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Relativistic mass of falling body

  1. Sep 21, 2011 #1

    zonde

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    Let's say that body with rest mass m0 is falling toward gravitating body from infinity.
    Near gravitating body it falls with speed [tex]v=\sqrt{\frac{2MG}{r}}[/tex]

    Is it right that relativistic mass of body falling at speed v near gravitating body is equal to the rest mass of the same body at infinite distance from gravitating body?
     
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  3. Sep 21, 2011 #2

    pervect

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    Most modern GR books dont even talk about "relatiivstic mass" anymore. it's mostly an outdated SR concept that didn't make it to "the big time", though you'll find the occasional vocal enthusiast.

    That said, you might find a few very old books (Tollman's Relativity, Thermodynamics, and Cosmology is the only one that comes to mind) that might find a way to use "relativistic mass" in a way that's compatible with GR.

    However, it seems safest to write that any correct esxpression for the equations of motion for a radial infall would be equivalent to finding two functions

    r(tau), t(tau)

    such that

    (dr/dtau)^2 + (1-2M/r) = constant = "energy at infinity" = E

    (dt/dtau) = E / (1-2M/r)

    (I've omitted the radial terms, see for example http://www.fourmilab.ch/gravitation/orbits/ for the full formula)

    I would suspect that if you try simple-minded substitutions for the "relativistic mass" using SR formulae that ignore the metric coefficients that you'd get an incorrect answer, but I haven't verified this.
     
  4. Sep 21, 2011 #3
    An observer sitting on your falling mass would not see anything unusual. He does not feel any acceleration and might assume he is floating free in open space. The actual rest mass M upon which he sits has not changed.

    An observer at somewhere approximating infinity would see the falling mass as having acquired a velocity (assuming he could actually see that far from approximately infinity). Any mass that acquires a velocity is subject to the normal Special Relativity formula M/(1-v^2/c^2)^1/2. In your example, this is a relativistic mass that could be tested in a laboratory near the surface of the planet before it got to the Schwarzschild Radius (for example, the falling mass could impact an object in the laboratory and the relativistic mass would be applicable to this collision per the above formula).
     
  5. Sep 21, 2011 #4

    pervect

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    The velocity given by [tex]\sqrt{\frac{2MG}{r}} [/tex] is not measured or defined via an observer at infinity.

    This velocity is the velocity measured by a "stationary" observer at the same location as the falling object.

    I suppose the same co-located observer could then use the SR relativistic mass formula to compute the relativistic mass , because you can treat space-time as being locally flat.

    The purist view (see http://math.ucr.edu/home/baez/einstein/node2.html) is that there isn't any good definition for relative velocity in GR for non-colocated observers, thus there isn't any good definition of the velocity of the falling object relative to the observer at infinity.

     
  6. Sep 22, 2011 #5

    zonde

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    Then let's say that distant observer sees such a sequence of events:
    Test mass falls toward large gravitating object and hits the surface.
    Kinetic energy of test mass is converted into heat and then radiated away.

    Now summary mass of gravitating object has increased by rest mass of falling object that it had at the moment when it hit the surface. But this rest mass should be less than initial rest mass when test body started to fall as we have to take into account energy that was radiated away and does not contribute to summary mass any more.

    Is the problem now well defined? If yes are my conclusions correct?
     
  7. Sep 22, 2011 #6
    I disagree with baez. Current GPS satelites calculate the relative time rates from their position in space relative to the time rates on the surface of the earth and have been experimentally shown to be accurate. To say that at any point in space you cannot possibly know what is happening at any other point is a statement of the invalidity of General Relativity. If the local stationary observer at the lesser radius is the only person who can know the true velocity of the falling mass, then how does this local observer know that the mass started out at zero velocity when it was at approximately infinity. The problem becomes insolvable.

    The mass starts falling from a position in space where he is next to the observer. This is the reference point from which his potential energy is converted into the kinetic energy which converts into his relativistic mass. This is the only valid reference point from which the assessment of his final kinetic energy can be made. How could anyone possibly know that the local observer at the lesser radius was stationary? You could only calculate the potential energy of a stationary observer at the lesser radius, but this would be done by converting his position back to the position approximating infinity. But that's circular logic.
     
  8. Sep 22, 2011 #7

    zonde

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    Have to say that I feel the same way. If we can't use GR to make statements about physical situation that spans area with different gravitational potentials then how is it theory about gravity?
     
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