91影视

Consider themaximum angle of radiation emitted is${胃}_{\text{max}}鈮�\sqrt{(\text{1}鈭�尾)/\text{2}}$.

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

$\mathbf{cos}\mathit{胃}\mathbf{=}\frac{\mathbf{卤}\sqrt{\mathbf{1}\mathbf{+}\mathbf{15}{\mathbf{尾}}^2}}{\mathbf{3}\mathbf{尾}}$鈥︹�� (1)

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

Write the formula ofultrarelativistic speed.

$\mathbf{cos}\mathit{胃}\mathbf{=}\frac{\sqrt{\mathbf{1}\mathbf{+}\mathbf{15}{\mathbf{尾}}^2}}{\mathbf{3}\mathbf{尾}}$ 鈥︹�� (2)

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

Write the formula of$畏$.

$\mathit{畏}\mathbf{=}{\mathbf{\left(}\frac{4}{5}\mathbf{\right)}}^5\frac{1}{\mathbf{2}{\mathbf{未}}^4}$ 鈥︹�� (3)

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

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

$\begin{array}{l}\frac{d}{d胃}\frac{dP}{d惟}=0\\ 鈬�\frac{d}{d胃}\frac{{\sin }^2胃}{{(1鈭�尾\cos 胃)}^5}=0\\ =\frac{2\sin 胃\cos 胃{(1鈭�尾\cos 胃)}^5鈭�5尾{\sin }^3胃{(1鈭�尾\cos 胃)}^4}{{(1鈭�尾\cos 胃)}^{10}}=0\\ 2\cos 胃鈭�2尾{\cos }^2胃=5尾{\sin }^2胃\end{array}$

Solve further as

$3尾{\cos }^2胃+2\cos 胃鈭�5尾=0$

Determine the angle at which the power distribution is maximum.

For $尾鈫�0$we expect$胃鈫�蟺/2$, so we want the RHS to tend to zero. We can only have that for the plus sign, so we have:

${胃}_m=\arccos \frac{\sqrt{1+15{尾}^2}鈭�1}{3尾}$

We have $尾鈮�1$ for the ultrarelativistic speeds, therefore$尾=1鈭�未$, $未<<1$. Increase the cosine:

Determine the value of ultrarelativistic speeds.

Substitute$(1鈭�未)$ for $尾$ into equation (2).

$\begin{array}{c}\cos 胃=\frac{\sqrt{1+15{(1鈭�未)}^2}}{3(1鈭�未)}\\ 鈮�\frac{\sqrt{1+15(1鈭�2未)}鈭�1}{3(1鈭�未)}\\ =\frac{\sqrt{16鈭�30未}鈭�1}{3(1鈭�未)}\\ 鈮�\frac{1}{3}(1+未)[4\sqrt{1鈭�\frac{15}{8}未}鈭�1]\end{array}$

Solve further as

$\begin{array}{c}鈮�\frac{1}{3}(1+未)[4(1鈭�\frac{15}{16}未)鈭�1]\\ =(1+未)(1鈭�\frac{5}{4}未)\\ 鈮�1鈭�\frac{1}{4}未\end{array}$

But:

$\begin{array}{c}\cos {胃}_m鈮�1鈭�\frac{1}{2}{胃}_m^2\\ {胃}_m=\sqrt{未/2}\\ =\sqrt{(1鈭�尾)/2}\end{array}$

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

$\begin{array}{c}{\left|\frac{dP}{d惟}\right|}_{rest}=S_{rad}鈰�\widehat{r}{r}^2\\ =\frac{{渭}_0{q}^2{a}^2}{16{蟺}^{2}c}{\sin }^2胃\end{array}$

Keep in mind that in this case, $胃$ refers to the spherical angle rather than the angle closed with the velocity vector. So, in this instance, $胃鈮�蟺/2$, and:

${\left|\frac{dP}{d惟}\right|}_{rest}=\frac{{渭}_0{q}^2{a}^2}{16{蟺}^{2}c}$

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

$\begin{array}{c}{\sin }^2{胃}_m鈮�{胃}_m^2\\ =\frac{未}{2}\\ {(1鈭�尾\cos {胃}_m)}^5={[1鈭�(1鈭�未)(1+\frac{1}{2}\frac{未}{2})]}^5\\ 鈮�{\left(\frac{5}{4}\right)}^5{未}^5\end{array}$

Determine the$畏$.

$\begin{array}{c}畏=\frac{{(dP/d惟)}_{UR}}{{(dP/d惟)}_{rest}}\\ =\frac{{\sin }^2{胃}_m}{{(1鈭�尾\cos {胃}_m)}^5}\\ ={\left(\frac{4}{5}\right)}^5\frac{1}{2{未}^4}\end{array}$

Now we need to determine the$未$ in terms of$纬$:

$\begin{array}{c}纬=\frac{1}{\sqrt{1鈭�{尾}^2}}\\ =\frac{1}{\sqrt{1鈭�{(1鈭�未)}^2}}\\ 鈮�\frac{1}{\sqrt{2未}}\\ 鈬�未=\frac{1}{2{纬}^2}\end{array}$

Determine the ratio is:

$\begin{array}{c}畏={\left(\frac{4}{5}\right)}^5\frac{1}{2{未}^4}\\ =\frac{1}{4}\frac{8^5}{5^5}{纬}^8\end{array}$