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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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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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