Maxwell’s equations in matter: boundary conditions

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Section 7.3.6.

Maxwell’s equations in matter are

\displaystyle \nabla\cdot\mathbf{D} \displaystyle = \displaystyle \rho_{f}\ \ \ \ \ (1)
\displaystyle \nabla\cdot\mathbf{B} \displaystyle = \displaystyle 0\ \ \ \ \ (2)
\displaystyle \nabla\times\mathbf{E} \displaystyle = \displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (3)
\displaystyle \nabla\times\mathbf{H} \displaystyle = \displaystyle \mathbf{J}_{f}+\frac{\partial\mathbf{D}}{\partial t} \ \ \ \ \ (4)

We can write these in integral form using the divergence theorem and Stokes’s theorem:

\displaystyle \int_{\mathcal{S}}\mathbf{D}\cdot d\mathbf{a} \displaystyle = \displaystyle Q_{f}\ \ \ \ \ (5)
\displaystyle \int_{\mathcal{S}}\mathbf{B}\cdot d\mathbf{a} \displaystyle = \displaystyle 0\ \ \ \ \ (6)
\displaystyle \int_{\mathcal{P}}\mathbf{E}\cdot d\boldsymbol{\ell} \displaystyle = \displaystyle -\frac{d}{dt}\int_{\mathcal{A}}\mathbf{B}\cdot d\mathbf{a}\ \ \ \ \ (7)
\displaystyle \int_{\mathcal{P}}\mathbf{H}\cdot d\boldsymbol{\ell} \displaystyle = \displaystyle \int_{\mathcal{A}}\mathbf{J}_{f}\cdot d\mathbf{a}+\int_{\mathcal{A}}\mathbf{D}\cdot d\mathbf{a}\ \ \ \ \ (8)
\displaystyle \displaystyle = \displaystyle I_{f}+\frac{d}{dt}\int_{\mathcal{A}}\mathbf{D}\cdot d\mathbf{a} \ \ \ \ \ (9)

Here, {Q_{f}} is the free charge enclosed by closed surface {\mathcal{S}}, {I_{f}} is the free current passing through closed loop {\mathcal{P}} and {\mathcal{A}} is any surface bounded by {\mathcal{P}}.

To work out the boundary conditions at the interface between two media, where medium {i} has permittivity {\epsilon_{i}} and permeability {\mu_{i}} we can look at the first equation for {\mathbf{D}}. Choosing a Gaussian pillbox of infinitesimal thickness straddling the the boundary, and taking the area of the top and bottom of the pillbox as {d\mathbf{a}}, we have

\displaystyle \mathbf{D}_{1}\cdot d\mathbf{a}-\mathbf{D}_{2}\cdot d\mathbf{a}=\sigma_{f}da \ \ \ \ \ (10)

where {\sigma_{f}} is the free charge density at the boundary. Any volume free charge doesn’t contribute since we can make the pillbox as thin as we like (for the same reason, the sides of the pillbox don’t contribute since they become vanishly small). The minus sign occurs because {d\mathbf{a}} points up on side 1 and down on side 2.

The product {\mathbf{D}_{1}\cdot d\mathbf{a}=D_{1}^{\perp}da} where {D_{1}^{\perp}} is the component of {\mathbf{D}_{1}} perpendicular to the surface, so we get

\displaystyle D_{1}^{\perp}-D_{2}^{\perp}=\sigma_{f} \ \ \ \ \ (11)

This is the same condition as the one for electrostatics and shows that the perpendicular component of {\mathbf{D}} is discontinuous at a boundary if there is free charge at the boundary. The same calculation for {\mathbf{B}} gives

\displaystyle B_{1}^{\perp}-B_{2}^{\perp}=0 \ \ \ \ \ (12)

so the perpendicular component of {\mathbf{B}} is continuous in all cases (unless magnetic monopoles exist!).

For the path integrals, we can choose a rectangular loop with a side length of {\ell} parallel to the surface and an infinitesimal length perpendicular to the surface. If we orient the loop so its face is perpendicular to the boundary and do the line integral around it, we get for {\mathbf{E}}:

\displaystyle \mathbf{E}_{1}\cdot\boldsymbol{\ell}-\mathbf{E}_{2}\cdot\boldsymbol{\ell}=-\frac{d}{dt}\int_{\mathcal{A}}\mathbf{B}\cdot d\mathbf{a} \ \ \ \ \ (13)

In the limit as the area of the loop goes to zero, the integral on the right goes to zero since the area {\mathcal{A}} over which the integral is done goes to zero. If we choose the boundary to lie in the {xy} plane, we can do this calculation for {\boldsymbol{\ell}} parallel first to the {x} axis and then to the {y} axis, so we get

\displaystyle E_{1x}-E_{2x} \displaystyle = \displaystyle 0\ \ \ \ \ (14)
\displaystyle E_{1y}-E_{2y} \displaystyle = \displaystyle 0 \ \ \ \ \ (15)

or with {\mathbf{E}_{i}^{\parallel}\equiv E_{ix}\hat{\mathbf{x}}+E_{iy}\hat{\mathbf{y}}} being the component of {\mathbf{E}_{i}} parallel to the surface:

\displaystyle \mathbf{E}_{1}^{\parallel}-\mathbf{E}_{2}^{\parallel}=0 \ \ \ \ \ (16)

For the {\mathbf{H}} integral, we can imagine the free current {I_{f}} as made up from a surface current {\mathbf{K}_{f}} that flows along the boundary. This current may not be perpendicular to the face of the loop, but since the face of the loop is normal to the boundary surface, if we have a unit vector normal to the loop’s face we can take the dot product of {\mathbf{K}_{f}} with this vector to find the component of the surface current density that flows through the loop. Griffiths does this by taking {\hat{\mathbf{n}}} to be a unit normal to the boundary surface, so that {\hat{\mathbf{n}}\times\boldsymbol{\ell}} is a vector with magnitude {\ell} normal to the face of the loop. The total current flowing through the loop is then

\displaystyle I_{f}=\mathbf{K}_{f}\cdot\left(\hat{\mathbf{n}}\times\boldsymbol{\ell}\right)=\left(\mathbf{K}_{f}\times\hat{\mathbf{n}}\right)\cdot\boldsymbol{\ell} \ \ \ \ \ (17)

since we can interchange the dot and cross in a vector triple product.

In the limit of an infinitesimally thin loop, the surface integral {\int_{\mathcal{A}}\mathbf{D}\cdot d\mathbf{a}} goes to zero, so we’re left with

\displaystyle \mathbf{H}_{1}\cdot\boldsymbol{\ell}-\mathbf{H}_{2}\cdot\boldsymbol{\ell}=\left(\mathbf{K}_{f}\times\hat{\mathbf{n}}\right)\cdot\boldsymbol{\ell} \ \ \ \ \ (18)

Since this must be true for arbitrary {\boldsymbol{\ell}} we again get a condition on {\mathbf{H}_{i}^{\parallel}}:

\displaystyle \mathbf{H}_{1}^{\parallel}-\mathbf{H}_{2}^{\parallel}=\mathbf{K}_{f}\times\hat{\mathbf{n}} \ \ \ \ \ (19)

If the media are both linear and homogeneous, then

\displaystyle \mathbf{D}_{i} \displaystyle = \displaystyle \epsilon_{i}\mathbf{E}_{i}\ \ \ \ \ (20)
\displaystyle \mathbf{H}_{i} \displaystyle = \displaystyle \frac{1}{\mu_{i}}\mathbf{B}_{i} \ \ \ \ \ (21)

so

\displaystyle \epsilon_{1}E_{1}^{\perp}-\epsilon_{2}E_{2}^{\perp} \displaystyle = \displaystyle \sigma_{f}\ \ \ \ \ (22)
\displaystyle B_{1}^{\perp}-B_{2}^{\perp} \displaystyle = \displaystyle 0\ \ \ \ \ (23)
\displaystyle \mathbf{E}_{1}^{\parallel}-\mathbf{E}_{2}^{\parallel} \displaystyle = \displaystyle 0\ \ \ \ \ (24)
\displaystyle \frac{1}{\mu_{1}}\mathbf{B}_{1}^{\parallel}-\frac{1}{\mu_{2}}\mathbf{B}_{2}^{\parallel} \displaystyle = \displaystyle \mathbf{K}_{f}\times\hat{\mathbf{n}} \ \ \ \ \ (25)

Finally, in cases where there is no surface charge or current at the boundary:

\displaystyle \epsilon_{1}E_{1}^{\perp}-\epsilon_{2}E_{2}^{\perp} \displaystyle = \displaystyle 0\ \ \ \ \ (26)
\displaystyle B_{1}^{\perp}-B_{2}^{\perp} \displaystyle = \displaystyle 0\ \ \ \ \ (27)
\displaystyle \mathbf{E}_{1}^{\parallel}-\mathbf{E}_{2}^{\parallel} \displaystyle = \displaystyle 0\ \ \ \ \ (28)
\displaystyle \frac{1}{\mu_{1}}\mathbf{B}_{1}^{\parallel}-\frac{1}{\mu_{2}}\mathbf{B}_{2}^{\parallel} \displaystyle = \displaystyle 0 \ \ \ \ \ (29)

Maxwell’s equations in matter

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 7.37.

Inside matter, we’ve seen that the polarization and magnetization give rise to bound charges and bound currents. The earlier results applied in the electrostatic and magnetostatic cases, respectively, so we’d like to generalize these to get the corresponding equations in electrodynamics.

For a medium with polarization {\mathbf{P}} and magnetization {\mathbf{M}}, the static cases are

\displaystyle   -\nabla\cdot\mathbf{P} \displaystyle  = \displaystyle  \rho_{b}\ \ \ \ \ (1)
\displaystyle  \mathbf{P}\cdot\hat{\mathbf{n}} \displaystyle  = \displaystyle  \sigma_{b}\ \ \ \ \ (2)
\displaystyle  \nabla\times\mathbf{M} \displaystyle  = \displaystyle  \mathbf{J}_{b}\ \ \ \ \ (3)
\displaystyle  \mathbf{M}\times\hat{\mathbf{n}} \displaystyle  = \displaystyle  \mathbf{K}_{b} \ \ \ \ \ (4)

where {\rho_{b}} and {\sigma_{b}} are the bound volume and surface charge densities and {\mathbf{J}_{b}} and {\mathbf{K}_{b}} are the bound volume and surface currents. These equations assume that nothing changes with time.

Now suppose that the polarization does change with time, so that {\mathbf{P}=\mathbf{P}\left(t\right)}. Considering a small cylindrical volume element with an axis parallel to {\mathbf{P}}, there is no bound surface charge on the sides of the cylinder, and on one end {\sigma_{b}=P} and on the other {\sigma_{b}=-P}. If {P} increases by {\Delta P} over time {\Delta t} then the surface charge density also increases (or decreases, at the negative end, but both ends increase in magnitude) by {\Delta P}. This change in charge density must be due to a current density {\mathbf{J}_{p}} parallel to {\mathbf{P}} of amount

\displaystyle  \mathbf{J}_{p}=\frac{\partial\sigma_{b}}{\partial t}\hat{\mathbf{n}}=\frac{\partial\mathbf{P}}{\partial t} \ \ \ \ \ (5)

This current density is called the polarization current. It is a current that is not present at all in the static case, so it’s a new current in addition to the bound current {\mathbf{J}_{b}} and free current {\mathbf{J}_{f}}.

If the magnetization changes with time, this will change the bound currents {\mathbf{J}_{b}} and {\mathbf{K}_{b}} but it doesn’t introduce any new currents into the system, so the only change we need to make in a dynamic system is the addition of the polarization current {\mathbf{J}_{p}}. The charge density {\rho} is thus still composed of two contributions:

\displaystyle  \rho=\rho_{f}+\rho_{b}=\rho_{f}-\nabla\cdot\mathbf{P} \ \ \ \ \ (6)

but the current density now has an additional term:

\displaystyle  \mathbf{J}=\mathbf{J}_{f}+\mathbf{J}_{b}+\mathbf{J}_{p}=\mathbf{J}_{f}+\nabla\times\mathbf{M}+\frac{\partial\mathbf{P}}{\partial t} \ \ \ \ \ (7)

Gauss’s law in matter can thus be written

\displaystyle   \nabla\cdot\mathbf{E} \displaystyle  = \displaystyle  \frac{\rho}{\epsilon_{0}}=\frac{1}{\epsilon_{0}}\left(\rho_{f}-\nabla\cdot\mathbf{P}\right)\ \ \ \ \ (8)
\displaystyle  \nabla\cdot\left(\epsilon_{0}\mathbf{E}+\mathbf{P}\right) \displaystyle  = \displaystyle  \rho_{f}\ \ \ \ \ (9)
\displaystyle  \nabla\cdot\mathbf{D} \displaystyle  = \displaystyle  \rho_{f} \ \ \ \ \ (10)

where {\mathbf{D}=\epsilon_{0}\mathbf{E}+\mathbf{P}}.
where {\epsilon} is the permittivity of the medium and is constant everywhere in the medium.

Ampère’s law with Maxwell’s correction must take into account the polarization current, so we have

\displaystyle   \nabla\times\mathbf{B} \displaystyle  = \displaystyle  \mu_{0}\mathbf{J}+\mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t}\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \mu_{0}\left(\mathbf{J}_{f}+\nabla\times\mathbf{M}+\frac{\partial\mathbf{P}}{\partial t}\right)+\mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t}\ \ \ \ \ (12)
\displaystyle  \nabla\times\left(\frac{1}{\mu_{0}}\mathbf{B}-\mathbf{M}\right) \displaystyle  = \displaystyle  \mathbf{J}_{f}+\frac{\partial}{\partial t}\left(\epsilon_{0}\mathbf{E}+\mathbf{P}\right)\ \ \ \ \ (13)
\displaystyle  \nabla\times\mathbf{H} \displaystyle  = \displaystyle  \mathbf{J}_{f}+\frac{\partial\mathbf{D}}{\partial t} \ \ \ \ \ (14)

The other two Maxwell equations don’t change, since they don’t refer explicitly to charge or current. Thus we get Maxwell’s equations in matter:

\displaystyle   \nabla\cdot\mathbf{D} \displaystyle  = \displaystyle  \rho_{f}\ \ \ \ \ (15)
\displaystyle  \nabla\cdot\mathbf{B} \displaystyle  = \displaystyle  0\ \ \ \ \ (16)
\displaystyle  \nabla\times\mathbf{E} \displaystyle  = \displaystyle  -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (17)
\displaystyle  \nabla\times\mathbf{H} \displaystyle  = \displaystyle  \mathbf{J}_{f}+\frac{\partial\mathbf{D}}{\partial t} \ \ \ \ \ (18)

For linear, homogeneous materials, the polarization and magnetization depend linearly on {\mathbf{E}} and {\mathbf{B}}, respectively, so

\displaystyle   \mathbf{D} \displaystyle  = \displaystyle  \epsilon\mathbf{E}\ \ \ \ \ (19)
\displaystyle  \mathbf{H} \displaystyle  = \displaystyle  \frac{1}{\mu}\mathbf{B} \ \ \ \ \ (20)

where

\displaystyle   \epsilon \displaystyle  = \displaystyle  \epsilon_{0}\left(1+\chi_{e}\right)\ \ \ \ \ (21)
\displaystyle  \mu \displaystyle  = \displaystyle  \mu_{0}\left(1+\chi_{m}\right) \ \ \ \ \ (22)

are the permittivity and permeability of the material. Maxwell’s original displacement current, {\epsilon_{0}\partial\mathbf{E}/\partial t}, becomes {\partial\mathbf{D}/\partial t}.

Example Suppose we have a large parallel plate capacitor embedded in sea water and maintain an AC voltage across the plates, so that

\displaystyle  V\left(t\right)=V_{0}\cos\left(2\pi\nu t\right) \ \ \ \ \ (23)

In terms of the conductivity {\sigma} of the sea water, the current density due to conduction is {\mathbf{J}=\sigma\mathbf{E}} and for a capacitor with plates separated by a distance {d}, {E\left(t\right)=V\left(t\right)/d}, so

\displaystyle  J=\frac{\sigma V_{0}}{d}\cos\left(2\pi\nu t\right) \ \ \ \ \ (24)

The displacement current is

\displaystyle   J_{d} \displaystyle  = \displaystyle  \frac{\partial D}{\partial t}\ \ \ \ \ (25)
\displaystyle  \displaystyle  = \displaystyle  \epsilon\frac{\partial E}{\partial t}\ \ \ \ \ (26)
\displaystyle  \displaystyle  = \displaystyle  -2\pi\epsilon\nu\frac{V_{0}}{d}\sin\left(2\pi\nu t\right) \ \ \ \ \ (27)

The ratio of the amplitudes is therefore

\displaystyle  \frac{J}{J_{d}}=\frac{\sigma}{2\pi\epsilon\nu}=\frac{1}{2\pi\epsilon\nu\rho} \ \ \ \ \ (28)

where {\rho} is the resistivity of sea water.

Plugging in the values given by Griffiths, we have {\rho=0.23\mbox{ }\Omega\cdot\mbox{m}}, {\epsilon=81\epsilon_{0}=7.17\times10^{-10}\mbox{ m}^{-3}\mbox{kg}^{-1}\mbox{s}^{4}\mbox{A}^{2}}, {\nu=4\times10^{8}\mbox{ s}^{-1}} and

\displaystyle  \frac{J}{J_{d}}=2.41 \ \ \ \ \ (29)

Maxwell stress tensor and electromagnetic waves

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.12.

The Maxwell stress tensor {\overleftrightarrow{\mathbf{T}}} gives the components of momentum flux density. That is, the component {-T_{ij}} is the momentum component {i} per unit area per unit time crossing a surface normal to the {j} direction. The formula is

\displaystyle T_{ij}\equiv\epsilon_{0}\left(E_{i}E_{j}-\frac{1}{2}\delta_{ij}E^{2}\right)+\frac{1}{\mu_{0}}\left(B_{i}B_{j}-\frac{1}{2}\delta_{ij}B^{2}\right) \ \ \ \ \ (1)

For a monochromatic plane wave with amplitude {A}, direction {\hat{\mathbf{z}}}, frequency {\omega} and phase {\delta} polarized in the {x} direction, we have

\displaystyle \mathbf{E} \displaystyle = \displaystyle E_{0}\cos\left(kz-\omega t+\delta\right)\hat{\mathbf{x}}\ \ \ \ \ (2)
\displaystyle \mathbf{B} \displaystyle = \displaystyle \frac{E_{0}}{c}\cos\left(kz-\omega t+\delta\right)\hat{\mathbf{y}} \ \ \ \ \ (3)

All the off-diagonal elements of {\overleftrightarrow{\mathbf{T}}} are zero, and the diagonal elements are, using {c=1/\sqrt{\epsilon_{0}\mu_{0}}}:

\displaystyle T_{xx} \displaystyle = \displaystyle \frac{1}{2}\epsilon_{0}E_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right)-\frac{1}{2\mu_{0}c^{2}}E_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right)\ \ \ \ \ (4)
\displaystyle \displaystyle = \displaystyle \frac{1}{2}\epsilon_{0}E_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right)-\frac{1}{2}\epsilon_{0}E_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right)\ \ \ \ \ (5)
\displaystyle \displaystyle = \displaystyle 0\ \ \ \ \ (6)
\displaystyle T_{yy} \displaystyle = \displaystyle -\frac{1}{2}\epsilon_{0}E_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right)+\frac{1}{2\mu_{0}c^{2}}E_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right)\ \ \ \ \ (7)
\displaystyle \displaystyle = \displaystyle 0\ \ \ \ \ (8)
\displaystyle T_{zz} \displaystyle = \displaystyle -\frac{1}{2}\epsilon_{0}E_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right)-\frac{1}{2\mu_{0}c^{2}}E_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right)\ \ \ \ \ (9)
\displaystyle \displaystyle = \displaystyle -\epsilon_{0}E_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right) \ \ \ \ \ (10)

Thus there is no momentum flux in the {x} or {y} directions, and the momentum flux density in the {z} direction is {\epsilon_{0}E_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right)}. Since the average of {\cos^{2}x} over a single cycle is {\frac{1}{2}}, the average momentum flux density is

\displaystyle \left\langle -T_{zz}\right\rangle =\frac{1}{2}\epsilon_{0}E_{0}^{2} \ \ \ \ \ (11)

Since the wave moves with speed {c}, a unit area sweeps out a volume {c} in unit time. Thus the average momentum density (that is, momentum per unit volume) is

\displaystyle \left\langle \mathfrak{p}\right\rangle =\frac{1}{c}\left\langle -T_{zz}\right\rangle =\frac{1}{2c}\epsilon_{0}E_{0}^{2} \ \ \ \ \ (12)

which agrees with our earlier result.

Also in the earlier post, we showed that rate per unit area of energy flow is given by the Poynting vector and is

\displaystyle \mathbf{S} \displaystyle = \displaystyle \epsilon_{0}cE_{0}^{2}\cos^{2}\left(kz-\omega t+\delta\right)\hat{\mathbf{z}}\ \ \ \ \ (13)
\displaystyle \displaystyle = \displaystyle -cT_{zz}\hat{\mathbf{z}} \ \ \ \ \ (14)

The energy density is thus {S/c=-T_{zz}} so the energy density is the same as the momentum flux density.

Average of product of two waves

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.11.

A common calculation that is required when analyzing any system that varies with a sinusoidal period is a time average over one cycle. For example, a monochromatic plane wave with amplitude {A}, direction {\mathbf{k}}, frequency {\omega} and phase {\delta} can be written as

\displaystyle   f \displaystyle  = \displaystyle  A\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta\right)=\Re\tilde{A}e^{i\left(\mathbf{k}\cdot\mathbf{r}-\omega t\right)}\ \ \ \ \ (1)
\displaystyle  \tilde{A} \displaystyle  = \displaystyle  Ae^{i\delta} \ \ \ \ \ (2)

Now suppose we have two waves with the same direction and frequency, but different amplitudes and phases. Then

\displaystyle   f \displaystyle  = \displaystyle  A\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{a}\right)\ \ \ \ \ (3)
\displaystyle  g \displaystyle  = \displaystyle  B\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{b}\right) \ \ \ \ \ (4)

The average of the product of these waves over a single cycle is then

\displaystyle  \left\langle fg\right\rangle =\frac{\omega AB}{2\pi}\int_{0}^{2\pi/\omega}\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{a}\right)\cos\left(\mathbf{k}\cdot\mathbf{r}-\omega t+\delta_{b}\right)dt \ \ \ \ \ (5)

We can transform this integral by defining

\displaystyle   \theta \displaystyle  \equiv \displaystyle  \mathbf{k}\cdot\mathbf{r}-\omega t\ \ \ \ \ (6)
\displaystyle  d\theta \displaystyle  = \displaystyle  -\omega dt\ \ \ \ \ (7)
\displaystyle  \left\langle fg\right\rangle \displaystyle  = \displaystyle  \frac{AB}{2\pi}\int_{0}^{2\pi}\cos\left(\theta+\delta_{a}\right)\cos\left(\theta+\delta_{b}\right)d\theta \ \ \ \ \ (8)

We’ve used the limits of 0 and {2\pi} since any interval of {2\pi} covers one complete cycle of {\theta}.

The two cosines have the same period and differ only in their phase, so we will get the same result from the integral if we replace them by

\displaystyle   \cos\left(\theta+\delta_{a}\right)\cos\left(\theta+\delta_{b}\right) \displaystyle  \rightarrow \displaystyle  \cos\theta\cos\left(\theta+\delta_{a}-\delta_{b}\right)\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  \cos^{2}\theta\cos\left(\delta_{a}-\delta_{b}\right)-\cos\theta\sin\theta\sin\left(\delta_{a}-\delta_{b}\right) \ \ \ \ \ (10)

We now have

\displaystyle   \left\langle fg\right\rangle \displaystyle  = \displaystyle  \frac{AB}{2\pi}\cos\left(\delta_{a}-\delta_{b}\right)\int_{0}^{2\pi}\cos^{2}\theta d\theta-\frac{AB}{2\pi}\sin\left(\delta_{a}-\delta_{b}\right)\int_{0}^{2\pi}\cos\theta\sin\theta d\theta\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}AB\cos\left(\delta_{a}-\delta_{b}\right)-0\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}AB\cos\left(\delta_{a}-\delta_{b}\right)\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2}\Re\left(fg^{*}\right)=\frac{1}{2}\Re\left(f^{*}g\right) \ \ \ \ \ (14)

Thus we can get the answer using complex notation without doing any integrals.

Electromagnetic waves: energy, momentum and light pressure

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.10.

We’ve seen that Maxwell’s equations in a vacuum predict that an electromagnetic field can propagate as a wave with the speed of light. We’ve also seen that any region where both an electric and magnetic field exists results in energy flowing through the region. The rate per unit area at which energy crosses a surface is given by the Poynting vector

\displaystyle  \mathbf{S}=\frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B} \ \ \ \ \ (1)

In an electromagnetic wave, {\mathbf{E}} and {\mathbf{B}} are perpendicular to each other, and also perpendicular to the direction {\mathbf{k}} of propagation of the wave. Thus their cross product is parallel to {\mathbf{k}}, indicating that an electromagnetic wave carries energy.

Electromagnetic fields also carry momentum, with the momentum density given by

\displaystyle  \mathfrak{p}=\epsilon_{0}\mu_{0}\mathbf{S}=\frac{1}{c^{2}}\mathbf{S} \ \ \ \ \ (2)

For a monochromatic plane wave travelling in the {z} direction

\displaystyle   \mathbf{E} \displaystyle  = \displaystyle  E_{0}\cos\left(kz-\omega t+\delta\right)\hat{\mathbf{x}}\ \ \ \ \ (3)
\displaystyle  \mathbf{B} \displaystyle  = \displaystyle  \frac{E_{0}}{c}\cos\left(kz-\omega t+\delta\right)\hat{\mathbf{y}}\ \ \ \ \ (4)
\displaystyle  \mathbf{S} \displaystyle  = \displaystyle  \frac{E_{0}^{2}}{\mu_{0}c}\cos^{2}\left(kz-\omega t+\delta\right)\hat{\mathbf{z}}\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  E_{0}^{2}c\epsilon_{0}\cos^{2}\left(kz-\omega t+\delta\right)\hat{\mathbf{z}} \ \ \ \ \ (6)

The average of cos or sin squared over a complete cycle is {\frac{1}{2}}, so the averages of energy and momentum density are

\displaystyle   \left\langle \mathbf{S}\right\rangle \displaystyle  = \displaystyle  \frac{1}{2}E_{0}^{2}c\epsilon_{0}\hat{\mathbf{z}}\ \ \ \ \ (7)
\displaystyle  \left\langle \mathfrak{p}\right\rangle \displaystyle  = \displaystyle  \frac{1}{2c}E_{0}^{2}\epsilon_{0}\hat{\mathbf{z}} \ \ \ \ \ (8)

The magnitude of {\left\langle \mathbf{S}\right\rangle } has the units of energy per unit area per unit time, or power per unit area, and is known as the intensity of the radiation:

\displaystyle   I \displaystyle  \equiv \displaystyle  \left\langle S\right\rangle =\frac{1}{2}E_{0}^{2}c\epsilon_{0}\ \ \ \ \ (9)
\displaystyle  \left\langle \mathfrak{p}\right\rangle \displaystyle  = \displaystyle  \frac{1}{2c}E_{0}^{2}\epsilon_{0}=\frac{I}{c^{2}} \ \ \ \ \ (10)

If an electromagnetic wave hits a surface, its momentum is either absorbed or reflected (or bits of both), so EM radiation actually exerts a pressure on a surface. Pressure is force per unit area, which is momentum transferred per unit time per unit area. The quantity {\left\langle \mathfrak{p}\right\rangle } is the momentum density in the wave, so the amount of momentum that falls upon an area {A} in time {\Delta t} is the volume of the wave that falls on the area times {\left\langle \mathfrak{p}\right\rangle }. The wave is travelling at speed {c} so this volume is {Ac\Delta t} and the momentum that falls on the area is

\displaystyle  \Delta p=\left\langle \mathfrak{p}\right\rangle Ac\Delta t \ \ \ \ \ (11)

The average pressure is the average force per unit area, which in turn is the momentum received by the area per unit area per unit time, so

\displaystyle  P=\frac{\Delta p}{A\Delta t}=\left\langle \mathfrak{p}\right\rangle c=\frac{1}{2}E_{0}^{2}\epsilon_{0}=\frac{I}{c} \ \ \ \ \ (12)

This assumes a perfect absorber, so the radiation is just absorbed and not reflected. A perfect reflector would experience a pressure twice as large, since the momentum in the fields is not just stopped, it is reversed.

To get an idea of how big this pressure is, suppose we’re outside on a sunny day at noon, when the solar radiation is at its maximum. At my latitude (Scotland) this intensity never gets very far above {1000\mbox{ Watts m}^{-2}} (if you’re interested in my weather data see here) but we’ll use Griffiths’s value of {1300\mbox{ Watts m}^{-2}}. If this sunlight strikes a perfect absorber then

\displaystyle  P=\frac{1300}{3\times10^{8}}=4.33\times10^{-6}\mbox{ N m}^{-2} \ \ \ \ \ (13)

For a perfect reflector, the pressure is just twice this, or {8.66\times10^{-6}\mbox{ N m}^{-2}}.

For comparison, standard atmospheric pressure is {101325\mbox{ N m}^{-2}} so it’s no surprise that the pressure exerted by sunlight isn’t noticeable.

Electromagnetic waves in vacuum

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.9.

Starting with Maxwell’s equations

\displaystyle   \nabla\cdot\mathbf{E} \displaystyle  = \displaystyle  \frac{\rho}{\epsilon_{0}}\ \ \ \ \ (1)
\displaystyle  \nabla\cdot\mathbf{B} \displaystyle  = \displaystyle  0\ \ \ \ \ (2)
\displaystyle  \nabla\times\mathbf{E} \displaystyle  = \displaystyle  -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (3)
\displaystyle  \nabla\times\mathbf{B} \displaystyle  = \displaystyle  \mu_{0}\mathbf{J}+\mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (4)

we can now investigate what happens if we have time-varying electric and magnetic fields in vacuum. In that case, there is no charge or current so {\rho=\mathbf{J}=0} and we get

\displaystyle   \nabla\cdot\mathbf{E} \displaystyle  = \displaystyle  0\ \ \ \ \ (5)
\displaystyle  \nabla\cdot\mathbf{B} \displaystyle  = \displaystyle  0\ \ \ \ \ (6)
\displaystyle  \nabla\times\mathbf{E} \displaystyle  = \displaystyle  -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (7)
\displaystyle  \nabla\times\mathbf{B} \displaystyle  = \displaystyle  \mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (8)

We can transform these equations into separate equations for {\mathbf{E}} and {\mathbf{B}} by taking the curl of the last two:

\displaystyle   \nabla\times\left(\nabla\times\mathbf{E}\right) \displaystyle  = \displaystyle  \nabla\left(\nabla\cdot\mathbf{E}\right)-\nabla^{2}\mathbf{E}\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  -\nabla\times\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  -\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{E}}{\partial t^{2}} \ \ \ \ \ (11)

Since {\nabla\cdot\mathbf{E}=0} in vacuum, we get

\displaystyle  \boxed{\nabla^{2}\mathbf{E}=\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{E}}{\partial t^{2}}} \ \ \ \ \ (12)

A similar calculation for {\mathbf{B}} gives us

\displaystyle   \nabla\times\left(\nabla\times\mathbf{B}\right) \displaystyle  = \displaystyle  \nabla\left(\nabla\cdot\mathbf{B}\right)-\nabla^{2}\mathbf{B}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \mu_{0}\epsilon_{0}\nabla\times\frac{\partial\mathbf{E}}{\partial t}\ \ \ \ \ (14)
\displaystyle  \displaystyle  = \displaystyle  -\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{B}}{\partial t^{2}} \ \ \ \ \ (15)

so we get

\displaystyle  \boxed{\nabla^{2}\mathbf{B}=\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{B}}{\partial t^{2}}} \ \ \ \ \ (16)

The wave equation can be generalized to higher dimensions. In two dimensions, we can consider the force on a patch of membrane held under tension (as in a drum), and the wave variable is the displacement of the membrane from equilibrium. In three dimensions, we can consider the change in some property of a 3-d substance. For example, we can think of the change in density of a fluid such as water as a sound wave passes through it. A proper derivation of the 3-d wave equation would take us a bit far afield here, so we’ll just quote the result. For a scalar field (that is, the quantity that ‘waves’) {f} the 3-d wave equation is

\displaystyle  \nabla^{2}f=\frac{1}{v^{2}}\partial_{t}f \ \ \ \ \ (17)

where {v} is the speed of the wave through the substance.

Given the 3-d wave equation, we can see that each component of {\mathbf{E}} and {\mathbf{B}} satisfies the wave equation, and that the speed of the wave is the same in both cases, namely

\displaystyle  v=\frac{1}{\sqrt{\mu_{0}\epsilon_{0}}} \ \ \ \ \ (18)

Experimentally, it was found that {v=c}, the speed of light. This result seems to me to be one of the most magical results in physics. It predicts that electric and magnetic fields, once produced, sustain each other and propagate as a wave. Not only that, it suggests (it doesn’t really predict) that light is itself an electromagnetic wave.

We can derive a few more properties of the electromagnetic wave by applying Maxwell’s equations to solutions of the wave equations. Since any solution of the wave equation can be expressed as a sum (or integral) over sinusoidal functions (that’s Fourier analysis), we can consider only sinusoidal solutions from now on. Considering waves that consist of only a single frequency {\omega} that travel in the {+z} direction and have no dependence on {x} or {y}, we can write the solutions as

\displaystyle   \tilde{\mathbf{E}} \displaystyle  = \displaystyle  \tilde{\mathbf{E}}_{0}e^{i\left(kz-\omega t\right)}\ \ \ \ \ (19)
\displaystyle  \tilde{\mathbf{B}} \displaystyle  = \displaystyle  \tilde{\mathbf{B}}_{0}e^{i\left(kz-\omega t\right)} \ \ \ \ \ (20)

where the tilde indicates we’re using complex notation, and that the physical wave is the real part. The parameters {\tilde{\mathbf{E}}_{0}} and {\tilde{\mathbf{B}}_{0}} are constants under the assumptions we’ve made here. Such a wave is called monochromatic (“one colour”) because it contains only one frequency (and hence, for visible light, only one colour) and plane because the wave is constant over any plane perpendicular to the direction of propagation.

We can now apply Maxwell’s equations to these solutions. First, an observation about the complex notation. For the fields above, the real parts depend on space and time through a term {\cos\left(kz-\omega t\right)} and the imaginary parts through a term {\sin\left(kz-\omega t\right)}. Maxwell’s equations involve only first derivatives with respect to space and time, and these derivatives will convert all cosines into sines and vice versa. Therefore, if the real part of {\tilde{\mathbf{E}}} or {\tilde{\mathbf{B}}} satisfies Maxwell’s equations (as it does), then applying the same equations to the imaginary parts just replaces all cosines by sines and thus the imaginary parts must also be solutions. So it’s safe to apply Maxwell’s equations to the full complex functions {\tilde{\mathbf{E}}} and {\tilde{\mathbf{B}}}.

In a vacuum, both {\nabla\cdot\mathbf{E}=0} and {\nabla\cdot\mathbf{B}=0} from which we get

\displaystyle   \nabla\cdot\tilde{\mathbf{E}} \displaystyle  = \displaystyle  ik\tilde{E}_{0z}e^{i\left(kz-\omega t\right)}=0\ \ \ \ \ (21)
\displaystyle  \nabla\cdot\tilde{\mathbf{B}} \displaystyle  = \displaystyle  ik\tilde{B}_{0z}e^{i\left(kz-\omega t\right)}=0 \ \ \ \ \ (22)

Since this must be true for all {z}, we must have

\displaystyle  \tilde{E}_{0z}=\tilde{B}_{0z}=0 \ \ \ \ \ (23)

That is, the wave has only {x} and {y} components, so it must be a transverse wave: a wave that oscillates in a plane perpendicular to the direction of propagation.

We can now apply {\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}}, which gives

\displaystyle   \nabla\times\tilde{\mathbf{E}} \displaystyle  = \displaystyle  \left(-ik\tilde{E}_{0y}e^{i\left(kz-\omega t\right)}\right)\hat{\mathbf{x}}+\left(ik\tilde{E}_{0x}e^{i\left(kz-\omega t\right)}\right)\hat{\mathbf{y}}\ \ \ \ \ (24)
\displaystyle  \displaystyle  = \displaystyle  \left(i\omega\tilde{B}_{0x}e^{i\left(kz-\omega t\right)}\right)\hat{\mathbf{x}}+\left(i\omega\tilde{B}_{0y}e^{i\left(kz-\omega t\right)}\right)\hat{\mathbf{y}} \ \ \ \ \ (25)

so

\displaystyle   \tilde{B}_{0x} \displaystyle  = \displaystyle  -\frac{k}{\omega}\tilde{E}_{0y}\ \ \ \ \ (26)
\displaystyle  \tilde{B}_{0y} \displaystyle  = \displaystyle  \frac{k}{\omega}\tilde{E}_{0x} \ \ \ \ \ (27)

which can be written in vector form as

\displaystyle  \tilde{\mathbf{B}}_{0}=\frac{k}{\omega}\hat{\mathbf{z}}\times\tilde{\mathbf{E}}_{0}=\frac{1}{c}\hat{\mathbf{z}}\times\tilde{\mathbf{E}}_{0} \ \ \ \ \ (28)

Therefore, not only are {\mathbf{B}} and {\mathbf{E}} transverse, they are also perpendicular to each other.

Now suppose we want a monochromatic, plane wave that travels in some arbitrary direction given by {\hat{\mathbf{k}}}. The value of the wave function at some point {\mathbf{r}} in 3-d space depends on the projection of {\mathbf{r}} onto the direction of propagation, since for a plane wave, the wave function depends only on the distance we’ve moved along this direction. This projection is {\mathbf{r}\cdot\hat{\mathbf{k}}}. We can therefore define the wave vector {\mathbf{k}=k\hat{\mathbf{k}}} that points along the propagation direction and replace the {kz} in the equations above by {\mathbf{k}\cdot\mathbf{r}}, so a monochromatic plane wave travelling in direction {\hat{\mathbf{k}}} is then

\displaystyle   \tilde{\mathbf{E}} \displaystyle  = \displaystyle  \tilde{\mathbf{E}}_{0}e^{i\left(\mathbf{k}\cdot\mathbf{r}-\omega t\right)}\ \ \ \ \ (29)
\displaystyle  \tilde{\mathbf{B}} \displaystyle  = \displaystyle  \tilde{\mathbf{B}}_{0}e^{i\left(\mathbf{k}\cdot\mathbf{r}-\omega t\right)} \ \ \ \ \ (30)

For such a wave, the directions given by {\tilde{\mathbf{E}}_{0}} and {\tilde{\mathbf{B}}_{0}} are fixed, and it’s conventional to take the direction of {\tilde{\mathbf{E}}_{0}} as the polarization direction {\hat{\mathbf{n}}}. Since {\tilde{\mathbf{B}}} is perpendicular both to {\tilde{\mathbf{E}}} and {\mathbf{k}} its direction is given by {\hat{\mathbf{k}}\times\hat{\mathbf{n}}} so

\displaystyle   \tilde{\mathbf{E}} \displaystyle  = \displaystyle  \tilde{E}_{0}e^{i\left(\mathbf{k}\cdot\mathbf{r}-\omega t\right)}\hat{\mathbf{n}}\ \ \ \ \ (31)
\displaystyle  \tilde{\mathbf{B}} \displaystyle  = \displaystyle  \tilde{B}_{0}e^{i\left(\mathbf{k}\cdot\mathbf{r}-\omega t\right)}\hat{\mathbf{k}}\times\hat{\mathbf{n}}\ \ \ \ \ (32)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{c}\hat{\mathbf{k}}\times\tilde{\mathbf{E}} \ \ \ \ \ (33)

Example 1 Suppose we have a monochromatic plane wave of (real) amplitude {E_{0}}, frequency {\omega} and phase angle {\delta=0}. The wave is travelling in the {-x} direction and polarized in the {z} direction. We then have

\displaystyle   \mathbf{k} \displaystyle  = \displaystyle  -k\hat{\mathbf{x}}\ \ \ \ \ (34)
\displaystyle  \hat{\mathbf{n}} \displaystyle  = \displaystyle  \hat{\mathbf{z}}\ \ \ \ \ (35)
\displaystyle  \mathbf{k}\cdot\mathbf{r} \displaystyle  = \displaystyle  -kx\ \ \ \ \ (36)
\displaystyle  \tilde{E}_{0} \displaystyle  = \displaystyle  E_{0}\ \ \ \ \ (37)
\displaystyle  \mathbf{E} \displaystyle  = \displaystyle  E_{0}\cos\left(-kx-\omega t\right)\hat{\mathbf{z}}\ \ \ \ \ (38)
\displaystyle  \mathbf{B} \displaystyle  = \displaystyle  \frac{E_{0}}{c}\cos\left(-kx-\omega t\right)\hat{\mathbf{y}} \ \ \ \ \ (39)

Example 2 Now we have the same wave, except that it is travelling in the direction {\left[1,1,1\right]} and is polarized parallel to the {xz} plane. This time

\displaystyle  \mathbf{k}=\frac{k}{\sqrt{3}}\left[1,1,1\right] \ \ \ \ \ (40)

Since {\hat{\mathbf{n}}} is parallel to the {xz} plane and {\mathbf{k}} is perpendicular to {\hat{\mathbf{n}}}, we have

\displaystyle   \hat{\mathbf{n}} \displaystyle  = \displaystyle  \left[n_{x},0,n_{y}\right]\ \ \ \ \ (41)
\displaystyle  \mathbf{k}\cdot\hat{\mathbf{n}} \displaystyle  = \displaystyle  0\ \ \ \ \ (42)
\displaystyle  \hat{\mathbf{n}} \displaystyle  = \displaystyle  \frac{1}{\sqrt{2}}\left[1,0,-1\right] \ \ \ \ \ (43)

Thus

\displaystyle   \mathbf{E} \displaystyle  = \displaystyle  \frac{E_{0}}{\sqrt{2}}\cos\left(\frac{k}{\sqrt{3}}\left(x+y+z\right)-\omega t\right)\left[1,0,-1\right]\ \ \ \ \ (44)
\displaystyle  \mathbf{B} \displaystyle  = \displaystyle  \frac{1}{c}\hat{\mathbf{k}}\times\mathbf{E}\ \ \ \ \ (45)
\displaystyle  \displaystyle  = \displaystyle  \frac{E_{0}}{\sqrt{6}c}\cos\left(\frac{k}{\sqrt{3}}\left(x+y+z\right)-\omega t\right)\left[-1,2,-1\right] \ \ \ \ \ (46)

or, since {\omega=ck}

\displaystyle   \mathbf{E} \displaystyle  = \displaystyle  \frac{E_{0}}{\sqrt{2}}\cos\left(\frac{\omega}{\sqrt{3}c}\left(x+y+z\right)-\omega t\right)\left[1,0,-1\right]\ \ \ \ \ (47)
\displaystyle  \mathbf{B} \displaystyle  = \displaystyle  \frac{1}{c}\hat{\mathbf{k}}\times\mathbf{E}\ \ \ \ \ (48)
\displaystyle  \displaystyle  = \displaystyle  \frac{E_{0}}{\sqrt{6}c}\cos\left(\frac{\omega}{\sqrt{3}c}\left(x+y+z\right)-\omega t\right)\left[-1,2,-1\right] \ \ \ \ \ (49)

Waves: polarization

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.8.

For the sinusoidal wave on a string, let’s assume that the wave travels in the {+z} direction and then define {x} and {y} axes in the usual way. We can produce the wave by shaking the string in the {xz} plane, in which case there is no component of the wave in the {y} direction, or by shaking the string in the {yz} plane, in which case there is no motion in the {x} direction. Or we could shake the string in some other plane intermediate between {xz} and {xy}, as long as that plane contains the {z} axis.

If the {x} axis points upwards and the {y} axis points horizontally, then a wave moving in the {xz} plane can be said to be vertically polarized, while a wave moving in the {yz} plane is horizontally polarized. (These terms aren’t really technical terms, but you get the idea.) Any transverse wave whose motion is restricted to a plane is linearly polarized. We can represent this in vector notation by writing

\displaystyle  f\left(z,t\right)=Ae^{i\left(kz-\omega t\right)}\hat{\mathbf{n}} \ \ \ \ \ (1)

where {\hat{\mathbf{n}}} is a unit vector perpendicular to the {z} axis and parallel to the plane of polarization. Thus for a wave moving in the {xz} plane, {\hat{\mathbf{n}}=\hat{\mathbf{x}}} and so on.

If {\theta} is the angle between {\hat{\mathbf{n}}} and {\hat{\mathbf{x}}}, then in general

\displaystyle  \hat{\mathbf{n}}=\cos\theta\hat{\mathbf{x}}+\sin\theta\hat{\mathbf{y}} \ \ \ \ \ (2)

and a wave linearly polarized in the {\hat{\mathbf{n}}} direction is given by

\displaystyle  f\left(z,t\right)=Ae^{i\left(kz-\omega t\right)}\left(\cos\theta\hat{\mathbf{x}}+\sin\theta\hat{\mathbf{y}}\right) \ \ \ \ \ (3)

More generally, we can have a wave that is the vector sum of a vertically and horizontally polarized wave:

\displaystyle  f\left(z,t\right)=A_{x}e^{i\left(kz-\omega t\right)}\cos\theta\hat{\mathbf{x}}+A_{y}e^{i\left(kz-\omega t\right)}\sin\theta\hat{\mathbf{y}} \ \ \ \ \ (4)

where the complex amplitudes are given by

\displaystyle  A_{i}=a_{i}e^{i\delta_{i}} \ \ \ \ \ (5)

with {i=x,y}. The real amplitudes {a_{x}} and {a_{y}} must be the same for both components in order for the polarization to be in the {\hat{\mathbf{n}}} plane.

For a wave to be linearly polarized, the phases of the two component waves must also be equal, so that {\delta_{x}=\delta_{y}=\delta}. In that case, the components of the wave at a fixed location {z=z_{0}} vary with time according to

\displaystyle   \mathbf{f}_{x}\left(z_{0},t\right) \displaystyle  = \displaystyle  ae^{i\left(kz_{0}-\omega t+\delta\right)}\cos\theta\hat{\mathbf{x}}\ \ \ \ \ (6)
\displaystyle  \mathbf{f}_{y}\left(z_{0},t\right) \displaystyle  = \displaystyle  ae^{i\left(kz_{0}-\omega t+\delta\right)}\sin\theta\hat{\mathbf{y}} \ \ \ \ \ (7)

Since it is only the real parts of these equations that are physically meaningful, we see that the time dependence is the same in both cases and is {\cos\left(kz_{0}-\omega t+\delta\right)}. Thus the horizontal and vertical components oscillate in phase, so that the wave remains in the {\hat{\mathbf{n}}} plane.

Now suppose that {\delta_{y}=\pi/2} and {\delta_{x}=0} so that the two components are out of phase. Then the real components of the wave are

\displaystyle   x=\Re\left(f_{x}\right) \displaystyle  = \displaystyle  a\cos\left(kz_{0}-\omega t\right)\cos\theta\ \ \ \ \ (8)
\displaystyle  y=\Re\left(f_{y}\right) \displaystyle  = \displaystyle  a\cos\left(kz_{0}-\omega t+\frac{\pi}{2}\right)\sin\theta\ \ \ \ \ (9)
\displaystyle  \displaystyle  = \displaystyle  -a\sin\left(kz_{0}-\omega t\right)\sin\theta \ \ \ \ \ (10)

In this case, the point on the string at position {z=z_{0}} follows a curve with equation

\displaystyle  \frac{x^{2}}{\left(a\cos\theta\right)^{2}}+\frac{y^{2}}{\left(a\sin\theta\right)^{2}}=1 \ \ \ \ \ (11)

which is the equation of an ellipse. In this case, the wave is elliptically polarized. If {\theta=\frac{\pi}{4}} then we get a circle and the wave is circularly polarized.

To see which direction the string moves, try a couple of values of {t} at {z_{0}=0}. For {t=0}, {x\left(0,0\right)=a\cos\theta}, {y\left(0,0\right)=0}. Then for {t=\pi/2\omega} we have {x\left(0,\pi/2\omega\right)=0}, {y\left(0,\pi/2\omega\right)=a\sin\theta}. Thus if {\theta} increases counterclockwise as we look down the {z} axis towards the origin, the point is rotating counterclockwise. The shape of the string at any given instant of time is a helix (either elliptical or circular, depending on {\theta}). We can generate this by shaking the end of the string in an ellipse or circle.

If we set {\delta_{y}=-\pi/2} then

\displaystyle   x=\Re\left(f_{x}\right) \displaystyle  = \displaystyle  a\cos\left(kz_{0}-\omega t\right)\cos\theta\ \ \ \ \ (12)
\displaystyle  y=\Re\left(f_{y}\right) \displaystyle  = \displaystyle  a\cos\left(kz_{0}-\omega t-\frac{\pi}{2}\right)\sin\theta\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  a\sin\left(kz_{0}-\omega t\right)\sin\theta \ \ \ \ \ (14)

and this time, a point on the string moves clockwise around the {z} axis.

Waves in a viscous fluid

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.7.

Another example of a transverse wave is that of a string embedded in a viscous fluid which adds a drag force on the string. The drag force is proportional to the string’s transverse speed, so we need to insert a term

\displaystyle \Delta F_{drag}=-\gamma\frac{\partial f}{\partial t}\Delta z \ \ \ \ \ (1)

where {\gamma} is a constant determined by the viscosity of the fluid, into our derivation of the original wave equation. This gives us a modified wave equation:

\displaystyle T\frac{\partial^{2}f}{\partial z^{2}}-\gamma\frac{\partial f}{\partial t}=\mu\frac{\partial^{2}f}{\partial t^{2}} \ \ \ \ \ (2)

where {T} is the tension in the string and {\mu} is the mass per unit length of the string.

If we assume that string’s frequency {\omega} is constant, then we can write

\displaystyle f\left(z,t\right)=e^{-i\omega t}F\left(z\right) \ \ \ \ \ (3)

and the wave equation reduces to an ODE:

\displaystyle TF''+i\omega\gamma F \displaystyle = \displaystyle -\mu\omega^{2}F\ \ \ \ \ (4)
\displaystyle F'' \displaystyle = \displaystyle -\frac{i\omega\gamma+\mu\omega^{2}}{T}F\equiv\alpha^{2}F \ \ \ \ \ (5)

The general solution is

\displaystyle F\left(z\right)=Ae^{\alpha z}+Be^{-\alpha z} \ \ \ \ \ (6)

However, {\alpha} is a complex number, so {F} contains both oscillatory and exponential parts.

Using Maple, we can find {\alpha}:

\displaystyle \alpha=\frac{1}{\sqrt{2T}}\left[\sqrt{\omega\sqrt{\omega^{2}\mu^{2}+\gamma^{2}}-\mu\omega^{2}}-i\sqrt{\omega\sqrt{\omega^{2}\mu^{2}+\gamma^{2}}+\mu\omega^{2}}\right] \ \ \ \ \ (7)

To keep {F\left(z\right)} finite for large {z} (we’re assuming the string extends from {z=0} to {z=+\infty}), we must have {A=0} and therefore

\displaystyle F\left(z\right)=A_{T}\exp\left\{ -\frac{z}{\sqrt{2T}}\left[\sqrt{\omega\sqrt{\omega^{2}\mu^{2}+\gamma^{2}}-\mu\omega^{2}}-i\sqrt{\omega\sqrt{\omega^{2}\mu^{2}+\gamma^{2}}+\mu\omega^{2}}\right]\right\} \ \ \ \ \ (8)

 


where {A_{T}} is the (complex) amplitude of the wave.

We can think of the imaginary part of the exponential as the wave number {k_{2}} and the real part as a spatial decay factor {\lambda}. That is

\displaystyle k_{2} \displaystyle \equiv \displaystyle \frac{1}{\sqrt{2T}}\sqrt{\omega\sqrt{\omega^{2}\mu^{2}+\gamma^{2}}+\mu\omega^{2}}\ \ \ \ \ (9)
\displaystyle \lambda \displaystyle \equiv \displaystyle \frac{1}{\sqrt{2T}}\sqrt{\omega\sqrt{\omega^{2}\mu^{2}+\gamma^{2}}-\mu\omega^{2}}\ \ \ \ \ (10)
\displaystyle \alpha \displaystyle = \displaystyle \lambda-ik_{2}\ \ \ \ \ (11)
\displaystyle F\left(z\right) \displaystyle = \displaystyle A_{T}e^{-\lambda z+ik_{2}z} \ \ \ \ \ (12)

Because of the negative real part in the exponent, the wave is attenuated (its amplitude falls off with distance) with a characteristic penetration distance {d} of

\displaystyle d=\frac{1}{\lambda}=\frac{\sqrt{2T}}{\sqrt{\omega\sqrt{\omega^{2}\mu^{2}+\gamma^{2}}-\mu\omega^{2}}} \ \ \ \ \ (13)

This is the distance at which the amplitude falls to {1/e} of its value at {z=0}.

Finally, we can consider the case of two strings joined by a massless knot at {z=0}, with the string on the right embedded in the viscous fluid. If the wave is sinusoidal, then we have

\displaystyle f\left(z,t\right)=\begin{cases} A_{I}e^{i\left(k_{1}z-\omega t\right)}+A_{R}e^{i\left(-k_{1}z-\omega t\right)} & z<0\\ e^{-i\omega t}F\left(z\right) & z>0 \end{cases} \ \ \ \ \ (14)

where {F\left(z\right)} is given by 8.

The boundary conditions require the wave function and its derivative to be continuous at {z=0}, so we get

\displaystyle A_{I}+A_{R} \displaystyle = \displaystyle A_{T}\ \ \ \ \ (15)
\displaystyle ik_{1}\left(A_{I}-A_{R}\right) \displaystyle = \displaystyle -\alpha A_{T}=-\alpha\left(A_{I}+A_{R}\right) \ \ \ \ \ (16)

Solving for {A_{R}} we get

\displaystyle A_{R} \displaystyle = \displaystyle \frac{\alpha+ik_{1}}{\alpha-ik_{1}}A_{I}\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle \frac{\lambda+i\left(k_{1}-k_{2}\right)}{\lambda-i\left(k_{1}+k_{2}\right)}A_{I} \ \ \ \ \ (18)

If {A_{I}} is real (that is, the incident wave has zero phase: {\delta_{I}=0}) then the magnitude of the amplitude of the reflected wave is

\displaystyle \left|A_{R}\right|=\sqrt{\frac{\lambda^{2}+\left(k_{1}-k_{2}\right)^{2}}{\lambda^{2}+\left(k_{1}+k_{2}\right)^{2}}}A_{I} \ \ \ \ \ (19)

We can convert 18 by multiplying top and bottom by the complex conjugate of the denominator:

\displaystyle A_{R} \displaystyle = \displaystyle \frac{\left(\lambda+i\left(k_{1}-k_{2}\right)\right)\left(\lambda+i\left(k_{1}+k_{2}\right)\right)}{\lambda^{2}+\left(k_{1}+k_{2}\right)^{2}}A_{I}\ \ \ \ \ (20)
\displaystyle \displaystyle = \displaystyle \frac{\lambda^{2}+k_{2}^{2}-k_{1}^{2}+2i\lambda k_{1}}{\lambda^{2}+\left(k_{1}+k_{2}\right)^{2}}A_{I} \ \ \ \ \ (21)

The phase of the reflected wave is

\displaystyle \tan\delta_{R} \displaystyle = \displaystyle \frac{\Im\left(A_{R}\right)}{\Re\left(A_{R}\right)}\ \ \ \ \ (22)
\displaystyle \displaystyle = \displaystyle \frac{2\lambda k_{1}}{\lambda^{2}+k_{2}^{2}-k_{1}^{2}} \ \ \ \ \ (23)

If {\gamma=0} (so there is no fluid surrounding string 2), then {\lambda=0} and

\displaystyle A_{R} \displaystyle = \displaystyle \frac{k_{2}^{2}-k_{1}^{2}}{\left(k_{1}+k_{2}\right)^{2}}A_{I}\ \ \ \ \ (24)
\displaystyle \displaystyle = \displaystyle \frac{k_{2}-k_{1}}{k_{2}+k_{1}}A_{I}\ \ \ \ \ (25)
\displaystyle \tan\delta_{R} \displaystyle = \displaystyle 0 \ \ \ \ \ (26)

which agrees with the equations in Griffiths for sinusoidal waves.

Waves: boundary condition with a massive knot

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.6.

In our earlier analysis of the boundary conditions for a one dimensional wave, we looked at the case of one string of mass per unit length {\mu_{1}} joined at {z=0} to another string with mass density {\mu_{2}}, and we assumed that the knot joining the two strings was massless. This led to the boundary conditions being that both the wave function and its first derivative are continuous at {z=0}.

Suppose we now take the knot to have a non-zero mass {m}. In that case, the derivative no longer needs to be continuous at the knot. We can analyze the situation the same way as we derived the original wave equation. Since the tensions on either side of the knot are equal in magnitude but can now point in different directions, we have the net transverse force on the knot

\displaystyle F_{t}=T\left(\left.\frac{\partial f}{\partial z}\right|_{0^{+}}-\left.\frac{\partial f}{\partial z}\right|_{0^{-}}\right) \ \ \ \ \ (1)

where {T} is the tension in the strings.

This force is equal to the knot’s mass times its transverse acceleration, so

\displaystyle T\left(\left.\frac{\partial f}{\partial z}\right|_{0^{+}}-\left.\frac{\partial f}{\partial z}\right|_{0^{-}}\right)=m\left.\frac{\partial^{2}f}{\partial t^{2}}\right|_{0} \ \ \ \ \ (2)

 

Now suppose we have an incident sinusoidal wave

\displaystyle I\left(z,t\right)=A_{I}e^{i\left(k_{1}z-\omega t\right)} \ \ \ \ \ (3)

using complex notation, so {A_{I}} is a complex amplitude with real magnitude {a_{I}} and phase {\delta_{1}}:

\displaystyle A_{I}=a_{I}e^{i\delta_{1}} \ \ \ \ \ (4)

(I’ll drop the tildes that Griffiths uses in his book, since we’ll be dealing with complex notation pretty well exclusively until we extract the physical meaning at the end.)

When this wave hits the knot, part of it is reflected and part is transmitted, so the overall wave function is

\displaystyle f\left(z,t\right)=\begin{cases} A_{I}e^{i\left(k_{1}z-\omega t\right)}+A_{R}e^{i\left(-k_{1}z-\omega t\right)} & z<0\\ A_{T}e^{i\left(k_{2}z-\omega t\right)} & z>0 \end{cases} \ \ \ \ \ (5)

The frequency {\omega} is the same on both sides of {z=0}, but the wave numbers {k_{1}} and {k_{2}} (and thus the wave speeds {v_{i}=\omega/k_{i}}) are different. The continuity condition gives us

\displaystyle A_{I}+A_{R}=A_{T} \ \ \ \ \ (6)

 

and the derivative condition 2 gives us

\displaystyle ik_{2}A_{T}-ik_{1}\left(A_{I}-A_{R}\right)=-\frac{m\omega^{2}}{T}A_{T} \ \ \ \ \ (7)

 

These two equations can be solved to give {A_{R}} and {A_{T}} in terms of {A_{I}}.

Example Suppose the string for {z<0} has a non-zero mass {\mu_{1}} so the wave’s speed {v_{1}} and wave number {k_{1}} are both finite and non zero. However, the string for {z>0} is taken to be very light (effectively massless). In this case, the speed is (as we saw in the derivation of the wave equation)

\displaystyle v_{2}=\sqrt{\frac{T}{\mu_{2}}}=\infty \ \ \ \ \ (8)

so

\displaystyle k_{2}=\frac{\omega}{v_{2}}=0 \ \ \ \ \ (9)

Then from 7 and 6 we have

\displaystyle \frac{Tk_{1}}{m\omega^{2}}i\left(A_{I}-A_{R}\right)=A_{T}=A_{I}+A_{R} \ \ \ \ \ (10)

With the definition

\displaystyle \alpha\equiv\frac{Tk_{1}}{m\omega^{2}} \ \ \ \ \ (11)

we can solve this to get

\displaystyle A_{R} \displaystyle = \displaystyle A_{I}\frac{\alpha i-1}{1+\alpha i}\ \ \ \ \ (12)
\displaystyle A_{T} \displaystyle = \displaystyle A_{I}\frac{2\alpha i}{1+\alpha i} \ \ \ \ \ (13)

Taking magnitudes we get the real amplitudes of the reflected and transmitted waves:

\displaystyle a_{R} \displaystyle = \displaystyle a_{I}\ \ \ \ \ (14)
\displaystyle a_{T} \displaystyle = \displaystyle \frac{2\alpha}{\sqrt{1+\alpha^{2}}}a_{I} \ \ \ \ \ (15)

To get the phases, we can find the ratio of the imaginary and real parts in 12 and 13 to get the tangents of the phases. This is a bit messy so it’s easiest to use Maple to do the algebra. We get

\displaystyle \tan\delta_{R} \displaystyle = \displaystyle \frac{2\alpha\cos\delta_{1}+\left(\alpha^{2}-1\right)\sin\delta_{1}}{\left(\alpha^{2}-1\right)\cos\delta_{1}-2\alpha\sin\delta_{1}}\ \ \ \ \ (16)
\displaystyle \displaystyle = \displaystyle \frac{2\alpha+\left(\alpha^{2}-1\right)\tan\delta_{1}}{\left(\alpha^{2}-1\right)-2\alpha\tan\delta_{1}}\ \ \ \ \ (17)
\displaystyle \displaystyle = \displaystyle \frac{\frac{2\alpha}{\alpha^{2}-1}+\tan\delta_{1}}{1-\frac{2\alpha}{\alpha^{2}-1}\tan\delta_{1}} \ \ \ \ \ (18)

We can simplify this equation by using the formula for the tangent of the sum of two angles:

\displaystyle \tan\left(u+v\right)=\frac{\tan u+\tan v}{1-\tan u\tan v} \ \ \ \ \ (19)

We get

\displaystyle \tan\delta_{R} \displaystyle = \displaystyle \tan\left(\arctan\frac{2\alpha}{\alpha^{2}-1}+\delta_{1}\right)\ \ \ \ \ (20)
\displaystyle \delta_{R} \displaystyle = \displaystyle \arctan\frac{2\alpha}{\alpha^{2}-1}+\delta_{1} \ \ \ \ \ (21)

For the transmitted wave, we have

\displaystyle \tan\delta_{T} \displaystyle = \displaystyle \frac{\cos\delta_{1}+\alpha\sin\delta_{1}}{-\sin\delta_{1}+\alpha\cos\delta_{1}}\ \ \ \ \ (22)
\displaystyle \displaystyle = \displaystyle \frac{\frac{1}{\alpha}+\tan\delta_{1}}{1-\frac{1}{\alpha}\tan\delta_{1}}\ \ \ \ \ (23)
\displaystyle \displaystyle = \displaystyle \tan\left(\arctan\frac{1}{\alpha}+\delta_{1}\right)\ \ \ \ \ (24)
\displaystyle \delta_{T} \displaystyle = \displaystyle \arctan\frac{1}{\alpha}+\delta_{1} \ \ \ \ \ (25)

It might seem odd that the reflected wave has the same amplitude as the incident wave, yet the transmitted wave still has a non-zero amplitude. Doesn’t this violate conservation of energy, since the kinetic energy of the string due to the reflected wave is equal to that of the incident wave, so how can there be any energy left over for a transmitted wave? The answer is that since the transmitted wave is in a massless string, it carries no energy, so energy can still be conserved.

 

Waves: boundary conditions

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 9.5.

For a one dimensional wave (on a string, say) suppose we now place a boundary at the point {z=0}. For a string, this could be a point at which one string is joined with another string of a different mass per unit length. If we send a wave down the string from large negative {z}, when this wave reaches {z=0}, there will be a reflected wave that returns towards negative {z} and a transmitted wave that proceeds beyond {z=0} towards positive {z}. We can get some idea of the nature of these reflected and transmitted waves if we impose some boundary conditions at the point {z=0}.

First, we require the wave function {f} to be continuous, for the simple reason that there is no break in the string at {z=0}. The second boundary condition requires that the derivative {\partial f/\partial z} is also continuous. The reason for this is a bit more subtle. Assuming there is no point mass (such as a knot) at the joining position, if the tangent to the string at that point were not continuous, then the second derivative would be infinite, meaning that there would be an infinite force at that point.

To see how the incident, reflected and transmitted waves are related, suppose we have an incident wave {I(z-v_{1}t)}, a reflected wave {R\left(z+v_{1}t\right)} and a transmitted wave {T\left(x-v_{2}t\right)}, where {I} and {T} are moving to the right, with {I} defined for {z<0} and {T} for {z>0}, and {R} moving to the left for {z<0}.

The continuity of the wave function at {z=0} gives us

\displaystyle  I\left(-v_{1}t\right)+R\left(v_{1}t\right)=T\left(-v_{2}t\right) \ \ \ \ \ (1)

The continuity of the derivative, if applied directly, just gives the same equation with each function replaced by its derivative, so doesn’t help much:

\displaystyle  \left.\frac{\partial I}{\partial z}\right|_{z=0^{-}}+\left.\frac{\partial R}{\partial z}\right|_{z=0^{-}}=\left.\frac{\partial T}{\partial z}\right|_{z=0^{+}} \ \ \ \ \ (2)

However, if we consider the original definition of a derivative as a limit, we can make some progress. Consider first the derivative of the incident wave just below {z=0}, at time {t=0}:

\displaystyle  \left.\frac{\partial I}{\partial z}\right|_{z=0^{-}}=\lim_{\Delta z\rightarrow0}\frac{I\left(0\right)-I\left(-\Delta z\right)}{\Delta z} \ \ \ \ \ (3)

The wave amplitude at the point {\left(z,t\right)=\left(-\Delta z,0\right)} will be at {z=0} after it travels the distance {\Delta z}, which takes a time {t=\Delta z/v_{1}}.

By a similar argument, the derivative of the reflected wave is

\displaystyle  \left.\frac{\partial R}{\partial z}\right|_{z=0^{-}}=\lim_{\Delta z\rightarrow0}\frac{R\left(0\right)-R\left(-\Delta z\right)}{\Delta z} \ \ \ \ \ (4)

This time, the wave amplitude at {\left(-\Delta z,0\right)} was at {z=0} at time {t=-\Delta z/v_{1}} since this wave is travelling to the left. Finally, for the transmitted wave

\displaystyle  \left.\frac{\partial T}{\partial z}\right|_{z=0^{+}}=\lim_{\Delta z\rightarrow0}\frac{T\left(\Delta z\right)-T\left(0\right)}{\Delta z} \ \ \ \ \ (5)

since this wave is defined for {z>0}. The wave amplitude at {\left(\Delta z,0\right)} was at {z=0} at time {t=-\Delta z/v_{2}} since the transmitted wave is travelling to the right with speed {v_{2}}.

We can now use the continuity condition 1 to eliminate either {R} or {T} from the limits. Start by eliminating {R} by evaluating everything at time {t=-\Delta z/v_{1}}:

\displaystyle  \lim_{\Delta z\rightarrow0}\frac{R\left(0\right)-R\left(-\Delta z\right)}{\Delta z}=\lim_{\Delta z\rightarrow0}\frac{1}{\Delta z}\left[T\left(0\right)-I\left(0\right)-\left(T\left(\frac{v_{2}}{v_{1}}\Delta z\right)-I\left(\Delta z\right)\right)\right] \ \ \ \ \ (6)

Now we can insert this into the continuity equation for derivatives 2, and we’ll leave off the limit and {1/\Delta z} to simplify the notation:

\displaystyle   I\left(0\right)-I\left(-\Delta z\right)+T\left(0\right)-I\left(0\right)-\left(T\left(\frac{v_{2}}{v_{1}}\Delta z\right)-I\left(\Delta z\right)\right) \displaystyle  = \displaystyle  T\left(\Delta z\right)-T\left(0\right)\ \ \ \ \ (7)
\displaystyle  \left(I\left(0\right)-I\left(-\Delta z\right)\right)+\left(I\left(\Delta z\right)-I\left(0\right)\right) \displaystyle  = \displaystyle  \left(T\left(\frac{v_{2}}{v_{1}}\Delta z\right)-T\left(0\right)\right)+\left(T\left(\Delta z\right)-T\left(0\right)\right) \ \ \ \ \ (8)

Restoring the limit and {1/\Delta z} we get

\displaystyle   2\frac{\partial I}{\partial z} \displaystyle  = \displaystyle  \left(\frac{v_{2}}{v_{1}}+1\right)\frac{\partial T}{\partial z}\ \ \ \ \ (9)
\displaystyle  T\left(z-v_{2}t\right) \displaystyle  = \displaystyle  \frac{2v_{1}}{v_{1}+v_{2}}I\left(z-v_{1}t\right)+K_{T} \ \ \ \ \ (10)

where {K_{T}} is a constant of integration and we used the chain rule in the form

\displaystyle  \lim_{\Delta z\rightarrow0}\frac{f(a\Delta z)-f\left(0\right)}{\Delta z}=\lim_{\Delta z\rightarrow0}a\frac{f(a\Delta z)-f\left(0\right)}{a\Delta z}=af'\left(0\right) \ \ \ \ \ (11)

For the reflected wave, we eliminate {T} using 1 at time {t=-\Delta z/v_{2}} :

\displaystyle  \lim_{\Delta z\rightarrow0}\frac{T\left(\Delta z\right)-T\left(0\right)}{\Delta z}=\lim_{\Delta z\rightarrow0}\frac{1}{\Delta z}\left[I\left(\frac{v_{1}}{v_{2}}\Delta z\right)+R\left(-\frac{v_{1}}{v_{2}}\Delta z\right)-I\left(0\right)-R\left(0\right)\right] \ \ \ \ \ (12)

Substitute into 2, again without the limit and {1/\Delta z} to simplify the notation:

\displaystyle   I\left(0\right)-I\left(-\Delta z\right)+R\left(0\right)-R\left(-\Delta z\right) \displaystyle  = \displaystyle  I\left(\frac{v_{1}}{v_{2}}\Delta z\right)+R\left(-\frac{v_{1}}{v_{2}}\Delta z\right)-I\left(0\right)-R\left(0\right)\ \ \ \ \ (13)
\displaystyle  \left(I\left(0\right)-I\left(-\Delta z\right)\right)-\left(I\left(\frac{v_{1}}{v_{2}}\Delta z\right)-I\left(0\right)\right) \displaystyle  = \displaystyle  -\left(R\left(0\right)-R\left(-\frac{v_{1}}{v_{2}}\Delta z\right)\right)-\left(R\left(0\right)-R\left(-\Delta z\right)\right) \ \ \ \ \ (14)

Restoring the limit and {1/\Delta z} we get

\displaystyle   \left(1-\frac{v_{1}}{v_{2}}\right)\frac{\partial I}{\partial z} \displaystyle  = \displaystyle  -\left(1+\frac{v_{1}}{v_{2}}\right)\frac{\partial R}{\partial z}\ \ \ \ \ (15)
\displaystyle  R\left(z+v_{1}t\right) \displaystyle  = \displaystyle  \frac{v_{1}-v_{2}}{v_{2}+v_{1}}I\left(z-v_{1}t\right)+K_{R} \ \ \ \ \ (16)

These results apply to any wave shape, not just to sinusoidal waves.

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