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I was researching online for different metrics of spacetime out of curiosity, and I found one that was said to be Schwarzschild metric in cylindrical coordinates:

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

I cannot remember from which site it was or find it in my history, but when I started to try to understand it, I found it to be wrong.

I was amazed to see that there such metrics out there. I never learned string theory, but I also discovered that cylindrical metrics are used to model the spacetime around cosmic strings.

So does the Schwarzschild metric in cylindrical coordinates describe cosmic strings? What is the Schwarzschild metric in cylindrical coordinates?

Edit: I should have mentioned this before, but I did not. I am asking for the Schwarzschild metric in cylindrical coordinates specifically for a cylindrical source/mass aligned/centered along the $z$ axis.

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    $\begingroup$ That metric isn’t even dimensionally consistent. $\endgroup$
    – Ghoster
    Jan 19, 2023 at 0:07
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    $\begingroup$ the $r$ component (proper length) increases larger $r$ is If you’re talking about $g_{rr}$, it approaches $1$ as $r\to\infty$. $\endgroup$
    – Ghoster
    Jan 19, 2023 at 0:09
  • $\begingroup$ Ah yes, I see that now. Thanks, @Ghoster I edited my post, correcting this. $\endgroup$
    – Tachyon
    Jan 19, 2023 at 0:12
  • $\begingroup$ The $r^2 dz^2$ term doesn't look right. Should it not just be $dz^2$? @Ghoster seems to be right. $\endgroup$
    – joseph h
    Jan 19, 2023 at 1:14
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    $\begingroup$ I was amazed to see that there such metrics out there. Why? Because it’s wrong or some other reason? $\endgroup$
    – Ghoster
    Jan 19, 2023 at 7:52

1 Answer 1

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As you noted, that's not the Schwarzschild metric in cylindrical coordinates.

In spherical coordinates, where the corresponding Cartesian coordinates would be $$(x,y,z) = r (\sin θ \cos φ, \sin θ \sin φ, \cos θ),$$ the metric is given by the line element $$\frac{r}{r - r_s} dr^2 + (r dθ)^2 + (r \sin θ dφ)^2 - \frac{r - r_s}{r} (c dt)^2.$$

To express the metric in cylindrical coordinates, where $$(x,y) = ρ (\cos φ, \sin φ),$$ set $$(ρ,z) = r (\sin θ, \cos θ).$$ Then $$ρ dρ + z dz = r dr, \hspace 1em z dρ - ρ dz = r^2 dθ, \hspace 1em ρ dφ = r \sin θ dφ, \hspace 1em r = \sqrt{ρ^2 + z^2},$$ and the line element becomes: $$\frac{\sqrt{ρ^2 + z^2}}{\sqrt{ρ^2 + z^2} - r_s} \frac{(ρ dρ + z dz)^2}{ρ^2 + z^2} + \frac{(z dρ - ρ dz)^2}{ρ^2 + z^2} + (ρ dφ)^2 - \frac{\sqrt{ρ^2 + z^2} - r_s}{\sqrt{ρ^2 + z^2}} (c dt)^2.$$

There is also the underlying question of what the cylindrical *generalization* of the Schwarzschild metric is. That's the Kerr metric which, when given in Boyer-Lindquist coordinates, is: $$ds^2 = Σ \left(\frac{dr^2}{Δ} + dθ^2\right) + \left(r^2 + a^2\right) (\sin θ dφ)^2 + \frac{r_s r}Σ \left(c dt - a \sin^2 θ dφ\right)^2 - (c dt)^2,$$ where the corresponding Cartesian coordinates are $$(x,y,z) = \left(\sqrt{r^2 + a^2} \sin θ \cos φ, \sqrt{r^2 + a^2} \sin θ \sin φ, r \cos θ\right),$$ and $$r_s = \frac{2GM}{c^2}, \hspace 1em a = \frac{J}{Mc}, \hspace 1em Σ = r^2 + (a \cos θ)^2, \hspace 1em Δ = r^2 - r_s r + a^2,$$ and this $r_s$ being the same as the $r_s$ for the Schwarzschild metric.

This describes a source with angular momentum $J$, mass $M$, in relativity, with $c$ being the in-vacuo light speed. You can work that what that comes out to in cylindrical coordinates, with the coordinates modified to the following form: $$ρ = \sqrt{r^2 + a^2} \sin θ, \hspace 1em z = r \cos θ,$$ and use the cylindrical version of the Schwarzschild metric to check this against the $J = 0$ (and $a = 0$) case.
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    $\begingroup$ The Schwarzschild metric you derived for cylindrical coordinates, does it have a spherical or cylindrical mass/source? I am asking because if it is spherical mass but in a cylindrical coordinate metric, then I have made a mistake. I wanted to see the Schwarzschild metric in cylindrical coordinates with a cylindrical source/mass aligned/centered along the $z$ axis. I should have mentioned it. I will edit my question. $\endgroup$
    – Tachyon
    Jan 19, 2023 at 22:33
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    $\begingroup$ If the angular momentum J is zero, and you're in the Schwarzschild metric, then it is spherically symmetric, even if written in cylindrical coordinates. (That also means that the direction you choose for the z coordinate is arbitrary.) For non-zero angular momentum, then you've generalized to the Kerr metric, which has only cylindrical symmetry about the axis of rotation, but not spherical symmetry. (Note: I also corrected my mistake for the expression for r_s.) $\endgroup$
    – NinjaDarth
    Jan 19, 2023 at 22:59
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    $\begingroup$ I see. But does it make a difference from what I said, that if the mass is spherical or cylindrical, will it make a difference in the metric? Will it be in a different form if it is a cylindrical mass and not a spherical one? $\endgroup$
    – Tachyon
    Jan 19, 2023 at 23:25
  • $\begingroup$ The mass, itself, doesn't have properties of being spherical or symmetric, the space-time and object do - with Schwarzschild being the spherical case, and Kerr its cylindrical generalization. Yeah, the Kerr and Schwarzschild solutions are definitely different. The difference shows up, not with any property of the mass itself, but by its angular momentum: whether it is 0 or not; i.e. by whether it's rotating or not. You can generalize this further by making it charged. Then Schwarzschild becomes Reissner-Nordström, and Kerr becomes Kerr-Newman. Kerr-Newman is the most general case of them all. $\endgroup$
    – NinjaDarth
    Jan 19, 2023 at 23:32
  • $\begingroup$ Okay. So if I understand this correctly, for the non-rotating and uncharged case, then the cylindrical Schwarzschild metric you have there should encode an arbitrary mass, spherical or cylindrical, or even any other mass that exhibits a cylindrical metric. Am I right? $\endgroup$
    – Tachyon
    Jan 19, 2023 at 23:41

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