If I have a 2d Schwarzschild metric $$ dS^2 = -(1-\frac{r_s}{r})dt^2 + \frac{dr^2}{1-\frac{r_s}{r}} $$ I want to find the relation between the time of an asymptotic observer $t$ and the proper time of a free infalling observer $\tau$, I know that due to redshift I have $$ d\tau = \frac{1}{\sqrt{1-r_s/r}}dt $$ But I found this following formula on some online lecture notes, with not much explanation attached $$ d\tau \sim e^{-t/r_s} dt $$ How can I derive it?

Edit: I will add some information about the context. We are considering a freely falling observer through the event horizon, and we are using Kruskal coordinates $$ UV = r_s(r_s-r)e^{r/r_s},\qquad \frac{U}{V} = -e^{-t/r_S} $$ giving the following metric $$ dS^2 = -\frac{4r_s}{r}e^{-r/r_s}dUdV $$ It says that a trajectory of the infalling observer is described by $V\sim$const and $U$ goes to zero linearly in their proper time $\tau$ (I don't get how to prove this also). The infalling observer proper time $\tau$ and the asymptotic observer time $t$ are related by $$ d\tau \sim e^{-t/r_s} dt $$ Why?

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    $\begingroup$ It would help to have more context. Please include some of the text surrounding the formulas in the online lecture notes you found. (The words in scientific documents aren't just decorations for the equations, after all!) $\endgroup$ Commented Nov 20, 2023 at 12:43
  • $\begingroup$ I have updated my question with all the information present on the online notes, sorry for that $\endgroup$ Commented Nov 20, 2023 at 23:02

1 Answer 1


Physics Koan wrote: "I know that due to redshift I have..."

That assumes $\rm v=0, \ \ dr/dt=0, \ $ but if you're talking about

Physics Koan wrote: "...a free infalling observer"

you also need to consider the $\rm v$ (the velocity relative to local and stationary observers) or $\rm dr/dt$. The stationary observer at infinity receives

$$\rm f=f_0 \ \sqrt{\frac{c+v}{c-v}} \ \sqrt{1-\frac{r_s}{r}}$$

from the falling observer, where negative $\rm v$ is away from and positive towards the observer at infinity (same if it's the other observer that is moving).

Physics Koan asked: "How can I derive it?"

For the time dilation of the falling observer in the frame of the stationary observer at infinity set $\rm ds=c \ d\tau$ and rearrange the line element

$$\rm c^2 \ d\tau^2 = \left(1-\frac{r_s}{r}\right) \ c^2 \ dt^2 - \frac{dr^2}{1-\frac{r_s}{r}}$$

to $\rm dt/d\tau$, then you get

$$\rm \frac{dt}{dτ}=\frac{\gamma}{\sqrt{g_{tt}}} = \frac{1}{\sqrt{\left(1-\frac{r_s}{r}\right)-\frac{dr^2}{dt^2} \frac{1}{c^2 \ \left(1-r_s/r\right)}}}=\frac{1}{\sqrt{\left(1-\frac{v^2}{c^2}\right)\left(1-\frac{r_s}{r}\right)}}.$$

A free faller from infinity has the negative escape velocity

$$\rm v=-c \ \sqrt{\frac{r_s}{r}} \ , \ \ \frac{dr}{dt}=v \ \sqrt{-\frac{g_{tt}}{g_{rr}}}=-c \ (r-r_s)\sqrt{\frac{r_s}{r^3}},$$

or if you start to fall from rest at some finite $\rm r_0$ then $\rm v=-c \ \sqrt{\frac{r_s \ (r_0-r)}{r \ (r_0-r_s)}}. $

Since the kinematic component of the time dilation is relative, if you want to calculate the time dilation of the bookkeeper in the frame of the free faller, you must use the inverse of the gammafactor instead (that cancels with the gravitational component if $\rm v=\pm v_{esc}$, therefore in raindrop coordinates the contravariant $\rm g^{tt}=1$). The gravitational component on the other hand is absolute. For some examples in different coordinates see here and in the links therein.

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    $\begingroup$ Just a quick question for my education (and I’ll delete it afterwards). $\gamma$ comes from special relativity in a flat space where remote speeds are uniquely defined. Here though you just throw $\gamma$ in, but none of the special relativity conditions hold. I am sure your are correct, but at least intuitively this doesn’t feel rigorous. Even the speed of light is not $c$ as observed remotely, which makes $\gamma$ itself unclear. Do you have a reference with a rigorous treatment of the velocity time dilation in general relativity? +1 $\endgroup$
    – safesphere
    Commented Nov 20, 2023 at 14:16
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    $\begingroup$ @safesphere - Locally you have Minkowski, so you calculate the time dilation relative to a local observer and then apply the factor the local observer has relative to the observer at infinity. I'd have to look for a reference for Schwarzschild, but I have a reference for Kerr at hand, you can simply set a=0 to reduce that to Schwarzschild, see here on page 5 in the text between equations (3.5) and (3.6). The difference to Minkowski is that v isn't the same as dr/dt, for radial Schwarzschild trajectories it's v=dr/dt·√|gᵣᵣ/gₜₜ|, which is v=c for light. $\endgroup$
    – Yukterez
    Commented Nov 20, 2023 at 14:57
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    $\begingroup$ I meant v=dr/dt·√|gₜₜ/gᵣᵣ|, in the previous comment I switched gₜₜ and gᵣᵣ. Now I can't edit it any more, but I added that to the answer so that there is no confusion. $\endgroup$
    – Yukterez
    Commented Nov 20, 2023 at 19:52
  • $\begingroup$ Sounds good. Thanks so much! $\endgroup$
    – safesphere
    Commented Nov 21, 2023 at 0:29

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