PHYS 208 Lecture 22

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Electromagnetic Waves

EM Spectrum

Mechanical Wave Equation

Forces on a string (fixed at both ends) when plucked:

• Tension to left and right
• Mass density per unit length of $\rho$, so $m=\rho L$

For length $\mathrm{\Delta }x$, the sum of these forces add to $F=ma=\rho \mathrm{\Delta }x\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}$:

Note: We use $y$ in the acceleration above because the string moves up and down.

Why haven't I learned this before in Calculus? $\text{slope}=\frac{\mathrm{d}y}{\mathrm{d}x}=\tan \theta$

$\begin{array}{r}T(\sin {\theta }_R-\sin {\theta }_L)=\rho \mathrm{\Delta }x\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}\\ (\tan {\theta }_R-\tan {\theta }_L)=\frac{\rho \mathrm{\Delta }x}{T}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}\\ \frac{\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}_R^2}-\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}_L^2}}{\mathrm{\Delta }x}=\frac{\rho \mathrm{\Delta }x}{T}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}\end{array}$

$\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{x}^2}=\frac{\rho }{T}\frac{{\mathrm{d}}^{2}y}{\mathrm{d}{t}^2}$

The solution for this thing is

$y(x,t)=A\sin (\omega t-kx)$

cosine could also be used, the terms inside the sine could be switched, etc.

Note the following:

• frequency [Hz]: $f$
• angular frequency: $\omega =2\pi f$
• amplitude (midline to crest/trough): $A$
• phase (offset): $\varphi$
• wavelength: $\lambda$
• period: $T=1/f$
• wave number: $k=\frac{2\pi }{\lambda }$
• velocity of propagation: $v=\lambda f$

Recall Maxwell's Equations

(See PHYS 208 Lecture 19#Maxwell's Equations→)

$\oint \overrightarrow{E}\cdot \mathrm{d}\overrightarrow{A}=\frac{q_{en}}{{ϵ}_0}$

$\oint \overrightarrow{B}\cdot \mathrm{d}\overrightarrow{A}=0$

$\oint \overrightarrow{E}\cdot \mathrm{d}\overrightarrow{\ell }=-\frac{\mathrm{d}{\mathrm{\Phi }}_m}{\mathrm{d}t}$

$\oint \overrightarrow{B}\cdot \mathrm{d}\overrightarrow{\ell }={\mu }_0(i_c+{ϵ}_0\frac{\mathrm{d}{\mathrm{\Phi }}_E}{\mathrm{d}t})$

When we are very far from charges and currents, $\frac{q_{en}}{{ϵ}_0}$ and $i_c$ terms will be zero. EM waves propagate through a vacuum... So what's the medium carrying EM waves? Under these conditions, the equations are highly symmetric.

EM Wave Equation

Consider a plane-wave with $\overrightarrow{E}$ in the $\hat{ȷ}$ direction and $\overrightarrow{B}$ in the $\hat{ı}$ direction, so the two vectors are perpendicular to each other.

Faraday's law says that $\oint \overrightarrow{E}\cdot \mathrm{d}\overrightarrow{\ell }=-\frac{\mathrm{d}{\mathrm{\Phi }}_m}{\mathrm{d}t}$, so after much complicated math, we arrive at the same equation for mechanical waves:

$\frac{\partial E(x,t)}{\partial x}=-\frac{\partial B(x,t)}{\partial t}$         and         $\frac{\partial B(x,t)}{\partial x}={\mu }_0{ϵ}_0\frac{\partial E(x,t)}{\partial t}$

$\frac{{\partial }^{2}E(x,t)}{\partial x^2}={\mu }_0{ϵ}_0\frac{{\partial }^{2}E(x,t)}{\partial t^2}$

The propagation velocity $\frac{1}{\sqrt{{\mu }_0{ϵ}_0}}=v=c$ is the speed of light! Direction of propagation is in $\overrightarrow{E}\times \overrightarrow{B}$ direction (perpendicular to both vectors)

Magnitude of each vector is given by

$E(x,t)=cB(x,t)$

Finally, we arrive at the solution to the wave equation:

$\overrightarrow{E}(x,t)=E_{max}\sin (\omega t-kx)\hat{ȷ}$

$\overrightarrow{B}(x,t)=B_{max}\sin (\omega t-kx)\hat{k}$

where $v=c=\lambda f=\frac{\omega }{k}$

Note: Interesting fact: Microwave ovens use standing waves

Energy and Momentum in EM Waves

Recall the energy densities of electric and magnetic fields:

• $u_e=\frac{1}{2}{ϵ}_0{E}^2$
• $u_b=\frac{B^2}{2{\mu }_0}$

$u_{tot}=\frac{1}{2}{ϵ}_0{E}^2+\frac{1}{2}\frac{B^2}{{\mu }_0}={ϵ}_0{E}^2$