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.

Let us return to the radial-equation for the Schwarzschild metric given by

\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…

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This entry was posted in Cosmology, General Relativity, Mathematics, Physics, Relativity, Universe. Bookmark the permalink.

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