# A particle discharge a photon

1. May 12, 2013

### buc030

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
$E = m{\gamma}c^2$
$P = m{\gamma}v$
$E_{photon} = pc$

3. 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.

2. May 12, 2013

### Staff: Mentor

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".

3. May 12, 2013

### 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...

4. May 12, 2013

### 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

Last edited: May 12, 2013
5. May 13, 2013

### buc030

Yeah you have a point there :)