In Schwarzschild coordinates (t,r,theta,phi) the Schwarzschild metric (or equivalently, the line element for proper time) has the form g=−c^2d pi^2=−(1−r_sr/)c^2dt^2+(1−rsr)^(−1) dr^2+r^2g Omega, where gOmega is the metric on the two sphere, i.e. g Omega=(d theta^2+sin^2 theta d phi^2).

nabakhi72

nabakhi72

Answered question

2022-08-13

In Schwarzschild coordinates ( t , r , θ , ϕ ) the Schwarzschild metric (or equivalently, the line element for proper time) has the form
g = c 2 d τ 2 = ( 1 r s r ) c 2 d t 2 + ( 1 r s r ) 1 d r 2 + r 2 g Ω ,
where g Ω is the metric on the two sphere, i.e. g Ω = ( d θ 2 + sin 2 θ d φ 2 ) .
Isn't ( t , r , θ , ϕ ) representing the spherical coordinates of an observer, looking at the black-hole from infinity?

Answer & Explanation

Emely English

Emely English

Beginner2022-08-14Added 16 answers

The word "observer" is used in totally different ways in special relativity, general relativity, and quantum mechanics. If you assume they're related, you'll get confused.
In special relativity, "observer" usually means "inertial reference frame."
In general relativity, "observer" usually means a scientist making observations from a particular vantage point (a particular location in spacetime).
In special relativity it happens to be the case that if a scientist is moving inertially, there's a unique (up to time translation) global inertial reference frame in which they're at rest at the origin. Often people call that "their" coordinate system, though this is very misleading since you can just as well use any other coordinate system to describe their motion if you want to. But that unique inertial frame does exist, and it can be defined mathematically. There's no analogous special coordinate system for a moving scientist in general relativity. You might, in some cases, find it useful to use coordinates in which the scientist is at rest at the origin, but that doesn't uniquely define what the coordinates look like away from the origin. And often that coordinate system isn't very useful anyway, because the metric has a complicated form with respect to it, and a simpler form with respect to some other coordinate system in which the scientist isn't at rest.
Schwarzschild coordinates ( t , r , θ , ϕ ) are just coordinates in which the Schwarzschild metric has a simple mathematical form. They don't represent any observer, in the special-relativistic or general-relativistic sense. You can use Schwarzschild coordinates to determine what an observer (general-relativistic) will see. You can also use totally different coordinates to work out what the same observer will see, and if you did it correctly you'll get the same answer, because the underlying physics is independent of the coordinates.

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