Some time ago, in this question (Proper Time of Fall in Schwarschild Metric), I asked how to find the proper time of fall of an observer in the Schwarzschild metric because I had found many different ways of deriving it and they did not seem to mean the same thing.

There @Yukterez said the proper acceleration of an infalling observer is: $$\frac{d^{2}r}{d\tau^{2}} = -\frac{GM}{r^{2}}$$ What I understand now. You can find this by solving the Geodesic Equation for radial fall. He said I could solve this differential equation by integrating from $\frac{d^{2}r}{d\tau^{2}}$ to $\frac{dr}{d\tau}$ and then the reciprocal, but I do not understand how this leads to the expression he gave afterwards. Which is:

$$\rm \tau=\frac{r_0^{3/2}arctan\left(\sqrt{r_0/r-1}\right)}{c \ \sqrt{r_s}}+\frac{r \ r_0\sqrt{r_s/r-r_s/r_0}}{c \ r_s}$$

After this, @samuel-adrian-antz helped me by showing a method of solving that differential equation that I was able to fully understand.

I thank them both for helping me, but here I'd like to ask how to reach Yukterez's answer, because I see many texts giving it as answer. But also, why there's so many different ways of stating the proper time of fall? And what all these aproaches mean? Are each valid only for some cases?

My biggest doubt might be why direct integration do not seem to work here propely when I try it. I know this differential equation is for $\frac{d^{2}r}{d\tau^{2}}$, and I need $\frac{d\tau}{dr}$, but then how to reach it since I do not know how $r$ depends on $\tau$?

  • $\begingroup$ My biggest doubt might be why direct integration do not seem to work here propely when I try it. What do you mean by “direct integration”? $\endgroup$
    – Ghoster
    Commented Mar 2, 2023 at 1:41
  • $\begingroup$ @Ghoster integrate both sides of the differential equation $\endgroup$ Commented Mar 2, 2023 at 9:15
  • $\begingroup$ You can’t integrate the right side with respect to $t$ because you don’t know $r(t)$. $\endgroup$
    – Ghoster
    Commented Mar 2, 2023 at 17:45
  • $\begingroup$ @Ghoster Yes I said in the question I did not know how to reach the answer because I do not know how r and $\tau$ are related. $\endgroup$ Commented Mar 3, 2023 at 2:27

1 Answer 1


Lets walk through the derivation, starting from $$\frac{d^{2}r}{d\tau^{2}} = -\frac{GM}{r^{2}}.$$

We can multiply both side with $\frac{dr}{d\tau}$, $$\frac{d^{2}r}{d\tau^{2}}\frac{dr}{d\tau} = -\frac{GM}{r^{2}}\frac{dr}{d\tau}.$$ We now apply the chain rule to rewrite each side as a total derivative w.r.t. $\tau$, $$\frac{1}{2}\frac{d}{d\tau} \left(\frac{dr}{d\tau}\right)^2 = \frac{d}{d\tau} \frac{GM}{r}.$$ Since both sides are total derivatives we can integrate once w.r.t. $\tau$ to get $$\frac{1}{2} \left(\frac{dr}{d\tau}\right)^2 = \frac{GM}{r}-\frac{GM}{r_0},$$ where have used the condition $\frac{dr}{d\tau}=0$ at $r=r_0$.

With some rearranging we get, $$ \frac{dr}{d\tau} = -\sqrt{ \frac{2GM(r-r_0)}{r_0r}}.$$

We can now apply the inverse function theorem, to get $$ \frac{d\tau}{dr} = -\sqrt{ \frac{r_0r}{2GM(r-r_0)}}.$$

We now have a first order ODE for $\tau$ as a function of $r$, with no explicit dependence on $\tau$ on the right hand side, so this can be solved by direct integration to get the answer.

  • $\begingroup$ +1 It’s worth mentioning that you are assuming that the fall starts from rest at $r_0$. $\endgroup$
    – Ghoster
    Commented Mar 2, 2023 at 17:48
  • $\begingroup$ How is integration w.r.t $\tau$ possible not knowing how r and $\tau$ are related? $\endgroup$ Commented Mar 3, 2023 at 3:31
  • $\begingroup$ @ViniciusAraujoRitzmann Because left and right are both total derivatives with respect to $\tau$. $\endgroup$
    – TimRias
    Commented Mar 3, 2023 at 6:02
  • $\begingroup$ @TimRias I got it now, but why does the term $\frac{-GM}{r_0}$ appear? The left side is the same as before just does not have the derivative anymore, but the right side now has this new term. I understand that we can add a constant there after the integration because if we differentiate again the constant vanishes, but why exactly that constant? $\endgroup$ Commented Mar 3, 2023 at 12:52
  • $\begingroup$ @ViniciusAraujoRitzmann As it says in the text, this comes from applying the condition that the particle starts at rest at $r=r_0$. $\endgroup$
    – TimRias
    Commented Mar 3, 2023 at 16:38

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