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According to general relativity, I know that time flows at a different rate in the presence of gravity which is nothing but the curvature of spacetime. And is called time dilation.

My doubt is, are there similar effects in space as well? I mean, when there is a curve in space-time, it's not just in time but also in space. So are there any effects like length contraction or 'Space dilation' under the effect of gravity? If one measures the distance between two points as 1m in a strong gravitational field, will it be different from the 1m at a place in weaker gravity?

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  • $\begingroup$ You are talking about spaghettification. $\endgroup$
    – PrawwarP
    Commented Dec 23, 2020 at 3:29
  • $\begingroup$ Possible duplicate here. Does that answer your question? $\endgroup$
    – joseph h
    Commented Dec 23, 2020 at 4:10

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If you are talking about gravitational time dilation and length contraction yes there are analogous formulations like in special relativity.

But in this case, they are rather more complicated since you are no longer working on flat spacetime. You need to define the metric. For simplicity if you choose a non-rotating black hole

$$ds^2 = -\Big(1-\frac{2GM}{rc^2}\Big)c^2 dt^2 + \Big(1-\frac{2GM}{rc^2}\Big)^{-1}dr^2 + r^2 d\Omega^2$$

where $d\Omega^2 = d\theta^2 + \sin^2 \theta d\phi^2$

you can calculate length contraction by taking $dt = 0$, $d\phi = 0$ and $d\theta = 0$ so

$$ds^2 = \Big(1-\frac{2GM}{rc^2}\Big)^{-1}dr^2$$

and this gives

$$ds = \frac{1}{\sqrt{\Big(1-\frac{2GM}{rc^2}\Big)}}dr$$

Similar relation for time dilation will hold, you can calculate it now pretty straightforwardly.

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  • $\begingroup$ probably worth explicitly stating that the r^2 dOmega term is the same as in the 0-field case, and so there is no circumferential contraction, only radial $\endgroup$
    – Tristan
    Commented Dec 23, 2020 at 13:37
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    $\begingroup$ @Tristan That isn't true. The $r$ coordinate is arbitrary. It happens to be defined so that the angular part of the metric is $r^2dΩ^2$, but you could just as well use an $r'$ for which the radial part is $dr'^2$. This answer doesn't make a whole lot of sense since it treats $r$ as the true radius against which "space dilation" should be measured. $\endgroup$
    – benrg
    Commented Dec 24, 2020 at 3:27

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