91Ó°ÊÓ

Consider themaximum angle of radiation emitted is${\mathrm{θ}}_{\text{max}}â‰\dots \sqrt{(\text{1}−\mathrm{β})/\text{2}}$.

Write the formula of angle at which the power distribution is maximum.

$\mathbf{cos}\mathit{θ}\mathbf{=}\frac{\mathbf{Â\pm }\sqrt{\mathbf{1}\mathbf{+}\mathbf{15}{\mathbf{β}}^2}}{\mathbf{3}\mathbf{β}}$…… (1)

Here,$\mathit{β}$ is an ultrarelativistic speed.

Write the formula ofultrarelativistic speed.

$\mathbf{cos}\mathit{θ}\mathbf{=}\frac{\sqrt{\mathbf{1}\mathbf{+}\mathbf{15}{\mathbf{β}}^2}}{\mathbf{3}\mathbf{β}}$ …… (2)

Here, $\mathit{β}$ is an ultrarelativistic speed.

Write the formula of$\mathrm{η}$.

$\mathit{η}\mathbf{=}{\mathbf{\left(}\frac{4}{5}\mathbf{\right)}}^5\frac{1}{\mathbf{2}{\mathbf{δ}}^4}$ …… (3)

Here, $\mathit{δ}$is intensity of the radiation.

Determine the angle at which the power distribution is greatest first:

$\begin{array}{l}\frac{d}{d\mathrm{θ}}\frac{dP}{d\mathrm{Î\copyright }}=0\\ ⇒\frac{d}{d\mathrm{θ}}\frac{{\sin }^2\mathrm{θ}}{{(1−\mathrm{β}\cos \mathrm{θ})}^5}=0\\ =\frac{2\sin \mathrm{θ}\cos \mathrm{θ}{(1−\mathrm{β}\cos \mathrm{θ})}^5−5\mathrm{β}{\sin }^3\mathrm{θ}{(1−\mathrm{β}\cos \mathrm{θ})}^4}{{(1−\mathrm{β}\cos \mathrm{θ})}^{10}}=0\\ 2\cos \mathrm{θ}−2\mathrm{β}{\cos }^2\mathrm{θ}=5\mathrm{β}{\sin }^2\mathrm{θ}\end{array}$

Solve further as

$3\mathrm{β}{\cos }^2\mathrm{θ}+2\cos \mathrm{θ}−5\mathrm{β}=0$

Determine the angle at which the power distribution is maximum.

For $\mathrm{β}→0$we expect$\mathrm{θ}→\mathrm{π}/2$, so we want the RHS to tend to zero. We can only have that for the plus sign, so we have:

${\mathrm{θ}}_m=\arccos \frac{\sqrt{1+15{\mathrm{β}}^2}−1}{3\mathrm{β}}$

We have $\mathrm{β}≈1$ for the ultrarelativistic speeds, therefore$\mathrm{β}=1−\mathrm{δ}$, $\mathrm{δ}<<1$. Increase the cosine:

Determine the value of ultrarelativistic speeds.

Substitute$(1−\mathrm{δ})$ for $\mathrm{β}$ into equation (2).

$\begin{array}{c}\cos \mathrm{θ}=\frac{\sqrt{1+15{(1−\mathrm{δ})}^2}}{3(1−\mathrm{δ})}\\ ≈\frac{\sqrt{1+15(1−2\mathrm{δ})}−1}{3(1−\mathrm{δ})}\\ =\frac{\sqrt{16−30\mathrm{δ}}−1}{3(1−\mathrm{δ})}\\ ≈\frac{1}{3}(1+\mathrm{δ})[4\sqrt{1−\frac{15}{8}\mathrm{δ}}−1]\end{array}$

Solve further as

$\begin{array}{c}≈\frac{1}{3}(1+\mathrm{δ})[4(1−\frac{15}{16}\mathrm{δ})−1]\\ =(1+\mathrm{δ})(1−\frac{5}{4}\mathrm{δ})\\ ≈1−\frac{1}{4}\mathrm{δ}\end{array}$

But:

$\begin{array}{c}\cos {\mathrm{θ}}_m≈1−\frac{1}{2}{\mathrm{θ}}_m^2\\ {\mathrm{θ}}_m=\sqrt{\mathrm{δ}/2}\\ =\sqrt{(1−\mathrm{β})/2}\end{array}$

Let's now compute the ratio. Starting with 11.69, the particle is immediately at rest.

$\begin{array}{c}{\left|\frac{dP}{d\mathrm{Î\copyright }}\right|}_{rest}=S_{rad}â‹\dots \widehat{r}{r}^2\\ =\frac{{\mathrm{μ}}_0{q}^2{a}^2}{16{\mathrm{Ï€}}^{2}c}{\sin }^2\mathrm{θ}\end{array}$

Keep in mind that in this case, $\mathrm{θ}$ refers to the spherical angle rather than the angle closed with the velocity vector. So, in this instance, $\mathrm{θ}≈\mathrm{π}/2$, and:

${\left|\frac{dP}{d\mathrm{Î\copyright }}\right|}_{rest}=\frac{{\mathrm{μ}}_0{q}^2{a}^2}{16{\mathrm{Ï€}}^{2}c}$

The angle ${\mathrm{θ}}_m$ for the UR situation is what was discovered in the previous phase. The ratio is thus:

$\begin{array}{c}{\sin }^2{\mathrm{θ}}_m≈{\mathrm{θ}}_m^2\\ =\frac{\mathrm{δ}}{2}\\ {(1−\mathrm{β}\cos {\mathrm{θ}}_m)}^5={[1−(1−\mathrm{δ})(1+\frac{1}{2}\frac{\mathrm{δ}}{2})]}^5\\ ≈{\left(\frac{5}{4}\right)}^5{\mathrm{δ}}^5\end{array}$

Determine the$\mathrm{η}$.

$\begin{array}{c}\mathrm{η}=\frac{{(dP/d\mathrm{Î\copyright })}_{UR}}{{(dP/d\mathrm{Î\copyright })}_{rest}}\\ =\frac{{\sin }^2{\mathrm{θ}}_m}{{(1−\mathrm{β}\cos {\mathrm{θ}}_m)}^5}\\ ={\left(\frac{4}{5}\right)}^5\frac{1}{2{\mathrm{δ}}^4}\end{array}$

Now we need to determine the$\mathrm{δ}$ in terms of$\mathrm{γ}$:

$\begin{array}{c}\mathrm{γ}=\frac{1}{\sqrt{1−{\mathrm{β}}^2}}\\ =\frac{1}{\sqrt{1−{(1−\mathrm{δ})}^2}}\\ ≈\frac{1}{\sqrt{2\mathrm{δ}}}\\ ⇒\mathrm{δ}=\frac{1}{2{\mathrm{γ}}^2}\end{array}$

Determine the ratio is:

$\begin{array}{c}\mathrm{η}={\left(\frac{4}{5}\right)}^5\frac{1}{2{\mathrm{δ}}^4}\\ =\frac{1}{4}\frac{8^5}{5^5}{\mathrm{γ}}^8\end{array}$