- #1

kalok87

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## 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**,

*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}##:*

**section 13,**$$\sin{\theta_{1\, max}} = \frac{m_{2}}{m_{1}}$$

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.