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I'm trying to solve the Kepler problem using the Lagrangian,

$$L = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\phi}^2) - U(r) $$

which after quite a bit of fiddling with, by noting that the angular momentum $M = mr^2 \dot{\phi}$ is a constant of motion and also $M = 2m\dot{f}$ where $\dot{f}$ s the sectorial velocity, leads to

$$\phi = \int{\frac{M dr/r^2}{\sqrt{2m(E - U(r)) - M^2 / r^2}}}{}$$

Now for the Kepler problem $U(r) \propto 1 / r$ and so $U(r) = \alpha / r$. Plugging that in, we get,

$$\phi = \int{\frac{M}{r^2\sqrt{2m(E + \alpha / r) - M^2 /r^2}}}{dr}$$

However, plugging that integration into WolphramAlpha gives an imaginary solution.

What am I doing wrong?

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    $\begingroup$ You may not be doing anything wrong. You may be looking at something like $\arcsin$ in disguise: recall that (on the appropriate branch) $\arcsin z=-\,i\,\log\left(i\,z+\sqrt{1-z^2}\right)$. Does that help? $\endgroup$
    – Selene Routley
    Commented Jan 16, 2015 at 6:31
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    $\begingroup$ What am I doing wrong? The answer is you are using Mathematica to do your homework for you. It's a great tool, but you have to take its answers with a grain of salt. $\endgroup$
    – David Hammen
    Commented Jan 16, 2015 at 7:31
  • $\begingroup$ @DavidHammen I wouldn't say that using Mathematica to evaluate an integral is doing it wrong. But I do agree that one should be ready to get the result in an unexpected form and have to adapt it to expectations. $\endgroup$
    – Ruslan
    Commented Jan 16, 2015 at 9:27
  • $\begingroup$ Additionally to Rod's comment, note that you are evaluating an indefinite integral. It's nothing wrong with it being imaginary, because it's defined up to an additive constant, which can be complex. $\endgroup$
    – Ruslan
    Commented Jan 16, 2015 at 9:36
  • $\begingroup$ @Ruslan - I didn't say that WA is doing it wrong. It does however do things stupidly. While it knows that $\int_0^x e^t\,dt = e^x - 1$, it somehow thinks that the best representation of $\int_0^x e^{-t}\,dt$ is $\sinh (x) - \cosh(x) + 1$. That is correct, but it's mindless and it's stupid. What's wrong with $1-e^{-x}$? $\endgroup$
    – David Hammen
    Commented Jan 16, 2015 at 9:42

1 Answer 1

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You can use $$ \frac{d}{dr} \cos^{-1} \left(f\left(r\right)\right) = -\left[1-\left(f\left(r\right)\right)^2\right]^{-1/2} \frac{df}{dr} $$ with $$ f\left(r\right) = \frac{M/r - m \alpha/M}{\sqrt{2 m E + m^2 \alpha^2 / M^2}} $$ to show that $$ \int dr \frac{M / r^2}{\sqrt{2 m E + 2 m \alpha / r - M^2 / r^2}} = \cos^{-1} \left(\frac{M/r - m \alpha/M}{\sqrt{2 m E + m^2 \alpha^2 / M^2}}\right) + C $$

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