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Why then is there no change in the angular coordinates in Schwarzschild metric? Why arent they unaffected by the mass?

Qmechanic
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steve
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  • Because the mass distribution is spherically symmetric (by assumption, this is what makes the metric Schwarzschild), so the spacetime curvature does not depend on the angle, only from the distance from the mass distribution. – Photon Feb 14 '17 at 12:57
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    @Photon that should be an answer! – John Rennie Feb 14 '17 at 13:06
  • Made it an answer. Was not sure if such a short explanation is worth an answer. :) – Photon Feb 14 '17 at 13:15

2 Answers2

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Because the mass distribution is spherically symmetric (by assumption, this is what makes the metric Schwarzschild), so the space-time curvature does not depend on the angle, only on the distance from the mass distribution.

Photon
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One of the aspects of general relativity that beginners find hard is what we mean by coordinates. In GR coordinates are just a way of labelling points in spacetime and they don't necessarily correspond to anything real like distances and times that would be measured by an observer.

You've already asked about the Schwarzschild radial coordinate, $r$, and has been explained before this does not mean the distance to the centre of the black hole. We define the value of the $r$ coordinate at a point as the circumference of the circle centred on the black hole and passing through that point divided by $2\pi$. This isn't the same as the distance we would measure if we let out a tape measure towards the black hole, but that doesn't matter because $r$ is just a label. We can use the $r$ coordinate to calculate things like the length of that tape measure if we need to.

In this question you're asking about the $\theta$ and $\phi$ coordinates. You are quite correct that these coordinates are not affected by the mass of the black hole, but that's because we've chosen them in such a way that they are unaffected by that mass. In flat spacetime the circumference of a circle of radius $r$ is given by:

$$ s = \int_0^{2\pi} r\,d\phi = 2\pi r $$

In the curved spacetime around a black hole the circumference of a circle of Schwarzschild radius $r$ centred on the black hole is also $2\pi r$, but remember that's because we defined our coordinate $r$ to make this true.

So the bottom line is that the $\theta$ and $\phi$ coordinates in the Schwarzschild metric are the same as in flat space because we've fiddled with the $r$ coordinate to make them the same. That is we've absorbed the curvature into the definition of the $r$ coordinate.

Community
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John Rennie
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