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Quantum physics

  1. Jan 27, 2010 #1
    will the energy liberated be same if the electron and proton are annihilated in the moving carriage of train and on the railway platform
  2. jcsd
  3. Jan 27, 2010 #2


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    Electron and proton don't annihilate... fortunately.

    But let's ask about electron and position. The answer in that case is... the energy released in annihilation is equal to the energy they had initially, and that will depend on their velocity, which is relative to the observer.

    For a bit more detail, here is an example. Consider a positron fired at 60% lightspeed to collide with with a stationary electron, and annihilating with the release of two gamma ray photons.

    Rest mass of an electron/positron is about 511 keV (using energy equivalents for mass).

    At 60% light speed, the positron energy is 1.25 times greater, which is 628.75 keV. Total energy released is 1149.75 keV. You'll get two gamma ray photons, adding up to that energy.

    Now consider the same collision in the rest frame of the center of mass of the initial particles. The center of mass is moving at 0.3c (faster than a train) and so each particle is approaching the center of mass at 0.3 c, in that frame. The energy of each particle is about 1.048 times greater, about 535.7 keV. You will get photons produced, each with that energy, and a total released energy of 1071.4 keV; which is a little bit less. Energy is conserved in all frames, of course.

    Cheers -- sylas
  4. Jan 29, 2010 #3
    how the moving and stationary frames are equivalent when the results of annihilation experiment using electron and positron are not same .By performing the experiment it can be decided which frame is really moving
  5. Jan 29, 2010 #4


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    If the the center of mass of an electron positron collision is at rest in one frame, then there's another collision in which the center of mass is at rest in the other frame. So there's no way to pick out one frame as being at absolute rest.
  6. Jan 29, 2010 #5


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    How? Describe the details of the experiment that allows one to distinguish this?
  7. Jan 29, 2010 #6
    Hi, sharma_satdev.

    Let us consider the case that electron mass m velocity v1 and positron mass m and velocity v2 collide and two photon frequency ω1, ω2 moving + direction, - direction each, appear. all the particles are moving on a line to make the case simple and essential.

    Conservation of momentum
    mv1/sqrt(1-v1^2/c^2) + mv2/sqrt(1-v2^2/c^2) = hbar ω1/c - hbar ω2/c

    Conservation of energy
    mc^2/sqrt(1-v1^2/c^2) + mc^2/sqrt(1-v2^2/c^2) = hbar ω1 + hbar ω2

    By these equations we can get ω1 and ω2 from v1 and v2 and vice versa in any frames of reference, including your cases ; the moving carriage of train and on the railway platform.

    If we observe v1=0, we are in the rest frame of the electron.
    If we observe v2=0, we are in the rest frame of the positron.
    If we observe v1=-v2, we are in the center of mass frame.

    If we observe v1=v2=0, gentle collision of the particles, we are in your case of the moving carriage I guess, liberated energy to photon is 2mc^2.
    If we observe v1=v2=V, we are in your case on the platform looking at the moving carriage of velocity V I guess, liberated energy to photon is 2mc^2/sqrt(1-V^2/c^2).

    Last edited: Jan 29, 2010
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