Special relativity - scattering angle

[Aleolomorfo]

Homework Statement

Finding the maximum scattering angle of a particle whose mass in $m_1$ which hits with relativistic velocity $v$ a particle at rest with mass $m_2<m_1$.

The Attempt at a Solution

I've written the 4-momenta (p before the collision, k after the collision and the z-axis is along the direction of the incident particle):
$$p_1=(m_1\gamma,0,0,m_1\gamma v)$$
$$p_2=(m_2,0,0,0)$$
$$k_1=(E,0,\sqrt{E^2-m^2_1}\sin{\theta},\sqrt{E^2-m^2_1}\cos{\theta})$$
For $k_2$ the components are not important, it's important that $k^2_2=m^2_2$
Then I've used the conservation of 4-momentum $p_1+p_2=k_1+k_2$, then $k_2=p_1+p_2-k_1$, then $k^2_2=p^2_1+p^2_2+k^2_1+2p_1p_2-2p_1k_1-2p_2k_1$. After calculations I've found:
$$\cos{\theta}=\frac{m^2_1+m_1m_2\gamma-2E(m_1\gamma+m_2)}{m_1\gamma v\sqrt{E^2-m^2_1}}$$.
Then I've taken the derivative $\frac{d(\cos{\theta})}{dE}$ and put it equal to 0. However, I've found an equation quite difficult to solve and I think it's wrong.
I think the way I set up the problem is not incorrect, but maybe there is a easier way or some trick to reduce calcus.

 

[TSny]

Then I've used the conservation of 4-momentum $p_1+p_2=k_1+k_2$, then $k_2=p_1+p_2-k_1$, then $k^2_2=p^2_1+p^2_2+k^2_1+2p_1p_2-2p_1k_1-2p_2k_1$. After calculations I've found:
$$\cos{\theta}=\frac{m^2_1+m_1m_2\gamma-2E(m_1\gamma+m_2)}{m_1\gamma v\sqrt{E^2-m^2_1}}$$.

Check the overall sign of the right hand side of the equation for $\cos \theta$ and also check if the factor of 2 in the numerator is correct.

Then I've taken the derivative $\frac{d(\cos{\theta})}{dE}$ and put it equal to 0. However, I've found an equation quite difficult to solve and I think it's wrong.
I think the way I set up the problem is not incorrect, but maybe there is a easier way or some trick to reduce calcus.

With the corrections mentioned above, it will work out if you slog through it. I don't know of a clever trick to reduce the algebra. Maybe someone else can show us a better way.

 

[Aleolomorfo]

Check the overall sign of the right hand side of the equation for $\cos \theta$ and also check if the factor of 2 in the numerator is correct.

With the corrections mentioned above, it will work out if you slog through it. I don't know of a clever trick to reduce the algebra. Maybe someone else can show us a better way.

Thank you for your help. The worst is when you make a mistake at the beginning. My result is $\cos{\theta_{max}}=\sqrt{1-\frac{m^2_2}{m^2_1}}$, I think it's correct or at least is plausible.

 

[TSny]

Thank you for your help. The worst is when you make a mistake at the beginning. My result is $\cos{\theta_{max}}=\sqrt{1-\frac{m^2_2}{m^2_1}}$, I think it's correct or at least is plausible.

Yes, that's the correct answer. It's even nicer when expressed in terms of $\sin \theta$. Since the answer is independent of the speed of the incoming particle, the result is the same as the nonrelativistic, Newtonian result.