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Work with the complex representation and take its real part to obtain the physical magnetic field. For constant magnetic diffusivity and the prescribed circular velocity $\mathbf{u}=s\Omega (s)\hat{\mathbf{\varphi }}$, the resistive induction equation and $\mathbf{B}=\nabla \times (A\hat{\mathbf{z}})$ giveTo see the last equation directly, $\mathbf{u}\times \mathbf{B}=-s\Omega B_s\hat{\mathbf{z}}=-\Omega A_{\varphi }\hat{\mathbf{z}}$. Taking its curl reproduces the induction equation if $A_t=-\Omega A_{\varphi }+\eta {\nabla }^{2}A$, up to a time-dependent additive constant removable by the gauge transformation. This is the scalar form of advection and magnetic diffusion in cylindrical coordinates.

Put $\psi =\varphi -\Omega (s)t$. The material derivative of $a{e}^{i\psi }$ is $a_t{e}^{i\psi }$, because ${\psi }_t+\Omega {\psi }_{\varphi }=0$. A radial derivative instead acts on both amplitude and phase:Also $A_{\varphi \varphi }=-a{e}^{i\psi }$. Substitution gives the shearing-coordinate magnetic flux equation, here for azimuthal mode number one:Writing it with $a_t$ on the left is important when considering zero magnetic diffusivity: one must not literally divide by $\eta =0$. In the ideal limit, $a_t=0$, so $A(s,\varphi ,t)=A(s,\varphi -\Omega (s)t,0)$. This is magnetic flux freezing: the scalar flux pattern is carried with each rotating fluid ring. It does not make the magnetic field stationary. Indeed,so differential rotation winds the field and can make its azimuthal component grow linearly in time. Rigid rotation, for which ${\Omega }^{\prime }=0$, only rotates the pattern.

For the specified interior angular velocity, ${\Omega }^{\prime }=-{\Omega }_0/d$ and ${\Omega }^{\prime \prime }=0$ away from the circulation edge. Let $T={\Omega }_{0}t$, $x=s/d$ and $\mathrm{Rm}={\Omega }_0{d}^2/\eta$. The amplitude equation becomesFor a smooth initial amplitude on the cell scale, at $x$ bounded away from zero and the edge, the $-T^{2}a$ term dominates when $T\gg 1$. Its accumulated damping is order one at $T=O({\mathrm{Rm}}^{1/3})$, whereas the integrated terms proportional to $T$ are only $O({\mathrm{Rm}}^{-1/3})$ and the terms independent of $T$ are $O({\mathrm{Rm}}^{-2/3})$. The resulting cubic-time resistive damping in a differentially rotating cell isThe magnetic field has this leading exponential decay envelope. Its components also contain the algebraic winding factor visible in $B_{\varphi }$, so the claim concerns the cubic exponent, not an exact amplitude with no prefactor. Equivalently the radial wavenumber grows as $|k_s|\simeq |{\Omega }^{\prime }|t$ and the integrated resistive damping is ${\int }_0^t\eta k_s^{2}dt=\eta ({\Omega }^{\prime }{)}^2{t}^3/3$.

This is phase mixing in magnetic flux expulsion. The circulation creates ever finer gradients, allowing weak magnetic diffusion to act much sooner than its unsheared time $t_{\eta }=d^2/\eta$:Rapid winding followed by resistive smoothing expels the imposed field from the cell interior. The maintained exterior magnetic field is redistributed into a circulation-edge boundary layer. The result is an interior high-magnetic Reynolds number transient, not a uniform assertion at the axis and the nonsmooth edge, nor the exact $t\to \infty$ solution with a continuously imposed far field. The diffusion length at ${\tau }_c$ is $\sqrt{\eta {\tau }_c}/d=3^{1/6}{\mathrm{Rm}}^{-1/3}$, identifying where a local bulk approximation can fail. The linear profile also needs smoothing at a regular rotation axis and at $s=d$ if used as a fully smooth global flow.

In photospheric magnetoconvection, circulating and diverging convection transports weak magnetic flux towards converging downflow regions. Solar granulation concentrates field in intergranular lanes, while larger-scale supergranulation helps form the solar magnetic network. Flux expulsion explains why a highly conducting photosphere can have relatively weak field in cell interiors and intermittent concentrations at their edges: high electrical conductivity delays diffusion on the original scale, but does not prevent diffusion on the fine scales created by advection. A long-lived circulation must last long enough for the expulsion time; rapidly changing cells need not reach this idealized state. Once the field becomes strong, the Lorentz force modifies the convection, so this prescribed-flow calculation alone neither determines the final field strength nor describes strong-field sunspots.

Choose a reference circular orbit at radius $r_0$, rotate the frame with $\mathbf{\Omega }=\Omega (r_0){\mathbf{e}}_z$, and introduce local Cartesian coordinates $x=r-r_0$, $y=r_0(\varphi -\Omega t)$ and vertical coordinate $z$. On scales small compared with $r_0$, neglect curvature of the coordinate directions but retain the leading differential rotation. The background velocity relative to the rotating frame isThus $A=3\Omega /4$ for a Keplerian disk. The shearing sheet retains Coriolis acceleration and the linear tidal acceleration. The radial shearing-sheet tidal potential is $-2\Omega A{x}^2$, whose force $4\Omega Ax{\mathbf{e}}_x$ balances the background Coriolis acceleration. If vertical tidal gravity is included, its equilibrium contribution is balanced by a background pressure $p_0(z)$ and disappears when that equilibrium is subtracted.

Write the total velocity as $\mathbf{u}={\mathbf{u}}_0+\mathbf{v}$. In the rotating-frame magnetohydrodynamic momentum equation, background advection is $({\mathbf{u}}_0\cdot \nabla )\mathbf{v}=-2Ax{\partial }_y\mathbf{v}$ and advection of the background by the perturbation is $(\mathbf{v}\cdot \nabla ){\mathbf{u}}_0=-2A{v}_x{\mathbf{e}}_y$. The uniform-viscosity diffusion of the linear background shear is zero. For a solenoidal magnetic field the Lorentz force per unit mass isAt constant density, absorb the magnetic-pressure gradient into $\psi =(p-p_0)/\rho +B^2/(2{\mu }_0\rho )$. Subtracting the background force balance then gives the stated momentum equation, with the remaining nonlinear advection $\mathbf{v}\cdot \nabla \mathbf{v}$, Coriolis acceleration $2\mathbf{\Omega }\times \mathbf{v}$, magnetic tension and viscous diffusion.

For constant finite electrical conductivity ${\sigma }_e$, put $\eta =({\mu }_0{\sigma }_e{)}^{-1}$. The resistive induction equation with uniform magnetic diffusivity is ${\partial }_t\mathbf{B}+\mathbf{u}\cdot \nabla \mathbf{B}=\mathbf{B}\cdot \nabla \mathbf{u}+\eta {\nabla }^2\mathbf{B}$. Substituting the background shear supplies the term $-2A{B}_x{\mathbf{e}}_y$ and the same background-advection operator as in the momentum equation. The two further conditions areThey express incompressibility and the absence of magnetic monopoles, respectively.

Now take real plane waves with common phase $\vartheta =\mathbf{k}(t)\cdot \mathbf{x}$ and real wavevector. The solenoidal conditions becomeEvery field depends on position only through $\vartheta$. Hence for either transverse vector field $\mathbf{P}$ and either wave field $\mathbf{Q}$, $(\mathbf{P}\cdot \nabla )\mathbf{Q}=(\mathbf{P}\cdot \mathbf{k}){\partial }_{\vartheta }\mathbf{Q}=0$. This proves that all the displayed quadratic advection and tension terms vanish pointwise, including products involving conjugate amplitudes. These waves are therefore exact solutions at finite amplitude, not merely a linearization.

Acting with $D_0={\partial }_t-2Ax{\partial }_y$ on the phase yieldsCanceling its position-dependent part requiresThe remaining amplitude equations areDifferentiating the velocity constraint determines the pressure amplitude:The resistive induction equation preserves its constraint because $\dot{\mathbf{k}}\cdot \tilde{\mathbf{B}}=2A{k}_y{\tilde{B}}_x$ cancels the corresponding shear term in $\mathbf{k}\cdot \dot{\tilde{\mathbf{B}}}$. The specified magnetic field has no spatially uniform part. Thus there is no background magnetic field producing an additional linear magnetic-tension coupling. The scalar $\psi$ is the modified pressure; the physical pressure can include a constant and a second spatial harmonic from $-B^2/(2{\mu }_0)$ while $\psi$ retains the stipulated plane-wave form. This is an exact zero-mean magnetic shearing wave.

For decay, take the real part of the scalar product of the velocity-amplitude equation with its complex conjugate. Pressure does no work because $\mathbf{k}\cdot \tilde{\mathbf{v}}=0$, and the real part of Coriolis work is zero. ThereforeThe resistive induction equation similarly givesusing $2|u_x{u}_y|\le |\mathbf{u}{|}^2$. Integrating these differential inequalities, defineThenA non-axisymmetric wave has $k_y\ne 0$. For nonzero differential shear and positive $\nu ,\eta$, the negative cubic terms dominate the linear upper bound on shear work. ConsequentlyThis cubic-exponent decay of a nonaxisymmetric shearing wave allows transient energy growth before the winding to short radial wavelengths makes diffusion dominant. The magnetic amplitudes can also be written explicitly aswhich confirms the damping independently. Positivity of the dissipative coefficients is necessary: with $k_z=0$, an inviscid vertical velocity amplitude can remain constant, and an ideal vertical magnetic amplitude can remain constant when $\eta =0$. The asserted decay applies to the viscous, finitely conducting model; taking an ideal limit before the long-time limit changes the conclusion.

In vacuum SI notation, Maxwell's equations areTaking the divergence of the last equation, using that the divergence of a curl vanishes and then the first equation, givesThis is local conservation of electric charge. For spatially uniform electrical conductivity, Ohm's law $J=\sigma E$ gives $\nabla \cdot J=\sigma \rho /{ϵ}_0$, soFor a homogeneous dielectric conductor with permittivity $ϵ$, replace ${ϵ}_0$ by $ϵ$. This charge relaxation is a statement about bulk charge; charge transported to a boundary is still accounted for by the conservation law.

Let $\ell$ be the characteristic radius over which the differential rotation changes, with angular velocity of order $\Omega$. The winding source has magnitude $F\sim B_0\Omega$. For times too short for appreciable magnetic diffusion, the zero initial toroidal magnetic field therefore grows as $B\sim B_0\Omega t$. In the steady balance, $F$ is opposed by $\eta B/{\ell }^2$, giving $B_{\mathrm{steady}}\sim B_0\Omega {\ell }^2/\eta$. Comparing these amplitudes gives the diffusive establishment time of a wound toroidal field:With $R_m=\Omega {\ell }^2/\eta$, this is $R_m$ rotation times and the steady toroidal amplitude is of order $R_m{B}_0$. Thus high electrical conductivity makes establishment very slow even though winding begins immediately.

More precisely, if $\mathcal{L}={\nabla }^2-s^{-2}$ on regular toroidal magnetic fields, the solution with zero initial value satisfies $B(t)=B_{\mathrm{steady}}-e^{\eta t\mathcal{L}}{B}_{\mathrm{steady}}$. The residual relaxes diffusively. In the unbounded domain, ${\ell }^2/\eta$ estimates equilibration of the rotating region; exact establishment at every radius is only an asymptotic statement, with diffusion on scale $r$ taking order $r^2/\eta$. As $\eta \to 0$, the time diverges and the prescribed-flow ideal solution continues its secular winding. In a real fluid, magnetic tension and Lorentz force feedback can modify the differential rotation before the very large kinematic steady amplitude is attained.

In a steady state, Maxwell's equations imply $\nabla \times \mathbf{E}=0$, since the magnetic field is time independent. The divergence of the steady Ampère equation gives $\nabla \cdot \mathbf{j}=0$. With uniform electrical conductivity, Ohm's law is $\mathbf{j}=\sigma \mathbf{E}$, soFor the usual straight-wire model, the cylindrical side is insulating and the two flat end faces are equipotentials. Put $\mathbf{E}=-\nabla V$. The electric potential solves Laplace's equation, with constant end values and zero normal derivative on the side. The linear axial potential $V(z)=V(0)-\Delta Vz/l$ satisfies all these conditions. It is unique: the difference $h$ of two solutions has zero end values and zero side normal derivative, and integration by parts gives $\int |\nabla h{|}^{2}dV=0$. ThusThis proves uniform current in a straight homogeneous wire with the usual contact conditions. Integrating the electric current density over a cross-section and using Ohm's law then givesThe second expression is the rate of Ohmic heating.

Strictly, steady current and uniform conductivity alone do not force uniformity; the contact conditions are implicit in the wire model. For example, let $\chi$ be a nonconstant transverse Neumann eigenfunction on the disk, satisfying ${\Delta }_{\perp }\chi =-{\lambda }^2\chi$ and zero normal derivative at its edge. Then $V=-E_{0}z+C\chi \cosh (\lambda (z-l/2))$ is harmonic and preserves an insulating cylindrical side, but its nonconstant end potentials produce nonuniform current. Its added axial current has zero cross-sectional mean. This supplies a counterexample if arbitrary contacts are permitted.

A bend generally redistributes the electric current density: the inner side of the bend carries more current than the outer side, as illustrated by $j_{\theta }\propto 1/r$ in an annular conductor. Exact resistance and local Ohmic heating can therefore change. For a slender, gently curved wire the straight-wire formulas remain the leading approximation; the identity $\dot{Q}=I^{2}R$ remains valid for the total steady dissipation.

For this question use the normalized magnetic scalar potential $\mathcal{U}$ defined by $\mathbf{B}=({\mu }_0/4\pi )\nabla \mathcal{U}$. The previous question instead includes ${\mu }_0/4\pi$ in its scalar potential; its $U$ equals $({\mu }_0/4\pi )\mathcal{U}$. This normalization change in the paper must be kept separate from the kernel calculation.

Take the radial component of the Geselowitz formula. On the spherical boundary, ${\mathbf{n}}^{\prime }$ is parallel to ${\mathbf{r}}^{\prime }$, andThus the surface-current term contributes no radial magnetic field. The primary Biot-Savart law term gives, along a fixed direction $\hat{\mathbf{r}}$,Choose $\mathcal{U}\to 0$ at infinity. Integrating this equation fixes the full potential along every ray, givingTo express it as a source-position derivative, defineWith $\mathbf{W}=\mathbf{Q}\times {\mathbf{r}}_0$, differentiating the integrand givesso $\mathcal{U}=\mathbf{W}\cdot {\nabla }_{{\mathbf{r}}_0}\varphi$ as required.

Let $b=|{\mathbf{r}}_0|$ and $P=|\mathbf{r}-{\mathbf{r}}_0|$. An elementary antiderivative gives the spherical current-dipole logarithmic kernelFor $r>a>b$, both logarithm arguments are positive. Its radial derivative is $-1/(rP)$ and its limit at infinity is zero, which verifies the preceding integral. At $b=0$ use the continuous limit $\varphi =1/r$.

For a direct check, differentiating along $\mathbf{W}$ leaves $b$ unchanged and gives $\mathbf{W}\cdot {\nabla }_{{\mathbf{r}}_0}P=-\mathbf{W}\cdot \mathbf{r}/P$. Sinceone obtainsIts gradient is the Sarvas formula. The sphere radius and electrical conductivity disappear from this exterior magnetic result because spherical symmetry removed the radial contribution of the return currents. The original derivation and spherical-conductor assumptions are discussed in Sarvas's paper.

Take the divergence of the Ampère-Maxwell equation, $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{ϵ}_0{\partial }_t\mathbf{E}$. Since the divergence of a curl is zero vanishes, and Gauss's law gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, the result isThis is the charge continuity equation.

At an internal point of a homogeneous Ohmic conductor, Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Since $\sigma$ is constant, the charge continuity equation becomes ${\partial }_t\rho =-(\sigma /{ϵ}_0)\rho$. ThereforeThis is charge relaxation. Here $\rho$ is the total charge appearing in the vacuum form of Gauss's law; in a homogeneous dielectric description of free charge the corresponding constant permittivity replaces ${ϵ}_0$.

For a finite isolated body, charge travels to the surface. The bulk formula applies at internal points, not across the interface where electrical conductivity jumps and surface charge density accumulates. Integrating the charge continuity equation over the body, including its surface charge, gives constant total electric charge because no electric current crosses the external boundary. Bulk decay and conservation of electric charge are therefore compatible.

Use Faraday's law ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{B}=0$. In the nonrelativistic, single-fluid approximation, neglect Hall and other nonideal electromotive terms and use the moving-conductor moving-conductor Ohm law, $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\mathbf{J}/\sigma$. Infinite electrical conductivity with finite current gives $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, henceThis is the ideal magnetohydrodynamic induction equation. For comparison, neglecting displacement current in Ampère-Maxwell equation gives $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$; with uniform finite electrical conductivity it produces magnetic diffusion ${\eta }_m{\nabla }^2\mathbf{B}$, where the magnetic diffusivity is ${\eta }_m=1/({\mu }_0\sigma )$. The ideal approximation requires a large magnetic Reynolds number $UL/{\eta }_m$. Dropping displacement current is useful for this finite-conductivity comparison, but Faraday's law and the ideal Ohm relation already suffice for the ideal induction equation.

Expanding the curl and using the solenoidal magnetic-field constraint gives the material derivative formCombine this with mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, to obtainNow parametrize a material curve by a fixed label $a$: $\mathbf{X}(a,t)$ obeys ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. Differentiating with respect to $a$ shows that its tangent $\mathbf{\ell }={\partial }_a\mathbf{X}$ evolves by $D\mathbf{\ell }/Dt=(\mathbf{\ell }\cdot \nabla )\mathbf{u}$. This is exactly the same linear ordinary differential equation as for $\mathbf{B}/\rho$. Initially parallel tangents remain parallel by uniqueness, with a label-dependent proportionality factor constant along each particle trajectory. Thus magnetic field lines are transported as material curves, wherever the field and fluid flow are smooth and the field is nonzero. This is the field-line form of magnetic flux freezing.

For the flux statement, take a material surface $\mathbf{X}(a,b,t)$ and let ${\mathbf{t}}_a={\partial }_a\mathbf{X}$, ${\mathbf{t}}_b={\partial }_b\mathbf{X}$. Both tangents obey $D{t}_i/Dt=({\partial }_j{u}_i)t_j$. Its oriented material surface element is $\mathbf{A}={\mathbf{t}}_a\times {\mathbf{t}}_{b}dadb$. Differentiating the cross product, rather than assuming its transport rule, givesEquivalently, $D\mathbf{A}/Dt=(\nabla \cdot \mathbf{u})\mathbf{A}-(\nabla \mathbf{u}{)}^T\mathbf{A}$, with $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Contracting this derived rule with the induction equation gives a pointwise cancellation:Integrating over the fixed material labels therefore proves conservation of flux through an open material surface:The surface need not be closed; its boundary is carried with the fluid. The result follows from material transport, rather than from the zero flux through a closed surface.

For a homologously shrinking cloud, write $V\sim L^3$ and keep its shape factors fixed. Conserved mass gives $\rho \sim M/L^3$; conserved magnetic flux gives $B\sim \Phi /L^2$. Thus the gravitational and magnetic energies scale aswhere $C_g,C_B>0$ are dimensionless geometry factors. Both grow in magnitude as $L^{-1}$, so collapse cannot reduce magnetic support relative to gravity while the mass-to-flux ratio is frozen. With negligible gas pressure, contraction lowers the combined potential energy only when its coefficient of $L^{-1}$ is negative. Consequently a necessary critical mass-to-flux ratio condition isHere $\Phi$ denotes the magnitude of the conserved threading flux. The numerical coefficient depends on geometry and boundary conditions; the scaling argument does not determine it or make the condition sufficient in the presence of other support.

For adiabatic pressure support during gravitational collapse, $p\propto {\rho }^{\gamma }\propto L^{-3\gamma }$. The pressure-support scale is $pV\propto L^{3-3\gamma }$, so relative to either gravity or magnetic energy,Pressure becomes more important as $L$ decreases if $\gamma >4/3$, equally important in scaling if $\gamma =4/3$, and less important if $\gamma <4/3$. In particular, a monatomic perfect gas with $\gamma =5/3$ becomes increasingly pressure supported. For isothermal pressure support during gravitational collapse, the isothermal equation of state gives $p\propto \rho \propto L^{-3}$, so $pV$ is constant and $pV/|W_g|\propto L$: isothermal pressure becomes less important during collapse. The same comparisons hold against magnetic support because its energy has the same $L^{-1}$ scaling as gravity.

A thermoelectric generator maintains hot and cold junctions of two materials with different Seebeck coefficients. Mobile charge carriers redistribute in the temperature gradient, producing an open-circuit voltage of magnitude $|V|=|S_p-S_n|\Delta T$ for constant coefficients. A load permits electric current; pairs of p-type and n-type legs are commonly connected electrically in series and thermally in parallel. Maintaining the output requires continuous heat transfer from the hot reservoir and rejection of heat to the cold reservoir. The thermoelectric figure of merit $ZT=S^2\sigma T/\kappa$ expresses the favorable combination of large Seebeck coefficient, large electrical conductivity and small thermal conductivity. Contact losses, Joule heating and parasitic heat leaks reduce practical efficiency.

A pyroelectric generator instead exploits a change of spontaneous electric polarization with temperature. If the relevant component of the pyroelectric coefficient is $p=d{P}_s/dT$, an electrode of area $A$ receives a short-circuit electric current approximately $I=pAdT/dt$. A heating stroke and a cooling stroke produce opposite charge changes. Rectification or a synchronized electrical cycle can transfer energy to storage. A spatial temperature difference drives the thermoelectric device; a temporal temperature change drives the pyroelectric device. Uniform, constant temperature gives no continuing current from pyroelectricity even if the material is hot. Reported nanowire generators deliberately use changing temperature, rather than infer continuous generation from a static hot specimen.^[ref]

Both approaches can be compact, silent and free of macroscopic moving parts. The thermoelectric generator gives continuous output under a sustained gradient, but needs two effective thermal contacts and sufficient heat flow; very small $\Delta T$ gives a small voltage and thermal contacts can consume much of the available gradient. Thin films exhibiting pyroelectricity can respond to intermittent heating without a permanently maintained hot/cold separation across their thickness, but require repeated thermal cycling, low electrical leakage and suitable charge extraction. Their thermal time constant limits useful cycling frequency, and sensible heating/cooling can dominate the heat budget. A steady hot pipe with an available cold sink favors a thermoelectric generator; periodic radiative heating can favor a pyroelectric generator. Neither converts equilibrium thermal motion at a single fixed temperature into sustained useful work.

The dimensional material figure of merit and its common dimensionless form areHere $S$ is the Seebeck coefficient, $\sigma$ the electrical conductivity, ${\rho }_e=1/\sigma$ the electrical resistivity, and $\kappa ={\kappa }_e+{\kappa }_l$ the total thermal conductivity. $Z$ has units ${\mathrm{K}}^{-1}$; $ZT$ is dimensionless.

Seek a large thermoelectric power factor $S^2\sigma$ and low thermal conductivity. Optimize carrier concentration and band structure to balance thermopower and conductivity. Alloy disorder, complex unit cells, grain boundaries or suitable nanoscale features can scatter phonons and reduce lattice heat transport.

The properties are coupled: increasing carrier density often reduces $|S|$, electronic heat conduction grows with electrical transport, and boundaries that scatter phonons may also scatter carriers. Evaluate the net change and the operating-temperature range, rather than maximizing each favorable term independently.^[ref]

Tune doping to the optimum carrier concentration; a rise in conductivity is useful only if it outweighs the accompanying loss of thermopower and extra electronic heat conduction. Band convergence, favourable density of states and energy-dependent transport can improve the power factor $S^2\sigma$, provided mobility is retained.

Reduce lattice thermal conductivity through alloy disorder, heavy or complex unit cells, grain boundaries and nanoscale inclusions that scatter phonons. Aim to scatter heat-carrying phonons much more strongly than charge carriers; indiscriminate disorder can reduce both thermal and electrical conductivity. Nanostructuring and interfaces can therefore help, but an improvement must be demonstrated in total $ZT$ at the operating temperature, not inferred from lower $\kappa$ alone.^[ref]

At low carrier concentration, $|S|$ can be large but electrical conductivity is too small. Increasing the carrier concentration usually increases $\sigma =ne\mu$, while $|S|$ decreases. Electronic thermal conductivity rises approximately as ${\kappa }_e=L\sigma T$ by the Wiedemann-Franz law; lattice conduction supplies an additional background. Thus neither an insulator nor an ordinary metal usually maximizes $Z$.

Heavily doped semiconductors and semimetals often give the useful compromise, near a maximum of $S^2\sigma /\kappa$ as doping is varied. The optimum is material- and temperature-dependent; there is no universal optimal concentration.

The Seebeck effect produces an open-circuit voltage when a material has a temperature gradient: $S=-dV/dT$ for the stated sign convention. A circuit containing dissimilar thermoelectric legs converts this into useful voltage and current. The intrinsic thermoelectric figure of merit isHere $\sigma$ is electrical conductivity and $\kappa ={\kappa }_e+{\kappa }_L$ total thermal conductivity. $Z$ has units ${\mathrm{K}}^{-1}$; $ZT$ is dimensionless. A large Seebeck coefficient and electrical conductivity, together with low thermal conductivity, favour efficient conversion.

A thermocouple uses two materials with different Seebeck coefficients. With their junctions at different temperatures, the net open-circuit voltage isUsing p-type and n-type legs makes the two contributions add. Many pairs can be connected electrically in series to add voltage while remaining thermally in parallel between the same hot and cold plates. A single homogeneous wire does not provide this same complete-circuit voltage when its return lead has the same thermopower.

The reciprocal Peltier effect transports heat when electric current passes through a junction. The Kelvin relation $\Pi =\alpha T$ connects the heat carried per unit charge to the Seebeck coefficient. The same combination of large thermopower, high electrical conductivity and low thermal conductivity gives a favorable dimensionless figure of meritThus materials useful for a generator are also useful for a cooler when electric current is driven in the appropriate direction. Joule heating and thermal leakage oppose cooling; device geometry, contact resistance and operating temperature must still be appropriate. Reversibility of the thermoelectric coupling does not make either device lossless.

For one homogeneous thermoelectric material, the dimensional thermoelectric figure of merit isThe dimensionless quantity is $ZT$. The Seebeck coefficient $S$ measures the open-circuit voltage response to a temperature difference and reflects the energy asymmetry of charge-carrier transport. The electrical conductivity $\sigma$ depends on carrier density and mobility and sets electrical loss; $\rho =1/\sigma$. The thermal conductivity is $\kappa ={\kappa }_e+{\kappa }_l$, with contributions from carriers and phonons, and represents unwanted heat leakage.

Increasing carrier concentration usually raises $\sigma$ while reducing $|S|$ once transport becomes strongly degenerate. The electronic part of $\kappa$ also tends to rise, approximately through the Wiedemann-Franz law where applicable. This competition produces an intermediate optimum, commonly in heavily doped semiconductors, rather than in a very resistive insulator or an ordinary high-carrier-density metal. The optimum depends on band structure and scattering; the schematic is not a universal measured concentration curve.

Nanowires can increase $Z$ by reducing lattice heat conduction more strongly than they degrade useful electronic transport. A wire diameter comparable to important phonon mean free paths introduces boundary scattering and reduces the lattice part of thermal conductivity. Interfaces, surface roughness and suitable nanostructure can suppress long-mean-free-path phonons while preserving shorter-mean-free-path carrier transport.

Electronic confinement or energy-dependent scattering can also change the Seebeck coefficient and thermoelectric power factor, but neither improvement is automatic. Surface depletion, disorder and contact resistance may instead reduce electrical conductivity; a smaller wire is not necessarily better. The necessary comparison is the complete ratio $S^2\sigma /({\kappa }_e+{\kappa }_l)$ over the operating temperature range. The experimentally established benefit of nanostructuring is especially clear when it suppresses phonon heat conduction while retaining a useful power factor.^[ref]

With a positive charge magnitude e, the relaxation-time equation for electrons is $m_e^{\ast }\dot{\mathbf{v}}=-e\mathbf{E}-m_e^{\ast }\mathbf{v}/\tau$. Steady drift is opposite the field, ${\mathbf{v}}_d=-e\tau \mathbf{E}/m_e^{\ast }$, so the positive electron mobility and electrical conductivity areThis is the weak-field Drude model; the mobility is a magnitude, not the signed velocity-to-field ratio.

For an intrinsic nondegenerate semiconductor, $n_i=p_i=\sqrt{N_c{N}_v}{e}^{-E_g/(2k_{B}T)}$. Equal electron and hole masses give $N_c=N_v$ and intrinsic chemical potential midway between the band edges. At300K, the exponential is $e^{-19.34}=3.98\times 10^{-9}$. Using the later supplied mass $0.015m_e$ givesThis is many orders below a typical metal's carrier density. Since ${\sigma }_i=e{n}_i({\mu }_e+{\mu }_h)$, ordinary finite carrier mobilities leave its intrinsic conduction negligible on a metallic scale; the gap alone does not specify an exact conductivity without the mobilities.

Under the explicitly stipulated full donor ionization, $n\simeq N_d=10^{24}{\mathrm{m}}^{-3}$ and the supplied mobility is $0.0100{\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1}$. ThereforeThe band diagram places the intrinsic chemical potential at midgap and a shallow isolated donor level just below the conduction edge. The doped chemical potential moves upward. Quantitatively, the degeneracy test for a fully ionized light-mass donor gas gives $N_d/N_c=21.7$, so a nondegenerate logarithmic chemical-potential formula is inconsistent. The parabolic-band zero-temperature estimate is $E_{F,d}-E_c={\hslash }^2(3{\pi }^2{N}_d{)}^{2/3}/(2m^{\ast })=0.243\mathrm{eV}$; its300K value is about0.241eV. The diagram accordingly shows the doped chemical potential inside the conduction band, not between an isolated donor level and the edge.

The hydrogenic donor state scaling using the supplied hydrogen values isThe binding is much smaller than $k_{B}T=25.9\mathrm{meV}$, indicating a shallow donor on the isolated-state estimate. However, the mean donor spacing is only $N_d^{-1/3}=10\mathrm{nm}$, well below the orbit radius. The donor-state overlap estimate invalidates an independent isolated-level ionization model at this concentration. Shallow binding supports abundant mobile carriers, but the full ionization versus isolated donor occupation distinction means full ionization cannot be justified solely by comparing binding with temperature while simultaneously retaining isolated levels and a degenerate chemical potential. The conductivity above follows the question's carrier-density assumption; an overlapping impurity/conduction-band description is needed for a self-consistent microscopic account. ^[ref]

A three-phonon normal process conserves crystal wavevector without a reciprocal-lattice shift, ${\mathbf{k}}_1+{\mathbf{k}}_2={\mathbf{k}}_3$. An Umklapp process instead hasThe crystal can absorb the reciprocal-lattice momentum; after reduction to the first Brillouin zone the phonon drift transporting heat is relaxed. A normal process redistributes phonons while preserving their total crystal momentum, so it is not by itself the same resistive mechanism. energy must also be conserved in either process.

At low temperature, large-wavevector phonons are rare, suppressing Umklapp scattering. Boundary and defect scattering then limit the mean free path; with a nearly constant boundary-limited path, ${\kappa }_{\mathrm{ph}}\simeq (C_{\mathrm{ph}}/V)v\ell /3$ grows as $T^3$. As temperature increases, heat capacity rises but Umklapp scattering shortens $\ell$, often producing a electrical conductivity maximum. At sufficiently high temperature a simple anharmonic model has $C_{\mathrm{ph}}$ nearly constant and ${\tau }_U^{-1}\propto T$, giving ${\kappa }_{\mathrm{ph}}\propto T^{-1}$. These are regime estimates, not a universal law for every material or size.

For degenerate carriers with approximately elastic scattering and a common transport relaxation time, $\sigma =n{e}^2\tau /m^{\ast }$ and ${\kappa }_e=(C_e/V)v_F^2\tau /3$. Using $C_e/V={\pi }^{2}n{k}_B^{2}T/(2E_F)$ and $v_F^2=2E_F/m^{\ast }$ givesThis is the Wiedemann–Franz law. The same mobile electrons carry charge and heat, but Pauli statistics set the electronic heat capacity and hence the Lorenz number. The relation applies to electronic thermal conductivity, not automatically to total electrical conductivity including phonons. Inelastic electron-phonon scattering can relax energy and charge currents differently; nondegeneracy and strong energy dependence of scattering also limit the simple form.

A meaningful test combines a four-terminal electrical measurement with a steady thermal-gradient measurement on the same sample and temperature range. Contact heat resistance, parasitic heat leaks, phonon conduction and sample geometry must be controlled before interpreting a measured Lorenz ratio as an electronic departure from the law.

The free electron model replaces the microscopic ionic potential and interacting electronic motion by independent mobile electrons in a uniform neutralizing background, with approximately parabolic dispersion $ϵ={\hslash }^2{k}^2/(2m^{\ast })$. Its quantum version fills states according to Fermi-Dirac statistics and the Pauli exclusion principle. Scattering is represented by a momentum relaxation time $\tau$, rather than derived from the model's ideal free trajectories. This model explains a Fermi energy much larger than room-temperature thermal energy, an electronic heat capacity proportional to temperature at low temperature, and a Drude electrical conductivity. With appropriate scattering assumptions it also yields the Wiedemann-Franz law. It cannot by itself explain band gaps, detailed band anisotropy, strong correlations, or the microscopic temperature dependence of $\tau$.

The inertial term is mass times the rate of change of mean drift velocity. The electric force is $eE$, taking the printed $e$ as the signed electron charge, and $-\gamma ⟨v⟩=-m^{\ast }⟨v⟩/\tau$ describes loss of directed momentum to scattering. electrons retain large microscopic velocities; this damping is of their mean directed motion. In a steady DC field,Both charge factors are negative for electrons, so the conductivity is positive. If $e$ instead denotes a positive charge magnitude, the force and current equations each require a minus sign and give the same result.

For AC response choose physical fields as real parts of phasors proportional to $e^{-i\omega t}$. Substitution givesThusThe real part decreases from ${\sigma }_0$ to zero. The imaginary part starts at zero, reaches ${\sigma }_0/2$ at $\omega \tau =1$, and falls as $1/\omega$. The real part falls as $1/{\omega }^2$ at high frequency. With an $e^{+i\omega t}$ convention, only the sign of the imaginary part changes. Positive $\mathrm{Re}\sigma$ gives positive cycle-averaged dissipation $⟨jE⟩=\mathrm{Re}\sigma |\hat{E}{|}^2/2$.

The supplied DC conductivity determines the transport time:Inertia matters when $\omega \tau \sim 1$, giving ${\omega }_c\simeq 4.02\times 10^{13}\mathrm{rad}{\mathrm{s}}^{-1}$, or ${\nu }_c={\omega }_c/(2\pi )\simeq 6.40\times 10^{12}\mathrm{Hz}$. Quoting the former number as hertz would introduce an erroneous factor of $2\pi$.