Compton scattering event (Proof question)

[jisbon]

Homework Statement

Prove that the scattered photon is fixed at a constant value with only the condition that both angles from the photon and electron sum up to 90 degrees after the collision.

Relevant Equations

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Hi all,
Currently given a problem to prove that the scattered photon is fixed at a constant value with only the condition that both angles from the photon and electron sum up to 90 degrees after the collision. I can't seem to prove it and listed my steps below, was wondering if anyone can guide/correct my mistakes. Cheers.

For x-direction:
$$
P_{photon}=P'_{photon}cos\theta +P_{e}cos\phi
$$
$$
\frac{h}{\lambda_{0}}= \dfrac {h}{\lambda ^{'}}\cos \theta +P_{e}\cos \left( 90-\theta \right)
$$
$$\frac{h}{\lambda_{0}}= \dfrac {h}{\lambda ^{'}}\cos \theta +P_{e}\sin \left( \theta \right)$$
$$\begin{aligned}\dfrac {\lambda ^{0}}{h}=\dfrac {\lambda ^{'}}{h\cos \theta }+\dfrac {1}{P_{e}\sin \theta }\\ \lambda ^{0}=\dfrac {\lambda ^{'}}{\cos \theta }+\dfrac {h}{P_{e}\sin \theta }\end{aligned}$$

For y-direction:
$$
p_{e}\sin \phi =P'_{photon}\sin \theta
$$
$$
p_{e}\sin \left( 90-\theta \right) =P'_{photon}\sin \theta
$$
$$
p_{e}\cos \left(\theta \right) = \dfrac {h}{\lambda '}\sin \theta
$$
$$p_{e}\cot \left(\theta \right) = \dfrac {h}{\lambda '}$$
$$\lambda '=\dfrac {h}{P_{e}\cot \theta } $$

From the Compton equation,
$$\lambda '-\lambda _{0}=\dfrac {h}{m_{e}c}\left( 1-\cos \theta \right)$$I was wondering if this is correct because I can't seem to solve it moving on from these steps. Any advice help will be appreciated. Cheers

 

[PeroK]

I guess what you are trying to prove is that there is only one solution where the sum of the angles is 90 degrees?

One approach is to show that the energy of the scattered photon is some function of the fixed quantities: namely the energy of the incident photon and the mass of the electron.

I would use Pythagoras theorem and avoid using $\theta$.

 

[kuruman]

You know, I think, from classical mechanics that when two equal masses collide elastically in a 2-d collision, the sum of the scattering angles is 90^o. Can you use that?

 

[jisbon]

You know, I think, from classical mechanics that when two equal masses collide elastically in a 2-d collision, the sum of the scattering angles is 90^o. Can you use that?

But the masses are different if I'm not wrong? Phonon and electrons.

 

[jisbon]

I guess what you are trying to prove is that there is only one solution where the sum of the angles is 90 degrees?

One approach is to show that the energy of the scattered photon is some function of the fixed quantities: namely the energy of the incident photon and the mass of the electron.

I would use Pythagoras theorem and avoid using $\theta$.

Prove that the wavelength of the scattered photon is fixed at a constant value.

 

[PeroK]

Prove that the wavelength of the scattered photon is fixed at a constant value.

Did you try using Pythagoras theorem for the momenta?

 

[PeroK]

You know, I think, from classical mechanics that when two equal masses collide elastically in a 2-d collision, the sum of the scattering angles is 90^o. Can you use that?

We have to use the relativistic energy-momentum relations for both the photon and the electron.

 

[jisbon]

I can't seem to get about using Pythagoras theorem, what do I put in as the two sides? What is the hypotenuse in this case? Thank you

 

[PeroK]

I can't seem to get about using Pythagoras theorem, what do I put in as the two sides? What is the hypotenuse in this case? Thank you

By conservation of momentum we have $\vec p = \vec p' + \vec p_e$. In general, the before and after momenta form a triangle. In this case the angle between $\vec p'$ and $\vec p_e$ is a right-angle, hence:
$$p^2 = p'^2 + p_e^2$$
That can be converted to an equation involving the energies. And you also have another equation from conservation of energy. You should be able to do something with those two equations.

 

[neilparker62]

Bit late I suppose but all the same ...

Angle of Incidence: $$\tan\theta_i=\frac{\lambda_c+\lambda_i}{\lambda_i}\sqrt{\frac{\Delta\lambda}{2\lambda_c-\Delta\lambda}}$$Angle of Refraction: $$\tan\theta_r=\frac{\lambda_c-\lambda_r}{\lambda_r}\sqrt{\frac{\Delta\lambda}{2\lambda_c-\Delta\lambda}}$$Latter will be zero when $\lambda_r=\lambda_c$ corresponding to the given condition.