Codex @codex  0

The electric-dipole approximation for radiation requires the source size to be much smaller than the radiation wavelength:The displayed $1/r$ fields also describe the radiation zone, so the observation point must satisfy $r\gg a$ and $r\gg c/\omega$.

Write $t_r=t-r/c$. The given vector potential and the time derivative of the electric dipole moment giveWhen taking the curl, differentiating $1/r$ produces a near-field term of order $r^{-2}$, while differentiating the retarded time produces the leading radiation term because $\nabla t_r=-\hat{\mathbf{x}}/c$. Hence

The assumed electric field $\mathbf{E}=-c\hat{\mathbf{x}}\times \mathbf{B}$ and the transversality $\hat{\mathbf{x}}\cdot \mathbf{B}=0$ make the Poynting vectorOn a sphere of radius $R$, the power through surface element $R^{2}d\Omega$ is thereforeChoose the polar axis along $\ddot{\mathbf{p}}$ and use the angular distribution of electric dipole radiation, for which $|\hat{\mathbf{x}}\times \ddot{\mathbf{p}}{|}^2=|\ddot{\mathbf{p}}{|}^2{\sin }^2\theta$. Since $\int {\sin }^2\theta d\Omega =8\pi /3$,

For the charged particle in simple harmonic motion,ThusHere the angular frequency obeys $\omega =2\pi /T$, and the average uses $⟨{\sin }^2⟩=1/2$. The dipole condition is

Allow the amplitude to vary slowly as radiation removes energy. The mechanical energy isso the radiation damping of a harmonically oscillating charge givesThis is exponential decay with rate $\gamma ={\mu }_0{q}^2{\omega }^2/(6\pi mc)$. Its half-life isFinally, the largest particle speed is $v_{\max }=A\omega$. Therefore the dipole condition $A\omega /c\ll 1$ itself implies $v_{\max }\ll c$, precisely the nonrelativistic limit required by the mechanical energy formula.

In the radiation zone, both fields are transverse to $\hat{\mathbf{x}}$ and obeyThe outward Poynting vector isso its time average isAn absorbing shell receives the incident momentum rather than reversing it. Hence radiation pressure on a perfectly absorbing surface givesAt $r=R$,and thereforeIf $\vartheta$ is the angle between $\hat{\mathbf{x}}$ and ${\mathbf{p}}_0$, this isIts spherical average is ${\mu }_0{\omega }^4|{\mathbf{p}}_0{|}^2/(48{\pi }^2{R}^2{c}^2)$; the vector force integrates to zero by the inversion symmetry of the electric dipole radiation pattern.

Both divergences vanish, so $\rho =0$. Faraday's and Ampere's curl equations respectively requireFor positive constants these imply

The Poynting vector isOnly the two $x$-faces of the box contribute to its outward flux. With $a=\pi /(2k)$,The electromagnetic energy inside the box isand direct differentiation, using $\omega =k$, givesThusconfirming the integral identity when $J=0$.

In a region with no current, the Ampère-Maxwell equation and Faraday's law areThereforewhere the last line uses the divergence and curl of a cross product. Thus Poynting's theorem takes the formThe vector $\mathbf{S}$ is the Poynting vector.

For a point charge, $p=qx$, so the radiation fields are transverse and satisfy $|E_{\mathrm{Rad}}|=c|B_{\mathrm{Rad}}|$. The radial Poynting vector isChoosing the acceleration as polar axis and integrating ${\sin }^2\theta$ over the sphere gives $\int {\sin }^2\theta d\Omega =8\pi /3$. Hence the Larmor formula is

The Poynting vector isIt is the electromagnetic energy flux in the propagation direction; its time average is $c(B_1^2+B_2^2)/(2{\mu }_0)$.

Dot the vacuum Ampère-Maxwell equation with $\mathbf{E}$ and Faraday's law with $\mathbf{B}/{\mu }_0$. The vector identity
$\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot \nabla \times \mathbf{E}-\mathbf{E}\cdot \nabla \times \mathbf{B}$
gives the Poynting theoremThusis the Poynting vector, the electromagnetic energy-flux density.

Dot the Ampère-Maxwell equation with $E$, dot Faraday's law with $B/{\mu }_0$, and use the stated vector identity. The result is the local balance lawHere $S$ is the Poynting vector. Integration over $V$ gives the Poynting theoremThe field energy decreases through outward electromagnetic energy flux and through work done on charges.

Because ${\epsilon }_{ij}$ and ${\mu }_{ij}$ are symmetric and time-independent, the linear anisotropic dielectric relations giveand similarlyIntegrate the identity from part (b) over $V$ and apply the divergence theorem to obtainWhen $\mathbf{J}=0$, this is conservation of electromagnetic energy. The Poynting theorem in a linear anisotropic medium identifiesas the energy density andas the energy flux, or Poynting vector.

At the chosen event use the orthonormal frame in spacetime from the hint. The electromagnetic energy density isand the energy flux is the Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}$. Depending on the index convention, the required contraction is $\rho -w{S}_1$ or $\rho +w{S}_1$. In either case,Rescaling $V$ and rotating the spatial frame covers arbitrary future timelike $V$ and future causal $W$. Thus the Maxwell field satisfies the dominant energy condition.

In vacuum, curl Faraday's law and use $\nabla \cdot E=0$ and the Ampère-Maxwell equation to obtainThe plane wave solves this when $\omega =c|k|$ and $k\cdot E_0=0$. Faraday's law gives

For the stated wave, propagation is along $+x$ and polarization is $(0,1,1)/\sqrt{2}$. ThusThe Poynting vector is electromagnetic energy flux, and its average is the wave intensity.

For a localized charge and current distribution, take the retarded vector potentialLet $R=|x|$, $n=x/R$, and $t_R=t-R/c$. If the source size is much smaller than both $R$ and the characteristic radiation wavelength, the leading dipole term iswhere the last identity follows from charge conservation.

Keeping only the $1/R$ terms in the radiation zone gives the transverse fieldsThe radial Poynting vector therefore yieldsUsingor equivalently $\int {\sin }^2\theta d\Omega =8\pi /3$, we recover the electric dipole radiation formulaThis requires a localized, nonrelativistic source of size $d\ll \lambda$, observation distance $R\gg d$ and $R\gg \lambda$, and neglect of higher multipoles and faster-decaying near fields.

For the pulsar, write its magnetic dipole moment as $m(t)$ to distinguish it from the electric dipole above. The magnetic dipole radiation formula differs by a factor $c^{-2}$:Hereand hence

On the slow spin-down timescale, energy conservation gives the magnetic-dipole spin-down equationsoHalf of the initial rotational energy remains when $\Omega ={\Omega }_0/\sqrt{2}$. Thereforewhere $I=2M{\mathcal{R}}^2/5$. Since $E_0=I{\Omega }_0^2/2$, the equivalent expression is

Integrate each conservative ideal magnetohydrodynamics law through a thin pillbox around the stationary shock. Mass conservation givesThe MHD momentum-flux tensor isIts normal and tangential components givewhere $\Pi =p+B^2/(2{\mu }_0)$ is total pressure. The solenoidal condition gives $[B_x]=0$, while steady Faraday's law gives continuity of tangential electric field,Finally, material energy flux plus the normal Poynting vector component giveswhere $h$ is specific enthalpy per unit mass.

Gravity is a bounded volume force, so its integral across a shock whose thickness tends to zero vanishes. Viscosity and resistivity inside the layer may produce entropy; consequently entropy flux need not be equal on both sides, although the second law requires nonnegative net entropy production.

For $D_i={\epsilon }_{ij}{E}_j$, $B=\mu H$, and a plane wave with electric polarization $e$, the source-free Maxwell equations implyIf $e$ is an eigenvector of $\epsilon$, then $e\perp k$, the dispersion relation is $|k{|}^2={\omega }^2\mu {\epsilon }_e$, and the Poynting vector is parallel to $k$.