Special Relativity collision problem

[subsonicman]

Homework Statement

A photon of Energy E_0 collides with a free particle of mass m_0 at rest. If the scattered photon flies off at angle θ, what is the scattering angle of the particle, β?

Homework Equations

The relevant equations are conservation of momentum in x and y direction and conservation of energy.

Conservation of energy:
E_0=E_s+m_0c^2γ

Conservation of momentum:
x direction: E_0/c=E_s*cosθ/c+m_0vγ*cosβ
y direction: E_s*sinθ=m_0vγ*sinβ

The Attempt at a Solution

Of course I need to first get rid of E_s and v (which is also inside of γ) and then solve for β. So first I get rid of E_s by plugging in E_0-m_0c^2γ in the two momentum equations. Then I solve for v in the y direction momentum equation. When I plug it in I get this ridiculous quadratic equation that, while it can be solved, becomes even more ridiculous when I plug it into the x direction momentum equation. And I can't solve it.

I was wondering if possibly there is some nicer way to solve this problem, possibly with better substitutions at the start or some other technique I'm not seeing. Anyways, thanks in advance!

 

[Curious3141]

Homework Statement

A photon of Energy E_0 collides with a free particle of mass m_0 at rest. If the scattered photon flies off at angle θ, what is the scattering angle of the particle, β?

Homework Equations

The relevant equations are conservation of momentum in x and y direction and conservation of energy.

Conservation of energy:
E_0=E_s+m_0c^2γ

Conservation of momentum:
x direction: E_0/c=E_s*cosθ/c+m_0vγ*cosβ
y direction: E_s*sinθ=m_0vγ*sinβ

The Attempt at a Solution

Of course I need to first get rid of E_s and v (which is also inside of γ) and then solve for β. So first I get rid of E_s by plugging in E_0-m_0c^2γ in the two momentum equations. Then I solve for v in the y direction momentum equation. When I plug it in I get this ridiculous quadratic equation that, while it can be solved, becomes even more ridiculous when I plug it into the x direction momentum equation. And I can't solve it.

I was wondering if possibly there is some nicer way to solve this problem, possibly with better substitutions at the start or some other technique I'm not seeing. Anyways, thanks in advance!

It's still a pretty ugly result (well, it's not pretty), but I don't get a quadratic. Here's how I did it:

I generally prefer not to introduce $\gamma$ into these problems because it complicates the algebra. Instead, I use $E^2 = m^2c^4 + p^2c^2$ for the energy conservation. Remember that this E represents the sum of the rest-mass energy plus the kinetic energy of a body. For a photon, it simply reduces to E = pc. Rewrite the photon momenta immediately in terms of energy.

You should get 3 simultaneous equations when you write down the conservation statements for energy, horiz. momentum and vert. momentum. There are three unknowns: $E'$, the energy of the photon post-collision, $p$, the momentum of the particle post-collision and $\beta$, what you need to solve for. I first solved for $E'$ in terms of the givens using the energy conservation equation, then substituted that into the equations for momentum conservation. A simple division of one equation over another should now give you the answer, if you remember that $\frac{\sin \beta}{\cos \beta} = \tan \beta$.