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

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

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

${\begin{aligned}T(\sin {\theta }_{R}-\sin {\theta }_{L})=\rho \Delta x{\frac {\mathrm {d} ^{2}y}{\mathrm {d} t^{2}}}\\(\tan {\theta }_{R}-\tan {\theta }_{L})={\frac {\rho \Delta x}{T}}{\frac {\mathrm {d} ^{2}y}{\mathrm {d} t^{2}}}\\{\frac {{\frac {\mathrm {d} ^{2}y}{\mathrm {d} x_{R}^{2}}}-{\frac {\mathrm {d} ^{2}y}{\mathrm {d} x_{L}^{2}}}}{\Delta x}}={\frac {\rho \Delta x}{T}}{\frac {\mathrm {d} ^{2}y}{\mathrm {d} t^{2}}}\end{aligned}}$

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

The solution for this thing is

$y(x,t)=A\sin {\left(\omega t-kx\right)}$

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): $\phi$
• 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→)

$\oint {\vec {E}}\cdot \mathrm {d} {\vec {A}}={\frac {q_{en}}{\epsilon _{0}}}$

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

$\oint {\vec {E}}\cdot \mathrm {d} {\vec {\ell }}=-{\frac {\mathrm {d} \Phi _{m}}{\mathrm {d} t}}$

$\oint {\vec {B}}\cdot \mathrm {d} {\vec {\ell }}=\mu _{0}\left(i_{c}+\epsilon _{0}{\frac {\mathrm {d} \Phi _{E}}{\mathrm {d} t}}\right)$

When we are very far from charges and currents, ${\frac {q_{en}}{\epsilon _{0}}}$ 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 ${\vec {E}}$ in the ${\hat {\jmath }}$ direction and ${\vec {B}}$ in the ${\hat {\imath }}$ direction, so the two vectors are perpendicular to each other.

Faraday's law says that $\oint {\vec {E}}\cdot \mathrm {d} {\vec {\ell }}=-{\frac {\mathrm {d} \Phi _{m}}{\mathrm {d} t}}$, 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         ${\frac {\partial B(x,t)}{\partial x}}=\mu _{0}\epsilon _{0}{\frac {\partial E(x,t)}{\partial t}}$

${\frac {\partial ^{2}E(x,t)}{\partial x^{2}}}=\mu _{0}\epsilon _{0}{\frac {\partial ^{2}E(x,t)}{\partial t^{2}}}$

The propagation velocity ${\frac {1}{\sqrt {\mu _{0}\epsilon _{0}}}}=v=c$ is the speed of light! Direction of propagation is in ${\vec {E}}\times {\vec {B}}$ 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:

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

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

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:

• $u_{e}={\frac {1}{2}}\epsilon _{0}E^{2}$
• $u_{b}={\frac {B^{2}}{2\mu _{0}}}$

$u_{tot}={\frac {1}{2}}\epsilon _{0}E^{2}+{\frac {1}{2}}{\frac {B^{2}}{\mu _{0}}}=\epsilon _{0}E^{2}$