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Relationship between gravitational time dilation and energy?

  1. Jun 25, 2015 #1
    The rate that a stationary clock slows down near a massive object, relative to one far away, can be read off from the Schwartzschild metric:
    $$c^2d\tau^2=\left(1-\frac{r_s}{r}\right)c^2dt^2-\left(1-\frac{r_s}{r}\right)^{-1}dr^2-r^2\left(d\theta^2+\sin^2\theta d\phi^2\right)$$
    by setting ##dr=d\theta=d\phi=0## to give:
    where the Schwartzschild radius ##r_s=2GM/c^2##.

    If the clock is running slowly compared to a distant clock is this equivalent to the clock having a lower energy compared to a distant clock?

    If the clock was an atomic system then the frequency of its oscillation would be less near the massive object. As energy is proportional to frequency for atomic systems then I would have thought that this would imply that the energy of the atomic system would be less near the massive object than it was far away.
  2. jcsd
  3. Jun 25, 2015 #2
    Let the energy at infinity of a clock be E.

    When we start lowering said clock into a gravity well, its energy at infinity starts decreasing, energy at infinity of the clock will be proportional to the gravitational time dilation of the clock.

    That is not controversial.

    On the other hand according to Wikipedia there is some controversy about the frequency of an atomic system in a gravity well:
  4. Jun 25, 2015 #3


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    Gravitational time dilation is directly related to the gravitational potential and thus potential energy.
  5. Jun 25, 2015 #4


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    Here's a nonrelativistic approximation (low speeds, weak gravitational fields) to the Schwarzschild metric for a clock of mass [itex]m[/itex] near a planet of mass [itex]M[/itex]:

    [itex]\frac{d \tau}{dt} = 1+ \frac{1}{mc^2} [- \frac{1}{2} mv^2 - \frac{GMm}{r}][/itex]

    The factor of [itex]m[/itex] cancels, so it's sort of pointless to put it in, except to compare with the Newtonian kinetic energy and potential energy:

    [itex]K = \frac{1}{2} mv^2[/itex]: Kinetic energy
    [itex]U = -\frac{GMm}{r}[/itex]: Potential energy

    So in terms of these:

    [itex]\frac{d \tau}{dt} = 1 + \frac{1}{mc^2} (U - K)[/itex]

    So what that means is that clocks with greater [itex]U[/itex] (higher potential energy) run faster, but clocks with greater [itex]K[/itex] (higher kinetic energy) run slower (because there is a minus sign in front of it. So kinetic energy and potential energy work in opposite directions, when it comes to time dilation.
  6. Jun 27, 2015 #5


    Staff: Mentor

    The energy of the system, relative to a similar system far away, is less near the massive object. Energy is relative. If you make a local measurement of the atomic system, you will find its energy unchanged. Only measurements made from far away, such as measuring the redshift of light from the atomic system near the massive object when the light is received far away, will show a difference.
  7. Jun 27, 2015 #6


    Staff: Mentor

    After reading this article, I'm not clear on exactly what is meant by "the frequency of an atomic system". What actual measurements are supposed to tell us this frequency?
  8. Jun 29, 2015 #7
    I don't know any quantum mechanics, but somehow we know the frequency of a standing wave of an electron in a box.

    Electron in a box is some kind of standing wave. Let's move the box some distance, this causes the standing wave to become a propagating wave for a while. After this operation the standing wave is supposed to be in different phase compared to the alternative where we let the box stay still. It's like we temporarily slowed down the standing wave.

    We bring a large mass near a box that contains an electron, this causes some energy to be sucked out of the standing wave of the electron, which causes a decrease of the frequency of the standing wave. Then we move the mass away, by doing so we increase the potential energy of the standing wave, which causes the frequency of the standing wave to increase to the original value. By this operation we affected the phase of the standing wave. We temporarily slowed down the standing wave.
  9. Jun 29, 2015 #8


    Staff: Mentor

    Which means you are leaving out a lot of complications. Not only that, you are trying to combine quantum mechanics with gravity, which adds even more complications.

    The short answer to your post is that, as far as GR is concerned, quantum systems are just like any other systems; gravitational time dilation does not show any local effects, it only shows up when you look at the system from a different position in the potential well. Local measurements made of an electron in a box deep in a gravity well will be the same as local measurements of an electron in a box far out in free space, away from the gravity well. If you make the electron in a box deep in a gravity well emit radiation somehow, that radiation will be redshifted if it is observed by someone far out in free space, away from the gravity well, just as any radiation would be. So as far as GR is concerned, this "phase of the standing wave" you are talking about is observer-dependent: it looks different to observers far away compared to local observers.

    A longer answer would require you to spend some time studying all those complications of QM and QM + gravity that I referred to above. I would recommend doing that if you are interested in questions like this one; the complications are not trivial.
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