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Gravitational redshift and time dilation

  1. Nov 14, 2009 #1
    It seems quite common to more or less equate gravitational redshift to gravitational time dilation. For example, the two might be explained as follows.

    We have two observers. A is standing on the surface of the earth, B on the top of a tower. A sends light up to B with some frequency, B receives the light with a lower frequency - the light has been redshifted because it's losing energy. Now B, after comparing the frequencies (maybe he has an identical light source), thinks that his units of time (derived from the frequency of his light source) are shorter than those of A (derived from the frequency of the light coming upwards). Thus there is time dilation.

    This seems like a silly argument to me. The red-shift of the light feels more like a Doppler effect than like something that is inherent to spacetime. If, classically, someone is creating a sound while moving away from me, I hear it at a lower frequency than they are transmitting it at. That doesn't mean that I think their clock must be going more slowly. The sound is just Doppler-shifted. The fact that you even know that the signal has been Doppler-shifted means that it is not equivalent to time dilation because if your units of time were adjusted to the frequency, you would be measuring the same frequency as the person sending out the signal.

    The gravitational redshift seems like the same kind of thing. Someone sends out a signal with a frequency and in the process of reaching you, the frequency is lowered. Then you receive it and measure the lowered frequency. It has nothing to do with time dilation, the light just lost energy. The only way you could call this time dilation is if you somehow demand that the signal was not changed along the way.

    This all in contrast to time dilation in SR, where it follows from the fundamental assumption that light moves at the same velocity for all observers, which has nothing to do with a signal being altered - the Doppler effect is a seperate concept with a seperate effect on things.

    All right, so the above is up for criticism. If I'm wrong about any of it, I'd be happy to hear it. However, it seems pretty clear cut and it is not what is puzzling me at the moment. The problem I'm having is how to reconcile all that with the fact that there does seem to be a time dilation equation in General Relativity, which just follows from the Schwarzschild metric. You can simply determine the interval between two events at the same location in a gravitational field and the metric will relate the proper time interval to the time interval as viewed by an observer at a different distance from the mass.

    [tex]\tau = t \sqrt{1-\frac{r_s}{r}}[/tex]

    This suggests that the time dilation must also be true for things other than light. For example, if your signal consists of light pulses at a certain interval, then that would need to be red-shifted as well, even though the energy in the signal has nothing to do with the frequency. This suggests a true time dilation.

    But we can't have it both ways. If there is this true time dilation, then light must be redshifted both by this effect and by the fact that it loses energy. Then the red-shift equation used (which is just the dilation equation above cast into a different form) is wrong - it's missing the loss-of-energy part.

    So now I'm stuck. I don't know how to reconcile all of this, or where the error in my thinking is. I welcome any insight you could offer.
     
  2. jcsd
  3. Nov 14, 2009 #2

    pervect

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    Staff Emeritus
    Science Advisor

    I'd encourage you to think of the situation geometrically. Let's draw a crude space-time diagram. With time running from left to right.

    *add* I finally got latex to work, it took me a while to remember the magic trick

    top of tower
    [tex]
    \]
    \begin{picture}(90,70)(0,0)
    \linethickness{0.3mm}
    \multiput(10,20)(0.12,0.2){250}{\line(0,1){0.2}}
    \put(40,70){\vector(2,3){0.12}}
    \linethickness{0.3mm}
    \multiput(60,20)(0.12,0.2){250}{\line(0,1){0.2}}
    \put(90,70){\vector(2,3){0.12}}
    \linethickness{0.3mm}
    \qbezier(0,20)(0,20)(0,20)
    \qbezier(0,20)(0,20)(0,20)
    \put(0,20){\vector(0,0){0.12}}
    \linethickness{0.3mm}
    \put(10,20){\line(1,0){50}}
    \put(60,20){\vector(1,0){0.12}}
    \linethickness{0.3mm}
    \put(40,70){\line(1,0){50}}
    \put(90,70){\vector(1,0){0.12}}
    \end{picture}
    \[
    [/tex]
    bottom of tower

    In the diagram, we have a certain time interval being measured by our clocks at the bottom of the tower. We send light signals upwards, to mark out an interval to be measured at the top of the tower. We find that the interval we measure on the top of the tower is longer than the interval we measure on the bottom of the tower.

    Because of the symmetry of the problem, we expect that the light signals travelling from bottom to top , emitted at different times, are parallel to each other - that they trace basically the same path through space-time, just offset by a certain amount of time.

    So, we have a parallelogram, where the lengths of the opposite sides are not equal. This basically implies a non-euclidean geometry - which is in fact the case.

    The geometrical viewpoint is that clocks always measure proper time, just as rulers always measure proper length. Time dilation in this view can be interpreted as just the fact that space-time doesn't have Euclidean geometry - loosely speaking, it's "curved". In fact, there are some papers that give us a 3-d surface on which we can draw our space-time diagrams to make them "come out right" for a gravitating body.

    So, where does time dilation fit into this? The expression itself is a bit confusing, since clocks in the geometric view always tick at one second per second. The time dilation argument relies on supplying something other than proper time. It needs the concept of some time coordinate.

    Time dilation, then, can be understood as the comparsion of coordinate time to proper time. The coordinate time doesn't advance at the same right as proper time does, the time that a clock actually measures.

    One can repeat the argument with light signals going the other way, from bottom to top to make the argument a bit more solid and rule out some other possibilities.
     
    Last edited: Nov 14, 2009
  4. Nov 14, 2009 #3
    Alright, so there is actual time dilation, which has nothing to do with your photons losing energy and thus being red-shifted. But then where are we leaving the energy-loss? If a photon loses energy when climbing out of a gravitational potential, shouldn't that give you an additional red-shift?
     
  5. Nov 14, 2009 #4

    pervect

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    Science Advisor

    Photons "loosing energy" (energy gets tricky in GR), or the difference in frequency between a local photon and a distant one from the same source (i.e. some specified atomic transition) isn't a separate phenomenon from time dilation that "adds on" - its just a different way of looking at the same thing.

    Note that the energy of a photon isn't a property of just the photon. You also have to specify the coordinate system that you use to measure the energy. If you look at a photon from different frames, you willl assign different energies to it depending on which frame you use. So the energy approach isn't coordinate independent or geometric. It is just an alternate way of comparing proper time and coordinate time, really.

    Another example might help. Suppose you do this experiment in an accelerating elevator, rather than on the Earth.

    If you look at the redshift phenomenon from an inertial frame, you will see only the doppler effect. If you look at the redshift phenomenon from the local frame of the accelerated observer (I oversimplify a bit here by not going into enough detail to really exactly specify what coordinate system I mean....), you[ll attribute it to "gravitational time dilation". But you'd get the wrong answer if you did both - you have to either choose the inertial frame, and the doppler shift explanation, or the non-inertial frame, and the gravitational time dilation explanation....
     
  6. Nov 14, 2009 #5
    This paradox goes away if you accept that photon's frequency is related to its *total* energy (rather than kinetic energy only). Indeed, when a photon is captured by a detector, the photon is fully destroyed, so the total energy of the photon gets transferred to the detector. This total energy always stays the same no matter whether the photon is "climbing" in or out of a gravitational potential. This also means that photon's frequency is also constant during the travel. Then the only reason for the red shift is the fact that sources that are deep in the gravitational potential emit photons whose frequency is lower than usual. There is no double counting.

    Eugene.
     
  7. Nov 15, 2009 #6
    Thanks to you both. I suspected that something like this might be the case, but I couldn't visualise how and why exactly.

    Ah yes. I think I understand what you're saying. If you've got an accelerating frame, you'd be inclined to call it a Doppler shift, because you're dealing with observers moving with respect to eachother. If you're dealing with gravitation, you'd be inclined to call it time dilation since the observers aren't moving. But since these things are equivalent in General Relativity, you're talking about the same phenomenon.

    Is this true because gravitation is no longer described as a force in General Relativity?
     
  8. Nov 15, 2009 #7
    This is simply the energy conservation law. The total energy (kinetic + potential) always stays constant. This conservation law applies to both a rock rolling down the hill and to a photon moving in the gravitational field.

    Eugene.
     
  9. Nov 15, 2009 #8
    I see what you mean. I tend to think of photons as having only kinetic energy, but I see where that leads to a problem.
     
  10. Nov 15, 2009 #9

    Dale

    Staff: Mentor

    The difference is that in the gravitational case the distance is not changing, so the transmission time is not changing. In the velocity case you don't need time dilation to explain it, you can say that the difference in frequency is due to the continually increasing time that the transmission takes. In the gravitational case you can't say that, so only time dilation is left.
     
  11. Nov 15, 2009 #10
    I think you are double counting here. The light loses energy because it is redshifted. To say that that the light must be further red shifted because it has lost energy, becomes a circular argument and can be applied ad infinitum. The further redshift caused by the loss of energy would cause even further loss of energy, causing further redshift and so on. The redshift due to gravitational time dilation is the cause and the loss of energy is the effect and it ends there.
     
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