If I have a light clock in the centre of the earth, and one at a distance from the earth, two observers at the clocks will agree that the inner-earth clock runs slower.

But if we'd look with a magic telescope at each-other's clock, we'd agree that the photons bouncing between the mirrors are travelling at c.

As far as I can work out, that implies that either:

  1. From the far-away clock's perspective, the inner-earth clock is larger, so that it takes photons longer to bounce between mirrors. I.e., dropping a clock to the centre of the earth makes it 'bigger' from the dropper's perspective.
  2. From the far-away clock's perspective, the inner-earth clock has mirrors which are constantly moving (like velocity time dilation).

I have seen the Schwarzschild metric quite often as an answer to this sort of question. I certainly understand that the inner-earth clock would tick less frequently. But that is not really the question.

If I looked with the magic telescope at the inner-earth clock the path of the photons must be different in some way for us to agree on c. Is either of my two implications above correct?

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    $\begingroup$ The speed of light changes with distance from the Earth. $\endgroup$ Oct 7, 2018 at 16:11
  • $\begingroup$ That resolves the issue. I was erroneously thinking I would observe the speed of light as the same everywhere, as opposed to merely within my own reference frame. Once again I am the victim of having read descriptions by people overly generalising. $\endgroup$ Oct 7, 2018 at 17:18
  • $\begingroup$ @JeroenDStout: Actually I don't think it has anything to do with this question. Coordinate velocities are not interesting, and discussing them generally leads nowhere. $\endgroup$
    – user4552
    Oct 7, 2018 at 17:59
  • $\begingroup$ @BenCrowell Given that my question was outside of an acceleration, the answer either had to be about me seeing the speed of light different or seeing the length of the clock as different. The answer linked showed it would be the former, i.e., my magic telescope would see the photons slow down (from my reference frame). $\endgroup$ Oct 7, 2018 at 18:55
  • $\begingroup$ @JeroenDStout: The answer linked showed it would be the former, i.e., my magic telescope would see the photons slow down (from my reference frame). No, GR doesn't give us a frame of reference big enough to allow you to say what this would mean. $\endgroup$
    – user4552
    Oct 7, 2018 at 21:26

2 Answers 2


You don't have to make up hypothetical experiments involving light clocks, clocks at the center of the earth, or magic telescopes that can see the center of the earth. An equivalent experiment was done by Pound and Rebka in 1959, using apparatus at the top and bottom of a tower. The results are the same as if the tower was accelerating upward at $g$. General relativity says that an inertial frame of reference is a free-falling frame. Therefore by the GR definition, the tower was accelerating upward at $g$. This is an example of the equivalence principle, and such experiments only test the equivalence principle. They do not test anything that would be more specifically predicted by GR as opposed to some other theory of gravity.

Your example involving the center and surface of the earth is more complicated because we can't construct a frame of reference big enough to cover both. Frames of reference are local in GR. However, we can make use of the fact that these differences in clock rates are transitive, in the sense that if clock A is slow by a factor $f_1$ relative to clock B, and B is slow by $f_2$ relative to C, then A is slow relative to C by $f_1f_2$. The result is qualitatively similar to the kind of local result seen by Pound-Rebka.

I don't think your proposed explanations in terms of the mirrors of the light clock are really workable, because gravitational time dilation is observed with all clocks, not just with one type of clock. GR doesn't interpret this sort of thing as a physical influence acting on the clock but as an effect on the signals that we use to compare them. Also, any physical effect on the clock would be expected to be proportional to the gravitational field or to tidal forces, but in reality time dilation depends on the difference in gravitational potential.

  • $\begingroup$ I think my example being more complicated is exactly why I did not use the tower example. There the acceleration is clear, whereas it is not between the centre of the earth (no accelerating as far as I can see) and space. My mistake was indeed assuming a "global" reference frame based on my own local reference frame. $\endgroup$ Oct 7, 2018 at 17:16

$\let\D=\Delta$ First of all I would like to state some facts (as opposed to interpretations).

It is not exactly true that gravitational effects like time dilation depend on gravitational potential. It is so only approximately, under two hypotheses:

  1. that the spacetime geometry is static
  2. that it deviates little from a flat spacetime.

Consider the simplest case of Schwarzschild spacetime. In the usuale coordinates ($G=1$, $c=1$) metric is $$d\tau^2 = \left(\!1 - {2 M \over r}\!\right) dt^2 - {r \over r - 2M}\,dr^2 - r^2\,(d\theta^2 + \sin^2\theta\,d\phi^2).\tag 1$$ The time dilation factor between two points at $r_1$ and $r_2$ is $${d\tau_1 \over d\tau_2} = \sqrt{1 - 2M/r_1 \over 1 - 2M/r_2}.\tag 2$$

Only if $2M/r_1\ll1$ and $2M/r_2\ll1$ we can approximate to $${d\tau_1 \over d\tau_2} \simeq 1 - {M \over r_1} + {M \over r_2}$$ which, reinserting $c$ and $G$, becomes $${d\tau_1 \over d\tau_2} \simeq 1 - {G M \over c^2 r_1} + {G M \over c^2 r_2} = 1 + {1 \over c^2}\,V(r_1) - {1 \over c^2}\,V(r_2)$$ where $V(r)=-GM/r$ is the gravitational potential. The approximation works well in Earth's gravitational field, and in the Sun's too. It would be unacceptable near a neutron star.

An analogous remark must be made about the equivalence principle (EP). It is not always possible to interpret GR effects as equivalent to those observed in an accelerated frame in flat spacetime. EP is a local property, and it does not hold for observations involving widely separated points of spacetime.

Let now examine statements like "the inner-earth clock runs slower", or, more generally "clocks down in a gravitational field run slower wrt to higher ones". Or, still worse: "time flows more slowly in a gravitational field".

Actually I will not even touch the latter, which I find absolutely unintelligible (my fault?). But consider the first or the second. What do they really mean? How can we test such a statement? As far as I can see, there is only one way: to send two signals fron one clock to the other, and to compare time interval at start, as measured by the first clock, with time interval at arrival, measured by second clock.

No doubt, if first clock stays lower, interval at arrival will be longer than interval at start. But to say that clocks tick at different rates is an interpretation, and one not unmistakably forced by experimental situation.

Consider eq. (1). We have two clocks, situated at $r_1$ and $r_2$ ($r_1<r_2$) with same $\theta$ and same $\phi$. Clock 1 sends two signals, separated by $\D\tau_1$ as clock 1 says. They arrive at $r_2$ and are registered with a separation $\D\tau_2$ (time of clock 2). But how do we compute the ratio $\D\tau_1/\D\tau_2$?

The first step is to compute $\D t$, setting $r=r_1$, $dr=d\theta=d\phi=0$ (clock 1 is stationary at $r=r_1$). We find $$\D t = {\D\tau_1 \over \sqrt{1 - 2M/r_1}}.\tag 3$$ Now we had to study radial propagation of signals, from $r_1$ to $r_2$. They will take some time, say from $t_1$ to $t_2$ for the first signal. And from $t_1+\D t$ to $\dots$ ?

Happily, we don't need the actual calculation, since Schwarzschild geometry is static. You see it by observing eq. (1): time coordinate $t$ never appears in the coefficients which multiply the coordinate differentials. Here is what this physically means. Let us conduct two experiments, involving widely separated spacetime points as well, the second experiment being the exact copy of the first apart for initial $t$-values: $t_1$ and $t_2$ precisely. Then both experiments will run in the same way, apart for a shift in $t$ at every phase of the experiment. (In physicists' parlance, this is said "invariance by time translations", where time here means coordinate $t$).

This gives the answer we were looking for: the $t$-time of arrival of the second signal is $t_2+\D t$. Then we can proceed using eq. (1) in the reverse, from $\D t$ to $\D\tau_2$: $$\D\tau_2 = \D t\,\sqrt{1 - 2M/r_2}.\tag 4$$ Multiplying eqs. (3) and (4) we get eq. (2).

Did we prove that clock 1 ticks slower than clock 2? Not at all. We studied the propagation of light signals between them, and showed that the distance (in local proper time) at start is different (less) than their distance at arrival. This a purely geometrical result, says something about spacetime, not about clocks' behaviour.

My answer to your question is as follows. No, neither of your implicatios is true. Simply, you are not allowed to measure sizes, times, velocities at-a-distance. Or rather, you may do it, but you should not call them time- size- or velocity-measurements. They are something entirely different and can give strange results, with no physically useful interpretation.


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