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Where does the energy go?

  1. Dec 14, 2011 #1
    I was thinking the other day, and came up with a scenario. Suppose you have two particles. The mass does not matter, but the distance between them is small enough for a gravitational field from one to have an effect on the other, but large enough that they do not immediately collide. Since they are separated, and in a gravitational field, they have a certain amount of potential energy given by, I believe, U=mgh. Suppose then that an anti-particle collides with one of the particles, completely annihilating both the particle and the antiparticle. Before the destruction of the particle, there was still a distance, still a mass, and still a gravitational field, so the other particle still has some potential energy. After the collision there is no gravitational field, so g=0 and the potential energy is 0. So where did this energy go? I asked my physics teacher, but he didn't know. Any thoughts?
     
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  3. Dec 14, 2011 #2

    Drakkith

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    Staff: Mentor

    The annihilation of the antiparticle and particle produce photons which do in fact contribute to gravity.
     
  4. Dec 14, 2011 #3
    Annihilation (like electron-positron) doesn't just make the energy disappear. Other particles come out of such collisions; for example, gamma rays (photons) are generated by electron-positron annihilation. These byproducts are where the energy goes.

    Now you may ask: gamma rays are massless and don't cause the gravitational attraction. Surely energy is lost. mgh = 0gh = 0, so where did it go.

    mgh is only correct for the non-relativistic limit. For relativistic particles, like the gamma rays (photons) generated by annihilation collisions, you need to consider relativity. In general relativity, gravitational "attraction" is not a "force" in the usual sense of the word, it is more like a fictitious force (such as centrifugal force) caused by the curvature of spacetime. Even massless particles cause such curvature. Curvature is related to the presence of what many textbooks call "mass-energy;" I'm sure you've seen the equation E=mc^2. I'm not exactly sure what kind of general relativistic effects happen during annihilation (surely they touch on quantum gravity). However it is observationally confirmed that (massless) photons' paths bend around gravitational bodies--they do feel the gravitational "fictitious forces" as they travel along geodesics in curved spacetime.
     
    Last edited: Dec 14, 2011
  5. Dec 16, 2011 #4
    Thanks! I think that clears it up.
     
  6. Dec 16, 2011 #5
    OK, I have another question for you about the same problem. The light wave is created, and it has enough gravitational pull to offset the loss in U by the missing particle, so no energy is lost. However, as the light leaves, the U decreases since the distance is increasing. This would be something along the lines of:

    F=Gm1m2/r^2 Substitutes into U=mgh

    So you have U=h*G*m1*m2/(r^2)

    Removing the constants gets U=h/(r^2) and h is equal to r so U=1/r

    So where does the U go this time? My best thought, and I'm doubtful of it, is that as the particle "catches up" with the light, it shortens the wave length of the light, increasing it's energy. But something just seems off about that. Thanks.
     
  7. Dec 17, 2011 #6
    I see what you are trying to say, but actually there is little need to overcomplicate the matter. Try to look at the particle as energy - the total energy is made up of its mass, but also of its gravitational energy, angular momentum and spin. The same is true for the anti-particle. When annihilation happens, the total energy is converted into photons of appropriate wavelengths and "radiated" off. If you do the maths you will find that there is no loss in energy, it all adds up perfectly. Also bear in mind that the other particle ( the one that is not annihilated and left behind ) retains its gravitation field, and therefore some gravitation energy.
    In general though, for a quantum particle, the gravitational forces involved are several orders of magnitude smaller than the other fundamental forces, and can normally be neglected in calculations.
     
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