I can't fill in the gaps in my solution to this and assistance or a reference would be appreciated.

The question begins with the straightforward derivation of the EoM for a massive particle orbiting in the equatorial plane, as

$$ \left( \frac{du}{d\phi}\right)^2 = \frac{c^2 k^2}{h^2} - \alpha \left( \frac{c^2}{h^2} + u^2 \right) $$ where $u = \frac{1}{r}$, $ h, k$ are constants arising as $ \alpha \dot{t} = k$ and $\dot{\phi} = h u^2$, and $\alpha = 1-\frac{r_s}{r} $ where $r_s$ is the Schwarzschild radius.

It then says a stationary experimenter at radius $a > r_s$ projects a massive particle with speed $v$ normal to the radial direction, and asks me to show that in the case $h^2 > 3 r_s^2 c^2$ the particle will be ejected if $v$ exceeds an escape velocity similar in form to the Newtonian one.

Clearly the above condition restricts to the case of three real roots, and I think that the condition I want is that the smallest root of the above cubic (there's an extra $u$ in the $\alpha$) is $\leqslant 0$, though I'm not entirely sure why that's necessary/sufficient. Given that, I obtain the result $ v \geqslant \sqrt{\frac{2GM}{a}} $.

Is this result correct? And could someone explain why that condition is the right one?


1 Answer 1


Let $f(u)$ be the third degree polynomial, so that $$\left(\frac{du}{d\phi}\right)^2 = f(u)\tag{*}$$ The experimenter starts at $u=1/a$ and must reach infinity, $u=0$. The crucial point is that if $f(u)$ is negative somewhere in the region $0<u<1/a$, then the equation of motion $(*)$ prevents crossing the negative region, so you can't reach infinity. In other words if $u(\theta)$ solves $(*)$ then $f(u)>0$; since $u$ is continuous, you can't connect $1/a$ to $0$ if $f$ is negative somewhere in between.

Now it's a matter of ordinary calculus to determine the shape of $f$: we learn that it has exactly one root $u_\ast$ in the range $u<r_s$ and that $f<0$ for $u<u_\ast$ and $f>0$ for $u_\ast<u<1/r_s$. Since $f$ must be positive for $0<u<1/a<1/r_s$, we must have $u_\ast<0$.

Whether that's the right equation depends on what you mean, i.e. see this question.


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