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by Miles Mathis

The Pound-Rebka Experiment took place at Harvard in 1959. It is said to be the first experiment to unequivocally prove gravitational blueshifts, and to be the experiment that began the age of precision tests of General Relativity. Is this true? Well, yes and no. Let us analyze the experiment a bit more closely than it has so far been analyzed.

Basically, Pound-Rebka set up an experiment to measure the blueshift of light traveling straight down. They emitted the light at the top of a four-story building and caught it in the basement. We are told this is high-tech, but it sounds pretty low-tech to me so far. They couldn't find a taller building on campus, or in Boston? But you will see why they kept the distance small in a moment.

To measure the amount of blueshift, they thought to offset it with redshift. This makes perfect sense to me, and I congratulate them for this part of the experiment. They took some iron57 that was emitting gamma rays, let some of it emit from the top of the building downwards, and let the rest re-absorb the same rays in the basement. We suppose they found it wouldn't re-absorb in the basement, because it was at a different gravitational level. The light had been blueshifted and wouldn't fit back into the hole it left. So they let their lower iron57 move down with the rays, in a little elevator. This redshift offset the blueshift, and they had “measured” it. Appears brilliant, so far (except for that short “tower”).

Let me say that I accept the initial assumptions of this experiment. I agree that the motion of the light down in a gravity field should cause a blueshift. I agree that the movement of the iron in its elevator should cause a redshift. I agree that these should be capable of offsetting. I agree with the main lines of both Special Relativity and General Relativity. I believe in time dilation and length contraction and the need for transforms. I am not here to disprove Relativity or blueshifts. I am here to analyze this experiment, and especially the math underneath it.

For the brilliance of the experiment fades very fast when we start looking at the equations they used. We are told at Wikipedia that “the redshift is predicted by Special Relativity,” and that it is expressed by this equation:

fr = fe[1 – (v/c)]
         [1 + (v/c)]

That is a very strange fudge, as I will now show. The redshift predicted by Special Relativity is called the relativistic Doppler Effect, and it is normally written with square roots, like this:

fr = fe[1 – (v/c)]
         √[1 + (v/c)]

That is the way it is written at Wiki, on the “Relativistic Doppler Effect” page. So why is the Pound-Rebka math done without the square roots? Maybe they think this redshift is not really relativistic, but if so, why did they just say that it was predicted by Special Relativity? And if the redshift is not relativistic, the “non-relativistic” equation for Doppler shift on frequency is

fr = fe[1 – (v/c)]

So in neither case is the first equation the current derived Doppler shift for frequency. You will say that they are ignoring the square roots in this case, since the velocity and time in the experiment is small, but it is still highly curious. You will find it even more highly curious in a moment.

The derivation of the relativistic Doppler Effect relies on Einstein's value for gamma. I can show you very quickly how the current derivation fails. We will follow the derivation at Wiki. We are given a velocity between emitter and receiver, then are given this equation for “the time observed between crests”:

t = λ/(c – v) = 1/[1 – (v/c)fe]

That is just our frequency transform in disguise.

fr = fe[1 – (v/c)]

Time is expressed in seconds and frequency is expressed in s-1, so the equation has just been flipped. This frequency transform has nothing to do with Special Relativity. It was known long before.

Then we are given this equation, where to is defined as the observed time (“the observer will measure this time to be”):

to = t/γ

That is the time transform of special relativity. But look what they just did: they just transformed a time that was already defined as “observed.” In other words, they just transformed an observed time into an observed time. You really have to watch your variables here, or they will fudge an equation before you know what hit you. Look above at their first variable assignment. The time t is already defined as observed. And now they are transforming that t with gamma into an observed time to. Incredible! What they should have done is label the first t as tr (time received). The frequency in that first equation is defined as f emitted, so the time should be t received. That would make the time transform,

tr = te

That would be the correct way to express it, supposing gamma were correct.

I will be told that the time in the first equation is really the emitted time, but that can't be right however you look at it. First of all, we can't observe an emitted time. If we could, we wouldn't need a frequency or time transform in the first place! Second, that first equation is just the classical frequency transform (from optics) flipped, as I already said. It should be written like this:

tr = 1/fr = 1/[1 – (v/c)fe]

That is this equation flipped

fr = fe[1 – (v/c)]

If you don't agree, than hang on to your hat, because Wiki admits it in the next step. The current derivation combines their equation with gamma with their first equation, obtaining

fo = 1/to = γ[1 – (v/c)] fe
fr = fe√[1 – (v/c)]
         √[1 + (v/c)]

See, they just admitted that fo = 1/to. That means that the observed frequency is one over the observed time. Which is just another way of saying the received frequency is one over the received time. The observed time and the received time are the same damn thing. They both stand for our data—the events on our end—as opposed to the events as seen from the other body or other system. And that means that they just substituted wrongly in their derivation. The derivation is a big fudge.

If you still can't see it, rewrite the time transform in the opposite way. If you don't like my notation, see how yours works. Let us write the transform like this

te = tr

It has to be one or the other, right? If I am wrong, then we have to write the transform like this. But that doesn't make any sense. You are transforming a received time into an emitted time? Why would you want to do that?

My critic will say, “So that we can discover the amount of redshift, of course.” But that is to miss the entire point of these transforms, both of frequency and time. You cannot observe emitted times and frequencies. If you could observe emitted times and frequencies, then there would be no shifts. You would be observing the unshifted waves. No, the observed data is shifted. The observed data is the received data. Which means the t variable in the first equation is already observed. We cannot transform an observed time into an observed time. The given time in the first equation is t, and it is not the emitted time. We know it is the observed time, both by the verbal assignment and by the form of the equation. Which means the equations should go like this, if we simply follow the mainstream math without fudging,

fr = 1/tr = γ/ te
1/[1 – (v/c)fe] = te
γ/[1 – (v/c)fe] = te
fr = γ/{γ/[1 – (v/c)fe]}
fr = fe[1 – (v/c)]

Yes, it reduces back down to the frequency transform from optics, since the frequency transform was already relativistic to start with. As this derivation showed in the first equation, the velocity is being subtracted from c just to find the frequency transform. t = λ/(c – v). That is already doing relativity. When you start subtracting a velocity from c, you are doing special relativity, by definition. Since the frequency transform was already a relative transform, we don't need to do a time transform on top of it. In fact, all the corrected first-degree transforms look like the frequency transform, as I have shown in many other papers3. I have shown that gamma is false and that the real time transform for one degree of relativity for a receding object is

tr = te [1 + (v/c)]

This means that the completely corrected derivation of the “relativistic” Doppler Effect goes like this:

fr = 1/tr = 1/te[1 + (v/c)]
te[1 + (v/c)] = 1/[1 - (v/c)fe]
te = 1/[1 - (v/c)][1 + (v/c)]fe
fr = {1/[1 + (v/c)]}{[1 + (v/c)][1 – (v/c)]fe
fr = fe[1 – (v/c)]

With or without gamma, the equation reduces the same way. But my derivation is the correct one, since gamma is not correct.

One final thing to notice before we move on to the blueshift equation. If the frequency is just 1 over the time, as Wiki admits, then why are they transforming the time once and the frequency twice? If you study their equations, it is easy to see that they have done two transforms on the frequency but only one on the time. My critic will answer that the first frequency transform 1 – (v/c) is not an SR transform, since it doesn't include gamma; but in fact it is. That differential 1 – (v/c) is a relativity transform, simply because it is found by subtracting v from c, as in λ/(c – v). Einstein found gamma exactly the same way: by subtracting v from c. Whenever you subtract v from c, you are already doing relativity. The frequency transform was already relativistic, and no one seems to understand that. It doesn't need to be made relativistic: it already is. It is just a simpler form of gamma. And it is correct, while gamma is incorrect. Gamma is just the outcome of a mistake Einstein made in his early variable assignments, as I have shown quite clearly elsewhere1.

But let us move on. Next, the Pound-Rebka derivation gives us this equation for the blueshift:

fr = fe {1 – [2GM/(R + h)c2}
            [1 – (2GM/Rc2]

Let us unwind it. Since a = GM/R2 and v2 = 2aR

The blueshift equation can be rewritten as

fr = fe[1 – (v22 /c2)]
         [1 – (v12/c2)]

And now we begin to see what a mess it is. Since it is some bastardized form of gamma, we know it is wrong from the beginning. It isn't even consistent with their derivation of relativistic Doppler, since if it followed that derivation, it would be

fr = 1/tr = γ[1 + (v/c)] fe = fe[1 + (v /c)]
                                             √[1 – (v2/c2)]

There is another major problem, since v2 = 2aR can't apply here. We have to ask for a definition or an assignment of that variable v. To get a blueshift, the surface of the Earth has to be taken as moving at the light during the entire experiment. That is the velocity assignment. I have no problem with that assignment. It is said to be allowed by Einstein's equivalence principle, and I agree with both the principle and the assignment. My problem is that the initial velocity is not zero. Neither the initial velocity nor the final velocity of the Earth toward the light can be zero, since the light is moving the entire time. The light does not start from zero, and neither does the Earth. Therefore, the equation v2 = 2aR can't apply. The equation for a situation with an initial velocity is

v2 = v02 + 2aR

I will be told that the light can't have a velocity before it is emitted, and it is emitted at t0, but that is to ignore how the equation v2 = 2aR must be applied. It is applied to an object that is accelerated from rest; an object whose velocity is due entirely to the acceleration a. But that does not apply to light. Light has its own velocity independent of the acceleration field. Therefore the velocity of the light must be integrated into every differential of acceleration.

A critic will say that the light cannot be accelerated, since it is already at c. It cannot go over c. But if that is the case, then this experiment should not be able to show any blueshift. If the velocity of the light cannot be changed by the field, then the field cannot cause a blueshift, by definition.

It is true that the light itself cannot go over c, but we are moving the surface of the Earth at the light, in order to show the shift. That is not disallowed, since it simply shrinks the distance light must go to get to the Earth. I have already shown this in my muon papers2, and the mainstream assumes it when it does any kind of math on this problem. Richard Feynman admitted this openly in his Lectures on Gravitation*. In Lecture 7.2, he allows the bottom of his box to move toward light emitted at the top, and his problem is absolutely equivalent to the one we are looking at here. It is just like subtracting v from c in the equation t = λ/(c – v). His math is the very math used in this problem, historically.

If we correct the blueshift as we did the redshift, we find

fr = 1/tr = 1/te[1 – (v/c)]
te[1 – (v/c)] = 1/[1 + (v/c)fe]
te = 1/[1 + (v/c)][1 – (v/c)]fe
fr = 1/tr = {1/[1 – (v/c)]}{[1 + (v/c)][1 – (v/c)]fe
fr = 1/tr = [1 + (v/c)]fe

In other words, it too reduces to the frequency transform. But since I agree that the gravity field of the Earth should cause a blueshift, I must come up with a new and better equation to express that.

To start with, the gravity transform should show more blueshift the longer the light is in the field. Gravity is an acceleration, so the shift must accelerate like anything else. It must change with time. The current equation expresses this in a way, but it does not express it clearly. In textbooks, the variables are always limited to hide this fact. In my paper on Feynman, I show this clearly, but Feynman himself blows by it without comment. His own equations show it, if we take an extended time, but in his Lectures he neither takes us there nor implies that it is true. If anything, he misdirects us, as the equations at Wiki and in other textbooks do.

Yes, light not only shifts blue, it must shifter bluer the longer it moves down in the field. And this means that it is not just the strength of the field that matters, it is the size of the field, and the time light spends in it.

The math for the Pound-Rebka experiment misdirects us very blatantly, since all the math above is stated and then thrown out. At Wiki, they set their two big equations equal to eachother, and then ditch them. They say that if h is much less than R,

v ≈ gh/c

They admit that if h is larger, this is not true, but we never get the full math for this experiment or any like it. Feynman does not give it to us, either, since he also limits the time and the velocity, finding the same equation.

The Pound-Rebka-Wiki math can't get the right answer, since it is full of gamma and upside-down substitutions, so let's look at Feynman's math. Feynman doesn't use the big blueshift equation above, he just substitutes gh/c into the "non-relativistic" frequency transform, like this

fr = fe[1 + (gh/c2)]

ds = dt[1 + (gh/c2)]

Then he says, “A more careful computation gives us an expression good for all velocities:”

ds = dt√[1 + 2(gh/c2)]

He doesn't bother to give us this careful computation, of course, since it isn't careful at all. How can it be, when the observed frequency becomes ds and the emitted frequency becomes dt? How is that logical? How can it be careful, when Feynman doesn't even understand how light or a clock would blueshift? In the very next section, Feynman teaches one of his famous puzzles. He asks himself how we would maximize the time change of a clock by moving it in a gravity field. The second sentence of this section is, “We know that a clock should run faster as we move it up, away from the surface of the Earth.” BUT HE JUST PROVED THE OPPOSITE IN THE PREVIOUS SECTION! He showed that either a clock or a ray of light would be blueshifted if it is moving down in the field. A blueshifted clock is running faster, by definition. The clock's period must be decreased, just like the light's. All we have to do is notice that the time between ticks is like the wavelength of the light. This wave gets compressed by the blueshift, just like the light's wave. The ticks are closer together, as seen or heard by the observer. If the ticks are closer together, the clock appears to be running faster.

My critic will say, “No, Feynman is talking about the SR shift in that sentence. In the previous section, he was calculating the GR shift.” That can't be, because in the next sentence after that one, he says, “On the other hand, as we move it, it should lose time because of the time dilation of special relativity.” So the first sentence is the GR effect and this sentence is the SR effect. That is why this sentence contains the clause “away from the surface of the Earth.” That is a gravitational clause.

He has the clock running faster moving up and running faster moving down, both caused by GR.

Feynman, like everyone else, has misunderstood the time period. He doesn't understand that time dilation is the redshift and that the blueshift is time compression. Time compression is a faster running clock. He also thinks that all motion in any direction causes time dilation, although he just proved the opposite in the previous section. To squirm out of this terrible misunderstanding, scientists now claim that Special Relativity causes redshifts in all cases, and that General Relativity is the cause of any blueshifts. But neither the equations nor the fields confirm that. It is totally illogical and mathematically asinine. All simple data from pulsars contradicts it. I have been told that when the pulsar is moving away, we see a redshift due to SR, and that when the pulsar is moving toward us, we see a blueshift due to GR. Am I the only one who sees how absurd that is? It is the same object: either the gravitational shifts or the Doppler shifts have to be stronger. If one is stronger than the other, it will be stronger in both directions. Besides, the Doppler shifts caused by motion have to overwhelm the gravitational shifts, otherwise gravitational shifts would have been easy to measure. It would not have required Pound-Rebka to prove them: we could have proved them even easier than Hubble shifts, supposing they existed with such strengths. Gravitational shifts are neither as large nor as easy to see or measure as Doppler shifts. What we are seeing with pulsars is Doppler shifts in both instances. In both cases, we are seeing SR, and in approach we see an SR blueshift. That destroys the current model completely, since a blueshift is not a dilation: it is a compression.

Now let me show you the correct math. I will show you both the short math to solve Pound-Rebka, and the long math to solve any possible experiment, over any distance. All Pound and Rebka were doing is trying to offset the motion of the Earth up by moving their iron down. We will not get into the question of whether the Earth's surface is actually moving. The equations must be done as if it were, and even Feynman admits that. This problem can only be solved by using the equivalence principle, and the mainstream knows that even though they like to hide it. If we want to offset the two motions, we have to recognize that the Earth is moving up with an acceleration, not a velocity. Therefore, our iron should have to move down with an acceleration as well. Of course, we only need to match the final velocities, so our solution will be a velocity, but to reach that final velocity, the Earth has to be accelerating during the time the light is moving down. If we don't admit that, we won't be able to express it in the equations.

This means that what we have here is an acceleration meeting a velocity. We have the acceleration of the Earth meeting the velocity of the light. Which means we have a cubed acceleration, y = t3. Since I have recently done the math for this in my muon papers, it makes this paper much easier. The mainstream has never bothered to notice this, and they never solve by noting or using a cubed acceleration. They try to use the old equation v = at, or some variation of it like v2= 2aR, but those equations don't work in this situation. They don't work because they don't take into account the fact that the light has a velocity uncaused by the field acceleration g. The light's velocity has to be integrated into the acceleration, creating a cubed acceleration, and no one has ever noticed that.

Now, the current derivation is correct in only one way: if the time is small, we can simplify the equations, like this

s = ct

That is the distance the light would travel in no field.

Δs = ct + gt2/2

That is the additional distance it would travel during each interval, due to the acceleration. Then we integrate:

2sΔs = 2ct(ct + gt2/2)

sf = 2c2t2 + gct3

If the time is small, we can ignore the second term as negligible.

If we give all the motion to the light, as we did in these equations, it will seem to have gone over c. But if we do not give all the motion to light, it won't.

vf = c + 2c2t
vav = (vf – c) /2
vav = c2t
h = c2t2
s = ct
h = s2

This just means that the light didn't have to go as far as we think it did. While the light was moving toward the Earth, the Earth was moving toward the light. So the light doesn't really have to travel h. It travels √h and the acceleration does the rest.

√h = √(c2t2) = ct

Since h is given as 22.5 meters in this experiment, t is equal to 1.58 x 10-8s, and gt is equal to 1.55 x 10-7m/s. Pound and Rebka found a value of 7.5 x 10-7m/s, so we are pretty close. But their value cannot be correct, because buried in their math is the assumption that the light traveled the entire distance h. Watch this:

h = ct
t = h/c
v = gt
v = gh/c

That is how their simplified final equation is derived. {Notice that v = gt. That was the secret info hidden here all along. We need all these equations just to find the time. If we can measure the time of the experiment accurately, we don't need all these equations.} But the first equation means that the light traveled h, all by itself. If the light traveled h, then the Earth could not have moved in the math. If the Earth does not move in the math, then we cannot find a blueshift. Even Feynman understands that and admits it in his math. The bottom of his box is moving up. To solve this problem, we have to assign a velocity to the Earth. Otherwise, what is the velocity of the iron offsetting? The iron is moving down, toward the Earth. If the Earth is stationary in the math, what is the iron offsetting, as a matter of kinematics? We are supposed to be offsetting a blueshift with a redshift. The motion of the iron causes the redshift. What motion causes the blueshift? The light cannot cause its own shift. We have to give the surface of the Earth a motion in the math. Even the Wiki math shows this. That is why the velocities are different in the numerator and the denominator of the blueshift equation, or why the distances are different. Something besides the light has to be moving to show a blueshift, by definition. You can give it to the field instead of the surface of the Earth, but the end is the same. If the field is moving during the equation, this must be represented in the math. If the light is traveling the entire distance h during the math, the field movement cannot be represented in the math.

For all these reasons, we know that the Pound-Rebka experiment has been pushed in both its math and its outcome. The simplified equation for small times should be v = (g/c)√h not v = gh/c. And all that larking about with time transforms was just an embarrassment, as I have shown. They had to throw out that equation because it is a fudge from beginning to end. They had best put it in the garbage as soon as possible. It is incredible that we continue to be assaulted with such things more than a century after 1905 and more than half a century after 1959.

This makes the frequency transform

fr = fe[1 + (v/c)]
v/c = (g/c2)√h
fr = fe[1 + (g/c2)√h]

That is what Feynman was trying to derive, without success.

Now for the fuller solution.

vf = c + 2c2t + gct2
vav = (vf – c) /2
vav = c2t + gct2/2
h = c2t2 + gct3/2
s = ct
h = s2 + (gct3/2)
s = √[h – (gct3/2)]
v = gs/c = (g/c)√[h – (gct3/2)]

Say we let the light move down for .00001 second. We find that h will be 9,000 kilometers. In that case, our equation is still not complete, because we have to include the variance in g. If we did this calculation near a red giant, for instance, the gravity field would be very large, so we could let the light travel down for quite a while. But the field would be changing, and we would have to monitor that also.

However, even at 9,000 km, the last term is still negligible. This means that the mainstream has still not fully understood this equation or this experiment, since Wiki tells us that if h ≈ R, the simplified equation doesn't work. We see that isn't true, since we are above R here. For the last term to become significant, t would have to go above 10,000s, which means the light would travel over 109km. That would have to be a very large gravitational field, a field larger than the Sun's field at the distance of Jupiter. Not impossible, but very large.

*I first looked at this math in my first paper on Feynman.

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