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Lecture 13: Deriving Electromagnetic Waves

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Lecture Topics

On a chalkboard pink and yellow curves sit below and above a straight line. The figure looks somewhat like a butterfly's wings.
  • Wave Equation from Maxwell's Equations
  • Wave Solutions to Maxwell's Equations
  • E and B in EM Waves
  • Polarization
  • Three-dimensional Propagation

Learning Objectives

By the end of this lecture, you should:

  • know how to combine Maxwell's equations and the "curl of a curl" to form a wave equation in vacuum.
  • identify the speed of light with a combination of permittivity and permeability.
  • relate the electric and magnetic fields of an EM wave.
  • describe an EM wave as seen at different places at one time.
  • describe an EM wave as seen at one place as time goes by.
  • combine solutions to form polarized waves.
  • write complete three-dimensional solutions for EM waves.
  • describe how wavefront projections may move faster than c.
  • relate polarization to antenna direction and describe polarizers.

Lecture Activities

Check Yourself

  • Verify, including checking the SI base units, that the product of the permittivity of free space and the permeability of free space gives the inverse of the square of the speed of light in vacuum.

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Since the lectures do not actually give the base units, a little external checking is needed. The permittivity of free space, also known as the vacuum permittivity, is \({\varepsilon _0} \approx \) 8.854187817620×10−12 [A2 s4 kg-1 m-3]  in SI base units (if looking this up on the internet seems like cheating to you, you could work it out using Coulomb’s Law). The vacuum permeability is defined as \({\mu _0} = 4\pi \times {10^{ - 7}}\)[NA-2] where the base units would be [kg m s-2 A-2]. In the SI system, the speed of light c is defined to be 299792458 [m s-1]. The numbers given work out to the fifth decimal place when converted to the speed of light. The SI base unit calculation is \[[{\varepsilon _0}{\mu _0}] = [{A^2}{s^4}k{g^{ - 1}}{m^{ - 3}}][kg{\kern 1pt}  m{\kern 1pt}  {s^{ - 2}}{\kern 1pt} {A^{ - 2}}] = [{s^4}{m^{ - 3}}][m{\kern 1pt} {s^{ - 2}}{\kern 1pt} ] = [{s^2}{m^{ - 2}}]\] which are the inverse of those of the square of velocity.

 

  • Electromagnetic waves in a vacuum can exist due to the displacement current term that replaces a "physical" current in Ampère’s Law, whose form in vacuum then becomes \[\vec \nabla \times \vec B = {\varepsilon _0}{\mu _0}\frac{{\partial \vec E}}{{\partial t}}\] Without the displacement current, the right hand side would be zero. Late in the classical physics period, people still wanted to think of waves as requiring a "medium" such as a string to travel in. Sometimes a physical concept that is incorrect leads to fairly extreme conclusions as a sign that something is wrong, and the properties of the "ether" that was hypothesized to support EM waves is no exception. To see this, assume first that EM waves propagate much as waves on stretched wires. If the wires had even a negligible mass density, say 10-6 kg/m, what would be the tension in them to give a speed of c? Conversely, if the tension was reasonable, say 100 N, what would the mass density be in such a wire? Comment on the results.

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The speed of waves on a wire is given by \(v = \sqrt {\frac{T}{\mu }}\), so if there is more tension it goes up, and if there is more mass "loading" each meter of wire, it goes down. If we assume an almost unimaginably small but possibly physically realizable linear mass density of 10-6 in SI units, then we need \[c = 3 \times {10^8} = \sqrt {\frac{T}{{{{10}^{ - 6}}}}}\] Squaring both sides and solving for T gives \(T = 9 \times {10^{22}}\)N. This is a huge tension, especially for a wire that is almost unimaginably thin and light. Conversely, if we try to imagine a reasonable tension of 100 N, we end up with \[c = 3 \times {10^8} = \sqrt {\frac{{100}}{\mu }} \] or \[\mu = \frac{{100}}{{9 \times {{10}^{16}}}} \approx {10^{ - 15}}kg/m. \] If we thought the initially proposed wire was thin and light, this is about a billion times more so. Although the various historic theories of the ether differed in detail from what we are trying to look at here, the common factor was that the ether had to have truly remarkable properties (and not limited to its mechanical properties). Usually, arriving at such extreme conclusions is a sign there is something wrong with one’s understanding.

 

  • The direction of polarization in EM waves is rather naturally chosen as the direction of the electric field because in many circumstances, the most basic interaction with matter is through acceleration of charges by the electric field, giving it more importance than the magnetic field that is also always present. In what circumstance might it make sense to talk about direction of polarization of waves on a string?

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Waves on a string can propagate along the string, causing transverse motion of the string in any plane containing the string and a perpendicular to it. (This is easier to physically demonstrate if the string is rather taut so that gravity does not affect it much.) The waves could pass easily through a narrow slot which was aligned exactly in the direction the string is vibrating. If the string is vibrating at an angle to such a slot, it would rub on the edges and the wave would be damped. This is a very similar situation to EM waves interacting with a polarizer, although the physical mechanisms are different. In this case it would be important to specify a direction of polarization for waves on a string.

 

  • Consider water waves coming in to a beach and their similarity to EM waves relative to some given plane. Again, one is torn in talking too much about water waves since in reality their physics is much more complex than that of EM waves, but on the other hand we can look at some very familiar situations and learn from them. The speed of waves whose wavelength is longer than the depth h (a common situation near shore) is \(v = \sqrt {gh}\). This means that waves slow down as they get nearer to shore. One of the implications of this is that waves almost always come in with their crests almost parallel to shore. Take the speed in 1 m deep water as typical, and calculate at what speed a wave front moving at that speed moves along the beach if it comes in at a 1º angle to shore. Do this by considering where a wave front, that hit the shore at the origin at time t = 0, will be 1 second later. You found the speed of the wave front; what was the speed of the water?

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The speed of the wave is taken as \(v = \sqrt {gh} = \sqrt {9.8(1)} = 3.1\)m/s without considering that it slows down even more even closer to shore. This leads to wave breaking and behavior we really are not equipped to discuss. In a coordinate system with the x axis parallel to the beach and y directed away from shore, the phase along the wave front is a constant: \(C = \vec k\bullet\vec x - \omega t\). Since the wave hits at the origin at t=0, in fact C=0. One second later, the same phase has moved toward the shore and the equation is \(C = \vec k\bullet\vec x - \omega(1)\). This leads to \(0 = kx\cos (89^\circ ) - \omega \) on the x axis (along the shore) since 89º is the angle between \(\vec k\), which is perpendicular to the wavefront, and the x axis (and on the x axis there is no y value to put into the dot product). The dispersion relation gives the wave (phase) speed as \(v = \frac{\omega }{k}\), so \(\omega = kv\) and the phase equation becomes \[0 = kx\cos (89^\circ ) - kv = k(x\cos (89^\circ ) - v)\] We do not need to solve for k, but we know it is not zero. Thus we must have that the other part is zero: \[x\cos (89^\circ ) - v = 0\] or \[x = \frac{v}{{\cos (89^\circ )}} = \frac{{3.1}}{{0.01745}} = 177m.\] However, x is the distance the phase front moved along the beach in 1 second: it is thus the phase velocity of the wave along the shore. This is quite a fast speed: more than 600 km/hour! Under ideal conditions, such very fast speeds can actually be seen. How fast did the water move? Hardly at all: the passing wave phenomena do not on average move much water around, with most of the motion being up and down and being gone after the wave passes. Note: compare this rather algebraic approach to that in the lecture, which is more geometric, to see which is clearer for you.

 

  • In the lecture, a polarization demonstration was done in which crossed polarizers let through no light, then another polarizer was inserted between them and light passed. Calculate how much light got through, ignoring any absorption in the polarizers, if the angle the third polarizer was placed at was 45º relative to either of the first two.

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The first polarizer took the initial unpolarized light and let through 50% of it, but that 50% was 100% polarized parallel to the axis of the polarizer. What is now the second polarizer (the inserted sheet) was at 45º to this, and by Malus’ Law would let through \({\cos ^2}(45^\circ ) = 0.5\) of this light, 100% polarized along its axis. Since the original two polarizers were crossed, they were at 90º to each other, so this light enters the final polarizer at 45º also, with a 50% reduction. The final intensity is (0.5)3=0.125 as intense as the original. This is of course infinitely more light than was getting through the crossed polarizers since that amount was zero!

 

  • The peak electric field associated with a typical radio signal is about 1 mV/m, while a laser spot may have a field about 1012 times higher. What is the wave magnetic field in each case?

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\(B = \frac{E}{c}\) relates the amplitudes of the in-phase electric and magnetic fields in a propagating EM wave (in SI units). In the radio wave this gives \[B = \frac{{.001}}{{3 \times {{10}^8}}} = 3.3 \times {10^{ - 12}}T.\] A laser spot would thus have a large field of about 3 T. However, this would not be directly measured as a magnetic field due to its very high frequency.

 

  • If the typical wavelength of visible light is 500 nm, what are the frequency and the wave number?

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All the waves stacked end to end in one second’s travel mean that \(\lambda f = c\) or \[f = \frac{c}{\lambda } = \frac{{3 \times {{10}^8}}}{{500 \times {{10}^{ - 9}}}} = 6 \times {10^{14}} Hz. \] The wave number is \[k = \frac{{2\pi }}{\lambda } = \frac{{6.28}}{{500 \times {{10}^{ - 9}}}} = 1.26 \times {10^7}m^{-1}.\]

 

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