I have a mathematical background and I have just derived the expression of the Schwarzschild metric. Now I was wondering what were the motivations and applications in physics of this metric. Any info would be welcome!
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I just meant I calculated the metric $ds^2$ = ... like in the answer below. – Sasha May 17 '15 at 09:47
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The Schwarzschild metric describes the geometry of the spacetime containing a time independant spherically symmetric mass and nothing else. In other words the spacetime has to have existed unchanged for an infinite time and continue to exist unchanged for an infinite time, and there must be nothing else in the universe. Obviously there is no object in the real universe that is exactly described by the Schwarzschild metric (which has led the loonier end of the physics fringe to claim that black holes don't exist).
However we expect the Schwarzschild metric to be an excellent approximation for spacetime geometry near the sorts of astronomical objects we see around us, like stars and planets as well as (of course) black holes. So for example we can use the Schwarzschild metric to calculate the corrections needed to the clocks on GPS satellites, without worrying that the spacetime geometry in and around the Solar System is a lot more complicated that the Schwarzschild metric describes.
As for specific applications, there must be hundreds of them. Top of the list would be the sorts of applications Stan mentions in his answer.
John Rennie
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I think I don't understand something. The Schwarzschild space time is meant to describe the geometry around an object which is spherically symmetric and static. But planets, stars... are not static ? I just read a paper where they say that the metric is static but not the source ?? Is this true ? Can we avoid assuming that the object is static and the metric remains static ? Thanks again ! – Sasha May 19 '15 at 07:45
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@Sasha: good point. If there is no inertial frame in which the object is static, for example because it's rotating or orbiting another body or both, the the Schwarzschild metric will at best be an approximation. My point is that in many cases it's a very, very good approximation. – John Rennie May 19 '15 at 07:48
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@Sasha A Schwartzschild solution can sometimes apply even if the star or planet isn't static. For instance if a star expands and contracts but in a totally spherically symmetric way then the region that is always outside the star follows a Schwartzschild solution the whole time. And if a real star already warped the spacetime outside its current boundary back when it formed then relatively small amounts of source out there now aren't going to change it much on average over regions large compared to the small stuff out there. So that's how it can be a good approximation to lots of situations. – Timaeus Jun 06 '15 at 01:57
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By "derived the Schwarzschild metric" I assume you mean calculating the exterior solution of the form
$$ds^2 = -\left(1-\frac{2M}{r}\right)dt^2 + \left(1-\frac{2M}{r}\right)^{-1}dr^2 +r^2\Omega^2$$
Applications:
- Describing deflection of light by the sun
- Precession of the perihelia of the orbits of the inner planets
- Schwarzschild singularity and blackholes
You can find info on these in Wald's General Relativity or Weinberg's Gravitation and Cosmology.
Stan Shunpike
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