Physics Asked on March 5, 2021
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
$$ left( frac{du}{dphi}right)^2 = frac{c^2 k^2}{h^2} – alpha left( frac{c^2}{h^2} + u^2 right) $$ 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?
Let $f(u)$ be the third degree polynomial, so that $$left(frac{du}{dphi}right)^2 = f(u)tag{*}$$ 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.
Answered by John Donne on March 5, 2021
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