This chapter starts with physical equations and proves their mathematical consequences. Chapters 2 and 3 define the derivatives, vector operations and integrals. Chapter 4 establishes wave propagation. No prior electromagnetism is assumed.
6.1. What is adopted, and what follows
Definition 6.1. The smooth Maxwell model
At position \(x=(x_1,x_2,x_3)\) and real time \(t\), the electric field \(E(x,t)\) and magnetic field \(B(x,t)\) are real three-component vectors. The charge density \(\rho(x,t)\) is charge per volume. The current density \(J(x,t)\) describes charge crossing an oriented surface: its rate is \(\int J\cdot n\,dA\). The positive constants \(\varepsilon_0,\mu_0\) are vacuum permittivity and permeability. Set \(c=(\mu_0\varepsilon_0)^{-1/2}\). In a fixed Cartesian frame the model adopts
\[\operatorname{div}E=\rho/\varepsilon_0,\quad \operatorname{div}B=0,\quad \operatorname{curl}E=-B_t,\quad \operatorname{curl}B=\mu_0J+c^{-2}E_t.\]A subscript \(t\) means differentiation in time. These are physical postulates, not theorems of real analysis. We assume joint \(C^2\) regularity whenever differentiating them twice: all derivatives through order two exist and are continuous on the open space-time region under discussion. A source-free vacuum region means \(\rho=0,J=0\) there; it need not mean the fields vanish.
Alphabeta definitions: laboratory geometry, SI conventions, sources, fields, and vacuum Maxwell postulate.
Point charges are not smooth densities. Chapter 5 separately gives their punctured-domain and integral interpretation. Also, a charged particle's force law \(F=q(E+v\times B)\) is a further physical law: Maxwell's equations contain no particle acceleration or mass and therefore cannot determine a particle trajectory. See the Lorentz coupling postulate. Media require further laws relating fields to material response; see material response.
Proposition 6.2. Charge conservation is a compatibility condition
Every smooth Maxwell solution satisfies \(\rho_t+\operatorname{div}J=0\). For a fixed bounded region \(V\) with the regularity required by the divergence theorem,
\[\frac d{dt}\int_V\rho\,dV=-\int_{\partial V}J\cdot n\,dA.\]Alphabeta: charge continuity.
Proof.
Take divergence of the last Maxwell equation. Chapter 2 proves \(\operatorname{div}\operatorname{curl}B=0\) by cancellation of mixed partials. Thus \(0=\mu_0\operatorname{div}J+c^{-2}\partial_t\operatorname{div}E=\mu_0(\operatorname{div}J+\rho_t)\). Divide by \(\mu_0>0\). Integrate over \(V\); differentiation under this bounded integral is justified by continuity of \(\rho_t\) on the compact integration region, as in chapter 3. Apply the divergence theorem.
□Incompatible data. Prescribing \(J=0\) and \(\rho=t\) on an open region violates this condition. There can be no smooth fields satisfying all four equations for those prescribed sources.
Proposition 6.3. Divergence constraints propagate
Suppose the two curl equations hold and \(\rho_t+\operatorname{div}J=0\). If the divergence equations hold initially, they hold at every later time along each time interval in the region.
Alphabeta: Gauss constraint propagation.
Proof.
The curl equations give \(\partial_t\operatorname{div}B=-\operatorname{div}\operatorname{curl}E=0\) and \(\partial_t(\operatorname{div}E-\rho/\varepsilon_0)=c^2\operatorname{div}\operatorname{curl}B-c^2\mu_0\operatorname{div}J-\rho_t/\varepsilon_0=0\), since \(c^2\mu_0=1/\varepsilon_0\). A function with zero time derivative is constant by the real mean value theorem. Initial zero remains zero.
□Proposition 6.4. Superposition includes the sources
If \((E_1,B_1,\rho_1,J_1)\) and \((E_2,B_2,\rho_2,J_2)\) are solutions with the same constants, their componentwise linear combination with real coefficients \(a,b\) is a solution with sources \(a\rho_1+b\rho_2,aJ_1+bJ_2\).
Alphabeta: linear superposition.
Worked example for Proposition 6.4
Counterexample: Superposition of fields does not mean addition of their energies
Proof.
Every derivative, divergence and curl is linear. For example \(\operatorname{curl}(aB_1+bB_2)=\mu_0(aJ_1+bJ_2)+c^{-2}\partial_t(aE_1+bE_2)\). The other three equations follow by the same substitution. No product of unknown fields occurs.
□If one solution for fixed sources exists, every other is obtained by adding a source-free solution: subtraction cancels the fixed sources, and addition preserves them. Simply doubling a field with nonzero sources doubles its sources.
6.2. Wave equations and energy
Proposition 6.5. Maxwell implies wave equations
Write \(\Delta=\partial_{x_1}^2+\partial_{x_2}^2+\partial_{x_3}^2\), acting on each vector component. Then
\[\Delta E-c^{-2}E_{tt}=\varepsilon_0^{-1}\nabla\rho+\mu_0J_t,\qquad \Delta B-c^{-2}B_{tt}=-\mu_0\operatorname{curl}J.\]In a source-free region each component satisfies \(u_{tt}=c^2\Delta u\).
Alphabeta: vacuum wave equations.
Proof.
Using the chapter 2 identity \(\operatorname{curl}\operatorname{curl}E=\nabla\operatorname{div}E-\Delta E\), curl Faraday's equation to get \(\nabla\rho/\varepsilon_0-\Delta E=-\partial_t\operatorname{curl}B=-\mu_0J_t-c^{-2}E_{tt}\). Rearrange. Curl the other equation: \(-\Delta B=\mu_0\operatorname{curl}J+c^{-2}\partial_t\operatorname{curl}E=\mu_0\operatorname{curl}J-c^{-2}B_{tt}\). Rearrange again. Commutation of the spatial and time derivatives uses the stated \(C^2\) hypothesis.
□The converse is false. For \(k>0\), let \(E=(0,0,\cos(kz-ckt))\), \(B=0\). Both vectors satisfy the vacuum wave equations componentwise. But \(\operatorname{div}E=-k\sin(kz-ckt)\) is not zero, and the second curl equation fails. Even imposing both divergence constraints is insufficient: \(E=(\cos(kz-ckt),0,0)\), \(B=0\) satisfies both wave equations and both divergence equations, but Faraday fails because \(\operatorname{curl}E=(0,-k\sin(kz-ckt),0)\). The first-order coupling must still be checked.
Proposition 6.6. Local Poynting identity
Define the nonnegative quadratic density and flux
\[u=\frac{\varepsilon_0}{2}|E|^2+\frac1{2\mu_0}|B|^2,\qquad S=\frac1{\mu_0}E\times B.\]Maxwell implies \(u_t+\operatorname{div}S=-J\cdot E\). Integrating on a fixed regular bounded region gives
\[\frac d{dt}\int_Vu\,dV=-\int_{\partial V}S\cdot n\,dA-\int_VJ\cdot E\,dV.\]The mathematical identity follows from Maxwell; interpreting \(u\) as electromagnetic energy, \(S\) as energy transport, and \(J\cdot E\) as power transferred to matter uses the energy and coupling model.
Alphabeta: vacuum energy and flux, Poynting balance.
Worked example for Proposition 6.6
Counterexample: Superposition of fields does not mean addition of their energies
Proof.
The product rule gives
\[u_t=\varepsilon_0E\cdot E_t+\mu_0^{-1}B\cdot B_t=\mu_0^{-1}\bigl(E\cdot\operatorname{curl}B-B\cdot\operatorname{curl}E\bigr)-J\cdot E.\]The required vector identity is \(\operatorname{div}(E\times B)=B\cdot\operatorname{curl}E-E\cdot\operatorname{curl}B\). To verify it, expand the three derivatives of \((E_2B_3-E_3B_2,E_3B_1-E_1B_3,E_1B_2-E_2B_1)\). The six terms differentiating \(E\) are precisely \(B_1(\partial_2E_3-\partial_3E_2)+B_2(\partial_3E_1-\partial_1E_3)+B_3(\partial_1E_2-\partial_2E_1)\). The six terms differentiating \(B\) are the negative of the analogous expression with \(E,B\) exchanged. Substitute the identity, then integrate and apply the divergence theorem.
□Flux bound. \(|E\times B|\le|E||B|\) by the cross-product norm identity in chapter 2. Put \(a=\sqrt{\varepsilon_0}|E|\) and \(b=|B|/\sqrt{\mu_0}\). Since \(2ab\le a^2+b^2\), proved by \((a-b)^2\ge0\), we have \(|S|\le cu\).
Proposition 6.7. Local domain of dependence
Two source-free \(C^2\) Maxwell solutions with the same \(E,B\) on the ball \(|x-x_0|\le R\) at time zero agree in \(|x-x_0|<R-ct\), for \(0\le t<R/c\), provided both solutions exist on that cone. This is a statement about finite propagation of changes, not a particle model.
Alphabeta: compatible Maxwell initial data, wave domain of dependence.
Proof.
Subtract the two solutions. By superposition the difference is source-free. The initial fields vanish throughout the ball. Their spatial curls therefore vanish in its interior, so the curl equations give \(E_t=B_t=0\) initially there. Every difference component satisfies the scalar wave equation of Proposition 6.5, with both initial value and initial time derivative zero. The proved shrinking-ball energy estimate in chapter 4 makes each difference component zero on the smaller ball. That estimate uses the three-dimensional divergence theorem and differentiation of a shrinking spherical integral; it requires no global finite-energy assumption. Hence all six components agree.
□For solutions on all of space, initially supported in \(|x|\le R\), apply this result at any point outside \(|x|\le R+ct\), comparing with the zero solution on a ball around that point disjoint from the initial support. This proves containment of the later support in \(|x|\le R+ct\). The result is conditional on existence of the smooth solution; it is not an existence theorem for arbitrary data. The initial data must satisfy both divergence constraints.
6.3. Complex notation without complex analysis
Definition 6.8. Phases and real fields
For real \(s\), \(e^{is}=\cos s+i\sin s\). The real part of a complex vector is taken componentwise. Complex fields here are a calculational representation; the physical fields are real. We use real spatial and time derivatives, taken componentwise on real and imaginary parts.
Alphabeta: complex plane-wave representation.
The needed construction from complex arithmetic is proved in the complex exponential definition, its convergence and derivative proof, and the trigonometric phase laws. Read these mathematical sections before continuing; none requires quantum physics. They prove the power-series definition \(e^z=\sum_{n\ge0}z^n/n!\), Euler's formula, \((\cos s)'=-\sin s\), \((\sin s)'=\cos s\), and the addition laws. No differentiation in a complex spatial variable or contour integration is used below.
If \(A=p+iq\) with real vectors \(p,q\), then direct multiplication gives \(\operatorname{Re}(Ae^{is})=p\cos s-q\sin s\). A complex constant \(C\) has \(\operatorname{Re}(Ce^{is})=0\) for every real \(s\) only if \(C=0\): evaluate at \(s=0\) and \(s=\pi/2\). Componentwise, this proves that vanishing real oscillations imply vanishing complex coefficients when the phase ranges over a whole cycle. Real constant-coefficient differential operators commute with taking real parts.
6.4. Every monochromatic vacuum plane wave
Definition and theorem 6.9. Nonzero-frequency plane waves
A monochromatic plane-wave ansatz is
\[E(x,t)=\operatorname{Re}(Ae^{i(k\cdot x-\omega t)}),\quad B(x,t)=\operatorname{Re}(De^{i(k\cdot x-\omega t)}),\]with fixed \(A,D\in\mathbb C^3\), real \(k\ne0\), and real \(\omega\ne0\). Monochromatic means one temporal frequency; plane means equal phase on the planes \(k\cdot x=\text{constant}\) at fixed time. Assume the field is nonzero. This ansatz solves all four vacuum Maxwell equations if and only if
\[k\cdot A=0,\qquad \omega^2=c^2|k|^2,\qquad D=\omega^{-1}k\times A.\]For \(\omega>0\), set \(n=k/|k|\); then \(D=c^{-1}n\times A\). Consequently the real fields obey \(B=c^{-1}n\times E\), \(n\cdot E=n\cdot B=E\cdot B=0\), and \(|B|=|E|/c\).
Alphabeta: real plane-wave definition, dispersion and transversality.
Worked example for Definition and theorem 6.9
Counterexample: A vacuum solution need not have perpendicular electric and magnetic fields
Proof.
Differentiate the ansatz, using \(\partial_j e^{i(k\cdot x-\omega t)}=ik_je^{i(k\cdot x-\omega t)}\) and \(\partial_t e^{i(k\cdot x-\omega t)}=-i\omega e^{i(k\cdot x-\omega t)}\). The four equations reduce, by the coefficient test above, to
\[k\cdot A=0,\quad k\cdot D=0,\quad k\times A=\omega D,\quad k\times D=-\omega c^{-2}A.\]Since \(\omega\ne0\), the third fixes \(D\). The chapter 2 triple-product identity gives \(k\times(k\times A)=k(k\cdot A)-|k|^2A=-|k|^2A\). Substitute into the fourth: \((\omega^2c^{-2}-|k|^2)A=0\). Nonzero fields imply \(A\ne0\), because \(A=0\) would force \(D=0\). Thus dispersion follows. Conversely the stated three conditions imply \(k\cdot D=0\) and both cross-product equations, using the same identity, so they imply every Maxwell equation.
For positive \(\omega=c|k|\), taking real parts in \(D=c^{-1}n\times A\) gives \(B=c^{-1}n\times E\), because \(n\) is real. For real vectors, \(n\cdot(n\times E)=E\cdot(n\times E)=0\). The norm identity \(|n\times E|^2=|n|^2|E|^2-(n\cdot E)^2=|E|^2\) proves the magnitude relation. These statements remain valid where both fields are zero; a zero vector has no direction.
□A sign check and curl calculation. Let \(k=k\hat z\), \(k>0\), \(\omega=ck\), and \(A=E_0\hat x\) with real \(E_0\). Then \(E=E_0\cos(kz-\omega t)\hat x\), \(B=(E_0/c)\cos(kz-\omega t)\hat y\), and \(\operatorname{curl}E=-kE_0\sin(kz-\omega t)\hat y=-B_t\). The curl peaks in magnitude at spatial zeros of \(E\), and vanishes at its extrema, by the sine and cosine phase laws. It is perpendicular to \(E\) in this example; a general curl need not be.
Three necessary qualifications. For \(\omega<0\), the propagation unit vector is \(-k/|k|\), and the formula with the actual direction remains \(B=n_{\mathrm{prop}}\times E/c\). Adding any constant \(E_s,B_s\) gives another source-free solution but can make the total fields longitudinal or nonperpendicular. For example \(E_s=\hat z,B_s=\hat z\) makes \(E\cdot B=1\) in the preceding linear-wave example. Faraday alone permits arbitrary time-independent additions to \(B\); within the pure nonzero-frequency ansatz it fixes the oscillatory amplitude, but outside that ansatz it does not fix the full magnetic field.
For complex amplitudes, \(a\cdot b=\sum_j a_jb_j\) here is the bilinear extension of the real dot product, not the Hermitian pairing \(\sum_j\overline{a_j}b_j\). The identity \(A\cdot(n\times A)=0\) holds algebraically but does not, by itself, prove a statement about real parts. For example \(A=(1,i,0)\), \(D=(-i,1,0)\) have bilinear dot product zero, whereas their Hermitian pairing is \(-2i\). Our real perpendicularity proof takes real parts of the cross-product relation first and then uses real geometry.
6.5. Polarisation, including the degenerate cases
The two-dimensional algebra and geometry used in the proof
A real two-by-two matrix \(M=\begin{pmatrix}u&v\\w&r\end{pmatrix}\) denotes the linear map \((x,y)\mapsto(ux+vy,wx+ry)\). Its transpose \(M^T\) exchanges rows and columns. Matrix multiplication means composition: \((MN)_{ij}=\sum_{\ell=1}^2M_{i\ell}N_{\ell j}\). The identity matrix \(I=\begin{pmatrix}1&0\\0&1\end{pmatrix}\) fixes every vector. An inverse is a matrix \(M^{-1}\) with \(MM^{-1}=M^{-1}M=I\). The determinant is \(\det M=ur-vw\). If it is nonzero, direct multiplication proves
\[M^{-1}=\frac1{ur-vw}\begin{pmatrix}r&-v\\-w&u\end{pmatrix}.\]If the determinant is zero, the columns are dependent: if one is zero this is immediate; otherwise dividing by a nonzero coordinate of the first column and using \(ur-vw=0\) shows the second is its scalar multiple. The image then lies on a line, or is zero, so an inverse cannot exist. For this two-dimensional setting we use rank zero for the zero map, rank one for a map whose image is a line through the origin, and rank two for an invertible map whose image is the whole plane. Thus these three cases exhaust all matrices, by the determinant test just proved.
An eigenvector of \(Q\) for the real eigenvalue \(\lambda\) is a nonzero real vector \(v\) satisfying \(Qv=\lambda v\). Such a vector exists precisely when \(\det(Q-\lambda I)=0\). Indeed a nonzero determinant gives an inverse and hence only the zero solution. For a zero determinant, if a row is nonzero, choose a nonzero perpendicular vector to that row; the other row is its multiple or zero, so the vector lies in the kernel of both rows. If both rows vanish, every vector lies in the kernel. The polynomial \(\det(\lambda I-Q)\) is called the characteristic polynomial; in two dimensions it equals \(\det(Q-\lambda I)\). The notation \(\operatorname{diag}(d_1,d_2)\) means the matrix with diagonal entries \(d_1,d_2\) and both off-diagonal entries zero.
Two unit vectors perpendicular to each other form an orthonormal basis of the plane: put them in the columns of \(U\); then \(U^TU=I\), so they span the plane because \(U\) is invertible. A matrix with \(U^TU=UU^T=I\) is orthogonal; it preserves lengths since \(|Uv|^2=v^TU^TUv=|v|^2\), and maps the unit circle onto itself using its inverse \(U^T\). Expansion of the four product entries gives \(\det(AB)=(A_{11}A_{22}-A_{12}A_{21})(B_{11}B_{22}-B_{12}B_{21})=\det A\det B\); exchanging rows and columns also gives \(\det A^T=\det A\). If \(NN^T=I\), these identities give \((\det N)^2=1\), so its determinant is nonzero, so \(N^T=N^{-1}\) and also \(N^TN=I\).
The unit circle is \(\{(x,y):x^2+y^2=1\}\). A nondegenerate ellipse centred at zero is a set \(\{s_1p\cos\theta+s_2q\sin\theta:\theta\in\mathbb R\}\), with an orthonormal pair \(p,q\) and positive numbers \(s_1,s_2\), its semiaxis lengths. When \(s_1=s_2\), this is a circle of that radius. A segment centred at zero is \(\{tv:-1\le t\le1\}\) for a fixed nonzero vector \(v\). The one-point set \(\{0\}\) is the remaining degenerate case. The proof below establishes that an invertible linear image of the unit circle has the stated ellipse form, rather than assuming that fact.
Alphabeta definitions: matrix product and identity, determinant, and eigenvalues and eigenvectors.
The proved polar form of complex numbers lets us write \(A_x=ae^{i\alpha_x}\), \(A_y=be^{i\alpha_y}\), with \(a,b\ge0\). For a zero component its phase can be chosen arbitrarily. Since \(\omega\ne0\), one full phase cycle takes time \(T=2\pi/|\omega|\), the period used below.
Proposition 6.10. The tip is an ellipse, a segment, or a point
For propagation along \(\hat z\), write the two transverse components as
\[E_x=a\cos\theta,\qquad E_y=b\cos(\theta+\delta),\quad \theta=kz-\omega t+\alpha_x,\quad \delta=\alpha_y-\alpha_x,\]with \(a,b\ge0\). At a fixed spatial point the locus of \((E_x,E_y)\) over one period is a nondegenerate ellipse exactly when \(ab\sin\delta\ne0\). It is a segment when the amplitude map has rank one, and a single point when \(a=b=0\). In particular:
- If \(a,b>0\), linear polarisation occurs exactly at \(\delta=0\) or \(\pi\) modulo \(2\pi\).
- If exactly one amplitude is zero, polarisation is linear regardless of \(\delta\); the phase of a zero component has no meaning.
- Nonzero circular polarisation occurs exactly when \(a=b>0\) and \(\delta=\pm\pi/2\) modulo \(2\pi\).
In the nondegenerate case the squared semiaxes are
\[\lambda_\pm=\frac{a^2+b^2\pm\sqrt{(a^2-b^2)^2+4a^2b^2\cos^2\delta}}2.\]The numbers \(a,b\) are the axis semilengths along \(x,y\) in the quadrature case \(\cos\delta=0\). In general the ellipse is rotated.
Alphabeta: polarisation definition.
Proof: Proposition 6.10. The tip is an ellipse, a segment, or a point
Proof.
The cosine addition formula writes the tip as
\[\binom{E_x}{E_y}=M\binom{\cos\theta}{\sin\theta},\qquad M=\begin{pmatrix}a&0\\b\cos\delta&-b\sin\delta\end{pmatrix}.\]The phase laws ensure \((\cos\theta,\sin\theta)\) traverses the unit circle. If \(\det M=-ab\sin\delta\ne0\), this invertible linear image is an ellipse. Explicitly, division by the nonzero \(a,b\) and elimination of \(\sin\theta\) give
\[\left(\frac{E_x}{a}\right)^2+\left(\frac{E_y}{b}\right)^2-2\frac{E_xE_y}{ab}\cos\delta=\sin^2\delta.\]If \(M=0\) its image is the origin. If \(M\ne0\) has rank one, its columns lie on a common line, so \(M(\cos\theta,\sin\theta)\) lies on that line. Write it as \(v(r\cos\theta+s\sin\theta)\) for a fixed nonzero \(v\); the scalar factor ranges through \([-\sqrt{r^2+s^2},\sqrt{r^2+s^2}]\), by the addition law after normalizing \((r,s)\) to a point of the unit circle. Thus the locus is the entire segment.
For axes, compute
\[Q=MM^T=\begin{pmatrix}a^2&ab\cos\delta\\ab\cos\delta&b^2\end{pmatrix}.\]Its characteristic polynomial is \(\lambda^2-(a^2+b^2)\lambda+a^2b^2\sin^2\delta\); the quadratic formula gives \(\lambda_\pm\). For completeness, a real symmetric two-by-two matrix has an orthonormal eigenbasis: if the eigenvalues differ, solve \((Q-\lambda I)v=0\) and normalize nonzero solutions; symmetry gives \((\lambda_+-\lambda_-)v_+\cdot v_-=0\). If they coincide, the displayed discriminant is zero, forcing the diagonal entries equal and off-diagonal zero, so every orthonormal basis works. Positivity follows from \(v^TQv=|M^Tv|^2>0\) for nonzero \(v\) when \(M\) is invertible. With the orthogonal eigenbasis matrix \(U\), put \(N=\operatorname{diag}(\lambda_+^{-1/2},\lambda_-^{-1/2})U^TM\). Then \(NN^T=I\), hence \(N\) is orthogonal and maps the unit circle onto itself. The tip in coordinates \(U\) therefore has semiaxes \(\sqrt{\lambda_+},\sqrt{\lambda_-}\). They coincide precisely when \((a^2-b^2)^2+4a^2b^2\cos^2\delta=0\). In the nonzero case this means \(a=b>0\), \(\cos\delta=0\), proving the circle classification.
□The article's example. For \(A=a\hat x+ib\hat y\), the real field is \(a\cos(kz-\omega t)\hat x-b\sin(kz-\omega t)\hat y\). The relative phase is \(\pi/2\), so its fixed-position tip follows an axis-aligned ellipse of semiaxes \(a,b\). For \(a=1,b=1.5\) this ellipse is noncircular. The arrow remains based at its spatial observation point; its tip moves in that point's transverse plane. At fixed time, joining the tips from different \(z\)-positions gives the parametrized curve \((a\cos(kz-\omega t),-b\sin(kz-\omega t),z)\), an elliptical helix. That curve is a display of field values, not a particle trajectory.
A rotated ellipse. If \(a=b=1\) and \(\delta=\pi/3\), the squared semiaxes are \(3/2\) and \(1/2\), not \(1,1\). Every amplitude and phase choice above gives a Maxwell solution by Theorem 6.9, with the corresponding magnetic amplitude. This proves admissibility in the source-free model; it makes no claim that all material sources or boundary conditions produce every polarisation equally.
6.6. Phase speed, rigid profiles, and packets
Proposition 6.11. What travels at \(c\)
For \(k>0,\omega=ck\), a surface of fixed phase \(kz-\omega t=\beta\) moves with \(dz/dt=\omega/k=c\). More generally, for any real transverse \(C^2\) profile \(f:\mathbb R\to\mathbb R^3\) with \(\hat z\cdot f=0\),
\[E(z,t)=f(z-ct),\qquad B(z,t)=\frac1c\hat z\times f(z-ct)\]is a source-free Maxwell solution independent of \(x,y\), and \(E(z+c\tau,t+\tau)=E(z,t)\), with the same equality for \(B\).
Proof.
Solving the phase equation gives \(z=(\beta+\omega t)/k\), proving its velocity. For the profile, \(E_t=-cf'\), \(\operatorname{curl}E=\hat z\times f'\), \(B_t=-\hat z\times f'\), and \(\operatorname{curl}B=c^{-1}\hat z\times(\hat z\times f')=-f'/c=c^{-2}E_t\). The last identity uses transversality and the triple-product identity. Both divergences vanish because their longitudinal components are zero. Translation leaves \((z+c\tau)-c(t+\tau)=z-ct\) unchanged.
□A packet here means a nonperiodic localized profile, rather than a single sinusoid. Its envelope or support translates at \(c\) in this one-direction vacuum construction. An elementary example of an envelope is \(f(s)=g(s)\cos(k_0s)\hat x\), where \(g\) is a smooth compactly supported real function: both oscillations and envelope depend on \(z-ct\). For a superposition of right-moving frequencies, every term depends on that same coordinate, so their sum translates. If an integral \(\int A(k)e^{ik(z-ct)}\,dk\) is used, sufficient hypotheses are compact \(k\)-support and continuous amplitudes; derivatives then pass under the integral by the compact-domain argument in chapter 3.
The group velocity in a differentiable dispersion relation is defined as \(d\omega/dk\). On the right-moving vacuum branch \(\omega=ck\), it equals \(c\). A two-frequency check gives \(\cos(k_1s)+\cos(k_2s)=2\cos((k_1-k_2)s/2)\cos((k_1+k_2)s/2)\), by the cosine addition laws, displaying the translating beat envelope. General dispersive media require their own relation \(\omega(k)\); these vacuum equalities must not be silently transferred to them.
An ideal nonzero monochromatic plane wave has infinite total energy on all of \(\mathbb R^3\): its density is independent of transverse coordinates. At any fixed time, moving along the nonzero wave vector makes the phase range over a whole cycle. The coefficient test in section 6.3 shows that a nonzero amplitude cannot give a real field identically zero along that line. The density is therefore positive at some point, and continuity bounds it below by some positive number on a smaller longitudinal interval. Integrating over that interval and increasingly large transverse squares gives unbounded energy. A profile compactly supported in \(z\) but uniform in \(x,y\) still has infinite total energy unless zero. The local finite-propagation theorem applies regardless. Finite-energy three-dimensional disturbances require spatial localisation in all directions and compatible Maxwell data.
For a pure right-moving plane wave, Proposition 6.6 and \(|B|=|E|/c\) give \(u=\varepsilon_0|E|^2\) and \(S=cu\,n\). Indeed \(E\times(n\times E)=n|E|^2\), and \(1/(\mu_0c)=\varepsilon_0c\). Scaling \(E,B\) by any real number scales \(u\) by its square. Maxwell therefore permits continuous classical energy values. Discrete photon energies are additional quantum physics; see chapter 7.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. A smooth sourced solution and propagated constraints
Applies to: Definition 6.1, Proposition 6.2, Proposition 6.3, Proposition 6.4.
Let \(a(t)\) be a real smooth function and set \(E=a(t)(x,y,z),B=0\), \(\rho=3\varepsilon_0a(t)\), \(J=-\varepsilon_0a'(t)(x,y,z)\). Then \(\operatorname{div}E=3a=\rho/\varepsilon_0\), \(\operatorname{curl}E=0=-B_t\), and \(\mu_0J+c^{-2}E_t=0=\operatorname{curl}B\) since \(c^{-2}=\mu_0\varepsilon_0\); \(\operatorname{div}B=0\) also holds. Thus all four Maxwell postulates hold with the declared sources. Continuity is \(\rho_t+\operatorname{div}J=3\varepsilon_0a'-3\varepsilon_0a'=0\).
On a fixed radius-\(R\) ball, charge is \(4\pi\varepsilon_0R^3a(t)\) and outward current flux is \(-4\pi\varepsilon_0R^3a'(t)\), verifying the integral conservation law. The Gauss residual \(\operatorname{div}E-\rho/\varepsilon_0\) stays zero. If instead one initially assigns \(\widetilde\rho=\rho+\rho_*\) for a nonzero constant \(\rho_*\), the curl equations and continuity still hold, but the residual stays \(-\rho_*/\varepsilon_0\), illustrating why initial satisfaction is required. Adding a second such solution with coefficient function \(b(t)\) gives the solution for \(a+b\), and scaling scales both sources, explicitly checking superposition.
Worked example 2. The wave equation and work term with nonzero current
Applies to: Proposition 6.5, Proposition 6.6.
Continue the preceding sourced solution. Its electric wave-equation left side is \(-c^{-2}a''(t)x\), where \(x=(x_1,x_2,x_3)\); the right side is \(\varepsilon_0^{-1}\nabla\rho+\mu_0J_t=0-\mu_0\varepsilon_0a''x\), the same expression. Both sides of the magnetic wave equation vanish since \(B=0\) and \(\operatorname{curl}J=0\). The energy and flux are \(u=\varepsilon_0a(t)^2|x|^2/2\), \(S=0\). Hence \(u_t=\varepsilon_0aa'|x|^2=-J\cdot E\), checking the local Poynting sign.
The ball integral is \(\int u=2\pi\varepsilon_0a^2R^5/5\); its derivative \(4\pi\varepsilon_0aa'R^5/5\) equals \(-\int J\cdot E\), with no boundary flux. Scaling the solution by a real \(\lambda\) scales this energy by \(\lambda^2\), checking the unboxed continuous classical energy-scaling claim.
Worked example 3. A travelling pulse and a silent cone
Applies to: Proposition 6.7, Proposition 6.11.
Let \(f(s)=\eta(s)e_x\), where \(\eta\) is a nonzero smooth bump supported in \((3,4)\). Define \(E=f(z-ct),B=c^{-1}\eta(z-ct)e_y\), independent of \(x,y\), with zero sources. Both divergences vanish, \(\operatorname{curl}E=\eta'(z-ct)e_y=-B_t\), and \(\operatorname{curl}B=-c^{-1}\eta'(z-ct)e_x=c^{-2}E_t\). Thus all four equations hold. Replacing \((z,t)\) by \((z+c\tau,t+\tau)\) leaves both fields unchanged, verifying rigid translation.
Compare this solution with zero. Their initial data agree on \(|x|\le1\), since there \(|z|\le1\). For \(0\le t<1/c\), \(|x|<1-ct\) gives \(z-ct<1\), so the pulse remains zero in the whole shrinking cone, exactly the local domain-of-dependence conclusion. For a phase \(kz-ckt=\beta\) with \(k>0\), solving gives \(z=ct+\beta/k\), independently checking the phase-velocity clause.
Worked example 4. A complex amplitude checked against all four equations
Applies to: Definition 6.8, Definition and theorem 6.9.
Choose \(k_0>0\), \(\omega=ck_0\), wave vector \(k=k_0e_z\), and \(A=2e_x+ie_y\). Then \(D=(k\times A)/\omega=c^{-1}(-ie_x+2e_y)\). With \(\theta=k_0z-\omega t\), the real fields are \(E=(2\cos\theta,-\sin\theta,0)\) and \(B=c^{-1}(\sin\theta,2\cos\theta,0)\). Their divergences vanish. Direct derivatives give \(\operatorname{curl}E=(k_0\cos\theta,-2k_0\sin\theta,0)=-B_t\), and \(\operatorname{curl}B=(2k_0\sin\theta,k_0\cos\theta,0)/c=c^{-2}E_t\). Thus every Maxwell equation is verified, with nonzero frequency and field and a transverse amplitude. Moreover \(B=e_z\times E/c\), \(E\cdot B=(2\cos\theta\sin\theta-2\sin\theta\cos\theta)/c=0\), and \(|B|=|E|/c\). This also illustrates the componentwise real-part convention rather than confusing a complex amplitude with a physical field.
Worked example 5. A rotated ellipse and its degenerate limits
Applies to: Proposition 6.10.
Take \(a=2,b=1,\delta=\pi/3\). Then \(E_x=2\cos\theta\), \(E_y=\tfrac12\cos\theta-\tfrac{\sqrt3}{2}\sin\theta\). The amplitude matrix from \((\cos\theta,\sin\theta)\) is \(M=\begin{pmatrix}2&0\\1/2&-\sqrt3/2\end{pmatrix}\), determinant \(-\sqrt3\ne0\); thus the tip traces a nondegenerate ellipse. Its squared semiaxes are the eigenvalues of \(MM^T=\begin{pmatrix}4&1\\1&1\end{pmatrix}\), namely \((5\pm\sqrt{13})/2\). The nonzero off-diagonal entry shows the principal axes are rotated, so component amplitudes 2 and 1 are not its semiaxes.
At \(\delta=0\) with the same amplitudes, the locus is \((2,1)\cos\theta\), a segment. With \(a=b=1,\delta=\pi/2\), it is \((\cos\theta,-\sin\theta)\), a unit circle; changing the phase sign reverses the traversal. With \(b=0,a>0\), it is a segment whatever \(\delta\), and with \(a=b=0\), it is the single origin. These calculations cover the rank-two, rank-one, rank-zero, and circular cases in the proposition.
Counterexamples and limits of the hypotheses
Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.
Counterexample 1. Superposition of fields does not mean addition of their energies
Related statements: Proposition 6.4, Proposition 6.6.
Claim refuted. The electromagnetic energy of a sum of solutions always equals the sum of their separate energies.
Proof.
Expanding the quadratic definition at each point gives
\[u(E_1+E_2,B_1+B_2)=u_1+u_2+\varepsilon_0E_1\cdot E_2+\mu_0^{-1}B_1\cdot B_2.\]The cross terms are not generally zero. Take two identical source-free plane-wave solutions \(E_1=E_2=a\cos(kz-ckt)e_x\), \(B_1=B_2=(a/c)\cos(kz-ckt)e_y\), with \(a\ne0,k>0\). Directly the curl equations require \(\omega=ck\) and both divergences vanish, so they are valid solutions with the same constants. Their sum has twice the amplitudes and energy density \(4u_1\), whereas the sum of the separate densities is \(2u_1\). Taking the second solution to be its negative instead gives zero total field and zero energy, although \(u_1+u_2=2u_1>0\) wherever the cosine is nonzero. This is classical coherent interference in a quadratic energy, consistent with linearity of Maxwell’s equations.
Counterexample 2. A vacuum solution need not have perpendicular electric and magnetic fields
Related statements: Definition and theorem 6.9.
Claim refuted. E·B=0 for every source-free Maxwell solution.
Proof.
Choose \(a,b,k\ne0\), \(k>0\), \(\theta=kz-ckt\), and set \(E=a\cos\theta\,e_x\), \(B=(a/c)\cos\theta\,e_y+b e_x\). Both divergences vanish because the components depend only on \(z,t\) and have zero \(z\)-component. Their curls are \(\operatorname{curl}E=-ak\sin\theta\,e_y=-B_t\) and \(\operatorname{curl}B=(ak/c)\sin\theta\,e_x=c^{-2}E_t\); the added constant field has zero derivatives. Thus all four source-free equations hold smoothly. Yet \(E\cdot B=ab\cos\theta\), nonzero at phases with \(\cos\theta\ne0\). The field is a monochromatic wave plus a static magnetic background, not the single nonzero-frequency ansatz of Definition and theorem 6.9. The perpendicularity conclusion requires that ansatz.
Comments