1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Dec 21, 2015 #1
    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. jcsd
  3. Dec 22, 2015 #2
    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.
  4. Dec 23, 2015 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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##.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted