Question

Black hole of Schwarzschild.
The Schwarzschild solution is an exact solution of Einstein's equations in the vacuum that is static and spherically symmetric.
The interval is given by: (1)

(a) they are conserved quantities: (2)
Due to an additional symmetry, the movement is confined to a plane, which can be chosen as the equatorial plane, θ = π / 2. Write (1) in terms of,A (r), (dr / dτ) 2 and dτ.

(b) Show that (3) is constant throughout geodesics.

(c) An observer falls radially towards a Schwarzschild black hole with initial velocity dr / dτ = Uo at a distance R from the center of the black hole. Express the constant in terms of M, R, and Uo.

(d) Calculate the 4-speed of the observer falling, as a function of, R and M.

(e) Calculate the own time that it takes the observer to reach the horizon, r = 2M.

dT' 3

Answer 1

(a)

$d\tau^2=Adt^2-A^{-1}dr^2-r^2d\varphi^2$

for

$\theta=\pi/2$.

(b) This is just the definition of interval in terms of metric. Thus,

$g_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau}=1$

always.

(c) NOT CLEAR, which constant?

(d)

The velocity in general is

$u^{\mu}=\left ( \frac{dt}{d\tau},\frac{dr}{d\tau},r\frac{d\theta}{d\tau},r\sin\theta\frac{d\varphi}{d\tau} \right )$

But for

$\theta=\pi/2$.

$u^{\mu}=\left ( \frac{dt}{d\tau},\frac{dr}{d\tau},0,r\frac{d\varphi}{d\tau} \right )$

Using (a) and 2,

$A^{-1}\epsilon^2-A^{-1}\dot{r}^2-l^2=1$

where

.

dr

So,

$\dot{r}=\pm\sqrt{\epsilon^2-A(l^2+1)}$

Thus,

$u^{\mu}=\left (\frac{\epsilon}{A},\pm\sqrt{\epsilon^2-A(l^2+1)},0,-l \right )$

where A is of course given by

2M A 1

(e)

Own time corresponds to the proper time. For inwards falling observer

Ve- A(12 +1)

dr

the integral is of the form

$\tau_o=\int_{2M}^R\frac{dr}{\sqrt{a+b/r}}$

where

$a=\epsilon^2-1-l^2,b=2M(1+l^2)$