Solution to problem 10.4.5

Return to Chapter 10, The Schwarzschild Black Hole

 c2dt2 = g00c2dt2  - (1-F/2c2)4(dr)2

c2 = c2 g00 (dt/dt)2 - (1-F/2c2)4(dr/dt)2

From equation 5.5.4b g00 dt/dt = g = constant

So

c2 = c2 g00 (g/g00)2 - (1-F/2c2)4(dr/dt)2

(1-F/2c2)4(dr/dt)2 = c2(g2/g00 - 1)

(dr/dt)2 = c2(g2/g00 - 1)/ (1-F/2c2)4

 

b.

For a weak potential

1/(1-F/2c2)4 ~ (1+ 2F/c2)

so

(dr/dt)2 = c2(g2/g00 - 1)(1+ 2F/c2)

(dr/dt)2 = c2(g2/g00)(1+ 2F/c2) - c2(1+ 2F/c2)

(dr/dt)2 = c2(g2/g00)(1+ 2F/c2) - c2 - 2F

(1/2)(dr/dt)2 + F = [c2(g2/g00)(1+ 2F/c2) - c2]/2

Again for a weak potential

2F/c2 < < 1

and

g00 » 1

so

(1/2)(dr/dt)2 + F = [g2 - 1]c2/2

(1/2)m(dr/dt)2 + mF = m[g2 - 1]c2/2

This takes the Newtonian conservation of energy equation form

KE + U = E

QED

Return to Chapter 10, The Schwarzschild Black Hole