Proper Time of Fall in Schwarschild Metric

Question

I've been trying to understand how the proper time of a radial fall into a black hole is finite, but I'm confused because I found many results and they do not seem to be the same.

Usually what I see is that the proper time depends on the radius in an equation that possesses a power of 3/2 like here ("Free falling into the Schwarzschild black hole: two times doubt" by riemannium). I did not find how this result is derived, but I see it everywhere.

Then here in one question ("Proper time of flight Schwarzschild metric being finite" by Axionlike particles) I found this pdf which gives a complete different result and now I do not know which is right.

Can someone explain me how to find this proper time of fall? If this pdf is right, what about the 3/2 power result?

Answer 1

If the motion is only along the $\rm r$ axis the coordinate acceleration is simply $\rm d^2 r/d \tau^2=-G M/r^2$ so in terms of the proper time $\tau$ there is no difference to Newton. If you also have tangential velocity there is an additional term different from Newton, but for a radial infall it is the same.

Answer 2

I will sketch how to solve the differential equation in the answer of Yukterez. It is nonlinear and of second order, but can be simplified with a little trick, the multiplication with $r'$ and applying the chain rule to both sides. We get: $$\left(\frac{1}{2}r'^2-\frac{GM}{r}\right)' =\left(r''+\frac{GM}{r^2}\right)r'=0.$$ Therefore there exists a constant $E$ (the total energy of the particle, but don't get confused with classical mechanics!) with: $$\frac{1}{2}r'^2-\frac{GM}{r}=E \Rightarrow r'=-\left(2E+\frac{2GM}{r}\right)^{1/2}.$$ We have to take the negative solution of the root since $r'<0$ as the particle is falling. Therefore the time difference $\Delta t=t_\mathrm{B}-t_\mathrm{A}$ for the radial fall from a radius $r_\mathrm{A}$ to a radius $r_\mathrm{A}$ is therefore given by: $$\Delta t=t_\mathrm{B}-t_\mathrm{A} =\int_{t_\mathrm{A}}^{t_\mathrm{B}}\mathrm{d}t =-\int_{r_\mathrm{A}}^{r_\mathrm{B}}\left(2E+\frac{2GM}{r}\right)^{-1/2}\mathrm{d}r =\int_{r_\mathrm{B}}^{r_\mathrm{A}}\left(2E+\frac{2GM}{r}\right)^{-1/2}\mathrm{d}r.$$ Notice that the negative sign introduce earlier fits that $t_\mathrm{A}<t_\mathrm{B}$, but $r_\mathrm{B}<r_\mathrm{A}$. This elliptic integral is usually difficult to solve, but with $E=0$ it is easy. (Notice that $E$ can be computed with the upper equation with the initial conditions $r(t_\mathrm{A})$ and $r'(t_\mathrm{A})$.) For the special case $E=0$, we get: $$\Delta t =\sqrt{2GM}\int_{r_\mathrm{B}}^{r_\mathrm{A}}\sqrt{r}\mathrm{d}r =3\sqrt{\frac{GM}{2}}[r^{3/2}]_{r_\mathrm{B}}^{r_\mathrm{A}} =3\sqrt{\frac{GM}{2}}(r_\mathrm{A}^{3/2}-r_\mathrm{B}^{3/2}).$$ Recently, a question about this differential equation was posted on MSE, which I also answered with a similar answer to this one. But there are other answers going other ways to compute the elliptic integral, which you can find here.