# Rewriting Central Force Problem of Black Hole Potential

Digital Honeycomb

## Homework Statement

From the homework:
In General Relativity it is found that the radial equation of an object orbiting a non-rotating black hole has the form $$\dot r^2 + (1 - 2 \frac {V_o} {r} ) (\frac {l^2} {r^2} + 1) = E^2$$ where ##r## is the radial coordinate, ##l## is the angular momentum, and ##E## is the total energy; the potential ##V_o = GM_o##, where ##M_o## is the mass of the black hole. Show, using the standard substitution ##u = \frac {1} {r}##, and rewriting ##\dot r## in terms of ##\dot \phi## and ##u' = \frac {du} {d\phi}##, that we can write a differential equation for ##u(\phi)## of form $$u'' + u - \frac {V_o} {l^2} - 3V_ou^2 = E^2$$​

All shown above.

## The Attempt at a Solution

First off I'm not entirely sure how to rewrite ##\dot r## in terms of ##\dot \phi##. In the notes it is described that the angular momentum ##l = mr^2\dot\phi##, so therefore this equation could be written in the form $$\dot r = \sqrt{E^2 - (1 - 2V_o u)(m\dot\phi +1)}$$ but it is not exactly like I can plug this back into the original equation. I can also try taking that since ##u = \frac {1} {r}##, that ##\frac {du} {dt} = \frac {-1} {r^2} \frac {dr} {dt} = \frac {du} {d\phi} \frac {d\phi} {dt}## and so ##\dot r = \frac {-1} {u^2} u' \dot \phi## but putting this into the first equation, I'm not exactly sure where the ##u''## is supposed to come from. Am I coming at this from the wrong direction entirely? Thank you!

Staff Emeritus
I'm not sure about the motivation for going from the equation in terms of $\dot{r}$ to the second-order equation in terms of $u$, but here's how it could be done:

First take a time derivative of your equation for $\dot{r}$. The right side of the equation is a constant, so that gives:

$2 \dot{r} \ddot{r} - F(r) \dot{r} = 0$

where $- F(r)$ is the derivative of that potential-like expression on the left side. Dividing through by $\dot{r}$ and rearranging gives you a second-order equation for $r$:

$\ddot{r} = \frac{1}{2} F(r)$

Now, at this point, you can use a trick from nonrelativistic orbital dynamics. You do a simultaneous change of dependent variable from $r$ to $u = \frac{1}{r}$ and a change of independent variable from $t$ to $\phi$.

Digital Honeycomb
Digital Honeycomb
Hmm, okay... so doing that I end up with the equation, $$\ddot r = \frac {l^2} {r ^3} - 3 \frac {V_o l^2} {r^4} - 2 \frac {V_0} {r^2},$$ substituting ##u = \frac {1} {r}## is simple enough for the RHS of the equation, but not for understanding exactly how to change ##\ddot r## into an equation of ##u(t)##. I know that ##\frac {d^2u} {dr^2} = \frac {2} {r^3}##, but I'm afraid I don't understand how to turn this into an equation in respect to time. I suppose I can take that ##\dot u = \frac {-1} {r^2} \dot r## and then ##\ddot u = \frac {2} {r^3} \dot r - \frac {1}{r^2}\ddot r##, but beyond that I'm not sure how to turn this purely into some ##u(t)## (beyond that, I realize I can take ##d\phi = \frac {L}{m}u^2 dt##, however).

Staff Emeritus
As I said, you have to simultaneously change $r$ to $u$ and $t$ to $\phi$.

$\frac{dr}{dt} = \frac{dr}{du} \frac{du}{d\phi} \frac{d\phi}{dt}$

$\frac{dr}{du} = -\frac{1}{u^2}$
$\frac{d\phi}{dt} = \frac{\mathcal{l}}{mr^2} = \frac{\mathcal{l}}{m} u^2$

So $\frac{dr}{dt} = -\frac{\mathcal{l}}{m} \frac{du}{d\phi}$

Then operate again by $\frac{d}{dt} = \frac{d\phi}{dt} \frac{d}{d\phi}$ to get:

So $\frac{d^2 r}{dt^2} = -(\frac{\mathcal{l}}{m})^2 u^2 \frac{d^2 u}{d\phi^2}$

Digital Honeycomb
Digital Honeycomb
Oh! That makes sense! And then plugging that into the equation gives $$- \frac {l^2} {m^2}u^2 u'' = l^2 u^3 - 3V_o l^2 u^4 - V_o u^2$$ and that turns into $$u'' + u - \frac {V_o} {l^2} - 3V_o u^2 = 0$$ if you take ##m = 1##! Thank you so much!