Time dilation for different observer in black hole metric

Question

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?

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.