# Magnetostatics

Consider a point charge $$q$$ that is moving at a velocity $$\vec{v}$$. The force felt by the charge is given by $$\vec{F}_m = q\vec{v}\times\vec{B}$$. We can use this relation to define what a magnetic field $$\vec{B}$$ is since we can measure all of the three other quantities. Motion parallel to the magnetic field yields no force and motion perpendicular to the field maximizes this force on the charge. Total electromagnetic force, Lorentz force, is $$\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$$. Einstein made the observation that this force is dependent on the reference frame, and so the electric and magnetic fields are reference frame dependent.

The archetipical motion of a charge in a magnetic field is circular motion. Cyclotron motion, the first type of particle accelerator.

Uniform circular motion about a perpendicular magnetic field. Then, $$qv_{xy}B = \frac{mv_{xy}^2}{R} \Rightarrow \frac{v_{xy}}{R} = \frac{qB}{m}$$. And $$v_z$$ is constant, we will set it to zero.

$$qB(v_y\hat{x} - v_x\hat{y}) = m\frac{d\vec{v}}{dt} \Rightarrow v_x(t) = v_{max}\cos(\omega_c t+\phi_0),v_y(t) = v_{max}\sin(\omega_c t+\phi_0)$$ with $$\omega_c = \frac{qB}{m}$$ and $$v_{max}^2 = v_x^2 + v_y^2$$.

Assume we have the initial conditions, $$x(0) = -R$$, $$y(0) = 0$$, $$v_x(0) = 0$$, and $$v_y(0) = v_{max}$$. So, $$v_x(t) = v_{max}\sin(\omega_ct)$$, $$v_y(t) = v_{max}\cos(\omega_ct)$$, $$x(t) = -R\cos(\omega_ct)$$, and $$y(t) = R\sin(\omega_ct)$$.

### Current

$$I = \frac{dQ}{dt}$$. The amount of charge passing through a well defined area over a given amount of time. $$Q = \int\rho dV$$.

Consider the current density $$\vec{J}$$.

$$I \leftarrow \vec{J}\leftrightarrow\vec{v}$$. $$\Delta I = \frac{\Delta Q}{\Delta t}$$.

$$\Delta I = \vec{J} \cdot\hat{n}\Delta A$$, for a small area.

$$I = \int_A\vec{J}\cdot\hat{n}da$$. Conventionally, we choose the current to be positive, so our surface normal must be selected to make the current positive.

From the work sheet, for a small area $$\Delta A$$, $$dQ = \rho\vec{v}\cdot\hat{n}dt\Delta A = \vec{J}\cdot\hat{n}dt\Delta A$$. $$\Delta I = \frac{dQ}{dt}=\vec{J}\cdot\hat{n}\Delta A$$. $$I = \int\vec{J}\cdot\hat{n}da$$.

Put simply, i.e. with a surface of area $$A$$ perpendicular to current density, $$I = JA$$. Remember we choose positive current.

Consider a closed surface, $$S$$. What is the current when a charge is converging to the inside of the surface? $$I = \oint_S\vec{J}\cdot\left(-\hat{n}\right)da$$, the negative to ensure positive current since we use $$\hat{n}$$ as the convential closed surface normal. Thus, $$I = \frac{dQ}{dt} = \frac{d}{dt}\int\rho dV = \int \frac{\partial \rho}{\partial t} dV = -\int_V\vec{\nabla}\cdot\vec{J}dV$$. Thus, $$\frac{\partial\rho}{\partial t} + \vec{\nabla}\cdot\vec{J} = 0$$.

$$\vec{\nabla}\cdot\vec{J} + \frac{\partial\rho}{\partial t} = 0.$$

$$\vec{\nabla}\cdot\vec{J} = 0$$. If this is not the case, then we get electromagnetic field propogation.

### Bio-Savart Law

C.f. Coulomb’s Law.

$$\vec{B} = \frac{\mu_0}{4\pi}\int\frac{Id\vec{\ell}\times\vec{x}}{|\vec{x}|^3}.$$

C.f. (TODO: Update this later)

\begin{equation*} \vec{E} = \frac{1}{4\pi\varepsilon_0}\int\frac{dq}{|\vec{x} - \vec{x}'|^3}. \end{equation*}

Consider that we have a steady current $$I$$ in a wire. Consider a segment of the wire $$d\vec{\ell}$$ and a point $$\vec{x}$$ from this segment. What is the magnetic field at that point due to the segment? The source is then $$Id\vec{\ell}$$ (c.f. $$dq$$).

$$\varepsilon_0\mu_0=\frac{1}{c^2}$$.

Then, for a line current,

$$\vec{B}(\vec{x}) = \frac{\mu_0I}{4\pi}\int_\tau\frac{d\vec{x}'\times(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3}.$$

$$Id\vec{\ell} = Jdad\vec{\ell} = \vec{J}dad\ell = \vec{J}dV = \vec{J}d^3x'$$.

For a volume current,

$$\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi}\int_V\frac{\vec{J}(\vec{x})\times(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3}d^3x'.$$

#### Example

Consider a line current on the z-axis of current $$I$$. What is the magnetic field $$\vec{B}$$ at a distance $$r$$ radially away from the current.

$$R\hat{x} - z\hat{z} = \vec{x} - \vec{x}'$$. Note $$z=R\tan\theta$$. Then, from the B-S law,

\begin{align*} \vec{B}(\vec{x}) &= \frac{\mu_0I}{4\pi}\int_{-\infty}^\infty \frac{dz\hat{z}\times(R\hat{x}-z\hat{z})}{(R^2+z^2)^{3/2}} \\ &= \frac{\mu_0IR}{4\pi}\hat{y}\int_{-\infty}^\infty \frac{dz}{(R^2+z^2)^{3/2}} \\ &= \frac{\mu_0IR}{4\pi}\hat{y}\int_{-\pi/2}^{\pi/2} \frac{d\phi}{(R^2+R^2\tan^2\phi)^{3/2}}R\sec^2\phi \\ &= \frac{\mu_0IR}{4\pi}\hat{y}\int_{-\pi/2}^{\pi/2} \frac{d\phi}{(R^2\sec^2\phi)^{3/2}}R\sec^2\phi \\ &= \frac{\mu_0I}{4\pi}\hat{y}\int_{-\pi/2}^{\pi/2} \frac{d\phi}{R\sec\phi}\sec^2\phi \\ &= \frac{\mu_0I}{2\pi R}\hat{\phi}. \end{align*}

#### Two Loops

Consider two loops, with currents $$I_1$$ and $$I_2$$ CCW.

$$d\vec{\ell}_1$$ and $$d\vec{\ell}_2$$ with $$\vec{x}_1$$ and $$\vec{x}_2$$ pointing to them from the same reference point.

Let $$\vec{x}_{12} = \vec{x}_1 - \vec{x}_2$$. Note: $$d\vec{F} = q\vec{v}\times\vec{B}$$. Considering a cylinder, $$da,d\ell$$, $$Id\ell = Jdad\ell = \rho V dad\ell$$. Then, $$d\vec{F} = \rho dad\ell\vec{V}\times\vec{B} = Id\vec{\ell}\times\vec{B}$$.

\begin{align*} \vec{F}_{12} &= \oint I_1d\vec{\ell}_1\times\vec{B}_{12} \\ &= \oint I_1d\vec{\ell}_1\times\left[\frac{\mu_0 I_2}{4\pi}\oint\frac{d\vec{\ell_2}\times\vec{x}_{12}}{|\vec{x}_{12}|^3}\right] \\ &= \frac{\mu_0I_1I_2}{4\pi}\oint\oint \frac{d\vec{\ell}_1\times(d\vec{\ell}_2\times\vec{x}_{12})}{|\vec{x}_{12}|^3} \\ &= A\oint\oint \frac{d\vec{\ell}_2(d\vec{\ell}_1\cdot\vec{x}_{12})-\vec{x}_{12}(d\vec{\ell}_1\cdot d\vec{\ell}_2)}{|\vec{x}_{12}|^3} \\ &= -A\oint\oint \frac{\vec{x}_{12}(d\vec{\ell}_1\cdot d\vec{\ell}_2)}{|\vec{x}_{12}|^3} \\ \vec{F}_{12} &= -A\oint\oint \frac{\vec{x}_{12}}{|\vec{x}_{12}|^3}(d\vec{\ell}_1\cdot d\vec{\ell}_2) = -\vec{F}_{21}. \end{align*}

From, $$\vec{x}_{12} = -\vec{x}_{21}$$.

### Force On Current

Consider a current $$I$$ along a wire. What is the force felt by the wire when a magnetic field $$\vec{B}$$ is applied to it. $$d\vec{F} = dq\vec{v}\times\vec{B} = \frac{dq}{dt}d\vec{\ell}\times\vec{B} = Id\vec{\ell}\times\vec{B}$$.

$$\vec{F} = I\int d\vec{\ell}\times\vec{B}.$$

Consider we have two closed surfaces with current $$I_1$$ and current $$I_2$$.

The force on circuit one from the induced magnetic field from current 2 is then, $$\vec{F}_{12} = I_1\oint d\vec{\ell}_1\times\vec{B}_2 = I_1\oint d\vec{\ell}_1\times\frac{\mu_0I_2}{4\pi}\oint\frac{d\vec{\ell}_{2}\times\vec{x}_{12}}{|\vec{x}_{12}|^3} = \frac{\mu_0I_1I_2}{4\pi}\oint d\vec{\ell}_1\cdot d\vec{\ell_2}\frac{\vec{x}_{12}}{|\vec{x}_{12}|^3} = -\vec{F}_{21}$$.

Consider two parallel line charges with currents $$I_1$$ and $$I_2$$ directed to the positive $$z$$ direction. $$\vec{F}_{12} = \frac{\mu_0 I_1I_2}{4\pi}\oint d\vec{\ell}_1\cdot d\vec{\ell}_2\frac{\vec{x}_{12}}{|\vec{x}_{12}|^3}$$. $$d\vec{\ell}_1 = \hat{z}dz$$.

Consider the first wire at $$y=0$$ and the second at $$y=d$$.

$$\frac{d\vec{F}_{12}}{dz} = \frac{d}{dz}I_1(\hat{z}dz)\times\left(\hat{x}\frac{\mu_0I_2}{4\pi}\right) = \frac{\mu_0 I_1I_2}{4\pi}(\hat{z}\times\hat{x}) = \frac{\mu_0 I_1I_2}{4\pi}\hat{y}$$. Parallel currents attract and anti-parallel are repulsed.

### Emergent Properties of B-S Law

Recall: From Coulomb’s law we derived the Maxwell laws $$\nabla\cdot\vec{E} = \frac{\rho}{\varepsilon_0},\nabla\times\vec{E} = 0$$.

$$\vec{B}(\vec{x}) = \frac{\mu_0}{4\pi}\int \vec{J}(\vec{x}')\times\frac{\vec{x}-\vec{x}'}{|\cdots|^3}d^3x = \frac{\mu_0}{4\pi}\int \nabla\left(\frac{1}{|\vec{x}-\vec{x}'|}\times\vec{J}(\vec{x}')\right) = \frac{\mu_0}{4\pi}\int \nabla\times\left(\frac{\vec{J}(\vec{x})}{|\vec{x}-\vec{x}'|}\right) = \nabla\times\left(\frac{\mu_0}{4\pi}\int \frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|}d^3x'\right) = \nabla\times\vec{A}$$.

My own aside (ansatz): Note: $$\nabla\times(\psi\vec{A}) = \nabla\psi\times\vec{A} + \psi\nabla\times\vec{A}$$. In our case, the second term is zero.

Therefore:

$$\vec{B} = \nabla\times\vec{A}.$$

$$\nabla\cdot\vec{B} = \nabla\cdot(\nabla\times\vec{A}) = 0$$.

$$\nabla\cdot\vec{B} = 0.$$

$$\nabla\times\vec{B} = \nabla\times\nabla\times\vec{A} = \nabla(\nabla\cdot\vec{A}) - \nabla^2\vec{A}$$. $$\nabla\nabla\cdot\vec{A} = \frac{\mu_0}{4\pi}\nabla\int\nabla\cdot\frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|}d^3x' = \frac{\mu_0}{4\pi}\nabla\int\left(\frac{\nabla\cdot\vec{J}}{|\vec{x}-\vec{x}'|} - \vec{J}(\vec{x}')\nabla'\frac{1}{|\vec{x}-\vec{x}'|}\right) = \frac{\mu_0}{4\pi}\nabla\int\left(- \vec{J}(\vec{x}')\nabla'\frac{1}{|\vec{x}-\vec{x}'|}\right) + \cdots = 0$$

From, $$\nabla'\cdot\frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|} = \frac{\nabla'\cdot\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|} + \vec{J}(\vec{x}')\cdot\nabla'\frac{1}{|\vec{x}-\vec{x}'|} = \vec{J}(\vec{x}')\cdot\nabla'\frac{1}{|\vec{x}-\vec{x}'|}$$

$$\nabla^2\vec{A} \Rightarrow \nabla^2 \frac{1}{|\vec{x}-\vec{x}'|} = -4\pi\delta(\vec{x}-\vec{x}') \Rightarrow \nabla^2\vec{A} = -\mu_0\vec{J}(\vec{x})$$

Thus, we get Ampere’s Law,

$$\nabla\times\vec{B} = \mu_0\vec{J}.$$ \begin{align} \int(\nabla\times\vec{B})\cdot\hat{n}da &= \mu_0\int \vec{J}\cdot\hat{n}da, \nonumber \\ \oint \vec{B}\cdot d\vec{\ell} &= \mu_0I. \end{align}

### Example

Consider a line current $$I$$ upward along the z-axis. Thus, we have cylindrical symmetry.

So, $$2\pi B_\phi = \mu_0 I$$. Thus, $$\vec{B} = \frac{\mu_0I}{2\pi}\hat{\phi}$$. Radial component is zero due to $$\nabla\cdot\vec{B} = 0$$. Z-component is zero due to infinite line current.

## Vector Potential

Coulomb Gague: $$\nabla\cdot\vec{A} = 0$$.

$$\vec{A}(\vec{x}) = \frac{\mu_0}{4\pi}\int \frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|}d^3x'$$.

From a circular (radius a) current $$I$$ on x-y plane, $$\vec{A}(\vec{x}) = \frac{\mu_0I}{4\pi}\oint \frac{d\vec{x}'}{|\vec{x}-\vec{x}'}$$. $$\vec{x}=(r\sin\theta,0,r\cos\theta)$$, $$\vec{x}'=(a\cos\phi',a\sin\phi',0)$$, $$d\vec{\ell} = (-a\sin\phi'd\phi',a\cos\phi'd\phi',0)$$. $$A_\phi(r,\theta) = \int_0^{2\pi}\frac{\cos\phi' d\phi'}{\left(r^2+a^2-2ar\sin\theta\cos\phi'\right)^{1/2}}$$. For $$r\ll a,r\gg a,\theta\approx 0$$, $$A_\phi \approx \frac{1}{\sqrt{r^2+a^2}}\int \cos\phi' d\phi' \left(1+\frac{ar}{a^2+r^2}\sin\theta\cos\phi'\right) = \frac{\mu_0I}{2}\frac{a^2\sin\theta}{\sqrt{a^2+r^2}^{3}}$$.

$$B_r = \frac{1}{r\sin\theta}\frac{\partial}{\partial\theta}(\sin\theta A_\phi) = \frac{\mu_0 Ia^2}{2}\frac{\cos\theta}{\sqrt{a^2+r^2}^3}$$. $$B_\theta = -\frac{\mu_0 Ia^2\sin\theta(2a^2-r^2)}{4(a^2+r^2)^{5/2}}$$.

For $$r\gg a$$, then $$A_\phi\approx \frac{\mu_0 I a^2\sin\theta}{4r^2} = \frac{\mu_0 m\sin\theta}{4\pi r^2} = \frac{\mu_0}{4\pi}\frac{\vec{m}\times\vec{x}}{r^3}$$. $$B_r \approx \frac{\mu_0}{4\pi}\frac{2m\cos\theta}{r^3}$$. $$B_\theta\approx -\frac{\mu_0}{4\pi}\frac{m\sin\theta}{r^3}$$.

$$r\ll a$$ then $$B_r\approx \frac{\mu_0}{4\pi}\frac{m\cos\theta}{a^3} = \frac{\mu_0I}{2a}\cos\theta$$. $$B_\theta \approx -\frac{\mu_0I}{2a}\sin\theta$$. $$\vec{B}\approx\frac{\mu_0I}{2a}\left(\cos\theta\hat{r}-\sin\theta\hat{\theta}\right)= \frac{\mu_0I}{2a}\hat{z}$$.

$$\frac{1}{r}\left(1+\frac{r'^2}{r^2}-2\frac{\vec{x}\cdot\vec{x}}{r^2}\right)^{-1/2} \approx \frac{1}{r}\left(1+\frac{\vec{x}\cdot\vec{x}'}{r}\right) \Rightarrow \vec{A}(\vec{x}) = \frac{1}{r}\int\vec{J}(\vec{x})d^3x' + \frac{\mu_0}{4\pi}\frac{1}{r^3}\int\vec{J}(\vec{x}')(\vec{x}\cdot\vec{x}')d^3x' = \frac{\mu_0}{4\pi}\frac{1}{r^3}\int\vec{J}(\vec{x}')(\vec{x}\cdot\vec{x}')d^3x' = \frac{\mu_0}{4\pi r^3} -\frac{1}{2}\int d^3x' \vec{x}\times(\vec{x}'\times\vec{J}(\vec{x}')) = -\frac{\mu_0}{4\pi r^3}\frac{1}{2}\vec{x}\times\int d^3x' (\vec{x}'\times\vec{J}(\vec{x}')) = -\frac{\mu_0}{4\pi r^3}\vec{x}\times\vec{m} = \frac{\mu_0}{4\pi r^3}\vec{m}\times\vec{x}$$.

$$\vec{m} = \frac{I}{2}\int\vec{x}'\times d\vec{x}' = m\hat{n} = I\pi a^2$$. Remember size of cross product is area of trapezoid, thus half the area is the area of the triangle.

Torque: $$\tau = \vec{m}\times\vec{B}$$.

For a magnetic dipole: $$\vec{A}(\vec{x}) = \frac{\mu_0}{4\pi}\frac{\vec{m}\times\vec{x}}{r^3}$$.

## Magnetic Fields in Matter

In matter, atoms could have microscopic magnetic dipoles, thus magnetization is given as: $$\vec{M} = \frac{\sum_i \vec{m}_i}{V}$$. 3 different types: dia-, para-, ferromagnetism. All diamagnetic fields are conductors and with a magnetic field applied gives a magnetization in the opposite direction of the magnetic field. Paramagnetic fields are parallel to the magnetic field. Ferromagnetic have a permanent magnetization that is felt when no field is applied.

B=0 B=0 B!=0
m M M
Dia m=0 M=0 M//-B
Para m!=0 M=0 M//B
Ferro m!=0 M!=0 Depends

So, $$\vec{A}(\vec{x}) = \frac{\mu_0}{4\pi}\int\frac{\vec{M}(\vec{x}')\times(\vec{x}-\vec{x}')}{|\vec{x}-\vec{x}'|^3}d^3\vec{x}' = \frac{\mu_0}{4\pi}\int\vec{M}(\vec{x}')\times\nabla'\frac{1}{|\vec{x}-\vec{x}'|}d^3\vec{x}' = \frac{\mu_0}{4\pi}\int\left(\nabla'\times\frac{\vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|} + \frac{\nabla'\times\vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|}\right)d^3\vec{x}' = -\frac{\mu_0}{4\pi}\int\frac{\hat{n}'\times\vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|}da' + \frac{\mu_0}{4\pi}\int\frac{\nabla'\times\vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|}d^3x' = \frac{\mu_0}{4\pi}\int\frac{\vec{M}(\vec{x}')\times\hat{n}'}{|\vec{x}-\vec{x}'|}da' + \frac{\mu_0}{4\pi}\int\frac{\nabla'\times\vec{M}(\vec{x}')}{|\vec{x}-\vec{x}'|}d^3x'$$.

C.f. $$\vec{J}(\vec{x})\to\vec{A}(\vec{x}) = \frac{\mu_0}{4\pi}\int\frac{\vec{J}(\vec{x}')}{|\vec{x}-\vec{x}'|}d^3x'$$. Thus, we have terms that are like sources.

The area term is the effective surface current, $$\vec{M}\times\hat{n}$$. The volume term is the effective volumetric current in the volume. Therefore,

\begin{align} \vec{K}_M(\vec{x}) &= \vec{M}\times\hat{n}, \\ \vec{J}_M(\vec{x}) &= \nabla\times\vec{M}, \\ \vec{A}(\vec{x}) &= \frac{\mu_0}{4\pi}\int\frac{\vec{K}_b(\vec{x}')}{|\vec{x}-\vec{x}'|}da' + \frac{\mu_0}{4\pi}\int\frac{\vec{J}_b(\vec{x}')}{|\vec{x}-\vec{x}'|}d^3x'. \end{align}

Recall: $$\nabla\times(f\vec{M}) = \nabla f\times\vec{M} + f\nabla\times\vec{M}$$.

Consider if we have a surface far away, then we only need to worry about the volumentric current density since the surface current term goes to zero. From Ampere’s law, $$\nabla\times\vec{J}_{total} = \mu_0\vec{J}_{free} + \mu_0\vec{J}_{M} = \mu_0(\vec{J}_f+\nabla\times\vec{M}) \Rightarrow \nabla\times\left(\frac{\vec{B}}{\mu_0}-\vec{M}\right)=\vec{J}_f\Rightarrow \vec{H} = \frac{\vec{B}}{\mu_0}-\vec{M}$$.

Macroscopically:

\begin{align} \nabla\cdot\vec{B}&=0, \\ \nabla\times\vec{H}&=\vec{J}_f. \end{align}

C.f. $$\nabla\cdot\vec{B}=\rho_f,\nabla\times\vec{E}=0$$ with $$\vec{D}=\varepsilon_0\vec{E}+\vec{P}$$.

$$\vec{B}$$ is called the magnetic induction and $$\vec{H}$$ the magnetic field, which is the old way of referring to these. We will use B-field and H-field to limit confusion since it now is the magnetic field and auxillary field.

Stokes theorem on this gives: $$\oint\vec{H}\cdot d\vec{\ell}=I_f$$.

For dia, para: $$\vec{M}\propto \vec{B}$$ hence $$\propto \vec{H}$$. Similar to linear dielectric media, $$\vec{P}\propto\vec{E}$$. The proportionality is $$\vec{M}=\chi_m\vec{H}$$. Remember the proportionalities are approximations, but typically very good. For ferro, $$\vec{H}=0,\vec{M}\neq0$$. Dia: $$\chi_m<0$$. Para: $$\chi_m>0$$. But, $$\chi_m$$ is pretty small in both cases $$\sim10^{-5}$$. $$n=\sqrt{1+\chi_e}\sim \mathcal{O}(1)$$. $$\chi_e\sim 1$$.

$$\vec{B} = \mu_0(\vec{H}+\vec{M})$$.

When light interacts with matter, the electric interaction is much more dominant than the magnetic term so the magnetic is neglected typically. $$n=\sqrt{\varepsilon_r\mu_r}= \sqrt{\frac{\varepsilon\mu}{\varepsilon_0\mu_0}},\mu\approx\mu_0$$.

## Interfaces

At an interface, from the divergence, $$\nabla\cdot\vec{B}=0$$, we know that $$B_{1,\perp} = B_{2,\perp}$$. From the curl, $$\nabla\times\vec{H}=J_f$$, we know that $$H_{1,||}-H_{2,||}=\vec{K}_f\times\hat{n}_{21}$$. Similarly, $$\vec{K} = \hat{n}_{21}(\vec{H}_2-\vec{H}_1)$$. And note that $$\hat{n}_{21}\times(\vec{H}_2-\vec{H}_1)_\perp=0$$. $$\hat{n}_{21}$$ points at region 2 with its tail pointing at region 1.

\begin{align} B_{1,\perp} &= B_{2,\perp}, \\ \vec{K} &= \hat{n}_{21}\times(\vec{H}_2-\vec{H}_1). \end{align}

### Boundary Value Problems for No Free Current

$$\nabla\cdot\vec{B}=0$$ and $$\nabla\times\vec{H}=0$$ with $$\vec{M}\neq 0$$.

$$\nabla\times\vec{H} = \nabla\times\left(\frac{\vec{B}}{\mu_0}-\vec{M}\right) = 0\Rightarrow \nabla\times\nabla\vec{A} = \mu_0\nabla\times\vec{M} = \mu_0\vec{J}_M$$. So, $$\nabla^2\vec{A}=-\mu_0\nabla\times\vec{M}$$. So**, $$\vec{A} = \frac{\mu_0}{4\pi}\left(\int \frac{\nabla'\times\vec{M}}{|\vec{x}-\vec{x}'|}d^3x' + \int\frac{\vec{M}\times\hat{n}}{|\vec{x}-\vec{x}'|}da\right)$$.

$$\vec{H} = -\nabla\Phi_M$$. $$\nabla\cot\vec{H} = -\nabla\cdot\vec{M}$$ so $$\nabla^2\Phi_M=\nabla\cdot\vec{M} = -\rho_M$$. And also we can define $$\sigma_M = \vec{M}\cdot\hat{n}$$.

#### Example - Cylindrical Magnet

Consider a bar magnet with azimutal symmetry aligned along the z-axis, $$\vec{M} = \mu\hat{z}$$. So, $$\nabla\cdot\vec{M} = 0$$ and $$\vec{M}\cdot\hat{n} = 0$$ on the sides and $$\vec{M}\cdot\hat{n} = \pm \mu$$ on the top and bottom, hence a dipole. $$\vec{J}_M = 0$$. $$\vec{K}_M = \vec{M}\times\hat{n} = M\hat{\phi}$$. So, it appears to be a solenoid.

#### Example - Ball Magnet

Let $$\vec{M} = M_0\hat{z}$$ and the radius of the ball to be $$a$$. So, $$\Phi_M=?$$. Then, $$\nabla\cdot\vec{M}=0$$ and $$\nabla\cdot\vec{B} = 0$$ so $$\nabla\cdot\vec{H}=0$$ and so $$\nabla^2\Phi_M=\nabla\cdot\vec{M}=0$$. For a spherical symmetric system with azimutal symmetry, $$\Phi_m(r,\theta,\phi) = \sum_\ell \left(A_\ell r^\ell + B_\ell\frac{1}{r^{\ell+1}}\right)P_\ell(\cos\theta)$$. Inside, we get $$\Phi_m(r,\theta,\phi) = \sum_\ell A_\ell r^\ell P_\ell(\cos\theta)$$ and outside we get $$\Phi_m(r,\theta,\phi) = \sum_\ell \frac{B_\ell}{r^{\ell+1}} P_\ell(\cos\theta)$$. NOTE: When the $$\vec{H}_t$$ is continuous at a boundary then the potential is continuous! So, $$A_\ell a^\ell = \frac{B_\ell}{a^{\ell+1}}\Rightarrow a^{2\ell+1}A_\ell=B_\ell$$. The other boundary condition we have is that $$B_{in,r}=B_{out,r}$$ so $$H_{in,r}+M_{in}\cdot\hat{r} = H_{in,r}+M_0\cos\theta = H_{out,r}$$. Then, $$-\frac{\partial\Phi_{m,in}}{\partial r}|_{r=a}+M_0\cos\theta = -\frac{\partial\Phi_{out}}{\partial r}|_{r=a}$$. So, $$-\sum_\ell \ell A_\ell a^{\ell-1}P_\ell(\cos\theta) + M_0 P_1(\cos\theta) = \sum_\ell (\ell+1)\frac{B_\ell}{a^{\ell+2}}P_\ell(\cos\theta)$$. For $$\ell=1$$, $$-A_1 + M_0 = \frac{2 B_1}{a^3}$$. For $$\ell\neq 1$$, $$\ell A_\ell a^{\ell-1} = (\ell+1)\frac{B_\ell}{a^{\ell+2}} = (\ell+1)\frac{a^{2\ell+1}}{a^{\ell+2}} = (\ell+1)A_\ell a^{\ell-1}\Rightarrow A_\ell=0$$. Then, $$A_1 a^3 = B_1$$ and $$-A_1 + M_0 = 2 \frac{B_1}{a^3} = 2A_1 \Rightarrow A_1 = \frac{1}{3}M_0$$ and $$B_1 = \frac{1}{3}M_0a^3$$.

$$\Phi_{M,in} = \frac{1}{3}M_0r\cos\theta$$ and $$\Phi_{M,out} = \frac{1}{3}M_0\frac{a^3}{r^2}\cos\theta$$. Thus, $$\Phi_M = \Theta(a-r)\frac{1}{3}M_0r\cos\theta + \Theta(r-a)\frac{1}{3}M_0\frac{a^3}{r^2}\cos\theta$$.

Alternatively: The bound charge is $$\sigma_M = \vec{M}\cdot\hat{n} = M\cos\theta$$. $$\Phi_M(r,\theta) = \frac{1}{4\pi}\int\frac{\sigma_m(\theta',\phi')}{|\vec{x}-\vec{x}'|}da' = \frac{M_0}{4\pi}\int \frac{\cos\theta'}{|\vec{x}-\vec{x}'|}a^2d\Omega' = \frac{M_0a^2}{4\pi}\int\cos\theta'\left(\sum_\ell \frac{4\pi}{2\ell+1}\frac{r_<^\ell}{r_>^{\ell+1}}\sum_m Y_\ell^{m*}(\theta',\phi')Y_\ell^m(\theta,\phi)\right) = \frac{1}{3}M_0a^2\frac{r_<}{r_>^2}\cos\theta$$.

Connection with magnetic dipole: $$\vec{m} = V\vec{M} = \frac{4\pi}{3}a^3\vec{M}$$. So, \$ΦM,out = \frac{1}{4\pi r^3}(3(\vec{m}⋅\hat{r})\hat{r}-\vec{m}) $$\vec{B} = \mu_{\0}\vec{H}$$.

$$\vec{H}_{in} = \frac{\vec{B}_0}{\mu_0} - \frac{1}{3}\vec{M}$$ and $$\vec{B}_{in} = \vec{B}_0+\frac{2}{3}\vec{M}$$.

$$\vec{B}_{in} = \mu\vec{H}_{in}$$ and $$\left(\frac{\mu}{3}+\frac{2}{3}\right)\vec{M} = (\mu_r-1)\vec{B}_0 \Rightarrow \vec{B}_0 = \frac{\frac{\mu}{3}+\frac{2}{3}}{\frac{\mu}{\mu_0}-1}\vec{M}$$.

Side Note: When $$\nabla\times\vec{H}=0$$ we can use the magnetic scalar potential, $$\vec{H} = -\nabla\Phi_M$$ with $$\nabla^2\Phi_M=\nabla\cdot\vec{M} = -\rho_M$$.

## EM Induction

Faraday’s Law. $$\mathcal{E} = -\frac{dF}{dt}$$. Note that $$\mathcal{E} = \oint \vec{E}'\cdot d\vec{\ell}$$ where $$\vec{E}'$$ is the electric field in the stationary frame of the circuit and $$F = \int_S\vec{B}\cdot\hat{n}da$$ is the magnetic flux.

Say we have a magnetic field and a circuit that is partially submerged circuit in the magnetic field. If you move the circuit then a current is induced. Same if you move the magnet.

In another case, if both are stationary but the magnetic field changes in time, a current is induced.

So, $$\oint \vec{E}'\cdot d\vec{\ell} = -\frac{d}{dt}\int_S\vec{B}\cdot\hat{n}da$$.

In the case that the circuit is stationary, $$\oint \vec{E}\cdot d\vec{\ell} = \int_S(\nabla\times\vec{E})\cdot\hat{n}da = -\frac{d}{dt}\int_S\vec{B}(t)\cdot\hat{n}da = -\int_S\frac{\partial}{\partial t}\vec{B}(t)\cdot\hat{n}da$$. So, $$\nabla\times\vec{E} = -\frac{\partial \vec{B}}{\partial t}$$.

For a varying current, $$\vec{B}\propto I$$ so $$\frac{dF}{dt} = \frac{dF}{dI}\frac{dI}{dt} = L\frac{dI}{dt}$$. So, $$\mathcal{E} = -L\frac{dI}{dt}$$.

Consider a solenoid of length $$L$$ and $$N$$ turns. $$\vec{B} = \frac{\mu_0NI}{\ell}$$ inside. The flux is $$F = N(\pi a^2B) = N\pi a^2 \frac{\mu_0NI}{\ell} = \frac{\mu_0 \pi a^2 N^2 I}{\ell}$$. So, $$B = \mu_0 \frac{NI}{\ell} = L$$, check notes on this one.

Say we have a circuit with a resistor, inductor, battery, and switch. Then, the loop around the circuit gives, $$V_0 - L\frac{dI}{dt}-IR = 0$$ hence $$V_0 = L\frac{dI(t)}{dt} + I(t)R$$. This gives a damped harmonic oscillator and so $$I(t) = I_0 + I_h(t)$$. When current is constant, $$I = \frac{V_0}{R}$$. For the homogeneous solution, $$\frac{dI}{dt} = - \frac{R}{L}I$$ hence $$I(t)=\frac{V_0}{R}\left(1-\exp\left(-\frac{R}{L}t\right)\right)$$.

Consider a loop current $$I(t)$$, then the self-EMF is $$\mathcal{E} = -\frac{dF}{dt} = -L\frac{dI}{dt}$$.

### Mutual Inductance

Suppose we have many circuits with currents, $$I_k$$.

The flux on circuit $$i$$ is, $$F_i = \sum_{j=1}^nF_{ij}$$. $$F_{ij} = \int_{S_i}\vec{B}_{ij}\cdot\hat{n}_ida_i = M_{ij}I_j \propto I_j$$. Then, $$F_i = \sum_{j=1}^n M_{ij}I_j$$ with $$M_{ji} = M_{ij}$$.

Proving this symmetry, $$F_{ij} = \int_S \vec{B}_{ij}\cdot\hat{n}_ida_i \int_S \nabla\times\vec{A}_{ij}\cdot\hat{n}_ida_i = \oint \vec{A}_{ij}\cdot d\vec{\ell}_i = \frac{\mu_0I_j}{4\pi}\oint\oint\frac{d\vec{\ell}_i\cdot d\vec{\ell}_j}{|\vec{x}_i-\vec{x}_j|}\equiv M_{ij}I_j$$. Also, we see $$M_{ij}$$ solely depends on the shape and size, i.e. only geometry dependent.

Then, the self inductance $$L_i$$ can be obtained with $$\vec{\ell}_i\to \vec{\ell}$$ and $$\vec{\ell}_j\to\vec{\ell}'$$.

$$\frac{dW}{dt} = -\vec{F}\cdot\frac{d\vec{x}}{dt} = -\vec{F}\cdot\vec{v} = -q\vec{E}'\cdot\vec{v}$$. So, $$\frac{dW}{dt} = \sum_i q_i\vec{v}_i\cdot\vec{E}' \Rightarrow -\int\rho\vec{v}\cdot\vec{E}'dV = -\int\vec{J}\cdot\vec{E}'dV\Rightarrow -I\int\vec{E}'\cdot d\vec{\ell} = -I\mathcal{E}$$.

Then, $$\frac{dW}{dt} = -I\mathcal{E} = I\frac{dF}{dt}$$.

For a single circuit, $$\delta W = I \delta F$$. For multiple, $$\delta W = \sum\limits_i I_i\delta F_i = \sum\limits_i I_i\int_S\delta \vec{B}_i\cdot\hat{n}_ida_i = \sum\limits_i I_i\int_S\nabla\times\delta\vec{A}_{i}\cdot\hat{n}_ida_i = \sum\limits_i I_i\oint \delta\vec{A}_{i}\cdot d\vec{\ell}_i \Rightarrow \int \delta\vec{A}\cdot \vec{J}dV$$. Using $$\vec{J} = \nabla\times\vec{H}$$, $$\delta W = \int \delta A\cdot(\nabla\times\vec{H})dV = \int \nabla\cdot(\vec{H}\times\vec{A})+\vec{H}\cdot(\nabla\times\delta\vec{A})dV = \int_S(\vec{H}\times\delta\vec{A})\cdot\hat{n}da + \int\vec{H}\cdot\delta\vec{B}dV$$. If we choose a surface very far away, $$\vec{H}\propto \frac{1}{r^2}$$ and $$\vec{A}\propto\frac{1}{r}$$ so, $$=\int\vec{H}\cdot\delta\vec{B}dV$$.

If we have a linear magnetic system, $$\delta W = \delta\int\frac{1}{2}\vec{H}\cdot\vec{B}dV\Rightarrow W = \int \frac{1}{2}\vec{H}\cdot\vec{B}dV$$. Then, $$w_m = \frac{1}{2}\vec{H}\cdot\vec{B}$$, recall $$w_e = \frac{1}{2}\vec{E}\cdot\vec{D}$$.

$$W = \frac{1}{2}\sum_{i}^N I_iF_i = \frac{1}{2}\sum_{i=1}^N\sum_{j=1}^N M_{ij}I_iI_j = \frac{1}{2}LI^2$$.

Say we have $$N$$ current loops with energy supplied by a battery. Then, $$dW_{battery}\equiv dW_b = dW_{m} + dW$$. $$dW_b = \sum_i I_idF_i$$.

If the current is constant then $$W = \frac{1}{2}\sum_i I_iF_i$$. So, $$dW = \frac{1}{2}\sum_i I_idF_i$$. Then half of the energy from the battery goes into the potential energy and the other half must go into the mechanical energy. I.e. a force was exerted on the circuit, exerting this mechanical energy. Hence, $$\vec{F} = (\nabla W)_I$$.

Say $$N=2$$. What is the force acting on circuit 2 due to circuit 1? Then, $$dW = I_1dF_1 + I_2dF_2$$ and so $$W = \frac{1}{2}\sum_{ij}^2M_{ij}I_j = \frac{1}{2}(L_1I_1^2 + 2MI_1I_2 + L_2I_2^2)$$. Then, $$F=(\nabla_2 W)_I = I_1I_2\nabla_2 M$$. Recall: $$M_12 = \frac{\mu_0}{4\pi}\oint\oint\frac{d\vec{\ell}_1\cdot\vec{\ell}_2}{|\vec{x}_2-\vec{x}_1|}$$. Then, $$\nabla_2M_{12} = \frac{\mu_0}{4\pi}\oint\oint d\vec{\ell}_1\cdot d\vec{\ell}_2\nabla_2\frac{1}{|\vec{x}_2-\vec{x}_1}$$. Hence $$\vec{F} = -\frac{\mu_0I_1I_2}{4\pi}\oint\oint d\vec{\ell}_1\cdot d\vec{\ell}_2\frac{(\vec{x}_2-\vec{x}_1)}{|\vec{x}_2-\vec{x}_1|^3}$$. Note we can write by massaging this expression as, $$F_{21} = I_2\oint d\vec{\ell}_2\times\vec{B}_{21}$$.

## Electromagnet

Suppose we have an iron bar (which is behaving linearly with $$\chi_m\sim 10^5$$ in a low field) in a solenoid of current $$I$$, $$N$$ turns, and area $$A$$. Then, in the solenoid, $$H=NI\hat{z}$$. So, $$F = \frac{\partial W}{\partial x}$$. Let $$x$$ be the length of insertion of magnet in solenoid.

Consider if we put it $$\Delta x$$ further in. Then, $$W_m = \frac{1}{2}\mu H^2$$. So, $$\Delta W = \left[\frac{1}{2}\mu H^2 - \frac{1}{2}\mu_0H^2\right] A\Delta x$$ Then, $$F \approx \frac{\Delta W}{\Delta x} = \frac{1}{2}(\mu-\mu_0)N^2I^2$$. Then the bar is attracted to go further in the solenoid.

Suppose we have copper, then $$\chi_m<0$$ hence we have a slight repelling force.

If we have a superconductor, $$\chi_m=-1$$.

### Modification of Ampere’s Law

$$\nabla\times\vec{H}=\vec{J}$$ is not consistent with conservation of charge: $$\nabla\cdot\vec{J} + \frac{\partial\rho}{\partial t} = 0$$.

Say we have a capacitor that accumulates charge on each terminal (equal and opposite) such that the current flow is constant. Then consider a surface that the wire that carries the current $$I$$ through it, $$\oint\vec{H}\cdot d\vec{\ell} = \int_S(\nabla\times\vec{H})\cdot\hat{n}_1da_1 = I$$. If we consider a new surface that goes through te capacitor, which has no current passing through it, then $$\oint\vec{J}\cdot d\vec{\ell} = 0$$. These should be the same!

Consider entire closed surface composed of both of these surfaces. Then, we get a net current into the region instead of zero net current. Specifically from the continuity equation: $$\int_V\nabla\cdot\vec{J}dV = 0 = -\int_V\left(\frac{\partial\rho}{\partial t}\right)dV \neq 0$$!

Therefore, the Maxwell differential equations,

\begin{align} \nabla\cdot\vec{D} &= \rho \\ \nabla\cdot\vec{E} &= \frac{\rho}{\varepsilon_0} \\ \nabla\times\vec{E} &= -\frac{\partial\vec{B}}{\partial t} \\ \nabla\cdot\vec{B} &= 0 \\ \nabla\times\vec{H} &= \vec{J} + \frac{\partial\vec{D}}{\partial t} \\ \nabla\times\vec{B} &= \mu_0\vec{J} + \frac{1}{c^2}\frac{\partial\vec{E}}{\partial t} \end{align}

From, $$\nabla\cdot\vec{D} = 0\Rightarrow \vec{B} = \nabla\times\vec{A}$$ and $$\nabla\times\left(\vec{E} + \frac{\partial A}{\partial t}\right)=0\Rightarrow \vec{E} + \frac{\partial\vec{A}}{\partial t} = -\nabla\Phi(t)$$.

Let $$c = \frac{1}{\sqrt{\epsilon_0\mu_0}}$$. Then, $$\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_0}\Rightarrow \nabla^2\Phi + \frac{\partial}{\partial t}(\nabla\cdot\vec{A}) = -\frac{\rho}{\varepsilon_0}$$ and $$\nabla^2\vec{A}-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\vec{A} - \nabla\left(\nabla\cdot\vec{A}+\frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2}\right) = - \mu_0\vec{J}$$ give two coupled partial differential wave equations.

\begin{align} \nabla^2\Phi + \frac{\partial}{\partial t}(\nabla\cdot\vec{A}) &= -\frac{\rho}{\epsilon_0} \\ \nabla^2\vec{A}-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\vec{A} - \nabla\left(\nabla\cdot\vec{A}+\frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2}\right) &= - \mu_0\vec{J}. \end{align} \begin{align} \nabla^2\vec{E} - \frac{1}{c^2}\frac{\partial^2\vec{E}}{\partial t^2} &= 0 \\ \nabla^2\vec{B} - \frac{1}{c^2}\frac{\partial^2\vec{B}}{\partial t^2} &= 0 \end{align}

#### Gague Freedom

1. Lorenz Condition: $$\nabla\cdot\vec{A} + \frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2} = 0$$ gives $$\nabla^2\vec{A} - \frac{1}{c^2}\frac{\partial^2\vec{A}}{\partial t^2} = -\mu_0\vec{J}$$ and $$\nabla^2\Phi - \frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2} = -\frac{\rho}{\varepsilon_0}$$ now decoupled equations. Note: $$\nabla^2 f = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2}\Rightarrow \exp(ikx-\omega t)$$ where $$v = \frac{\omega}{k}$$. In special relativity, we get the 4-vectors $${\nabla\choose\frac{1}{c}\frac{\partial}{\partial t}}\leftrightarrow{\frac{\Phi}{c}\choose \vec{A}}\leftrightarrow{c\rho\choose\vec{J}}$$.
2. Lorentz Transformation and Lorentz Gauge. For a Gauge Function $$\Lambda$$ then the transformation: $$\vec{A}' = \vec{A} + \nabla\Lambda$$ and $$\Phi' = \Phi - \frac{\partial\Lambda}{\partial t}$$ are invariant $$\vec{B} = \nabla\times\vec{A}' = \nabla\times\vec{A}$$ and $$\vec{E} = -\nabla\Phi'-\frac{\partial\vec{A}'}{\partial t} = -\nabla\Phi-\nabla\frac{\partial\Lambda}{\partial t}-\frac{\partial\vec{A}}{\partial t} + \frac{\partial}{\partial t}\nabla\Lambda = \vec{E}$$. Lorentz gague gives the Lorenz condition on the Gague function $$\Lambda$$, $$\nabla^2\Lambda - \frac{1}{c^2}\frac{\partial^2\Lambda}{\partial t^2}=0$$.
3. Coulomb Gague: $$\nabla\cdot\vec{A}=0$$. Then, $$\nabla^2\Phi=-\frac{\rho}{\varepsilon_0}$$ and $$\nabla^2\vec{A}-\frac{1}{c^2}\frac{\partial^2\vec{A}}{\partial t^2} -\nabla\frac{1}{c^2}\frac{\partial^2\Phi}{\partial t^2}=-\mu_0\vec{J}$$. So, $$\Phi(\vec{x},t) = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\vec{x}',t)}{|\vec{x}-\vec{x}'|}d^3x'$$. Let $$\vec{J}=\vec{J}_\ell + \vec{J}_t$$ for longitudinal and transverse. With, $$\mu_0\vec{J}_\ell = \frac{1}{c^2}\nabla\frac{\partial\Phi}{\partial t}$$. We have the conditions: $$\nabla\times\vec{J}_\ell = 0$$ and $$\nabla\cdot\vec{J}_t=0$$. Then, $$\nabla^2\vec{A}-\frac{1}{c^2}\frac{\partial^2\vec{A}}{\partial t^2} = -\mu_0\vec{J}_t$$. Therefore: $$\vec{B}=\nabla\times\vec{A}$$ and $$\vec{E} = -\frac{\partial\vec{A}}{\partial t}$$. $$\vec{J}_\ell(\vec{x},t) = \int \vec{J}_\ell(\vec{x}',t)\delta(\vec{x}-\vec{x}')d^3x' = \int \vec{J}_\ell(\vec{x}',t)\left(-\frac{1}{4\pi}\nabla^{'2}\frac{1}{|\vec{x}-\vec{x}'|}\right)d^3x' = -\frac{1}{4\pi}\int \vec{J}_\ell(\vec{x}',t)\nabla^{'2}\frac{1}{R}d^3x'$$. Using vector identity on integrand, $$-\nabla\cdot(\vec{J}_\ell\nabla\frac{1}{R}) + \nabla\cdot\vec{J}\nabla\frac{1}{R}$$ and the second term is the only one that survives. $$\nabla'\cdot\vec{J}\nabla'\frac{1}{R} = -\nabla\cdot\vec{J}\nabla\frac{1}{R}$$. Thus, $$\frac{-1}{4\pi}\nabla\int\frac{\nabla'\cdot\vec{J}}{|\vec{x}-\vec{x}'|}d^3x' = \frac{1}{4\pi}\nabla\int\frac{\frac{\partial\rho}{\partial t}}{|\vec{x}-\vec{x}'|}d^3x' = \frac{1}{4\pi}\frac{\partial}{\partial t}\nabla\int\frac{\rho}{|\vec{x}-\vec{x}'|}d^3x'$$. Therefore, we show $$\mu_0\vec{J}_t = \frac{1}{c^2}\nabla\frac{\partial\Phi}{\partial t}$$. Also, $$\vec{J}(\vec{x},t) = \frac{-1}{4\pi}\int\vec{J}(\vec{x}',t)\nabla^2\frac{1}{R}d^3x' = \frac{-1}{4\pi}\int\frac{\nabla'\cdot\vec{J}}{R}d^3x'+\frac{1}{4\pi}\nabla\times\nabla\times\int\frac{\vec{J}}{R}d^3x' = \vec{J}_\ell + \vec{J}_t$$. Hence, we can find the transverse current and thus solve $$\nabla^2\vec{A} - \frac{1}{c^2}\frac{\partial^2\vec{A}}{\partial t^2}=-\mu_0\vec{J}_t$$. General wave equation:

$$\nabla^2\psi - \frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2} = -4\pi f(\vec{x},t)$$

Using Green’s functions: $$\nabla^2G(\vec{x},t;\vec{x}',t') - \frac{1}{c^2}\frac{\partial^2G}{\partial t^2} = -4\pi\delta(\vec{x}-\vec{x}')\delta(t-t')$$. Using a Fourier transform, $$\Psi(\vec{x},\omega) = \frac{1}{2\pi}\int_\mathbb{R}dt\Psi(\vec{x},t)\exp(i\omega t)$$. The inverse fourier transform, $$\Psi(\vec{x},t) = \int_\mathbb{R}d\omega\Psi(\vec{x},\omega)\exp(-i\omega t)$$. $$\mathcal{F}[\nabla^2G(\vec{x},t;\vec{x}',t') - \frac{1}{c^2}\frac{\partial^2G}{\partial t^2} = -4\pi\delta(\vec{x}-\vec{x}')\delta(t-t')] \Rightarrow \nabla^2G(\vec{x},\omega;\vec{x}',\omega') +\frac{\omega^2}{c^2}G(\vec{x},\omega;\vec{x}',\omega') = -4\pi\delta(\vec{x}-\vec{x}')$$. So, $$\nabla^2\tilde{G} + k^2\tilde{G} = -4\pi\delta(\vec{x}-\vec{x}')$$. Note that when $$k=0$$ we get the static case.

Let $$\vec{R} = \vec{x} - \vec{x}'$$. So, $$\tilde{G}_k(R)\Rightarrow \frac{1}{R}\frac{d^2}{dR^2}(R\tilde{G}_k(R)) + k^2\tilde{G}_k(R) = -4\pi\delta(\vec{R})$$. Hence, $$\frac{d^2}{dR^2}(R\tilde{G}_k(R)) + k^2(R\tilde{G}_k(R)) = 0\rightarrow \tilde{G}_k(R) = \frac{\exp(\pm ikR)}{R}$$. So we get waves diverging and converging to the source.

Then the general solution, $$\tilde{G}_k(R) = A\tilde{G}_k^+(R) + B\tilde{G}_k^-(R)$$ for the diverging and converging waves. Our time-dependent green function is then, $$\nabla^2 G(\vec{x},t;\vec{x}',t') - \frac{1}{c^2}\frac{\partial^2G}{\partial t^2} = -4\pi\delta(\vec{x}-\vec{x}')\delta(t-t')$$. Using our inverse transform, $$\int dt\exp(i\omega t)$$: $$\nabla^2G(\vec{x},\omega;\vec{x}',t') + k^2G(\vec{x},\omega;\vec{x}',t') = -4\pi\delta(\vec{x}-\vec{x}')\exp(i\omega t')$$. Hence, $$G(\vec{x},\omega;\vec{x}',t') = G_k^{\pm}(R)\exp(i\omega t')$$. Then,

\begin{align} G(\vec{x},t;\vec{x}',t') &= \frac{1}{2\pi}d\omega \frac{\exp(\pm i\frac{\omega}{c}R)}{R}\exp(i\omega t')\exp(i\omega t) \nonumber \\ &= \frac{1}{R} \int d\omega \exp\left(i\omega\left(t' -\left(t \mp \frac{R}{c}\right)\right)\right) \nonumber \\ &= \frac{1}{R}\delta\left(t'-t\pm \frac{R}{c}\right) \end{align}

So,

$$G^{\pm} = \frac{1}{|\vec{x}-\vec{x}'|}\delta\left(t'-\left(t\mp\frac{|\vec{x}-\vec{x}'|}{c}\right)\right)$$

Hence, $$\Psi(\vec{x},t) = \int d^3x'dt' f(\vec{x}',t')G^{\pm}$$. For the outgoing wave, $$\Psi(\vec{x},t) = \int d^3x'dt' f(\vec{x}',t')G^{+} + \Psi_{homogeneous}(\vec{x},t)$$. Where $$\nabla^2\Psi_{homogeneous} - \frac{1}{c^2}\frac{\partial^2\Psi_{homogeneous}}{\partial t^2} = 0$$. $$G^+$$ is the retarded Green’s function $$\left(t=t'+\frac{R}{c}\right)$$ and $$G^-$$ is the advanced Green’s function$$\left(t=t'-\frac{R}{c}\right)$$.

NOTE: Fourier Transform of General wave equation: $$\mathcal{F}[\nabla^2\psi - \frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2} = -4\pi f(\vec{x},t)] \Rightarrow \nabla^2\psi(\vec{x},\omega) + \frac{\omega^2}{c^2}\psi(\vec{x},\omega)= -4\pi f(\vec{x},\omega)$$.

Lets consider the retarded solution for no homogeneous wave, $$\Psi(\vec{x},t) = \int d^3x' \left.\frac{f(\vec{x}',t')}{|\vec{x}-\vec{x}'|}\right|_{t'=t-\frac{|\vec{x}-\vec{x}'|}{c}}$$. Hence, $$\Phi(\vec{x},t) = \frac{1}{4\pi\varepsilon_0}\int d^3x'\frac{[\rho(\vec{x}',t')]_{ret}}{|\vec{x}-\vec{x}'|}$$ Also, $$\vec{A}(\vec{x},t) = \frac{\mu_0}{4\pi}\frac{[\vec{J}(\vec{x},t)]_{ret}}{|\vec{x}-\vec{x}'|}d^3x'$$.

MIDTERM MATERIAL ENDS HERE

## Next Topic

Using double curl, $$\left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\vec{B} = -\mu_0\nabla\times\vec{J}$$. Also, $$\left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\vec{E} = \frac{1}{\varepsilon_0}\left(\nabla\rho+\frac{1}{c^2}\frac{\partial\vec{J}}{\partial t}\right)$$.

$$\vec{E} = \frac{1}{4\pi\varepsilon_0}\int\frac{1}{R}\left[-\nabla'\rho-\frac{1}{c^2}\frac{\partial\vec{J}}{\partial t'}\right]_{ret}d^3x'$$

$$\vec{B} = \frac{\mu_0}{4\pi}\int\frac{1}{R}\left[\nabla'\times\vec{J}\right]_{ret}d^3x'$$

Note that $$[\nabla'\rho]_{ret}\neq\nabla'[\rho]_{ret}$$ since $$\rho(\vec{x}',t')$$ has $$t'=t-\frac{|\vec{x}-\vec{x}'|}{c}$$. Specifically, $$[\nabla'\rho]_{ret} + \left(\frac{\partial \rho}{\partial t'}\right)_{ret}\nabla'\left(t-\frac{R}{c}\right) = \nabla'[\rho]_{ret}$$. Hence: $$[\nabla'\rho]_{ret} = -\left(\frac{\partial \rho}{\partial t'}\right)_{ret}\hat{R} + \nabla'[\rho]_{ret}$$. Then, $$\vec{E}(\vec{x},t) = \frac{1}{4\pi\varepsilon}\int\left([\rho]_{\text{ret}}\frac{\hat{R}}{R^2} + \left(\frac{\partial \rho}{\partial t'}\right)_{ret}\frac{\hat{R}}{cR} - \left(\frac{\partial \vec{J}}{\partial t'}\right)_{ret}\frac{1}{c^2R}\right)dV$$. $$\vec{B}(\vec{x},t) = \frac{\mu_0}{4\pi}\int\left([\vec{J}]_{ret}\times\frac{\hat{R}}{R^2} + \left(\frac{\partial \vec{J}}{\partial t'}\right)_{ret}\times\frac{\hat{R}}{cR}\right)dV$$.

Note if we have a static field, no time dependence, we have the Colomb and Biot-Savart laws.

For near-field: $$R\ll\lambda$$.

For far-field: $$R\gg\lambda$$. Then we can ignore the columb terms. Additionally, the second term in the electric field typically is much smaller.

So, $$c\vec{B}=\hat{R}\times\vec{E}$$.

Microscopic:

\begin{align} \nabla\cdot\vec{e} &= \frac{\eta}{\epsilon_0} \\ \nabla\cdot\vec{b} &= 0 \\ \nabla\times\vec{b} &= \mu_0\vec{j} + \frac{1}{c^2}\frac{\partial \vec{e}}{\partial t} \\ \nabla\times\vec{e} &= -\frac{\partial \vec{b}}{\partial t} \end{align}

Note, $$\nabla\cdot\vec{E} = \frac{1}{\varepsilon_0}\langle\eta_f + \eta_b\rangle = \frac{1}{\varepsilon_0}(\rho_f + \rho_b)$$

Consider Ampere’s law. $$\nabla\times\frac{\vec{B}}{\mu_0} = \langle\vec{j}\rangle + \varepsilon_0 \frac{\partial\vec{E}}{\partial t} = \langle\vec{j}_f + \vec{j}_b\rangle + \varepsilon_0 \frac{\partial\vec{E}}{\partial t}$$. $$dI = \frac{\partial Q}{\partial t} = \frac{\partial (Pda)}{\partial t} \Rightarrow \vec{J}_{p} = \frac{dI}{da} = \frac{\partial P}{\partial t}$$ is the polarization current, for when the polarization is changing in time. So, $$\nabla\times\frac{\vec{B}}{\mu_0} = \langle\vec{j}_f + \vec{j}_b + \vec{j}_p\rangle + \varepsilon_0 \frac{\partial\vec{E}}{\partial t}$$. Hence, $$\nabla\times\left(\frac{\vec{B}}{\mu_0}-\vec{M}\right) = \vec{J}_f + \frac{\partial}{\partial t}(\epsilon_0\vec{E} + \vec{P}) \Rightarrow \nabla\times\vec{H} = \vec{J}_f = \frac{\partial\vec{D}}{\partial t}$$.

## Poynting Theorem

Starting from Maxwell’s equation (for macroscopic materials). Consider Faraday’s law $$\nabla\times\vec{E} + \frac{\partial\vec{B}}{\partial t} = 0$$ and Ampere’s law $$\nabla\times\vec{H} - \frac{\partial \vec{D}}{\partial T} = \vec{J}$$.

Multiplying Ampere’s law by $$\vec{E}$$ and subtracting, we get the exact equation,

\begin{align} \nabla\cdot(\vec{E}\times\vec{H}) + \vec{H}\cdot\frac{\partial\vec{B}}{\partial t} + \vec{E}\cdot\frac{\partial\vec{D}}{\partial t} = -\vec{J}\cdot\vec{E} \nonumber \\ \nabla\cdot\vec{S} + \vec{H}\cdot\frac{\partial\vec{B}}{\partial t} + \vec{E}\cdot\frac{\partial\vec{D}}{\partial t} = -\vec{J}\cdot\vec{E}. \end{align}

For linear media, $$\vec{E}=\varepsilon\vec{E},\vec{B} = \mu\vec{H}$$. So,

\begin{align} \nabla\cdot\vec{S} + \mu\vec{H}\cdot\frac{\partial\vec{H}}{\partial t} + \varepsilon\vec{E}\cdot\frac{\partial\vec{E}}{\partial t} = -\vec{J}\cdot\vec{E} \nonumber \\ \nabla\cdot\vec{S} + \frac{\partial}{\partial t}\left(\frac{1}{2}\vec{E}\cdot\vec{D} + \frac{1}{2}\vec{H}\cdot\vec{B}\right) = -\vec{J}\cdot\vec{E} \nonumber \\ \nabla\cdot\vec{S} + \frac{\partial}{\partial t}\left(u_E + u_M\right) = -\vec{J}\cdot\vec{E} \nonumber \\ \nabla\cdot\vec{S} + \frac{\partial u}{\partial t} = -\vec{J}\cdot\vec{E}. \end{align}

For no current, we get $$\nabla\cdot\vec{S} + \frac{\partial u}{\partial t} = 0$$ gives a continuity equation. Hence currents become a source and $$\vec{S}$$ is an energy flux since $$u$$ is an energy density.

Consider the integral form of the Poynting vector,

\begin{align} \oiint\vec{S}\cdot\hat{n}da &= -\frac{d}{dt}\int udV - \int \vec{J}\cdot\vec{E}dV \\ \oiint\vec{S}\cdot\hat{n}da &= -\frac{d}{dt}U - \int \rho\vec{v}\cdot\vec{E}dV \\ \oiint\vec{S}\cdot\hat{n}da &= -\frac{d}{dt}(U + T) \end{align}

So, if the energy in a region is not changing, then no radiation is emitting. If energy in a region is changing, radiation must be coming in or leaving.

### Momentum of E&M Wave

Classically: $$\vec{F} = q(\vec{E} + \vec{v}\times\vec{B}) = \frac{d\vec{p}_{mech}}{dt}$$. Doing a volume integration of the system, $$\frac{d\vec{p}_{mech}}{dt} + \frac{d\vec{p}_{em}}{dt} = \vec{F}$$ with $$\frac{d\vec{p}_{em}}{dt} = \int\vec{g}dV$$ where $$\vec{g} = \frac{1}{c^2}\vec{S}$$.

A photon has energy $$\hbar\omega$$ then the energy density is $$\frac{N\hbar\omega}{V}$$ then the flux is $$\frac{N\hbar\omega}{V}\vec{c} = \vec{S}$$. So, the momentum is then (say we have a single photon in a unit volume), $$S = \hbar\omega c$$ so, $$g = \frac{\hbar\omega}{c}$$ so the momentum is $$\frac{\hbar\omega}{c} = \frac{\hbar 2\pi c}{c\lambda} = \hbar k$$.

Consider the harmonic field: $$\vec{F}(x,t) = \Re[\vec{F}(x)\exp(-i\omega t)] = \frac{1}{2}(\vec{F}(x)\exp(-i\omega t) + \vec{F}^*(x)\exp(i\omega t))$$

Then the product of two fields, $$\vec{F}(x,t)\cdot\vec{G}(x,t) = \frac{1}{4}(\vec{F}^*\vec{G} + \vec{F}\cdot\vec{G}^* + \vec{F}\cdot\vec{G}\exp(-2i\omega t) + \vec{F}^*\vec{G}^*\exp(2i\omega t))$$.

If we time-average this, $$\langle \vec{F}(x,t)\cdot\vec{G}(x,t)\rangle = \frac{1}{4}\langle\vec{F}^*\vec{G} + \vec{F}\cdot\vec{G}^*\rangle = \frac{1}{2}\Re(\vec{F}^*\cdot\vec{G})$$.

So, the Complex Poynting Theorem can be derived. $$\nabla\times\vec{E} = i\omega\vec{B}$$ and $$\nabla\times\vec{J} = \vec{J}-i\omega\vec{D}$$. Taking the complex conjugate of the second equation and multiplying by $$\vec{E}$$ and multiplying the first by $$\vec{H}$$,

\begin{align*} \vec{H}^*\cdot\nabla\times\vec{E} &= i\omega\vec{B}\cdot\vec{H}^* \\ \vec{E}\cdot\nabla\times\vec{H}^* &= \vec{E}\cdot\vec{J}^* -i\omega\vec{D}^*\cdot\vec{E}. \end{align*}

Subtracting,

$$\nabla\cdot(\vec{E}\times\vec{H}^*) &= -i\omega(\vec{E}\cdot\vec{D}^* - \vec{B}\cdot\vec{H}^*) - \vec{J}^*\cdot\vec{E}$$

So, $$\vec{S} = \frac{1}{2}\vec{E}\times\vec{H}^*$$. Then, $$\nabla\times\vec{S} = -\frac{i\omega}{2} (\vec{E}\cdot\vec{D}^* - \vec{B}\cdot\vec{H}^*) - \frac{1}{2}\vec{J}^*\cdot\vec{E}$$. So, $$\int\vec{S}\cdot\hat{n}da = -\frac{i\omega}{2}\int dV\left(\vec{E}\cdot\vec{D}^* - \vec{B}\cdot\vec{H}^*\right) - \frac{1}{2}\int dV\vec{J}^*\cdot\vec{E}$$.

So, $$\Re\int\vec{S}\cdot\hat{n}da = -\frac{1}{2}I^*V = -\frac{1}{2}|I|^2R$$.

Note that the half comes from the time-average.

Then, $$I_{rms}^2R$$ is the energy loss of the system.

## Monochromatic Waves

It is useful to write $$\vec{E},\vec{B},\vec{D},\vec{H}\propto\exp(-i\omega t)$$.

Then the field, $$\vec{F}(x,t) = \vec{F}(x)\exp(-i\omega t)$$.

$$\langle F(x,t)G(x,t)\rangle_{time} = \frac{1}{2}\Re\left[F^*(x)G(x)\right]$$.

$$\nabla\cdot(\vec{E}\times\vec{H}^*) = \nabla\cdot\vec{S} = -i\omega\left(\vec{E}\cdot\vec{D}^*-\vec{B}\cdot\vec{H}^*\right) - \vec{J}^*\cdot\vec{E}$$. $$u_E = \frac{1}{4}\vec{E}\cdot\vec{D}^*$$, $$u_M = \frac{1}{4}\vec{B}\cdot\vec{H}^*$$.

Taking the real part, $$\Re\left[\nabla\cdot\vec{S}=-i\omega(\vec{E}\cdot\vec{D}^*-\vec{B}\cdot\vec{H}^*) - \vec{J}^*\cdot\vec{E}\right]$$.

When we have $$R,C,L$$ components in a circuit, we have $$V_i=ZI_i$$. Where $$Z = R - iX$$ and $$X$$ relates to both the inductance and capacitance.

Note that $$-i\omega(\vec{E}\cdot\vec{D}^* - \vec{B}\cdot\vec{H}^*)$$ relates to the reactance and $$-\vec{J}^*\cdot\vec{E}$$ relates to resistance. $$\Re,\Im$$ relates to reactance and resistance.

Created: 2023-06-25 Sun 02:35

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