# Why is the Schwarzschild radius of a Schwarzschild black hole $\frac{2GM}{c^2}$?

$d = \frac{2GM}{v^2}$ is the formula for computing in terms of any speed the distance at which the escape velocity is that speed in Newtonian physics. Why is it also true that for a Schwarzschild black hole, $r = \frac{2GM}{c^2}$? I see why for some positive real number $k$, the formula is $r = \frac{kGM}{c^2}$ but why is $k$ exactly 2?

I'm guessing $k$ = 2 because the Gullstrand–Painlevé were chosen in such a way that $k$ = 2.

Here are my 2 questions.

1. Is it 2 in the Gullstrand–Painlevé coordinate system or is that only in another coordinate system?
2. In which ever coordinate system it is, why were the coordinates chosen that way?

I know space-time is curved so we could have chosen a coordinate system where $k$ is not exactly 2. I think the Gullstrand–Painlevé coordinates were chosen to satisfy certain intuitive properties, such as the property that the time dilation and length contraction of an object hovering outside the event horizon both vary as the minus half power of the distance from the event horizon. Maybe they were specially chosen to make $k$ be exactly equal to 2.

• What kind of explanation do you want, i.e. what assumptions do you want to start form? The $2$ can be derived just be crunching equations; are you looking for something more intuitive than that? – knzhou Aug 19 '18 at 21:17
• @knzhou I fixed up my question to address your concerns. – Timothy Aug 19 '18 at 21:26
• The Schwarzschild radius applies in Schwarzschild coordinates, not G-P as far as I'm aware. – Javier Aug 19 '18 at 21:41
• What a useful heuristic and intuitive way to think about the Schwarzschild radius. I would not have naively thought of the concept that a black hole is an object with an Newtonian escape velocity equal to or greater than the maximum possible Newtonian escape velocity given special relativity, but it is. – ohwilleke Aug 19 '18 at 21:53

If we assume a nearly flat spacetime, we can expand around Minkowski space in the linear limit $$g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$$
Our line element is now $$ds^2 = (1 + h_{00})dt^2 - (1 - h_{ii})dx_i^2$$
We can now plug this into the equations of motion using Einsteins equations $$m\left(\ddot{x}^k + \Gamma^k_{\mu\nu}\dot{x}^\mu \dot{x}^\nu\right) = 0$$ We want to ultimately get the Newtonian limit of the Schwarzchild solution, so we will assume spherical symmetry and a static solution. This means that, for $k = 0$, we find $$m\ddot{x}^0 = 0$$ The spatial part, however, looks like $$m\left(\ddot{x}^i +\Gamma^i_{\mu\nu}\dot{x}^\mu \dot{x}^\nu\right) = 0$$ Assuming spherical symmetry, and knowing the equation for the Christoffel symbol, we can perturbatively define $$\Gamma^i_{00} = \frac12\nabla h_{00}$$ In the Newtonian limit, all of the $\dot{x}^i$ terms are much smaller than the $\dot{x}^0$ terms, so we can ignore them. Plugging in $\Gamma$, we find $$m\ddot{x}^i = -\frac12m\nabla h_{00}$$ We recognize the left hand side as Force, $F = ma$, and deduce that the right hand side must be the force of a particle with mass $m$ moving in a gravitational field. Comparing this with newtons equation $F = -\frac{GMm}{(x^i)^2}$, we deduce that $h_{00} = -\frac{2GM}{x^i}$. Plugging this into the line element then exactly gives the Schwarzchild metric in the weak field limit $$ds^2 = \left(1 - \frac{2GM}{x^i}\right)dt^2 - \left(1 + \frac{2GM}{x^i}\right)dx_i^2$$