This question is mainly inspired after watching the movie known as Interstellar

We knew that for time dilation caused by relativistic motion between A and B. A will measure B's clocks slowing down, and B will measure A's clock slowing down by the same rate, while they both measure their own clocks ticking normally

In interstellar, there's a planet that experience severe gravitational time dilation relative to earth's frame due to being in close proximity to the black hole Gargantula, therefore for every hour spent there measured by the astronaut's frame, 7 years will have passed in Earth's frame

Thus this means to Earth's frame, they will measure the astronaut's clocks to be ticking slower relative to them

But what about earth's clock as measured by the astronauts? Will they measure Earth's to be ticking faster than theirs (at the same magnitude it is slowed in earth's frame measuring theirs)?

Because if they both measured the other's clocks as running slower, then how could the twin paradox like aging difference be possible?

  • $\begingroup$ NB This question was posted by mistake in the maths stack exchange site, thus the one there is now deleted and the question is moved to the correct place $\endgroup$
    – Secret
    Dec 19, 2014 at 9:29
  • $\begingroup$ Yes, however, the depiction of the effect in "Interstellar" is not physically realizable as shown. It's a plot device, and if you allow me a little bit of film criticism here, it's an awfully unnecessary one. The writers were simply throwing the scifi kitchen sink in there, hoping that something awesome would bubble up... and it didn't. $\endgroup$
    – CuriousOne
    Dec 19, 2014 at 9:47

3 Answers 3


Yes, the observers on Earth will measure the astronaut's clocks to be be running slow while the astronauts will measure the clocks on Earth to be running fast. So the situation is asymmetric.

The situation is asymmetric because the two sets of clocks are in different environments. Specifically there is a (gravitational) potential energy difference between them i.e. you get energy out going from Earth to the planet but you have to put energy in going from the planet to Earth. In fact you can directly relate the time dilation to the potential energy difference using the weak field equation:

$$ \frac{\Delta t_1}{\Delta t_2} = \sqrt{1 - \frac{2\Delta\Phi}{c^2}} \tag{1} $$

This asymmetry does not exist for observers moving at different speeds in flat spacetime.

  • $\begingroup$ But flipping the sign on ΔΦ won't invert the value of Δt_1/Δt_2. It will work approximately for small values of ΔΦ, but I want the equation for big values of ΔΦ. $\endgroup$
    – DanielLC
    Jun 2, 2015 at 19:31
  • $\begingroup$ @DanielLC: when $\Delta\Phi$ is large you need to use the metric that describes the curvature of your spacetime. The spacetime around the Earth is approximately described by the Schwarzschild metric, and you'll find lots of answers on this site explaining how to do the calculation. $\endgroup$ Jun 3, 2015 at 5:06

Relativistic time dilation and gravitational time dilation are two different coexisting phenomena. At speeds much less than $c$, the gravitational time dilation effect is much more pronounced, and is the reason that clocks on satellites must run slightly slower than earth time. At speeds approaching $c$, the relativistic time dilation effect is more pronounced than the gravitational effect.

Earth's clock as measured by the astronauts will be running quickly as measured by an observer near a black hole: the gravitational potential is an absolute quantity, not a relative one. Similarly, the clocks on earth measured by an observer on a satellite run slightly slower than real time.


For an intuitive understanding of this, let me remind you that we know from special relativity that the proper time of an uniformly accelerated observer is "absolutely" slower than the proper time of an inertial guy (supposing clock hypothesis). The more the (proper) acceleration, the slower proper time of an accelerated observer is.

Now, thanks to strong equivalence principle, a guy that is not free falling is actually accelerating, which means that on our chair we are all accelerated +1g upward right now. This, in particular implies that a stationary observer higher in the atmosphere is less accelerated (still in terms of proper acceleration) than us here on the ground, and consequently measure that our time is passing slower than his own (and reciprocally, there's a symmetry here).

The same reasoning can be done in the case you consider. The guys far away from the earth have a higher proper acceleration than us here, and thus, their proper time is much slower.


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