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The only possible difference between two suspended objects of different make up would be their tangential velocity in the rotating field, since all other motions are either caused by the gravitational field (g) or are a direct response to it (c or f). These latter motions must be assigned to “gravitational mass”, obviously. This means that the difference in 2g t = √c^{2} + 2gh) - cg In this time, the box has gained this much more velocity v = gt = √c ^{2} + 2gh) - cFeynman said that to a first approximation that would have been v = gh/c In his first equation above, t is c/h. In his second t is h/c. I have shown that the second is correct. Perhaps the first was a misprint. Now if φ = gh, then v = √c ^{2} + 2φ) - c= √1 + 2φ/c ^{2}) - cIn order to find the final transformation equation, Feynman says that all we have to do is substitute the last equation into this one: f = f'(1 + v/c) This equation defines the apparent change in frequency due only to relative motion. It has nothing to do with gravity, obviously, since there is no acceleration or gravity variable in it. Substituting we obtain: f = f'[1 + √1 + 2φ/c^{2}) - c ]c f = f'√1 + 2φ/c ^{2}That is the change in frequency from the top of the box to the bottom. Since a clock behaves like a wave, Feynman implies that the equation also applies to the period of a clock at the top of the box seen from the bottom. f = frequency measured at the bottom f' = frequency emitted at the top Since f > f' time appears to go faster at the top of the box. First of all, I find it very interesting that Feynman uses the equation f = f'(1 + v/c). This equation means that because the bottom of the box is moving toward the ray of light, the frequency will increase. The relative motion by itself, with no acceleration, causes a blue shift. That is what this equation must mean. But he does not seem to see that a signal from a clock will act just like a ray of light, as it moves from top to bottom. The light has a frequency of 10^{16}/s, say, and a clock is defined as 1/s, but they are both waves. If the light is blue-shifted, then the clock will be too, as he admits. He says, "the clock at the top looks bluer" (p.94, 7.2.1). A blue clock will be ticking faster. By the equation f = f'(1 + v/c), the clock will be ticker faster even without the gravitational field. If the bottom of the box was moving at a constant velocity, the equation would still apply, and the f would describe the period of the clock. This is important since it contradicts Special Relativity. SR states that all relative motion causes time dilation, which is to say slowing of time. But here Feynman is assuming just the opposite in order to prove that GR speeds up time. Feynman assumes that motion toward in SR causes time to speed up in order to prove that motion toward in GR causes time to speed up. By carrying his equation for v into the equation f = f'(1 + v/c) he is calculating a sort of double blue-shift. Let's try to work the problem out another way. Let's say the clock at the top has a period of t'. We want to calculate the period as seen from the bottom, t. We are given g and h. The bottom of the box receives the first tick at T _{1} = T_{1}' + Δt_{1}. Δt _{1} is the time it takes for light to travel from top to bottom, after tick #1.Δt _{1} = x_{1}/cx _{1}= h - Δh_{1} where Δh_{1} is the distance the bottom travels up while light is traveling down Δh _{1} = v_{av}Δt_{1} = (v_{1} + v_{2}) Δt_{1}/2v _{av} stands for v averageWe need a velocity variable in there instead of g, so that we can vary it from one time to the next. g does not change, and so it will show no increase over time, if such exists. Δt _{1} = 2h - [(v_{1} + v_{2}) Δt_{1}]2c 2cΔt _{1} = 2h - [(v_{1} + v_{2})Δt_{1}]2cΔt _{1} + v_{1}Δt_{1} + v_{2}Δt_{1} = 2hΔt _{1}(2c + v_{1} + v_{2} ) = 2hΔt _{1} = 2h/(2c + v_{1} + v_{2}) The bottom receives the second tick at T _{2} = T_{2}' + Δt_{2}Δt _{2} = x_{2}/cx _{2}= h - Δh_{2}Δh _{2} = vΔt_{2} = (v_{2} + v_{3}) Δt_{2}/2Δt _{2} = 2h - [(v_{2} + v_{3}) Δt_{2}]2c 2cΔt _{2} = 2h - [(v_{3} + v_{2}) Δt_{2}]2cΔt _{2} + v_{2}Δt_{2} + v_{3}Δt_{2} = 2hΔt _{2}(2c + v_{2} + v_{3} ) = 2hΔt _{2} = 2h/(2c + v_{3} + v_{2} )The observed period is then t = T _{2} - T_{1} = (T _{2}' + Δt_{2}) - (T_{1}' + Δt_{1})= T _{2}' + Δt_{2} - T_{1}' - Δt_{1})But T _{2}' - T_{1}' is defined as 1s.So, t = 1s + (Δt _{2} - Δt_{1})= 1s + [2h/(2c + v _{3} + v_{2})] - [2h/(2c + v_{1} + v_{2})]I don't see immediately how to reduce this, but it doesn't really matter. Because what it means is clear. For one thing, it means that the clock is not just blue shifted—the clock at the top is seen by the bottom to be increasing its rate. The clock does not have a constant rate. I think you can see that T _{3} - T_{2} is going to be less than T_{2} - T_{1}, so the period is going to be steadily shrinking. I did not need the equations of SR to find this out though. I did not use the transformation for frequency, like Feynman did. I did need to assume that the clock was ticking a normal rate in its own vicinity, and that we could know that rate, as did Feynman. He says (p.94), "How much is the time difference at various points in space? To calculate it we compare the time rates with an absolute time separation, defined in terms of the proper times ds." This is one of the clearest statements I have found in favor of my interpretation of Relativity. Feynman seems to recognize at times like this that time differences must be measured against a standard, which standard is local time. His assumption of absolute time separation is astonishing, really, since it contradicts the current interpretation of Relativity, which states that the local field is either an illusion or an impossibility. Regardless, I hope you can see that just as with SR, one must assume all fields are equivalent locally in order to calculate how they will be seen differently from a distance. Feynman understands that he can't solve the problem without "absolute time separation" and he makes that explicit in this problem. Let's try one more time to get a simpler transformation equation. Let us try to obtain an x variable instead of a v variable. That will also allow us to solve for different times. Δt _{1} = x_{1}/c the t variable here designates the time it takes light to travel to the bottom of the box. x _{1} = h - Δh_{1} Δh _{1} is the distance the bottom travels up while light is traveling downΔh _{1} = gΔt_{1}^{2}/2 the t variable here is the time it takes for the bottom to travel up Now, set the two times equal to eachother, since the time it takes the light to meet the bottom is the same as the time it takes the bottom to meet the light. x _{1}^{2}/c^{2} = 2Δh_{1}/g Δh _{1} = -gx_{1}^{2}/2c^{2}Δt _{1} = h/c + (gx_{1}^{2}/2c^{3})The first part of this equation is Feynman's first approximation from above. You see that it should have been h/c not c/h. Just like before, we find that Δt _{2} = h/c + (gx_{2}^{2}/2c^{3})t _{1} = T_{2} - T_{1} = (T _{2}' + Δt_{2}) - (T_{1}' + Δt_{1})= 1s + h/c + (gx _{2}^{2}/2c^{3}) - h/c - (gx_{1}^{2}/2c^{3})t _{1} - 1s = g(x_{2}^{2} - x_{1}^{2}/2c^{3})x _{1} is bigger than x_{2}, so t_{1} will be smaller than 1 second. We have proven a blue shift. Butt _{2} = g(x_{3}^{2} - x_{2}^{2}) + 1s2c ^{3}x _{2} > x_{3}, so we still have a blue shift, but x_{2} < x_{1}, which means the blue shift is increasing with time. Notice that as the box accelerates, Δh is going to get larger and therefore x is going to get smaller. Each successive light signal from the top is going to have a shorter distance to go before meeting the bottom. With each passing moment, the clock will get bluer and bluer. Feynman and Einstein don't ever recognize this, and so they don't see it as problematical. Feynman thinks that GR implies simple (and constant) time contraction, as shown by his equations, and that SR always implies time dilation—as in the standard interpretation of Einstein. This gives him an opportunity to pursue an "amusing puzzle" in the next section (see below). But the truth is that it is quite problematical for an acceleration to imply an increasing blueshift, since we do not see such things in reality. According to Einstein's theory of equivalence, we cannot tell if we are accelerating or are in a gravitational field. But the fact is we are in a gravitational field at all times, and yet do not see things the way this experiment tells us. I feel a force pulling me down right now that we call gravity. But according to this thought experiment, I should be able to say that I am in a rocket accelerating upwards. In which case I should see a clock on the ceiling gaining time with each passing second. Even at a relatively small acceleration, I should, after a few years, be going really fast, in which case even a small initial blue shift would become obvious. And after a few decades at this acceleration, people in my rocket (people on this earth) should be approaching c. After millions of years, we should be so close to c as to not matter at all. That clock on the ceiling should be going so fast I can't read it. Why isn't it?For instance, if the gravitational field of the earth is 9.8m/s ^{2} near the surface, and I am 40 years old, then even if the my rocket was going zero at my birth, it would now be going,
9.8m/s/s x 40 x 365 x 24 x 60 x 60 = 1.24 x 10^{10}m/sThat is, I would be going faster than the speed of light, if not for mass considerations. According to Feynman, I should have been gaining mass now for at least a decade, just to slow me down. All incoming light should be so blue-shifted it would have become invisible years ago, unless I had learned to see cosmic rays. If I were Feynman, I would make a mathematical game of it, and figure out how many years and months old I was when I began packing on the pounds and seeing x-rays, but I find such things juvenile, especially when they involve calculating from false theories. I have real work to do—a large part of which is becoming correcting all the mistakes of Feynman. Current physicists might answer that we do not see blue-shifts like this because we are not really accelerating. We are in fact in a field. But if this is the case then Einstein's equivalence fails, and we cannot use math derived from blueshifts to define our field. Feynman has just used acceleration equivalence in order to derive his transformation equation. But the situation is either equivalent or it is not. If we derive equations from blueshifts, then we must explain the lack of blueshifts. If there are no blueshifts, then there is no equivalence, and we must derive equations some other way. The whole reason Einstein postulated equivalence was so that he could calculate the potentials in just this way, and that is why Feynman quotes his method half a century later, calling it marvelous. Feynman's amusing puzzle is to calculate the best movement of a clock in order to make it gain the most time. He assumes that greater potentials give faster clocks (GR), but that motion slows down the clock (SR). Since I have shown in another paper that SR is wrong about motion toward an observer, as with Feynman's frequency increase equation, this whole exercise is moot. Besides, I just showed how Feynman assumes that SR causes a blueshift. He uses the equation f = f'(1 + v/c) above, which is nothing more than a blueshift caused by relative motion. He assumes SR causes blueshifts on one page and that it causes redshifts on another page. This is what is called a contradiction. Not a paradox. A mistake. The Hafele-Keating experiment obviously took its cue from Feynman, though. That experiment is precisely the working out of Feynman's amusing puzzle. [Remarks on the Lectures on Gravitation to be continued] For the completed equations on Feynman's box, see my newer paper on the Pound-Rebka experiment. Similar equations are solved in my paper on the muon and in my paper on the equation v = v _{0} + at.
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