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Relativistic fall equation.

  1. Mar 21, 2013 #1
    Say I have a special type of black hole, an infinetly small one. I am standing still in deep space and i let a point mass to fall. What is it's equation of motion? I'm interested in the equation itself not just an explanation. It should be sth like this x=f(t).
     
  2. jcsd
  3. Mar 21, 2013 #2

    mfb

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    A black hole with a Schwarzschild radius of 0 has a mass of 0, so it is just vacuum. Any black hole with a finite mass has a size - this is not the size of a matter concentration, it is just the Schwarzschild radius.
    To get x=f(t), you have to define a coordinate system for spacetime first. In GR, this is not trivial, and the solution will depend on your choice.
     
  4. Mar 21, 2013 #3
    It will look just like the Newtonian fall equation until you are very close to the hole.
     
  5. Mar 21, 2013 #4
    A matter of geodesics

    It doesn't matter how big the black hole is, the solution is still that of the Schwarzschild metric. Standing still is a matter of context, as, how would you propose an observer stands still with respect to his/her 4-velocity in the vicinity of the black hole.

    However, in general, the point mass as long as it is only moving under gravity and inertia moves according to the geodesic equation. The geodesic equation will give you the set of equations that give the particle's motion. However, you will need to first calculate the Christoffel symbols. You will, in general, obtain a coupled system of nonlinear ordinary differential equations which can be integrated to give you what you seek.
     
  6. Mar 21, 2013 #5
    You can find a full analysis here: http://mathpages.com/rr/s6-04/6-04.htm
     
  7. Mar 21, 2013 #6

    Bill_K

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    It's a lot simpler than that - you don't even need Christoffel symbols. The Lagrangian for geodesics in the Schwarzschild equatorial plane is

    F = (1 - rS/r)-1(dr/dτ)2 + r2(dφ/ds)2 - (1 - rS/r)(dt/dτ)2

    with first integrals L = r2(dφ/dτ) [angular momentum], E = (1 - rS/r)(dt/dτ) [energy], and F = 0 for a photon or -1 for a particle.

    In our case, radial motion starting at r = ∞ we have L = 0 and E = 1. The geodesic equation reduces to just

    (dr/dτ)2 = rS/r

    which can be easily integrated to give τ = const x r3/2. (Note Kepler's Law! period squared ∝ distance cubed!) Thus, we see the remarkable fact that, expressed in terms of the Schwarzschild radius r and the proper time τ, the equation for radial infall from infinity is exactly the same as in the Newtonian case.
     
  8. Mar 21, 2013 #7
    Doesn't an infinitely small black hole evaporate infinitely fast?
     
  9. Mar 21, 2013 #8
    You actually have given me the Christoffel symbols

    You actually find the Christoffel symbols from minimizing the action of geodesics. They are read off the Lagrangian you have provided. The Newtonian potential method you have alluded to is generally not a good idea, as it suggests that this method is applicable to any metric, which is not true.
     
  10. Mar 21, 2013 #9

    Bill_K

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    :bugeye: Read it again, ikjyotsingh. I am not using any "Newtonian potential method". I am solving the exact geodesic equations, and the method is completely general. What I am doing is taking advantage of the two Killing vectors. It is a waste of time to work with the second-order equations when first integrals are available.
     
  11. Mar 21, 2013 #10

    pervect

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    The "infinitely small" part is confusing me, along with everyone else. Let's just take the equation for a point particle falling into a black hole of mass m.

    The equations I have aren't in the form you requested, they are i the form

    r = f(tau)
    t = f(tau)

    Here tau is the "proper time", the time that a wristwatch on the infalling point particle would measure.

    r is the Schwarzschild r coordinate, and t is the Schwarzschild time coordinate. Note that that r is not a distance, rather r is defined by the fact that the circumference is 2*pi*r.

    Then the equations of motion for a free-fall from a particle at rest at infinity are

    r = [ (9*m/2)*(tau^2) ] ^ (1/3)

    Here tau ranges from -infinity to zero. At tau = -infinity, the point particle is at infinity. At tau=0, the point particle is at r=0, i.e. at the singularity in the center of the black hole.

    ooops - looks like I'll have to look up t(tau). My notes aren't clear enough, I'd have to go back to the source.

    However, you'll find that the Schwarzschild "t" coordinate goes to infinity before r reaches 2m, even though "tau" is finite.
     
  12. Mar 21, 2013 #11
    Nope

    So, when you defind E = (1 - rS/r)(dt/dτ) [energy], you are saying that this is completely general?? You can define "Energy" in ALL spacetimes?

    Indeed, this method is not general at all. The reason you can do this for the Schwarzschild metric is that you have 4 killing vectors, one timelike, and 3 spacelike. Recall from classical mechanics what a timelike vector, and hence a time-symmetry indicates, obviously it indicates a static solution (which is one part of the S-vacuum metric), but it generally implies that there is conservation of energy. For spacetimes such as the Bianchi metrics which only have 3-dimensional isometry groups, of which none of the killing vectors are timelike, you absolutely, positively CANNOT define such an energy quantity, and even if you could, if would not be conserved. In fact, for the FLRW metrics, for which there exists a 6-dimensional isometry group, there are no time-like killing vectors either!

    So, the method is anything BUT general.
     
  13. Mar 21, 2013 #12

    Bill_K

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    Same as what I gave;
    ----
    t(τ) can be integrated explicitly but is not pretty!
     
  14. Mar 21, 2013 #13

    Bill_K

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    I said I was making use of the two Killing vectors. Sorry if you have a problem with that.
     
  15. Mar 21, 2013 #14
    Geodesics

    You can do this because the geodesics are planar (as this problem represents a central force problem), so I don't have a problem with that. I had a problem with the incorrect statement you made, "and the method is completely general", which as I just showed is not true.
     
  16. Mar 21, 2013 #15

    pervect

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    Yes :-)

    I'd go so far as to say that it's ugly. I know I posted t(tau) before, along with some conversions to some better behaved coordinates, but I don't recall the name of the thread anymore.

    My Maple worksheet has a note which leads me to believe that Maple got a bit confused carrying out the integration (Maple, at least my old version, often gives expressions for the integral that don't evaluate to real numbers in the desired domain. I don't know if other programs do better). Hence the need to look it up to be sure the answer is correct.

    I'm not sure if r(t) has a closed form solution or not - I rather suspect not. r(tau) is by far the simplest expression to write down. I hope that it will be sufficient for the OP's curiosity, r(t) (which is what I think he requested) if it has a closed form expression at all, will be a lot messier.
     
    Last edited: Mar 21, 2013
  17. Mar 21, 2013 #16

    pervect

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    The method Bill K used is a general and well-known for space-times which have Killing vectors.

    I did not see anyone claim that all space-time's have Killing vectors.
     
  18. Mar 22, 2013 #17
    Unless I am missing the point, r(t) is given in MTW, and also in the reference I posted earlier (a full derivation by the mysterious Kevin Brown). I do not consider it pretty, but not sure if I 'd call it ugly ;)
     
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