A student claims: 'The fact that electromagnetic waves are transverse — that E and B are perpendicular to the propagation direction — is an independent physical postulate required to complete Maxwell's theory.' How would you respond?
AThe student is correct; transversality must be stated separately as an experimental observation
BTransversality follows directly from applying Gauss's law (∇·E = 0) to the plane wave ansatz — it is derived, not assumed
CTransversality is only approximate; at very high frequencies, E develops a small component along k
DTransversality follows from the wave equation alone, without needing any of Maxwell's other equations
Transversality is not a separate postulate — it is derived. Substituting E = E₀e^{i(k·r − ωt)} into Gauss's law ∇·E = 0, the divergence acts on the exponent to produce ik·E₀ = 0, which means k·E₀ = 0: the electric field amplitude is perpendicular to k. Option D is wrong because the wave equation alone doesn't enforce transversality — you specifically need the zero-divergence equations (Gauss's laws for E and B).
Question 2 Multiple Choice
Why does a glass prism separate white light into colors, while a vacuum does not?
AGlass absorbs some frequencies more strongly, removing them selectively from the beam
BIn glass the dispersion relation is no longer ω = ck — different frequencies travel at different speeds, so they refract at different angles
CThe plane wave approximation breaks down in glass, making light travel in curved paths
DGlass introduces a phase delay that is the same for all frequencies, shifting but not separating them
In vacuum, ω = ck holds exactly for all frequencies — all EM waves travel at c regardless of frequency. In glass, interactions with the medium produce a frequency-dependent index of refraction n(ω), so different frequencies travel at c/n(ω) and refract at different angles at the glass surface. This dispersion is absent in vacuum, which is why vacuum propagation is non-dispersive and wave packets travel without spreading.
Question 3 True / False
In vacuum, gamma rays travel faster than radio waves because their much higher frequency gives them greater energy and therefore greater speed.
TTrue
FFalse
Answer: False
The dispersion relation ω = ck holds exactly for all electromagnetic frequencies in vacuum. The phase velocity ω/k = c is the same for every EM wave, regardless of frequency or energy. This is a fundamental consequence of Maxwell's equations in vacuum — there is no medium to interact with and produce frequency-dependent propagation. All electromagnetic radiation travels at exactly c in vacuum.
Question 4 True / False
The relationship B = k̂ × E/c between the electric and magnetic fields of a plane wave follows from applying Faraday's law to the plane wave ansatz.
TTrue
FFalse
Answer: True
Applying Faraday's law ∇ × E = −∂B/∂t to the plane wave: the curl of E brings down ik × E₀, and the time derivative of B gives iωB₀. Setting ik × E₀ = iωB₀ and using ω = ck yields B₀ = (k × E₀)/ω = k̂ × E₀/c. This confirms that B is perpendicular to both k and E, with magnitude |E|/c — entirely derived from Maxwell's equations, not assumed.
Question 5 Short Answer
Explain how applying Gauss's law (∇·E = 0) to a plane wave E = E₀e^{i(k·r − ωt)} proves that the electric field must be transverse.
Think about your answer, then reveal below.
Model answer: Taking the divergence: ∇·E = ∇·(E₀e^{i(k·r − ωt)}). The spatial derivatives act on the exponent and pull down ik, giving ik·E₀ e^{i(k·r − ωt)}. Setting this equal to zero (Gauss's law in vacuum) yields k·E₀ = 0, which is exactly the condition that E₀ is perpendicular to k — the wave is transverse.
The key step is that differentiating a plane wave with respect to position replaces ∇ with ik algebraically. This converts the differential equation (Gauss's law) into an algebraic dot-product condition on the amplitude vector, directly implying transversality.