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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

    vela

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    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##.
     
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