Ohm’s Law

  • $\vec{J} = \sigma (\vec{E}+v\times \vec{B}) \approx \sigma \vec{E}$

    • $J$ is the current density
    • $\sigma$ is the conductivity
    • $\vec{E}$ is the electric field
    • Microscopic description
    • typically, the velocities are so slow that we can drop the magnetic field term

  • $\vec{I} = \int \vec{J}\cdot \vec{dA}$

  • $V = IR = I \frac{L}{\sigma A}$

    • Macroscopic description
    • $ \frac{1}{\sigma} = \eta$ is called the resistivity
    • Can think of resistance R as water flowing through pipe: Higher cross section means less resistance. Longer length means more resistance

  • conductivity varies dramatically across materials. (Fun Fact: Silver, at room temperature, has the highest conductivity)

Drude Model

  • Assumes that electrons are accelerated by electric field,but frequent collisions allow steady state velocity (think skydiver reaching terminal velocity)
  • There is an average distance of $\lambda$ between collisions, and the speed of the electrons follows a Boltzmann distribution. Hence, the collision time $\tau = \frac{\lambda}{v}$
  • The average velocity is then $\frac{a \tau}{2}$, where $a = \frac{qE}{m}$
  • Hence, $J = nqv_{avg} = \frac{nq^2\lambda}{2m v_{thermal}}E = \sigma E$

Flux

  • $\Phi = \int \vec{B}\cdot d\vec{a}$

  • $\frac{\partial \Phi}{\partial t} = -\mathcal{E} = - \int \vec{E}\cdot \vec{dl}$

    • In words, a change in flux creates an electromotive force (EMF ie. $\mathcal{E}$) that creates an opposing flux
    • Recall that $\mathcal{E} = \int_{a}^{b} \frac{\vec{F}(r)}{q}\cdot \vec{dr}$.
      • In words: EMF is the integrated force per charge summed along the the path between a and b
    • Lenz’s law: nature abhors a change in flux
      • ie. current will be produced to counteract magnetic flux to counteract a change in flux
    • One caveat to this is is that only a single loop can be under consideration
      • If the current has multiple defined paths, flux rule breaks down

  • Demonstration is pendulum in magnetic field. Moving pendulum in field creates an emf that slows the pendulum down

    • Cutting slots in the pendulum greatly reduces enclosed area, which reduces the possible change in flux

Maxwell’s Equations

  • $\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_{0}}$

  • $\nabla \cdot \vec{B} = 0$

  • $\nabla \times \vec{E} = -\frac{d \vec{B}}{dt}$

  • $\nabla \times \vec{B} = \mu_{0}\vec{J}+\frac{1}{c^2}\frac{d \vec{E}}{dt}$

    • $\frac{1}{c^2} = \mu_{0}\epsilon_{0}$
    • For low frequency stuff, can occasionally drop $\frac{1}{c^2}\frac{d \vec{E}}{dt}$ term (quasi-static process )
      • $J_{d} = \epsilon_{0} \frac{d \vec{E}}{dt}$ is called the displacement current

Mutual Inductance

  • $\Phi_{2} = M_{12} I_{1}$

    • In words, the flux induced in the second loop is proportional to the current in the first loop
    • $M_{21} = \frac{\mu_{0}}{4 \pi}\int \int \frac{dl_{1} \cdot dl_{2}}{|\mathcal{R}|}$
      • $\mathcal{R} = r-r^{’}$
      • By symmetry, $M_{12} = M_{21}$

  • $M_{12} = \frac{\mu_{0}}{4\pi} \oint \oint \frac{dl_{1}\cdot dl_{2}}{(r_{1}-r_{2})^{0.5}}$

    • Neumann’s formula for self inductance
    • Derived from substituting in vector potential into flux equation, using Stokes, and rearranging

Self Inductance

  • $\Phi = IL$

    • L has units of $\frac{V\cdot s}{A}$ (H for henries)
    • Inductance of loop where it interacts with it’s own field
    • Assuming a fixed geometry: $V = -L \frac{d I}{dt}$

  • Adding in self inductance to the two loop problem:

    • $\Phi_{1} = L_{1}I_{1}+M$
    • $\Phi_{2} = MI_{1}+L_{2}I_{2}$
    • Can solve for eigenvalues and vectors to find stable currents

Energy in Fields

  • Energy stored in an inductor is $W = \frac{1}{2}LI^{2}$

  • Energy stored in an capacitor is $W = \frac{1}{2}\frac{Q^{2}}{C}$

  • $W_{mag} =\frac{1}{2\mu_{0}}\int\int\int B^{2} dV = \int \frac{V \rho}{2} d\tau$

    • You derive the above by taking the vector potential version, substituting in Ampere’s law, doing integration by parts with vector identities, and then taking the volume to be all of space

  • $W_{elec} = \frac{\epsilon_{0}}{2}\int\int\int E^{2} dV = \int \frac{\vec{A} \cdot \vec{J}}{2} d\tau$

Maxwell’s Equation in Matter

  • $\nabla \cdot \vec{D} = \rho_{f}$

  • $\nabla \cdot \vec{B} = 0$

  • $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$

  • $\nabla \times \vec{H} = \vec{J_{f}}+\frac{\partial \vec{D}}{\partial t}$

Electric

  • $\rho_{b} = -\nabla \cdot \vec{P}$

    • P is the polarization

  • $\nabla \cdot \vec{D} = \rho_{free}$

    • $\vec{D} = \epsilon \vec{E} +\vec{P} = \epsilon \vec{E}$
      • $\epsilon = \epsilon_{0}(1+\chi_{e})$ (assuming linear material)
      • Called permittivity of material
    • $\vec{D}$ is the displacement current
    • $\vec{P} = \epsilon_{0}\chi_{e}\vec{E}$ (linear)

Magnetic

  • $\vec{J_{b}} = \nabla \times \vec{M}+ \frac{\partial \vec{P}}{\partial t}$

    • M is the magnetization
    • The partial derivative term is the polarization current

  • $\nabla \times \vec{B} = \mu_{0}(\vec{J_{f}}+\nabla\times \vec{M}+ \frac{\partial \vec{P}}{\partial t})+\mu_{0}\epsilon_{0}\frac{\partial \vec{E}}{\partial t}$

    • Possible sources of magnetic fields: free currents, bound currents, polarization current, and displacement current

  • Rearranging to get: $\nabla \times \vec{H} = J_{f}+ \frac{\partial D}{\partial t}$

    • $H = \frac{B}{\mu_{0}}-M$
      • $H = \frac{1}{\mu}B$ (linear material)
      • $\vec{M} = \chi_{m}\vec{H}$ (linear)

Boundary Conditions

  • $D_{1}^{\perp}-D_{2}^{\perp} = \sigma_{f}$

  • $B_{1}^{\perp}-B_{2}^{\perp} = 0$

  • $E_{1}^{\parallel} - E_{2}^{\parallel} = 0$

  • $H_{1}^{\parallel}-H_{2}^{\parallel} = \vec{K_{f}}\times \vec{n}$

Continuity Equations

Charge

  • $\frac{\partial}{\partial t} \rho = -\nabla \cdot \vec{J}$
    • J is constrained by charge density, and vice versa

Energy

  • Total energy density is $u = \frac{\epsilon_{0}}{2} E^{2} + \frac{1}{2 \mu_{0}} B^{2} $ were we integrate over all of space

  • To calcuate conservation equation:

    • Start with Lorentz law and take dot product with $\vec{dl} = \vec{v} dt$
      • $\vec{F}\cdot \vec{dl} = q(\vec{E}+\vec{v}\times \vec{B}) \cdot \vec{v} dt = dW$
    • $\frac{dW}{dt} = \vec{E}\cdot q\vec{v} \rightarrow \frac{\partial W}{\partial t} = \int \vec{E}\cdot \vec{J} d\tau$
    • Write $\vec{E}\cdot \vec{J}$ in terms of E and B:
      • $\mu_{0} \vec{J} = \nabla \times \vec{B} - \epsilon_{0}\mu_{0} \frac{\partial \vec{E}}{\partial t}$ (Ampere’s law)
      • Substitute into above, use identity $\vec{E}\cdot \frac{\partial \vec{E}}{\partial t} = \frac{1}{2}\frac{\partial }{\partial t}(\vec{E})^{2}$
      • Use one of the front cover identities to massage other term, apply the above identity to the B field, and applying divergence theorem to cross product term, we find that:
        • $\frac{dW}{dt} = \frac{-d}{dt}\int \frac{B^2}{2\mu_{0}}+\frac{\epsilon_{0}E^2}{2} - \oint_{S} \frac{\vec{E}\times\vec{B}}{\mu_{0}} = \frac{-d}{dt}\int u d\tau - \oint_{S} \vec{S} \cdot d\vec{a} $
          • $\vec{S} = \frac{\vec{E}\times\vec{B}}{\mu_{0}}$ is called the Poynting vector. This states that flow of energy is perpendicular to both electric field and magnetic field
        • In a region without work being done on a charge (say, in a region were there is no charge), the equations reduce to:
          • $\frac{\partial u}{\partial t} = -\nabla \cdot \vec{S}$
    • The energy of the fields alone is NOT conserved. You need to take into account field and matter energy together

Momentum

  • Momentum is much worse to derive (Similar to above. Bigger pain to do). Final result:
    • $f = \nabla \cdot T - \epsilon_{0}\mu_{0} \frac{\partial \vec{S}}{\partial t}$
      • f is the force per unit volume
      • $T_{ij} = \epsilon_{0}(E_{i}E_{j}-\frac{1}{2} \delta_{ij} E^{2}) +\frac{1}{\mu_{0}} (B_{i}B_{j} -\frac{1}{2}\delta_{ij}B^{2})$
    • Momentum density is defined as $\vec{g} = \epsilon_{0}\mu_{0} \vec{S}$
    • $\frac{\vec{g}}{\partial t} = \nabla \cdot T$
    • Therefore, momentum is just integrating $\vec{g}$ over all of space

Angular Momentum

  • $\vec{l = \vec{r}\times\vec{g}}$ defines the angular momentum density associated with an EM field

Waves

  • $\frac{\partial^{2} f}{\partial z^{2}} = \frac{1}{c^2}\frac{\partial^{2} f}{\partial t^2}$ were c is the speed of propagation of the wave

    • Solutions take the form of $f = Re[A e^{i(\vec{k}\cdot\vec{r}-\omega t)+\phi}]$
      • $k_{i} = \frac{2\pi}{\lambda_{i}}$
      • $\omega = 2\pi f$
      • Period T is $\frac{2\pi}{kc}$
      • The dispersion relation is is $\frac{\omega}{k} = c$

  • When light passes between interfaces, the frequency stays the same while the wavelength get’s modulated

Basic Waves

  • $n = \sqrt{\frac{\epsilon \mu}{\epsilon_{0} \mu_{0}}} = \frac{c}{v}$

Polarization

  • Refers to the geometry in which the oscillations are restricted to

  • Suppose that we have a wave traveling in the $\hat{z}$ direction. A polarized wave would have the form

    • $f = A e^{i(kz-\omega t)}\hat{n}$
    • $\hat{n} = \cos \theta \hat{x}+ \sin\theta \hat{y}$ $\hat{n}$ refers to the plane in which the wave is restricted to

EM Waves

  • Can tease out two wave equations via some vector calculus kung-fu. Then end result is:

    • $\nabla^{2} \vec{E} = \frac{1}{c^2}\frac{\partial^{2} \vec{E}}{\partial t^2}$
    • $\nabla^{2} \vec{B} = \frac{1}{c^2}\frac{\partial^{2} \vec{B}}{\partial t^2}$
    • Maxwell’s equations imposes the additional constraints:
      • that $\vec{E}\cdot\vec{B} = 0$ (ie. the fields are transverse to each other)
        • Equivalently: $\hat{n}\cdot\hat{k} = 0$
      • The E and B fields are in phase with each other
      • $\vec{B} = \frac{k}{\omega}(\hat{k}\times \vec{E}) = \frac{k}{\omega}(\hat{k}\times\vec{E})$
        • For a monochromatic wave, this means that $B_{0} = \frac{1}{c} E_{0}$

  • From the relation $B = \frac{1}{c} E$, it can be shown that the energy density is evenly split between the E and B fields (ie. $\frac{1}{\mu_{0}}B_{0}^{2} = \epsilon_{0}E_{0}^{2}$)

  • Since waves carry momentum, we can define a radiation pressure as follows:

    • $P = \frac{1}{A}\frac{\delta p}{\delta t} = \frac{\epsilon_{0}E^{2}_{0}}{2} = \frac{I}{c}$
      • $I = \frac{\epsilon cE^2}{2}$ where I is the intensity
    • If this is a perfect reflector, double this, since the momentum switches direction

Reflection and Transmission

  • Assume linear media (ie. that $D=\epsilon E$ and $H = \frac{B}{\mu}$) and no free charge

  • Let $E(\vec{r},t) = E_{0} exp(i(\vec{k}\cdot \vec{r}-\omega t))$

  • Let $B(\vec{r},t) = \frac{1}{v_{1}}(\vec{k}\times \vec{E})$

  • The phase of the incident, transmitted, and reflected waves must be the same

    • $k_{i}\cdot r-\omega t = constant$
    • As a result of this:
      • The transmitted and reflected waves are in the same plane as the incident wave
      • The incident angle is the same as the reflected angle
      • $n_1 \sin \theta_{1} = n_{2} \sin \theta_{2}$

  • Let $\alpha = \frac{\cos \theta_{T}}{\cos \theta_{I}}$ and $\beta = \frac{\mu_{1} \nu_{1}}{\mu_{2}\nu_{2}} = \frac{\mu_{1}n_{2}}{\mu_{2}n_{1}}$

  • Using the boundary conditions in free space at the boundary $z=0$, can derive relation between reflected and transmitted to incident electric fields

  • The Brewster angle is where there is no reflected electric field

    • $\tan \theta_{B} = \frac{n_{2}}{n_{1}}$

  • Intensity is given by $I_{I} = \frac{\epsilon_{1}v_{1}E_{o}^{2}}{2}\cos(\theta)$

  • R is reflection coefficient and is given by $\frac{I_{R}}{I_{I}}$

  • T is transmission coefficient and is given by $\frac{I_{T}}{I_{I}}$

Absorption and Dispersion

Absorption

  • Combining Ohm’s Law ($J = \sigma E$), Gausses Law ($\nabla \cdot E = \frac{\rho_{f}}{\epsilon}$) and the continuity equation ($\nabla \cdot J = -\frac{\partial \rho_{f}}{\partial t}$) yields that, for homogeneous linear medium

    • $\rho_{f}(t) = e^{\frac{-\sigma}{\epsilon}t}\rho_{f}(0)$
      • where $\tau = \frac{\epsilon}{\sigma}$ is the characteristic dissipation time of the conductor

  • After applying Ohm’s law, one can construct the following modified wave equations

    • $\nabla^{2} E = \mu \epsilon \frac{\partial^{2} E}{\partial t^{2}}+\mu\sigma \frac{\partial E}{\partial t}$
    • $\nabla^{2} B = \mu \epsilon \frac{\partial^{2} B}{\partial t^{2}}+\mu\sigma \frac{\partial B}{\partial t}$
    • Solutions to these are still monochromatic waves, but now the wave vector is complex
      • $\tilde{k}^{2} = \mu \epsilon\omega^{2} + i\mu \sigma \omega$
      • Taking the square root: $\tilde{k} = k+i\kappa$
        • $k = \omega\sqrt{\frac{\epsilon \mu}{2}}\sqrt{\sqrt{1+(\frac{\sigma}{\epsilon\omega})^{2}}+1}$
        • $\kappa = \omega\sqrt{\frac{\epsilon \mu}{2}}\sqrt{\sqrt{1+(\frac{\sigma}{\epsilon\omega})^{2}}-1}$
      • The imaginary part results in attenuation. Define skin depth to be $d = \frac{1}{\kappa}$ as a measure of how far the wave penerates into the conductor
      • The magnetic field also lags behind the electric field

Dispersion

  • permittivity, permeability and conductivity are have some frequency dependence that was ignore until now

    • Each component of a wave travels at a phase velocity of $v = \frac{\omega}{k}$, while the packet as a whole travels at the group velocity $v_{g} = \frac{d\omega}{dk}$

  • A simplified mechanism to explain frequency dependency of $\epsilon$ on $\omega$:

    • Treat the system as a driven harmonic oscillator with damping
      • $m\frac{d^{2}x}{dt^{2}}+m\gamma \frac{dx}{dt}+m\omega_{0}^{2}x = q E_{0}\cos(\omega t)$
      • Using phasors, can tease out that dipole moment is $\tilde{p(t)} = q\tilde{x}(t) = \frac{\frac{q^{2}}{m}}{\omega_{0}^{2}-\omega^{2}-i\gamma\omega}E_{0}e^{-i\omega t}$
        • The imaginary part means that p is out of phase with E
    • The polarization is then: $P = \frac{Nq^{2}}{m} (\Sigma_{j}\frac{f_{j}}{\omega_{j}^{2}-\omega^{2}-i\gamma \omega}) E = \epsilon_{0}\tilde{\chi_{e}} \tilde{E}$
      • $f_{j}$ represents the number of electrons that have frequency $\omega_{j}$ and damping of $\gamma_{j}$
    • If we define $D = \tilde(\epsilon) E$ where $\tilde{\epsilon} = \epsilon_{0}(1+\tilde{\chi_{e}})$, we can see that
      • $\epsilon_{r} = \frac{\tilde{\epsilon}}{\epsilon_{0}} = 1+\frac{Nq^{2}}{m\epsilon_{0}}\Sigma_{j} \frac{f_j}{\omega_{j}-\omega^{2}-i\gamma\omega}$
    • Define $\tilde{k} = \sqrt{\tilde{\epsilon}\mu_{0}}\omega$
      • Absorption coefficient is $\alpha = 2\kappa$ where $\kappa$ is imaginary part of above

Wave Guides

  • Assume that at the boundaries, the wave guide is a conductor, which means that $E_{\parallel} = 0$ and $B_{\perp} = 0$

    • Assume monochromatic waves travelling in the z direction

  • Assuming that the interior of the wave guide is a vacuum, can expand maxwell equations into x,y,z components of E and B, utilize Faraday’s and Ampere’s laws to solve for x and y components of E and B, and then plug into Gauss’s laws to yield two uncoupled equations for z components:

    • $[\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+(\frac{\omega}{c})^{2}-k^{2}]E_{z} = 0$
    • $[\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+(\frac{\omega}{c})^{2}-k^{2}]B_{z} = 0$
      • If $E_{z} = 0$, this is called TE waves
      • If $B_{z} = 0$, this is called TM waves
      • If both are zero, this is called TEM waves
        • Not physically possible in an empty wave guide due to no local maxima allows for Laplace’s equation (but if there is a conductor down the middle, it works)

  • Use separation of variables to some $E_{z}$ and $B_{z}$ equations

Potentials

  • We can write the fields in terms of potentials.
    • Using magnetic Guass’s law: $B = \nabla \times A$
    • Applying the above to Faraday’s Law and rearranging yields that $E = -\nabla V -\frac{\partial A}{\partial t}$

Gauge Freedom

  • Assume that we can change the potentials in such a way that the physical fields remain unchanged. It can be shows that, for any scalar function $\lambda(r,t)$, we have that
    • $A’ = A+\nabla \lambda$
    • $V’ = V-\frac{\partial \lambda}{\partial t}$

Coulomb Gauge

  • Let $\nabla \cdot A = 0$

    • $V(r,t) = \frac{1}{4\pi \epsilon_0} \int \frac{\rho(r’,t)}{r} d\tau'$
    • $\nabla^{2} A - \mu_{0}\epsilon_{0} \frac{\partial ^{2} A}{\partial t^{2}} = -\mu_{0} J + \mu_{0} \epsilon_{0} \nabla (\frac{\partial V}{\partial t})$

  • Potential easy to calculate, vector potential hard to calculate

Lorenz gauge

  • $\nabla \cdot A = -\mu\epsilon_{0}\frac{\partial V}{\partial t}$

    • $\nabla^{2} A -\mu_{0}\epsilon_{0} \frac{\partial^{2} A}{\partial t^{2}} = -\mu_{0} J$
    • Similar for V, with J swapped for p and $\mu$ for $\epsilon$

  • Both A and V are on equal footing

  • Solutions to equations are the retarded potentials

    • $V(r,t) = \frac{1}{4\pi \epsilon_{0}}\int \frac{\rho(r’,t_{r})}{R} d\tau$
      • $t_{r} = t - \frac{R}{c}$
      • $R = r’-r$
    • Can think of this conceptually as calcuating the distance between where you are measuring the potential at (r) to where the in space the charge is located at (r’). You then turn back time by R/c and read use the charge at this earlier time as the contribution from that region of space. Continue on for all r’ over all space

  • Hence, given the potentials, you can calculate the fields (Jefimenko’s equations)

    • $E(r,t) = \frac{1}{4\pi \epsilon_{0}}\int [\frac{\rho}{R^{2}}\hat{R}+\frac{\dot{\rho}}{cR}\hat{R}-\frac{\dot{\vec{J}}}{c^{2}R}] d\tau'$
    • $B(r,t) = \frac{\mu_{0}}{4\pi} \int [\frac{J}{R^{2}}+\frac{\dot{J}}{cR}] \times \hat{R} d\tau'$

Leinard-Weichart Potentials

  • What is V and A for a point charge q that follows some trajectory w(t)?

    • You might guess that since $V(r,t) = \frac{1}{4\pi\epsilon_{0}} \int \frac{\rho(r,t)}{R} d\tau’$, you can just use the electrostatic potential
      • Since the charge only exists at on point in time,can’t do this
      • Use delta function to figure ou at what retarded time the charge was at
      • End results
        • $V(r,t) = \frac{1}{4\pi\epsilon_{0}}\frac{qc}{Rc-R\cdot v}$
        • $A(r,t) = \frac{\mu_{0}}{4\pi}\frac{qcv}{Rc-R\cdot v}$

  • Taking many derivative of potentials yields that

    • $E(r,t) = \frac{q}{4\pi\epsilon_{0}} \frac{R}{(R\cdot u)^{3}}[(c^{2}-v^{2})u+R\times(u\times a)]$
      • $u = c\hat{R}-v$
      • First term is generalized Coulumb field, second is radiation field

Radiation

  • Define radiation of a local charge distribution as $P_{rad}(t_{0}) = lim_{r\rightarrow \infty} P(r, t_{0}+\frac{r}{c})$
    • Since surface area increase like $r^{2}$, we need a field that drops off slower than $\frac{1}{r^{2}}$
      • the radiation field is the only part of the E/B field that has this dependency

Dipoles

  • For a perfect electric dipole with dipole moment $p(t) = p_{0}\cos(\omega t)$, we make 3 key assumptions

    • that $d « r$ or that the size of the dipole is much smaller than the distance at which we are measuring the radiation
      • This allows use to utilize the binomial expansion with respect to d
    • that $d « \frac{c}{\omega}$ or that the dipole distance is much smaller than the wavelength of the frequency of oscillation of the dipole
      • Allows us to utilize $\sin\theta \approx \theta$ and $\cos \theta \approx 1$

  • $r » \frac{c}{\omega}$ or that the radiation distance is mny times larger than the wavelength of oscillation

    • Allows us to drop terms of higher order thant $\frac{1}{r}$

  • End result is that

    • $E = \frac{-\mu_{0}p_{0}\omega^{2}}{4\pi}(\frac{\sin\theta}{r})\cos(\omega(t-\frac{r}{c}))\hat{\theta}$
    • $B = -\frac{\mu_{0}p_{0}\omega^{2}}{4\pi c}\frac{\sin \theta}{r} \cos(\omega(t-\frac{r}{c}))\hat{\phi}$
    • $S = \frac{1}{\mu_{0}}E\times B = \frac{\mu_{0}}{c}(\frac{p_{0}\omega^2}{4\pi}\frac{\sin\theta}{r}\cos(\omega(t-\frac{r}{c}))\hat)^{2}{r}$

  • Importantly, you get no radiation along the axis of the dipole, and maximum radiation perpendicular to the dipole

  • For a magnetic dipole, we have that $m(t) = m_{0} \cos(\omega t)$

  • Using the same assumptions as the electric dipole, one can deduce the following fields

    • $E = \frac{\mu_{0}m_{0}\omega^{2}}{4\pi c}(\frac{\sin\theta}{r})\cos(\omega(t-\frac{r}{c}))\hat{\phi}$
    • $B = -\frac{\mu_{0}m_{0}\omega^{2}}{4\pi c^{2}}\frac{\sin \theta}{r} \cos(\omega(t-\frac{r}{c}))\hat{\theta}$
    • $S = \frac{1}{\mu_{0}}E\times B = \frac{\mu_{0}}{c}(\frac{m_{0}\omega^2}{4\pi c}\frac{\sin\theta}{r}\cos(\omega(t-\frac{r}{c}))\hat)^{2}{r}$
      • The magnetic dipole power is much smaller than the electric dipole power

Point Charges

  • Only the acceleration field contributes to radiation
    • $E_{rad} = \frac{q}{4\pi \epsilon_{0}}\frac{R}{(R\cdot u)^{3}}(R\times (u\times a))$
      • $u = c\hat{R}-v$