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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.

The mode bases are related by a Bogoliubov transformation,The last identities follow from conservation of the Klein-Gordon inner product and are the canonical identities for a bosonic Bogoliubov transformation. Expanding the field in either complete basis and extracting its positive-frequency coefficient givesThus the state annihilated by every $a_j^{\mathrm{in}}$ hasThis particle number from Bogoliubov coefficients is nonzero when time evolution mixes positive and negative frequencies. It describes one state as vacuum in the early particle basis and populated in the late basis; it does not require an arbitrary choice of a vacuum at each intermediate time. For a finite implementable transformation the state remains a squeezed pure quantum state. In infinitely many modes, a common unitary bosonic Fock space implementation requires additional conditions, such as the beta map being a Hilbert-Schmidt operator; finite wave packet observables need not share every global divergence of an ideal continuum calculation.

For gravitational collapse, the future is not globally Minkowski: it contains a black hole. The relevant late out-basis includes modes reaching future null infinity together with modes entering the future event horizon. Modes on infinity alone are not a complete Cauchy basis. Nevertheless the same mode-mixing calculation determines the outgoing particle flux. Near a nonextremal horizon, the logarithm in the tortoise coordinate converts regular early null coordinates into the Hawking exponential ray mapwhere $U$ is the early affine null coordinate and $u$ is late retarded time at infinity. Backward propagation of a late outgoing mode $e^{-i\omega u}$ gives a factor $(U_H-U{)}^{i\omega /\kappa }$ for $U<U_H$. Its positive- and negative-frequency Fourier integrals differ by analytic continuation around the logarithmic branch. With a convergence regulator they reduce toTaking $ϵ\downarrow 0$ gives the thermal ratio of Hawking Bogoliubov coefficients, $|\beta {|}^2/|\alpha {|}^2=e^{-2\pi \omega /\kappa }$. Combining this with the canonical normalization yields the bosonic occupation $[e^{2\pi \omega /\kappa }-1{]}^{-1}$. Late-time wave packets turn formal continuum coefficients into a finite number flux.

A Schwarzschild black hole has $\kappa =1/(4M)$, so this spectrum has $T_H=1/(8\pi M)$. The exterior curvature potential partly reflects the outgoing modes. Its transmission probabilities are the greybody factors, giving, for one massless scalar species in the stationary late-time approximation,Thus the horizon temperature is universal, while the spectrum received at infinity is not a perfect featureless blackbody radiation spectrum. The derivation assumes the near-horizon quantum state inherited from a regular collapse vacuum and ignores rapid backreaction over the timescale of the packets. Exponentially blueshifted precursor frequencies indicate the usual short-distance assumption in the semiclassical calculation, not an independently demonstrated quantum-gravity description of the endpoint.

The outgoing positive energy flux drives black-hole evaporation. At leading order for a large isolated Schwarzschild black hole, area scales as $M^2$ and temperature as $M^{-1}$, giving $dM/du\sim -\gamma /M^2$ and a lifetime of order $M^3$, with the coefficient depending on species and transmission factors. Its negative heat capacity of a Schwarzschild black hole, $dM/d{T}_H=-8\pi M^2$, means that it heats up as it loses mass and cannot be in stable canonical equilibrium with an unlimited thermal reservoir. The approximation fails when quantum-gravitational scales are reached and does not fix whether the endpoint is complete evaporation, a remnant, or something else.

Area loss does not contradict the classical area theorem, because the quantum stress-energy tensor need not obey the classical null energy condition; negative horizon energy accompanies the positive outgoing flux. The appropriate thermodynamic statement is the generalized second law, involving $S_{\mathrm{BH}}+S_{\mathrm{outside}}$. Finally, Hawking radiation raises the black hole information paradox: if a pure quantum state completely evaporates into an exactly thermal mixed quantum state and its partners disappear, the result conflicts with ordinary unitary time evolution. Early outgoing radiation can be mixed simply because of its entanglement with interior modes while the complete state remains a pure quantum state; that fact alone is not information loss. Resolving the fate of all correlations through the evaporation endpoint requires more than the leading semiclassical flux calculation.

Let $\rho =a/\sqrt{3}$ be the distance from the origin to each fixed charge, and denote their position vectors by ${\mathbf{R}}_j$. For a small planar displacement $\mathbf{r}$, the Taylor expansioncan be summed using the threefold symmetry relationsThe electric potential due to the fixed charges is consequentlyThe moving charge has electrostatic potential energy of point charges $U=q\Phi$. Because the quadratic coefficient is positive, the origin is a strict local minimum of $U$, and the equilibrium is stable.

Use $\mathbf{E}=-\nabla \varphi$ and Gauss's law $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$. The divergence theorem and the decay at infinity giveMultiplying by $1/2$ proves that the two expressions for the electrostatic energy agree.

The uniform volume charge density in the thick shell isSpherical symmetry and the integral form of Gauss's law giveThus $|\mathbf{E}|$ is zero inside, rises continuously through the charged region, reaches $Q/(4\pi {ϵ}_0{b}^2)$ at $b$, and then decays as $r^{-2}$. Taking $\varphi (\infty )=0$ and integrating $E=-d\varphi /dr$ givesThe electric potential is continuous, constant inside, decreases smoothly across the charge, and continues as $1/r$ outside.

As $b\to a$, the charge becomes a surface shell. ThenThe potential remains continuous, but the normal electric field jumps by $Q/(4\pi {ϵ}_0{a}^2)$ across the surface charge density.

The field-energy expression givesThe charge-potential expression gives the same result directly because the entire charge lies where the potential equals $\varphi (a)$:This is the electrostatic energy of a uniformly charged spherical shell.

Since $\varphi (a)=Q/(4\pi {ϵ}_{0}a)$ is proportional to $Q$,to first order. This is the work required to bring the additional charge $\delta Q$ from infinity to the potential created by the charge already present.

The free-space Green function of the Laplacian for the Poisson equation gives the electric potentialFor $r=|\mathbf{x}|\gg R$, its electric multipole expansion begins withDefine the total electric charge and electric dipole moment byThen

The analogous solution for the magnetic vector potential isIts apparent $r^{-1}$ coefficient is ${\mu }_0(4\pi {)}^{-1}\int \mathbf{J}{d}^3{x}^{\prime }$. For each component, the supplied identity and the divergence-free vector field condition giveThe divergence theorem turns this into a surface integral outside the compact support of the current density, where $\mathbf{J}=0$. Hence $\int \mathbf{J}{d}^3{x}^{\prime }=0$, so the $1/r$ term vanishes.

The electric potential of a point charge $Q$ at the origin isFor the two charges forming the dipole,The first-order Taylor expansion at large $|\mathbf{x}|$ isand therefore, with the electric dipole moment $\mathbf{p}=Q\mathbf{d}$,to leading order. Taking the interaction of a second dipole with the corresponding electric field gives the stated electric dipole-dipole interaction

Let the lattice spacing be $d$ and write the central dipole asIts two horizontal neighbours have moment $p(\cos \theta ,-\sin \theta )$. For either one, the expression in parentheses, after extracting $d^{-3}$, isIts two vertical neighbours have moment $p(-\cos \theta ,\sin \theta )$, and each contributes insteadAdding all four nearest-neighbour interactions and using the Pythagorean trigonometric identity givesThe angle has cancelled, so the energy is independent of $\theta$.

Let $r$ be distance from the common centre. By Gauss's law, the electric field is radial and equalsTaking the electric potential to vanish at infinity and requiring continuity across each shell gives

The electrostatic energy is the integral of the electric-field energy density:Only the two nonzero-field regions contribute, soTherefore

Let $\theta$ be the angle between $r$ and the positive $z$-axis. By superposition of the electric potential of point charges,For $r>d$, the Legendre polynomial expansion givesThereforeThe three displayed terms are respectively the monopole, dipole, and quadrupole terms of the electric multipole expansion. In particular, the total charge is $Q(N-M-1)$ and the electric dipole moment is $Qd(M-1)\hat{z}$. This is the multipole expansion of three collinear charges.