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Energy conservation in particle decay

  1. Feb 26, 2008 #1
    two massive particles a finite distance from each other are bound to each other gravitationally, so if one dissapeared the other one would have its energy altered, and the conservation of mass-energy would be broken.

    so what happens when a massive particle decays to a less massive particle with the emission of say a gamma ray? how would the energy of another particle in the vicinity be altered?
     
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  3. Feb 27, 2008 #2

    vanesch

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    Consider two heavenly bodies, say the earth and the moon. Now, consider that the moon breaks up into two pieces in an explosion. What do you think would happen gravitationally ?
     
  4. Feb 27, 2008 #3
    eh the center of mass would still be at the same point so not much i would guess? but there's still the same amount of mass, it's just been seperated. i'm tsalking about when the net mass of the system is reduced after the event, i dont see how the two relate.
     
  5. Feb 27, 2008 #4

    vanesch

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    Well, the center of mass, even in the decay of a particle, will still remain the same, and the total energy also. That is to say, if you add the 4-vectors (E,px,py,pz) of the decay products, you will obtain the original (E,px,py,pz) of the original particle before decay. Now, what counts gravitationally is the E (in fact, the energy-momentum tensor, but we can do with the E here).

    So imagine that you have a pi-0, gravitationally bound to, say, a proton (quite hypothetic, I know). If the pi-0 decays into two photons (each initially of energy about the mass of the pi-0 divided by 2), then those two photons will have to "climb out of the gravitational potential well" of the proton, and loose energy (shift to red). In the end, the two photons will have a total energy which is less than the energy of the pi-0, with exactly the amount of gravitational energy they needed to overcome the gravitational potential energy.

    Of course, in practice this is ridiculous, because the gravitational energy of a proton and a pion are so terribly tiny as compared to the mass-energy of a pion, that you will never be able to measure this.
     
  6. Feb 27, 2008 #5
    If photons could travel against the gravitational field without loosing energy, then energy conservation would be violated. We can use this to calculate the effect of gravity on the energy (frequency) of photons.
    Imagine this experiment:

    A particle with mass m decays into two photons, which travel upwards for a hight z in gravitational field g. Then they assemble back to original particle, which travels back to the original location. If gravity did not affect photons, then this (cyclical) proces would be a perpetuum mobile, since we would get out work done by gravitational force on every cycle.
    For energy to be conserved, a photon traveling in gravitational field must loose exactly as much energy as a massive particle with m=E/c^2=h*f/c^2 (f=photon frequency).

    This must also be true for infinitezimal hight dz:

    m*g*dz=h*df
    E*g*dz/c^2=h*df
    f*g*dz/c^2=h*df

    Consequence:

    df/dz=f*g/c^2
     
  7. Feb 28, 2008 #6
    ok so the photons are still affected by the gravitational potential of the other particle after decay, but what happens to the other particle after this? is the energy lost by the photons making up for the energy gained by the other particle by losing its original graitational well?
     
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