Scattering angle in relativistic kinematics

[kalok87]

Homework Statement

Considering 2 scattering particles with momenta $p_{1}, p_{2}$, where $p_{2} = 0$ in the Lab reference. The momenta of these 2 particles after elastic collision are $p_{1}', p_{2}'$, respectively. Due to the 4-momentum relation, we have $p_{1i}p_{1}^{'i} = e_{1}e_{1}' - p_{1} \cdot p_{1}' = e_{1}e_{1}' - p_{1} \cdot p_{1}' \cos{\theta_{1}}$, here $\theta_{1}$ is the angle of scattering of the incident particle $m_{1}$, and the $e_{1}$ is the energy of particle $m_{1}$. Finally we can derive the equation for scattering angle $\theta_{1}$:

$$\cos{\theta_{1}} = \frac{e_{1}'(e_{1} + m_{2}) - e_{1}m_{2} - m_{1}^{2}}{p_{1}p_{1}'}$$

In Landau's book The Classical Theory of Fields, section 13, a function of $\theta_{1}$ is derived as follows if the rest mass of incident particle $m_{1}$ is larger than the mass of another particle $m_{2}$:

$$\sin{\theta_{1\, max}} = \frac{m_{2}}{m_{1}}$$

But how could we derive this result?

Homework Equations

The Attempt at a Solution

If we don't consider the relativity, an equation for $\theta_{1}$ can be derived:

$$\cos{\theta_{1}} = \frac{(1 + \alpha) \beta^{2} + 1 - \alpha}{2\beta}$$

where $\alpha = \frac{m_{2}}{m_{1}}$ and $\beta = \frac{v_{1}'}{v_{1}}$. Since $\alpha$ is a constant, we can find the extreme of $\theta_{1}$ by treating $\beta$ as an independent variable.:

$$ \frac{\mathrm d}{\mathrm d \beta}\cos{\theta_{1}} = 0$$

From above equation we obtain $\cos{\theta_{1\, min}} = \frac{2\beta}{1 + \beta^{2}}$ for $\alpha = \frac{1 - \beta^{2}}{1 + \beta^{2}}$. Finally we get $\sin{\theta_{1}}_{max} = \frac{m_{2}}{m_{1}}$.

But if we are considering the relativity, the equation for $\theta_{1}$ is hard to simplify. How can I resolve this? Thank you for the advice.

 

[TSny]

$$\cos{\theta_{1}} = \frac{e_{1}'(e_{1} + m_{2}) - e_{1}m_{2} - m_{1}^{2}}{p_{1}p_{1}'}$$
But if we are considering the relativity, the equation for $\theta_{1}$ is hard to simplify. How can I resolve this? Thank you for the advice.

It is messy, but it will work out. It might be helpful to write the above equation as $$\cos{\theta_{1}} = \frac{ae_{1}' -b}{p_{1}\sqrt{e_{1}'^2-m_1^2}}$$where $a$ and $b$ are certain constants. Show that the derivative of the right hand side equals zero when $e_{1}' = \large \frac{a m_1^2}{b}$.