## The escape velocity for the Schwarzschild metric (2).

In the other post The escape velocity for the Schwarzschild metric. we have found the escape velocity

$\displaystyle u = c \sqrt{ \displaystyle \frac{2\phi}{c^2} - \left( \frac{2\phi}{c^2} \right)^2 }$,

however – this was based on the limit $r \rightarrow \infty$ and $\dot{r} \rightarrow 0$.

$\displaystyle \frac{1}{2} \frac{\dot{r}^2}{\psi} + \frac{1}{2} \psi c^2 = \textit{constant}$,

where $\displaystyle \psi = 1 - \frac{r_s}{r}$. At $r \rightarrow \infty$ the metric becomes Minkowski, thus we may assume that $\dot{r} = \beta c$, where $-1 \le \beta \le +1$. It is clear that $\psi \rightarrow 1$ for $r \rightarrow \infty$, then we obtain

$\displaystyle \displaystyle \frac{1}{2} \frac{\dot{r}^2}{\psi} + \frac{1}{2} \psi c^2 = \frac{1}{2} \left( 1 + \beta^2 \right) c^2$,

so the general escape velocity becomes

$\displaystyle u = c \sqrt{\beta^2 + \left(1-\beta^2\right) \frac{2\phi}{c^2} - \left(\frac{2\phi}{c^2}\right)^2}$.

For light we have $\beta=1$ thus

$\displaystyle u_\textit{light} = c \sqrt{1 - \left(\frac{2\phi}{c^2}\right)^2}$.

For matter we have the critical case $\beta=0$ thus

$\displaystyle u_\textit{matter} = c \sqrt{\frac{2\phi}{c^2} - \left(\frac{2\phi}{c^2}\right)^2}$

Here are some plots…

For $\beta=0$ (matter) we have

For $\beta=\frac{1}{4}$ (matter) we have

For $\beta=\frac{1}{2}$ (matter) we have

For $\beta=1$ (light) we have

Sofar part 2…