## The escape velocity for the Schwarzschild metric.

The Schwarzschild metric is given by

$\displaystyle ds^2 = \psi c^2 dt^2 - \frac{1}{\psi} dr^2 - r^2 \left( d\theta^2 + \sin^2(\theta) d\phi^2 \right)$

where $\displaystyle \psi = 1 - \frac{r_s}{r}$ and $\displaystyle r_s = \frac{2 GM}{c^2}$.

The geodetic equation is given by

$\displaystyle \frac{d^2x^\kappa}{ds^2} + \Gamma^\kappa_{\mu\nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds} = 0$

The path for escaping is given by $\theta = \textit{constant}$ and $\phi = \textit{constant}$, so we only need

$\left[\begin{array}{rcl} \Gamma^r_{rr} &=& \displaystyle \frac{1}{2} \frac{\psi'}{\psi} \\\\ \Gamma^t_{rt} = \Gamma^t_{tr} &=& \displaystyle \frac{1}{2} \frac{\psi'}{\psi} \\\\ \Gamma^r_{tt} &=& \displaystyle \frac{1}{2} \psi \psi' c^2 \end{array}\right.$

And we obtain the equations

$\left[\begin{array}{rcl} \displaystyle \frac{d^2r}{ds^2} + \Gamma^r_{rr} \left(\frac{dr}{ds}\right)^2 + \Gamma^r_{tt} \left(\frac{dt}{ds}\right)^2 &=& 0\\\\ \displaystyle \frac{d^2t}{ds^2} + 2 \Gamma^t_{tr} \frac{dr}{ds} \frac{dt}{ds} &=& 0 \end{array}\right.$

So

$\left[\begin{array}{rcl} \displaystyle \frac{d^2r}{ds^2} + \frac{1}{2} \frac{\psi'}{\psi} \left(\frac{dr}{ds}\right)^2 + \frac{1}{2} \psi \psi' c^2 \left(\frac{dt}{ds}\right)^2 &=& 0\\\\ \displaystyle \frac{d^2t}{ds^2} + \frac{\psi'}{\psi} \frac{dr}{ds} \frac{dt}{ds} &=& 0 \end{array}\right.$

Now it is clear that $\displaystyle \frac{d^2r}{ds^2} = \frac{d}{ds} \left( \dot{r} \frac{dt}{ds} \right) = \ddot{r} \left(\frac{dt}{ds}\right)^2 + \dot{r} \frac{d^2t}{ds^2}$. So we obtain the equation

$\displaystyle \left( \ddot{r} - \frac{1}{2} \frac{\psi'}{\psi} \dot{r}^2 + \frac{1}{2} \psi \psi' c^2 \right) \left( \frac{dt}{ds} \right)^2 = 0$

Then

$\displaystyle \ddot{r} - \frac{1}{2} \frac{\psi'}{\psi} \dot{r}^2 + \frac{1}{2} \psi \psi' c^2 = 0$

Note that for $c \rightarrow \infty$ we obtain $\ddot{r} + \phi' = 0$ – which is the Newton-limit, where we use $\displaystyle \psi = 1 - \frac{2 \phi}{c^2}$.

Now

$\displaystyle \frac{d}{dt} \left( \frac{1}{2} \frac{\dot{r}^2}{\psi} + \frac{1}{2} \psi c^2 \right) = \frac{\dot{r}}{\psi} \left( \ddot{r} - \frac{1}{2} \frac{\psi'}{\psi} \dot{r}^2 + \frac{1}{2} \psi \psi' c^2\right) = 0$

So

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

Escaping means that there is some limit $r \rightarrow \infty$ and $\dot{r} \rightarrow 0$. Thus we obtain

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

Then we obtain

$\displaystyle \left( \frac{\dot{r}}{c} \right)^2 = \psi - \psi^2$

Let the escape velocity be denoted as $u$ then we can write

$u = c \sqrt{\psi - \psi^2}$

As $\displaystyle \psi = 1 - \frac{r_s}{r} = 1 - \frac{2\phi}{c^2}$ we obtain

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

The only solutions are given by $\displaystyle \frac{r_s}{r} \ge 0$ and $\displaystyle \frac{r_s}{r} \le 1$ – thus $r \ge r_s$. Or $\displaystyle \frac{2\phi}{c^2} \le 1$ thus $\displaystyle \frac{2GM}{rc^2} \le 1$ so $\displaystyle r \ge \frac{2GM}{c^2}$

The escape velocity for the Schwarzschild metric is given by

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

Hope you liked this post!

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### 11 Responses to The escape velocity for the Schwarzschild metric.

1. Xavier Terri says:

Excellent article. The formula for escape velocity is the correct formula from the point of view of General Relativity. I would like to comment it.
It is easy to show that this formula is the same that is obtained in Newton’s Universal Gravitation, but multiplied by a factor which is equal to (1 – (Schwarzschild radius) / r). (see page 9 of http://vixra.org/abs/0912.0031 )
Then, according to General Relativity, the escape velocity at the critical radius is not equal to the speed of light ‘c ‘, but… zero!!!. How then defines general relativity a black hole? ( http://21physics.blogspot.com/p/new-lorentz-transformation.html )

• Hello Xavier,

there is a difference between the speed of escape and what you reach… What is important – what is the time required for a particle/light to move away from the Schwardzschild radius.

As for you link to that article (it seems to be Spanish) – the modified metric (page 5) is not a solution of the Einstein Field equation – therefore not valid within General Relativity – neither are the conclusions.

– Johannes

• Xavier Terri says:

Obviously is not the same speed (of escape) than space (what you reach…). But what is the escape speed at the critical Schwarzschild radius? In old Newton’s theory is equal to ‘c ‘. In Einstein’s theory is equal (as you can see in the last formula of your own post) to zero!!!
I agree with the last paragraph. The ‘modified metric’ is no solution of the Einstein Field equation, therefore is not valid within General Relativity Is a solution of a new theory: http://www.bubok.com/libros/6346/Extracto-de-la-Teoria-Conectada

• Hello Xavier, about the escape velocity…

First, as $g_rr \rightarrow \infty$ for $r \rightarrow r_s$ and as $ds$ does not go to infinity, we must have $dr=0$ at the Schwarzschild radius.

Second, the escape velocity has been solved for $r \rightarrow \infty$ and $\dot{r} \rightarrow 0$.

I guess that I need t write another post with more analysis in detail…

• Hi Xavier, I have tried to look at the links – but I am not very good at spanish – so reading those documents is somehow difficult. Do you have English versions as well?

2. Xavier Terri says:

Hi Johannes. Thank you for asking for my work.
I share your penultimate comment, including the conclusion dr = 0.
Regarding your last comment, I have to say that I have very few texts in English. Three papers on viXra.org: 1, 8 and 10 (the number 8 is an automatic translation by Google Translator). ( http://vixra.org/author/Xavier_Terri_Casta_ntilde_eacute )
The quickest way to understand what I try to say is at this link, which I have the intention to translate into English as soon as possible: http://www.slideshare.net/xaterri/conferencia-de-xavier-terri-para-unt-tres-teoras-sobre-la-gravedad (Since this article is very sketchy, the language problem is perhaps less important … )
Sorry I can not offer, for now, most texts in English

3. An impressive share, I just given this onto a colleague who was doing a little analysis on this. And he in fact bought me breakfast because I found it for him.. smile. So let me reword that: Thnx for the treat! But yeah Thnkx for spending the time to discuss this, I feel strongly about it and love reading more on this topic. If possible, as you become expertise, would you mind updating your blog with more details? It is highly helpful for me. Big thumb up for this blog post!

4. Starting from the formula for escape velocity, witch is the speed for a particle that came from
infinity with an initial speed Uo ??

5. Amid says:

I thought that the Schwarzschild radius was already the radius corresponding to the escape of light: if you put c as an escape velocity, than in the equation c = sqrt(2GM/r) you can find the Schwarzschild radius. Am I wrong?

• The escape velocity of light is the classical approax. Within general relativity the speed of light at the Schwardzschild radius is actually zero.

6. voltar says:

The christoffel symbol for rrr should be the negative of what you have.