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For the wave equation with d'Alembert operator $\square =-{\partial }_t^2+\Delta$, the wave speed is one. The domain of dependence of $(t,x)$ is the initial ball $\overline{B}(x,t)$. More precisely, two solutions whose Cauchy data agree on a neighbourhood of that ball have the same value at $(t,x)$. Thus, if both Cauchy data have support in $K$, finite propagation speed givesThe Strong Huygens principle is sharper in three spatial dimensions: only the initial data on the sphere $|y-x|=t$ contribute, including the first spatial derivatives of the initial displacement there. The Kirchhoff formula makes this precise:In particular, if the Cauchy data have compact support in $\overline{B}(0,R)$, thenFor $t>R$, the solution therefore vanishes behind the inward edge $|x|=t-R$ as well as outside $|x|=t+R$. The absence of an interior tail is the additional content of the Strong Huygens principle.

We prove finite propagation speed using a shrinking cone energy argument. Fix $T>0$ and $x_{\ast }$, and suppose the Cauchy data vanish on $B(x_{\ast },T)$. For $0\le s<T$, define the local wave energyThe homogeneous wave equation gives the local conservation lawDifferentiate the integral over the moving ball. Its boundary moves inward with speed one, so the divergence theorem givesHere $n$ is the outward unit normal, ${\partial }_n$ the normal derivative, and ${\nabla }_{\tan }$ the component of the gradient tangent to the boundary. Since $E(0)=0$ and $e\ge 0$, we have $E(s)=0$. Hence both ${\varphi }_t$ and $\nabla \varphi$ vanish inside the backward light cone. Integrating ${\varphi }_t(s,x_{\ast })=0$ from the zero initial displacement gives $\varphi (s,x_{\ast })=0$; continuity then gives $\varphi (T,x_{\ast })=0$.

Applying the same energy estimate to the difference of two solutions proves the domain of dependence assertion. If $\mathrm{dist}(x_{\ast },K)>T$, the initial ball misses $K$, and the preceding argument proves the stated support bound. No disturbance propagates faster than one.

For a radial function, the Laplacian is $\Delta \varphi ={\varphi }_{rr}+2{\varphi }_r/r$. Introduce the retarded and advanced null coordinates and the radial reduction of the three-dimensional wave equation:The chain rule gives ${\partial }_t={\partial }_u+{\partial }_v$ and ${\partial }_r=-{\partial }_u+{\partial }_v$. Consequently the inhomogeneous wave equation isMultiplying the radial wave equation by $r$ removes its first-order radial derivative:Thus the particularly useful retarded and advanced null coordinates form isThese formulas hold for $r>0$. At the axis, smooth spherical symmetry imposes ${\varphi }_r(t,0)=0$ and $w(t,0)=0$; equivalently $w$ has a smooth odd extension across $r=0$.

Use the radial reduction of the three-dimensional wave equation $w=r\varphi$ and extend $w$ oddly in $r$. Write $w=w_h+w_F$, where $w_h$ is the homogeneous wave equation solution with the prescribed Cauchy data, and $w_F$ has zero Cauchy data. With $w_0(r)=r{\varphi }_0(r)$ and $w_1(r)=r{\varphi }_1(r)$, the D'Alembert formula givesBoth odd initial profiles have compact support, soThe Duhamel principle, applied to $w_{tt}-w_{rr}=-rF$, givesNow write $r=t+a$ with $1/2\le a\le 1$. The lower integration limit is $s+a>0$, so this domain of dependence never crosses the axis. Intersecting its integration interval with the support of the source leaves a subset of $[s+a,s+1]$, of length at most one. On this interval,The outgoing shell source estimate for a radial wave therefore yieldsFinally $r\ge t+1/2\ge (1+t)/2$, and thusThis constant depends only on the Cauchy data and the fixed unit bound for the source. The logarithm comes from integrating the shell's accumulated forcing, whereas the factor $(1+t{)}^{-1}$ comes from division by $r$.

One version of the Klainerman-Sobolev inequality, for a sufficiently decaying smooth function $\psi$ on ${\mathbb{R}}^{1+3}$, isHere $I$ denotes a word of length $|I|$ in the following eleven commutation vector fields for the wave equation:The first four are spacetime translation vector fields; the next three are spatial rotation vector fields; the next three are Lorentz boost vector fields; and $S$ is the scaling vector field. The vector field method for wave equations uses these vector fields because their commutators with the d'Alembert operator obeyIn particular the Klainerman-Sobolev inequality impliesThe displayed L2 norms are spatial norms at fixed time; the spacetime vector fields can contain time derivatives. No wave equation assumption is needed for the Klainerman-Sobolev inequality itself.

Use signature $(+---)$ and take the covariant spatial components $A_i$ as canonical coordinates. The canonical quantization of the electromagnetic field begins with the canonical momentaThus the primary momentum constraint of the electromagnetic potential is ${\Pi }^0=0$: $A_0$ has no independent velocity. The Hamiltonian obtained by the Legendre transform in mechanics, up to a boundary term, isPreserving the primary constraint requires the Gauss law constraint in gauge theory, ${\partial }_i{\Pi }^i=0$. It also follows by varying $A_0$. These are two first-class constraints; they generate the gauge freedom and remove two canonical pairs from the four potential components. The reduced phase space has four dimensions per spatial mode, hence two propagating photon degrees of freedom.

Impose Coulomb gauge, ${\partial }_i{A}_i=0$. With no charges, Gauss's law then gives ${\nabla }^2{A}_0=0$; vanishing boundary conditions set $A_0=0$. This is radiation gauge. The remaining components are transverse and obey the massless wave equation. Let ${ϵ}_i^{(r)}(\mathbf{k})$, $r=1,2$, be orthonormal transverse polarization vectors. Their photon polarization completeness relation isThe canonical transverse photon field is the Hermitian operatorwhereThe field and its conjugate momentum have the transverse equal-time commutatorThis is the quantized reduced bracket, or equivalently the Dirac bracket after imposing the constraints and gauge conditions. The normal-ordered Hamiltonian is ${\sum }_r\int d^{3}k{\omega }_{\mathbf{k}}{a}_r^{†}{a}_r/(2\pi {)}^3$. Its excitations are photons; circular combinations of the two transverse polarizations have helicity $+1$ and $-1$. The scalar and longitudinal potential components do not create additional physical photons.

The Feynman propagator is the vacuum expectation of a time-ordered product. The mode expansion directly gives the radiation-gauge photon propagator, with $z=x-y$:The first expression comes from the creation-annihilation commutator; the second is its contour-integral representation. The positive-energy pole lies below the real axis and the negative-energy pole above it. In this reduced free-field description, the temporal operator is zero. A photon propagator must specify its gauge; the spatial transverse propagator is not the same tensor as the covariant four-potential propagator.

For the commonly used covariant form, add the gauge fixing term $-({\partial }_{\mu }{A}^{\mu }{)}^2/(2\xi )$. The resulting Fourier-space kinetic operator isThe inversion of the gauge-fixed Maxwell kinetic operator gives $K^{\mu \nu }{D}_{\nu \rho }^F=i{\delta }^{\mu }{}_{\rho }$. In Feynman gauge, $\xi =1$, the photon propagator isEquivalently, $\square D_{\mu \nu }^F=i{\eta }_{\mu \nu }{\delta }^4(x-y)$ with the vacuum pole prescription. For general $\xi$, the momentum-space numerator is ${\eta }_{\mu \nu }-(1-\xi )k_{\mu }{k}_{\nu }/(k^2+i0)$.

A covariant canonical realization uses four polarization oscillators with $[a_r,a_s^{†}]=-(2\pi {)}^3{\eta }_{rs}{\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })$. The resulting indefinite inner product is auxiliary. In Gupta-Bleuler quantization, impose $({\partial }_{\mu }{A}^{\mu }{)}^{(+)}|\mathrm{phys}⟩=0$ and take the Gupta-Bleuler null-state quotient. The scalar-longitudinal combination is thereby removed from the physical state space, leaving the same two transverse photon states. Thus the four-component Feynman-gauge numerator does not imply four physical polarization states.

Contract the printed Riemann curvature tensor using $R_{ab}=R^{\mu }{}_{a\mu b}$. Only the derivative terms can contain second derivatives of the metric tensor:Substituting the Christoffel symbol formula, its second derivative part isSet $H_b=g_{b\lambda }{H}^{\lambda }=g^{\mu \nu }{\partial }_{\mu }{g}_{\nu b}-\frac{1}{2}{g}^{\mu \nu }{\partial }_b{g}_{\mu \nu }$. The first, second and fourth terms are the second derivative terms of $\frac{1}{2}({\partial }_a{H}_b+{\partial }_b{H}_a)$. Derivatives of the inverse metric tensor produce only products of first derivatives. Thuswhere $Q$ contains at most first metric derivatives. In harmonic coordinates, $H^{\lambda }$ vanishes throughout the chart, so $H_b$ and its derivatives vanish. The vacuum equations have the wave principal partThese are quasilinear partial differential equations with the principal part of a wave equation, because $g^{\mu \nu }$ has Lorentzian signature. This is the harmonic reduction of the vacuum Einstein equations.

Use the Dirichlet Laplacian eigenvalues and eigenfunctions on $(-1,1)$:An expansion $u={\sum }_j{q}_j(t)e_j$ reduces the wave equation to independent ordinary differential equationsWhen $\alpha <{\lambda }_1$, put ${\omega }_j=\sqrt{{\lambda }_j-\alpha }$; each coefficient isFor finite-energy data $\varphi \in H_0^1(-1,1)$ and $\psi \in L^2(-1,1)$, a uniform spatial bound follows from the energy method, rather than from unjustified absolute summation of the Fourier series. The conserved quantity isThe Poincare inequality gives $\Vert u{\Vert }_2^2\le {\lambda }_1^{-1}\Vert u_x{\Vert }_2^2$. Hence $2E\ge c_{\alpha }\Vert u_x{\Vert }_2^2$ with $c_{\alpha }=1-\alpha /{\lambda }_1>0$ for $0\le \alpha <{\lambda }_1$, and with $c_{\alpha }=1$ for $\alpha <0$. Since the Dirichlet boundary condition gives $u(x,t)={\int }_{-1}^x{u}_x(s,t)ds$, the Cauchy-Schwarz inequality yieldsAt $\alpha ={\lambda }_1$, the first coefficient is $q_1(t)={\varphi }_1+t{\psi }_1$, which is unbounded for some admissible data. For $\alpha >{\lambda }_1$, that coefficient has an exponentially growing component for some admissible data. Thus all-time boundedness of a wave equation with a reaction term requiresAt equality, boundedness for a particular data set requires ${\psi }_1=0$; the remaining modes are bounded by their spectral gap. Above the threshold, every unstable mode must have its growing component canceled, and any zero-frequency mode must have zero initial velocity. There is no condition on $\alpha$ alone for arbitrary specially chosen data.

The printed reference to a limit needs qualification. Bounded oscillations generally have no limit as $t\to \infty$. If actual existence of that limit for every admissible initial datum is required, no real $\alpha$ works: for every $\alpha$, choose $j$ with ${\lambda }_j>\alpha$ and nonzero pure oscillatory data in that eigenfunction. The boxed inequality answers the intended long-time boundedness question.

Put $v=e^{t/2}u$. The cancellation of the damping and mass terms gives $v_{tt}=v_{xx}$, with $v(x,0)=u_0(x)$ and $v_t(x,0)=u_1(x)+u_0(x)/2$. For the wave equation, $w_{+}=v_t+v_x$ satisfies $({\partial }_t-{\partial }_x)w_{+}=0$, and $w_{-}=v_t-v_x$ satisfies $({\partial }_t+{\partial }_x)w_{-}=0$. Thus these combinations are transported on straight characteristics. Integrating them and matching the initial displacement yields the D'Alembert formula, and henceHere the initial functions are extended periodically to the real line. Direct differentiation verifies both initial conditions and the equation, and arbitrary repeated differentiation shows the solution is smooth and periodic.

This characteristic derivation also proves finite propagation speed from first principles: the value at $(x,t)$ depends only on the two endpoint displacements and initial velocity data in $[x-t,x+t]$. If two data sets agree on that interval, their solutions agree at that point. Equivalently initially supported data have no influence outside the unit-speed characteristic cone, interpreted on the periodic covering line. Multiplication by $e^{-t/2}$ does not enlarge this domain of dependence.

The Neumann boundary conditions select the spatial eigenfunctions $\cos nx$, $n=0,1,\dots$, of the separated wave equation. Indeed separation gives $X^{\prime \prime }+\lambda X=0$ with $X^{\prime }(0)=X^{\prime }(\pi )=0$. Integration by parts gives $\lambda \int X^2=\int X^{\prime 2}\ge 0$. The zero value yields the constant mode; for positive $\lambda$, the first boundary condition removes the sine term and the second requires $\sqrt{\lambda }=n\in \mathbb{N}$. For $n>0$, the temporal factor is $\cos nct$: the sine factor vanishes because the initial velocity is zero. The zero mode has constant displacement, with no term linear in time for the same reason.

The constant Fourier cosine series coefficient is the mean ${\pi }^{-1}{\int }_0^{\pi }bxdx=b\pi /2$. The remaining coefficients areOnly odd modes survive. Thus the displacement isThe series for displacement converges absolutely and uniformly. The initial velocity is zero in the finite-energy sense, and the initial displacement is $bx$. There is an endpoint compatibility qualification: when $b\ne 0$, the initial slope does not satisfy the Neumann boundary conditions at $t=0$. The answer is therefore a finite-energy weak solution, piecewise classical between the propagating corners, rather than a globally twice continuously differentiable solution up to the initial endpoints. The mode expansion implements reflection at those endpoints.

For the wave equation, introduce characteristic coordinates $\xi =x-ct$, $\eta =x+ct$ when $c\ne 0$. The equation becomes $u_{\xi \eta }=0$, so $u=F(x-ct)+G(x+ct)$. The initial conditions imply $F+G=\varphi$ and $-c{F}^{\prime }+c{G}^{\prime }=0$. Constants can be absorbed into $F,G$, giving the D'Alembert formulaIf $\varphi$ vanishes outside $[-\alpha ,\alpha ]$, then $u(x,t)=0$ whenever $|x|>\alpha +|c|t$. Thus Property P fails: the wave equation has finite propagation speed $|c|$. No arbitrarily short time can transmit the disturbance arbitrarily far. If $c=0$, the zero initial velocity instead gives $u(x,t)=\varphi (x)$, and the conclusion still holds.

Transform in time and use the zero interior initial conditions. The Laplace transform of a derivative formula turns the wave equation into $c^2{Y}_{xx}=p^{2}Y$. Its transformed endpoint conditions are $Y(0,p)=a/p$ and $Y(L,p)=0$. A homogeneous solution satisfying the right condition is a multiple of $\sinh [p(L-x)/c]$; the left condition fixes that multiple. Thus, for $\mathrm{Re}p>0$,Here $L,c>0$ and the denominator has no zeros in this half-plane. The initial-boundary corner at $(0,0)$ is deliberately discontinuous: the applied step launches propagating fronts, so one should not demand a globally smooth solution across them.

Use the geometrical optics conventionBecause ${\partial }_t=\epsilon {\partial }_T$ and ${\partial }_x=\epsilon {\partial }_X$, differentiating $e^{i\theta /\epsilon }$ gives an order-one physical angular frequency and order-one physical wavenumbers. Absorb a constant leading amplitude phase into $\theta$ and take $A_0$ real; for a complex amplitude the corresponding quadratic density is $|A_0{|}^2$.

Substitution into the wave equation gives the eikonal equation for a variable-speed wave at leading order:Choose the positive-frequency branch $\omega =\Omega =cK$, $K=\sqrt{k^2+{\ell }^2}$; the negative branch can be treated in exactly the same way. The imaginary first-order terms giveMultiplying by $A_0$ puts this wave-action transport for a variable-speed wave in conservation form:Its density is $\omega A_0^2$ and its flux is that density times the group velocity. Independently, equality of mixed wave phase derivatives givesThis phase compatibility in geometrical optics is essential when converting the differentiated dispersion relation into equations along rays.

For completeness let $\omega =\Omega (\mathbf{k};\mathbf{X},T)$ be a general local dispersion relation. Define ray tracing by $d\mathbf{X}/dT={\mathbf{c}}_g={\nabla }_{\mathbf{k}}\Omega$. Using the chain rule in ${\nabla }_{\mathbf{X}}\omega$ and the symmetry ${\partial }_{X_i}{k}_j={\partial }_{X_j}{k}_i$ givesLikewise ${\omega }_T={\Omega }_T+{\mathbf{c}}_g\cdot {\mathbf{k}}_T$ cancels its advective term because ${\mathbf{k}}_T=-\nabla \omega$. Hence the Hamiltonian ray equations for a local dispersion relation areThe derivatives on the right hold the wavevector fixed. For this wave equation they reduce to

When $c=c(Y)$, the ray invariants are $k$ and $\omega$, while $d\ell /dT=-K{c}_Y<0$. For a monochromatic component with these fixed invariants write $\theta =kX-\omega T+S(Y)$, soUsing the stated independence of $A_0$ from $X,T$, the wave-action transport for a variable-speed wave becomes $(c^2\ell A_0^2{)}_Y=0$. ThereforeThis is the stationary wave amplitude in a stratified wavespeed; $\mu$ is the constant for that component or ray family.

At the proposed simple turning point of a variable-speed wave, take $c_s=c(Y_s)=\omega /k$ and $c_s^{\prime }=c_Y(Y_s)>0$. On the chosen positive branch this entails $k>0$. DefineAs the ray approaches from below, $\ell \sim [q(Y_s-Y){]}^{1/2}$ andThe amplitude diverges and the vertical group velocity $c^2\ell /\omega$ vanishes. The WKB approximation also fails its slow-variation condition $\epsilon |{\ell }_Y|/{\ell }^2\ll 1$ when $Y_s-Y=O({\epsilon }^{2/3})$. The turning-point enhancement of wave amplitude shows that this divergence is a failure of the outer ray approximation, rather than an actual infinite solution.

To derive the inner wavefield directly, separate the conserved horizontal wave phase and frequency in the original differential equation:Dividing by $c^2$ gives ${\epsilon }^2{\Psi }_{YY}+2{\epsilon }^2(c_Y/c){\Psi }_Y+Q(Y)\Psi =0$. Set $Y-Y_s=\delta \eta$. The second derivative is of size ${\epsilon }^2/{\delta }^2$, whereas $Q\sim -q\delta \eta$, so the Airy scaling at a variable-speed wave turning point isThe first-derivative term is smaller by $O(\delta )$, and higher coefficients in the expansion of $Q$ give the same relative order. Thus the general leading inner profile is a linear combination of the two Airy functions $\mathrm{Ai}(\eta )$ and $\mathrm{Bi}(\eta )$. For decaying Airy continuation and ray reflection, if the physical continuation is bounded and decays into $Y>Y_s$, its growing $\mathrm{Bi}$ coefficient is zero. In that usual evanescent continuation the wavefield has the formWithout a condition in the evanescent region the corresponding $\mathrm{Bi}$ term remains admissible locally; the given existence of a wavefield below the turning point alone does not select its coefficient.

Finally, the outer amplitude at distance $\delta$ is $O({\delta }^{-1/4})=O({\epsilon }^{-1/6})$ for $\mu =O(1)$ and fixed nonzero $q$. The oscillatory Airy turning-point connection formula has exactly the same $(-\eta {)}^{-1/4}$ envelope on the allowed side. Therefore the inner wavefield has magnitude $O({\epsilon }^{-1/6})$. The decaying Airy continuation matches an incident and reflected pair of rays, so a single upward $\ell >0$ branch must be supplemented by its reflected branch near the turning point; no formal coefficient matching is needed to identify this magnitude.

Choose $X=x-2y$ and $T=x$, so $(\alpha ,\beta ,\gamma ,\delta )=(1,-2,1,0)$. The Jacobian determinant of the coordinate change is two, so it is an invertible linear map. By the chain rule, ${\partial }_x={\partial }_X+{\partial }_T$ and ${\partial }_y=-2{\partial }_X$. ConsequentlyThis is the wave equation with $c=1$. The D'Alembert formula is $F(X,T)=U(X+T)+V(X-T)$, with arbitrary twice differentiable functions $U,V$. Returning to the old coordinates and absorbing factors of two into the arbitrary functions givesDirect differentiation checks the cancellation. Equivalently, first integrating ${\partial }_x(f_x+f_y)=0$ produces the same two-function family.

For smooth rapidly decaying initial data, apply the Fourier transform in $x$ and differentiate under the integral. The Fourier transform of a derivative turns the wave equation intoSolving this constant-coefficient ordinary differential equation givesAt $k=0$ the quotient means its continuous limit $t$, so $\tilde{v}(0,t)=\tilde{f}(0)+t\tilde{g}(0)$.

Since $\cos kt=(e^{ikt}+e^{-ikt})/2$, the Translation property of the Fourier transform inverts the first term as $(f(x+t)+f(x-t))/2$. Also $K_t(x)=\frac{1}{2}{1}_{[-t,t]}(x)$ has Fourier transform $\sin kt/k$, including its limiting value at zero. The convolution theorem therefore inverts the second term asThis proves the D'Alembert formulaThe formula itself extends beyond the rapidly decaying class: $f\in C^2$ and $g\in C^1$ suffice for a classical solution, verified by differentiation and the two initial conditions. The decay assumption above justifies the ordinary transform derivation without distributional qualifications.

For the wave equation, the principal symbol is $-{\tau }^2+|\xi {|}^2$, so its value on the conormal $dt$ is $-1$. Consequently the Cauchy-Kovalevskaya theorem applies to real analytic displacement and velocity data on this non-characteristic hypersurface:In addition, this Cauchy problem for a partial differential equation has Hadamard well-posedness for nonanalytic data in the usual energy spaces. For example, data in $H^1({\mathbb{R}}^n)\times L^2({\mathbb{R}}^n)$ give a unique solution in $C_t{H}^1\cap C_t^1{L}^2$ on bounded time intervals, with continuous dependence; conservation of $\int (|u_t{|}^2+|\nabla u{|}^2)$ is its basic estimate. Thus real analytic existence and energy-space stability are different assertions, and in this example both are available.

Work first in units $G=c=\hslash =k_B=1$. Black-hole thermodynamics starts with the laws of black-hole mechanics. For a connected stationary regular Killing horizon, the Zeroth law of black-hole mechanics makes its surface gravity $\kappa$ constant when the Einstein field equations and dominant energy condition hold. This parallels uniform equilibrium temperature. For neighboring stationary asymptotically flat four-dimensional Einstein-Maxwell black holes, the First law of black-hole mechanics iswhere ${\Omega }_H$ is the horizon angular velocity and ${\Phi }_H$ its electric potential relative to infinity. The last terms are rotational and electromagnetic work, analogous to the work terms in $\delta E=T\delta S+\text{work}$. The second law of black-hole mechanics, expressed by Hawking's area theorem, makes the total future horizon area nondecreasing for classical matter obeying the null energy condition and global assumptions such as strong asymptotic predictability. The third law of black-hole mechanics is an unattainability statement: subject to the usual regularity assumptions, the weak energy condition, and a bounded stress-energy tensor, no finite physical process reduces the surface gravity of a regular horizon to zero. It does not assert that extremal black holes have zero area or zero entropy.

Classical geometry alone fixes an analogy, not a nonzero physical temperature. Quantum field theory supplies Hawking temperature $T_H=\kappa /(2\pi )$. Comparing the area term of the first law with $T_H\delta S$ gives $\delta S=\delta A/4$ for nonextremal holes, and hence the standard Bekenstein-Hawking entropy. Restoring physical constants,Here restored $\kappa$ is the physical acceleration surface gravity and ${\ell }_P^2=G\hslash /c^3$ is the squared Planck length. The first-law comparison determines the area coefficient, leaving an additive entropy constant unspecified; the displayed formula is the usual convention. Its area scaling, rather than ordinary volume scaling, signals that the available thermodynamic degrees of freedom of a gravitating system are constrained by the horizon geometry.

Particle production explains the quantum input. For a real scalar field obeying the massless covariant wave equation, the Klein-Gordon inner product on solutions isThe wave equation makes its current divergence-free, so the product is independent of a Cauchy hypersurface when the boundary flux vanishes. Use suitably normalized wave packets or the appropriate continuum distributions. Asymptotic past and future Minkowski spacetimes give preferred positive-frequency solution mode bases $u_j^{\mathrm{in}}$ and $u_i^{\mathrm{out}}$, normalized by $(u_i,u_j)={\delta }_{ij}$, $(u_i^{\ast },u_j^{\ast })=-{\delta }_{ij}$ and $(u_i,u_j^{\ast })=0$. Between those regions, a nonstationary geometry generally has no preferred positive-frequency splitting.