The textbook from which I teach physics at the end of secondary school calculates the escape velocity at the surface of a non-rotating neutron star Newton-style using $\frac{1}{2}\cdot m\cdot v^2=\frac{G\cdot M\cdot m}{R^2}$. The mass of the star is $M_{\text{sun}}=2\cdot10^{30}~\text{kg}$, the radius is $10^4~\text{m}$. This gives $v_{\text{esc}}=0.54\cdot c$.

Even if this would (coincidentally?) be the correct value, how can I calculate the escape velocity from the surface of the neutron star, according to an observer at rest at the surface, in general relativity?


Let's say you assume that the neutron star is spherically symmetric, e.g., ignore the effects of rotation. Then for a radial trajectory in the resulting Schwarzschild spacetime, the calculation is actually not quite wrong, although you must be careful in interpreting it.

The reason is that orbits in a Schwarzschild spacetime have an effective potential that conserves an analogue of specific orbital energy: $$\mathcal{E} = \frac{1}{2}\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2 - \frac{GM}{r} + \frac{l^2}{2r^2} - \frac{GMl^2}{c^2r^3}\text{,}$$ where $l$ is specific angular momentum. The difference with the Newtonian case is that $r$ is the Schwarzschild radial coordinate, not the radial distance, and $\tau$ is the proper time of the orbiting particle, not absolute time, and finally, the very last term is absent in Newtonian gravity.

For a radial orbit, $l = 0$. Therefore, if by $v$ you mean the rate of change in Schwarzschild radial coordinate with respect to proper time of the particle, then the Newtonian calculation is actually correct: $$\left|\frac{\mathrm{d}r}{\mathrm{d}\tau}\right| = \sqrt{\frac{2M}{r}}\text{.}$$ However, if $v$ means something else, such as the rate of change in Schwarzschild radial coordinate with respect to Schwarzschild time $t$, then $$\left|\frac{\mathrm{d}r}{\mathrm{d}t}\right| = \left(1-\frac{2M}{r}\right)\sqrt{\frac{2M}{r}}\text{.}$$ Finally, the velocity measured by a stationary observer at a given Schwarzschild radial coordinate can be calculated via an inner product between the stationary observer's four-velocity $\left(1/\sqrt{1-2M/r},0\right)$ and the particle's four-velocity $\left(1/(1-2M/r),\pm\sqrt{\frac{2M}{r}}\right)$, but in this case it happily enough turns out to be equal to $\mathrm{d}r/\mathrm{d}\tau$: $$\left|v_s\right| = \sqrt{\frac{2M}{r}}\text{.}$$

  • $\begingroup$ Thank you. Could you please explain where the G has gone in the expression $v_{\text{esc}}=\sqrt{\frac{2 M}{r}}$? Is this a choice of units? $\endgroup$ – gamma1954 Apr 19 '15 at 21:30
  • $\begingroup$ @gamma1954 Yes, I lapsed into $G = 1$ units because they're more convenient. ... Also, if you're wondering about the effective potential, I gave an outline of a derivation here. $\endgroup$ – Stan Liou Apr 19 '15 at 21:36
  • $\begingroup$ @gamma1954: Stan has also used units where $c = 1$. With the constants the equation is $\left|v\right| = c\sqrt{2GM/c^2r}$. Note that at the event horizon $2GM/c^2 = 1$ so the escape velocity is $c$. $\endgroup$ – John Rennie Apr 20 '15 at 6:16
  • $\begingroup$ @John and Stan Thank you for your help. There is an explanation on en.wikipedia.org/wiki/Geometrized_unit_system I'm looking forward to reading my first book on GR. Hartle, Schutz, Zee, Glendenning? $\endgroup$ – gamma1954 Apr 20 '15 at 8:48
  • $\begingroup$ @gamma1954: if you're doing this for fun start with Schutz. I'm not sure it would be ideal for someone wanting to do research in GR, but it's the most approachable GR book I've ever read. $\endgroup$ – John Rennie Apr 20 '15 at 11:20

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