I've been reading about GR recently and I can follow the derivation of a Schwarzschild solution to it's final and well known form in Schwarzschild coordinates.

The orbit stability argument (for a massive test particle) is also clear - no stable circular orbit can exists for $r<6M$.

What usually follows after that is a calculation for the Earth:

$r = 6GM/c^2 = 0.03m$

radius of the Earth $= 6300km$.

So comparing them one notes that it is not a problem for the Earth because 0.03m is well below the surface.

My question is - how can we make such a comparision? Radius of a planet is measured in spherical coordinates but $r$ in $r=6M$ is in Schwartzschild coordinates - while deriving Schwartschild solution one starts with spherical coordinates but makes a lot of coordinates transformations so the resulting $r$ is really a very complicated function of a spherical radius and comparing their values seems wrong.

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    $\begingroup$ Much more simply: within Earth spacetime geometry is not Schwarzschild. It is Schwarzschild outside Earth if three simplifying assumptions are done: 1) Earth's mass distribution is spherically symmetric 2) Earth's rotation is neglected 3) Earth is alone into space (no Sun, no Moon, etc.). The value 0.03 m << Earth radius only means that spacetime geometry around Earth is very slightly affected by its mass - it departs very little from being Lorentzian. $\endgroup$
    – Elio Fabri
    Sep 23, 2018 at 13:07
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    $\begingroup$ while deriving Schwartschild solution one starts with spherical coordinates but makes a lot of coordinates transformations This sounds to me like a misunderstanding, although it's hard to be certain without seeing what source you're reading. There is no underlying, flat-space system of spherical coordinates that later gets transformed to Schwarzschild coordinates. $\endgroup$
    – user4552
    Sep 23, 2018 at 14:46

2 Answers 2


The Schwarzschild coordinate $r$ is defined so that the area of an $r=const.$ surface is $4\pi r^2$ with the area being evaluated using the metric at fixed $t$. This means that one can regard our (to a very good approximation) flat space-time radius $r$ as coinciding with the Schwarzschild coordinate $r$ once we are outside the body of the earth. (The Schwarzschild metric does not apply inside he earth)

  • $\begingroup$ Thanks for answering but it is not exactly what I have a problem with. Yes, I know about S2-foliation and that it is an exterior solution. The thing I cannot understand is if we do a coordinate transformation: r -> r' = f(r) then it is not meaningful to do a comparision of the form: r < r' and we do a lot of tranformations when deriving Schwartzschild solution $\endgroup$
    – shull
    Sep 23, 2018 at 13:31

The comparison is still valid, although what it means is a little hidden. The Schwarzschild r coordinate is defined as the square root of the surface area of a sphere at that distance divided by $4 \pi$. In other words $A=4\pi r^2$.

So saying that the radius of the earth is greater than $0.03 m$ is saying that the surface area of the earth is greater than $0.01 m^2$, which is clearly true. And saying that the radius of the Earth is $6300 km$ is saying that it’s surface area is $4\pi (6300 km)^2$ which if we are approximating the earth as spherically symmetric and non rotating is approximately true.


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