Rewriting Central Force Problem of Black Hole Potential

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1. Feb 19, 2017

Digital Honeycomb

1. The problem statement, all variables and given/known data
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 $$u'' + u - \frac {V_o} {l^2} - 3V_ou^2 = E^2$$​
2. Relevant equations
All shown above.

3. 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!

2. Feb 19, 2017

stevendaryl

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 $\phi$.

3. Feb 19, 2017

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).

4. Feb 19, 2017

stevendaryl

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

5. Feb 19, 2017

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!