1
$\begingroup$

With studying Schwarzschild metric geodesics one can easily come up with the following differential equation \begin{equation} \dfrac{dr}{d\tau} = - \sqrt{C^2-\left( 1-\dfrac{2GM}{r}\right)} \end{equation} which relates the radial coordinate and the proper time outside the event horizon $r_H=2GM$ (I'm using, of course, $c=1$).
Imagine that an observer is falling from the initial position $r_0=2GM+h$ (with $h>0$ in order to be outside the event horizon) all the way down to the event horizon $r_H=2GM$. The proper time of flight is, thanks to the differential equation given above, by \begin{equation} \Delta \tau= \int_{\tau_0}^{\tau_H}d\tau = \int_{r_0=2GM+h}^{r_H=2GM} -\;\dfrac{dr}{\sqrt{C^2-\left( 1-\dfrac{2GM}{r}\right)}} \end{equation} However, the coordinate time of flight is given by the relation $dt=\frac{C}{(1-2GM/r)}d\tau$ (where $C>0$, see for example http://gfm.cii.fc.ul.pt/events/lecture_series/general_relativity/gfm-general_relativity-lecture4.pdf), resulting \begin{equation} \Delta t= \int_{t_0}^{t_H}dt = \int_{r_0=2GM+h}^{r_H=2GM} -\; \frac{C}{1-\dfrac{2GM}{r}}\;\dfrac{dr}{\sqrt{C^2-\left( 1-\dfrac{2GM}{r}\right)}} \end{equation} From this (see for example the reference given above) it is said that '' It is easy to see, by direct evaluation of the integrals, that the proper time of flight is finite, while the coordinate time of flight is infinite, the orbit observer will never see the infalling observer reach the event horizon, except asymptotical''.

I've tried to calculate the first integral using Mathematica and the result I get is absurd, giving a complex proper time. Here is the code I used:
Integrate[1/Sqrt[C^2 - (1 - 2*G*M/r)], {r, 2*G*M, 2*G*M + h}]
And the output is:
ConditionalExpression[(G M (2 Sqrt[C^2 - C^4] - I Log[2] - I Log[((-I + 2 I C^2 + 2 Sqrt[C^2 - C^4]) G M)/Sqrt[1 - C^2]]))/(1 - C^2)^(3/2) + (-Sqrt[1 - C^2] (h + 2 G M) Sqrt[C^2 - h/(h + 2 G M)] + I G M Log[2] + I G M Log[(-I h + I C^2 h - I G M + 2 I C^2 G M + Sqrt[1 - C^2] h Sqrt[C^2 - h/(h + 2 G M)] + 2 Sqrt[1 - C^2] G M Sqrt[C^2 - h/(h + 2 G M)])/Sqrt[1 - C^2]])/(1 - C^2)^(3/2), ((G M)/h != 0 && Re[(GM)/h] >= 0) ||Re[(G M)/h] < -(1/2) || (G M)/h \[NotElement] Reals]

What I am missing up here? Thanks

$\endgroup$
5
  • $\begingroup$ The best way to approach integrals is to perform some nice substitutions to simplify the integrands as much as possible (as suggested in the second answer), leaving so many variables tends to muck up the CAS because it has to make assumptions about those quantities. It doesn't know that M is a mass and must be positive for instance as you can see in the conditional expression. One way to help this is to set assumptions using refine as Refine[ expression, Assumptions->{...}], Where the braces hold all assumptions needed to clean up the expression. $\endgroup$
    – Triatticus
    Commented Mar 12, 2021 at 15:11
  • $\begingroup$ The analytical result is given here: physics.stackexchange.com/questions/426143/427025#427025 - The second chart also shows that $t$ (in blue) diverges at the horizon while $\tau$ (in green) does not. $\endgroup$
    – safesphere
    Commented Mar 12, 2021 at 16:11
  • $\begingroup$ @Triatticus Thanks! I've tried with your answer but the result given, although it is indeed real and makes sense, doesn't agree with the expected value. $\endgroup$
    – ALPs
    Commented Mar 12, 2021 at 19:15
  • $\begingroup$ @safesphere Thanks! The results you give are correct, those are the results I was trying to check with Mathematica. However, I still don't know why the command doesn't work. $\endgroup$
    – ALPs
    Commented Mar 12, 2021 at 19:17
  • $\begingroup$ Is there a physics question here ? $\endgroup$
    – ProfRob
    Commented May 2, 2022 at 8:38

2 Answers 2

1
$\begingroup$

I also do not quite know how this is "easy to see" from that.
However, you can solve the integral for proper time by first noting that for $C^2=\text{const.}$ we have (with $r_s = 2GM$)

$$C^2 = 1 - \frac{r_s}{r_0} $$

which follows from plugging in $(4.2)$ into $(4.1)$ and considering an infalling particle which is at rest at $r = r_0$, meaning $\dot{r}\vert_{r=r_0} = 0$.
This you can use to write your first equation as

$$\dot{r} = -r_s^{1/2} \left(\frac{r_0 - r}{r_0 r}\right)^{1/2}$$

From this, the proper time integral will turn out to be

$$\Delta \tau = -r_s^{-1/2} \int_{r_0}^{r_s} \left(\frac{r r_0}{r_0 - r}\right)^{1/2}\text{d}r $$

which you can solve using Mathematica or also by introducing the parametrization

$$r(\eta) = \frac{r_0}{2} (1+\cos\eta)$$

with $\eta \in [0,\pi]$.

$\endgroup$
3
  • $\begingroup$ Thanks! Yeah, you know, scientific papers are plenty of words like ''trivial'', ''obvious'', ''very clear'', ... in moments where it is not even close to be trivial or easy haha. Thanks for your answer, but putting the command in Mathematica with your expression for $\Delta\tau$ the program doesn't give a result, I don't know what happens. $\endgroup$
    – ALPs
    Commented Mar 12, 2021 at 19:21
  • $\begingroup$ Have you tried it using the parametrization as well? With that it's actually quite straightforward and can be done with pen and paper. I don't have access to Mathematica myself and thus can't really help you with the software part unfortunately. $\endgroup$
    – P-A
    Commented Mar 13, 2021 at 0:19
  • $\begingroup$ Yes, I did it with the parametrization as well and it doesn't compute a result. Don't worry! Thanks for your answer. I know it can be done analytically but I wanted to know what's going on with the software. $\endgroup$
    – ALPs
    Commented Mar 13, 2021 at 10:28
1
$\begingroup$

the result I get is absurd, giving a complex proper time.

Your result has I in it, which I assume is Mathematica's notation for $i=\sqrt{-1}$, but that doesn't mean the result is complex. It's common with certain techniques of integration to get results that look complex but actually turn out to be real when you evaluate them.

You're making your results unnecessarily complicated by taking $G$ and $M$ as parameters. When doing this sort of integral, the thing to do is to convert as many variables as possible to unitless quantities, which generally has the effect of eliminating unnecessary constants like these. Here you want to change to the variable $x=r/2GM$. Don't just feed it into the Computer Algebra System (CAS) without doing this prep. CAS's are stupid and produce complicated output, so give it a fighting chance by doing the same setup you would normally do if working without a CAS.

I use the open-source CAS maxima, rather than mathematica. It has no problem producing a result with no explicit $i$'s in it.

$ maxima -q --batch-string="assume(h>0 and (c^2-1)<0 and c>0); integrate(-(c^2-(1-1/x))^(-1/2),x,1+h,1);"

(%i1) assume(h > 0 and c^2-1 < 0 and c > 0)
                                     2
(%o1)                       [h > 0, c  < 1, c > 0]
(%i2) integrate(-(c^2-(1-1/r))^((-1)/2),r,1+h,1)
                                   2                     2           2
                 2       sqrt(1 - c ) sqrt(h + 1) sqrt((c  - 1) h + c )
(%o2) (sqrt(1 - c ) atan(----------------------------------------------)
                                        2           2
                                      (c  - 1) h + c  - 1
     2                         2           2     4      2
 + (c  - 1) sqrt(h + 1) sqrt((c  - 1) h + c ))/(c  - 2 c  + 1)
                                 2
             2       c sqrt(1 - c )     3
   sqrt(1 - c ) atan(--------------) + c  - c
                          2
                         c  - 1
 - ------------------------------------------
                  4      2
                 c  - 2 c  + 1

You might want to try setting it up the same way in mathematica. It will certainly be simpler and easier to interpret after the change in variables.

Actually, there's no guarantee that this result comes out real just because there are no $i$'s in it. It does have one square root in it that looks like it could be imaginary. The way we know that it should be real is that it's a definite integral of a real integrand. If you then want to check whether it really is real, try just using some random numbers for $C$ and $h$. If the result comes out to be real, then that's not a coincidence. If it comes out complex, then one of the following may hold: (1) you've coded it incorrectly, (2) there is a bug in the CAS, or (3) there's an issue with branch cuts for the inverse tangent.

$\endgroup$
1
  • $\begingroup$ I've tried to use your substitution into dimensionless units, but the result is the same given above and still has the terrible explicit $I=i=\sqrt{-1}$. I don't understand what happens, maybe I try to use Maxima or other software and check it there. Thanks! $\endgroup$
    – ALPs
    Commented Mar 12, 2021 at 19:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.