# Two particle collision - One in excited state

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1. Jan 9, 2015

### unscientific

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

(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

2. Relevant equations

3. The attempt at a solution

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:

$$Mc^2 = E + E_1$$
$$pc = E$$

Solving gives us:

$$E = \frac{M^2 - m^2}{2M} c^2$$

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:
$$(Mc + \gamma_r mc)^2 - (\gamma_r m \vec{v_r})^2 = (Mc + \gamma_r' mc)^2 - (\gamma_r' m \vec{v_r'})^2$$

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?

2. Jan 10, 2015

### Staff: Mentor

How did you find out that $\gamma_r = \gamma_r'$ is the only meaningful result?

You can use the result of (a) to calculate the velocity change of particle 1 after the process, and find the system where the photon has the right energy to get absorbed by particle 2.

3. Jan 10, 2015

### unscientific

I know I had to use part (a) somehow, but I didn't know how. After emitting the photon, the momentum gained by the atom is $p = \frac{E}{c} = \frac{M^2 - m^2}{2M}c$.

What do you mean by 'the system where the photon has the right energy to get absorbed by particle 2. ' ?

For part (b):

1. Start off with rest frame of particle 1. Particle 2 is moving towards 1 with speed $v_2$.
2. Particle 1 emits photon, moving backwards with momentum $p = \frac{M^2 - m^2}{2M}c$. Now photon with energy $E = \frac{M^2 - m^2}{2M}c^2$ is moving towards particle 2.
3. Particle 2 absorbs photon, slows down?
4. Go into rest frame of particle 1 and find relative velocity of particle 2 using relativistic addition.

But how do I find the velocity of particle 2 after absorption?

Last edited: Jan 10, 2015
4. Jan 10, 2015

### Staff: Mentor

Absorption is like emission, just with the opposite time direction. You can calculate the energy of the photon in the frame of the ground state atom, and you can calculate the energy of the emitted photon in every frame. Now you have to find the frame where the photon energy is right for the absorption process: the second atom will be at rest in this frame.

5. Jan 10, 2015

### unscientific

From part (a), I understand that the 'right frame' for absorption is when both the photon and atom are moving towards each other with momentum $p = \frac{M^2 - m^2}{2M}c$. Then they combine to form a stationary, excited atom of rest mass $M$..

6. Jan 10, 2015

### Staff: Mentor

... and the second atom will need a specific velocity to make that possible.

7. Jan 10, 2015

### unscientific

Ok, let's think about the relative velocities.

1. Before emission in the rest frame of particle 1, particle 2 had to be travelling towards particle 1 with momentum $p = \frac{M^2 - m^2}{2M}c$ for proper absorption.

2. After particle 1 emits the photon and the photon is absorbed by particle 2, particle 2 comes to rest while particle 1 recoils back with momentum $p = \frac{M^2 - m^2}{2M}c$.

3. Now in the rest frame of particle 1, particle 2 is now moving away from it at momentum $p = \frac{M^2 - m^2}{2M}c$, instead of towards it as before.

To find the speed, we start off by:

$$p = \frac{M^2 - m^2}{2M}c = \gamma_v m v$$
$$(\frac{M^2 - m^2}{2M}c )^2 (1 - \frac{v^2}{c^2}) = m^2 v^2$$
$$v = \frac{\frac{M^2 - m^2}{2M}}{\sqrt{m^2 + \frac{M^2 - m^2}{2M} }}c$$

Last edited: Jan 10, 2015
8. Jan 13, 2015

### unscientific

Part (b)

Using this expression

$$v_r = \frac{E}{\sqrt{m^2c^2 + \frac{E^2}{c^2}}}$$

and $m = 40u = 40 \times 1.67 \times 10^{-27}~kg$ and $E = \frac{hc}{\lambda} = 2.73 \times 10^{-19} J$, I get the relative speed to be $v_r = 0.014~ m s^{-1}$. Is something wrong?

I realized that $m^2c^2 >> \frac{E^2}{c^2}$.

Last edited: Jan 13, 2015
9. Jan 13, 2015

### unscientific

Is there an easier way to do part (c)?

Part (c)

Before collision in the rest frame of particle 1, particle 2 is moving towards particle 1 at speed $v_r$ as we have found. The momentum of particle 2 is simply $p_r = \frac{E}{c}$ from part (b). The It's 4-vector is $= (\frac{Mc^2 + \frac{E^2}{c^2}c^2 + m^2c^4 }{c},- \frac{E}{c})$.

In the CM frame, particle 1 and particle 2 are moving towards each other at speeds $v$ and $v_2$ respectively. It's 4-vector is (E_{cm},0).

The energy of the CM frame is found by taking invariance of the square and found to be:

$$E_{cm} = \left( M^2c^4 + m^2c^4 + 2Mc^2 \sqrt{E^2 + m^2c^4} \right)^{\frac{1}{2}}$$

For the velocity, I boost the frame leftwards by speed $v$.

$$v_2 = \frac{v_r - v}{1 - \frac{v v_r}{c^2}}$$

For overall momentum to be zero,

$$\gamma_2 m v_2 = \gamma_v M v$$

Eliminating $v_2$ from these equations will allow us to find $v$, but it seems to be extremely tedious.

Am I missing something here?

10. Jan 13, 2015

### Staff: Mentor

v<<c, you can ignore all relativistic effects for the atoms.

11. Jan 13, 2015

### unscientific

That's brilliant!