A particle of mass m collides elastically with an identical particle at rest. Classically, the outgoing trajectories always make an angle of $\mathbf{90}\mathbf{°}$. Calculate this angle relativistically, in terms of$\mathit{\varphi }$ , the scattering angle, and v, the speed, in the center-of-momentum frame.

Write the expression for outgoing 4-momenta.

$\begin{array}{c}{r}^{\mu }=\left(\frac{E}{c},p\cos \varphi ,p\sin \varphi ,0\right)\\ s^{\mu }=\left(\frac{E}{c},-p\cos \varphi ,-p\sin \varphi ,0\right)\end{array}$

Write the expression for the Lorentz transformation equation.

$\begin{array}{c}{\overline{\mathbf{r}}}_{\mathbf{x}}\mathbf{=}\mathbf{\gamma }\mathbf{(}{\mathbf{r}}_{\mathbf{x}}\mathbf{-}{\mathbf{\beta r}}^{\mathbf{0}}\mathbf{)}\\ {\overline{\mathbf{r}}}_{\mathbf{y}}\mathbf{=}{\mathbf{r}}_{\mathbf{y}}\\ {\overline{\mathbf{s}}}_{\mathbf{x}}\mathbf{=}\mathbf{\gamma }\mathbf{(}{\mathbf{s}}_{\mathbf{x}}\mathbf{-}{\mathbf{\beta s}}^{\mathbf{0}}\mathbf{)}\\ {\overline{\mathbf{s}}}_{\mathbf{y}}\mathbf{=}{\mathbf{s}}_{\mathbf{y}}\end{array}$

It is known that:

$\begin{array}{c}{r}_x=p\cos \varphi \\ \beta =-\frac{pc}{E}\\ r^0=\frac{E}{c}\end{array}$

Hence, the value of ${\overline{r}}_x$will be,

$\begin{array}{c}{\overline{r}}_x=\gamma \left(p\cos \varphi +\frac{pc}{E}\frac{E}{c}\right)\\ {\overline{r}}_x=\gamma p(1+\cos \varphi )\end{array}$

The value of ${\overline{r}}_y$,${\overline{s}}_x$and${\overline{s}}_y$will be,

$\begin{array}{c}{\overline{r}}_y=p\sin \varphi \\ {\overline{s}}_x=\gamma p(1-\cos \varphi )\\ {\overline{s}}_y=-p\sin \varphi \end{array}$

Write the equation for $\cos \theta$.

$\cos \theta =\frac{\overline{\mathbf{r}}\cdot \overline{\mathbf{s}}}{\overline{r}\cdot \overline{s}}$ …… (1)

Here,$\overline{\mathbf{r}}={\overline{r}}_x\widehat{\mathbf{x}}+{\overline{r}}_y\widehat{\mathbf{y}}$ and $\overline{\mathbf{s}}={\overline{s}}_x\widehat{\mathbf{x}}+{\overline{s}}_y\widehat{\mathbf{y}}$.

Substitute $\overline{\mathbf{r}}={\overline{r}}_x\widehat{\mathbf{x}}+{\overline{r}}_y\widehat{\mathbf{y}}$, $\overline{\mathbf{s}}={\overline{s}}_x\widehat{\mathbf{x}}+{\overline{s}}_y\widehat{\mathbf{y}}$, ${\overline{r}}_x=\gamma p(1+\cos \varphi )$, ${\overline{r}}_y=p\sin \varphi$, ${\overline{s}}_x=\gamma p(1-\cos \varphi )$and ${\overline{s}}_y=-p\sin \varphi$.

$\begin{array}{c}\cos \theta =\frac{({\overline{r}}_x\widehat{\mathbf{x}}+{\overline{r}}_y\widehat{\mathbf{y}})\cdot ({\overline{s}}_x\widehat{\mathbf{x}}+{\overline{s}}_y\widehat{\mathbf{y}})}{\overline{r}\cdot \overline{s}}\\ \cos \theta =\frac{(\left(\gamma p(1+\cos \varphi )\right)\widehat{\mathbf{x}}+\left(p\sin \varphi \right)\widehat{\mathbf{y}})\cdot (\left(\gamma p(1-\cos \varphi )\right)\widehat{\mathbf{x}}+(-p\sin \varphi )\widehat{\mathbf{y}})}{\sqrt{[{\gamma }^2{p}^2{(1+\cos \varphi )}^2+p^2{\sin }^2\varphi ][{\gamma }^2{p}^2{(1-\cos \varphi )}^2+p^2{\sin }^2\varphi ]}}\\ \cos \theta =\frac{{\gamma }^2{p}^2(1-{\cos }^2\varphi )-p^2{\sin }^2\varphi }{\sqrt{[{\gamma }^2{p}^2{(1+\cos \varphi )}^2+p^2{\sin }^2\varphi ][{\gamma }^2{p}^2{(1-\cos \varphi )}^2+p^2{\sin }^2\varphi ]}}\\ \cos \theta =\frac{({\gamma }^2-1){\sin }^2\varphi }{\sqrt{[{\gamma }^2{(1+\cos \varphi )}^2+{\sin }^2\varphi ][{\gamma }^2{(1-\cos \varphi )}^2+{\sin }^2\varphi ]}}\end{array}$

Substitute $\cos \theta =\frac{({\gamma }^2-1){\sin }^2\varphi }{\sqrt{[{\gamma }^2{(1+\cos \varphi )}^2+{\sin }^2\varphi ][{\gamma }^2{(1-\cos \varphi )}^2+{\sin }^2\varphi ]}}$in equation (1).

$\begin{array}{c}\frac{({\gamma }^2-1)}{\sqrt{\left[{\gamma }^2{\left(\frac{1+\cos \varphi }{\sin \varphi }\right)}^2+1\right]\left[{\gamma }^2{\left(\frac{1-\cos \varphi }{\sin \varphi }\right)}^2+1\right]}}=\frac{\overline{\mathbf{r}}\cdot \overline{\mathbf{s}}}{\overline{r}\cdot \overline{s}}\\ \frac{\overline{\mathbf{r}}\cdot \overline{\mathbf{s}}}{\overline{r}\cdot \overline{s}}=\frac{({\gamma }^2-1)}{\sqrt{\left({\gamma }^2{\cot }^2\frac{\varphi }{2}+1\right)\left({\gamma }^2{\tan }^2\frac{\varphi }{2}+1\right)}}\end{array}$

Let $\omega ={\gamma }^2-1$

$\begin{array}{c}\cos \theta =\frac{\omega }{\sqrt{\left(1+{\cot }^2\frac{\varphi }{2}+\omega {\cot }^2\frac{\varphi }{2}\right)\left(1+{\tan }^2\frac{\varphi }{2}+\omega {\tan }^2\frac{\varphi }{2}\right)}}\\ \cos \theta =\frac{\omega }{\sqrt{\left({\mathrm{cosec}}^2\frac{\varphi }{2}+\omega {\cot }^2\frac{\varphi }{2}\right)\left({\sec }^2\frac{\varphi }{2}+\omega {\tan }^2\frac{\varphi }{2}\right)}}\\ \cos \theta =\frac{\omega \sin \frac{\varphi }{2}\cos \frac{\varphi }{2}}{\sqrt{\left(1+\omega {\cos }^2\frac{\varphi }{2}\right)\left(1+\omega {\sin }^2\frac{\varphi }{2}\right)}}\\ \cos \theta =\frac{\frac{1}{2}\omega \sin \varphi }{\sqrt{\left[1+\frac{1}{2}\omega (1+\cos \varphi )\right]\left[1+\frac{1}{2}\omega (1-\cos \varphi )\right]}}\end{array}$

On further solving,

$\begin{array}{c}\cos \theta =\frac{\sin \varphi }{\sqrt{\left[\left(\frac{2}{\omega }+1\right)+\cos \varphi \right]\left[\left(\frac{2}{\omega }+1\right)-\cos \varphi \right]}}\\ \cos \theta =\frac{\sin \varphi }{\sqrt{{\left(\frac{2}{\omega }+1\right)}^2-{\cos }^2\varphi }}\\ \cos \theta =\frac{\sin \varphi }{\sqrt{\frac{4}{{\omega }^2}+\frac{4}{\omega }+{\sin }^2\varphi }}\end{array}$

Here,${\tau }^2=\frac{4}{{\omega }^2}+\frac{4}{\omega }$

$\cos \theta =\frac{1}{\sqrt{1+{\left(\frac{\tau }{\sin \varphi }\right)}^2}}$

Now, consider the following figure,

Apply the Pythagoras theorem,

role="math" localid="1655907747852" $\begin{array}{c}{\left(PR\right)}^2={\left(PQ\right)}^2+{\left(QR\right)}^2\\ {\left(\sqrt{1+{\left(\frac{\tau }{\sin \varphi }\right)}^2}\right)}^2={\left(PQ\right)}^2+{\left(1\right)}^2\\ {\left(PQ\right)}^2=1+{\left(\frac{\tau }{\sin \varphi }\right)}^2-1\\ PQ=\frac{\tau }{\sin \varphi }\end{array}$

Now, calculate the value of $\tan \theta$.

$\begin{array}{c}\tan \theta =\frac{\tau }{\sin \varphi }\\ \tan \theta =\frac{\sqrt{\frac{4}{{\omega }^2}+\frac{4}{\omega }}}{\sin \varphi }\\ \tan \theta =\frac{\sqrt{\frac{4}{{({\gamma }^2-1)}^2}+\frac{4}{({\gamma }^2-1)}}}{\sin \varphi }\\ \tan \theta =\frac{2\gamma }{({\gamma }^2-1)\sin \varphi }\end{array}$ .....(2)

Here,$\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

Hence, equation (2) becomes,

$\begin{array}{c}\tan \theta =\frac{2\gamma }{\left({\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\right)}^2-1\right)\sin \varphi }\\ \tan \theta =\frac{2c^2}{\gamma v^2\sin \varphi }\end{array}$

Therefore, the angle subtended by two particles is$\tan \theta =\frac{2c^2}{\gamma v^2\sin \varphi }$ .