Question

Find the angle ${\mathit{\theta }}_{\max }$ at which the maximum radiation is emitted, in Ex. 11.3 (Fig. 11.13). Show that for ultra relativistic speeds ( $\mathit{\upsilon }$close to$\mathrm{c}$), ${\mathit{\theta }}_{\max }\mathbf{\cong }\sqrt{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{\beta }\mathbf{)}\mathbf{/}\mathbf{2}}$. What is the intensity of the radiation in this maximal direction (in the ultra relativistic case), in proportion to the same quantity for a particle instantaneously at rest? Give your answer in terms of$\mathit{\gamma }$.

Short answer

The value ofangle at which the power distribution is maximum at .

${\theta }_m=\text{arc}\cos \frac{\pm \sqrt{1+15{\beta }^2}-1}{3\beta }$

The value of ultrarelativistic speeds is ${\theta }_m=\sqrt{(1-\beta )/2}$.

The ratio is$\eta =\frac{1}{4}\frac{8^5}{5^5}{\gamma }^8$ .

Step-by-step solution

Write the given data from the question.

Consider themaximum angle of radiation emitted is${\theta }_{\text{max}}\cong \sqrt{(\text{1}-\beta )/\text{2}}$.

Determine the formulaof angle at which the power distribution is maximum, ultrarelativistic speeds and ratioη .

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

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

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

Write the formula ofultrarelativistic speed.

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

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

Write the formula of$\eta$.

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

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

Step 3:Determine thevalue of angle at which the power distribution is maximum, ultrarelativistic speeds and ratioη .

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

$\begin{array}{l}\frac{d}{d\theta }\frac{dP}{d\Omega }=0\\ \Rightarrow \frac{d}{d\theta }\frac{{\sin }^2\theta }{{(1-\beta \cos \theta )}^5}=0\\ =\frac{2\sin \theta \cos \theta {(1-\beta \cos \theta )}^5-5\beta {\sin }^3\theta {(1-\beta \cos \theta )}^4}{{(1-\beta \cos \theta )}^{10}}=0\\ 2\cos \theta -2\beta {\cos }^2\theta =5\beta {\sin }^2\theta \end{array}$

Solve further as

$3\beta {\cos }^2\theta +2\cos \theta -5\beta =0$

Determine the angle at which the power distribution is maximum.

For $\beta \to 0$we expect$\theta \to \pi /2$, so we want the RHS to tend to zero. We can only have that for the plus sign, so we have:

${\theta }_m=\arccos \frac{\sqrt{1+15{\beta }^2}-1}{3\beta }$

We have $\beta \approx 1$ for the ultrarelativistic speeds, therefore$\beta =1-\delta$, $\delta <<1$. Increase the cosine:

Determine the value of ultrarelativistic speeds.

Substitute$(1-\delta )$ for $\beta$ into equation (2).

$\begin{array}{c}\cos \theta =\frac{\sqrt{1+15{(1-\delta )}^2}}{3(1-\delta )}\\ \approx \frac{\sqrt{1+15(1-2\delta )}-1}{3(1-\delta )}\\ =\frac{\sqrt{16-30\delta }-1}{3(1-\delta )}\\ \approx \frac{1}{3}(1+\delta )[4\sqrt{1-\frac{15}{8}\delta }-1]\end{array}$

Solve further as

$\begin{array}{c}\approx \frac{1}{3}(1+\delta )[4(1-\frac{15}{16}\delta )-1]\\ =(1+\delta )(1-\frac{5}{4}\delta )\\ \approx 1-\frac{1}{4}\delta \end{array}$

But:

$\begin{array}{c}\cos {\theta }_m\approx 1-\frac{1}{2}{\theta }_m^2\\ {\theta }_m=\sqrt{\delta /2}\\ =\sqrt{(1-\beta )/2}\end{array}$

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

$\begin{array}{c}{\left|\frac{dP}{d\Omega }\right|}_{rest}=S_{rad}\cdot \widehat{r}{r}^2\\ =\frac{{\mu }_0{q}^2{a}^2}{16{\pi }^{2}c}{\sin }^2\theta \end{array}$

Keep in mind that in this case, $\theta$ refers to the spherical angle rather than the angle closed with the velocity vector. So, in this instance, $\theta \approx \pi /2$, and:

${\left|\frac{dP}{d\Omega }\right|}_{rest}=\frac{{\mu }_0{q}^2{a}^2}{16{\pi }^{2}c}$

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

$\begin{array}{c}{\sin }^2{\theta }_m\approx {\theta }_m^2\\ =\frac{\delta }{2}\\ {(1-\beta \cos {\theta }_m)}^5={[1-(1-\delta )(1+\frac{1}{2}\frac{\delta }{2})]}^5\\ \approx {\left(\frac{5}{4}\right)}^5{\delta }^5\end{array}$

Determine the$\eta$.

$\begin{array}{c}\eta =\frac{{(dP/d\Omega )}_{UR}}{{(dP/d\Omega )}_{rest}}\\ =\frac{{\sin }^2{\theta }_m}{{(1-\beta \cos {\theta }_m)}^5}\\ ={\left(\frac{4}{5}\right)}^5\frac{1}{2{\delta }^4}\end{array}$

Now we need to determine the$\delta$ in terms of$\gamma$:

$\begin{array}{c}\gamma =\frac{1}{\sqrt{1-{\beta }^2}}\\ =\frac{1}{\sqrt{1-{(1-\delta )}^2}}\\ \approx \frac{1}{\sqrt{2\delta }}\\ \Rightarrow \delta =\frac{1}{2{\gamma }^2}\end{array}$

Determine the ratio is:

$\begin{array}{c}\eta ={\left(\frac{4}{5}\right)}^5\frac{1}{2{\delta }^4}\\ =\frac{1}{4}\frac{8^5}{5^5}{\gamma }^8\end{array}$