Previous Update....... Updates Index.......My Post-Trib Book


November 28 - December 4, 2023

Discovery: Sky Rainbows Have Light Source Behind Them
Similarity Between an Empty Atom and an Evolutionist's Skull
Spectral Lines Are Not Light Gone Missing

Videos/articles on sky rainbows teach that each raindrop forms a full compliment of rainbow colors, but this may be misleading. First off, each drop forms such little color, if any at all, that this method of forming sky color seems a stretch. It stretches my mind to see a full rainbow of colors formed in the width of a mere rain drop when it takes an inch or so through glass to get the job done. Some would argue that while each drop produces a minute faintness of all the colors, an abundance of drops can make the colors rich. Is this really true? Can one drop produce even faint colors?

Everyone describing the formation of a sky rainbow always has the sun coming down on the droplets from above the head, then deflecting into and out of raindrops so that the light comes FURTHER DOWN to your eyes. But the sky rainbow exists also to both sides of a person AT/NEAR THE GROUND, and when the sun shines on raindrops near the ground, the light can't deflect to your eyes in the same way as it does with how they describe it for a droplet high in the sky. If the same is happening to droplets near the ground, as they say for the sky, the rainbow colors would deflect into the ground, not into your eyes.

Let me put it this way, that the sun is so far away that it enters sky-high droplets at the same angle as it enters droplets near the ground. There's not even a degree of difference. Therefore, if sunlight indeed enters droplets and then deflects off their backsides, the red color, for example, would come out from the same part of the droplet, IN THE SAME DIRECTION, in which case the red light can't go to the eye from both the sky-high and near-ground drops. It's just impossible.

To keep the people from realizing this problem, those who present rainbow angles don't tell you what I just did, that the angle of the sun into all raindrops at the rainbow is identical. Instead, they tell you what the angle is from the raindrop, between the line of the sun and the line to the eyes of the person who sees the rainbow. In that scenario, there's about one degree of difference between the red and blue colors, but this in no way is the difference of the angle of sunlight into the drops. The top line, from the sun, is the same into the red and blue regions of the rainbow. You need to know that.

Plus, in the way they describe the angles, they get more deceptive yet, putting two degrees between red and blue so that you might not catch the problem that tends to make their explanation of the rainbow incorrect.

Plus, they use perfectly ROUND raindrops in their drawings whenever they intricately show how they think sunlight enters and exits raindrops. But falling raindrops will tend to flatten to oval shapes, making their intricate explanations look foolish. Here's a sample of what they teach:

They have the sunlight deflecting DOWNWARD from their primary-rainbow droplet, and must, to the contrary, maintain that the same light out of droplets deflects HORIZONTAL with the ground, out of droplets, near the ground; otherwise the light can't get to your eyes from the drops at/near the ground. That's not consistent. I've considered it, and I don't think it works; I can't see how light could deflect horizontally out of the drop. If they find a way, they then need to say that light exits horizontally from drops in the sky, but that will miss your eyes.

In short, they cannot draw light exiting the raindrops from multiple angles, or the rainbow would then not form a circular shape. If colored light goes out also horizontally to the eye from the ends of the rainbow, then colored light should also come horizontally to the eye from on/near the ground directly in front of the observer, but it does not do that.

An alternative theory for colored-light formation is from sunlight entering and exiting many raindrops. As the light goes though one drop and exits with a little spread of sunlight, the drop's rainbow colors are absolutely minimal, hardly colored yet. But as this spreading light enters and exits more drops toward the eyes, the colors become more distinct from one another, and richer/deeper too. But in order for this theory to be correct, the light must originate behind the rainbow, and the only option is scattered sunlight off of the atoms and molecules in the air.

Each raindrop is a prism / lens because it's roughly spherical in shape.... I've just realized how the rainbow works, struggling as I write. I've got it. I'm sure I've got it. Let me tell you. Some of the light from behind the rainbow comes to the eyes as color, and the rest is typically invisible. While I find it nuts to try and find where exactly the colored light is sourced, IT DOESN'T MATTER where, IT'S THERE. That's the realization. It's there, and can form all the colors. I've now got to work this out into an explanation.

I tried and I tried and I tried to understand how the scattered back-of-rainbow sunlight could possibly create VARIOUS colored lights such that a HALF-CIRCLE appears in the sky. It seemed a mind-boggling task, even though my last update was on prism light, making me presently familiar with this type of light. But then I realized, IT DOESN'T MATTER where the source light is located that creates the colors, because there will simply exist such spots from which all colors have their source, while adjacent spots form colors that don't quite go to the eye. Instead, the adjacent spots send invisible light to the eyes through the same raindrops that bring color to the eyes from the SOURCE SPOTS.

I don't need to prove this theory correct by finding how the light comes to the eye from the bottom of the rainbow, and how it comes from the top of the rainbow, and in between too. That complication was too much when using the sun as the only source. But after realizing that scattered light behind the rainbow could be the color source, I realized, IT DOESN'T MATTER where the light sources are...because they will be back there somewhere. Somewhere, behind every sliver of the rainbow circle, there will be white light -- reflected off the FRONT of raindrops as white light -- sending colored light to the eye THROUGH raindrops.

Before you get confused: the colors forming in raindrops, in this theory, don't bounce and deflect around, inside the raindrops, as the diagrams everywhere have them. In this theory, the light goes straight through the drops, in and out without a bounce within.

We make a straight line from the eye to the red-colored drops, and then, instead of continuing straight back, we go at a slight angle to the white scattered light that is the red-light source. We can't call this the red-light spot, because the same spot behind every sliver of rainbow circle produces all the colors (per sliver).

Let's take the track starting at THE SPOT. If we draw a line from the center of The Spot (you can view it as spherical) of raindrops, to the raindrops in the red region of the rainbow, the light forms a full rainbow within the raindrops such that only the red light angles to the eye. But if we then draw a line from the center of the same spot to the yellow region, the angle of the line has changed such that, now, only the yellow light of all raindrops goes to the eye. Ditto with a line from the center of the spot to the region below the yellow, only the blue part of the rainbow within all raindrops goes to the eye.

From within the air directly beneath the violet of the circle, or directly above the red of the circle, only the ultra-violet or infrared (invisible) light of all drop rainbows goes to the eyes. From within the air higher or lower from the regions sending infrared and ultraviolet, plain white light goes to the eyes.

For every line of colored light coming to the eye, there is a specific angle upward into the sky, and further back from the visible colors, from which its white source light originates. The angle back the The Spot is always the same no matter where one line of colored light comes from on the rainbow circle.

Every line of yellow to the eye originates from the same angle behind the rainbow circle, but different than the angle for a red-region line of light. Ditto for all the colors. There will be source light behind the colored circle somewhere, even if I can't figure out where exactly it is! That was the key realization. I don't need to prove where the source light is, back there, because its ALL POTENTIAL SOURCE LIGHT back there. There's rainy drops back there everywhere that become the source light for each color. There will be ONE MAGIC SPOT per color per sliver of rainbow circle.

Looking into this further, I realized that one magic spot of source light will produce ALL the colors per sliver of rainbow circle! That is how the rainbow circle is formed. There cannot be a rainbow circle if every different color, per sliver, has a different source spot. That is now the biggest key to understanding.

You can view a sliver at any thickness you wish, a centimeter, an inch, or a foot, but not too wide. If you view a sliver as an inch wide, then a source spot an inch wide above and behind the colors is the magic spot for that particular sliver. It's works beautifully.

However, "above" is literal only at the top of the rainbow. In this discussion, "above" also means to the far side of both legs of the rainbow on/near the ground.

The reason The Spot needs to be higher in the sky than the colors is that rainbow light within each drop spreads to very minutely as it traverses the drop. It means that the angle back to The Spot is only slightly different than a straight line of sight to the colored regions.

The colored regions begin immediately at The Spot, and yet the coloring through the first many drops will be so minimal as not yet to be considered color. When the colors finally begin to appear, they deepen in richness progressively to a maximum because they afterward gradually disappear as the angle to the eye changes, as the light gets closer to the eye (from any one color region).

The more I get into this, the more the situation presents itself. As it turns out, the ONE SPOT per sliver forms a circle when all One Spots are combined. The ONE SPOT is a circle itself. It has the same shape, only wider, than the colored rainbow circle. This makes it very easy to understand what is taking place. Invisible light in a circle from above and behind the rainbow colors forms the colors.

IT'S THERE. There is scattered light coming from everywhere, in every direction, from behind the rainbow, and in some of those parts there will be ALL THE ONE SPOTS. Who would have thought? The mind is tricked into thinking that all colors originate within the raindrops in the midst of the colors, but this cannot work if every drop produces all the colors. It simply cannot work that way

If we shine white light at a different angle into a raindrop, the colors will not exit the drop at the same place. You need to have the light entering the drop at a certain angle (from the droplet) to get any one color into any one direction, such as the eye. When all colors exit raindrops in the red region of the sky rainbow, only the red comes to the eye, and when all colors exit the drops in the blue region of the sky rainbow, only the blue comes to the eye. It's due just to a different angle that red radiates in the one case, and the blue in the other. I've drawn this on paper, and it works. All the colors through the drops that don't come to your eye are going out to some other person's eyes.

To put it another way, when a One Spot is forming red light into the observer's eyes, it NECESSARILY creates all the other colors of the circle because they are directly beside the sky's red region, and because the other colors within the droplets are right beside the red within the droplets. So, as the angle of droplets changes toward the eye as one looks closer to the ground, the next color shows (orange/yellow) in the sky that is beside the red within the droplets. As orange/yellow is beside the red within the droplets, orange/yellow in the sky shows right beside the red in the sky. As the eye looks lower or over from the yellow to the blue region, the angle to the droplets has changed accordingly.

The One Spot rules.

I'm extremely happy with this theory, so happy that I can't see how it's not the reality. It's a perfect-working theory, and there's nothing left unexplained.

There are other rainbows above the one that a person on the ground sees, but these other rainbows cannot be seen unless one goes a little higher into the sky. It's all just in the angles between the raindrops and the locations of observers.

To explain the so-called "secondary rainbow," I suggest that each raindrop can possess an internal rainbow-making mechanism by another method than the one under discussion thus far. The internal rainbow from this other method shines to a different place than the internal rainbow under discussion thus far. A person can't see the colors of both internal rainbows, per raindrops, by standing at the same spot.

A person can see both internal rainbows, however, from standing at the same spot, when the second sky rainbow (the secondary rainbow) is seen from different raindrops. The secondary and primary sky rainbows are seen at the same time by one person, and the secondary rainbow must be the one shining the second type of internal rainbow (from each raindrop). That's a fact, we need only discover how the internal rainbows are formed that form the secondary rainbow.

On the webpage above, you can see that the secondary rainbow has a color sequence in reverse order from the color sequence of the primary rainbow. The blue region is now at the top of the rainbow circle. The secondary rainbow must possess its own One Spots, behind the secondary rainbow. But in order to have the color scheme in reverse, each One Spot sends light bouncing off of an internal surface of each raindrop, and only then comes back out toward the observer.

That internal, bouncing light is the theory of the goofs. It's the theory that I thought couldn't make rich enough color to form a rainbow, but then the secondary rainbow is very ghost-like (faint) at times, if not most of the time. It rarely even shows. This argues in favor of the secondary rainbow's formation from internal deflection. We don't need to know the angle of the internal deflections, because there is scattered-light EVERYWHERE behind the secondary rainbow. The light SOMEWHERE there will bounce off the internal walls of droplets such that colored light comes to the eyes.

It's interesting what I've read, that secondary rainbows form mainly when the sun is close to the horizon, suggesting that the One Spots are formed behind and BELOW secondary rainbows.

I can thank the goofs for developing this type of color formation in the sky. I suppose it's correct after all, though I can't believe it causes the primary rainbow. Light that strikes the internal side of a raindrop goes mainly out the raindrop, and therefore deflects VERY-LITTLE light as color once it exits the drop. Yet this picture well explains the dimmer secondary rainbow.

The goofs invent excuses to explain why the secondary rainbow has less than half the brightness of the primary. They need to invent an excuse because they have the primary rainbow formed from the same type of internally-deflected light.

As I said, their theory is impossible because they don't have the One Spots. Instead, they have colored light going from sunlight to each raindrop, to the observer, and while this works from the top of the rainbow, it doesn't work from the bottom, explaining why all diagrams of this theory always show the internal deflections from the top of the rainbow. The One Spots are needed to arrange the proper colors at EVERY SLIVER, top and bottom, of the sky rainbow.

Okay, so I've drawn my secondary rainbow on paper with its own One Spot in exactly the same spot as the One Spot for the primary rainbow...not in the same location of the sky, but at the same spot relative to the colored region of the sky (for both rainbows). That is, I've tried it with the One Spot above and behind the secondary rainbow, because one can make the light deflections work with that picture. You're likely going to lose me without drawings, but I don't have a draw package yet.

In order for all the One Spots to be behind and BELOW the secondary rainbow, the light within the drops has to do a loop, deflecting twice within each drop before exiting. I suppose it's possible, but the light may predictably get too dim in this case.

With the One Spots above the secondary rainbow, the light can form two internal rainbows, one on each and opposite side of a raindrop. Both internal rainbows flow roughly toward the viewer, though only one color from one of the two internal rainbows enters the observer's eyes while all colors of the other internal rainbow miss the eyes.

From my ventures in the last update, I can make this work even though modern science always puts the blue line of a rainbow in the same positioning relative to the red line. The last update taught me that the blue light always forms where the prism allows a LONGER path through glass than for the red light. It should be the same with paths through water. If we go by that rule, the color sequence of one internal rainbow will get to an observer's eyes successfully to form the secondary rainbow, while the rainbow on the side of the drop closest to the ground will not because it will not give the reverse order of colors possessed by the secondary rainbow.

Both internal rainbows hug a side of the droplet. One of the two hugs the side closer to the sky (top of the drop), and the other internal rainbow hugs the side closer to the bottom of the drop. The latter has the red line hugging closest to the drop's side i.e. with the blue line closer (but not close) to the center of the drop. This arrangement wont get the reverse order for the secondary rainbow, and neither will the internal rainbow on the opposite side of the drop, if we go by modern science, because it thinks the red versus blue lines should be in the same arrangement on both sides of the drop (i.e. if red is on the right, blue's always on the left).

In case you're confused, the lines of the internal rainbows move downward but diagonally through the drop, and it's the diagonal direction that makes for an upper and lower side to each drop. The rainbow lines on the bottom side take a path at roughly 4 o'clock of the drop, and the rainbow lines at the top side take a path at about 10 o'clock.

On the page above, the red and blue lines they show (bottom side of the drop) for their primary rainbow droplet are the two lines I'm talking about here for the bottom side of the drops. However, what I'm saying here has the light originating from outside the drop, not as they have it bouncing within the drop.

A line of light goes through less water content the closer it traverses to the side of the spite of all drops becoming a little flattened (oval shapes) as they fall through the air. Modern science would put the blue line closest to the outside of the top side of the drop, but it's necessary to have the red line closest there because it then correctly forms a rainbow circle with colors in reverse order (to the colors of the primary rainbow).

Again, they would put the blue line closer to the outside of the drop's top side, but I need the red line closer there to explain the secondary rainbow. However, the last update taught me that the violet-blue side of the rainbow forms where the light traverses the most material, and it just so happens that the most water exists NOT closest to the outside of the drop. Therefore, I do declare: the red line at the drop's top side will form closest to the drop's exterior, and thus this arrangement will make a secondary rainbow.

As it works, I'm very happy, as if I've discovered the real cause for the rainbow. On the other hand, God said the real cause is to show His Grace, no more drowning of a wicked humanity in one swoop.

In the photo at the page below, you see primary and secondary rainbows together, and so just realize that they form where the light happens to come to the camera with colored light from droplets. It's just the nature of the reflective beast. Nowhere else amongst the droplets will colored light come to this camera. There must be two different kinds of reflections/deflections, through raindrops, to form the two rainbows. As the colors are vertical near the ground, the mechanism of the goofs, for colored-light formation, cannot work. Their mechanism works only where the colors are horizontal across the sky.

The Goofs Who Invented Empty Mass

In the video below, the atom of the first minute is never fully covered with electrons, but there is more disservice done to the viewer because, in reality, there needs to be far more than one layer of electrons around a proton. Electrons inter-repel so that they would spread fully around the proton, with no proton "skin" showing, as shown in the first minute.

The atomic-physics goofs started to pile up, and formed a gang assigning the atom the impossible orbiting electron. That's what happened, to the shame of science. And here we are at the brink of 2024, and the world of physics is still so utterly devoid of integrity that it won't advertise the obvious impossibilities of the orbiting electron. It doesn't inspire me to praise this establishment, let's put it nicely like that. We need a clean break with evolutionism, because it's ultimately responsible for this nuttiness, because they created an atomic model that makes the big-bang viable. That's what they wanted most.

Rutherford was an idiot, and those who push this "discovery" of his are idiots too, uninclined to come up with an alternative explanation for what his experiment showed. He was shooting alpha particles...defined as positively-charged helium atoms, but this is crock. Alpha particles must be positive atoms SPIT out of radioactive materials, but how can helium be in radioactive materials...are they still nuts today? At least they give me a laugh once in a while.

If correct that alpha-particles are positive in charge, they won't stay that way very long, does anyone in the establishment tell you that part? If they come out of materials positively charged, they must be spit out by electrically-charged regions in the material. It's as though they are clinging to atoms until a point arrives that they are unlocked from whatever has them clinging, and out they fly dressed with fewer captured electrons due to the jolt each receives. That is, they are expected to lose a few electrons from off of their butts when something boots them hard out of the material surface. But helium material THEY ARE NOT.

Let's give Rutherford the benefit of the doubt, and say that alpha-particles went through the gold foil. Is this any reason to suggest that the gold foil is almost all empty space? Only a lunatic makes such a claim, one not fit to be a scientist.

Here's the reality: every atom anywhere has outer electrons (not orbiting) barely clinging to protonic attraction. The deeper into the atom, the closer the electrons are to protonic attraction, and therefore the more-strongly they are held to the proton. It's a no-brainer. But the most-outer electrons are held on the weakest, and barely held on because they are under as much inter-repulsion forces OUTWARD as they are under attraction forces INWARD (to the proton). The outer electrons are repelled outward by all other electrons beneath them.

The outer layer is defined as the one receiving as much outward force as inward force. This equilibrium of forces is what determines how wide the atom will be. Every atom, regardless of size or shape, is roughly at equilibrium at the outer layer. The stronger the protonic attraction, the more electrons it can load, and therefore the wider / larger the atom will be.

The point is, the outer electrons are hovering over the protons in equilibrium such that they can easily be plowed through by bullets such as Rutherford was shooting. As the bullets go deeper into the atom, they find more stress in plowing through so that, when they slow down enough, their tendency is to snake / weave through materials along paths of least resistance, which is along the outer layers of electrons, which can almost be like flying through empty space due to electrons hovering in near-perfect balance between positive and negative forces. Eventually, the bullets can exit the foil, but they will come out altered in course because the tightly-held, deeper electrons deflect them from their original paths.

The electron bodies of all atoms, which make up a solid, are merged into one another. When merged, their outer electrons are attracted by the proton of every atom involved in the merger; it's what causes atomic bonding, it's what causes liquids and solids to form. A bullet could, if fast enough, plow through protons, moving them aside while moving through. But when slowed down enough by successive collisions, the bullets are more prone to following the paths of least resistance along the peripheries of atoms. But by the time they reach a slow speed, they could ricochet in any directions possible, and therefore they can exit the foil in any direction.

Why couldn't alpha-particle bullets move aside hovering electrons, and ram through? It's such an easy, logical conclusion to make, any magnet-savvy kid could come up with it. Electrons hovering above protons means that empty space does exist in gold foil, but not to the massive extent claimed by those who invented orbiting electrons, who see very few electrons per atom. That view is nuts and unnecessary. It's unnecessary because the alternative, to the contrary, is logical: gold foil is packed with electrons such that they likely take up more space than the protons. There is easily as much matter in foil as there is empty space. Doesn't that sit better with you than the claim that a solid is in the range of 99-percent space?

Why did he use gold foil? Would the alpha particles not go through aluminum foil? If not, why not, if indeed aluminum is almost all empty space? There's all kinds of foils and other material just as thin. Did he try those before making his irrational mad dash? What ailed him to rush rashly into the arms of his orbiting-electron magic? His fellows have had a century to espouse an alternative, sane theory, but they are all involved in a conspiracy to shelter the irresistible, orbiting electron. What comparison is there between merely grandma's plum pudding to dazzling, dizzy deceit, it's just so much fun to play with orbiting electrons.

He may have been using gold foil because gold is a soft material, easy to bend compared to other metals. The atomic bonds are weaker. The outer layers of atoms hold each other more weakly, and thus passage of bullets through gold can be predicted to be easier. But if the reason for penetration is truly 99-percent space, then aluminum, iron, silver, tin, copper would have that too. Let us see what happens in the same experiment using foils of those materials in the same thickness used for the gold foil. I can tell you this, that if this experiment could be repeated with several other metals, we would have heard about it.

Rutherford was a zealot, and his zeal got the better part of his common sense. Electrons cannot orbit protons, it's just so easy to figure. The entire human race must be hopelessly stupid. Had he never seen the moon in orbit? Does an electron at the speed of light look like a normal orbit? Did he even ask, what puts the electron into orbit? Did he even ask: aren't orbiting electrons going to crash in clouds of smoking vapors when they hit something? Who's going to put them back together again? Rutherford was in fantasyland, and all the world of science went with him. Why?

There must have been a secret motive, because nobody adopts anything this crazy without one, and nobody tells what the motive is when adopting something this crazy. Why did they want an impossible, fast electron? The big bang, their speed baby. They want the big bang to live on in perpetual speed to this day and beyond. That is what they imagine: big-bang speed everywhere.

A version of Thomson's atom, which you saw at the beginning of the video above, had a proton much too large, I think, and definitely did not have enough electrons covering the proton, though this is debatable depending on how he imagined the atom. He probably had more than one theory going through his mind on how the electrons and the proton were built together. He had the electrons IMBEDDED in the positive field of the proton, a genius guess, as it turned out. Yes, protons are imbedded in a positive field, and they hover in that positive field, making it easy to move some of them aside. Apparently, what Thomson didn't have was a hard atomic core, but to be fair, he may not have ventured a guess as to how hard it is.

Rutherford, by counting the deflections from the gold foil, and comparing the number with the shots that went forward after penetrating the foil, thought he could estimate how much hard protonic material there was as compared to empty space. What he didn't allow into his teaching is that some percentage of bullets could have plowed right through gold protons, and out to his detection surface in spite of collisions with them. Therefore, he really could not know how many of the shots frontward into his detection surface were proton misses versus proton hits.

Besides, there's no way for bullets to miss all the protons from one side of the gold foil to the other. Rutherford was wrong because the bullets did strike many protons and yet came forth roughly in the frontward direction in which they were shot.

The way in which people tell of his experiment, he apparently assumed that every proton hit resulted in a deflection of the bullet. But he could not know how many gold protons were plowed through, each one a strike, per alpha particle, by the time the latter came forth from the gold.

To push the empty-space theory with falsifications, Rutherford's proponents would be prone to minimizing the number of gold atoms across the thickness of his foil. For example, a google offering: "...the gold leaf would be around 490 atoms thick". That's ridiculous. Five hundred atoms is nothing. Another google offering is more like it: "Depending on how thick the hair is, it might be 300,000 to 1,000,000 atoms from one side to the other."

What I can tell you with certainty is that an alpha-particle is larger than the gold atom, for gold has one of the smallest atoms of all the elements. The goofs have it backward, thinking that helium is the second-smallest atom while gold has one of the largest. Rutherford's experiment does better for the empty-space theory if alpha-particles are very tiny in comparison with gold atoms, and so this need can possibly explain why the physics establishment chose metals to have the largest atoms while choosing hydrogen and helium as the smallest. I don't think they can prove the size of atoms as they teach them, but I can offer evidence that hydrogen has the largest atom, and that metals have the smallest.

Okay, so if the gold leaf was 100,000 atoms thick, each alpha particle penetrating the gold foil could strike 1,000 or more gold protons, depending on how wide the alpha-particle is in comparison to gold atoms. In that case, the deflections of particles backwards and sidewards from the foil may have included products from the gold foil: clusters of electrons and/or individual gold atoms.

I don't know the basis / reasoning for their claim that the alpha-particle issuing from a heavy metal needs to be equivalent to their perception of a helium atom, but, guaranteed, these goofs have the atomic model so wrongly modeled that it's not likely good reasoning. I don't know enough about the alpha-particle's origins to know what tricks they may have played with it, but some evolutionists of that time period were willing to pull off hoaxes to advance evolution. It was their nature to falsify science. I think they enjoyed misinformation, to see how far they could trick the public.

If the alpha-particle can zip to the detection target after it has penetrated past 1,000 atoms, then it can also bring with it some gold atoms. Right? Did Rutherford include that idea? It may have crossed his mind, but he chose to publicize something else: all strikes on his detection surface were from alpha particles.

You've probably seen motionless metal balls lined up in a row; when a moving ball strikes the row from one direction, it stops dead in its tracks, but the last one in the row simultaneously starts moving with the same velocity and force as the striker had before it was halted. Why couldn't that sort of motion transfer be possible in Rutherford's experiment? That is, the bullets may have struck one side of the gold, but only gold atoms came shooting out on the other side. And these gold atoms would probably have been positively charged because the jolt of knocking them out flung some of their captured electrons away.

I've just read at Wikipedia's Alpha Particle article that these bitties supposedly move at a quarter the speed of light, hee-hee, which is how you can easily know that the goofs are science lunatics. They've driven themselves mad with errors piled on errors.

Is there no other way to propose how electrons are captured on atoms besides orbiting at the near-speed of light? Did Rutherford and others conspire to ruin Thomson's atomic model because the latter's electrons were stationary while cosmic evolutionism favored unimaginable speeds from the big-bang explosion? Looks like. In those days, evolutionists were still seeking to become dominant. They weren't yet, and so they had to work secretly to conform science departments to whatever ideas would favor their developing, mainly-secret beliefs.

Do you know how itty-bitty small the circle of an orbiting electron is? Do you realize what the speed of light is around that itty-bitty circle? Do you really think that these ittie-bitties could hold to an orbit at that speed? Do you realize what would happen when these itty-bitties crash into one another, and they go zinging at the speed of light off of every street corner throughout the unsuspecting atomic neighborhood? There's going to be some bad crunch sounds in the machinery alright. How do you spell, DESTRUCTION?

An orbit is a perfect balance between a moving object and gravitational force. The proton has a positive, gravitational force. The moon cannot be placed anywhere around the earth where it would stay in orbit at zillions of revolutions per second. How did Rutherford expect, or how did he prove to his peers, that the electron could stay in orbit? It's not scientific. It's not true to reality.

The fast electron can be marked out as so absolutely necessary to the God killers that they refuse to abandon it to this day.

Inventing Neutrons

When they went forward with an erroneous atomic model, there were more erroneous inventions needed to patch things up when observations spoke to the arising problems. They opted to devise an atomic model where all atoms were made of hydrogen atoms at their core. All atomic cores were deemed identical in the sense that all had differing numbers of hydrogen atoms. And they called these hydrogen atoms, protons. Can you think of just one reason why grown men should not have chosen multiple protons as the core of atoms? Just one.

Why did they choose this atom? Why didn't they choose atoms all having just one proton, but with every atom type having a different proton, each with their own characteristics, abilities, and quirks?

There must have been a secret reason for the choosing of their protonic core, because only a circus buffoon decides that atoms have more than one proton at their cores. Only a science buffoon, asking to be ridiculed and spurned by his peers, would advance the idea that particles repelling each other are joined at the hip in the middle of all atoms. But as the whole gang ran with this foolery, it's obvious to me that there was a conspiracy to effect a secret mission. They wanted an atomic model that only the big-bang could love. It's as simple as that. To hell with reality; they were on a mission to kill God using (pseudo-)science, deceptions, and wits as their weapons.

Here's the next video telling why neutrons were invented: due to their atomic model not jibing with their observations:

At 40 seconds in the video, you can hear the speaker tell that "most atoms were found to be about twice as heavy as the mass of their protons," and while this is said to justify the invention of neutrons (added weight), the fact is, they did NOT know the true weight of atoms in the first place. You need to take that into consideration, asking, how did they find the weights of atoms? Is that an easy thing to do? You can't find the weight of atoms, even today, how possibly did they do it a century ago? Go ahead, find the weight of an atom, call me up when you find it, I'll be waiting right by the phone.

It's clear that they devised a way(s) to weigh atoms by error or deception. The way they did this was to first pass off a theory that later turned magically to fact when the public had no interest in disputing the theory. The science books were written, and they ran with the theory that all gases at STP had the same number of atoms, wherefore the relative weights of atoms were in accordance (proportional) to the weights of the gases at STP. Look at how weasily they found the relative weights of all atoms, so lickety-split by first assuming fact from a cheap error.

It's not true that all gases have the same number of atoms. By what cosmic coincidence could that be true? Therefore, if its not true, they invented neutrons for nothing. They invented as many neutrons, per atom, as was necessary to patch up their already broken atomic model.

They all wrote jargon-filled volumes on this atomic model, creating mountains of buried mathematics to describe it, craftily inventing experiments to prove it, inviting royal magicians to explore it, and they dissected a sheer phantom upon their table in bright lights, naming all of its parts and labeling all of its bones which did not exist. They operated like physicians on a cold fantasy already dead. They reached into a carcass of make-believe and birthed many more mirages all milked at their breasts so preciously. These men were lunatic-breeding lunatics, and their whole game was to deceive subsequent generations. Alas, its today, and the world is still infected from the rot in the brains yestercentury.

So here's what they did. As they knew that hydrogen gas weighs twice as much as helium gas at STP, they wrongly assigned the helium atom with twice as much weight as the hydrogen atom. You can see right away how asinine and amateurish this is. But they then went and made the further mistake of defining a helium atom as two hydrogen protons stuck together. The one hydrogen proton, and the two helium protons, constituted roughly the full weight of both atoms because electrons were not assigned any weight to speak of.

Why didn't they count the weight of electrons? Because, they could gather from experiments that they didn't weigh anything. And so, rather than realizing that gravity repels electrons, hee-hee, they went and made their third mistake: they made atoms almost naked, giving them few electrons. They gave only one electron to the hydrogen atom, poor thing. Can you imagine anything more stupid as to assign a particle, that attracts electrons, with just one electron? Just one lousy electron, hahahaha. There's room there for a city, but they gave it just one wee-wee electron, hahahahah.

What possible argumentation could they have had to deny the proton the ability to attract more than one electron? They then made another mistake by assigning the electron with as much negative charge exactly as the proton has positive charge, and they crossed their fingers hoping that the world would accept this as sufficient evidence that H atoms can only gather in one electron. And guess what? The world and the science pupils were together stupid enough to let the science bosses get away with it. Some future generation has a big mess to clean up.

They were then forced to make another mistake when they needed to define a hydrogen gas as having "diatomic" atoms, meaning that each atom was two hydrogen atoms bonded together. Yup, that means that they stuck two protons together that repel one another, just like an idiot would, and, don't worry, they'll invent another error to make "glue" to keep those protons together. I can't remember why they invented the diatomic hydrogen atom, but they may have needed the extra weight for the hydrogen gas to jibe with some fact(s).

The new problem was, they had defined the helium atom likewise as two protons stuck together, yet helium gas weighed twice as much as hydrogen gas, which flew into the face of their theory: all gases at STP have the same number of atoms. In that fantasy, helium atoms needed to weigh twice as much as the diatomic hydrogen atom, yet by creating the diatomic H atom, they made it weight the same as helium. So they needed to invent neutrons for extra weight, and gave two of them (free of charge) to the helium atom because the goofs conveniently assigned each neutron with the same weight as a proton, what a laughable farce.

How many of you still think the atomic-physics pioneers were the gods? They spent half their time, and half their money, fixing their mistakes by tacking on new mistakes.

And they thought: as long as we're inventing neutrons to go into the atomic nucleus with the protons, why don't we invent a way to make the neutrons keep the inter-repelling protons from flying away from each other. That way, the people won't think we're altogether law breakers.

Lithium metal is a problem for the goofs because they assign it an atomic weight of seven, and if this is correct, lithium gas must weigh seven times as much as hydrogen gas. If it doesn't, it flies in the face of their theory: all gases at STP have the same number of atoms (or molecules, if they apply).

Instead of giving the lithium atom seven protons so that it weighs seven times as much as the hydrogen atom, they give it three protons and four neutrons, child's play. So, when they need to, they can give the atom's nucleus more neutrons than protons. As the need is, use the magic wand.

Does it sound right to you that the lithium atom, which forms a metal, is not much different than a helium or hydrogen atom? Doesn't anybody in this world ask questions like this? Is everyone totally devoid of think-for-yourself power?

Beryllium metal was assigned four protons and five neutrons, but I have no idea why they assign it four protons. I've never wanted to know...because it's just trash. It's worse than trash because it's got the whole world deceived.

Near before the 2:00 point of the video above, we see alpha-particles shot at lithium and beryllium to result in something passing through an electromagnetic field. It doesn't say whether the metals are in foil form. The mystery thing could be hard light waves, or it could be beryllium / lithium atoms, but the video wants to give the impression that neutrons are being shot through the magnetic field, because the goofs were hard-pressed to prove the existence of neutrons. The experiment is a perfect set-up for inventing neutrons.

In this experiment, the mystery thing does not bend through the magnetic field, i.e. it goes straight through. They can tell if it bends by making the particle strike a detection surface past the field.

The interpretation of a straight-through particle in a magnetic field is that it is neutral in charge, and so you can see what they are insinuating: it must be neutrons moving through. However, it is possible for particles to travel fast enough through the magnetic field that they do not bend through it.

They used three metals to get the straight-through track, beryllium, lithium and boron, all classified as light metals. Lithium even floats in water. Did they do the experiment with heavy metals? If not why not? Is it possible that when the atoms of these metals pass through the field, they go neutral in charge?

If one argues that beryllium was used because it doesn't get attracted by a magnet, it seems like a good argument. However, Boron was used too, and it's attracted by a magnet. All materials can develop a positive charge, but only few are attracted by a magnet.

Why might the alpha-particles knock the atoms of these light metals through the field but not of other metal atoms? Can metal lightness indicate that the atoms are not deeply merged, and therefore easily dislodged from one another? Sounds reasonable. However, this can be tricky because, the more shallow they are merged, the less shock they receive from bombardments. The bombardment energy hits "cushions" when the protons of atoms are far apart, and covered with more electrons. The latter are indeed cushioning to protect the protons from the full energy of the strikes.

When alpha-particles struck one side of the beryllium / lithium / boron metal, sending energy through it, it could have knocked metal atoms off the other side, into the field, due to the GREAT DEPTH of atomic merger. With protons thus closer together, more shock coming through the material can hit them. If we put a row of metal balls side-by-side with a little air between them, a striker hitting a ball on one side of the row won't send a ball at the opposite side moving very fast. The closer the balls, the faster the last ball in the row moves because it gets more shock force.

Then, with metal atoms only slightly merged, they don't loose many electrons, for atoms lose more electrons the deeper they are merged. If beryllium and boron atoms are deeply merged, they have lost many of their outer electrons, which are the loosest electrons i.e. the ones offering the most cushioning. I'm not including lithium in this discussion in case it's porous, in which case shock force to its atoms gets complicated.

However, if the metal targets (in the experiment) have deeply-merged atoms, they are expected to go through the magnetic field highly positive in charge, yet we are told that the mystery material went with zero charge. The best explanation is, therefore, that beryllium and boron get such great shock force that they go too fast through the field to register as a positive force.

It's a no-brainer that the nature of atomic bonding is such that it releases half the electrons (as heat particles) within the merged regions, for the protons cannot keep (attract and trap) a double-density region of electrons; half must go free (into the air as a gas becomes a liquid). The proton cannot keep a double-density region because the atom is already fully loaded with electrons prior to merger. Electrons of atoms hover, forming literal atmospheres (i.e. have space between electrons), and thus one atmosphere can sink (merge) into another. It's child play when one goes through the simple steps of atom formation correctly.

It works easy-to-figure like this: a naked proton's gravity attracts many electrons. The deepest electrons come closest to one another because the positive gravity is greater on the proton's surface, same situation exactly as a higher density of air atoms on the earth's surface. The higher up from the protonic surface, the more there will form outgoing negative charge from lower electrons (closer to the surface). This increasing negative charge counteracts the proton's positive gravity, allowing the higher electrons to create more space between each other (because they resist contact with each other)...until an outer layer of electrons is formed where the positive force of the proton equals the force of the earth's gravity. Say what? What's earth's gravity got to do with it?

The formation of the outer layer is the atom's fully-loaded condition. If you merge two or more, heat will come out from their merged regions, because freed electrons define heat. The goofs have an impossible definition of heat.

As earth gravity is the negative charge from heat particles (free electrons) within the planet, or heat within the sun, another no-brainer can be had. An atom's core can load electrons only as far as its outer surface gets positive proton attraction equal to the negative charge of earth gravity upon the atom. Can you grasp that? Gravity won't allow more proton-captured electrons upon the atom, but instead causes them to drift away from all atoms in the upward direction as rising heat, for earth gravity repels atmospheric electrons upward, duh. Air atoms are always surrounded by heat particles, but they are all drifting past the atoms (the sun replenishes them continuously)...when the wind don't blow.

What you've just read is a huge eye-opener, but your eye probably missed it. As gravity arranges for all atoms to load electrons to the point that the outer atoms get a positive charge equal to the charge of gravity, then, regardless of what specific force gravity has, the outer layers of ALL atoms have the same level of positive charge, which is how I discovered that ALL atoms weigh the explain Galileo's experiments Pisa. This is how I know that the list of atomic weights of the goofballs is a farce. ALL ATOMS WEIGH THE SAME. Earth gravity arranges, and continually demands, it.

And so where helium gas weighs twice as much as hydrogen gas, it's because there's twice as many helium atoms in that gas than hydrogen atoms.

Rutherford and his group of conspirators opted to make the atom like a planet revolving around the sun, so much fun, but the reality is that God made it like the earth's atmosphere. It's like a plum pudding if the pudding is the empty but positively-charged space between electrons. The only thing Rutherford had to do was put a hard plum seed at the center of this pudding, but instead opted for the impossible...because he was a goof.

At 2:18 of the video above, the claim is that a ray of the mystery material was shot through some wax (thickness not given). The speaker says that protons came out of the wax once shot by the mystery material, but this is impossible. Bare protons cannot exist because they instantly load electrons. If bare protons were coming out, all the men had to do was to let them pile up (keep the gun turned on) until they become visible / detectable as atomic material.

What they needed to do, to pass this experiment off as thorough, was to prove that these protons were indeed from the wax, but they did not do that, when they could have. Nobody tells us what the protons at the end of the path amounted to. The experiment is therefore NO GOOD. Blow the whistle. Penalty.

Yes, for the only way to know that they were protons was to make them land on a detection surface long enough, and so tell us what that surface looked like by doing so, otherwise the explanation for this experiment is NO GOOD. Penalty, major misconduct for failure to cross-check. Red card for knee-jerk reaction.

If instead of bare wax protons there were full wax atoms shot out having some positive charge, the goofs could have let them pile up for several days to see if wax became detectable. Do we think they at least tried to do it, as a test, or is what they ever wanted was to deceive the people into believing in the existence of neutrons?

What the video is implying is that the neutrons in the alpha-particles went though the lithium / beryllium metals, and then into the wax but not through the wax. Achem, isn't wax softer than metal? Yes, yet they claim that only protons (= something positively charged) came out of the wax, nothing else. Doesn't it seem more logical that the mystery material penetrated fully through the wax as a positively-charged material? Yes. In that case, it wasn't neutrons.

In that case, the mystery material looks like positive beryllium or positive lithium. No neutrons came through those metals in the first place, but, rather, the alpha particles knocked the atoms of those metals through the magnetic field either fast enough that there was no bending through it, or through a field that had very little electrical power for the purpose of deceiving.

Did the men allow the mystery material to pile up on the detector, then test to see what builds up there? If it takes a month to get the job done, so be it, but they cannot claim neutrons there without testing to see what builds up. As we hear nothing of what built up at the landing site, the experiment looks like a mountain of cow patties.

There were two magnetic fields in this experiment (Juliot and Curie version), though the video shows only one. In the second field is where they claim to detect protons, and this is where I stand tall above the goofs for accepting the obvious fact that bare protons do not exist. I can be confident when interpreting this experiment because I stick to logic.

The video's argument is that gamma light waves can't knock protons out of the wax, and that this is proof that the mystery striker material was neutrons (at the core of the alpha-particles). But the striker material through wax could have been something else besides gamma rays, and since the striker material didn't produce bare protons after entering the wax, the striker material becomes feasibly discerned as positively-charged metal atoms. And why did they use wax, of all materials, to prove that this was a neutron stream? Is it because they were more trickster magicians than scientists?

Logically, the positively-charged metal atoms giving the appearance of a neutral charge, through the first magnetic field, lost more electrons when penetrating fully through the wax, and therefore registered with a positive charge through the second field. Again, we could conjecture that the metal atoms went positively charged through the first magnetic field, but registered as neutral due to their great speed (made their paths unbendable). Then, they lost speed through the wax so as to make their positive charge more detectable, this time, through the second magnetic field. The wax just slowed them down enough to register their positive charge, but the tricksters claimed protons there instead, to hide the fact that metal atoms were coming through.

So, yes, the wax could both have made the metal atoms more positive, and slower too. The only thing we need to find to make this experiment lame is whether the old school knew that particles could go through a magnetic field fast enough not to bend. Have fun trying to find such a thing in the college textbooks, for it would be a guarded secret, to protect this experiment.

Besides, they think that wax is almost all empty space, just like gold leaf, but now suddenly the neutrons alone of the alpha particles can knock protons out of wax??? In Rutherford's experiment, the whole alpha particle (neutrons and protons in their eyes) was deemed incapable of knocking gold protons out, otherwise Rutherford's empty-space claim was made a non-starter.

It should be imperative that we see the final detection surface, where they claim the landing of protons. Did all particles strike at the same spot? If not, how did they strike? In Rutherford's gold-leaf experiment, the landings created a large circle, not a small spot. The particles from the leaf were spitting out, not rifling straight out.

What did the detection surface look like when they allowed the mystery ray (from the beryllium / lithium metal) through only one magnetic field? If the mystery material was neutrons, did they deflect out of the metal in all directions, same as in Rutherford's experiment? If the metal was much thicker than the gold leaf, there should have been more deflections.

On the other hand, if the mystery material landed at one small spot on the detector, couldn't it suggest that the alpha-particles were sending only motion energy through the metal, and putting the metal atoms to flight from the opposite side of the metal?

The atomic model of the goofs didn't allow them to know the atomic make-up of the light metals, but here's the true model: the lighter a material per unit volume, the fewer the atoms, because all atoms weigh the same. If a cubic inch of one material weighs a pound while a cubic inch of another material weighs a half-pound, there are half as many atoms in the latter material. But why are there a different numbers of atoms per unit volume? There's two reasons: 1) the atoms are of different sizes, and: 2) atoms can be merged by different depths. These two factors determine the density (weight) of materials.

All atoms weigh the same and thus all carry the same punch power when striking, yet the bigger they are, the harder it is to penetrate materials when shot. Larger atoms have the potential to be more-deeply merged, but thus far I haven't been able to discover the thing(s) determining depth of atomic merger. It's a vastly-complicated task. One first needs to discover how deeply merged various atoms are before asking why it's so, and molecules in merger only makes it more complicated if not impossible to know.

My thinking is that atoms more-deeply merged have a higher atomic-bond strength. Seems reasonable. To discover the force level of atomic bonds, one can use the evaporation points of materials, but can only use it partially because the evaporation points (when atoms dislodge) depends on both, the level of atomic bond, and the size of the atom. The bigger it is, the more lift power heat has to dislodge it from the liquid surface...same as a shot of upward air lifting a two-inch ball more easily than a one-inch ball where both balls weigh the same.

BEHOLD. Beryllium is more than four times lighter than iron, yet both have boiling points in the same ballpark. Iron boils at 2,862 C, and beryllium at 2,469 C. Although the evaporation points of materials are at some degree below their boiling points, we might assume for the moment that beryllium has the lower evaporation point. As this point is when heat finally gets the upper hand, when it unmerges a liquid atom (to send it into the air as a gas atom), I can tell that iron and beryllium have atoms of roughly the same size.

KEEP ON BEHOLDING. As beryllium is more than four times less dense per cubic inch, it reveals that iron has atoms more deeply merged by far simply because both are roughly the same size. This deeper merger of iron atoms is probably why iron has the higher evaporation point, because deeper merger needs more heat lift (higher temperature) to tear atoms apart.

IN SHORT: beryllium has slightly-merged atoms so that when one is knocked out of another, through a magnetic field, it doesn't have a strong positive charge. That is, it's almost fully-loaded and NEUTRAL (!) even before it's unmerged, and it moreover captures some electrons immediately upon exit from the chunk of metal! That's why its BEHOLD time. This can explain what the goofs insist to be neutral neutrons.

But by forcing beryllium atoms through wax, you strip some of its electrons off, same as when the goof dives into water and his swimming trunks come off...he gets all emBARE-ASSed. The result is not bare-naked protons, but beryllium protons stripped of many electrons.

Boron was used in the same experiment, which has a much-higher boiling point (4,000 C) than iron, and a density of 3.4 times less than iron. If the evaporation point of boron is higher than that of iron, then boron atoms are substantially smaller than iron atoms, yet the lower density, by far, of boron atoms means that they too are not deeply merged, though considerably more merged than beryllium atoms. This is an area of science I like to play in, but haven't got the time to make tables and recordings to see how well my atomic models checks out.

A difficulty I've encountered is in how shallow-merged atoms create weakness in material structure. Beryllium is stronger than iron in one way, but I'm reading: "Beryllium's brittleness is the down side of its advantageous stiffness." So, it's stiffer than iron, yet breaks apart while bending, possibly due to its shallow atomic mergers. But if it's got shallow merger, what makes it stiffer than iron? Is there something besides depth of merger that creates atomic and material strength? Yes, or predicting the strengths of all materials would be fairly straight-forward, much proportional to depth of merger. Shapes of protons/atoms can alter atomic bonds. Imagine protons shaped as crosses or forks.

A factor in brittleness could be porosity. I'm reading, "...two types of porous beryllium..." I can't find how porous it is, and assume it refers to air/gas pockets. I'm reading that many metals have porosity.

I'm trying to figure out how likely it is for alpha-particle bombardment to dislodge beryllium atoms in comparison to other metals. As I argued above, the more-deeply merged, the more shock-force the cores of atoms ought to receive when the material they comprise gets bombardment. On the other hand, depth of merger can make it harder to dislodge atoms.

Perhaps beryllium is just so shallow in atomic merger that this alone explains why it goes SPEEDILY through the magnetic field, for the bombardment force first needs to overcome the atomic bond, and if much energy is spent on that, less energy will be left to produce velocity of the knocked-out atom. Making sense.

The goofs probably tried many materials to find a neutral stream through the magnetic field, and when they found one with beryllium, they conspired doing the winkie-winkie to discover the neutron.

If indeed the neutrons of alpha-particles fully penetrated the beryllium, as they claim, and then continued on through the field, it requires some concocted excuse for why the protons of the alpha-particles didn't also go through the beryllium and into the field. Or, they need to concoct an excuse to keep the beryllium atoms from likewise flying through the field, for if the neutrons went fully through, why couldn't they bring beryllium atoms along in their train? It looks like a pick-whatever-you-want game from the goofs. They picked neutrons.

We now address why the beryllium atoms went neutral through the field while, to be assumed, alpha-particles shot direct through the same field would show a positively-charged curvature. That is, if the beryllium went neutral at a speed predicted to be SLOWER than the alpha-particles, and if the beryllium couldn't show a positively-charged curve due to its high speed, why should a faster alpha-particle show a positively-charged curve? Because, it's got more positive force than beryllium atoms.

This is the key: beryllium has shallow-merged atoms that don't come out of merger with high, positive force. My atomic model reveals the shallow merger of beryllium atoms, I'm so positively electrified.

There is a claim online that Rutherford shot alpha-particles into nitrogen gas to create a little oxygen from the merging of alpha-particles with nitrogen atoms, which is either a hoax, or an error as stated/understood. Looking further into that claim, I found another claim that Rutherford was splitting nitrogen atoms with alpha-particles and creating bare-naked protons to boot. You see, in their atomic model, combining helium with nitrogen adds up close to the number of protons and neutrons needed to form the oxygen atom, and so this might be how they came to define the alpha-particle as a helium nucleus.

As nitrogen gas weighs about 14/16ths as much as oxygen gas, and as nitrogen and oxygen make up almost all of the atmosphere, it had me wondering whether nitrogen is a non-combusting, modified form of oxygen. If so, then bombarding nitrogen atoms could possibly convert them to oxygen atoms, but this would be the case with or without helium involved. It's not mainstream physics that oxygen is created from nitrogen bombardment.

Again, the erroneous atomic model of the establishment is what necessitated neutrons. The model did not work, and needed extra weight without extra charge at the atomic core. So, they invented the extra weight as neutrons. That's all there is to it. However, they gave the neutron more than this one role, as the patchwork needs arose, if they could get away with it.

To explain why the alpha-particle was positive in charge, Rutherford proposed that it had more protons than electrons, an impossibility that only a BIG GOOF would stand on, in public. He assigned the alpha-particle just two electrons, which is another way to know that he was a very stupid man scientifically. He gave the nucleus of the alpha-particle four massive protons that could only capture two wee-wee- electrons, just like the most-stupid man in all of science. He was a hijacker, the villain.

Would you bet your house on one electron having as much charge as one proton? Why were the goofs so stuck on inventing an atom that had one electron per one proton? What absolute need did this set of hijackers possess that they would go this route to tarnation when all they had to do, to fly to paradise instead, was give the plum pudding a plum seed at its center? Just one plum seed. Only one, that didn't break the law. Instead, they opted for the law-breaking, multi-protonic core because they were law-breaking hijackers. It's that simple. The devil had their minds.

Thompson's Gas Atoms Play a Trick on Him

In a video below on the same beryllium experiment under discussion, I caught wind that James Chadwick "discovered" the neutron thanks to a so-called "charge to mass ratio," which had me suspicious because it sounds foreign to logic, a perfect thing to be bounced around by physics magicians in their crazy skulls. And so I went fishing to find a video that could explain this animal to me, and came to this one below (I'll go over it with you below). Excuse the speaker's voice, who, together with his music, sounds like hypnotist playing a trick on us:

This gadget is simple, but only after someone discovered that electrons could be coaxed out of metals with some positive force.

Thomson hooked up some electricity to his metal "cathode," a fancy word for electron shooter. The electrons from the cathode came from the wire having electricity at the ready. To get the little bitties to shoot, he ran high voltage to the cathode, but he also needed the anode, a piece of metal that would attract the electrons clear off of the cathode. He didn't want all the electrons, just the ones going in a straight line, and to accomplish this, he put a hole in the anode. Then, to make sure he got the straightest of the straight ones, he put another piece of metal further down from the anode, and it too had a hole. The only electrons he studied were the ones going through both holes. This entire outfit is inside the light-green gas that you see in the second half of the video.

What the mystifying speaker fails to tell you is that the "low pressure," midway into the video, is a low-pressure gas, the light-green color. As they fire through, electrons strike some gas atoms. Why do we think Thomson had the gas atoms in the path of the electrons? Why doesn't Mr. Mystify mention the gas-atom bombardments that inevitably occur? My guess is that he's stealing this set-up from other video producers, and doesn't himself fully know what's going on.

What effect did the gas atoms have as electrons struck them? That's easy: gas atoms released electrons. What did that do? That's easy: they increased the density of free electrons in the tube. Did the density increase when Thomson raised the voltage to the cathode? That's easy: yes. What happened as the density of free electrons increased? That's easy: the gas atoms became more positive because they lost more electrons. Where did the positive gas atoms go? That's easy: to the cathode. How long did they stay there? Until they re-loaded with electrons. Can any of this affect Thomson's calculations? That's not so easy (to answer), but this video and others don't mention any of this. I thought you might like to know.

Plus, nothing is said on whether Thomson closed off the part of his chamber holding the cathode and anode from the larger chamber holding the electric field. I recall reading that the cathode wouldn't shoot electrons without a low-pressure gas in the chamber, and I'm also quite sure that the larger chamber had low-pressure gas, suggesting no need to separate the two chambers. Thomson discovered the electron, but perhaps didn't yet know that shooting gas atoms with them released both free electrons and turned atoms positive. Can you see how this could have ruined his calculations?

The second half of the video tells of three requirements to get the electron "beam" to bend MORE. The first requirement is the charge of the particle, defined as its magnetic force. As we are dealing with the electron here, that charge remains the same.

The second requirement for greater bending of the beam is particles having less mass. Mr. Misty fails to tell us which particles he's referring to, but the cathode shoots only electrons, which always come with the same mass, and so we assume that Thomson always used the same number in every math calculation when he needed to plug the mass of the electron into it.

The third requirement for greater bend of the beam is simply turning up the power to the electric and/or magnetic field, making perfect sense.

But there could be another means to make the beam bend by different amounts which Misty fails to mention, I wonder why. The higher the voltage to the cathode, the faster the electrons fly, the less they bend. More voltage to the cathode pushes electrons closer together in the electrical wire, and in the cathode metal, so that electrons fly faster once they get pulled into the gas.

Ahh, the higher the voltage, the more the chaos in the gas due to more electrons flying free. Them bitties always fly away from each other, the very things that cause explosions. Doesn't it suggest that the electrons in the beam slowed down with higher cathode voltage, even though they have the capacity to fly faster? Can a bird fly faster in pouring rain than in the air alone? Not even if it had waterproof wings.

The storm of free electrons slowed the electron birds through the electric field. As Thomson calculated the charge of one electron using their changing speeds in various sessions of experimentation, his calculation was cheap-cheap wrong if he thought higher voltage made the birds fly faster.

Secondly, it's known that positively-charged particles take some time to reload and return to the neutral condition. As flying electrons create positive gas atoms, some of them could have migrated to the glass beside the negative plates of his electric and/or magnetic fields. These atoms contributed to a positive force toward the beam, no doubt about it. Did Thomson figure on this? What fraction of beam bending was due to positive atoms versus the magnetic fields alone? Did his time yet know that firing electrons at atoms made them positive?

Thomson's calculations would have been amiss, for he calculated the charge of the electron using the various charges of his electric and magnetic fields...which were measured by some sort of electricity-measuring device that did not take into consideration additional positive charge from some gas atoms trapped invisibly within the field(s).

Plus, some of the positive atoms would have covered his cathode, which could have further slowed electrons "jumping" out of it.

On the other hand, the electrons freed from the atoms would not have migrated to the glass nearest his positive plates, if true, what they say, that magnets don't attract stationary (or slow) electrons. Chances are, there's not a pile of electrons on a the positive pole, or they could be revealed in a couple of ways. The reason a magnet has a positive side is that this side's metal atoms resist electrons, and keep a depletion of electrons in order to become positively charged. Therefore, firing an electron beam through a gas will make the negative poles of Thomson's electromagnet and magnet less negatively charged.

I'm reading erroneously: "Cathode rays come from the cathode, because the cathode is charged negatively. So those rays strike and ionize the gas sample inside the container. The electrons that were ejected from gas ionization travel to the anode." "Ions" here refer to positively-charged atoms. It is not true that freed electrons go to the anode.

First of all, the goofs have trained the world NOT to see that a vacuum is filled with free electrons. The heat of a vacuum is free electrons. If you stick an anode in the vacuum (or thin gas), the electrons are not going to gravitate to it because they all inter-repel and thus seek equa-distance, meaning they spread out in the vacuum as much as possible. When a few more electrons are freed in the thin gas, they will only spread in the vacuum (adding to its density), and, as was said, magnets don't attract stationary electrons. An anode is the positive terminal of an electric magnet.

A bird cannot fly in a vacuum because electrons weigh zero. Unless particles are trapped downward by gravity, the bird's wings have nothing by which to get lift. If a bird or fly tries to fly in a vacuum, the electrons just move out of the way with zero work done on them. Work on air atoms is what makes for wing flight.

I was under the impression that a little gas was needed in the cathode-ray tube in order to form the electron beam, but maybe this is misinformation. I can't see how a few gas atoms in the tube could be needed to get the electrons to fly from cathode toward the anode, can you? It makes more sense that the gas was needed to slow the electron beam, because a beam too fast cannot be bent by magnets. The force of any object's path bending (or curving) through a field is done by magnetic force x time. The shorter the time that the object is in the field, the less the magnetic force can shift its path. Did Thomson know this? If not, his calculations would have gone off of true.

In short, it may be that Thomson needed gas in his tube to get the the electron beams to bend, not to get the electrons to shoot. If the beam would not bend without some gas in the tube, then Thomson and the entire establishment has egg on their faces if they did not add this "minor detail" to their reports.

Just to show how far and thorough atoms can migrate when charged with either positive or negative force, a cheap and rudimentary outfit to make hydrogen gas by passing electricity through a watery acid causes the water molecules to separate into individual oxygen and hydrogen atoms AT ONE BATTERY TERMINAL, yet one only or the other, I'm not sure which, the oxygen atoms or hydrogens atoms, are instantly pulled, through the water, to the other battery terminal by electromagnetic attraction. If charged gas atoms can migrate more than a foot through water, then they also migrated a few inches to Thomson's negative plates.

The proof that charged atoms migrate to a battery terminal is that hydrogen gas is collected in a chamber directly above one terminal while oxygen gas is collected above the other terminal. If any substantial oxygen gas got into the hydrogen chamber, it would be dangerously explosive, but it's known that pure hydrogen collects there, while pure oxygen collects at the other chamber. Even though both gases rise in water, it's known that at least one of the gases (= individual atoms) can be pulled DOWNWARD in the water (in a funnel) before moving across to the other terminal. DEFINITELY, gas atoms are attracted (and probably repelled) by battery terminals.

Thomson knew that having too much gas in the chamber wouldn't allow the electron beam to form, and he also knew that not having the gas thin enough, he couldn't get the beam to bend. Yet he didn't evacuate the entire gas. Why not? Why wouldn't a vacuum work for him? I've just asked google why the gas is needed, and we get this: "The cathode ray tube is evacuated to a low pressure so that the electrons can move freely through the tube. In the presence of normal pressure, the electrons are obstructed by the air molecules."

They make it sound as though the only purpose of a thin gas is to let electrons fly as trouble-free as possible, and yet a perfect vacuum would do much better for both a trouble-free flight, and a faster flight. The statement above is like a motorcycle racer saying: "I'm so smart I put only a few nails on the road only to make it safer." But stupid, how about zero nails?

One guy in his youtube video, "DEMO: Electron Beam in a Magnetic Field", says that the cathode tube "also has some background gas that ionizes to keep everything stable and it doesn't charge up too much." Apparently, that's the idea that was set into his noggin, which makes no sense. Apparently, everyone is being lied to about the purpose of the thin gas. How can positive atoms and higher-density electrons in the tube "stabilize" things, and how can he say that the thin gas keeps the charge level down when ionization is exactly an extra, charged environment that only complicates the path of the electron beam?

If electron speed is their game, why not use a vacuum?????????? Why am I reading that they prefer to use a thin hydrogen gas in the tube??? WHY NOT A VACUUM??????????????????????

I maybe can explain: hydrogen atoms have more captured electrons (largest electron atmosphere) than any other gas, meaning hydrogen atoms allow for more chaos in the tube, slowing the beam more efficiently than smaller atoms. It seems that the voltage necessary to get electrons to fly at all across the tube gets them to fly too fast, wherefore they need to be slowed some. Only a small fraction of the voltage level used for the tube can get electrons to fly as sparks, but in the tube, they need the electrons to fly a good distance.

Why haven't other scientists debunked Thomson? I assume it's because they are altogether too willing to play goofball. Many of them love the excitement of fantastic quests of science. They love their beloved fathers of science struggling at the oars with invisible bitties, exposing their hide-outs, conquering them. Nobody in the establishment is permitted to topple the everlasting statues of these semi-gods without repercussions. And the evolutionists make sure they make renewed shrines to these fathers lest anyone dare attack them.

Although the beam is made of countless electrons, one can treat it as one electron because each one gets treated the same by the field. Each electron gets the same curvature, one after the other. Thomson would at times use only the electric field, sometimes only the magnetic field, and sometimes both. The trick for finding the charge or mass of the electron is to find how much the curve changes per doubling or quadrupling of the magnetic force. It can be used in math, but only if one finds the speed of the electron, for speed counteracts the magnetic force. That is, you need more magnetic force to overcome added electron speed.

PLUS, you can't get the correct calculation if positive atoms are hiding at the negative plate laughing at you. They would have laughed hardest at Thomson because he wore a straight face. As he found the spread of electrons to be a huge fraction of the speed of light, I already know that his calculations were bogus.

Thomson's Mathology

From all of this, can you yet see what "charge to mass ratio" means? I cannot. Such a phrase seems to imply that mass and charge are proportional, the sort of magical hoopla that only the goofs would devise for each other's ongoing entertainment. I'll need to view more videos to figure out what they meant by that phrase, but it gives the impression that it befits the multi-protonic atomic model wherein all atoms are made of the same type of mass i.e. different combinations of the same protons.

In reality, where all atoms can be expected to have protons each with a different charge level, how is charge related to mass at all? Couldn't God create charge level not proportional to the proton's mass? But Thomson was very apparently seeking to advance the multi-proton model of the atom.

Here's Thomson's result: "Where, m = mass of an electron in kg = 9.10938356 10-31 kilograms[,] e = magnitude of the charge of an electron in coulombs = 1.602 x 10-19 coulombs." That latter figure is 10 to the power of 19. I can tell you immediately that Thomson was wrong because he calculated the WEIGHT of an electron, using his gadget under discussion, to figure its mass, but the electron has no weight. The positive atoms were not the only bitties laughing at him.

He either allowed his gadget to deceive him into thinking that the electron had weight, or he used his gadget deceptively to arrive to the charge of the electron in ways nobody in the world would/could seek to test. His math alone looks too monstrous to approach. Plus, I've looked at his math formula, but it makes no sense.

As the charge of his electron was "found" to be exactly the charge of one hydrogen nucleus, which is one proton in the view of the goofs, that's how the goofs arranged one proton per electron in every atom! Thomson must have been part of the hijacking conspirators. In fact, Rutherford was one of his "classmates." They probably got together on full moons to drink blood (kidding).

We read: "The specific charge of an object is its charge divided by its mass." It means that they view charge in direct relation to the mass of the particle, idiotic. This is like Newtonian gravity, erroneous, where every atom has a graviton so that gravity force is equal to the volume of mass. It's just plain wrong, and can be shown wrong where electrons are repelled by gravity (i.e. gravity is not sourced in atoms, but in electrons).

Online: "The charge of an electron is -1.610-19 and the charge of proton is +1.610-19 coulomb (C).". It hasn't changed since Thomson got his figure. It suggests that nobody else was able to figure the charge / mass of an electron, unless he/she too rigged an experiment to prove Thomson correct. Nobody else has been able to change the charge of an electron because it's impossible to figure, but Thomson got it because it was needed to prop up the multi-proton atomic Frankenstein. That's what this situation looks like. If anyone else was able to find the mass of the electron, it couldn't be Thomson's figure because he sorely misinterpreted the things going on in his gadget.

Can you see how necessary Thomson's number is to "prove" that each proton has only one electron? Can you see how rigged it is that he merely "stumbled" on an electron having the same charge as a hydrogen nucleus?

The speed of the electrons resist bending through a field, but the electrons are both pulled and pushed in the same direction (when one of the fields is on and the other off). Herein is the conflict: the force to curve the path versus the forward force to resist curvature. The forward force isn't alone; it has a partner. The forward force is speed x time. The wider the plates, the longer any electron is acted upon. Thomson did include both the time and length of the field by adding the following sub-formula to his math: time = length of field / velocity through the field.

However, that bit of math only gave him the time it took for any one electron to pass the field, providing he correctly had the velocity. It didn't tell him how much curvature the electron was undergoing per unit time, or over the whole time within the field. To find that, he needed the acceleration of the curvature over the whole time in the field. Let me explain this acceleration because it's easy to get lost here. It's not longitudinal acceleration, but how fast the curve curves more than it did before.

As one electron passes the field from left to right, all magnet atoms, in both plates, move the atom toward one plate in the first unit of time. While MOVING ALREADY, all magnet atoms move it more in the second unit time...which amounts to an acceleration of the curvature. By "one atom," I mean one on the positive plate, and its counterpart on the negative plate (both are curving the electron).

But that's not all. The electron is first moved before it reaches the plates, for the plates act at a distance. But that's not all, for the magnetic power is less in the first unit of time than in the second unit of time, and even more in the third unit of time, etc., until the electron reaches the middle of the plates, where the magnetic force is strongest. Then, while going forward further, the magnetic force decreased per unit of time. Did Thomson reckon with this factor?

Thomson plugged the acceleration of curvature into his math with this sub-formula: electromagnetic force on the electron / mass of electron. He knows neither of the two, and needs to solve other things in other sub-formula's to find them both. In order to solve this sub-formula, he also needs to express the electromagnetic force on the electron in other terms that are found in other places of the whole formula. This sub-formula above only tells that the acceleration on any particle is twice as much when there is half as much mass. It doesn't tell him anything of what the level of acceleration was for the electron. Where any of his numbers in the full formula are wrong, especially in the electron's velocity, ditto for a number(s) in his acceleration sub-formula...the key part of the formula because it defines the shape of the beam's curvature.

I can now show how his sub-formula for acceleration, electromagnetic force on the electron / mass of electron, is incomplete or even misleading. In this formula, substitute electromagnetic force on electron with gravity force (on any object), because electromagnetic force and gravity act the same for creating acceleration (same level of acceleration for both). Yet the formula for finding acceleration due to gravity is: gravity x mass / radius of travel squared. Therefore, his sub-formula for finding the acceleration of the electron ought to be: electromagnetic force on electron x mass of electron / radius of travel squared. Why isn't it that? Why does he have force DIVIDED by mass whereas gravitational acceleration has force TIMES mass? The difference could make a big difference in the final results.

Most readers won't want to spend two seconds at the page where I'm getting these formula's, but for anyone interested in delving, it's here:

Next, I can show another error in Thomson's formula, where he substitutes his sub-formula for acceleration above with: charge of the electron x strength of the electric field / mass. He therefore viewed his magnetic force on the electron the same as the strength of the electric field upon the electron, which isn't a big change in terms except that we can now discern a problem, an omission: he didn't have differences in his strength of the electric field, which he denoted with, E. He always used, E, yet the strength of the field changes upon the atom as if crosses the field.

Just a reminder in case you got confused: the acceleration under discussion is not for the electron beam's forward direction, but for curvature of one and each electron, meaning the curved path gets more curvey per unit distance in the forward direction. The curve starts slight, then gets more curvey fast. The curve doesn't stay the same. Thompson could see the basic shape of the curve (per session) because he knew the original path was down the middle of the tube, and he could also see the final landing spot of the electrons shortly after leaving the field, but he had to explain this amount of curvature by the unknown magnetic force (charge) that the electron offered. That is, how much electron charge could explain such-and-such a curve, per session?

Each unit of time across the field can be one magnet atom wide. Each magnet atom continues to accelerate the curved path, and the strength of the magnet is not as strong when the electron first enters it as when the electron is in the middle of it. Therefore, Thomson used (for his math) the full strength of the magnet (whatever was on his dial) for the entire path of the electron, and that resulted in an incorrect charge for the electron. I'm not yet sure as I write, but this can maybe go a long way in explaining how he got a charge for this wee-wee particle that is as great as big-daddy proton.

The page above shows that his E is equal to: voltage applied / distance between two plates, which does not recognize a change in voltage applied across the lengths of the plates. This sub-formula, still part of his quest for the acceleration of electron curvature, simply says that the voltage applied is reduced with greater distance between the magnetic plates. Dandy, but again, no matter what voltage was applied to form the specific strength of the electromagnetism, this strength acted more weakly to both sides of the plates than half way across them.

He needed to change E to the middle ground between E and zero, because the force on the electron starts at the first number after zero, and E (the full voltage power) acted only half way across the plates. His E needs to be 1/2 of E, and if you study his acceleration formula, you'll see that 1/2 of E results in half the acceleration as compared to what E would find. I have no idea how half the acceleration changes the curve, nor can I off-hand tell whether it will increase or decrease the math result for the electron's charge. I don't know whether half the acceleration will cause the electrons to strike half as high at Thomsons striker material.

On the same page above, there is a zero beneath his sub-formulas for acceleration, where he apparently changed the acceleration to an average velocity (toward the positive plate) called, velocity-y, to distinguish from velocity-x, the latter being the longitudinal velocity through the field. I don't know why this zero exists, but it doesn't appear to relate to the need to change E to 1/2 of E, for the zero does nothing in the formula. Where this zero exists, the only things that are added to it is his acceleration rate combined with time spent within the field, or total acceleration.

The next part of his formula is over my head to verify, where he finds an angle using both the upward and longitudinal velocities of each electron. He's apparently converting the curved path to a straight-line path to make further calculation possible. If he goofed here, he throws the whole out. The very bottom of this page defines the "specific charge" of his electron as charge / mass.

The bottom of the page also has a link to the next page, which claims that they were shooting electrons through gold foil. This page starts with this clue that Thomson was way off-track of reality: "In his experiment, J.J. Thomson had found a charged particle that had a specific charge two thousand times greater than that of the hydrogen ion, the lightest particle known in 1897. Once the charge on the particles was measured he could say with certainty that they were two thousand times lighter than hydrogen." It's malarkey. He was partnering with the impositions of the goofballs who had a multi-protonic core. I don't know what his part was in establishing that quackery, but it looks like he was one of the court jesters, in the least.

It adds to the evidence of a conspiracy to seat the orbiting electron as king, though Thomson didn't front the orbiting electron. It seems that the conspirators had differences, competing on the fine points of a theory having a multi-proton foundation. It had to be a conspiracy, to produce mainly empty space between atoms, because grown men can't be so stupid as to think that one big-daddy proton attracts only one electron. Just go ahead and picture a watermelon beside a pea. The watermelon can only attract one pea. Were they nuts? Or were they evil?

The next page again says: "Find the velocity the cathode rays by balancing the electric and the magnetic field. When the deflection angle is zero, the two fields are of equal strength, and the cathode rays' velocity can be found." I understand that. He played around with voltage levels until the electron beam went straight ahead though the field. I suppose that Thomson needed to know what "voltage" his magnet (two ends) had in the positions that he arranged for them, for while he had a way to measure volts for his electric field, magnets don't come with a dial or optional volts. I can't see how velocity could be found by balancing the two magnetic systems, but this page has all the math that, maybe, I can understand;

The following on the page above troubles me: "The force (F) on a charged object [the electron] in a magnetic field depends on the strength of the magnetic field (B), multiplied by both the charge (q) and velocity (v) of the object. F = Bqv" What does the longitudinal velocity of the electron have to do with force applied to it horizontally toward the magnetic plate(s)? I've gone over and over this, but no light bulb turns on in my head. The velocity of the electron only determines how much time the magnetic force can be applied to it, not the level of its force. Is this a trick, therefore? Didn't Thomson understand this? Did his fellow goofs mislead him with one of their erroneous ideas?

The formula is saying that, the faster the electron or any particle through a magnetic field, the less force that's applied. The only way this can be true is when seeking TOTAL force over the entire path, not specific force at any one time. There is a different total force level on the electron with changing speeds because changing speeds changes the duration of the path, but if this is what the formula was trying to nail down, it would include "time" too. But there's no time in this formula. Therefore, Thompson may have goofed here for acquiring the velocity.

I'm going to repeat the statement above immediately after quoting from the same page with a similar quote that's driving me hairy:

The force (F) on a charged object in an electric field depends on the strength of the electric field (E), multiplied by the charge (q) on the object. F = Eq

The force (F) on a charged object in a magnetic field depends on the strength of the magnetic field (B), multiplied by both the charge (q) and velocity (v) of the object. F = Bqv

The two quotes are not only almost identical, but the two scenarios they describe are perhaps exactly identical, yet the second scenario for the MAGNETIC FIELD claims to require a velocity factor while the first scenario for an ELECTRIC FIELD has no velocity factor. Both fields are created by electromagnetic plates. The magnetic field, the mechanics of which are not easy to understand as it affects the electron beam, ultimately pushes the electron beam, same as does the negative plate of the electric field.

Apparently, the fathers of the empty-space model of the atom decided that the magnetic field gets INTO THE TRAFFIC of the electron beam, to DEFLECT it, for which Thomson thought he could enter the velocity factor when calculating the magnetic field's force on the beam. But in the video below, showing a magnet changing the direction of a cathode ray, it seems that the magnet only pushes the ray up or down, not necessarily getting into the traffic to deflect it, but same as when the negative plate of the electric field pushes it:

If Thomson was wrong about the magnetic field getting in front of the beam, then his whole project in determining the charge of the electron is ruined. He was basing his project on the velocity of the beam, and thought it could be discovered thanks to the effects of the magnetic field. I don't think anyone understand how a magnet works in the mechanical sense. We see what it does, but we don't know how it does it.

Thomson thought that the electric field did not alter the velocity of the beam. Just in case you're wondering, the discussion above on the acceleration of the curve of the beam was in an electric field.

The problem I'm having is in how Thomson dealt with the velocity factor in his magnetic field. The page where the sub-formulas above were taken claims that the two formulas are equal in their assault against the electrons. Both fields exert the same force on the electrons in his arrangement, which we can't dispute, yet he claims that the magnetic field assaults, or goes against, electron velocity to alter its path so that its straightens out what the electric field bent.

I need to get that point across. He first curved the beam with one of the two fields, so that the curvature was either upward with the electric field, or downward with the magnetic field. Let's assume he first made it curve downward with the magnet. He then adjusted the voltage to the electric field until he ironed out the full curve and made the beam a straight-through one. It's hard to argue against this being a balanced situation in which both fields are doing the same work on the electrons.

However, not only may he have been wrong in assuming that the magnetic field gets into the path of the electrons, but the negative plate of the electric field was covered in positive atoms which he arranged with the gas in his tube. Yet the assault (or pull effect) of these atoms are not a part of his formula.

Imagine a flat piece of wood getting in front of water out of a hose such that it deflects a water flow at 45-degrees upward to a horizontal flow. This is about all that the magnetic field does to the electron flow. You don't say that the piece of wood is acting against the velocity of the water to describe the deflection because you're not cancelling all the velocity. The latter is not irrelevant, but how do you use it, as a number, in the formula, if it's not all canceled?

The better way to describe the situation is: how much force from the piece of wood was used to curb the flow from angle A to angle B? Thomson was feverish, it appears, to use velocity in his formula, because he needed to find the velocity of the electron stream, but could not find it. He instead invented a means that was not true to the situation. That's what it looks like to me.

When he had the beam balanced in the straight-through flow, with both fields applying the same force, he described it as a combination of F = Eq, and F = Bqv. The latter letter is the velocity of the magnetic-field situation. As both formulas begin with F on the one side, he broke it down to Eq = Bqv, and from that he got v = E/B (q cancels q, ignore both)...of which I haven't got the ability to verify. But what it maintains is that the force (E) of the electromagnet is NOT EQUAL equal to the force (B) of the magnet, but rather E is larger than B, and B needs to be combined with the velocity of the electron for both to equal E. If you can understand that, you must be a lunatic, the same sort of lunatic that can understand space-time.

The page ends by saying of v = E/B: "The velocity was equal to the electric field strength divided by the magnetic field strength [B]. Thomson could measure these field strengths and use them to calculate the velocity of the rays."

What he means by, v = E/B, when taken from Eq = Bqv, is that the velocity is equal to whatever B lacks to make E. So, for example, if E is twice as strong as B, then he makes the velocity twice as strong as B. It troubles me, I cannot grasp it. If the electromagnets are twice as strong as the magnetic field, twice the velocity of electrons is needed to make up for the weakness of the magnet in comparison to the electromagnets. How can this make any sense? He's got v proportional to E. If E is three times stronger than B, v is three times faster.

I have no confidence in this method for finding the speed of the electrons.

A youtube video, "Cathode Ray Tube Experiment and Charge To Mass Ratio of an Electron," says that the velocity of the beam is proportional to the magnetic force, which doesn't help to understand the mechanics of what he's describing. It apparently means that, when the magnetic force is increased by .3, or by 100-percent, for examples, the velocity will be .3 faster, or 100-percent faster. I still cannot understand that.

Perhaps it's to be taken as the magnet utilized first, before the electromagnets are turned on. In that picture, the beam first comes straight though without anything to curve it. All the velocity is there. Then the magnet is put in place with its two ends, one on either side of the beam, which curve the beam downward...but how does this relate to velocity? I assume that both ends of the magnet contribute to the downward curve, and it's possibly done by push-down power without the force getting into the path of the electrons.

The pages I've been quoting from turn out disappointing in not showing how Thomson finally calculated the velocity, or what velocity figure he calculated, nor do they show how he finally calculated the mass or charge of the electrons.

If I recall from one video, Thomson thinks the electron was moving at something like 60,000 miles per second. Where in tarnation did he get that? Today, the claim at google is that the electron in a cathode ray is one tenth the speed of light (19,000 miles per second), way to high for reality. Does anyone in his right mind think that merely a wee-wee electron coming out of a hot piece of steel, by the coaxing of a mere anode, flies almost once around the earth in one second? Give your heads a hang of shame. If that were true, Thomson would have needed an electron-proof vest to keep from getting plugged full of holes.

I've found the youtube video with speaker saying, "Thomson found speeds up to 60,000 miles per hour, or up to a third of the speed of light." She apparently meant, 60,000 miles per second. It's at the 5th minute of the video title: "JJ Thomson Cathode Ray Tube Experiment: the Discovery of the Electron" (I played it at 3/4 speed because she talks too fast; she only gives the basics.)

If correct that he calculated it at 60,000 mps, who changed it to 19,000 without also changing the charge of the electron? Is this just a minor question? Thomson uses his electron-velocity figure to find the charge on the electron. No, it's not a minor question.

It was through the use of the electric field that he claimed to find the weight of the electron, which he gave as its mass, but as it weighs nothing, he must have been imagining things when discovering its mass. I've yet to come across a video/article telling how he discovered its weight, though the video above gives some brief mention and math without explanations. It says that he needed to know the electron's velocity before he could find the mass, meaning he was way-wrong on the mass by at least as much as he was way-wrong with the velocity. <> You need to be logical. The electron is just a bittie thing. The electron in his heated cathode was almost spilling out on account of the voltage alone. One video tells he was using only 15,000 volts at the cathode, which only makes one heck of a painful spark if you put a finger near it. It's not exactly a nuclear bomb. The anode gave the electron a little help, and out it came, free from protonic attraction. Do you think that such a small thing, flying out almost wholly due to its MERE REPULSION FORCE against other MERE electrons on a MERE atom, and with the little tug from a mild-mannered anode, could screech at one-tenth the speed of light? Give your head a shameful face if you believe this.

Let's explain it in more detail. The electrons flew out from the outer realm of metal atoms at the surface of the metal. The electrons that flew out were smack on the balcony overlooking the gas of the tube, almost all pure-empty space. There wasn't even any railings on the balcony. They were just about falling off. They were lucky because their next-door-neighbor electrons were trying to push them off with only so much push-power. With the proton that held them on the balcony by an elastic thread being half-way across town, the anode could lasso them with a rope and just pull with one hand, not even the whole arm. Do you think that, when the thread snapped, their next-door neighbors had enough push-power to make them go 19,000 miles in ONE SECOND? From the United States to Europe over the Pacific ocean? In one second? Them bittie things????? Them little tiny nothins?

Thomson! They can't even get past a couple of inches of normal air.

Big Discovery on Spectral Lines

There is no doubt in my mind that all electron ejections from cathodes also form light waves. It's impossible not to. Thomson was maybe measuring fast speeds due to the instantaneous light waves in his tube. I don't know how he got those fast speeds, so I can't comment but to say, it wasn't the electrons going that fast. In such a small tube, electrons at 100 mph would be as-instant, to the eye, from cathode to field.

His electron beams were known to light up the gas atoms that the electrons collided with, because light is caused from the emission / jolting of electrons from/upon atoms. Do you think that those gas atoms had photons on the ready? Like, uh, maybe the atoms had the photons in their torpedo launchers, but didn't want to shoot them out until they themselves got bombarded by electrons? The fact is, atoms don't have photons, but rather, as soon as the electrons of the gas atoms were jilted off, or jolted around while still on the atoms, light waves were formed through the electron aether in the tube.

You can't teach photons unless you can explain where they come from when electrons fly in the tube. They are not in the tube when there's no voltage to the cathode. They are not in the cathode when there's no voltage to the cathode. You can send voltage to the cathode in the dark, and still there's no photons, and you can let the electrons fly in a dark room where there are no photons, and then bango, as soon as they fly, light appears in the gas. Duh, I wonder what happened.

I just proved that light is a wave through an electron aether when electrons strike the aether electrons. It's the only logical explanation. And this is by far not the only proof.

Sooner or later, while messing around with electricity, it was going to happen. Somebody was going to learn that electricity shoots out some light-forming material, and it was Thomson who realized, thanks to his work with cathodes, that no matter what kind of steel he used for the cathode, the same beam came forth having the very same properties when his magnets had the same settings. He realized all atoms had the same material, learned that it had an electromagnetic charge, and so the electromagnetic atom was suddenly in the cradle.

The video above with female speaker tells that Hertz could not get his beam to bend by using the same gadget (cathode-ray tube) that Thomson used. Apparently, Hertz was trying to copy or test Thomson's experiment. The video said that Thomson tackled Hertz's problem and found that Hertz had too much air in his tube, but that when the gas was thinned, the beam did bend for Thomson.

Hertz did have light shining on his detection surface at the end of the tube, but, no matter how much magnetic force he provided between the beam, the light remained at the center of the tube i.e. would not curve.

However, that doesn't make sense because, if the beam can get through the magnetic field to the detection surface, it should also bend. The only apparent explanation is that Hertz had only light going through his set-up. Hertz's air may have caused the electron beam to come to a stop early in its trip, due to striking too many air atoms, but light waves (which cannot be pulled by magnetic force) continued through the magnetic field to his light-detecting surface.

Then, when Thomson got the beam to bend with fewer atoms in his gas, he succeeded in making the beam reach across the field, which caused the beam to rip through atoms in the field, creating positive atoms which migrated to the negative plate(s), then regathered their electrons from the aether, and rejoined their gaseous pals, on-the-ready for another electrifying blast.

If the new abundance of free electrons in the gas caused the unstruck gas atoms to take on some negative charge, it wasn't equal to the charge of struck, positive atoms because most of the freed electrons would spread out in the empty space, and its these that crowd around and saddle themselves upon atoms to make them negatively charged. The positive force level of the struck atoms increased with every electron lost, but the negative atoms could not each gain anywhere near one electron for the one lost by the positive atoms...meaning that the positive atoms at a negative plate would far out-muscle any negative atoms at a positive plate, assuring that the electron beam bent harder toward the negative plate than would otherwise have been the case.

This can perhaps explain why the old school wrongly labeled the terminals of the battery. It's widely known that they labeled the negative terminal as the positive, and vice-versa. The true negative is where electrons flow out. As positive gas atoms migrated to the true-negative terminal, and as Thomson and others all figured that the negative electron beam needs to be attracted by the true-positive terminal, perhaps the cathode experiments made the old school realize that they had the terminals tagged backward. In those days, as it is to this day, they hooked up the "positive" terminal to the cathode...because it's really the negative terminal.

Below is the history shortly before Thomson, when light in cathode tubes was a rave. Nobody disputes that the light is from struck atoms. The pioneers began to realize that "corpuscles" (bullet-like) formed light, and Thomson's erroneous velocity of the electron may have conceived the soon-to-be-king, the impossible photon, an imposter.

The leading pioneers were either unable to see that the light was formed from the electrons exiting atoms, or some ringleaders amongst them forbade them to think along those lines. They knew that the fast electron beams caused the light, but it's as though they couldn't get it into their heads that the fast electrons caused the light by knocking the electrons of atoms into the aether in the tubes. It was such a shame because they already believed in an aether in those days, and so Thomson, of all people, should have proclaimed that free electrons in his tubes were the aether. He would have been super-famous for such a discovery, in time to sink the photon.

Late in the last update, I began on spectral light as science sees it from stars. I told that one could destroy their red-shift / blue-shift theory which they use partially to decide on the distance of stars / galaxies. Spectral light (formed through a prism) always comes with black lines, they assure us, and every light shining through a particular gas always comes with the gas' "signature," the same arrangement of thin, black lines spread across the rainbow of a prism. Each gas has the same set of black lines.

The big question is how these black lines form in the glass prism, and I'm after an alternative explanation because I disagree that the rainbow exists in light prior to entering the prism. I say the glass produces the rainbow, and offered some evidence in the last update. I've come up with an explanation since then. But first see the second minute of this video where there are two kinds of spectral lines per spectral light source:

The first type of spectral lines per gas has black lines, and, they say, the same lines appear but with colors, this time, when they let light reflect from a gas into the prism (without a light source behind the gas). The same lines are black with going into the prism from a light source behind the gas i.e. it penetrates the gas first, then enters the prism. From this, it seems very certain that the gas produces the lines, and from that it's got a high probability that the lines are formed from the gas atoms as opposed to something else within the gas.

Their theory is that atoms absorb certain "frequencies" of light so that whatever is absorbed goes on to become black in the prism, which is incomprehensible to me on a mechanical level. Then, oops, the same light has colors when put through the prism from reflected light. That makes no sense and already busts their light mechanics. Besides, I don't think light frequency has anything to do with these lines.

The first order of business here is to show the difference of the light. In the case of the black lines, the light waves penetrate around the perimeters of atoms, atom-to-atom and across the space (aether) between them, and, in the case of transparent gases, most of these waves come out of each atom before entering the prism. A small portion of the wave energy is absorbed by the atoms, explaining why light dims, from white to yellow to red to invisible, as it goes though more gas. Light never dims through the violet > blue > green colors.

The sun never takes on those latter colors as it approaches the horizon, suggesting that they do not exist in sunlight. Or, at least, someone needs to explain why the sun never shines with those colors if indeed they pre-exist in sunlight.

In the case of colored spectral lines (emission spectrum), there's at least two possible explanations: 1) scattered light is reflecting from gas atoms; 2) the light waves first absorbed by the gas atoms are reflecting out. The latter does seem like a potent argument in favor of rainbow colors all pre-existing in white light. However, I would argue that the black lines are not due to the absence of colors though the prism, but due to having light too strong for the eye to register as light.

The star gazers think that, because gas atoms absorb some light, these are the pre-existing colors that don't arrive to the prism, and so they form the black lines. But they know that the same atoms are reflecting their absorbed light, wherefore what ails them that they maintain zero/weak light to explain the black lines? The same atoms that they say do not put colored light through the prism in fact do put colored light into the prism, and they admit it in their so-called "emission spectrum."

Therefore, I do declare, the black lines of the so-called "absorption spectrum" are strong-light waves, a combination of both their colored light and the white light coming through the gas. The total spit-power, as I called it in the last update, amounts to invisible light, which appears black to the eye. "Absorption" spectrum is wrongly named, therefore, because it implies that there's no light at the black lines.

The 1) option above can be differentiated from the 2) type white light passing through a gas, entering atoms and jiggling (exciting) their electrons, and coming back out, due to the jiggling, in all directions as colored light.

The 1) option is white light first bouncing off of gas atoms before getting absorbed within them, same as white light bounces off of glass without entering through the glass. Some white light apparently strikes the hard center of the glass atom, and bounces like a ball off of a wall. When this light re-strikes a gas atom, it can, this time, enter to become absorbed as the 2) type light.

What the experts are telling us is that type 2) reflected light, from gas atoms, comes with distinct and spaced-out, spectral colors of various colors. Yet, from what we can gather in all objects we see, all atoms jiggle in one color only. Your green wall doesn't come with spectral lines from its atoms, yet the light experts tell us that all atoms, wall (or paint) atoms included, absorb colors when reflecting light. Why, therefore, is it only light from gases that produce the black or colored spectral lines?

How is the light from gas atoms different from the light of solid objects? Why doesn't light from (cold) solid objects do what gas atoms do if atoms of solid objects absorb various colors and reflect the rest? Is the latter even true? I don't think so. Absorbed light must come out; it doesn't just disappear in gas atoms. If it slips off of an atom into the aether, it then re-enters other atoms as one of two different kinds of scattered light.

Why couldn't it be possible that one atom, upon receiving white light, mixes up all of its various waves, of various strengths, to produce wholly different strengths that come out combined as one color to the eye? When we are dealing with gas atoms, I don't think "frequency" is the correct term to describe colors, but rather "strength" is much better.

Atoms struck by light get tumultuous, we may say, sending out waves of all strengths, each of which becomes a different color in the prism, and each of which could become a different color in the eye than in the prism. I'm proposing that light, before entering the prism, has various spit-power strengths each registering as color to the eye that the latter sees differently after the spit-powers are modified in the prism.

I would argue similarly to light experts that mixing of various "frequencies" (their word) inside the atom is the very definition of absorption of various "colors," but I say that those colors are spit-power instead, and that they vanish once mixed in the atom. Electrons have no color. If you mix them up, they still have no color. Only when they send out new waves do they come with a color for the eye to see. It's possible that the goofs have a trick / faked experiment or two to "prove" otherwise if they need their view badly enough, and they do need it badly enough with red-shift cosmology.

The goofs agree that light falling on atoms cause the electrons to deflect light. And I can now argue that they chose against an atom covered in countless electrons because such a thing mixes colors and shoots them out again in different colors. The task may have been given to Einstein to "prove" that whatever color going into an atom is the same color coming back out, but one should take a close look at such a claim, and the experiment used to back it up. It doesn't work because the outer atom is an atmosphere, or a sea.

When the source light produces a certain level / characteristic of chaotic jiggling action from the full combination of its received light, the jiggling comes with outward jolts which REPRODUCE (like sons and daughters having different mixed genes) different spit-power strengths because it doesn't seem logical that light splashing into an electron atmosphere will result in forthcoming jolts all having exactly the same spit-power force. There is no need for photons of different colors here, but only jolts of various spit-power strengths.

How can their photons produce different spit-powers if they all have the same speed? Jolts produce more spit-power with different jolt speeds, and each jolt is from one electron only, not more than one that would them look like a wave of water having many molecules. One jolt, one jolt speed, that is all (aside from what God may have added to the scene that we know nothing about). Exchange the photon for an electron jolt, and light mechanics will become sane. Do it, leave the goofs behind to suffer their fate. Jesus is coming with a new science. Start now to grasp it.

The more speed a jolt has, the higher it can jolt, though not necessarily. This is akin to a photon coming from a higher electron orbit, but there can be no orbit. Leave that idea to the goofy past of wicked magicians. An electron jolt can be straight up, or horizontal, or anything in between (just common sense). A jolt is a weak "ejection" that does not leave the atom. Full ejections, per atom type, have the stronger spit-powers. Full ejections cause rainbows in the prism, and jolts cause spectral light.

The reality here depends on the mechanics of the landing of light into the atom. A solid ball landing in water will remain a solid ball, but what happens to a ball of water (with no balloon skin) landing in water? Disappeared. A photon will remain a photon when landing in the atom, and this may be a chief reason that the goofs opted for the photon instead of the electron atmosphere, so that they could claim that light color doesn't change when striking the atom.

The goofs define "frequency" of a photon, but you need more than one photon to have frequency. It is STUPID to say that one photon has frequency. Then they define their frequency as the "energy level," which is my split-power level. They have red and blue photons, I have red and blue jolts. Which of the two models seems more like reality to you?

If a ball of water strikes water, it's energy gets dispersed into the water particles. The ball vanishes, unable to bounce out of, or leave, the water as it had entered. One may argue that its energy will eventually come out of the water to the air, but it won't be a ball of water flying out. That's how I view light landing in the sea surface that is the outer atom. White source light lands with both different ejection spit-powers and different jolt spit-powers, and when they reproduce spit-powers from transparent gas atoms as sons and daughters, they are weaker than the spit-powers in the light source because gas atoms absorb a small portion of source light only. Most of the light through a gas goes straight through; ask the crisp circle shape seen of the sun about that if you don't believe it.

I'll agree that sunlight has different waves with different frequencies, but when this light enters atoms, it's a whole new ball game. The frequencies are no longer, but mix into one chaotic soup. Plus, "frequency" here means the number of straight-through waves per second, and may even be irrelevant in creating colors because the thing that creates color, I feel sure, is the strength of the wave. Gas atoms don't create straight-through waves and so they don't form literal frequency. Frequency of waves needs straight-through waves like upon one track. High-frequency waves may cause a blur in the eye rather than colors.

If you shoot a million drops of water into water so that a half-million drops splash out of the water, the drops that splash out are not the original drops. The other half-million will just become part of the water body as absorbed drops, and they create water waves instead. In the atom, the powerful out-splashing of "drops" become color, and the mild waves probably end up as invisible infrared, though all light absorbed by atoms contributes to the total jiggling and jolting. It seems like a no-brainer, I'm not trying to appear genius here, just emphasizing the truer atomic model using different illustrations.

"Jiggle" is not necessarily the best term to use. Deeper electrons that are held tightly enough by the proton will do literal jiggling, but outer-surface electrons are more free to move around in currents that can reach up to different heights and splash down again. That sounds just like waves of a lake / ocean, but produced by the "winds" of passing light waves.

The up-and-down motion is best described as, jolting, like four-foot wave chops on a lake, or ten-foot chops, depending on how much source-light energy a gas atom can take on, but also depending on how tightly or loosely different atom cores hold their electrons.

Situation: the same source light lands on green paint as on yellow paint. The two paints are made of a different kind of atom. It seems logical that one or the other, green paint or yellow, holds electrons tighter to the proton so that they don't both send out the same sorts of color-forming jolts. My tentative claim is that, in the prism, yellow gets more wave-train density and weaker spit-power, which can suggest that yellow paint has the weaker-held electrons i.e. with more jolts per square area, but weaker jolts as a result.

One way to provide more weakly-held electrons is where electron atmospheres are larger, which I would interpret as stronger protons, which at first glance seems a contradiction to the above with electrons held less tightly. I see stronger protons accumulating more electrons so that each general region, from top down, is thicker so that the outer layers are thicker in weakly-held electrons, and the inner layers are thicker in strongly-held electrons. The stronger proton obviously holds the inner electrons more strongly.

There should be an entire science field in physics predicting the various mechanics of atoms depending on proton size and strength, but the goofs fail to understand that atoms have electron atmospheres. Proton size, strength and shape can result is a large assortment of changes in behaviors from atom to atom.

As the violet-blue-green side of the prism gets more-compressed electrons, as is my claim at the moment, it can jibe with the above in which it suggests that protons of green paint/objects hold electrons more strongly than yellow. In this picture, especially as mixed paints jibe with the color sequence in a prism (e.g. blue and yellow paint mixed = middle-ground green), it can reveal that violet and blue paints have the tightest electrons with lower jolts when excited, and also more stringent, if that's the right word. More rigid could be better.

Just think about it. Blue paint has shallower jolts than green, and green shallower jolts that yellow, and when we mix yellow with blue paint, we get the average jolt level of green. In the mixed paints, the eye sees HALF blue jolts and HALF yellow jolts, and registers them in the brain as the same color as when it see ALL green jolts. Why? Apparently, it's simply because green objects have a mix of jolt levels that are roughly half blue and half green. Or, atoms having one-third violet jolts, two-thirds blue jolts, and half yellow jolts register as the same color as atoms having all green jolts by some other mix such as one-third orange, two-thirds yellow and half blue.

I would not interpret any color out of the prism (landing on a white wall) as a mix of its neighboring colors, however. I would interpret them as pure colors that happen to jiggle the electrons of wall atoms with the same total force as yellow atoms from various colors mixes due to a mix of diffeent types of atoms (= different products). Paints are made of different products to get different colors.

White light doesn't have green until it lands into atoms that produce it. It's even debatable as to whether white light has yellow or red within it. A light bulb's filament can glow yellow or red when colder, but it's debatable whether those colors exist in the filament when the voltage is turned up to make brilliant white. The jolting that produced the red and yellow light become stronger jolts with more voltage pumped into the filament. If no red or yellow light is made by the filament, then it can't enter prism. The white light into a prism needs to create red and yellow from scratch.

The spectral lines from gases may be able to reveal whether colored light pre-exists in white light. Already, I can tell you that light entering a gas atom is remixed, no doubt about it. Pre-existing light does not enter orbits of differing heights. So far as I know, that's all that the goofs have to explain colors from atoms, differing orbital heights. The electrons throw out color from different orbital heights. Does anyone even really want to know about how this impossible picture works? If light is absorbed so that it can't go into a prism, where does it go when an orbiting electron catches it? If the electron eats it, doesn't it need to come out somewhere? It's easy to say that the electron can bounce to a different orbit so that the color gets changed, but it's not easy to explain an orbit, or a change in orbit. It's impossible.

Why should the photon thrown out by an electron come with a specific color just because the electron goes up and down in orbit heights? You can see that the dopes are trying to mimic the mechanics of a jolting electron atmosphere with orbiting electrons, because they are speed freaks. They want the speed so badly, they'll cling dizzy to the orbiting electron and photon partnership to their deaths.

How can the black spectral lines form after a mixing of all source waves in gas atoms? To where do the orbiting electrons take the colors "missing" in the spectral lines? But the colors are not missing there. They want you to think the colors are missing so that they can stick to the concept of pre-exiting colors in white light. Why? Because all photons carry a color, but they don't have a sea of photons in the atom so that they all mix therein and change colors by that method. They say that, whatever colors of photons enter a gas atom, they all come back out again except for the absorbed light.

If there's going to be black lines, shouldn't they be in different parts of the prism each time that light through a gas is put through a prism, if there is a chaotic mixing of light frequencies prior to putting the light through the prism? Yet the black lines, per gas, always appear in the same place within the prism, and ditto for the colored lines of the emission spectrum of the same gas. What is going on? When the black lines are in the midst of all three, the violet, green, and orange rays of the prism rainbow, the colored lines of the emission spectrum are violet, green and orange (same location of the prism) the midst of a black background this time; no rainbow colors are showing, just the colored lines.

I showed above that type 1) light, the bounce-off-wall SCATTERED light, will inevitably become reflected light upon striking and entering gas atoms, and thus becoming absorbed by them. Absorbed light is supposed to become the black lines. Scattered light is essentially identical with source light until its absorbed. Only a portion of scattered light will get absorbed, per atomic strike; the rest remains scattered light and moves on to the next atom it strikes, and there it loses more of its energy as absorbed light, and so on.

Light off the mirror that shows your face is that scattered light barely altered at all (only dimmed in brightness). It becomes type 2) reflected light only when it enters the atoms of your face. Before that, it bounces off the glass like a ball off of a wall.

What might be the difference, in the prism, between the light that was once scattered light in the gas, and the straight-moving source light that enters atoms directly and transfers atom-to-atom through the entire gas, and finally into the prism? Source light cannot travel atom-to-atom without exiting each atom. It moves through the aether toward the next atomic strike, and when it strikes the next atom in its path, it becomes partially scattered light, with only a small portion becoming absorbed as it passes through that next atom. The scattered waves goes about their business of bouncing off of atoms in all directions, whereas the straight-through waves maintain their straight lines to the prism. Can this picture possibly expose the method of producing spectral lines? How much scattered light gets to the prism, and how will it differ when it gets there?

More importantly, how much scattered light gets into a prism in the lab, where they create spectral lines to be compared to the spectral lines of stars? How much scattered starlight gets to the prism? Probably near zero. How much of the prism spectrum do the black lines from stars fill? Lots. It's therefore not looking good to define spectral lines from scattered light, yet this is the light source, they say, for the colored emission spectra.

Stellar spectrums can have an enormous number of lines that can be misinterpreted, I wonder how many tricks the star gazers play when they want something but the lines don't quite give it to them. Do they concoct reasons for accepting the lines, as they want them to be, even though they are not quite or nearly the same set of lines? It could be exactly like how the same evolutionist scum make a whole dinosaur out of a couple of bones without telling the public.

Even distant starlight is white after getting through the air. There is no way that white light from the sun can get to our eyes without ALL of it cutting across the perimeters of air atoms. They way the cosmologists put it, and they may be correct, the colors absorbed by hydrogen atoms at stellar atmospheres explain why these colors are black on the prism. And so they will inevitably show how their electron model can explain the absorption of those colors, and I can bet my life that they are flat wrong with that explanation(s).

I have tried my best to understand how a color of light can be absorbed by a sane atom, such that the same color does not appear on a prism. I can understand how the color doesn't appear if indeed all colors in the source light remain unchangeable in color, but what I can't understand is how the atom can absorb one, two or three colors alone such that it no longer sends out those colors. This is especially difficult where the goofs tell us that scattered light in a gas sends those very colors to the prism that are not there with the straight-through light through the same gas.

Okay, so how can straight-through light absorb the colors while scattered light sends the colors? Impossible, apparently, for they are both white light from the same source. It makes more sense that the straight-through light sends BOTH the "black" and the colored lights. Do you understand me? Scattered light goes in all directions so that the so-called "absorption spectrum," which creates black lines, is also receiving the colored (emission) lines simultaneously. How can that double-whammy situation produce black lines?

Easy: by being too strong of a color for the eye to detect. In that case, I don't need to explain how the gas atoms absorb colors, because they don't. Hurray, let's have some popcorn.

It doesn't explain, however, why gas atoms send colored lines, to the SAME PLACES only, of a prism spectrum. But first thing's first, let';s nail down what I just said in case some readers missed it. To put it another way, there are two light sources in a gas, the straight-through normal light, and the scattered light. The straight-through forms the rainbow, and zero black lines. And MEANWHILE the scattered light sends the EXTRA colors to put make invisible / black any parts of the spectrum that they arrive to. It's such a relief to come to this LOGICAL explanation.

I showed in the last update that when two paint colors alone turn to black, it's not, as we may have expected, from an absorption of colors, but too much color becoming invisible, and thereby looking black because it's ON PAPER. The same two colors (for example purple and yellow) combined in the air will turn white or near-white, however. The stellar spectrums are "on paper" because they don't show inside the prism. The prism light needs to land on an atomic surface to be seen.

Scattered light is laughing at the goofs, they can't seem to get anything right. God is working toward a checkmate, setting up a trap for these proud peacocks who do nothing better than propagate falsifications. That's called propaganda on behalf of their dark god.

The star wizards call scattered light, "re-emission." They say that emission spectrums are from re-emission of light. That sounds correct where straight-through is the first emission, and scattered is the second. But as re-emission is taking place ON TOP of the "absorption spectrums," why don't they inform the people about that? Why did I need to hunt this fact down? BECAUSE, if re-emission happens on the absorption lines, then the goofs can't claim that the atoms absorb the light. LAUGHING AT THEM.

They've been hiding this from us. I was almost to the point of confessing that they are correct, when the solution dawned on me to show why they are incorrect. The date is December 1, 2023, the night I discovered that absorption lines aren't.

It seems impossible to me that sun or stars could produce something so weak as red light, and especially not infrared light. Therefore, I contend that these colors, and yellow too, are formed by the glass prism. The prism forms these colors partially because it spreads and therefore THINS the white light. This situation seems to match the yellowing and reddening light with the sun on the horizon, which is not yellow / red light from the sun, but due to the accumulation of air atoms weakening the light.

I now have a big problem in continuing to contend that SOURCE LIGHT doesn't spread in a prism according to its pre-existing strengths or frequencies, but rather, I contend that the glass spreads the light into rays each having a different combination of characteristics that themselves create the colors in the glass. It seems clear from emission spectrums that pre-existing colors of REFLECTED LIGHT go through the prism exactly where colored rays form in the prism from white source light.

Asking google what color green light becomes in a prism gets the expected comments such as that green light is already separated and therefore stays green. But, no, green light from a source is not already spread through a prism. And when working on prism light in the last update, I didn't come across videos showing colored lights passing through a prism. Why not? It seems like a natural thing for novices to do when they show rainbow colors through a prism.

I shared one photo (last update) of red light only, through a prism, yet rainbow colors appeared in the result, spreading from red to blue. Here's a google offering:

Why does green laser turn red in a prism?

1 Answer. The reddish colour produced by a green laser pointer passing through common glass objects is due to excitation of luminescence centres [blah-blah probably trash] produced by the particular composition and matrix structure [more blah-blah inventiveness] of the glass. Fluorescence is when light excites a system that then almost immediately emits a lower energy photon.

You can believe that explanation if you want, but I never trust the establishment when it's trying to cover for its errors. They first tell us that colored light through a prism can't change, then give us blah-blah to explain why it's not reality, or at least not always reality. Just know that there are different sources for colored light, some being reflected light from colored things in glass/plastic. Such colored light may not behave the same in a prism as pure colored light.

They came to call the photon a flying "electromagnetic wave," but they don't tell us where the electromagnet is. They're hiding it somewhere. They don't give it a medium to fly through either. It's just a flying wave, what goofs. You're allowed to imagine it in any shape you wish because nobody has a clue about that. If it's electromagnetic, how does it use magnetism to fly? What goofs. It can't fly using attraction force, and so it's got to use repulsion force. But how? Easy: it "flies" via electron-to-electron bumps in the electron aether. Whaddaya know, it's not really a flying wave.

Plus, if light can be electromagnetic, and the sun is the electromagnet, why not also gravity? As the sun is indeed electromagnetic, why can't that feature be its gravity source? If they put their minds to it, they could realize how gravity attracts all atoms. It took me five minutes to do so, not because I'm a genius, but because it's easy. All you need to do is find a way that negatively-charged gravity can make all electrons positively charged. Piece of cake. Positively-charged atoms lose electrons. How do you suppose negatively-charged gravity can make atoms lose electrons. Duh.


Maybe I shouldn't be celebrating yet. I haven't yet got an explanation on why scattered light creates the colored spectral lines. This thing tends to agree with their theory that light color is pre-existing prior to entry into the prism. I don't disagree with that idea outright except that I view "color" within white light as varying waves of frequency that can be altered in the prism. So, where the signature colors of hydrogen gas atoms traverse the prism, they can be changed to another color if another color comes through the same path, and that's why the emission spectrum superimposed on the absorption spectrum turns invisible.

I'm not saying that the prism will show red after red light traverses where the green had first traversed, but, rather, the final prism color will be red light added to green light...though the final hue also depends on how strong each are when combined. Can red and green through a prism form black? Yes.

In the last update, I explained, and showed evidence for, why all prism colors have higher frequencies than anything in white light, for passage through glass increases frequencies of all colors by compressing light waves (in the glass). Therefore, it doesn't take much extra prism light added to another prism color to get the combination into the ultra-violet zone.

I don't know whether we're going to get black by shining both red and green prism light on the same spot of white paper. It seems doubtful, and failure can be explained in that the prism lights become uncompressed when exiting the prism. It's the compression that increases colors toward ultra-violet.

But when red light goes through the same prism track as green light, there's only one color exiting the prism to become uncompressed, which should show more compressed on a white paper than when the two different color streams got uncompressed independent of one another. If, for example, there is a red air hose with 30 psi pressure, and a green at 40 psi pressure, and they both released air on the same spot of some white paper, the total pressure won't go above 40 psi. Even if we connected the two hoses, the pressure will drop below 40 psi to about 35. But if we had only one red hose with green stripes having 70 psi to begin with, it's going to spray on the white paper with twice the pressure. I think that's a good analogy for combining red to green light in one "hose." The two hoses get yellow, and the one hose gets ultra-violet = black. That's what I'm proposing. That's my explanation for black spectral lines.

Now if I could just explain why spectral lines exist in the first place, we (anybody) might be able to re-interpret what the red and blue shifts are on stellar spectrums. Our reward might be to give a good boot to the backsides of evolutionist scum. I'm driven.

The red / blue shifts are the spectral lines of stars having a whole set of lines (for example, of hydrogen gas) either closer to the red color, or closer to the blue color, they being the two ends of the rainbow. The magicians call it "shift," but is it? They say that the a star moving away from earth will have the lines shift toward the red end of the rainbow.

BUT WAIT. If the star is moving away from earth, so is it's hydrogen atmosphere, meaning that if the hydrogen atmosphere is causing a shift in the prism light, the rainbow colors produced by the star should also be shifting by the same amount and in the same direction, resulting in zero net-shift of the black lines (as compared to the shift of all colors). MAGICIANS PLAYING TRICKS ON US.

That debunker has never dawned on me before (that I can recall). December 1 must be my lucky day. I've only got an hour and a half to go before the day's over.

Remember, there's two kinds of absorbed light in the gas atoms, one from the straight-through light, and one from the scattered light. Do the emission lines come from both kinds of absorbed light, or only from the absorbed light resulting from scattered light? This discussion is Greek to the goofs because they don't think absorbed light can get to the prism, but STUPIDS, all atoms absorb light and then re-emit it. The goofs themselves will tell you that you can't destroy energy. If light energy goes into atoms, it has to come out.

I now know that the emission lines are from scattered light, and so what difference does its absorbed light have as compared to straight-through light? For one, colors. Straight-through light is white and brighter (more wave-train density). Absorbed light shoots out colors from jolting electrons, and so to explain the four main spectral colors of jolting hydrogen atoms: they jolt with four main colors, rendering most others weakly invisible. They say that the hydrogen spectrum radiates red, cyan, blue and violet colors, yet there's many more than four lines, several in each of the violet, cyan and red regions (others show lines in the green instead of cyan). I'll call them four basic lines to make it easier to follow.

The page below shows the emission spectrum for hydrogen gas made from a white light in a tube filled with the gas, but how is that an emission spectrum if there is white source light between the gas and the prism? That's supposed to be an absorption spectrum. No white light source, according to my understanding, is to make it to the prism to form an emission spectrum:

The best word to describe the goofball explanation of colored-light transmission is: incomprehensible. They say that a photon strikes an electron, but instead of knocking it out of orbit, it flies higher from the proton and orbits there like nothing unusual takes place. Planets have less velocity = less energy when more distant from the sun, otherwise they would spin out of orbit, but the goofs, backward as ever, say the electron has more energy when in an orbit further from the proton.

. It has higher energy because the photon is piggy-backing upon it for some instant of time, and not until the photon flies off will the electron return to its lower orbit as if nothing unusual took place. Over and over again, this electron gets knocked by photons, but never goes out of orbit, and always goes to a higher orbit. If you still think that this theory deserves credibility, you are not fit to learn physics. The photons comes from every direction, the electron can never know from where at any time, and yet the result is always a perfect higher orbit as if it's the most natural thing in the universe.

Back to the hydrogen spectrum with its four main colors. I suppose the full numbers of colors depends on how the light is made, and who makes it. I can certainly imagine the magicians simplifying things if eradicating key information is to their advantage, and complicating things if need be to salvage their theories.

I'm going to propose that hydrogen gas produces only four main colors after it absorbs white scattered light. I'll then propose that these four colors enter the prism in accordance with some quality they possess that bends their light to where they end up upon exit of the prism (see diagram page above). In this latter point, I at first agreed that light enters the prism at an angle in accordance with their various dispositions to curve at the entry point. However, I'm not necessarily in agreement with the foes on what the bending factor is.

Nor am I ready to concur that the four colors at the prism's exit are the four colors pre-existing within the hydrogen atoms. The latter could be of any spit-power levels that, only after passage through the prism, become red, cyan, blue and violet spit-powers.

Sticking to logic and the rules of physics, I cannot agree with my foes that violet light bends most in a prism, and red the least, due to their having the highest and lowest frequencies. Logic tells me that the more frequent wave trains strike glass, the more their shared path is straight in, rather than curved, where the glass seeks to curve it.

Spectral light has no waves on the same track because that belongs to stronger source light. This is to be viewed as one poke at a time at the same place upon the prism, like train after train non-stop to the same place. The stronger the train, the higher frequency it has i.e. more pokes per second. But spectral light has no such thing. Instead, it makes pokes of upon different spots on the glass, each poke a wave. Each poke is a disappearing train followed by trains that likewise disappear the moment they land on the prism. A wave has a short life from source to landing, yet it then reproduces sons and daughters who look different than the parent.

The point is: spectral-line trains come with different strengths, not different frequencies, and the weakest wave is, I think, most susceptible to curvature upon entry into the glass.

Imagine a ray of light, of a square centimeter in size, coming through hydrogen gas to the prism. At first, I thought that the only way to form only four basic lines of colored light at the exit is for the same four lines to enter the glass at only four locations along the one-centimeter of width. But how possibly could light accomplish such an entry? How can light from anywhere be brought, as a ray, to a prism or anything else, so as to have light of different characteristics at a few spots only, across the width of the ray? Impossible.

The light fully across the width of the ray must be roughly identical in characteristics. This is what has bothered me for years concerning spectral light. I can accept that only four colors come to the prism from the gas, but each of four colors must spread evenly across the width of the ray, and thus it seems impossible for them to form spectral lines.

EUREKA!!! It's shortly before 2 am December 2, and I have solved it. It was due to the two paragraphs above, all spelled out. I was looking at my drawing while writing it, and realized, zikers, the colors at the entry, all spread evenly across the width of the ray, each bend with different amounts upon entry, BUT THEY CONVERGE, not spread out!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

AHHHHHHHHHHHHHHHHHHHHH, relief. Thanks to the last update, I did notice converging of light in the prism even though almost nobody ever draws it converging. We get tricked into thinking that light always spreads out in the prism. Not so.

SO, what probably happens is that the red line of the hydrogen gas starts out at the full thickness of the entry ray. The light is white at that point, I will maintain, and it then bends on a path that takes it to the part of the prism that forms red light. The light becomes thinner through passage of the prism until it's a thin line upon exit. Ditto for all the colors. They pass through becoming thinner too, but moving toward different parts of the prism.

For this theory to be correct, the spectral lines will not show if the prism glass is too wide, for each of the four converging rays will then meet inside the prism, and thus not show outside the prism. It's not clear to me that the black lines showing as a spectrum outside the prism will ever meet, because there's a good chance that the lines will spread out once they exit the glass, due to the glass re-directing the light, as glass is known to do.

So, for spectral lines to show, the waves within the prism converge to a near-point at the exit. If they spread out at the exit, it could be a long ways from the prism before they become bars (fat) rather than lines. Some lines can be fat right out of the exit.

You can see here that converging of light waves will become brighter color. Transparent gas atoms emit very little color, but convergence can explain why emission-spectra colors are bright, or even existent at all.

It's a good thing that I delved into different types of light for this discussion. The reason that spectral lines can be due to converging waves in prisms is that these lines are from reflected light. Or if you don't like "reflected," use "atomic light" to distinguish if from source light. However, source light is from atoms too. And even reflected light travels across atoms and aether. How else? In other words, both the source light and secondary light take the same mode of transportation to the prism.

Source light from the sun is a lot hotter / stronger than scattered light in the air, and aside from that, I can't see any difference between the two, though this energy-level difference is what makes colored light versus white light. Reflected light, especially through transparent gas atoms, comes with much-weaker waves. I don't like using "energy level" because that's their phrase. I use "strength" or "power" instead.

For hydrogen atoms in light, they seem to be undergoing four main modes of reflection, each causing a different strengths. For example, light in hydrogen atoms can cause swishing of electrons to form one strength level, jolting to cause another, swerving of electrons to cause another, and jiggling to cause the forth. Or, there's just things about this atom that nobody know about.

Why should source light spread out, while reflected light converges, in prisms? What is it at the entry point that causes either one? It seems very odd that light of the same strength, striking different parts of the entry point, should converge in the glass because we would rather expect identical rays (= same strength) to go parallel into the glass, i.e. stick to the tracks they have in their approach to the glass.

The theory of the goofs is impossible where they say that the light from the gas is already colored, and then goes through the prism in those colors due to every color being bent the same through a prism. It's a trick because it's impossible for light to enter the prism at four spots only. It needs to enter equally all across the width of the ray. It's a realization that has never occurred to me before...because it was never taught to me by the goofs, magicians. They were hiding it. People who spend countless hours on prism light were hiding it, not disseminating it.

Nitrogen and oxygen gas have many, major spectral lines. We can't say that each line arrives to the entry point at exactly the same positions in which they exit the glass, it's IMPOSSIBLE.

Air atoms make the air blue, from reflected light, and so this blue would arrive to the prism as faint light because air is mainly transparent. Would this blue light converge in a prism? Shouldn't it converge to form the emission spectral lines of nitrogen and oxygen? YES. And if we include direct sunlight along with air, the colored spectral lines become black, incomprehensible unless the extra mega-light from the sun makes the colors invisible to the eye. Without the sunlight going direct into the prism, most of each emission spectrum is black due to light being too weak to register in the eye. Logical.

The goofs need to explain the blue color of the sky so as to jibe best, if possible, with their mechanics for the emission lines of oxygen and nitrogen. If the sky is blue, air atoms are emitting light waves of different spit-powers that, altogether combined, look blue, but when this light is run through a prism, it doesn't necessarily stay blue.

It seems the prism does separate all the colors coming from the gas atoms, but they can change through the prism. Yes, literal separation is possible if each color converges in the prism. Each color arrives separately and distinct to the prism entry with a different spit-power level, but each color also has a distinct and different density of wave trains. Which of the two, or both, determines the initial direction through the glass?

The trains all go through the glass at different angles, not only per different color, but each color sends trains through that come closer together -- brighter, more dense -- with distance through the prism.

Chew-Chew Over This

Just become the train director. You get into the prism at the entry, and you study traffic there. What might be happening as the trains come through? Get your self into the glass. You see that some white trains all set out evenly spread all along the entry point, but then get painted red along the way as they end up in the red region of the prism. The passengers all had tickets to Red Beach.

You then note that white other trains, likewise setting out while evenly spread at the entry, got painted blue as they all ended up in the blue region of the prism. They were all headed for the train station at Blue Mountain.

You then scratch your head because: when all trains ended up at Red Beach, the bottom trains angled more than the top trains. And the top trains of those headed to Blue Mountain were angled more than the bottom trains. Hmmm, now what could be making that difference at the starting line? You note that the trains to Blue Mountain all had train cars closer together than the trains to Red Beach. Hmm. But then you get confused because, when you counted the trains, there were more going to Red Beach than to Blue Mountain. You call the train manager: "what's going on"?

The manager responds: "check out what happens to the yellow-green trains." Er, the top yellow-green trains angle exactly as much as the bottom yellow-green trains, and so they end up exactly midway between Red Beach and Blue Mountain, at Yellow-Green Plains. What made them both go at the same angle? You call the manager again, but he just says, "think harder, longer."

You note that most of the blue trains angled more at the entry point than most of the red trains, and so most of the blue trains went against the grain, uphill. Most of the red trains went less against the grain, down hill to the beach. That means that half the yellow-green trains angled perfectly reciprocal to the other half of the yellow-green trains. That is, for as much greater angle as half the yellow trains that went, the other half went at a lesser angle. These trains stayed on level ground all across the plains.

Why did the blue trains, with more spit-power, go more against the grain of the glass, by taking the longer route than both the red and yellow trains? Shouldn't it be the other way around? You are sure that, with more spit-power, it amounts to more engine power, so that the trains should go straighter though the glass, at less of an angle.

You call the manager: "so why do the stronger engines turn sharper into the glass?"

"Congestion," he replies. "Stronger engines create more congestion at the entry.

At first, you thought he meant more congestion of train cars. But then you entertain more congestion of glass atoms. Say what? Light pushes glass atoms closer to each other? No, light isn't strong enough to do that. But light pushes the electrons of atoms closer together, and it's through those electrons that light moves in trains Okay.

Plus, light waves add electrons to everything they strike. It results in heat. Everything in sunlight heats up, right? That's because each striker of a light wave, which is last aether electron in the wave, enters an atom (at the surface of where the light strikes).

All wave trains, from all colors, at the bottom of the ray striking the prism, enter the glass first, followed by the center of the ray, followed by the top of the ray. That's because the ray hits the glass at an angle, not dead-on at 90 degrees (rainbow effect won't happen with the latter).

You start to ponder: the bottom of the ray compresses glass electrons in both the horizontal and UPWARD directions as it enters. When the next fraction of the ray (the one most-directly above this initial entry point) arrives to the glass, the atoms have been compressed for its entry, and so the change in entry angle could be a little different as compared to a no-compression scenario.

Compression of glass / water electrons can explain why light always bends more when it approaches either one at a sharper angle. The sharpness of the angle determines the level of glass-electron compression in the direction semi-perpendicular to the flow of light. The mechanics works as described here, with compression in view, for both a prism and a flat pane of glass. .

The compression process above occurs progressively toward the top of the ray; the electrons at the glass entry become progressively more compressed toward the top (semi-perpendicular to flow of light).

You were trained for train traffic jams. You begin to ponder: compression of glass atoms makes more congestion, making it harder for trains to get through. The blue trains create more glass congestion and bend steeper into the glass, in spite of their bigger engines, while the wee-wee red trains create less congestion and bend straighter into the glass. The yellow-green trains with medium-sized engines are the middle ground. Bigger engine size causes more compaction of electrons, more congestion.

You then realize: the more the congestion, the less the trains can go straight ahead because compacted electrons act more the wall that DEFLECTS the path. The congestion forces them more sideways. The blue trains that enter first, at the bottom of the ray, have less congestion than the blue trains entering higher up, and so the blue trains at the bottom bend less through the glass than the blue trains at the top, and all blue trains converge at Blue Mountain (i.e. the same spot exactly) because all blue trains enter the prism with PROGRESSIVELY greater congestion! Every blue train higher (closer to the apex) than another gets bent a little more, and, to boot, on a compounding level.

The reverse is true for the red trains. The ones at the bottom of the ray enter the glass straighter so that they end up climbing upward in the glass at a greater angle than the red trains entering at the top of the ray, and consequently they all meet at Red Beach. Not meaning to confuse you, but let's add that the top red trains angle more deeply into the glass i.e. in the downward direction than the top red trains. That is, the lower trains angle more upward, and the top red trains angle more downward. It's the reverse with the blue trains.

I've said that gas atoms are to be distinguished from source light in that the atoms radiate colors instead, but that was in the context of spectral lines. We could instead say that all incident light from gases is white with different spit-powers. We could then ask whether the weakest waves of gas-based light turns to red prism light while the strongest gas-based waves turns to violet. Etc.

To the best of my ability to interpret this situation, it appears that the red spectral line originates from the weakest, visible waves...which science departments would interpret as original red light staying red through the prism. However, let's throw a monkey wrench into this in case, per chance, it's wrong that the strongest incident light bends the most. If it instead bends the least, then red spectral lines are from strong input light, and the blue / violet is from weak input light. I don't think the prism can turn the strongest light waves into the weakest unless the strongest are thinned out into more trains i.e. more tracks, and vice-versa. Note how red paint is mixed with blue to get a violet tone. Could it be that violet originates from red, and red from violet?

There is a problem that will hopefully be a solution. The to-be-red and to-be-blue waves that strike the prism don't do so independent of one another. With hydrogen gas, there are four main proto-colors striking the glass simultaneously, all causing the same constant electron compression for each proto-color to get deal with. Why, then, should proto-red bend less while proto-blue bends more at that same location, if the compression is identical for both proto-colors? It suggests that the entry angle of different colors has nothing to do with differing levels of electron compression, but rather due to something else?

But wait. Each proto-color can produce its own compression level while ignoring (being unaffected by) the compression levels of the other proto-colors. This seems very convenient of me, to make the electron-compression theory be true, and it even seems like a ridiculous idea. But wait. Millions of light waves can and do go through each other virtually unscathed, for example in communication wires. It's as though each wave is non-existent to other waves. How to explain this?

Each wave acts in its own TIME, and time is divisible by extremely-short units that we cannot imagine. A wave always causes compression of electrons. When you speak on the phone, you are compressing electrons in the transmission cable. Other people talking on the phone are compressing electrons in the same cable, in the same direction, and the electrons all bring each person's voice across that wire unscathed because each electronic impulse (= sound in your voice) creates its own electron compression that, regardless of how we explain the mechanics, carries as a wave for miles and miles down the wire. Why can't electrons in prism glass do the same over a few inches with only four colors speaking at the same time?

I'm proposing that, when a proto-red wave strikes the glass, it produces its own compression levels for the instant of time that it strikes, and then the other colors strike and produce their own compression levels, in their own instants of time. In this picture, it was wrong to say above that compression is "constant" upon the glass surface. If the strikes come with time between them, the total compression can be non-constant as far as the waves are concerned. In our human economy of time, it's definitely constant.

It seems very difficult to explain why BOTH top and bottom trains should follow different angles, toward one another, when the rainbow is formed by an opposite scenario, trains away from one another. Why don't the rainbow colors converge as do the light waves from gases?

You get on the phone, "Manager, baby, just one more question. Where's the hotdogs?"

"At the end of the rainbow, in the pot."

"But I want them grilled."

"Try the other end of the rainbow."

So, off you go, and, just your luck, you went to the boiled dogs first. Across you go to the other end, and you get a good lay-out of the land. You walk over Yellow-Green Plains, way up there, then walk down over Blue Mountain, and you see the end of the rainbow right at the home of Violet, your aunt. "I don't want hot dogs that bad," you think.

Over yonder, some violet-painted trains at the depot, you're sure that the trainsvestites move out to these parts. No wonder they use the rainbow symbol. You start to puke, but hold it in just in time, feeling a hot sensation in your throat. You make a bee-line toward the boiled hotdogs, and notice than no trains converge at any train station under the rainbow. All the train tracks get further from each other the further they travel from the rainbow, and the tracks come in all colors to match the colors of the trains. It gives you pause, and you stop at the crest of the bow, and look toward the entry point in the glass. What do you see?

You see red tracks closer together than all other colors, and the violet tracks have the most distance between them. You get the manager on the phone: "Boss, sorry but, why are their fewer pink tracks?"

"It's more repulsive on the south side of Blue Mountain. All the fags vacation there."

That explains it. In the last update, I concluded, tentatively anyway, that the rainbow tracks form as the wave trains at the bottom of the ray push other wave trains upward by inter-repulsion force, causing the colored tracks as a whole to spread-out upward, from the bottom up, and arranging fewer wave-train tracks at the blue-violet end of the rainbow. The idea was that, the more repulsive the violet tracks, the more distance they put between each other, and consequently the red tracks should be the closest together. The yellow-green tracks are again the middle ground between the two situations.

That theory was based on the variations of glass-atom compression in the longitudinal (horizontal) direction from entry to exit. The more compressed the waves were longitudinally, the more train cars the waves got, so to speak, per unit distance along the tracks. And the train cars are a symbol of captured electrons that repel each other, forcing trains to take tracks further apart where the density of train cars is greater.

So, there was a push-up force from the bottom, gradually weakening with distance upward (toward the prism's apex) as train cars became less dense. But, now, in hoping to find an explanation for light-wave convergence, the possibility, with probability, is that there is simultaneously an opposite, top-down push from the UPWARD compression (think crunch) of glass atoms at the entry point. Compression force is upward at an angle, not straight up, because the light source comes up at an angle.

This sets up a conflict, between the two mechanisms, one formed by the upward compression of glass atoms, and the other formed by the longitudinal compression. The conflict is to determine whether the rays will ultimately spread (diverge) or converge. They can't do both even while both mechanisms are in play simultaneously. And so it seems that, the more powerful the waves, the more the out-spreading wins the conflict. One can then conclude that convergence is indeed due to weakness of the waves.

The mechanism for convergence is yet active even though the wave trains spread out. If it were not active, the wave trains would spread out more. As the convergence mechanism minimizes the out-spreading, it predicts that stronger source light into the prism will increase the out-spreading, and vice-versa. Laser light should get a wider rainbow than weaker bulb light.

Again, solid atoms don't make spectral lines. Why not? I think the solution is in the fact that only gases having at least some transparency produce spectral lines. The key is that such atoms produce very little color...while solid atoms produce much color. Then, as solid atoms produce light with strengths between those of gas atoms and source light, the waves of solid atoms can be predicted to barely converge or spread out in a typical prism. That is, the colors move more parallel in a prism. Should they slightly converge, they would not nearly meet just outside of the prism, and so there's no spectral lines.

If the light of solid objects gets brighter/deeper through a prism, then the lines of all their colors are converging. One can peer though a prism and see solid objects coming through. The less distortion to their shapes, the more parallel are the light waves from those solids. Heated solids (e.g. molten metal) create a rainbow from the light of heat (a source light), not from the atoms.

Let's Go Surfing

The red-shift / blue-shift of the star gazers cannot be interpreted as they do because they compare the rainbow colors of a star with the spectrum of the star's atmosphere. This comparison is done when viewing one photo image; the rainbow and lines appear together in one image. The gazers don't apply the rainbow superimposed upon a separate image of the gas' spectral lines.

Therefore, when they show that the spectral lines are shifted toward Red Beach or Blue Mountain as compared to the rainbow, it cannot have anything to do with the star's direction from earth, for all rainbow colors would then be shifted exactly as much as the spectral lines. The claim is a non-starter that only trickster magicians would use on the unsuspecting public.

Instead, the shift of some spectral lines, from star to star, and always compared to the spectral lines of a gas on earth, must be due either to something pertaining to the gas, or in the something pertaining to how the light arrived to earth. The best way to solve variations in spectral lines is to first discover the mechanism behind them.

I can front theories, but none with any means to serve evidence. I can be sure that differences in spit-powers from gas atoms are from motions of electrons. There's "one" kind of electron motion as source light passes across electrons, but how can we know how many spit-powers this one process offers that become spectral lines? Impossible to know. There must be near-infinite number (so to speak) of spit-powers amongst the near-infinite number of waves, and so how do any of them, IN PARTICULAR, end up as spectral lines? One possible answer: none of the spit-powers make it visibly to the prism but the ones that cause the visible lines. Why them?

What about wave different spit-powers formed from white, scattered light hitting the atoms like balls off of a wall? They are going to cause different electron motion, more like jolts, as compared to white-light waves passing across the atomic surface that might only cause mild jiggling.

This scattered light then enters as "absorbed" light into atoms, upon subsequent atomic collisions, and this will cause yet another type of electron motion, we may assume, harder in gas types where the atoms absorb more.

How can we possibly trace spectral lines to any of these different motions? Maybe a computer program can help, but it's only as good as its human (or goofball) programmer.

It seems predictable that the scattered light of the bounce-off-wall type will be a punch to the center of the atom, where the proton provides bone. If it doesn't hit the center bone, it won't be much of a punch. Per each set of scattered waves, the first scattered wave that arrives to the gas is a backward wave, because it hits the atom in the back, and bounces out toward the light source. After it punches a second atom, it goes in random directions, partially punching atoms in their bones, partially becoming surfers who cross atoms from one side to the other on their surface waters, and partially becoming absorbed light.

The life of a puncher doesn't last long, but he has sons that carry the torch forward. When a puncher becomes a surfer, he goes out of the atom and punches another atom, and thus the son is born, who becomes a surfer and births a son...

Every puncher is just one wave, and every surfer is just one wave circling upon the electron waters of an atom. As the puncher weakens, he becomes a finger-tapper, and finally he disappears into the mist of aether. Every son is just one wave that produces a number of waves of various but weakening waves. They can never be as strong as when a son is born, and the son is never a wave as strong as the prism's ultra-violet light. When the goofs speak of the sun's ultra-violet light giving us a sun tan, they err. Ultra-violet is born in a prism, whenever a puncher or surfer goes into it. The puncher can go in fighting mad, or with lazy punches, or with taps.

Punchers, and their children everywhere, the surfers, are Caucasians of different spit-powers. They all have white skin. Different spit-powers are not colors aside from through the prism. White light comes with different spit-powers and frequencies of waves. Scattered light is from many different strengths of light, but you never see it in colors. It doesn't shine into your windows with violet, blue or orange color. As you turn the voltage down to your light bulb, it never becomes violet, blue or green.

When the portion of scattered light that doesn't act as the puncher crosses the outer electrons as a surfer, it does exactly what straight-through source light does. However, the latter is big-daddy surfer who can surf from the sun all the way to the prism without becoming a puncher. They are both Caucasian.

Scattered light is a copy-cat. It's going to be milder on the atom when it apes the straight-through light because scattered light loses energy by punching atoms around. Scattered light gets tired, and grows old, then dies. Straight-through light, which stays robust all the way to the prism, never turns black through the prism. When scattered light goes through the prism, it makes a black rainbow. It's the weakling-light son of the puncher, not fit for the boxing ring. Part of the black is the knocked-out surfer. But big-daddy surfer never goes black through the prism.

A punch is not going to result in the same type of electron motion, upon the atom, as the surfer. The entire gamut of punches to an atom raises its water level with chops. Yet punchers deprive the atom of surfer energy, and not always by the same amount. There is an average puncher-energy that's exactly midway between the hardest puncher and the died-off puncher. There is an average surfer-energy that's exactly midway between the highest surfer and the drowned surfer.

If the puncher bounces out of Adam with one-percent of the energy that it went in with, the surfer carries onward with 99-percent, minus what it gives to the atom in absorbed energy. Into another atom, the bounced out puncher then becomes a weakling-surfer wave with almost 100 times less strength than the surfer upon Adam. The punch waves keep coming into Adam, from the light source, at the same rate (= frequency) as there are surfer waves. Both have the same frequency, but the surfer waves in Adam have much more spit-power than the puncher waves bouncing out of Adam.

Therefore, says Adam: unlike a light source that increases in spit-power with higher frequency (or viced-versa) of shooting light waves, scattered light from the source can have waves of identical frequency but different levels of spit-power, all of them white-colored waves...until they go through a prism. However, I can now realize that FREQUENCY of scattered light waves exists only from the punchers, not from absorbed-and-reflected light.

The light-infiltrated atom itself becomes a "light source" shooting out different waves of identical frequency with the straight-forward waves, but the atom in this case always shoots the light backward, away from the prism and toward the sun, because there's constant punchers hitting the backs of atoms as the straight-through waves continue. This makes me realize that light on the opposite side of the sun sends these backward waves through the sun, and into the prism...and therefore through the atmosphere on both sides of the sun. Could these backward-shooting waves explain spectral lines? I don't think so, as they would be very weak.

A punch to a whole atom could produce mild atom-to-atom waves that will ultimately send energy to the aether. This looks like more scattered light, but might just produce still a different kind of electron motion such that it sends the weakest waves, because a whole atom being nudged forward by a single punch is going to put into motion countless aether electrons, meaning the entire punch is shared by countless light carriers, and that to me looks like invisible, infrared light. It means that I may have been "wrong" to say that sunlight can't produce weak light in the red zone, because sunlight through the air can punch and stab air atoms. But then that's not source light.

When electricity invades a filament, the metal atoms probably release strong white waves at generally the same frequency, because electrons are ejecting (fully emitting) from the atoms at generally the same rate, because the electricity is pushing at generally the same force. If the force (voltage) of the electricity is lowered, the light turns red because, I think, we are now seeing jolting (not ejecting) electrons. I have a hard time believing that atoms on the sun can jolt with that weak color. When a bulb is red, there's almost no light filling the room.

Every transfer of straight-through light to an atom involves a punch and a stab. Stabbing creates the spectral lines. A stab is the absorbed light from the punch. The stab can go deep to the bone, but always comes fully out of the atom as surface jolting. Straight-through light struggles to cut across the jolting surface, but may also cut across in the milder waters under the jolting. The latter reaches a maximum, per light source, and is expected to be very mild in highly-transparent gas atoms.

Absorbed light creates MANY jolting electrons, MILD, and the "many" become frequency, don't they? Yes, but they become frequency of wave trains, not the same as a high-frequency wave. It's tricky: high-frequency waves are not the same as a high frequency of waves. See the difference? The high-frequency waves are on the same, straight-through track, and they pass one-at-a-time across/around one gas atom that simultaneously jolts with wave trains each of different strengths. The latter is what forms the spectral lines, and the high-frequency waves is what forms the full rainbow. Both are white light, and although the gas atoms do jolt some faint visible colors, are they the colors that form the spectral lines?

How can faint colors, not picked up by the eye in a foot of width of a gas, in the lab, cause spectral lines? The men have a small container of gas, with a bulb in it, and they form colored spectral lines with this. Are the colored lines from the color of the gas that occurs to the eye only when the gas is miles thick? Maybe not. I don't have knowledge on the faint colors of various gases.

The different strengths of jolting trains -- each train = one wave -- forms different colors in the emission spectrum, most of them too weak for the eye to see. I'm proposing that the trains of highest strength pile up at a train station on one side of the spectrum, and that all the trains of the lowest, visible strength pile up at a train station on the opposite end of the spectrum.

If every jolt off of an atom forms a wave while not followed up by a jolt in the same direction, how can that form a frequency of waves in one direction? As each jolt is a train, this type of scattered light results instead in a frequency of trains (plural). For frequency to apply to a wave train (singular), we need straight-through light that has one train (wave) after the other in the same direction. A high-frequency wave is one with waves on the same track, and with waves closer together in time than a wave on one track with waves further apart in time. How will these two, different situations show on the prism?

I'm proposing that the prism FILTERS OUT (or separates) waves of different spit powers. The prism's front tunnel has redirection tracks sending all trains, with the same spit-power, through the same tunnel to the same train station. Across the other side of the prism mountain, there are numerous tunnel exits each with train stations of different colors. As the gas must send many trains with very weak engines to the front tunnel, they must go through the prism too, but rather than all going to the infrared station, might there be skinny, colored trains too faint to see that go to colored stations? That is, could there be FEW yellow and green trains, for example, of such low density amongst the total density of incoming trains, that the eye doesn't pick them up, because they make extremely-thin colored lines on the spectrum?

Or, they make a very small pile at the prism exit because trains of this particular spit-power are few. I don't see a problem with this view, begging why the large spectral lines are big piles, or loaded with trains.

I'd like to know why gas atoms have a high density of trains with the same engine powers? The secret must be in the way gas atoms jolt. But then it's not a secret anymore. The answer is plain: the power of the electron jolts makes the size of the engines. This secret isn't even exciting, such a disappointment, especially for the goofs with elastic brains who crave addictively to illusionary ideas.

So, we can conclude that the hydrogen atom jolts with three main colors, plus a little cyan/green. The atom just happens to form chops of a few different heights when its showered in white light.

But, just as backward-shooting waves on the opposite side of the sun must penetrate the sun to the prism, so the chops on the "dark side" of the hydrogen atom, the side facing away from the prism, can arrive to the prism by first moving through-and-around the atom. These chops don't go up only, but also down, and when they come down on the dark side, some waves (not all) will form toward the prism. Could these dark-side waves contribute something like sister lines to some of the spectral lines? If, for example, chops on the front of the atom produce a violet line, might the same chops on the dark side, due to penetration of the atom, form a second violet line beside the one above?

What about the chopping of electrons that sends energy down to the proton, then bounce away, off of it, to the prism? Could that form an altogether different color of spectral line?

Only the full rainbow is formed with trains of high-frequency waves, each of a different frequency in unit time, and of corresponding (proportional) spit-power. Superimposed upon that rainbow are the spectral-line trains of different and milder spit-powers, where the spit powers are not proportional to their frequency in unit time. Why should frequency of wave trains produce the rainbow instead of spectral lines?

It's not enough to say that they are more powerful, if the power level of waves causes convergence though the prism and a piling-up at certain spots. It's unlikely that high-frequency waves have more wave-train density than reflected light. The only difference seems to be in pokes upon the prism glass: high-frequency waves poke on the SAME PLACE of the glass, and so, along with being the most-powerful waves, they compress electrons CONSTANTLY along the same track, longitudinally-crushing trains so that, with train cars (electrons) coming closer, trains repel trains perpendicular to their tracks and therefore spread the waves out i.e. diverge through the glass instead of converging. I'll probably not remember all of this the next time I delve into it.

As reflected light doesn't poke the glass constantly at the same place, it acts differently, without the ability to compress trains, and therefore absent of spread-out force. Instead, the wave trains in the glass disappear as soon as they pass through, and get replaced only intermittently. It's more like a bunch of arrows or pins flying through instead of a steady train.

Frequency of train cars (arrows and pins) is not proportional to wave strength. There can be a low density of arrows and pins with high power, or low power. Their density of arrows / pins determines how bright/visible the spectral line is, but we can also add that higher density is more likely to create a fatter line. If you magnify the spectral lines, more lines should become visible.

Molten materials like the sun have so many waves that they train each other straight forward even while waves are passing through them in all directions. It's remarkable. Even if it's debatable on whether there is such a thing as a wave frequency, since one can view waves as individual entities (two or more waves needed for frequency), the fact is now exposing itself that source light from sun, other molten materials, and bulb filaments become the same colors (in the prism) as the much-weaker reflected light BECAUSE THE LATTER GETS COMPACTED by convergence. Otherwise, if reflected light were to spread out instead, it would become something different.

How much different? Maybe a different color altogether. The green spectral line might not become fainter green, but probably red. If one shines green light into a prism by first shining white light through green-tinted glass, that's reflected light, and so I'd expect the green color to get compacted in the prism, and therefore it should stay green, even a deeper green. But if one shines a green source light i.e. having wave frequency, I'd expect that light to spread out, and so it could change color altogether.

Shining green prism light through a prism is not the same as sending green source light through, because the green prism light has been made weaker by spreading. It won't compact glass electrons as much as green-source light, and may therefore not spread as much, or not at all. If it doesn't spread, it stays green, is the prediction. Try sending it through different amounts of prism glass to find what the difference might be.

The straight-through waves are big-daddy-white-potent, the whistle-happy conductor. He drives the white, high-frequency train cars, yet he cannot unload electrons from metal atoms for to set-off the photo-electric effect, whereas ultra-violet from a prism can. The latter must have more train-car density than big-daddy. That's because, when big-daddy-white-potent drives through a prism, his train becomes compressed so that the train-car density / frequency increases.

But big-daddy-white-potent is not part of the spectral lines. Instead, he becomes the shining spectrum, a partner with the sun of God's goodness and genius. God is GGG, Goodness and Genius and Glory. He will not share his glory with Einstein.

The first video below has the man claiming that the red line on the spectrum of water vapor is the red line of hydrogen, yet oxygen's spectrum, seen in the second video, also has the red line. In the first video he claims that the two blue lines of water vapor are from its oxygen atoms, yet the oxygen spectrum also has an orange and green line, absent from water vapor. It would be easy to make false / misinterpreted claims in this way where the lines could be coincidentally the same color(s). The third video shows orange and green lines too, for nitrogen this time, and even nitrogen has the blue lines and some additional red to boot. One could make the case that nitrogen is a modified version of an oxygen atom:

I assume that the bulbs used in the videos are not high-frequency white because they don't make rainbows.

An additional problem the goofs have is that all colors, as per their view, strike the glass fully across the thickness of the incoming white ray. They cannot say that violet enters only at the bottom of the ray, and from there bends, because they have violet also at the top of the ray. They have all colors of light striking roughly equal at both top and bottom of the ray, and in between. They pretend that red enters at the top of the ray only, but then they believe, without saying, that red hits the glass also at the bottom of the ray. How does this situation in its totality form the rainbow we see at the exit of the glass? Go ahead and try to draw it on paper with merely a ruler needed, so what you get.

When you draw the lower violet line and the top red line, you have made the top and bottom of the rainbow. Contrary to diagrams, the rainbow at the exit is very-slightly wider than the white ray that hits the glass. You might draw the exit only two or three times wider than the entry ray.

You then need to draw a line for the blue light starting at the same place as the violet line does. Of course, this blue line ends a little higher, at the other side of the prism, than the violet line. BUT, you then need to draw another line for the blue light that starts where the red line starts, because their blue light hits the prism there too. This upper blue line needs to be parallel with the bottom blue line because blue, according to their theory, enters the glass at the same angle no matter where it strikes at the same angle.

On my drawing with the exit three times wider than the entry, the top blue line ends up reaching past the full green section, all the way to the start of the yellow color, at the exit. It means that green is ruined with overlapping blue, becoming cyan (blue-green) instead. Then, the top of the green line goes as far as the start of the orange color, meaning that there can be no pure yellow, but in its place will be yellow-green. This ruination of colors gets worse when the exit is less than three times the entry width, and better when it's more than three times.

The thing I cannot find an explanation for is the directional shift of light as it exits glass. I can't even offer a theory. It's almost as though light is a rubber ball packed with internal energy, and once it finds flight back into the air, that energy sends it off in a different direction. But this theory has no basis so far as I can see in the way I view light-wave transmission.

I must be lucky. If I were very smart, I'd be able to spot the extravagances that the goofballs are trying to lure us into, and then I'd be trapped into looking at things as their minds do. But as I'm rather slow of mind, I can't grasp it, so I don't even try to see things their way. I can see things from low-down only, and so rather than trying to get into their minds, I'm seeing only their feet. And I can see where they're walking, stinky, and what they are really trying to accomplish, the evil, and their feet don't match what their mouths are saying. And I can see everything they're doing under the table. Wicked. Thank God for my being so dumb.

Blessed are the dumb, for they shall inherit the Kingdom of God. Woe to the wise of this age, who are wise in their own eyes.

The reason that God made space is that we can't now say that nothing would exist if there were no God, no Creation. For if we say that nothing would exist, we know that space is something, and we know that by eliminating space, nothing can't exist. NOTHING CANNOT EXIST. Try to imagine nothing without even space. You can't do it. Space tells you that nothing(ness) can't exist because space is nothing, yet there it is in your midst, right in front of your nose. God has the evolutionists in checkmate. The space in front of their eyes is evidence of a Creator. He just was. And He was intelligent. He was not out of nothing, but was out of was. He was that He was, which is why he said, "I am that I am." Before God was, was was. Don't fight it, just believe. God was in the was before the was was. God was, God rules. Before Abraham, Jesus was. Just let Him be the King, because you wasn't always, you pip-squeak.

Imagine a brain five inches wide contending with God on a mortal level, to the death. Such a fool has been the evolutionist. He's a has been, and will be an is not. It's past-due time to take the land from him.


This is probably the most important message today:

I, with others, proposed that Tamara Lich, so-called "organizer" of canada's trucker convoy, was a fake, a plant in some way, though the truckers who came out to Ottawa were not supposed to be as numerous as they were. I'll add that her 31st day in court, to date this week, is not yet at its end, and she's been charged for nothing. It seems impossible that anybody could be in court for this long unless the government is faking this to scare canadians into never protesting again. This is an identical tactic to the faked January-6 arrests.

Look at it this way. Even if she organized the truckers, ALL of them, she's not responsible for anything unlawful that any of them did. It is not unlawful to organize a protest. She didn't drive the trucks to the capital. She didn't honk their horns and make noise. Therefore, 31 days and counting is an obvious hoax. That's my take. And for good measure, for scaring the people, she supposedly spent time in jail.


Here's all four Gospels wrapped into one story.

For Some Prophetic Proof for Jesus as the Predicted Son of God.
Also, you might like this related video:

Pre-Tribulation Preparation for a Post-Tribulation Rapture