Question: In Ex. 11.3 we assumed the velocity and acceleration were (instantaneously,

at least) collinear. Carry out the same analysis for the case where they are

perpendicular. Choose your axes so that v lies along the z axis and a along the x axis

(Fig. 11.14), so that$\mathbf{v}\mathbf{=}\mathbf{v}\hat{\mathbf{z}}\mathbf{,}\mathbf{a}\mathbf{=}\mathbf{a}\hat{\mathbf{x}}\mathbf{,}\mathbf{and}\hat{\mathbf{r}}\mathbf{=}\mathbf{sin\theta cos}\varphi \hat{\mathbf{x}}\mathbf{+}\mathbf{sinv}\mathbf{=}\mathbf{v}\hat{\mathbf{z}}\mathbf{,}\mathbf{a}\mathbf{=}\mathbf{a}\hat{\mathbf{x}}\mathbf{,}\mathbf{and}\hat{\mathbf{r}}\mathbf{=}\mathbf{sin\theta cos}\varphi \hat{\mathbf{x}}\mathbf{+}\mathbf{sinsin}\varphi \hat{\mathbf{y}}\mathbf{+}\mathbf{cos\theta }\hat{\mathbf{z}}$Check that P is consistent with the Lienard formula.

$\frac{\mathbf{dP}}{\mathbf{d\Omega }}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{16}{\mathbf{\pi }}^{\mathbf{2}}\mathbf{c}}\frac{\left[{\left(\mathbf{1}\mathbf{-}\mathbf{\beta cos\theta }\right)}^{\mathbf{2}}\mathbf{-}\left(\mathbf{1}\mathbf{-}{\mathbf{\beta }}^{\mathbf{2}}\right){\mathbf{sin}}^{\mathbf{2}}{\mathbf{\theta cos}}^{\mathbf{2}}\varphi \right]}{{\left(\mathbf{1}\mathbf{-}\mathbf{\beta cos\theta }\right)}^{\mathbf{5}}}\mathbf{,}\mathbf{P}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}{\gamma }^{\mathbf{4}}}{\mathbf{6}\mathbf{\pi c}}$[Answer: .For relativistic velocities ( $\mathbf{\beta }\approx \mathbf{1}$) the radiation is again sharply peaked in the forward

direction (Fig. 11.15). The most important application of these formulas is to circular motion-in this case the radiation is called synchrotron radiation. For a relativistic electron, the radiation sweeps around like a locomotive's headlight as the particle moves.]

Consider the formula for the angular distribution of the power as:

$\frac{\mathbf{dP}}{\mathbf{d\Omega }}\mathbf{=}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{16}{\mathbf{\pi }}^{\mathbf{2}}{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{{\left|\hat{\mathbf{r}}\mathbf{\times }\left(\mathbf{u}\mathbf{\times }\mathbf{a}\right)\right|}^{\mathbf{2}}}{{\left(\hat{\mathbf{r}}·\mathbf{u}\right)}^{\mathbf{5}}}$

Here, the terms are:

$$\begin{gathered} \mathbf{u}=x\hat{r}-v\hat{z} \\ \mathbf{a}=a\hat{x} \\ \hat{\mathbf{r}}=\sin \theta \cos \varphi \hat{x}+\mathrm{sinsin}\varphi \hat{y}+\cos \theta \hat{z}\backslash \end{gathered}$$

Determine the terms in the second fraction as:

$$\begin{gathered} d\hat{r}·u=c-v\cos \theta \\ =c\left(1-\beta \cos \theta \right) \end{gathered}$$ …. (1)

Solve further as:

$$\begin{gathered} {\left|\hat{r}\times \left(u\times a\right)\right|}^2={\left|u\left(\hat{r}·a\right)-a\left(\hat{r}·u\right)\right|}^2 \\ {\left|\hat{r}\times \left(u\times a\right)\right|}^2={\left|ua\sin \theta \cos \varphi -ac\left(1-\beta \cos \theta \right)\hat{x}\right|}^2 \\ {\left|\hat{r}\times \left(u\times a\right)\right|}^2=a^2{c}^2{\left(1-\beta \cos \theta \right)}^2-\left(1-\beta \right){\sin }^2\theta {\cos }^2\varphi \end{gathered}$$ …… (2)

From equation (1) and (2) determine the formula for the angular power as:

$$\begin{gathered} \frac{dP}{d\Omega }=\frac{q^2}{16{\pi }^2{\epsilon }_0}\frac{a^2{c}^2\left[{\left(1-\beta \cos \theta \right)}^2-\left(1-{\beta }^2\right){\sin }^2\theta {\cos }^2\varphi \right]}{c^3{\left(1-\beta \cos \theta \right)}^5} \\ \frac{dP}{d\Omega }=\frac{{\mu }_0{q}^2{a}^2}{16{\pi }^{2}c}\frac{\left[{\left(1-\beta \cos \theta \right)}^2-\left(1-{\beta }^2\right){\sin }^2\theta {\cos }^2\varphi \right]}{{\left(1-\beta \cos \theta \right)}^5}\left\{{\mu }_0{\epsilon }_0=\frac{1}{c^2}\right\} \end{gathered}$$

In order to determine the total power integrate over the space angle

$$\begin{gathered} d\Omega =d\theta \sin \theta \varphi asfollows \\ P=\frac{{\mu }_0{q}^2{a}^2}{16{\pi }^{2}c}{\int }_0^{2\pi }{\int }_0^{\pi }\frac{\left[{\left(1-\beta \cos \theta \right)}^2-\left(1-{\beta }^2\right){\sin }^2\theta {\cos }^2\varphi \right]}{{\left(1-\beta \cos \theta \right)}^5}\sin \theta d\theta d\varphi \end{gathered}$$

Using ${\int }_0^{2\pi }{\cos }^2\varphi =\pi$integrate over as follows:

$$\begin{gathered} P=\frac{{\mu }_0{q}^2{a}^2}{16{\pi }^{2}c}{\int }_0^{\pi }\frac{\left[2{\left(1-\beta \cos \theta \right)}^2-\left(1-{\beta }^2\right){\sin }^2\theta \right]}{{\left(1-\beta \cos \theta \right)}^5}\sin \theta d\theta \\ =\frac{{\mu }_0{q}^2{a}^2}{16{\pi }^{2}c}\left(J\right) \end{gathered}$$

Solve for the integral J as follows:

$$\begin{gathered} J={\int }_{1-\beta }^{1+\beta }\frac{\left[2u^2-\left(1-{\beta }^2\right)\left(1-\left(1-2u+\frac{u^2}{{\beta }^2}\right)\right)\right]}{{\left(1-\beta \cos \theta \right)}^5}\sin \theta d\theta \left\{1-\beta \cos \theta =u,{\sin }^2\theta =1-\frac{1-2u+u^2}{{\beta }^2},\backslash \mathrm{hfill} \\ du=\beta \sin \theta d\theta \right\} \\ =\frac{1}{{\beta }^3}{\left[-{\left(1-{\beta }^2\right)}^2\frac{1}{4u^4}+2\left(1-{\beta }^2\right)\frac{1}{3u^3}-\left(1+{\beta }^2\right)\frac{1}{2u^2}\right]}_{1+\beta }^{1+\beta } \end{gathered}$$

Solve for the terms in the bracket as:

First term:

$$\begin{gathered} {\left.-\left(1+{\beta }^2\right)\frac{1}{2u^2}\right|}_{1+\beta }^{1+\beta }=-\frac{1+{\beta }^2}{2}\frac{-4\beta }{{\left(1-\beta \right)}^2{\left(1+\beta \right)}^2} \\ =\frac{2\beta {\left(1+\beta \right)}^2}{{\left(1-{\beta }^2\right)}^2} \end{gathered}$$

Second term:

$$\begin{gathered} {\left.2\left(1-{\beta }^2\right)\frac{1}{3u^3}\right|}_{1+\beta }^{1+\beta }=\frac{2}{3}\left(1-{\beta }^2\right)\frac{-6\beta -2{\beta }^3}{{\left(1-\beta \right)}^3{\left(1+\beta \right)}^3} \\ =\frac{4\beta {\left(3+\beta \right)}^2}{3{\left(1-{\beta }^2\right)}^2} \end{gathered}$$

Third term:

$$\begin{gathered} {\left.-\left(1-{\beta }^2\right)\frac{1}{4u^4}\right|}_{1+\beta }^{1+\beta }=\frac{\left(1-{\beta }^2\right)}{4}\frac{8\beta +8{\beta }^3}{{\left(1-\beta \right)}^4{\left(1+\beta \right)}^4} \\ =\frac{2\beta \left(1+{\beta }^2\right)\beta }{{\left(1-{\beta }^2\right)}^2} \end{gathered}$$

Rewrite the integral as:

$$\begin{gathered} J=\frac{1}{3{\left(1-{\beta }^2\right)}^2}\left[2\left(1+{\beta }^2\right)\beta -\frac{4}{3}\beta \left(3+{\beta }^2+2\beta \left(2+{\beta }^2\right)\right)\right] \\ =\frac{8}{3{\left(1-{\beta }^2\right)}^2} \\ =\frac{8}{3}{\gamma }^4\left\{\gamma ={\left(1-{\beta }^2\right)}^{\frac{-1}{2}}\right\} \end{gathered}$$

Therefore, the expression for the power is written as:

$$\begin{gathered} P=\frac{{\mu }_0{q}^2{a}^2}{16\pi c}\left(\frac{8}{3}{\gamma }^4\right) \\ =\frac{{\mu }_0{q}^2{a}^2}{6\pi c}{\gamma }^4 \end{gathered}$$ ….. (3)

From the Lienard formula solve as:

$$\begin{gathered} P=\frac{{\mu }_0{q}^2{\gamma }^6}{6\pi c}\left(a^2-{\left|\frac{v\times a}{c}\right|}^2\right) \\ =\frac{{\mu }_0{q}^2{\gamma }^6}{6\pi c}\left(a^2-a^2{\beta }^2\right)\left\{\left|v\times a\right|=va\right\} \\ =\frac{{\mu }_0{q}^2{a}^2}{6\pi c}{\gamma }^4 \end{gathered}$$$$\begin{gathered} P=\frac{{\mu }_0{q}^2{\gamma }^6}{6\pi c}\left(a^2-{\left|\frac{v\times a}{c}\right|}^2\right) \\ =\frac{{\mu }_0{q}^2{\gamma }^6}{6\pi c}\left(a^2-a^2{\beta }^2\right)\left\{\left|v\times a\right|=va\right\} \\ =\frac{{\mu }_0{q}^2{a}^2}{6\pi c}{\gamma }^4 \end{gathered}$$ …… (4)

Since, the velocity and acceleration are perpendicular to each other so the equation (3) and (4) agrees:

Therefore, the Lienard formula is

$\frac{dP}{d\Omega }=\frac{{\mu }_0{q}^2{a}^2}{16{\pi }^{2}c}\frac{\left[{\left(1-\beta \cos \theta \right)}^2-\left(1-{\beta }^2\right){\sin }^2\theta {\cos }^2\varphi \right]}{{\left(1-\beta \cos \theta \right)}^5}$

and the power found by integrating and the distribution agree to each other.