Two particle collision - One in excited state

  • #1
1,728
13

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]
2wqesns.png


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?
 

Answers and Replies

  • #2
35,268
11,537
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
1,728
13
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.
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:
  • #4
35,268
11,537
What do you mean by 'the system where the photon has the right energy to get absorbed by particle 2. ' ?
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
1,728
13
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.
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
35,268
11,537
... and the second atom will need a specific velocity to make that possible.
 
  • #7
1,728
13
... and the second atom will need a specific velocity to make that possible.
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:

[tex] p = \frac{M^2 - m^2}{2M}c = \gamma_v m v [/tex]
[tex](\frac{M^2 - m^2}{2M}c )^2 (1 - \frac{v^2}{c^2}) = m^2 v^2 [/tex]
[tex] v = \frac{\frac{M^2 - m^2}{2M}}{\sqrt{m^2 + \frac{M^2 - m^2}{2M} }}c[/tex]
 
Last edited:
  • #8
1,728
13
Part (b)

Using this expression

[tex] v_r = \frac{E}{\sqrt{m^2c^2 + \frac{E^2}{c^2}}}[/tex]

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:
  • #9
1,728
13
... and the second atom will need a specific velocity to make that possible.
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:

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

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

By velocity addition formula,

[tex] v_2 = \frac{v_r - v}{1 - \frac{v v_r}{c^2}}[/tex]

For overall momentum to be zero,

[tex]\gamma_2 m v_2 = \gamma_v M v [/tex]

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
35,268
11,537
v<<c, you can ignore all relativistic effects for the atoms.
 
  • Like
Likes unscientific
  • #11
1,728
13
v<<c, you can ignore all relativistic effects for the atoms.
That's brilliant!
 

Related Threads on Two particle collision - One in excited state

Replies
2
Views
950
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
2
Views
5K
Replies
9
Views
4K
  • Last Post
Replies
6
Views
722
Replies
4
Views
689
Replies
1
Views
1K
  • Last Post
Replies
4
Views
827
Replies
0
Views
2K
Top