Two particle collision - One in excited state

• unscientific
In summary,In summary, according to the homework statement, the relative velocities after de-excitation are reversed. The expression for CM frame energy and momentum of either particle in CM frame can be found by solving for these values in the rest frame of particle 1 after the process.

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]

The Attempt at a Solution

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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?

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.

mfb said:
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?

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unscientific said:
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.

mfb said:
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##..

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

mfb said:
... 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 traveling 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:

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

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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}##.

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mfb said:
... 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:

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

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

By velocity addition formula,

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

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?

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

unscientific
mfb said:
v<<c, you can ignore all relativistic effects for the atoms.
That's brilliant!

1. What is a two particle collision?

A two particle collision is when two particles, such as atoms or subatomic particles, interact with each other and exchange energy and/or momentum.

2. What does it mean for one particle to be in an excited state?

A particle in an excited state means that it has absorbed energy and its electrons have moved to higher energy levels. This can happen through collisions with other particles or by absorbing energy from an outside source.

3. How does a collision affect the excited state of a particle?

A collision can cause the excited state particle to transfer some of its energy to the other particle, causing it to become excited as well. This can result in both particles emitting energy in the form of light or heat, or they may continue to collide and exchange energy until one or both return to their ground state.

4. What types of particles can experience an excited state collision?

Any type of particle that can be in an excited state can experience an excited state collision. This includes atoms, molecules, and subatomic particles such as electrons and protons.

5. What is the significance of studying two particle collisions in excited states?

Studying two particle collisions in excited states can provide valuable information about the properties and behavior of particles. It can also help us understand how energy and momentum are exchanged between particles, and how this can impact larger systems such as chemical reactions and nuclear reactions.