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When orbiting a massive object, perhaps a neutron star, you maintain your altitude while at the same time you are close to a very strong gravitational field. Being in orbit, and being in free fall is indistinguishable for the object in question. The object is not affected by gravity, locally.

Another free falling object can't be "less affected" than this first object, since the first object is in fact not noticing any gravity. Logically, both object should age equally fast if we ignore the relativeness of aging.

There is no doubt that the gravitational field is strongest around the first object, but neither object experiences any gravitational field. Object one looks at object two is in free fall, and vice versa.

Of object one records a video of object two, and then later plays that video - but adjust the playback rate to account for all doppler shift effects - what would object one see?

Is doppler the only effect causing the difference in relative aging in this case?


marked as duplicate by user108787, user36790, Jon Custer, Rococo, John Rennie Sep 24 '16 at 19:23

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Yep! Still applies.

Set $c=1$ and use the Schwarzschild metric. This metric is the unique solution for any vacuum, spherically symmetric spacetime which behaves like a Newtonian $\frac{1}{r}$ potential in the large $r$ limit. If the Schwarzschild radius is $R$, we just have have:

$${d\tau }^{2}=\left(1-{\frac {R}{r}}\right)dt^{2}-r^{2}d\theta ^{2}$$

Let's say the velocity from an asymptotic observer is $v=\frac{r d\theta}{dt}$. Then $r^2 d\theta^2=v^2 dt^2$, and:

$${d\tau }^{2}=\left(1-{\frac {R}{r}}\right)dt^{2}-v^{2}dt^{2}$$

$$\frac{d\tau }{dt}=\sqrt{ 1-{\frac {R}{r}}-v^{2}}$$

The prediction in special relativity is just

$$\frac{d\tau }{dt}=\sqrt{ 1-v^{2}}$$

so indeed time passes by more slowly due to the gravitational well.

Note that the "speed of light" here is reached when $1-{\frac {R}{r}}-v^{2}=0$, which gives $v_{max}<1$. This is due to the fact that $v$ defined this way is some sort of weird coordinate speed.


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