Total Energy of a Particle After Discharging a Photon in a Moving Frame

[buc030]

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

E = m{\gamma}c^2
P = m{\gamma}v
E_{photon} = pc

The Attempt at a Solution

Let's mark the new mass after the discharge: m'
And the new \gamma after the discharge : \gamma'
And the energy of the particle after the discharge relative to the lab: E_{particle}The energy before the discharge:
E_i = mc^2

The momentum before the discharge:
P_i = 0

After the discharge the energy and the momentum are same, so:

E_{photon} = 1/2mc^2
E_f = mc^2 = E_{photon} + E_{particle} = m'{\gamma'}c^2 + 1/2mc^2
so:
1/2m = m'{\gamma'}

since E_{photon} = pc we can say that:
P_{photon} = 1/2mc
and:
0 = P_i = P_f = 1/2mc + m'{\gamma'}v.
but
m'{\gamma'} = 1/2m
so:
0 = P_i = P_f = 1/2mc + 1/2mv.
so:
v = -c.
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:
Ans = mc^2-1/2mc^2=1/2mc^2.

However this is not correct (by my reference), does anyone know why?
By my reference the right answer is:
Ans = {3/2}mc^2.
Thanks,
Shai.

 

[mfb]

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:

The particle has to be massless, I agree (which is a bit strange, but possible). This gives a speed of c in all reference frames, but the energy is still frame-dependent.

By the way, I never heard "discharge" for those processes. Usually, they are called "emission", if the result is another massive particle. If a particle decays to two photons (as in this example), I would simply call it "decay".

 

[buc030]

You are right!
After I apply

E'={\gamma}(E-pv)
I get the correct result!
Thank.
Regarding the name of the process you are probably right, I translated this question from Hebrew so...

 

[tiny-tim]

<h2><strong>welcome to pf!</strong></h2>

hi buc030! welcome to pf!

a more direct way would be to use the red-shift formula to calculate the energy of the photon, and the initial energy of the particle, in the .8c frame, and just subtract

 

[buc030]

Yeah you have a point there :)