91Ó°ÊÓ

Write the expression for outgoing 4-momenta.

$\begin{array}{c}{r}^{\mathrm{μ}}=\left(\frac{E}{c},p\cos \mathrm{ϕ},p\sin \mathrm{ϕ},0\right)\\ s^{\mathrm{μ}}=\left(\frac{E}{c},−p\cos \mathrm{ϕ},−p\sin \mathrm{ϕ},0\right)\end{array}$

Write the expression for the Lorentz transformation equation.

$\begin{array}{c}{\stackrel{\mathbf{¯}}{\mathbf{r}}}_{\mathbf{x}}\mathbf{=}\mathbf{γ}\mathbf{(}{\mathbf{r}}_{\mathbf{x}}\mathbf{-}{\mathbf{β°ù}}^{\mathbf{0}}\mathbf{)}\\ {\stackrel{\mathbf{¯}}{\mathbf{r}}}_{\mathbf{y}}\mathbf{=}{\mathbf{r}}_{\mathbf{y}}\\ {\stackrel{\mathbf{¯}}{\mathbf{s}}}_{\mathbf{x}}\mathbf{=}\mathbf{γ}\mathbf{(}{\mathbf{s}}_{\mathbf{x}}\mathbf{-}{\mathbf{β²õ}}^{\mathbf{0}}\mathbf{)}\\ {\stackrel{\mathbf{¯}}{\mathbf{s}}}_{\mathbf{y}}\mathbf{=}{\mathbf{s}}_{\mathbf{y}}\end{array}$

It is known that:

$\begin{array}{c}{r}_x=p\cos \mathrm{ϕ}\\ \mathrm{β}=−\frac{pc}{E}\\ r^0=\frac{E}{c}\end{array}$

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

$\begin{array}{c}{\stackrel{¯}{r}}_x=\mathrm{γ}\left(p\cos \mathrm{ϕ}+\frac{pc}{E}\frac{E}{c}\right)\\ {\stackrel{¯}{r}}_x=\mathrm{γ}p(1+\cos \mathrm{ϕ})\end{array}$

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

$\begin{array}{c}{\stackrel{¯}{r}}_y=p\sin \mathrm{ϕ}\\ {\stackrel{¯}{s}}_x=\mathrm{γ}p(1−\cos \mathrm{ϕ})\\ {\stackrel{¯}{s}}_y=−p\sin \mathrm{ϕ}\end{array}$

Write the equation for $\cos \mathrm{θ}$.

$\cos \mathrm{θ}=\frac{\stackrel{\mathbf{¯}}{\mathbf{r}}â‹\dots \stackrel{\mathbf{¯}}{\mathbf{s}}}{\stackrel{¯}{r}â‹\dots \stackrel{¯}{s}}$ …… (1)

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

Substitute $\stackrel{\mathbf{¯}}{\mathbf{r}}={\stackrel{¯}{r}}_x\widehat{\mathbf{x}}+{\stackrel{¯}{r}}_y\widehat{\mathbf{y}}$, $\stackrel{\mathbf{¯}}{\mathbf{s}}={\stackrel{¯}{s}}_x\widehat{\mathbf{x}}+{\stackrel{¯}{s}}_y\widehat{\mathbf{y}}$, ${\stackrel{¯}{r}}_x=\mathrm{γ}p(1+\cos \mathrm{ϕ})$, ${\stackrel{¯}{r}}_y=p\sin \mathrm{ϕ}$, ${\stackrel{¯}{s}}_x=\mathrm{γ}p(1−\cos \mathrm{ϕ})$and ${\stackrel{¯}{s}}_y=−p\sin \mathrm{ϕ}$.

$\begin{array}{c}\cos \mathrm{θ}=\frac{({\stackrel{¯}{r}}_x\widehat{\mathbf{x}}+{\stackrel{¯}{r}}_y\widehat{\mathbf{y}})â‹\dots ({\stackrel{¯}{s}}_x\widehat{\mathbf{x}}+{\stackrel{¯}{s}}_y\widehat{\mathbf{y}})}{\stackrel{¯}{r}â‹\dots \stackrel{¯}{s}}\\ \cos \mathrm{θ}=\frac{(\left(\mathrm{γ}p(1+\cos \mathrm{Ï•})\right)\widehat{\mathbf{x}}+\left(p\sin \mathrm{Ï•}\right)\widehat{\mathbf{y}})â‹\dots (\left(\mathrm{γ}p(1−\cos \mathrm{Ï•})\right)\widehat{\mathbf{x}}+(−p\sin \mathrm{Ï•})\widehat{\mathbf{y}})}{\sqrt{[{\mathrm{γ}}^2{p}^2{(1+\cos \mathrm{Ï•})}^2+p^2{\sin }^2\mathrm{Ï•}][{\mathrm{γ}}^2{p}^2{(1−\cos \mathrm{Ï•})}^2+p^2{\sin }^2\mathrm{Ï•}]}}\\ \cos \mathrm{θ}=\frac{{\mathrm{γ}}^2{p}^2(1−{\cos }^2\mathrm{Ï•})−p^2{\sin }^2\mathrm{Ï•}}{\sqrt{[{\mathrm{γ}}^2{p}^2{(1+\cos \mathrm{Ï•})}^2+p^2{\sin }^2\mathrm{Ï•}][{\mathrm{γ}}^2{p}^2{(1−\cos \mathrm{Ï•})}^2+p^2{\sin }^2\mathrm{Ï•}]}}\\ \cos \mathrm{θ}=\frac{({\mathrm{γ}}^2−1){\sin }^2\mathrm{Ï•}}{\sqrt{[{\mathrm{γ}}^2{(1+\cos \mathrm{Ï•})}^2+{\sin }^2\mathrm{Ï•}][{\mathrm{γ}}^2{(1−\cos \mathrm{Ï•})}^2+{\sin }^2\mathrm{Ï•}]}}\end{array}$

Substitute $\cos \mathrm{θ}=\frac{({\mathrm{γ}}^2−1){\sin }^2\mathrm{ϕ}}{\sqrt{[{\mathrm{γ}}^2{(1+\cos \mathrm{ϕ})}^2+{\sin }^2\mathrm{ϕ}][{\mathrm{γ}}^2{(1−\cos \mathrm{ϕ})}^2+{\sin }^2\mathrm{ϕ}]}}$in equation (1).

$\begin{array}{c}\frac{({\mathrm{γ}}^2−1)}{\sqrt{\left[{\mathrm{γ}}^2{\left(\frac{1+\cos \mathrm{Ï•}}{\sin \mathrm{Ï•}}\right)}^2+1\right]\left[{\mathrm{γ}}^2{\left(\frac{1−\cos \mathrm{Ï•}}{\sin \mathrm{Ï•}}\right)}^2+1\right]}}=\frac{\stackrel{\mathbf{¯}}{\mathbf{r}}â‹\dots \stackrel{\mathbf{¯}}{\mathbf{s}}}{\stackrel{¯}{r}â‹\dots \stackrel{¯}{s}}\\ \frac{\stackrel{\mathbf{¯}}{\mathbf{r}}â‹\dots \stackrel{\mathbf{¯}}{\mathbf{s}}}{\stackrel{¯}{r}â‹\dots \stackrel{¯}{s}}=\frac{({\mathrm{γ}}^2−1)}{\sqrt{\left({\mathrm{γ}}^2{\cot }^2\frac{\mathrm{Ï•}}{2}+1\right)\left({\mathrm{γ}}^2{\tan }^2\frac{\mathrm{Ï•}}{2}+1\right)}}\end{array}$

Let $\mathrm{Ó\neg }={\mathrm{γ}}^2−1$

$\begin{array}{c}\cos \mathrm{θ}=\frac{\mathrm{Ó\neg }}{\sqrt{\left(1+{\cot }^2\frac{\mathrm{Ï•}}{2}+\mathrm{Ó\neg }{\cot }^2\frac{\mathrm{Ï•}}{2}\right)\left(1+{\tan }^2\frac{\mathrm{Ï•}}{2}+\mathrm{Ó\neg }{\tan }^2\frac{\mathrm{Ï•}}{2}\right)}}\\ \cos \mathrm{θ}=\frac{\mathrm{Ó\neg }}{\sqrt{\left({\mathrm{cosec}}^2\frac{\mathrm{Ï•}}{2}+\mathrm{Ó\neg }{\cot }^2\frac{\mathrm{Ï•}}{2}\right)\left({\sec }^2\frac{\mathrm{Ï•}}{2}+\mathrm{Ó\neg }{\tan }^2\frac{\mathrm{Ï•}}{2}\right)}}\\ \cos \mathrm{θ}=\frac{\mathrm{Ó\neg }\sin \frac{\mathrm{Ï•}}{2}\cos \frac{\mathrm{Ï•}}{2}}{\sqrt{\left(1+\mathrm{Ó\neg }{\cos }^2\frac{\mathrm{Ï•}}{2}\right)\left(1+\mathrm{Ó\neg }{\sin }^2\frac{\mathrm{Ï•}}{2}\right)}}\\ \cos \mathrm{θ}=\frac{\frac{1}{2}\mathrm{Ó\neg }\sin \mathrm{Ï•}}{\sqrt{\left[1+\frac{1}{2}\mathrm{Ó\neg }(1+\cos \mathrm{Ï•})\right]\left[1+\frac{1}{2}\mathrm{Ó\neg }(1−\cos \mathrm{Ï•})\right]}}\end{array}$

On further solving,

$\begin{array}{c}\cos \mathrm{θ}=\frac{\sin \mathrm{Ï•}}{\sqrt{\left[\left(\frac{2}{\mathrm{Ó\neg }}+1\right)+\cos \mathrm{Ï•}\right]\left[\left(\frac{2}{\mathrm{Ó\neg }}+1\right)−\cos \mathrm{Ï•}\right]}}\\ \cos \mathrm{θ}=\frac{\sin \mathrm{Ï•}}{\sqrt{{\left(\frac{2}{\mathrm{Ó\neg }}+1\right)}^2−{\cos }^2\mathrm{Ï•}}}\\ \cos \mathrm{θ}=\frac{\sin \mathrm{Ï•}}{\sqrt{\frac{4}{{\mathrm{Ó\neg }}^2}+\frac{4}{\mathrm{Ó\neg }}+{\sin }^2\mathrm{Ï•}}}\end{array}$

Here,${\mathrm{Ï„}}^2=\frac{4}{{\mathrm{Ó\neg }}^2}+\frac{4}{\mathrm{Ó\neg }}$

$\cos \mathrm{θ}=\frac{1}{\sqrt{1+{\left(\frac{\mathrm{τ}}{\sin \mathrm{ϕ}}\right)}^2}}$

Now, consider the following figure,

Apply the Pythagoras theorem,

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

Now, calculate the value of $\tan \mathrm{θ}$.

$\begin{array}{c}\tan \mathrm{θ}=\frac{\mathrm{Ï„}}{\sin \mathrm{Ï•}}\\ \tan \mathrm{θ}=\frac{\sqrt{\frac{4}{{\mathrm{Ó\neg }}^2}+\frac{4}{\mathrm{Ó\neg }}}}{\sin \mathrm{Ï•}}\\ \tan \mathrm{θ}=\frac{\sqrt{\frac{4}{{({\mathrm{γ}}^2−1)}^2}+\frac{4}{({\mathrm{γ}}^2−1)}}}{\sin \mathrm{Ï•}}\\ \tan \mathrm{θ}=\frac{2\mathrm{γ}}{({\mathrm{γ}}^2−1)\sin \mathrm{Ï•}}\end{array}$ .....(2)

Here,$\mathrm{γ}=\frac{1}{\sqrt{1−\frac{v^2}{c^2}}}$

Hence, equation (2) becomes,

$\begin{array}{c}\tan \mathrm{θ}=\frac{2\mathrm{γ}}{\left({\left(\frac{1}{\sqrt{1−\frac{v^2}{c^2}}}\right)}^2−1\right)\sin \mathrm{ϕ}}\\ \tan \mathrm{θ}=\frac{2c^2}{\mathrm{γ}{v}^2\sin \mathrm{ϕ}}\end{array}$

Therefore, the angle subtended by two particles is$\tan \mathrm{θ}=\frac{2c^2}{\mathrm{γ}{v}^2\sin \mathrm{ϕ}}$ .