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 $$u'' + u - \frac {V_o} {l^2} - 3V_ou^2 = E^2$$​

Homework Equations

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!

 

[stevendaryl]

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]

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).

 

[stevendaryl]

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]

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!