Non-relativistic speed $\beta\ll1$

💼 Case: accelerating point charge

$\vec E$ field $\perp$

$\vec B$ field out of the page

$\vec S$ is outwards

• Lets explicitly calculate the pointing vector $\vec S$, first we find $\vec E \times \vec B$

  $$ \begin{aligned} \vec E \times \vec B &= \frac 1c \vec E \times (\hat R \times \vec E)=\frac 1c [\hat R \overbrace{(\vec E \cdot \vec E)}^{E^2}-\overbrace{\vec E (\vec E \cdot \hat R)}^{=\;0} ] \\ \Rightarrow \qquad \hat S_\text{rad}&=\frac{1}{\mu_0 c} E^2 [\hat R ]_\text{ret} \end{aligned} $$

  • we used the vector identity (expand) and $_\text{rad}$ means the part of $\vec S$ associated with radiation

  🗒️ Note: the radiation is centered on the retarded position for acceleration

Why does radiation appear when $\dot \beta \ne 0$

💼 Case: $\dot \beta =0$ and $0<\beta\ll1$

Since the potential fields move along with the charge then even if $\vec S=\frac 1 {\mu_0} \vec E \times \vec B \ne 0$ for one frame then there exists a frame in which the charge is stationary and the Poynting vector would be $0$

• More quantitively lets look at the total flux of energy though a spherical surface
  • For $\dot \beta =0$ we have $E,B \propto \frac{1}{R^2}$ so $S\propto \frac{1}{R^4}$
  • The flux is thus $\oint \vec S\cdot \text d \vec a\propto \oint \frac{1}{R^4}R^2\,\text d \Omega \to 0$ as $R\to \infin$

💼 Case: $\dot\beta \ne 0$ and $0<\beta \ll 1$

• Again looking at the total flux of energy through a spherical surface
  • For $\beta\ne 0$ we have $E,B\propto \frac{1}{R}$ so $S\propto \frac{1}{R^2}$
  • The flux is thus $\oint \vec S\cdot \text d \vec a\propto \oint \frac{1}{R^2}R^2\,\text d \Omega \to \text{constant}$ as $R\to \infin$

💎 Conclusion: here we expect radiation of photons from the charge

Calculate the radiation

🧠 Remember: our equation for the $\vec E$ field

$$ E_T=\frac{q}{4\pi \epsilon_0 c} \frac{\dot {\vec \beta} \sin \theta }{R} $$

• Using this on a pointing vector we have

  $$ \begin{aligned} \vec S & = \frac{1}{\mu_0 c} E^2 [\hat R]\text{ret}=\frac{1}{\mu_0 c} \frac{q}{16\pi ^2 \epsilon_0^2}\frac{\sin^2 (\theta) \dot \beta^2}{c^2 R^2} [\hat R]\text{ret}\\ & = \frac{\mu_0 c q^2}{16 \pi^2} \frac{\sin^2 (\theta) \dot \beta ^2}{R^2} [\hat R]_\text{ret} \end{aligned} $$

  $\vec S$ gives us the radiated power (flux) per unit area

• $R^2 \vec S$ gives power per unit solid angle

  $$ \begin{aligned} R^2 \vec S=\frac{\text d P}{\text d \Omega}=\frac{\mu_0 c q^2}{16 \pi^2}\sin^2 (\theta )\dot \beta ^2[\hat R]_\text{ret} \end{aligned} $$

  which is independent of $R$ as expected

  Plotting this result in 2D gives the picture, and since it is symmetric to the direction of acceleration it looks like a torus or donut in 3D

• We can also calculate the total rate with which energy is radiated ie the power

  $$ \begin{aligned} &\;P=\int \frac{\text d P}{\text d \Omega} \,\text d \Omega =\frac{\mu_0cq^2 \dot \beta^2 }{16\pi^2}\overbrace{\int^{2\pi}_0 \,\text d \phi}^{=\;2\pi}\; \overbrace{\int ^\pi_0 \sin^2 \theta \sin \theta \,\text d \theta}^{=\;\frac{4}{3}} \\ &\boxed{P=\frac{\mu_0 c q^2 \dot \beta^2}{6\pi}} \end{aligned} $$

  which is the Larmor Formula

Relativistic speeds $\beta\sim1$

To convert our current equations of $P$ and $\frac {\text dP}{\text d \Omega}$ which are valid in there rest frame $S'$ into relativistic equations we can imagine a frame $S$ moving at relativistic speed and lorentz transform the equations to get what an observer in $S$ would experience, and thanks to the symmetry of the situation it would give us the general relativistic equations of $P$ and $\frac{\text dP}{\text d \Omega}$

Total radiated power

$$ P'=\frac{\text d \mathcal E'}{\text dt'}=\frac{\mu_0 c q^2}{6\pi} (\dot \beta')^2 $$