# Compton scattering problem

1. Jun 23, 2008

### ajhunte

Can Someone look over this and tell me if the work is correct.

1. The problem statement, all variables and given/known data
An object comes in with known velocity (v) and known mass (m) moving directly along the x-axis. It strikes another object of the same mass (m) which is stationary. Find the change in energy of the incoming object based on the angle at which it scattered ($$\theta$$) and the velocity of the target after the collision (u). The collision only occurs in 2 dimensions. (Note: The angle at which the target object is moving can be removed algebraically.)

$$E_{i}$$=Energy of Incoming Object before collision
$$E_{f}$$=Energy of Incoming Object after collision
$$E_{2}$$=Energy of Target Object after collision

$$p_{i}$$= Momentum of incoming object before collision.
$$p_{f}$$=Momentum of Incoming object after collision.
$$p_{2}$$=Momentum of Target object after collion.

$$\phi$$= Arbitrary Angle of Target object scattering (should not matter based on the note.

2. Relevant equations
$$E=1/2*m*v^{2}$$
p=m*v
$$p^{2}/(2*m)=E$$

3. The attempt at a solution

Energy Balance:
$$E_{i}=E_{f}+E_{2}$$

X-Momentum Balance:
$$p_{i}=p_{f}*Cos(\theta) +p_{2}*Cos(\phi)$$

Y-Momentum Balance: (This should be a zero momentum system in y-direction)
$$p_{f}*Sin(\theta)=p_{2}*Sin(\phi)$$

Squaring only the Momentum Equations and adding them together.

Y-Balance:
$$p_{f}^{2}*Sin^{2}(\theta)=p_{2}^{2}*Sin^{2}(\phi)$$

X-Balance: (after getting $$\phi$$ isolated on one side then squaring)
$$p_{2}^{2}*Cos^{2}(\phi)=p_{i}^{2} - p_{i}*p_{f}*Cos(\theta)+p_{f}^{2}*Cos^{2}(\theta)$$

Adding the two momentum Equations and using $$Sin^{2}+Cos^{2}=1$$
$$p_{2}^{2}=p_{i}^{2}-p_{i}*p_{f}*Cos(\theta)+p_{f}^{2}$$

Relating back to energy, using the relationship defined in section 2.
Since all masses are the same I divide the newly found momentum equation by 2m in order to get to energy:
$$E_{2}=E_{i}+E_{f}-\sqrt{E_{i}}*\sqrt{E_{f}}*Cos(\theta)$$

(Note: $$\sqrt{2m}*\sqrt{2m}=2m$$ and $$p/\sqrt{2m}=\sqrt{E}$$ )

Combining with the original Energy Balance Equation to Eliminate $$E_{2}$$. This involves subtracting the equation I just solved for and the original equation.
$$0=2*E_{f}-\sqrt{E_{i}}*\sqrt{E_{f}}*Cos(\theta)$$

$$\sqrt{E_{f}}=\sqrt{E_{i}}*Cos(\theta) +\- \frac{\sqrt{\sqrt{E_{i}}^{2}*Cos^{2}(\theta)-0}}{4}$$

$$E_{f}=\frac{E_{i}*Cos^{2}(\theta)}{4}$$

Is this the correct solution, or is there a step that I made a mistake or false assumption? It seems to me that this is wrong, because even a grazing trajectory decreases the initial energy by 3/4.

2. Jun 23, 2008

### Mindscrape

Re: Scattering

This is essentially compton scattering. Why do you have to angles? Also, are you assuming it is inelastic, or were you told that it is inelastic? I would probably assume an elastic collision.

3. Jun 24, 2008

### ajhunte

Re: Scattering

Your right, this is basically compton scattering without the relativisitic effects. That is why the first 4 steps are the same as the Compton Scattering derivation. The difference is that the since this does not use relativistic effects:
$$E=\frac{p^{2}}{2m}$$

For Compton Scattering, the conversion between energy and momentum for Photons and relativistic electron is different:

For Relativistic Electrons: $$pc=\sqrt{E_{0}^{2}-m_{0}^{2}c^{4}}$$

For Photons: $$\frac{E}{c}$$

This is where the difference is. That is why I asked the question, because this answer does not seem intuitive that if I have a grazing trajectory (i.e. $$\theta$$ goes to 0) I still lose 3/4 of the energy.

4. Jun 24, 2008

### alphysicist

Re: Scattering

Hi ajhunte,

You are missing a factor of two here. The factor of two will keep the four from appearing in the denominator of the final answer.

5. Jun 25, 2008

### ajhunte

Re: Scattering

Thank You, That is exactly what I was missing.