When I tried to answer the SE question, Clock on a pendulum, I faced an ambiguity regarding time dilation in the gravitational field. To make the argument clearer, I designed the simpler thought experiment below:

Two similar clocks (1 & 2) are located side-by-side close to each other at point $A$ in a uniform gravitational field $g^\prime$ at rest with respect to the lab observer ($A$). The clocks are synchronous and are simultaneously set in motion at $u^\prime$ in opposite directions perpendicular to the field. After reaching two equidistant points from $A$, say, $E$ and $F$, the clocks are read by the lab observer. It is anticipated that both of the clocks, regardless of how much they have been dilated, show similar numbers as measured by the lab observer. See the attached Figure.


Now assume an observer $B$ approaches the lab observer at constant $v$ perpendicular to the field direction. In the absence of gravitation, it is easy to prove that both clocks, despite undergoing different time dilation due to different relative velocities as seen by $B$, show similar times as passing points $E$ and $F$. In this case, however, $B$ asserts that clock 1 first reaches point $E$, and then it's clock 2 that passes point $F$. In other words, the arrival of the clocks at the assigned points is no longer simultaneous, yet the clocks show similar times as passing the points.

However, when we consider the scenario in the presence of the gravitational field $g^\prime$, circumstances become odd from the perspective of $B$. Using acceleration transformation [Eqs. (1c)], $B$ measures two different accelerations $g_1$ and $g_2$, respectively, for clock 1 and clock 2 as follows:

$$g_1=\frac{g^\prime}{{\gamma_v}^2 \left( 1-\frac{uv}{c^2} \right)^2}$$

$$g_2=\frac{g^\prime}{{\gamma_v}^2 \left( 1+\frac{uv}{c^2} \right)^2}$$

So that we always have $g_1>g_2$. But, as we know, the greater the gravitational field, the slower the clocks run, and vice versa. Therefore, it is anticipated that, from $B$'s viewpoint, clock 1 runs slower due to greater G-field ($g_1$); and clock 2 runs faster due to smaller G-field ($g_2$). On the other hand, as mentioned earlier, $B$ sees that clock 1 (the slower one) reaches point $E$ sooner than clock 2 (the faster one) does for point $F$. Therefore, it seems that the clocks do no longer show similar times as passing over the points! Where is the problem?

Recall that if the obtained $g_1$ and $g_2$ were replaced by each other so that we had $g_1<g_2$, it would become more plausible that the reading of the clocks would be similar as calculated by $B$, however, this is not the case!

Moreover, assume that observer $B$ is outside (away from) the gravitational field all time. However, if one assumes that $B$ moves at $v$ sliding onto the surface of the earth/planet inside the G-field, the final result won't be affected much. In this case, assume that the clocks and observer $A$ are all located on the surface of the earth/planet contrary to what is shown in the figure.

All in all, is it true to say that the more G-field, the slower the clocks run? Or, maybe there are other parameters that can indicate the reverse, at least from the viewpoint of some specific inertial observers.

  • $\begingroup$ Is B inertial or is B also accelerating to counteract the uniform gravitational field? $\endgroup$
    – Dale
    Commented Oct 18, 2019 at 18:54
  • 1
    $\begingroup$ I'm editing the post. Consider B inertial yet. $\endgroup$ Commented Oct 18, 2019 at 18:59

3 Answers 3


You're making your life unnecessarily difficult by using three-acceleration and its unique transformation formula. It's easier to use four-vectors for everything, since they transform in a consistent way. But then, you're making your life unnecessarily difficult by solving the same problem in two different frames to begin with – one is enough, since they're all equivalent – so maybe you want to do the extra work, just to convince yourself it's all consistent.

There's no rule in general relativity that clocks undergoing a higher acceleration run slower. The rule is the same as in special relativity: an ideal clock measures the length of its worldline. When you apply that rule to certain patterns of motion in spacetimes with certain symmetries, you get gravitational time dilation.

The Euclidean analogue of gravitational time dilation is that if you have two runners on a circular track in different lanes, the runner in the outer lane has to run faster to keep up. Circular motion in Euclidean space is analogous to constant acceleration in Minkowski space, and distance along a path in Euclidean space is analogous to elapsed proper time (that is, the length of the worldline).

If you carefully work out the elapsed proper time of each clock in the primed frame, and don't add an extra factor for the gravitational time dilation that's already accounted for in the geometry of the problem, you should find that the readings on the clocks are the same when they reach E and F respectively.

  • $\begingroup$ If it is really complicated to answer this simple question using 4-vectors or whatsoever as precise calculations, we have not learned our GR lessons, or GR is hardly applicable to such simple examples! $\endgroup$ Commented Oct 19, 2019 at 10:50

I'd like to elaborate a little on the answer given by contributor benrg.

GR does not make a distinction between gravitational time dilation and velocity time dilation.

That is where the comparison with a circular track comes in.

Imagine an O'Neill cylinder, with pairs of atomic clocks at various distances to the central axis of rotation.

The principle of equivalence asserts: if you are limited to local measurement only then as a matter of principle there is no difference.

When two clocks are at different distance to the axis of rotation they are at different gravitational potential. For the clock closer to the axis of rotation more proper time will elapse. You can measure the difference in rate of proper time elapsing and infer a velocity difference. The principle of equivalence demands that the equations of GR must be such that you can also evaluate the case in terms of difference in gravitational potential and everything should remain consistent.

As benrg points out, difference in time elapsing is attributed to difference in gravitational potential.

You asked: "is it true to say that the more G-field, the slower the clocks run?" Presumably your intention was to compare with a clock that is so far away from the gravitational well that for the far-away clock gravitational time dilation is negligable. The thing is: the only way to assess elapsed time is to compare with another clock, therefore it is better to always express the physics in terms of comparison.


The usual way to analyze a uniform gravitational field is through the Rindler coordinates on Minkowski spacetime. Since the gravitational field is uniform, the spacetime is flat and gravity is simply an accelerating coordinate chart on Minkowski spacetime.

So in this problem the primed frame is a Rindler frame with acceleration parameter $g'$ meaning that at a position of $y'=1/g'$ the gravitational acceleration is $g'$. In units where c=1 metric in these coordinates is $$ds^2=-d\tau^2=-(g'y')^2 dt'^2+dx'^2+dy'^2+dz'^2$$

In these coordinates the worldline of the clocks is given by $$ X'=(t',x',y',z') = (t',\pm u' t', 1/g', 0)$$ so the four-velocity is $$U'=\frac{dX'}{d\tau}=\left(\frac{1}{\sqrt{1-u'^2}},\frac{\pm u'}{\sqrt{1-u'^2}},0,0\right)$$ and the four-acceleration is $$A'=\frac{dU'}{d\tau}=\left( 0, 0, \frac{g'}{1-u'^2}, 0 \right)$$

Transforming to the inertial coordinates of B we find the worldline $$X=(t,x,y,z)=\left(\frac{\sinh(g' t')\pm g' t' u' v}{g' \sqrt{1-v^2}},\frac{v \sinh(g' t') \pm g' t' u'}{g'\sqrt{1-v^2}} ,\frac{\cosh(g' t')}{g'},0\right)$$ and the four-velocity $$U=\left(\frac{\cosh(g't')\pm u' v}{\sqrt{1-u'^2}\sqrt{1-v^2}},\frac{v \cosh(g't')\pm u'}{\sqrt{1-u'^2}\sqrt{1-v^2}},\frac{\sinh(g' t')}{\sqrt{1-u'^2}},0 \right)$$ and the four-acceleration $$A=\left(\frac{g' \sinh(g' t')}{(1-u'^2)\sqrt{1-v^2}},\frac{g'v \sinh(g' t')}{(1-u'^2)\sqrt{1-v^2}},\frac{g' \cosh(g' t')}{1-u'^2},0 \right)$$

Now, there is no closed form solution for $t=\frac{\sinh(g' t')\pm g' t' u' v}{g' \sqrt{1-v^2}}$, so if you want to calculate it explicitly you need to use numerical methods. However, one thing that we can notice is that because of the $\cosh$ terms the four-velocity components are not symmetric. The time dilation can be calculated by $$\gamma = \frac{dt}{d\tau} = \sqrt{\frac{\cosh(g't')\pm u' v}{(1-u'^2)(1-v^2)}}$$ Notice that this is not the same for the two clocks, meaning that one clock will be more time dilated than the other. This explains why one clock registers more time than the other, they are simply more time dilated.

Therefore, it is anticipated that, from B's viewpoint, clock 1 runs slower due to greater G-field (g1); and clock 2 runs faster due to smaller G-field (g2).

Since B is an inertial observer in flat spacetime then from B's viewpoint there is no G field at all and the time dilation (as calculated above) is purely kinematic. Because the time dilation is kinematic and because the velocities are not the same, it is unsurprising that the time dilation is different and it is clear that clock 2 is more time dilated.

Remember that in GR the analog of the Newtonian gravitational field is frame-variant and is given by the Christoffel symbols, and that the Christoffel symbols are zero in an inertial frame.

All in all, is it true to say that the more G-field, the slower the clocks run?

No, this is not true. In fact, this is a common misunderstanding. The gravitational time dilation is related to the gravitational potential, not the gravitational acceleration. In the center of a planet the time dilation is the largest even thought the gravitational acceleration is zero.

  • $\begingroup$ This explains why one clock registers more time than the other, they are simply more time dilated. You mean that if clock 1 registers #6 as reaching point $E$, clock 2 would register, say, #5 as passing point $F$ due to different gamma factors as viewed by $B$?! $\endgroup$ Commented Oct 20, 2019 at 21:22
  • $\begingroup$ Almost. Clock 1 passing E reads the same as clock 2 passing F, but clock 2 passes F later than clock 1 passes E. So the greater time dilation of clock 2 explains why it reads the same at a later time. Or equivalently, at any point in time clock 1 registers more time than clock 2 due to the different gamma $\endgroup$
    – Dale
    Commented Oct 20, 2019 at 21:32

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