# Monochromatic Plane Waves

$$\rho=0,\vec{J}=0$$.

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

Free space: $$\square\vec{E} = 0,\square\vec{B} = 0$$.

In a medium: $$\nabla^2\vec{E} = \varepsilon\mu\frac{\partial\vec{E}}{\partial t^2}$$, $$\nabla^2\vec{B} = \varepsilon\mu\frac{\partial\vec{B}}{\partial t^2}$$.

$$\vec{E},\vec{B}\propto\exp(-i\omega t)$$.

For a plane wave, $$\nabla^2\vec{E} = \frac{\partial^2\vec{E}}{\partial z^2} + \varepsilon\mu\omega^2\vec{E} = 0$$. So, $$\vec{E} = \vec{\varepsilon}_-\exp\left(-i(\omega t-kz)\right) + \vec{\varepsilon}_+\exp\left(i(kz-\omega t)\right)$$ with $$k^2 = \epsilon\mu\omega^2$$. In free space, $$k = \frac{\omega}{c}$$. In medium, $$k = \sqrt{\varepsilon_r\mu_r}\frac{\omega}{c} = n\frac{\omega}{c}$$. Note the index of refraction is $$n = \sqrt{\frac{\varepsilon\mu}{\varepsilon_0\mu_0}}$$. In most mediums, we have a dielectric and so $$n=\sqrt{\varepsilon_r}$$. $$v = \frac{c}{n}$$. In medium, $$n>1$$ so speed of light is reduced below the speed of causality and note that index of refraction in a medium is usually a function of frequency. Note, if you know everything about the dispersion relation, $$k^2 = f(\omega)^2\Rightarrow k = f(\omega)$$, then you know everything about the wave.

In general, $$\exp(i(\vec{k}\cdot\vec{x} - \omega t))$$. Note that this dispersion is isotropic, i.e. has no explicit directional dependence. The wave goes in the direction of $$\hat{k}$$. Note also $$\frac{2\pi}{\lambda} = k$$. The phase velocity is $$v_\psi= \frac{\lambda}{T}$$. $$k = n(\omega)\frac{\omega}{c}$$.

In most media the permeability is roughly the same as the vacuum permeability, $$n=\sqrt{\varepsilon_r}$$.

From $$\nabla\cdot\vec{D}=0$$ and $$\nabla\cdot\vec{B}=0$$ we see that $$i\vec{k}\cdot\vec{E} =0$$ and $$\vec{k}\cdot\vec{B} = 0$$ so the wave is propogating normal to both the electric and magnetic field in a linear medium. Note, $$\vec{k}\times\vec{E} = \omega\vec{B}$$. Then, $$\vec{B} = \frac{\vec{k}}{\omega}\times\vec{E}$$ and $$\vec{E} = -\frac{c}{n}\hat{k}\times\vec{B}$$

Energy density: $$u = \frac{1}{2}\vec{E}\cdot\vec{D} + \frac{1}{2}\vec{H}\cdot\vec{B}$$ so $$u = \frac{1}{4}\vec{E}\cdot\vec{D}^* + \frac{1}{4}\vec{B}\cdot\vec{H}^*$$.

$$\vec{S} = \frac{1}{2}\vec{E}\times\vec{H}^* = \frac{1}{2\mu}\vec{E}\times\vec{B}^* = \frac{1}{2\mu}|\vec{E}|\cdot|\vec{B}|\hat{k} = u\vec{v}$$.

Note we have the continuity equation then to be $$\nabla\cdot\vec{S} + \frac{\partial u}{\partial t} = 0$$ to be the continuity equation.

$$\vec{E}(\vec{x},t) = \vec{\mathcal{E}}\exp(i(\vec{k}\cdot\vec{x}-\omega t)) = e_1\hat{\epsilon}_1\exp(i(\vec{k}\cdot\vec{x}-\omega t)) + e_2\hat{\epsilon}_2\exp(i(\vec{k}\cdot\vec{x}-\omega t))$$. The real part gives, $$\Re\vec{E}(\vec{x},t) = \Re(e_1)\hat{\epsilon_1}\cos(\omega t) + \Re(e_2)\hat{\epsilon_2}\cos(\omega t - \delta)$$.

Linear Polarization, the polarization is in phase: $$\vec{E}_r(t) = \hat{\epsilon}_1a_2\cos\omega t + \hat{\epsilon}_2a_2\cos(\omega t-\delta)\Rightarrow \delta = 0$$.

Consider $$\vec{x}=0$$. Then, $$\vec{E}_r(t) = (\hat{\ep}_1a_1+\hat{\ep}_2a_2\exp(i\delta))\exp(-i\omega t)$$ Then $$\vec{E}_r = \hat{ep}_1a_1\cos\omega t + \hat{\ep}_2a_2\cos(\omega t-\delta)$$. For linear polarizaiton, $$\delta = 0,\pi$$ or $$a_1\vee a_2=0$$, then $$\vec{E}_r(t) = (a_1\hat{\ep}_1 + a_2\hat{\ep_2})\cos\omega t$$.

If $$\delta = \pm\frac{\pi}{2}$$ and $$a_1=a_2=a$$. Then, $$\vec{E}(t) = a(\hat{\ep}_1+\exp\left(\pm i\frac{\pi}{2}\right)\hat{\ep}_2)\exp(-i\omega t) = a(\hat{\ep}_1\pm i\hat{\ep}_2)\exp(-i\omega t)$$. Also, $$\vec{E}_r(t) = (\hat{\ep}_2\cos\omega t\pm\hat{\ep}_2\sin\omega t)a$$. Hence, we get circularly polarized light. $$\delta=\frac{\pi}{2}$$ gives $$+$$ helicity (left-circular, ccw). $$\delta=-\frac{\pi}{2}$$ gives $$-$$ helicity (right-circular, cw). Then,

\begin{align*} \vec{E}(t) &= a(\hat{\ep}_1\pm i\hat{\ep}_2)\exp(-i\omega t) \\ &= \sqrt{2}a\frac{1}{\sqrt{2}}(\hat{\ep}_1\pm i\hat{\ep}_2)\exp(-i\omega t) \\ &= \sqrt{2}a\hat{\ep}_{\pm}\exp(-i\omega t) \end{align*}

This helicity is related to the fact that light’s magnetic quantum number, angular momentum, $$m\in\{-1,1,0\}$$. By relativity, $$0$$ is not possible. And so $$\pm1$$ relates to helicity. Thus polarized light carries angular momentum.

The intensity: $$I\propto |\vec{E}|^2$$.

Polarizers can select specific linear polarizations or circular polarizations. By using these polarizers, you can determine the field components.

Right or left circularity can be determined by using thumb but pointing away rather than toward you.

Created: 2023-06-25 Sun 02:28

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