# Homework Help: Radially infalling object - when to use + or - sqrt?

1. Dec 21, 2015

### liron

1. The problem statement, all variables and given/known data
An object falls radially towards a non rotating black hole from infinity (i.e. the velocity is the same as the escape velocity except negative). The black hole has a Schwarzschild radius of Rs = 2GM/c2 where G is the gravitational constant, M is the black hole's mass and c is the speed of light in free space. The object has mas m and velocity v = dr/dt where r is distance from the centre of the black hole and t is time.

When I calculate the momentum, I can get either a positive or negative result, depending on how I do it, since a square root is involved. I know it should be negative so in this case and in general,how can I be sure I have the right root?

2. Relevant equations
According to a website I looked up, the velocity should be:

dr/dt = (+/-) sqrt(M/r)(1-Rs/r)

However i believe that M=1 & c=1 in their cauculations. Adding them back in gives:

dr/dt = (+/-) sqrt(2GM/rc2)(1-Rs/r) = Rs/r

Since radius decreases as time increases, the negative root is the correct one:

dr/dt = -c sqrt(Rs/r)(1-Rs/r)

I calculated that the momentum (not 4-momentum) in this situation should be:

p = m v sqrt(1 - Rs/r - v2/(c2(1-Rs/r))

3. The attempt at a solution
When I plug in the velocity into the momentum equation, I get:
p = -mc sqrt(Rs/r)(1-Rs/r) / (sqrt(1 - Rs/r - c2(Rs/r)(1 - Rs/r)2/(c2(1-Rs/r)))

This reduces to:

p = -m c sqrt(Rs/r)(1-Rs/r) / sqrt((1 - Rs/r)(1 - Rs/r))

This reduces to p = -m c sqrt(Rs/r)

However, if I change the second last equation to be:

p = m c sqrt(Rs/r)(Rs/r-1) / sqrt((Rs/r-1)(Rs/r-1))

This reduces to p = m c sqrt(Rs/r)

So depending on how I do my calculations, I can get the answer or the negative of the answer, even when I start out with the correct root.

Last edited: Dec 21, 2015
2. Dec 22, 2015

### Brian T

Since momentum is a vector, you would just need to specify its magnitude |dr/dt| and its direction. If I have a vector like <2, - 3>, it doesn't really make sense saying whether its positive or negative.

3. Dec 23, 2015

### vela

Staff Emeritus
I'll assume the rest of your work is correct. You just need to recognize that $\sqrt{x^2} = \lvert x \rvert$, so If $r>R_s$, then $\sqrt{\left(1-\frac{R_s}{r}\right)^2} = \sqrt{\left(\frac{R_s}{r}-1\right)^2} = 1-\frac{R_s}{r} > 0$.