How to calculate the geodesics in polar coordinates?

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Brute-force method

From your second equation:

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

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

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

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

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

From your first equation (you have a typo):

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

Introduce substitution:

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

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

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

Now replace (5) into (4):

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

...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 :)

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Updated on August 01, 2022

Comments

  • Fernando Franco Félix
    Fernando Franco Félix over 1 year

    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

    • Narasimham
      Narasimham about 5 years
      Should n't that be $\ddot{r} = r\dot{\phi}^2 $ at start?
    • Oldboy
      Oldboy about 5 years
      I assume you are talking about a central force?
    • Oldboy
      Oldboy about 5 years
      ...and do you have any initial conditions? Find a general solution might be too complicated
  • Fernando Franco Félix
    Fernando Franco Félix about 5 years
    wow, thank you so much, now I wonder, how can I come up with a substitution like $r=1/u$?
  • Oldboy
    Oldboy about 5 years
    @FernandoFrancoFélix I still remember that lession from my university professor. It's often used when you deal with central forces, like gravity.