Neutron star: escape velocity

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?

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 $\mathrm{d}r/\mathrm{d}\tau$: $$\left|v_s\right| = \sqrt{\frac{2M}{r}}\text{.}$$
• 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? – gamma1954 Apr 19 '15 at 21:30
• @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. – Stan Liou Apr 19 '15 at 21:36
• @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 $c$. – John Rennie Apr 20 '15 at 6:16