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Pinning two test particles at two different points in space, how can I calculate their spatial distance, when the geometry is given by the Schwarzschild metric?

Let's say particle 1 is pinned at $r=R$, $\theta=\frac \pi 2$, $\varphi = 0$ where $R$ is a positive radius bigger than the Schwarzschild radius and particle 2 is pinned at $r=R+L$, $\theta=\frac \pi 2$, $\varphi = 0$ where $L$ is also a positive constant.

What is now the spatial distance between the two particles?

Do I have to calculate the length of a geodesic from one particle to the other? Is this equal to the distance?

thyme
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2 Answers2

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I'm guessing that when you say:

What is now the spatial distance between the two particles?

You mean the proper distance. The coordinate distance is of course just $L$. The proper distance is the distance you would measure if you sat at radius $R+L$ and let out a tape measure until it reached radius $R$.

To calculate the proper distance start with the Schwarzschild metric:

$$ ds^2 = -\left(1-\frac{r_s}{r}\right)dt^2 + \frac{dr^2}{\left(1-\frac{r_s}{r}\right)} + r^2 d\Omega^2 $$

Presumably you want the proper distance along the radial path from $r = R$ to $r = R+L$, in which case $dt = d\theta = d\phi = 0$ so the metric simplifies considerably to:

$$ ds = \frac{dr}{\sqrt{1-\frac{r_s}{r}}} $$

and to get the proper distance just integrate:

$$ s = \int_R^{R+L} \frac{1}{\sqrt{1-\frac{r_s}{r}}} dr $$

According to Wolfram this integrates to:

$$ s = \left[ r\sqrt{1-\frac{r_s}{r}} + \frac{r_s}{2} log \left( 2r \left( \sqrt{1-\frac{r_s}{r}} + 1 \right) - r_s \right) \right]_R^{R+L} $$

I'll leave you to finish the algebra because it's straightforward but messy and unrewarding.

John Rennie
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  • So is $s$ less than or greater than $L$? In other words, do lengths contract or expand in a gravitational field? Does this only apply to radial lengths? – Gumby The Green Mar 16 '22 at 05:49
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    @GumbyTheGreen For the Schwarzschild metric the proper distance is greater than the coordinate distance i.e. $s > L$. – John Rennie Mar 16 '22 at 06:55
  • Thanks. I guess that's also answered by the fact that $\frac{1}{\sqrt{1-\frac{r_s}{r}}} > 1$. Now the question asked for the spatial distance but I just noticed that you've given the distance in spacetime, which I assume could contain a temporal component. Do we somehow know that it doesn't in this case? – Gumby The Green Apr 03 '22 at 23:03
  • @GumbyTheGreen to do the calculation I used $dt = 0$ i.e. the distance is the length of the tap[e measured at the same (coordinate) time. It is a displacement only in $r$. – John Rennie Apr 04 '22 at 04:27
  • Thank you for this VERY interesting discussion and thanks to John Rennie for the answer that the proper distance is greater than the coordinate distance. I can fully understand that as well as the idea that, for points well separated in the gravitational field we integrate the coordinate distance but simply add the proper distances. Then I suspect again the total proper distance will be greater than the total coordinate distance. Assuming this is correct, what do we mean when we say that the coordinate distance "has no metrical significance"? I can understand that the only true measure of the – Paul Hinrichsen Feb 13 '23 at 17:29
  • distance is the proper distance. Do we mean that the coordinate distance depends on the observer in much the same way that, in special relativity, there can be a multitude of observers and hence a multitude of improper length measures. – Paul Hinrichsen Feb 13 '23 at 17:29
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Distance is relative to the observer. You probably want the distance as measured by static observers (particularly $r=\textrm{const}$, $\theta=\pi/2$, $\phi=0$), then it is $\int(1-2M/r)^{-1/2}dr$. But if you want the distance as measured by "raindrop" observers (who fell from rest far away from the black hole), then it is simply $L$.

Colin MacLaurin
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