## 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, $$\vec{E}=\ve\vec{E},\vec{B} = \mu\vec{H}$$. So,

\begin{align} \nabla\cdot\vec{S} + \mu\vec{H}\cdot\frac{\partial\vec{H}}{\partial t} + \ve\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.

## Stoke’s Paramters

• $$S_0 = |E_1|^2 + |E_2|^2 \propto I$$
• $$S_1 = |E_1|^2 - |E_2|^2$$
• $$S_2 = 2\Re(E_1^*E_2^*) = 2a_1a_2\cos\delta$$
• $$S_3 = 2\Im(E_1^*E_2) = 2a_1a_2\sin\delta$$
• $$S_0^2 = S_1^2 + S_2^2 + S_3^2$$.

Say we have $$E_1=a_1,E_2=a_2\exp(i\delta)$$. Then, $$S_0 = a_1^2 + a_2^2$$ and $$S_1 = a_1^2 -a_2^2 \propto I_1-I_2$$. From these, $$a_1$$ and $$a_2$$ are fully determinable. $$S_2$$ can be determine by putting polarizers at 45$$^\circ$$ above and below the polarization basis vector $$\hat{\epsilon}_1$$, and measure the intensities, $$\frac{I_+}{I_-} = \cos\delta$$, hence you get $$|\delta|$$. $$S_3$$ can be measured with far away plate, $$\lambda/2$$, $$\frac{I_R}{I_L}=\sin\delta$$.

Light-matter interaction: atom absorbs light then eventually radiates (thus light travels slower in a medium). For transparent materials, the loss of the light through the medium is very small.

When light hits a superconductor, light gets reflected. For a conductor, light gets mostly reflected.

## Ohm’s Law

$$\vec{J} = \sigma\vec{E}$$.

For $$\vec{J} = 0 \Rightarrow \nabla^2\vec{E} = \ve\mu\frac{\partial^2\vec{E}}{\partial t^2}$$. For $$\vec{J}\neq 0 \Rightarrow \nabla^2\vec{E} = \ve\mu\frac{\partial^2\vec{E}}{\partial t^2} + \mu\sigma \frac{\partial\vec{E}}{\partial t}$$.

## Losses in Complex Refractive Media

For a monochromatic wave, $$\vec{E}(\vec{x},t) = \vec{\mathcal{E}}\exp(i(\vec{k}\cdot\vec{x} - \omega t))$$ but in the medium $$\vec{k}$$ becomes complex and thus introduces a loss. Then, $$k^2 = \ve\mu\omega^2 + i\mu\omega\sigma$$. Thus, $$k=\sqrt{\ve\mu\omega^2+i\mu\omega\sigma} = k_r+ik_i = (n_r + in_i)\frac{\omega}{c}$$. Then we have the decaying wave, $$\exp(ikz) = \exp(ikr)\exp(-k_iz)$$.

Note, when the electron absorbs the light, it gets accelerated. From the defects in the material, it gets kicked around and the energy gets turned into thermal energy. Thus, the wave decays as it progresses through the medium.

$$\ve_r = n^2 = \frac{1}{\ve_0\mu_0}\left(\ve\mu + \frac{i\mu\sigma}{\omega}\right)$$ Then, $$n^2 = n_r^2 - n_i^2 + 2n_rn_ii$$. So, $$k_{r,i} = \frac{n_{r,i}\omega}{c}$$.

We can write this as, $$\exp(-k_iz) = \exp(-z/\delta)$$ were delta is called the transmission depth.

Suppose that $$\frac{\mu\sigma}{\omega}\gg\ve\mu$$, then $$k^2 = i\mu\sigma\omega$$ so $$k=\sqrt{\mu\sigma\omega}\frac{1}{\sqrt{2}}(1+i)$$. Thus, the imaginary and real parts have the same magnitude.

For copper, $$\rho = \frac{1}{\sigma} = 1.7\cdot10^{-8}\Omega\cdot m$$. Then, $$\delta = \frac{1}{k_i} = \sqrt{\frac{2}{\mu\sigma\omega}} \approx \sqrt{\frac{2}{\mu_0\sigma\omega}} = \frac{6.5\cdot10^{-2}}{\sqrt{\nu}}$$ m. For visible light, $$10^{14}-10^{16}$$. Then, $$\delta\sim10^{-9}-10^{-10}$$ m. The wave has a wavelength of $$10^{-6}$$, so most get reflected. The penetration depth is much shorter than the wavelength.

## Polarizations

For s-polarization, $$\vec{E},\vec{E}',\vec{E}''\|-\hat{y}$$. $$E_0 + E_0'' = E_0'\Rightarrow 1 + r_s = t_s$$ and $$cB_{0t} + cB_{0t}'' = cB_{0t}'\Rightarrow 1-r_s = \frac{n'}{n}\frac{\cos\theta_t}{\cos\theta_i}t_s$$. So, $$t_s = \frac{2n\cos\theta_i}{n\cos\theta_i + n'\cos\theta_t}$$ $$r_s = 1-t_s$$.

For p-polarization, $$\vec{B},\vec{B}',\vec{B}''\|\hat{y}$$. $$(E_0-E_0'')\cos\theta_i = E_0'\cos\theta_t$$ and $$c(B + B'' = B')\Rightarrow nE + nE'' = n'E'$$. Thus, $$t_p = \frac{2n\cos\theta_i}{n\cos\theta_t + n'\cos\theta_i}$$ $$r_p = 1-t_p$$.

Recall $$n\sin\theta_i = n'\sin\theta_t$$ hence $$\theta_t$$ depends on $$\theta_i$$. Therefore, we have fully determined the system: $$\cos\theta_t = \sqrt{1 - \frac{n^2}{n'^2}\sin^2\theta_i}$$.

## Special Angles

If $$\theta_i=0$$ then, $$\cos\theta_t = 1$$, $$t_s = \frac{2n}{n+n'} = t_p$$ and $$r_s = \frac{n'-n}{n+n'} = -r_p$$. For s-pol, $$\vec{E}\|-\hat{y}$$. For p-pol, $$\vec{E}\|\vec{E}'\|-\vec{E}''\|\hat{x}$$.

The total reflectance is then,

The brewster angle, when p-reflectance goes to zero, $$\tan\theta_B = \frac{n'}{n}$$. There is no reflectance since the dipoles are oscillating parallel to the reflected wave, which means it does not show up.

Total internal reflection: $$\theta_i = \theta_c$$ at $$\theta_t = 90^\circ$$. That is, there is only reflected light: $$R = 1$$. We get a decaying wave and from phase matching $$k_x=k_x'$$. And $$\vec{k}' = k_x\hat{x} + k_z'\hat{z}$$. We have the dispersion relation, $$k'^2 = n'^2\frac{\omega^2}{c^2} = k_x^2 + k_z'^2$$. So, $$k_z'$$ must be imaginary to make the phase matching possible above the critical angle. So, if we let $$k_z' = \frac{i}{\delta}$$. Then we get the decaying wave of $$\exp\left(-\frac{z}{\delta}\right)$$. Further, $$k_z' = ik\sqrt{\sin^2\theta_i - \sin^2\theta_c}$$. So, $$\delta = \frac{\lambda}{2\pi\sqrt{\sin^2\theta_i - \sin^2\theta_c}}$$.

For $$n'\theta_c$$, $$\cos\theta_t = \sqrt{1-\frac{n^2}{n'^2}\sin^2\theta_c} = \sqrt{1 - \frac{\sin^2\theta_i}{\sin^2\theta_c}} = i\sqrt{\frac{\sin^2\theta_i}{\sin\theta_c}-1}$$. Then, $$r = \frac{a-ib}{a + ib} \Rightarrow R = |r|^2 = 1$$.

## Frequency Dispersion

$$k = \frac{n\omega}{c} = \frac{\omega}{v_{ph}}$$.

### Drude-Lorentz SHOM

• Lorentz - Dielectric medium

Consider if you have an electron and atom. Apply a wave. The electron is initially in roughly a harmonic well, $$V(x) = \frac{1}{2}kx^2$$. So for an ideal dielectric, $$F(x) = m\frac{d^2x}{dt^2} = -e\vec{E} - k\vec{x}$$. Realistically, we have a damping force, $$F(x) = m\frac{d^2x}{dt^2} = -e\vec{E} - k\vec{x} - \gamma\frac{d\vec{x}}{dt}$$. For a 1d system, $$\frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + \omega_0^2x = -\frac{e}{m}E_x$$, where $$\omega_0^2 = \frac{k}{m}$$. Using an electric dipole approximation for our wave, $$E = E_0\exp(i(kz-\omega t)) = E_0\exp(-i\omega t)$$. Then, for a 1d system, $$\frac{d^2x}{dt^2} + \gamma \frac{dx}{dt} + \omega_0^2x = -\frac{e}{m}E_0\exp(-i\omega t)$$, where $$\omega_0^2 = \frac{k}{m}$$. Then, $$x(t) = x_0\exp(-\omega t)$$ with $$x = -\frac{e}{m}\frac{E}{\omega_0^2 - \omega^2) - i\omega\gamma}$$. So, $$\vec{p} = -e\vec{x} = \frac{e^2}{m}\frac{E}{\omega_0^2 - \omega^2 -i\omega\gamma}$$ Then, $$\vec{P} = N\vec{p} = \varepsilon_0\chi_e\vec{E}$$. So, $$\chi_e(\omega) = \frac{Ne^2}{m\varepsilon_0}\frac{1}{\omega_0^2 - \omega^2 -i\omega\gamma}$$.

For a collection of these oscillators, $$\sum_{j=1}^z = \frac{f_i}{\omega_0^2-\omega^2-i\omega\gamma}$$.

For $$\gamma\ll\omega_0$$, $$\chi_e(\omega)\approx \frac{Ne^2}{2m\varepsilon_0\omega_0}\frac{1}{\omega_0-\omega-i\frac{\gamma}{2}} = \frac{A}{\omega_0 - \omega - i\Gamma}$$.

$$k = k_r + ik_i = \beta + i\frac{\alpha}{2}$$. Then, $$I\propto |E_0|^2\exp(-\alpha z)$$. Also, $$k^2 = \alpha^2 + \frac{\beta^2}{4} + i\alpha\beta = n^2\omega^2/c^2 = \varepsilon_r\frac{\omega^2}{c^2}$$. So, $$\alpha^2 + \frac{\beta^2}{4} = \frac{\omega^2}{c^2}\Re[\varepsilon_r]$$. $$\alpha\beta =\frac{\omega^2}{c^2}\Im[\varepsilon_r]$$.

For a gaseous medium. $$\varepsilon_r = 1 + \chi_e \Rightarrow n \approx 1 + \frac{1}{2}\chi_e$$. Then, $$\alpha = \frac{\omega}{c} = \frac{2\pi}{\lambda}$$. So, $$n_R \approx 1 + \frac{1}{2}\Re(\chi_e) \approx 1$$ and $$n_I \approx \frac{1}{2}\Im \chi_e$$ Then, $$\beta = \frac{\omega}{c}\Im\chi_e$$.

### Drude Model

$$\vec{F} = m\frac{d\vec{v}}{dt} = -e\vec{E} - m]gamma\vec{v}$$ $$\vec{E}\propto\exp(-i\omega t), \vec{v} = \vec{v}_0\exp(-i\omega t)$$. $$\vec{v} = -\frac{e}{m}\frac{1}{\gamma-i\omega}\vec{E}$$

$$\vec{J} = \rho\vec{v} = -N_ee\vec{v} = \sigma\vec{E}$$. $$\frac{N_ee^2}{m}\frac{1}{\gamma-i\omega} = \sigma$$. $$\sigma(\omega) = \frac{N_e e^2}{m}\frac{1}{\gamma-i\omega} = \frac{N_e e^2}{m}\frac{\gamma+i\omega}{\gamma^2 + \omega^2}$$ Then the scattering time is, $$\tau_0 = \frac{1}{\gamma}$$. $\sigma(\omega) = \frac{N_e e^2}{m}\frac{\tau_0+i\omega\tau_0^2}{1 + \omega^2\tau_0^2}$

In DC limit, $$\omega = 0$$, $$\sigma_0 = \frac{N_ee^2}{m}\tau_0$$. For $$\gamma\sim 10^{14}$$ Hz, waves even as high as $$10^{12}$$ Hz are still behaving with a DC ’constant’ resistance. $$\nabla\times\vec{H} = \vec{J} + \frac{\partial\vec{D}}{\partial t} = \sigma\vec{E}-i\omega\varepsilon_b\vec{E} = -i\omega\left(\varepsilon_b + i\frac{\sigma}{\omega}\right)\vec{E}$$. $$\varepsilon_r = \frac{\varepsilon_b}{\varepsilon_0} + i\frac{\sigma}{\varepsilon_0\omega}$$.

At high-frequency, $$\sigma(\omega) = \frac{N_ee^2}{m}\frac{i\omega\tau_0^2}{\omega^2\tau_0^2} = \frac{N_ee^2}{m}\frac{i}{\omega}$$ $$\varepsilon_r(\omega) = \frac{\varepsilon_b}{\varepsilon_0} + i\frac{N_ee^2}{m\omega\varepsilon_0}\frac{1}{\gamma-i\omega} = \frac{\varepsilon_b}{\varepsilon_0} - \frac{N_0e^2}{\varepsilon_0 m\omega^2} \approx 1 - \frac{N_0e^2}{\varepsilon_0 m\omega^2} = 1-\frac{\omega_P^2}{\omega^2}$$ Where this is called the plasma frequency.

For $$\omega<\omega_p$$ $$\varepsilon_r < 0$$ so it is purely imaginary $$\varepsilon_r$$ and so we get total reflectance. For $$\omega>\omega_p$$ $$\varepsilon_r > 0$$.

### Gaussian Pulse

$$E(t) = E_0\exp(-at^2)\cos(\omega_0 t)$$. $$I_0(t) = \frac{1}{2}\varepsilon_0 cnE_0^2\exp(-2at^2)$$. Since if you Fourier transform a gaussian, you get a guassian. A well localized packet has a wide frequency spectrum.

When the group velocity is equal to the phase velocity, the pulse remains the same and just translates.

For $$v_g < v_{ph}$$. then the gropu velocity stays with the peak but the fast oscillation moves past the peak at a faster speed. For $$v_g > v_{ph}$$. then the gropu velocity stays with the peak but the slow oscillation moves past the peak backwards.

### Beats Phenomena

$$\tilde{E}_{tot}(t) = \tilde{E}_0\exp(i\omega_1 t) + \tilde{E}_0\exp(i\omega_2 t)$$. $$\omega_{ave} = \frac{\omega_1 + \omega_2}{2}, \Delta\omega = \frac{\omega_1 - \omega_2}{2}$$. Then, $$\tilde{E}_{tot}(t) = 2\tilde{E}_0\exp(i\omega_{ave}t)\cos\Delta\omega t$$. The cosine gives the envelope.

Similarly with the wavenumbers, $$E_{tot}(x, t) = 2E_0\cos(\Delta k x-\Delta\omega t)\exp(i(k_{avg}x-\omega_{avg}t))$$. Group velocity: $$v_{g} = \frac{d\omega}{dk}$$. From the dispersion relation, $$v_g = \left(\frac{dk}{d\omega}\right)^{-1} = \left(\frac{1}{c}\left(n + \omega\frac{dn}{d\omega}\right)\right)^{-1} = \frac{v_{ph}}{1 + \frac{\omega}{n}\frac{dn}{d\omega}}$$ Phase velocity: $$v_{ph} = \frac{\omega_{avg}}{k_{avg}}$$.

Created: 2024-05-30 Thu 21:20

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