I'm a new member on StackExchange. I'm french and my english is awfull, so I beg you to excuse me and I hope you can understand my question anyway…

I'm looking for a very precise equation for free fall in Schwarzschild coordinates (r;t) but with an initial speed superior at release speed.

I already have the formula of the speed with a parameter K


Where Rs is de Schwarzschild radius. This is not a local speed but the "slope" of the path I'm looking for.

For exemple, if K=0 it gives the speed at r for free fall from infinity, if K=-∞ it gives the speed of light (Shapiro's one). I'm interrested in the case K belongs to ]-∞;0[ wich means that a particle could follow a geodesic from an initial r=Ro with an initial speed superior at release speed.

The subject :

This formula can be written v(r). If I had v(t) I could intégrate it and find the equation of the path r(t), but this is not the case. Is anyone here able to write the formula t(r) or r(t) I need to draw the path of this kind of particles in Schwarzschild coordinates (r;t) ?

Either you can find a way to integrate this, either you already have a "ready made" formula without the K parameter. I just want the result, not the whole explanation… waste of time because I can't understand it. I'm not a good mathematician but I'm interrested in black holes and I like to "draw formulas" because it's the only way for me to understand somethis about relativistic équations.

Thank you for your help and sorry for my english…


  • 1
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Ben Crowell Oct 21 '19 at 23:31
  • $\begingroup$ Thank you for the link. But there are several formulas in this page... can you specify which one you think I have to pick ? Keep in mind that I'm not good in maths (even worse than in english) I only use formulas do draw it. $\endgroup$ – Mailou75 Oct 22 '19 at 23:51
  • $\begingroup$ There was an answer and a plot this morning. Who has delated them ? And why... ? Ok it was not exactly the good answer but it’s better than nothing... $\endgroup$ – Mailou75 Oct 24 '19 at 18:05
  • $\begingroup$ My guess is that the answer and plots were deleted because the person who posted them thought that another “answer” (actually just a comment), posted by you in response but deleted by a moderator, was rude. $\endgroup$ – G. Smith Oct 28 '19 at 5:06

I haven't checked whether your equation is correct, but I can tell you how to integrate it to get $t(r)$. (Sorry, you can't get $r(t)$.) All you need to do is write your differential equation in the separated form


where $t$-stuff is on one side and $r$-stuff is on the other, and then integrate both sides:

$$\int dt=\sqrt{\frac{1}{R}-K}\int\frac{dr}{\sqrt{\frac{1}{r}-K}(1-\frac{R}{r})}.$$

The $t$ integral is trivial, and Mathematica can do the messy $r$ integral. The result, slightly rewritten for easy calculation in the case $K<0$, is

$$t=\sqrt{\frac{1}{R}-K} \left(-\frac{(2 K R+1) \tanh ^{-1}\left(\frac{\sqrt{-K}}{\sqrt{\frac{1}{r}-K}}\right)}{(-K)^{3/2}}-\frac{2 R^{3/2} \tanh ^{-1}\left(\frac{\sqrt{R} \sqrt{\frac{1}{r}-K}}{\sqrt{1-K R}}\right)}{\sqrt{1-K R}}-\frac{r \sqrt{\frac{1}{r}-K}}{K}\right)+C$$

for $r>R$ and

$$t=\sqrt{\frac{1}{R}-K} \left(-\frac{(2 K R+1) \tanh ^{-1}\left(\frac{\sqrt{-K}}{\sqrt{\frac{1}{r}-K}}\right)}{(-K)^{3/2}}-\frac{2 R^{3/2} \tanh ^{-1}\left(\frac{\sqrt{1-K R}}{\sqrt{R} \sqrt{\frac{1}{r}-K}}\right)}{\sqrt{1-K R}}-\frac{r \sqrt{\frac{1}{r}-K}}{K}\right)+C$$

for $r<R$.

Here $C$ is a constant of integration determined by the initial conditions.

You should be able to verify that this solution satisfies your differential equation by differentiating this to compute $dt/dr$; the reciprocal $dr/dt$ should simplify to the right-hand-side of your equation.

By the way, if it had been impossible to obtain an analytic solution, it is always possible to solve a differential equation numerically.

I suspect that your equation has at least one mistake: shouldn’t $dr/dt$ be negative? If so, simply negate my solution.

Two final comments: This is called radial free-fall, and what you mean by “superior at release speed” is unclear because this is not a phrase used in English.


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