I know that:

$$ \ddot{r}^2 = r\dot{\phi}^2 $$


$$ \ddot{\phi}=-\frac{2}{r}\dot{r}\dot{\phi} $$

and I know that using this I should be able to get the geodesic equations, but for my life I just can't, I keep trying but I end up with things I cannot solve.

I tried getting $r$ from the first equation and replacing it in the second, getting $\dot{\phi}$ from the second and replacing it in the first, I even found that:

$$ r^2\dot{\phi} = constant $$

but that didn't help at all, I am completely stuck and I can't find any derivation of the geodesic equations anywhere in the internet, please help I want to understand

  • 1
    $\begingroup$ Should n't that be $\ddot{r} = r\dot{\phi}^2 $ at start? $\endgroup$ – Narasimham Oct 25 '18 at 11:35
  • $\begingroup$ I assume you are talking about a central force? $\endgroup$ – Oldboy Oct 25 '18 at 13:09
  • $\begingroup$ ...and do you have any initial conditions? Find a general solution might be too complicated $\endgroup$ – Oldboy Oct 25 '18 at 13:51

Brute-force method

From your second equation:

$$r\ddot{\varphi}+2\dot r\dot\varphi=0$$

$$r^2\ddot{\varphi}+2r\dot r\dot\varphi=0$$




From your first equation (you have a typo):

$$\ddot r=r\dot\varphi^2\tag{4}$$

Introduce substitution:

$$r=\frac 1u\implies\frac{dr}{du}=-\frac{1}{u^2}$$

$$\dot r = \frac{dr}{dt}=\frac{dr}{d\varphi}\dot\varphi=\frac{dr}{du}\frac{du}{d\varphi}\frac{C}{r^2}=-\frac{1}{u^2}\frac{du}{d\varphi}Cu^2=-C\frac{du}{d\varphi}$$

$$\ddot r=\frac{d\dot r}{dt}=\frac{d\dot r}{d\varphi}\dot\varphi=\frac{d}{d\varphi}(-C\frac{du}{d\varphi})\frac{C}{r^2}=-C^2u^2\frac{d^2u}{d\varphi^2}\tag{5}$$

Now replace (5) into (4):


...Which leads to:




or, finally:


But this is all baloney :)

Smart method

Your first equation simply says that the radial component of accelaration $a_r=\ddot r-r\dot\varphi^2$ is equal to zero.

Your second equation says that the circular component of accelaration $a_c=r\ddot\varphi+2\dot r\dot\varphi$ is also equal to zero.

So the total acceleration is zero, velocity is constant and trajectory must be a straight line:



$$r=\frac{b}{\sin\varphi - a\cos\varphi}\tag{8}$$

...which is equivalent to (7), just with a different constants :)

  • $\begingroup$ wow, thank you so much, now I wonder, how can I come up with a substitution like $r=1/u$? $\endgroup$ – Fernando Franco Félix Oct 25 '18 at 19:57
  • $\begingroup$ @FernandoFrancoFélix I still remember that lession from my university professor. It's often used when you deal with central forces, like gravity. $\endgroup$ – Oldboy Oct 25 '18 at 20:01

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