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Einstein energy versus thermodynamic energy

  1. Jun 8, 2006 #1
    Dear friends,
    I have asked the following question before, but I am not still satisfied with the answer given by the experts of this forum.
    In Einstein equation E=m/c2
    ¿ Does E equal to the thermodynamic internal energy of matter plus kinetic energy plus potential energy of a system?. ¿ or what energy does this equation refers?
     
  2. jcsd
  3. Jun 8, 2006 #2

    Andrew Mason

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    [itex]E = mc^2[/itex]

    E represents the energy that would be released if all of the matter (of mass m) were to be converted into energy. (ie. 1/2 of the energy that would be released if the mass were to encounter and equal amount of anti-matter).

    A very very tiny portion of this energy (mass) is contributed by the thermal energy of the atoms and the potential and kinetic energy of the body as a whole.

    In other words, "Einstein energy" (which I take to mean mc^2) and "thermal energy" are different forms of energy. Both forms of energy contribute intertia, but the thermal energy contributes such a small amount (E/c^2) that it can be ignored for most purposes.

    AM
     
    Last edited: Jun 8, 2006
  4. Jun 9, 2006 #3

    pervect

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    If you have a small isolated thermodynamic system, the following relationship will be true

    E^2 - (pc)^2 = (mc^2)^2

    In this equation, E is the total energy of the system. It includes the energy due to the rest mass, kinetic energy of pieces of the system, and energy due to non-gravitational fields (electric and magnetic fields, for instance). This would include chemical and nuclear binding energy, for instance.

    E can be computed by integrating the T_00 component of the stress-energy tensor of the system over the volume of the system. P can be computed by integrating the T_01 component of the stress-energy tensor of the system over the volume of the system. m is computed from E and P, and is an invariant of the system when it is isolated.

    In order for the system to be isolated, the fields emanating from the system must be zero. Furthermore the system must be in a vacuum (the pressure at the system boundary is zero).

    If the system is not isolated, the above simple relationship will not be true.

    Note that E = mc^2 only if p=0.

    So, in conclusion, if E=mc^2

    1) E is the total energy (including kinetic energy, field energy, etc)

    2)The system must be isolated. (The system must not be interacting with its environment, the pressure on the system must be zero, the system must not have external fields (internal fields are OK but must be included in the energy).

    3) The system must have zero momentum, otherwise the more comlete relationship E^2 - (pc)^2 = (mc^2)^2 must be used.

    4) The system must be small enough not to have significant gravitaitonal self-energy. The case where the system has significant gravitational self-energy can be handled if the system is static, or if the system is in an asymptotically flat space-time, but I'm not going to do it today in this post. The static case is the easiest to handle, and leads to the Komar mass of the system. The asymptotically flat case leads to the Bondi and ADM mass of the system, which differ only in whether or not gravitational radiation terms are included in the "mass".

    5) If the system is both not static and also is not asymptotically flat, there is no general defintion for the "mass" of the system in GR at all.
     
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