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Mass loss through gravitational radiation

  1. Nov 21, 2009 #1
    A system emitting gravitational radiation loses mass. But how is this explained in terms of the system. Take for example a neutron star with non-radial oscillations - this will emit gravitational waves, and lose mass. So does the number of particles in the star decrease? Or does the binding energy in the star decrease?
     
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  3. Nov 21, 2009 #2

    Janus

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    The gravitational waves will not cause the system to lose mass. The waves are created at the expense of the kinetic energy of the system, not its mass. The oscillations will dampen, the mass stays the same.
     
  4. Nov 21, 2009 #3
    From Gravitational Waves: Theory and Experiment by Michele Maggiore, page 108:

    "Of course, a physical system that radiates GWs loses mass. The conservation of the mass M of the radiating body, expressed by eq. (3.43), is due to the fact that in the linearized approximation the back action of the source dynamics due to the energy carried away by the GWs is neglected"
     
  5. Nov 21, 2009 #4

    Jonathan Scott

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    Yes, the effective gravitational mass of the system as seen from a distance, taking into account all its internal kinetic energy and binding energy, decreases when it emits gravitational waves. This doesn't however mean that the system loses any massive particles, but mainly that the constituent parts of the system lose kinetic energy, as Janus already said.
     
  6. Nov 21, 2009 #5
    What do you mean here by gravitational mass? In most modern books on special relativity, mass means invariant rest mass, which doesn't include any contributions from kinetic energy. Does this change in general relativity?
    Janus said "the gravitational waves will not cause the system to lose mass".
    The textbook said "Of, a physical system that radiates GWs loses mass".

    Are you saying these are both right?
     
  7. Nov 22, 2009 #6

    Jonathan Scott

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    The invariant rest mass of a system, in both special and general relativity, includes internal kinetic energy (for example rotational energy and thermal energy) and binding energy.

    It does not include the overall kinetic energy due to the movement of a system as a whole.
     
    Last edited: Nov 22, 2009
  8. Nov 22, 2009 #7
    Here is a quote from the same textbook which seems to imply that rest mass and kinetic energy are independent:

    "Dimensionally [tex] \frac{T^{00}}{c^2} [/tex] is a mass density but of course, besides the contribution due to the rest mass of the source, it also contains all contributions to [tex] T^{00} [/tex] coming from the kinetic energy of the particles which make up the source, contributions from the potential energy, etc. For sources that generate a strong gravitational field, such as neutron stars, the gravitational binding energy will also be important. Only for weak field sources, and in the non-relativistic limit, [tex] \frac{T^{00}}{c^2} [/tex] becomes the mass density."
     
  9. Nov 22, 2009 #8

    Jonathan Scott

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    This seems to be rather unhelpful in the way it uses the word "mass" in three different ways.

    The phrase "the rest mass of the source" is clearly intended to refer to the rest mass of the component particles which make up the system. The overall effective mass density also includes all other forms of internal energy.

    In Special Relativity (and loosely speaking in General Relativity too) the effective rest mass of an overall system is the invariant magnitude of the total four-vector energy and momentum added up (as a four-vector) for all components of the system (with appropriate factors of c for unit conversion):

    [tex](mc^2)^2 = E_{total}^2 - (p_{total}c)^2[/tex]

    The rest mass of a system of two or more particles is not exactly equal to the sum of the rest masses of the individual particles except when they are at rest relative to one another and there is no binding energy between them.
     
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