Strange Posted October 24, 2016 Share Posted October 24, 2016 If for 24h, as measured with a system on the ground (which is assumed to be in rest), a certain number of cycles is sent upward to a detector (which is also in rest), then - according to the ground system - that number of cycles must also be received by the detector in the same time interval. Else cycles would go missing "in mid air"! That excludes a change of light frequency "in transit". So you were saying that the number of cycle counted in the observers reference frame will be the same as the number of cycle counted in the observers reference frame. That is completely tautological so I don't see how it can be used as an argument for (or against) anything. It would be true if there was a change of frequency in transit, or not, or if photons briefly turned into chocolate mousse in transit. It can't not be true. Whatever happens. Link to comment Share on other sites More sharing options...
Sriman Dutta Posted October 25, 2016 Author Share Posted October 25, 2016 Thanks now I have learnt the velocity addition formula in SR. Thanks for all your help. Link to comment Share on other sites More sharing options...
Tim88 Posted October 25, 2016 Share Posted October 25, 2016 That's great Sriman. And I may have narrowed down the issue of Strange, it's probably a matter of definition. - frequency = number of cycles per second. - to change = to become altered or modified. For example, if I measure a length a stretch of water in nautical miles, I will find a different distance than if I measure it in sea miles. Nevertheless, the length of that stretch of water cannot be altered or modified by my measuring of it with different length standards; it certainly did not change. By definition, if a frequency changes in transit, this means that the number of cycles that is received in one second is different from the number of cycles that is emitted during the same time, according to the same time standard. But that is impossible in a stationary situation - just as impossible as photons turning into chocolate mousse. And as E=hf, also the energy cannot change. I think that that's just what Sriman already understood in the first post. Link to comment Share on other sites More sharing options...
Strange Posted October 25, 2016 Share Posted October 25, 2016 By definition, if a frequency changes in transit, this means that the number of cycles that is received in one second is different from the number of cycles that is emitted during the same time, according to the same time standard. The only way you can know how many cycles were emitted in another frame of reference is to count the number of cycles received. (So, trivially, you will always get the same number for "emitted" and "received"). Any mechanism you use to count the number of cycles emitted will be affected in the same way as the frequency of the photon and so, again, you will always get the same number. This tells you nothing. Link to comment Share on other sites More sharing options...
swansont Posted October 25, 2016 Share Posted October 25, 2016 That's great Sriman. And I may have narrowed down the issue of Strange, it's probably a matter of definition. - frequency = number of cycles per second. - to change = to become altered or modified. For example, if I measure a length a stretch of water in nautical miles, I will find a different distance than if I measure it in sea miles. Nevertheless, the length of that stretch of water cannot be altered or modified by my measuring of it with different length standards; it certainly did not change. By definition, if a frequency changes in transit, this means that the number of cycles that is received in one second is different from the number of cycles that is emitted during the same time, according to the same time standard. But that is impossible in a stationary situation - just as impossible as photons turning into chocolate mousse. And as E=hf, also the energy cannot change. I think that that's just what Sriman already understood in the first post. This points back to a terminology issue we've seen in other discussions. Change vs difference. The number of cycles per unit time is different in the two frames, i.e. they do not see the same frequency. But nothing changes in a single frame. Link to comment Share on other sites More sharing options...
Tim88 Posted October 25, 2016 Share Posted October 25, 2016 (edited) emphasis mine: The only way you can know how many cycles were emitted in another frame of reference is to count the number of cycles received. (So, trivially, you will always get the same number for "emitted" and "received"). Any mechanism you use to count the number of cycles emitted will be affected in the same way as the frequency of the photon and so, again, you will always get the same number. This tells you nothing. Once more, we use only a single reference system for a "global" analysis; otherwise it's meaningless and one is almost sure to mess up, as I explained. As a matter of fact GPS uses a single reference system with detectors at different altitudes and speeds. If there was a change of frequency in transit due to altitude (or if photons would turn into chocolate mousse), GPS systems would be messed up. The conservation of cycles is not really different from for example conservation of total current in a river or a conductor. One doesn't need to "count in different frames" for those laws of physics; in fact one should not do that. This points back to a terminology issue we've seen in other discussions. Change vs difference. The number of cycles per unit time is different in the two frames, i.e. they do not see the same frequency. But nothing changes in a single frame. Yes, exactly. A difference is not a change. Maybe I explained this less well than Einstein; and as he explained it in a very clear manner, I'll cite his explanation next as I now found it back: "[..] it is difficult to discover whether the inferred influence of the gravitational potential really exists.On a superficial consideration equation (2) [f1 = f2 (1 + gh/c2] [..] seems to assert an absurdity. If there is constant transmission of light from S2 to S1, how can any other number of periods per second arrive in S1 than is emitted in S2? But the answer is simple. [..] we must certainly define the time in K in such a way that the number of wave crests and troughs between S2 and S1 is independent of the absolute value of time ; for the process under observation is by nature a stationary one. [..] If we measure time in S1 with the clock U, then we must measure time in S2 with a clock which goes 1 + gh/c2 times more slowly than the clock U when compared with U at one and the same place." - Einstein 1911, "On the influence of gravitation on the propagation of light" (in the last sentence I substituted gh back for "phi"). Edited October 25, 2016 by Tim88 Link to comment Share on other sites More sharing options...
Strange Posted October 25, 2016 Share Posted October 25, 2016 Once more, we use only a single reference system for a "global" analysis; otherwise it's meaningless and one is almost sure to mess up, as I explained I agree that you can only use a single reference frame, which is why saying that a number is equal to itself is beyond meaningless. Perhaps you don't understand the conservation of cycles I have never heard of such a thing before. And, as it is not generally true, it doesn't seem a very useful concept. Link to comment Share on other sites More sharing options...
swansont Posted October 25, 2016 Share Posted October 25, 2016 emphasis mine: Once more, we use only a single reference system for a "global" analysis; otherwise it's meaningless and one is almost sure to mess up, as I explained. As a matter of fact GPS uses a single reference system with detectors at different altitudes and speeds. If there was a change of frequency in transit due to altitude (or if photons would turn into chocolate mousse), GPS systems would be messed up. And yet GPS systems in orbit are set to run at a different frequency than the frequency that at which a ground-based clock would run. You do analysis in one reference frame, that's true. Mixing frames is a common way to mess everything up. But you can do the analysis in another frame, and it will be valid, and the numbers will be different. Link to comment Share on other sites More sharing options...
DimaMazin Posted October 25, 2016 Share Posted October 25, 2016 Thanks now I have learnt the velocity addition formula in SR. Thanks for all your help. And don't forget to study fraquency/energy change due to velocity. Janus is very simple here: "Now for light,traveling in a vacuum which does not require a medium and adjusting for Relativity, we use where v/c is the relative speed of source and receiver measured as a fraction of the speed of light and is positive when they are approaching each other. With this formula, it doesn't matter which(source or receiver), is considered to be moving." Edited by Janus, 31 December 2014 - 08:45 PM. Link to comment Share on other sites More sharing options...
Sriman Dutta Posted October 25, 2016 Author Share Posted October 25, 2016 Hi DimaMazin I see that the formula you posted is quite identical to the Doppler's Effect equation. 1 Link to comment Share on other sites More sharing options...
Tim88 Posted October 26, 2016 Share Posted October 26, 2016 (edited) I agree that you can only use a single reference frame, which is why saying that a number is equal to itself is beyond meaningless. I have never heard of such a thing before. And, as it is not generally true, it doesn't seem a very useful concept. This is not about "a number is equal to itself"; indeed that would be meaningless! Here both Einstein and I already explained why the idea that frequency might change "in mid air" is absurd, and why that fact is useful for correctly interpreting gravitational time dilation - as a matter of fact, it was an important step on the road to GR. But never mind we'll leave it at that, just as you said. And yet GPS systems in orbit are set to run at a different frequency than the frequency that at which a ground-based clock would run. You do analysis in one reference frame, that's true. Mixing frames is a common way to mess everything up. But you can do the analysis in another frame, and it will be valid, and the numbers will be different. Yes, that's how to tune everything into a single system - and as cited, this procedure was already foretold by Einstein in 1911. Hi DimaMazin I see that the formula you posted is quite identical to the Doppler's Effect equation. That equation includes the time dilation effect, which makes it invariant between reference systems. See here for the classical Doppler effect equation: https://en.wikipedia.org/wiki/Doppler_effect#General Edited October 26, 2016 by Tim88 Link to comment Share on other sites More sharing options...
Sriman Dutta Posted October 26, 2016 Author Share Posted October 26, 2016 Thanks Tim.... I think after such a long discussion, my concept of SR has improved a lot. Link to comment Share on other sites More sharing options...
Tim88 Posted October 27, 2016 Share Posted October 27, 2016 (edited) That's good Sriman - but be aware that your original question was about GR! Einstein found the approximate gravitational time dilation equation of posts # 13 and #31 with the equivalence principle. The equivalence principle is already known from Newton's mechanics: if you are inside a free falling elevator (let's hope it never happens!), the effects of acceleration compensate the effects of gravity - it will look as if you are floating. Inversely, if you are far away from Earth in a fast accelerating rocket, the effect that you feel due to acceleration at 1g is just as if you are at rest on Earth. Einstein reasoned that this also should hold for measurements of clock rates: the effect from gravity should be the same as the effect from acceleration. When a rocket accelerates, there is a Doppler effect due to the changing speed. In that way it was possible to predict the effect of gravity on observed frequencies between two clocks at a distance. Now I cannot find a link to the derivation, but it's not so difficult to work out - maybe someone else will fill that in, or else I can look it up later. After he had found the gravitational frequency equation, he reasoned how that equation should be understood. With only clocks in rest, there cannot be a Doppler effect. And wave crests cannot get lost in transit. There was only one logical solution possible: a clock should run at a faster rate when it has been placed higher up in a gravitational field. And as he had visited a polytechnic high school, he immediately thought about how one should deal with that in practice. It's no good to do measurements with clocks that run at different rates! So, he already considered that very precise clocks that are put at a high elevation, should be corrected for their height so that they will run at the same rate as clocks at sea level. In this way the clocks can be used as part of a single reference system. Edited October 27, 2016 by Tim88 Link to comment Share on other sites More sharing options...
DimaMazin Posted October 27, 2016 Share Posted October 27, 2016 That's good Sriman - but be aware that your original question was about GR! Einstein found the approximate gravitational time dilation equation of posts # 13 and #31 with the equivalence principle. The equivalence principle is already known from Newton's mechanics: if you are inside a free falling elevator (let's hope it never happens!), the effects of acceleration compensate the effects of gravity - it will look as if you are floating. Inversely, if you are far away from Earth in a fast accelerating rocket, the effect that you feel due to acceleration at 1g is just as if you are at rest on Earth. Einstein reasoned that this also should hold for measurements of clock rates: the effect from gravity should be the same as the effect from acceleration. When a rocket accelerates, there is a Doppler effect due to the changing speed. In that way it was possible to predict the effect of gravity on observed frequencies between two clocks at a distance. Now I cannot find a link to the derivation, but it's not so difficult to work out - maybe someone else will fill that in, or else I can look it up later. After he had found the gravitational frequency equation, he reasoned how that equation should be understood. With only clocks in rest, there cannot be a Doppler effect. And wave crests cannot get lost in transit. There was only one logical solution possible: a clock should run at a faster rate when it has been placed higher up in a gravitational field. And as he had visited a polytechnic high school, he immediately thought about how one should deal with that in practice. It's no good to do measurements with clocks that run at different rates! So, he already considered that very precise clocks that are put at a high elevation, should be corrected for their height so that they will run at the same rate as clocks at sea level. In this way the clocks can be used as part of a single reference system. Posted by Janus on 28 May 2016 - 09:13 PM "The time dilation factor for a clock in circular orbit ( as measured by a distant observer) is found by: where r is the orbital radius. This is what you get when you take the combined effect of gravitational and relative velocity time dilation effects found by. And substitute , the orbital velocity at r, for v" You can define different time like Janus defines, but how defined different time can help to define energy change of photon? Link to comment Share on other sites More sharing options...
Sriman Dutta Posted October 28, 2016 Author Share Posted October 28, 2016 Different times has to do something with the energy of photon DimaMazin. If t' = ty, y is Lorentz factor => 1/t' = 1/ty => f' = f/y => hf' = hf/y => E' = E/y Link to comment Share on other sites More sharing options...
Tim88 Posted October 28, 2016 Share Posted October 28, 2016 (edited) Bold face added: That's good Sriman - but be aware that your original question was about GR! Einstein found the approximate gravitational time dilation equation of posts # 13 and #31 with the equivalence principle. The equivalence principle is already known from Newton's mechanics: if you are inside a free falling elevator (let's hope it never happens!), the effects of acceleration compensate the effects of gravity - it will look as if you are floating. Inversely, if you are far away from Earth in a fast accelerating rocket, the effect that you feel due to acceleration at 1g is just as if you are at rest on Earth. Einstein reasoned that this also should hold for measurements of clock rates: the effect from gravity should be the same as the effect from acceleration. When a rocket accelerates, there is a Doppler effect due to the changing speed. In that way it was possible to predict the effect of gravity on observed frequencies between two clocks at a distance. Now I cannot find a link to the derivation, but it's not so difficult to work out - maybe someone else will fill that in, or else I can look it up later. [..] On a second look, my phrasing in bold was imprecise (and it somewhat mismatches with the rest). SR only took care of effects due to speed; eventual effects from gravitation were not yet included in optics, and new corrections to mechanics were also foreseeable. The research program of that time was not yet completed. So, Einstein used the equivalence principle to predict the effects of gravity on optical phenomena and related physical measurement tools. As I could not quickly find the derivation, I now re-derived it myself as follows.(Funny enough, I next found Einstein's derivation which was more complex, as he started from SR in combination with light-based simultaneity; but next he simplified so much that he effectively didn't use SR!).We can think of a rocket in "deep space", accelerating from v=0 to v= 9.8 m/s in 1s. Inside, the phenomena should look just as on Earth. Also, special relativistic effects are negligible for this case.At t0=0 a light pulse is sent from the floor upwards to a receiver at the ceiling, at a height h. We'll assume that the ceiling will not move much before the light arrives. Then the light distance = h.The light pulse arrives at time t. The receiver is then moving with speed vr:vr = at (1)The light pulse travels the distance ct:h = ct => t = h/c (2)(1) + (2) -> vr = a h/c (3)For this, classical Doppler is good enough:fr/f0 = (c+vr)/c (4)(3) + (4) => fr/f0 = (c + ah/c)/cfr/f0 = 1+ah/c2 (5)The same observation should be made on earth with gravitational acceleration g:fr/f0 = 1+gh/c2 (6) After he had found the gravitational frequency equation, he reasoned how that equation should be understood. With only clocks in rest, there cannot be a Doppler effect. And wave crests cannot get lost in transit. There was only one logical solution possible: a clock should run at a faster rate when it has been placed higher up in a gravitational field. And as he had visited a polytechnic college, he immediately thought about how one should deal with that in practice. It's no good to do measurements with clocks that run at different rates! So, he already considered that very precise clocks that are put at a high elevation, should be corrected for their height so that they will run at the same rate as clocks at sea level. In this way the clocks can be used as part of a single reference system. If some part is not understandable, please tell; it's all simple math. Edited October 28, 2016 by Tim88 Link to comment Share on other sites More sharing options...
DimaMazin Posted October 28, 2016 Share Posted October 28, 2016 Different times has to do something with the energy of photon DimaMazin. If t' = ty, y is Lorentz factor => 1/t' = 1/ty => f' = f/y => hf' = hf/y => E' = E/y Lorentz factor reduces time there.But lorentz factor can increase or reduce frequency.Time doesn't tell about fraquency at motion in gravitational field. Link to comment Share on other sites More sharing options...
Sriman Dutta Posted October 28, 2016 Author Share Posted October 28, 2016 Thanks Tim. Link to comment Share on other sites More sharing options...
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