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buc030
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Homework Statement
A particle of mass m is at rest in the lab frame.
The particle discharge a photon with energy 1/2mc^2 to the direction of x+.
A spaceship is moving at v = 0.8c in the direction x+ (in the same direction the photon is moving).
What is the total energy of the particle after it discharge the photon relativity to the spaceship frame?
Homework Equations
[itex]E = m{\gamma}c^2[/itex]
[itex]P = m{\gamma}v[/itex]
[itex]E_{photon} = pc [/itex]
The Attempt at a Solution
Let's mark the new mass after the discharge: [itex]m'[/itex]
And the new [itex]\gamma[/itex] after the discharge : [itex]\gamma'[/itex]
And the energy of the particle after the discharge relative to the lab: [itex]E_{particle}[/itex]The energy before the discharge:
[itex]E_i = mc^2[/itex]
The momentum before the discharge:
[itex]P_i = 0[/itex]
After the discharge the energy and the momentum are same, so:
[itex]E_{photon} = 1/2mc^2[/itex]
[itex]E_f = mc^2 = E_{photon} + E_{particle} = m'{\gamma'}c^2 + 1/2mc^2[/itex]
so:
[itex]1/2m = m'{\gamma'}[/itex]
since [itex]E_{photon} = pc [/itex] we can say that:
[itex]P_{photon} = 1/2mc[/itex]
and:
[itex]0 = P_i = P_f = 1/2mc + m'{\gamma'}v[/itex].
but
[itex]m'{\gamma'} = 1/2m[/itex]
so:
[itex]0 = P_i = P_f = 1/2mc + 1/2mv[/itex].
so:
[itex]v = -c[/itex].
So the particle is seen from all frames with the same velocity and there for
the energy is the same that observed in the lab frame:
[itex]Ans = mc^2-1/2mc^2=1/2mc^2[/itex].
However this is not correct (by my reference), does anyone know why?
By my reference the right answer is:
[itex]Ans = {3/2}mc^2[/itex].
Thanks,
Shai.