So looking over this answer: Does gravitational time dilation happen due to height or difference in the strength of the field?

I am not sure if I am following completely. My issue relates to time dilation being multiplicative, though perhaps there I'm already wrong somehow.

Let's say there are three observers at different heights - A, B, and C. The gravitational field (or acceleration) is constant, and small, and the observers are relatively close. I'd like to think about the dilation from A to C in terms of from A to B and from B to C. I expect that:

$$\frac{dt_A}{dt_C} = \frac{dt_A}{dt_B}\frac{dt_B}{dt_C} \tag{1}$$

In his answer, John Rennie used the Rindler Metric to show the veracity of the approximation

$$\frac{dt_A}{dt_B} \approx \sqrt{ 1 + \frac{2\Delta\phi_{AB}}{c^2}} \approx 1 + \frac{\Delta\phi_{AB}}{c^2} \tag{2}$$

Because the metric indicates

$$\frac{d\tau}{dt} = 1 + \frac{gx}{c^2}\tag{3}$$

Which at least looks similar.

However, I'm not quite getting the connection between those. If the approximation is reasonably accurate, then since

$$\Delta\phi_{AC} = \Delta\phi_{AB}+\Delta\phi_{BC} \tag{4}$$

We would get

$$\frac{dt_A}{dt_C} \approx { 1 + \frac{\Delta\phi_{AB}+\Delta\phi_{BC}}{c^2}} \tag{5}$$

Which is not the same as this, via $(1)$:

$$\left( 1 + \frac{\Delta\phi_{AB}}{c^2}\right)\left( 1 + \frac{\Delta\phi_{BC}}{c^2}\right)\tag{6}$$

The latter has an additional $\frac{\Delta\phi_{AB}\Delta\phi_{BC}}{c^4}$ term. Now, I am happy to believe that this term is small enough to ignore in any single calculation. However, if we keep subdividing the distance, we end up with more such product terms and we have an infinite product to evaluate, which is a Volterra product integral:

$$\prod \limits_{0}^h\left(1+\frac{g(x)}{c^2}dx\right) = e^{\int \limits_{0}^h\frac{g(x)}{c^2} \, dx}\tag{7}$$

For a constant $g$ this evaluates to


I understand that for small exponent values, this is close to $(1+\frac{gh}{c^2})$ but it's not the same. And if I start with equation 8 instead, and do the same subdivision exercise, I end up back at equation 8 once again.

So I'm in the position of: if I think of using the approximation, I can prove to myself that it should actually be something else, and that something else no longer looks like the equation extracted from the Rindler Metric, so I no longer see the connection between $(2)$ and $(3)$.

  • $\begingroup$ I think perhaps my title isn't so great, it's more about reconciling the approximation with the metric. I admit I'm significantly out of my depth when it comes to fully understanding the implications of a spacetime metric. $\endgroup$
    – geshel
    Apr 14, 2021 at 1:18

1 Answer 1


I think John Rennie's answer is needlessly confusing.

Gravitational time dilation is not due to differences of gravitational potential energy. It isn't an "effect" at all. It's just a geometric property of worldlines in certain manifolds.

If two ships separated by 2 km on the equator both sail north to the 30th parallel, they'll be about 1 km apart when they arrive. If they instead both sail northeast – or at any other compass heading as long as it's the same for both – they'll still be about 1 km apart when they arrive. The reason is that their paths are equivalent under a rotation around the polar axis, and the 30th parallel rotates at half the rate of the equator.

The same thing happens in spacetimes with a time-translation symmetry. The 2 km and 1 km separations correspond to separations in time of emitted and received signals – so, in this case, to a Doppler shift factor of 2. That geometric relationship between worldlines is what gravitational time dilation is. It only exists (can only be defined) in spacetimes with an appropriate time-translation symmetry.

In Rindler coordinates $$ds^2 = (gx)^2 dt^2 - dx^2 - dy^2 - dz^2$$ the redshift factor for a signal emitted from rest at $x_1$ and received at $x_2$ is simply $gx_1/gx_2=x_1/x_2$. If it's absorbed and reemitted at $x_{1.5}$, then the total redshift over the two legs is $(x_1/x_{1.5})(x_{1.5}/x_2)=x_1/x_2$. This is all exact.

When the redshift factor is close to $1$, you can write it as $1+2\phi$ and use the approximation $(1+2\phi)(1+2\phi')\approx 1+2(\phi+\phi')$, but there's obviously no need to do so here since the exact formula is so simple. You could also take the log of the factor to get an exact additive representation of it, but that isn't typically done.

John Rennie used $(1+gx)^2$ rather than $(gx)^2$ in the metric. This merely corresponds to a different choice of origin for $x$. With this off-center origin, the Doppler shift factor is $(1+gx_1)/(1+gx_2)$, which looks like an approximation of some sort but is actually exact. You could of course approximate it as $1+g(x_1-x_2)$ for small $x$. This form of the metric obscures the fact that Rindler coordinates are merely a kind of polar coordinates for Minkowski space. (The advantage of this choice of origin is that an object at rest at $x=0$ experiences a proper acceleration of $g$, but I think that's poor compensation for the disadvantages.)


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