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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?

Vinicius Araujo Ritzmann
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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.

Yukterez
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  • But how this leads to the proper time? – Vinicius Araujo Ritzmann Jul 13 '22 at 20:11
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    you integrate, first from $\rm d^2r/d \tau^2$ to $\rm dr/d \tau$ and then integrate the reciprocal $\rm d \tau/dr$ over $ \rm r$, then you get $\rm \tau=\frac{r_0^{3/2}arctan\left(\sqrt{r_0/r-1}\right)}{c \ \sqrt{r_s}}+\frac{r \ r_0\sqrt{r_s/r-r_s/r_0}}{c \ r_s}$ if your initial velocity at $\rm r_0=0$, with initial velocity the equation is a little longer, see http://notizblock.yukterez.net/viewtopic.php?p=554#p554 and/or the link in the answer – Yukterez Jul 13 '22 at 20:23
  • And how do you find that d^2r/dtau^2 is -G M/r^2 ? I tried to solve the Geodesic Equation for it and got a really big expression. – Vinicius Araujo Ritzmann Jul 13 '22 at 21:21
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    I found this answer here: https://www.mathpages.com/rr/s6-04/6-04.htm. It uses a very neat trick to solve that Geodesic Equation. – Vinicius Araujo Ritzmann Jul 13 '22 at 21:37
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    I like that book a lot - it contains many derivations like the one you are asking about, which MTW simply states without proof. I particularly like the Lorentz Transform derivation in https://www.mathpages.com/rr/s1-07/1-07.htm His real tour de force IMO (in another book, Physics in Space and Time) is "From the Field Equations to the Kruskal Metric" https://www.mathpages.com/home/kmath655/kmath655.htm which does not go via Schwarzschild. Sadly the author seems to have been harassed off the internet during the nineties . . . either that or Kevin Brown is a pseudonym. – m4r35n357 Jul 14 '22 at 08:58
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    This appears to be the only web site that mentions him. https://www.numericana.com/fame/#brown, although there are a couple of threads on Physics Forums asking about him. – m4r35n357 Jul 14 '22 at 10:59
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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.

Samuel Adrian Antz
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  • By the Geodesic Equation dt/dτ of 1- Rs/r is a constant. If I take this and put on the metric, as dτ^2 = g00dt^2 - g11dr^2, I can find an answer very similar to Yukterez's, and I sort of understand yours as well, but are they all telling the same story? – Vinicius Araujo Ritzmann Jul 13 '22 at 21:14
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    I will think again about that part of question. – Samuel Adrian Antz Jul 13 '22 at 21:30
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    I found this website https://www.mathpages.com/rr/ which has a lot of content on General Relativity. Part 6. has a lot to add to this doubt, I'm reading it and thought you all would like to know about this website. – Vinicius Araujo Ritzmann Jul 13 '22 at 21:40