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## Homework Statement

(a)Find energy of photon emitted

(b)Show that relative velocities after de-excitation are reversed

(c)Find an expression for CM frame energy and find momentum of either particle in CM frame

(d)Consider now, for a nuclear process and describe the initial and final conditions in the CM frame[/B]

## Homework Equations

## The Attempt at a Solution

[/B]

**Part (a)**

Doing everything in the lab frame:

4-vector before collision: ##(Mc,0) ##

Let ##E## be energy of photon emitted, ##p## be momentum of de-excited particle.

4-vector after collision: ## (\frac{E_1}{c},-p) + (\frac{E}{c},\frac{E}{c})##.

By conservation of momentum and energy:

[tex]Mc^2 = E + E_1 [/tex]

[tex]pc = E [/tex]

Solving gives us:

[tex]E = \frac{M^2 - m^2}{2M} c^2 [/tex]

**Part (b)**Before collision in the rest frame of particle 1:

4-vector before collision: ## (Mc,0) + (\gamma_r mc, \gamma_r m \vec{v_r}) ##

After collision in the rest frame of particle 2:

4-vector after collision: ## (\gamma_r' mc, \gamma_r' m \vec{v_r'}) + (Mc,0) ##

Invariance of 4 vector gives:

[tex](Mc + \gamma_r mc)^2 - (\gamma_r m \vec{v_r})^2 = (Mc + \gamma_r' mc)^2 - (\gamma_r' m \vec{v_r'})^2 [/tex]

The only meaningful result here is ##\gamma_r = \gamma_r'## and the relative velocities in the frame of particle 2 reverses upon absorption, i.e. ##v_r = v_r'##.

How do I find an expression for ##v_r## without knowing their initial speeds?