How to calculate the geodesics in polar coordinates?

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

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

Brute-force method

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

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

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

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

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

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):

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

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

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 :)

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