Plane Waves, Transversality and Polarization
~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with vector calculus identities, specifically the curl of a gradient and the cross product, and you must have derived the wave equation for electromagnetic fields in vacuum in the previous lecture.
Why this matters
Understanding the geometric constraints on plane waves is the foundation for analyzing light propagation, optical devices, and radiation patterns.
While the wave equation tells us that electromagnetic disturbances propagate at speed \( c \), it does not fully specify the relationship between the electric and magnetic fields. Maxwell's equations impose strict geometric constraints: the fields must be transverse to the direction of propagation, mutually orthogonal, and in phase. These constraints define the concept of polarization, which is central to modern optics, telecommunications, and quantum information science. Without this structural understanding, one cannot correctly interpret how light interacts with matter or how antennas radiate energy.
The Geometry of Propagation
Imagine a wave travelling in a straight line. In a mechanical wave on a string, the displacement is perpendicular to the direction of travel. For electromagnetic waves in vacuum, the situation is even more rigid. The electric field vector \( \mathbf{E} \) and the magnetic field vector \( \mathbf{B} \) do not just wiggle randomly; they are locked into a specific geometric configuration relative to the direction the wave is moving, defined by the wave vector \( \mathbf{k} \). Neither field has a component along the direction of travel. This is what we mean by "transverse." Furthermore, \( \mathbf{E} \) and \( \mathbf{B} \) are always at right angles to each other. If you point your right thumb in the direction of propagation (\( \mathbf{k} \)), your fingers curl from \( \mathbf{E} \) towards \( \mathbf{B} \). This right-handed triad is a direct consequence of Faraday's law and Ampère's law in the absence of sources. It means that knowing the direction of travel and the electric field completely determines the magnetic field; there is no freedom to choose them independently.
Transversality and Field Relations
For a monochromatic plane wave, the fields take the form \( \mathbf{E}(\mathbf{r}, t) = \mathbf{E}_0 e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)} \) and \( \mathbf{B}(\mathbf{r}, t) = \mathbf{B}_0 e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)} \). Substituting these into Maxwell's equations in vacuum yields two critical constraints. First, Gauss's law for electricity, \( \nabla \cdot \mathbf{E} = 0 \), implies that the divergence of the electric field is zero. For a plane wave, this reduces to the dot product of the wave vector and the electric field amplitude being zero: \( \mathbf{k} \cdot \mathbf{E}_0 = 0 \). This confirms that \( \mathbf{E} \) is perpendicular to \( \mathbf{k} \). Similarly, Gauss's law for magnetism, \( \nabla \cdot \mathbf{B} = 0 \), implies \( \mathbf{k} \cdot \mathbf{B}_0 = 0 \), confirming that \( \mathbf{B} \) is also perpendicular to \( \mathbf{k} \).
The relationship between the magnitudes and directions of \( \mathbf{E} \) and \( \mathbf{B} \) is derived from Faraday's law. The result is that the magnetic field is determined by the electric field and the direction of propagation:
where \( \hat{\mathbf{k}} \) is the unit vector in the direction of propagation. This equation enforces the right-handedness of the triad. Taking the magnitude of both sides, and noting that \( \hat{\mathbf{k}} \) is perpendicular to \( \mathbf{E} \), we find the ratio of the field amplitudes:
This is a fundamental result: in vacuum, the electric field is always \( c \) times larger than the magnetic field in SI units. The fields are in phase, meaning they reach their maxima and minima simultaneously.
Polarization States
Polarization describes the time evolution of the electric field vector at a fixed point in space. Since \( \mathbf{E} \) must lie in the plane perpendicular to \( \mathbf{k} \), we can decompose \( \mathbf{E}_0 \) into two orthogonal components, say along \( \hat{\mathbf{x}} \) and \( \hat{\mathbf{y}} \), assuming propagation along \( \hat{\mathbf{z}} \). The general form is \( \mathbf{E}_0 = a_x \hat{\mathbf{x}} + a_y e^{i\delta} \hat{\mathbf{y}} \), where \( a_x \) and \( a_y \) are real amplitudes and \( \delta \) is the phase difference between the components.
If \( \delta = 0 \) or \( \pi \), the components are in phase or anti-phase. The tip of the electric field vector traces a straight line. This is linear polarization. If \( a_x = a_y \) and \( \delta = \pm \pi/2 \), the components have equal magnitude and are \( 90^\circ \) out of phase. The tip of the vector traces a circle. This is circular polarization. The sign of \( \delta \) determines the handedness (left or right). For any other combination of amplitudes and phase differences, the tip traces an ellipse, known as elliptical polarization. Linear and circular polarizations are special cases of elliptical polarization.
Superposition and Coherence
Maxwell's equations are linear, so the superposition principle applies. If \( \mathbf{E}_1 \) and \( \mathbf{E}_2 \) are solutions, their sum is also a solution. This is the basis for interference and the creation of complex polarization states. However, for stable interference patterns, the waves must be coherent, meaning they maintain a constant phase relationship over time. In practice, this requires the sources to be phase-locked. When two plane waves with the same frequency and propagation direction but different polarization states are superimposed, the resulting field is generally elliptically polarised. The degree of polarization can be quantified using the Stokes parameters, which provide a complete description of the polarization state without explicitly referencing the phase.
The derivation
Structure of Monochromatic Plane Waves
In a source-free, linear, homogeneous, isotropic, non-dispersive medium (vacuum unless stated), every monochromatic plane-wave solution of Maxwell's equations, E(r,t) = E0 ei(k·r − ωt) and B(r,t) = B0 ei(k·r − ωt), is strictly transverse: k·E = k·B = 0. The triad (E, B, k) is mutually orthogonal and right-handed, the two fields oscillate in phase, their amplitudes obey |E| = c|B|, and the complex vector E0 encodes the polarization state — linear, circular, or elliptical.
A common misconception is that the magnetic field is "caused" by the electric field in a temporal sense. In reality, they are coupled components of a single electromagnetic field tensor. In a different inertial frame, what appears as a pure electric field in one frame may appear as a mixture of electric and magnetic fields in another. The distinction between \( \mathbf{E} \) and \( \mathbf{B} \) is frame-dependent, but the invariants \( \mathbf{E}^2 - c^2\mathbf{B}^2 \) and \( \mathbf{E} \cdot \mathbf{B} \) are not. For a plane wave in vacuum, both invariants are zero, which is why the wave is always transverse and \( |\mathbf{E}| = c|\mathbf{B}| \) in any inertial frame. C
Worked example
Problem. A monochromatic plane wave propagates in the \( +\hat{\mathbf{z}} \) direction in vacuum. The electric field is given by \( \mathbf{E}(\mathbf{r}, t) = (3.0 \hat{\mathbf{x}} + 4.0 \hat{\mathbf{y}}) \cos(kz - \omega t) \ \mathrm{V/m} \). Determine the magnetic field vector \( \mathbf{B}(\mathbf{r}, t) \) and the intensity of the wave.
- Identify the direction of propagation: \( \hat{\mathbf{k}} = \hat{\mathbf{z}} \).
- Use the relation \( \mathbf{B} = \frac{1}{c} (\hat{\mathbf{k}} \times \mathbf{E}) \). Calculate the cross product: \( \hat{\mathbf{z}} \times (3.0 \hat{\mathbf{x}} + 4.0 \hat{\mathbf{y}}) = 3.0 (\hat{\mathbf{z}} \times \hat{\mathbf{x}}) + 4.0 (\hat{\mathbf{z}} \times \hat{\mathbf{y}}) = 3.0 \hat{\mathbf{y}} - 4.0 \hat{\mathbf{x}} \).
- Divide by \( c = 3.0 \times 10^8 \ \mathrm{m/s} \): \( \mathbf{B} = \frac{1}{3.0 \times 10^8} (-4.0 \hat{\mathbf{x}} + 3.0 \hat{\mathbf{y}}) \cos(kz - \omega t) \ \mathrm{T} \).
- Calculate the intensity \( I \) using the time-averaged Poynting vector magnitude: \( I = \frac{1}{2} c \epsilon_0 E_0^2 \), where \( E_0 = \sqrt{3.0^2 + 4.0^2} = 5.0 \ \mathrm{V/m} \).
- Substitute values: \( I = \frac{1}{2} (3.0 \times 10^8) (8.85 \times 10^{-12}) (5.0)^2 \).
Units check. The magnetic field is in Tesla (\( \mathrm{V \cdot s / m^2} \)), which is consistent with \( \mathrm{V/m} / (\mathrm{m/s}) \). The intensity is in \( \mathrm{W/m^2} \), consistent with \( (\mathrm{m/s}) (\mathrm{F/m}) (\mathrm{V/m})^2 = (\mathrm{m/s}) (\mathrm{C/(V \cdot m)}) (\mathrm{V^2/m^2}) = \mathrm{C \cdot V / (s \cdot m^2)} = \mathrm{J / (s \cdot m^2)} = \mathrm{W/m^2} \).
Three quick tests
- Is the magnetic field in a plane wave in vacuum parallel or perpendicular to the electric field?
Answer
Perpendicular. They are mutually orthogonal and both are perpendicular to the direction of propagation. - If the electric field of a plane wave is linearly polarised along the x-axis, what is the direction of the magnetic field if the wave propagates along the z-axis?
Answer
Along the y-axis (specifically \( +\hat{\mathbf{y}} \) if \( \mathbf{E} \) is along \( +\hat{\mathbf{x}} \)), determined by the right-hand rule \( \hat{\mathbf{k}} \times \mathbf{E} \). - What is the ratio of the magnitude of the electric field to the magnitude of the magnetic field in a vacuum plane wave?
Answer
The ratio is \( c \), the speed of light. \( |\mathbf{E}| = c |\mathbf{B}| \).
Problem set for PU-204 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes checked line by line by the editor; every number in the worked example was recomputed.