return to homepage AN EXPLOSION OF THE 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.
[1 – (2GM/Rc ^{2}]
Let us unwind it. Since a = GM/R ^{2} and v^{2} = 2aR
The blueshift equation can be rewritten as f_{r} = f_{e}[1 – (v_{2}^{2} /c^{2})][1 – (v _{1}^{2}/c^{2})]
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
f_{r} = 1/t_{r} = γ[1 + (v/c)] f_{e} = f_{e}[1 + (v /c)]√[1 – (v ^{2}/c^{2})]There is another major problem, since v ^{2} = 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 v^{2} = 2aR can't apply. The equation for a situation with an initial velocity is
v ^{2} = v_{0}^{2} + 2aR
I will be told that the light can't have a velocity before it is emitted, and it is emitted at t _{0}, but that is to ignore how the equation v^{2} = 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 papers ^{2}, 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 f_{r} = 1/t_{r} = 1/t_{e}[1 – (v/c)]t _{e}[1 – (v/c)] = 1/[1 + (v/c)f_{e}]t _{e} = 1/[1 + (v/c)][1 – (v/c)]f_{e}f_{r} = 1/t_{r} = {1/[1 – (v/c)]}{[1 + (v/c)][1 – (v/c)]f_{e}f_{r} = 1/t_{r} = [1 + (v/c)]f_{e}
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
f_{r} = f_{e}[1 + (gh/c^{2})]
ds = dt[1 + (gh/c ^{2})]
Then he says, “A more careful computation gives us an expression good for all velocities:” ds = dt√[1 + 2(gh/c ^{2})]
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 = t ^{3}. 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 v^{2}= 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 + gt ^{2}/2
That is the additional distance it would travel during each interval, due to the acceleration. Then we integrate: 2sΔs = 2ct(ct + gt ^{2}/2)
s _{f} = 2c^{2}t^{2} + gct^{3}
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. v _{f} = c + 2c^{2}t v _{av} = (v_{f} – c) /2v _{av} = c^{2}t h = c ^{2}t^{2} s = ct h = s ^{2}
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 = √(c ^{2}t^{2}) = ct
Since h is given as 22.5 meters in this experiment, t is equal to 1.58 x 10 ^{-8}s, and gt is equal to 1.55 x 10^{-7}m/s. Pound and Rebka found a value of 7.5 x 10^{-7}m/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 f_{r} = f_{e}[1 + (v/c)]v/c = (g/c ^{2})√h f_{r} = f_{e}[1 + (g/c^{2})√h]
That is what Feynman was trying to derive, without success. Now for the fuller solution. v _{f} = c + 2c^{2}t + gct^{2}v _{av} = (v_{f} – c) /2v _{av} = c^{2}t + gct^{2}/2h = c ^{2}t^{2} + gct^{3}/2s = ct h = s ^{2} + (gct^{3}/2)s = √[h – (gct ^{3}/2)]v = gs/c = (g/c)√[h – (gct ^{3}/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 10 ^{9}km. 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.
^{1}http://milesmathis.com/adp.html^{2}http://milesmathis.com/muon.html^{3}http://milesmathis.com/long.html*I first looked at this math in my first paper on Feynman. If this paper was useful to you in any way, please consider donating a dollar (or more) to the SAVE THE ARTISTS FOUNDATION. This will allow me to continue writing these "unpublishable" things. Don't be confused by paying Melisa Smith--that is just one of my many |