2
$\begingroup$

I know that:

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

and

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

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

$\endgroup$
  • 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
3
$\begingroup$

Brute-force method

From your second equation:

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

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

$$\frac{d}{dt}(r^2\dot{\varphi})=0$$

$$r^2\dot{\varphi}=C$$

$$\dot{\varphi}=\frac{C}{r^2}\tag{3}$$

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):

$$-C^2u^2\frac{d^2u}{d\varphi^2}=\frac{1}{u}(\frac{C}{r^2})^2=C^2u^3$$

...Which leads to:

$$\frac{d^2u}{d\varphi^2}+u=0\tag{6}$$

or:

$$u=\frac{1}{r}=C_1\cos\varphi+C_2\sin\varphi$$

or, finally:

$$r=\frac{1}{C_1\cos\varphi+C_2\sin\varphi}\tag{7}$$

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:

$$y=ax+b$$

$$r\sin\varphi=ar\cos\varphi+b$$

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

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.