Falling at rest from infinity to black hole.

I calculated the velocity when an object fall from infinity to a black hole or vicinity of one.

$c\Big(1-\frac{2GM}{c^2r}\Big)\sqrt{\Big(\frac{2GM}{c^2r}\Big)}=\frac{dr}{dt}$, but this was assumed that the object was at rest $\frac{E}{mc^2}=1$. If the object has a velocity at infinity, then $\frac{E}{mc^2}=\frac{1}{\sqrt{(1-v^2/c^2)}}$. I can not just multiply the equation above by this factor right? I suppose the lorentz factor changes over time, so the solution is horrible.

  • $\begingroup$ This - physics.stackexchange.com/questions/170502/… - might help you find your answer. What do you get when you integrate that equation? I think writing it in function of proper time might simplify it, then you won't have to bother with the $\gamma$ term. $\endgroup$ – WarreG Aug 16 '18 at 15:32

With the object starting from rest at infinity the initial energy is $mc^2$ and becaase the metric is static this is a conserved quantity. Some messing around later we get the equation you mentioned:

$$ v(r) = -c\left(1 - \frac{r_s}{r}\right) \sqrt{\frac{r_s}{r}} $$

If the object starts off moving with some radial velocity the initial energy becomes $\gamma_0mc^2$, where $\gamma_0$ is the Lorentz factor at infinity. The same working then gives us:

$$ v(r) = -c\left(1 - \frac{r_s}{r}\right) \sqrt{1 - \frac{1}{\gamma_0{}^2}\left(1 - \frac{r_s}{r}\right)} $$

  • $\begingroup$ If I implement the lorentz factor I get the following: \begin{align*}c\Big(1-\frac{2GM}{c^2r}\Big)\sqrt{\Big(1-\frac{1}{\gamma^2}(1-\frac{\gamma^2 r_s}{r})\Big)}=\frac{dr}{dt}\end{align*} $\endgroup$ – Jordy Molenaar Aug 16 '18 at 20:45
  • $\begingroup$ Oops, my mistake , it was only dividing by $\gamma^2$ and then substract $-1$ and multiply the entire equation by $-1$. Thanks. $\endgroup$ – Jordy Molenaar Aug 16 '18 at 21:13

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