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Rewriting Central Force Problem of Black Hole Potential

  1. Feb 19, 2017 #1
    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. jcsd
  3. Feb 19, 2017 #2

    stevendaryl

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

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

    [itex]2 \dot{r} \ddot{r} - F(r) \dot{r} = 0[/itex]

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

    [itex]\ddot{r} = \frac{1}{2} F(r)[/itex]

    Now, at this point, you can use a trick from nonrelativistic orbital dynamics. You do a simultaneous change of dependent variable from [itex]r[/itex] to [itex]u = \frac{1}{r}[/itex] and a change of independent variable from [itex]t[/itex] to [itex]\phi[/itex].
     
  4. Feb 19, 2017 #3
    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).
     
  5. Feb 19, 2017 #4

    stevendaryl

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    As I said, you have to simultaneously change [itex]r[/itex] to [itex]u[/itex] and [itex]t[/itex] to [itex]\phi[/itex].

    [itex]\frac{dr}{dt} = \frac{dr}{du} \frac{du}{d\phi} \frac{d\phi}{dt}[/itex]

    [itex]\frac{dr}{du} = -\frac{1}{u^2}[/itex]
    [itex]\frac{d\phi}{dt} = \frac{\mathcal{l}}{mr^2} = \frac{\mathcal{l}}{m} u^2[/itex]

    So [itex]\frac{dr}{dt} = -\frac{\mathcal{l}}{m} \frac{du}{d\phi}[/itex]

    Then operate again by [itex]\frac{d}{dt} = \frac{d\phi}{dt} \frac{d}{d\phi}[/itex] to get:

    So [itex]\frac{d^2 r}{dt^2} = -(\frac{\mathcal{l}}{m})^2 u^2 \frac{d^2 u}{d\phi^2}[/itex]
     
  6. Feb 19, 2017 #5
    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!
     
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