general relativity - Where 2 comes from in formula for Schwarzschild radius?

In general theory of relativity I've seen several times this factor:

$$(1-\frac{2GM}{rc^2}),$$

e.g. in the Schwarzschild metric for a black hole, but I still don't know in this factor where 2 comes from?

Answer

On the one hand, for nonrelativistic particle moving in an external gravitational field, the Lagrangian has the form: $$ L=-mc^{2}+\frac{m\mathbf{v}^{2}}{2}-m\phi,\quad\quad(1) $$ where $m$ is a mass of particle, $\phi$ is a gravitational potential. On the other hand, general relativity requires the following action for the particle: $$ S=-mc\int ds,\quad\quad\left( 2\right) $$ where $$ ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu},\quad dx^{\mu}=\left( cdt,d\mathbf{r} \right) , $$ thus for a static field ($g_{i0}=0$) $$ ds^{2}=g_{00}c^{2}dt^{2}+g_{ij}dr^{i}dr^{j}=\left( g_{00}c^{2}+g_{ij} v^{i}v^{j}\right) dt^{2}. $$ Let's find the leading correction to the metric in the nonrelativistic limit, i.e., in the $c\rightarrow\infty$ limit: $$ g_{00}=1+h_{00}+O\left( c^{-4}\right) ,\quad g_{ij}=-\delta_{ij}+O\left( c^{-4}\right) , $$ so that $h_{00}=O\left( c^{-2}\right) $, hence \begin{align*} ds & =dt\sqrt{g_{00}c^{2}+g_{ij}v^{i}v^{j}}=cdt\sqrt{1+h_{00}-\mathbf{v} ^{2}/c^{2}+O\left( c^{-4}\right) }\quad\quad\quad\left( 3\right) \\ & =cdt\left( 1+h_{00}/2-\mathbf{v}^{2}/2c^{2}+O\left( c^{-4}\right) \right) . \end{align*} Using the action (2) we obtain the following Lagrangian: $$ S=\int Ldt=\int dt\left( -mc^{2}-mc^{2}h_{00}/2+m\mathbf{v}^{2}/2+O\left( c^{-4}\right) \right) . $$ Comparison with the Lagrangian (1) yields: $$ \frac{c^{2}h_{00}}{2}=\phi,\quad\Rightarrow\quad h_{00}=\frac{2\phi}{c^{2} }=-\frac{2GM}{c^{2}r}.\quad\quad\left( 4\right) $$ Hence, one can see that the factor $2$ appears due to the square root operation in the interval (3).

It is worth noting here that it is not directly connected with the metric singular points. The expansion (4) is so called gauge independent, but the position of singularity is not. For example, the expression of the Schwarzschild metric in the harmonic coordinates has the form:(see, e.g., S. Weinberg, Gravitation and Cosmology, eq. (8.2.15)): $$ ds^{2}=\frac{1-GM/rc^{2}}{1+GM/rc^{2}}c^{2}dt^{2}-\frac{1+GM/rc^{2} }{1-GM/rc^{2}}\,\frac{G^{2}M^{2}}{r^{4}c^{4}}\left( \mathbf{r}\cdot d\mathbf{r}\right) ^{2}-\left( 1+\frac{GM}{rc^{2}}\right) ^{2} d\mathbf{r}^{2}, $$ which has the singularities in the points: $$ r=\pm\frac{GM}{c^{2}}. $$ However the expansion always has the form of eq. (4): $$ g_{00}=\frac{1-GM/rc^{2}}{1+GM/rc^{2}}=1-\frac{2GM}{rc^{2}}+O\left( c^{-4}\right) . $$ The Schwarzschild gauge is the simplest gauge such that $g_{00}$ has exactly the form of the expansion (4): $$ g_{00}=1-\frac{2GM}{rc^{2}}. $$ $$ $$