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First construct the periodic response required in the introductory clause. At ${\omega }_0^2=1/(LC)$, the terms $L{I}_0^{\prime \prime }+I_0/C$ cancel for a sinusoid, leaving $R{I}_0^{\prime }={\omega }_0\cos ({\omega }_{0}t)$. ThusThis resonant series RLC response dissipates energy through Joule heating. Over one period,The Q factor, using the maximum magnetic energy $L{I}_M^2/2$, is consequently

Hot-Jupiter radius inflation is the excess radius of many irradiated gas giants relative to standard cooling models of the same mass, composition and age. A young gas giant starts with high entropy and a large radius, but ordinarily contracts through Kelvin-Helmholtz contraction. An old inflated object must either retain that thermal reservoir unusually well or receive power that affects the deep interior.

The first class is delayed cooling of an inflated giant planet. Two specific mechanisms are increased atmospheric opacity, which makes internal radiation escape less readily, and inhibited interior convection caused by composition gradients, potentially with layered transport. Both slow entropy loss. Stellar irradiation can also maintain an outer radiative blanket, but an appropriate irradiated boundary is already included in many baseline models and does not alone explain every extreme radius.

The second class is heating of an inflated giant planet. Two mechanisms are tidal heating from eccentricity, obliquity or other time-dependent tidal forcing, and Joule heating from wind-driven electric currents in an ionized atmosphere coupled to the magnetic field. Mechanical energy carried inward from atmospheric circulation is another candidate. Heating must be deposited sufficiently deeply and with sufficient power; superficial absorption that is promptly reradiated is not equivalent to deep interior heating. A circular synchronized isolated orbit does not automatically provide a persistent tidal source.

The classes can coexist and are tested through radius-age-irradiation trends, orbital properties and the required energy budget. Merely changing an observed transit altitude by a few atmospheric scale heights is generally distinct from inflating the bulk giant-planet interior.

Exoplanet transit photometry discovers a giant planet through a periodic flux decrement and measures $R_p/R_{\star }\simeq \sqrt{\delta }$. Stellar-radius estimates turn this into a planetary radius. The radial-velocity method discovers the stellar orbital reflex motion and constrains $M_p\sin i$. Follow-up exoplanet transit photometry is required to measure a geometric radius when such a planet also transits; a radial-velocity detection alone gives no direct size. Alternatively, exoplanet direct imaging finds young luminous giants, whose sizes are inferred less directly from luminosity, temperature, distance, and a planetary mass-radius relation or atmosphere model.

The two broad explanations for hot-Jupiter radius inflation are retention of primordial heat and addition of new interior power.

Delayed cooling of an inflated giant planet can arise from enhanced atmospheric opacity, which slows radiative escape; an irradiation-maintained radiative blanket, which insulates the convective interior; or compositional stratification and inefficient layered convection, which inhibit the outward transport of heat. These alter the rate of Kelvin-Helmholtz contraction.

Heating of an inflated giant planet can arise from tidal heating maintained by eccentricity or obliquity; Joule heating of currents driven by atmospheric winds through a magnetic field; or downward transport and dissipation of atmospheric mechanical energy generated by irradiation. To affect radius, the energy must be deposited at a depth and rate that changes the interior cooling balance. Simply absorbing starlight high in the atmosphere does not automatically supply deep heating. These are proposed mechanisms with different efficiencies, not six universally established contributions in every inflated planet.

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.

A Hertzian dipole is an ideal localized oscillating source much smaller than its radiation wavelength. Its electric version is an infinitesimal current element; its magnetic version is a small current loop with moment $\mathbf{m}=IA\mathbf{n}$. Such elements model the far-field radiation and radiation resistance of electrically small antennas. The approximation requires source dimensions much smaller than $c/\omega$.

The supplied magnetic field has a quasistatic $[m]/r^3$ term, an induction $[\dot{m}]/(c{r}^2)$ term and a transverse radiation $[\ddot{m}]/(c^{2}r)$ term. The electric field has induction and radiation terms. Brackets mean evaluation at the retarded time $t_r=t-r/c$, not a time average. Near terms mainly exchange stored energy; only the $r^{-1}$ fields give a nonvanishing outgoing power integral as $r\to \infty$.

For a fixed dipole axis the radiation fields satisfy $E_{\varphi }=-c{B}_{\theta }$. The radial Poynting flux is thereforeUsing $\int {\sin }^2\theta d\Omega =8\pi /3$ yieldsFor a general changing dipole direction, replace the squared scalar acceleration by $|\ddot{\mathbf{m}}{|}^2$.

Radiation resistance is the equivalent input resistance whose mean power equals radiated power, not ohmic heating. For a fixed small loop carrying $I(t)=I_0\cos \omega t$, $m=AI$, soThis uses peak current $I_0$; with rms current the power is $I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$.

The satellite loop needs its changing orientation as well as its induced current. Put the spin axis along z, the external field along x, and let $\mathbf{n}(t)=(\cos \omega t,\sin \omega t,0)$. With $A=\pi b^2$ and resistance $R=10^{-4}\Omega$, the resistance-only approximation givesWrite $m_0=B{A}^2\omega /R$. The full magnetic moment isIt has a static component and a rotating component at $2\omega$. Consequently $|\ddot{\mathbf{m}}{|}^2=4m_0^2{\omega }^4$, andHolding the dipole axis fixed while inserting the sinusoidal current would instead give $B^2{A}^4{\omega }^6/(12\pi {ϵ}_0{c}^5{R}^2)$, eight times smaller. That is the fixed-axis-loop result; it does not describe the stated rotating loop. This is frequency doubling of an induction-driven rotating loop; the changing vector direction is essential.

The resistance-only model is conditional: loop inductance is not supplied. If self-inductance L matters, solve $L\dot{I}+RI=BA\omega \sin \omega t$. In the steady state $I_0=BA\omega /\sqrt{R^2+{\omega }^2{L}^2}$. Its phase changes the static moment direction but not the twice-frequency component's magnitude, so the power becomesThe first expression is the requested resistive idealization and an upper estimate within this passive RL correction. Radiation reaction is negligible compared with the stipulated ohmic resistance. The radiated energy comes from rotational kinetic energy: the induced magnetic torque brakes the loop. Most lost rotational energy is Joule heating; a small part escapes as radiation. Maintaining a constant spin would require external mechanical power.

For scale only, take $B=50\mu \mathrm{T}$, a representative near-Earth field rather than a supplied orbital measurement. The terrestrial surface range reported by NOAA is roughly $25$–$65\mu \mathrm{T}$; the actual orbital value depends on position.^[ref] With $b=1\mathrm{m}$ and $\omega =2\pi (500{\mathrm{s}}^{-1})$, the resistance-only model givesInductance would reduce this power. The kilohertz signal is exceptionally weak and a metre-scale receiving antenna is electrically tiny. Ionospheric absorption/reflection and strong natural or instrumental low-frequency backgrounds make ordinary ground reception impractical. This is not a theorem that the ionosphere blocks every low-frequency mode: field-guided propagation and condition-dependent leakage are real, as the VIPER mission's measurements were designed to test.^[ref]

At a separation $r=100\mathrm{km}$, $k_{\mathrm{rad}}r\simeq 2.1$, so the observer is not securely in the far field. The radiation term alone has at most the characteristic magnetic amplitudein this optimistic resistance-only vacuum model. Near and induction fields must be retained for a full prediction; a simple far-field inverse-square power estimate is not an exact received signal here. Another satellite avoids the whole ground ionospheric path, but plasma propagation, receiving sensitivity, bandwidth, orientation, integration time and spacecraft noise remain unspecified. Detection is not forbidden in principle, but routine practical detection is implausible at this scale; a definitive instrument-level yes/no does not follow from the supplied data.

antenna quantities. radiation resistance represents emitted power by $P_{\mathrm{rad}}=I_{\mathrm{rms}}^2{R}_{\mathrm{rad}}$ at the feed. It is not Joule heating in the antenna; a separate loss resistance accounts for that. antenna gain is the directional far-field intensity relative to an isotropic radiator supplied with the same input power:It includes radiation efficiency and equals antenna directivity only for a lossless antenna. antenna effective area is defined by the power available to a matched load, $P_{\mathrm{av}}=A_{\mathrm{eff}}{S}_{\mathrm{inc}}$, for incidence from a specified direction and matched polarization. It need not equal a literal geometric collecting area.

Gain–area relation. To derive the antenna gain-effective-area relation, take a passive reciprocal antenna. Its electromagnetic reciprocity makes the receive pattern proportional to the transmit pattern; write $A_{\mathrm{eff}}(\Omega )=KG(\Omega )$. The normalization follows from thermal equilibrium. In the Rayleigh-Jeans law limit, blackbody radiation radiance per unit frequency per unit solid angle is $2k_{B}T/{\lambda }^2$ for two polarizations. A single antenna polarization couples to one half, so the available incident noise power isThe open-circuit Johnson voltage spectrum of the antenna resistance is $4k_{B}T{R}_{\mathrm{rad}}d\nu$. A matched load receives half the open-circuit voltage, giving available power $k_{B}Td\nu$. In a lossless antenna at thermal equilibrium these powers agree; hence $\int A_{\mathrm{eff}}d\Omega ={\lambda }^2$. Since $\int Gd\Omega =4\pi$,The same efficiency factor multiplies receiving area and gain if reciprocal dissipative losses are included consistently. Impedance mismatch or polarization mismatch reduces delivered power further. The factor of two for thermal polarization modes and the factor of four between open-circuit voltage variance and matched-load power are separate effects.

Normalize the radar beam. Interpret the azimuth as signed about the beam centre at $\varphi =0$, so both sides of the narrow lobe are included. Put $u=\theta -\pi /2$. To leading order in the small angular width, $\sin \theta \simeq 1$ and the angular integrals can be extended over the real line:The Gaussian antenna beam solid angle gives the peak on the positive $x$-axis consequently hasHere $N$ is power per solid angle, equal to $r^2{S}_r$ in the far field. The leading neglected solid-angle correction is relative order ${\alpha }^2$.

Return using the supplied electric-dipole response. The electric-dipole radar cross section of a small conducting sphere determines this return. At target range $D$, the incident intensity is $S_t=C/D^2$. Write its root mean square electric field as $E_t$, with $S_t={ϵ}_{0}c{E}_t^2$. The induced electric dipole is $p_{\mathrm{rms}}=4\pi {ϵ}_0{a}^3{E}_t$. A dipole's angular radiation pattern, normalized by $\int {\sin }^2\chi d\Omega =8\pi /3$, gives $d{P}_{\mathrm{sc}}/d\Omega =(3P_{\mathrm{sc}}/8\pi ){\sin }^2\chi$. The backward direction is perpendicular to the induced dipole, so the supplied total dipole power givesThe received intensity is this angular power divided by $D^2$. Using the monostatic radar equation, or directly multiplying by the receive area, givesThe receiver threshold is a voltage across the matched load, so the detectable power is $V_m^2/R$, without another matching factor:This is the electric-dipole result following from the response data provided. It assumes $ka\ll 1$, far-field propagation on each leg, alignment at the beam maximum and no attenuation or additional receiver losses.

Perfect-conductor qualification. perfect-conductor sphere dipole interference arises because the sphere also excludes the oscillating magnetic field. Solving the magnetic boundary problem gives $\mathbf{m}=-2\pi a^3{\mathbf{H}}_t$ in addition to the electric dipole. The magnetic polarizability of a perfectly conducting sphere follows from its induced surface currents.^[ref] In the backward direction the magnetic contribution has half the electric scattering amplitude and the same sign, so the returned intensity is $9/4$ of the electric-only value. Equivalently, the monostatic radar cross section is ${\sigma }_{\mathrm{back}}=9\pi k^4{a}^6$, rather than $4\pi k^4{a}^6$. Including both leading dipoles therefore givesBoth results have been distinguished because the supplied auxiliary data describe only the electric dipole, while the literal perfect-conductor boundary conditions also generate the magnetic one. The backward interference is a leading-order effect, not a higher-order correction in $ka$.^[ref]

At terminal velocity, the upward magnetic force balances the weight. Substitute the electric current from the preceding solution:The geometric factors cancel, givingHere $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $B=1\mathrm{T}$, ${\rho }_m=8960\mathrm{kg}{\mathrm{m}}^{-3}$ and ${\rho }_c=1.7\times 10^{-8}\mathrm{\Omega }\mathrm{m}$. As numerical checks, the specified $a=0.08\mathrm{m}$ and $d_w=0.001\mathrm{m}$ give $M=2.252\times 10^{-3}\mathrm{kg}$, $R=6.926\times 10^{-3}\mathrm{\Omega }$ and $I_s=0.276\mathrm{A}$.

The gravitational power lost, $Mg{v}_s$, equals the Joule heating $I_s^{2}R$, consistent with energy conservation. This terminal velocity applies only while the top arm remains in the magnetic field; after it exits, the magnetic braking ceases, apart from any brief inductive transient.

Assume a passive linear electrical circuit in sinusoidal steady state, with transients absent and $V_0$ the peak voltage. Write $Z=|Z|e^{i\varphi }=R+iX$. The electric current is $I(t)=I_0\cos (\omega t-\varphi )$, with $I_0=V_0/|Z|$. Average the instantaneous electric power $V(t)I(t)$ over a full period. The double-frequency term averages to zero, leavingThereforeEquivalently, root mean square values give $⟨P⟩=V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi$. The reactance exchanges stored electric field or magnetic field energy with the source, whereas the real part produces Joule heating. A purely imaginary complex impedance dissipates no mean electric power under these ideal assumptions.

The magnetic dipole moment is $\mathbf{\mu }=IA\hat{\mathbf{n}}$ and its magnetic torque is ${\mathbf{\tau }}_{\mathrm{mag}}=\mathbf{\mu }\times \mathbf{B}$. Its component along the rotation axis opposes the imposed rotation. For positive $\omega$, the external driving torque needed to keep the angular velocity constant isAveraging over a complete rotation, with $⟨{\sin }^2\omega t⟩=1/2$, givesAs an independent sign and dimension check, $\omega ⟨{\tau }_{\mathrm{drive}}⟩=R⟨I^2⟩$. Thus the mechanical input power equals the mean Joule heating; a passive loop cannot supply a sustained driving torque. This average assumes the same negligible-self-inductance approximation as the current expression.

During switch-on, the Joule heating in $R_2$ has instantaneous power $P_2=R_2[V_0/(R_1+R_2){]}^2{e}^{-2t/\tau }$. Integrating the finite transient givesAlthough the source continues supplying power to $R_1$ indefinitely, $R_2$ is eventually shorted and its heat integral converges. It would be incorrect to identify this heat with the final stored magnetic energy, because the source and $R_1$ also participate during switch-on.

Opening the switch removes the source and $R_1$ but leaves a closed path through the inductor and $R_2$. Set the new time origin at switch-off. Inductor current continuity gives $i_L(0)=V_0/R_1=10\mathrm{A}$. Since the two parallel branches are now isolated, Kirchhoff's current law gives $I_2=-i_L$, and $L{\dot{i}}_L=-R_2{i}_L$. ThusAll the initially stored magnetic energy is ultimately converted to Joule heating in $R_2$:Direct integration of $R_2{i}_L^2(t)$ gives the same result. The initial shared voltage is $v(0^{+})=-R_2{i}_L(0)=-1000\mathrm{V}$; its reversal drives the return current. The ideal circuit permits this transient voltage; real switching would also be limited by stray capacitance, breakdown and other parasitic effects.

Orient the ring's area normal with positive external magnetic flux. Faraday's law gives the applied electromotive forceThe ring's self-inductance opposes changes in its electric current. With positive electrical resistance $R$, Kirchhoff's voltage law therefore gives $L\dot{I}+RI={\mathcal{E}}_{\mathrm{ext}}$. Define $\varphi ={\tan }^{-1}(\omega L/R)$. The periodic steady state isThe second result follows from Joule heating, $P=R{I}^2$, and $⟨{\sin }^2⟩=1/2$. For unspecified initial conditions, add a transient $I_h=C\exp (-Rt/L)$; the displayed time average is over the periodic steady state after that transient has decayed. The inductor temporarily stores and returns energy, with no net storage change over one complete cycle.

The component of magnetic force along the slope has magnitude $IBs\cos \varphi$. The rate at which the sliding bar does work against this magnetic braking isBy Ohm's law, $\mathcal{E}=IR$, soThis exactly equals the Joule heating in the resistor in the negligible-self-inductance approximation. The normal force does no work because it is perpendicular to the velocity.

The gravitational power is instead $mgv\sin \varphi$. By the work-energy theorem,It equals the Joule heating only at constant speed. For a concrete counterexample, during the initial acceleration the kinetic energy increases, so some gravitational power remains mechanical.

With phasors proportional to $e^{i\omega t}$, the resistor, inductor, and capacitor have complex impedances $R$, $i\omega L$, and $1/(i\omega C)$. In a series circuit they carry the same electric current, so their voltage drops add:The real part describes Joule heating; the imaginary part changes sign between capacitive and inductive response, passing through zero at ${\omega }_0=1/\sqrt{LC}$.

For the imposed electric current $I=I_0\cos \omega t$, Faraday's law gives the self-induced electromotive forceThe externally applied terminal voltage, with the passive sign convention, has the opposite sign:Thus the terminal voltage leads the electric current by $\pi /2$. The instantaneous input power isAn ideal inductor alternately stores magnetic energy and returns it to the source. Real winding electrical resistance would add mean Joule heating $I_0^2{R}_w/2$; the ideal zero result assumes that loss is neglected.

The single secondary turn links the primary magnetic flux ${\mu }_{0}n\pi a^2{I}_p$, so its mutual inductance with the primary isChoose positive winding directions so that positive secondary electric current would reinforce the primary magnetic flux. Faraday's law and Kirchhoff's voltage law then giveAfter the transient proportional to $e^{-Rt/L_s}$ has decayed, put $\alpha ={\tan }^{-1}(\omega L_s/R)$. The steady electric current isIts wave amplitude is $\omega M{I}_0/\sqrt{R^2+(\omega L_s{)}^2}$ and it lags the induced electromotive force by $\alpha$. Relative to the primary $I_0\cos \omega t$, its phase shift is $-\pi /2-\alpha$, modulo $2\pi$. Reversing the chosen secondary winding direction adds $\pi$ to this signed phase shift and leaves all power predictions unchanged.

The lamp's mean Joule heating isFor $\omega L_s\ll R$ the response is nearly in phase with the induced electromotive force; for $\omega L_s\gg R$ the secondary electric current nearly opposes the primary electric current in the stated winding convention.

Position A disconnects the battery and closes the resistor-inductor loop. Taking clockwise electric current as positive, Kirchhoff's voltage law for this RL circuit givesThe electric current through an ideal inductor remains continuous across a switch with no impulsive voltage. Thus $I(0^{+})=I_0$, and separation of variables givesIf the switch remains at A, the same result holds for every $t>0$. The time constant is $L/R$; the stored magnetic energy is dissipated by Joule heating.

Measure an unwanted effect separately or design a comparison in which it cancels. Calibration against reference standards at several settings tests zero offset, gain and nonlinearity. Blanks reveal background. Reversal is useful when the wanted response is odd and the nuisance is even: $D_{\pm }=\pm S+b$ gives $S=(D_{+}-D_{-})/2$ and $b=(D_{+}+D_{-})/2$. For example reversing current in a resistance measurement cancels a stable thermoelectric voltage, but does not cancel the current-scale error or even-in-current Joule heating.

Interleave reversals to detect drift; interchange sensors or leads to expose asymmetry. Null or bridge methods reduce dependence on detector gain. Alignment checks, shielding and temperature control target specified physical couplings. Independent measurement methods provide a stronger test than repetition of the same setup. Repeated readings reduce independent random errors, not a shared bias. Correct significant identified effects and propagate the uncertainty of the correction and standards; complete removal of an unknown systematic error cannot generally be certified.^[ref]

A systematic error is a reproducible displacement caused by the apparatus, procedure or measurement model. Repeating the same experiment can determine its mean very precisely while retaining that displacement. Reducing random measurement error therefore does not establish accuracy.

Start with a measurement model listing the response, zero offset, calibration scale and relevant environmental corrections. Compare the instrument with traceable standards at several points across its working range; a single zero check cannot reveal a gain error or nonlinearity. Measure blanks and backgrounds using the same acquisition chain as the signal. Repeat calibration before and after the run to detect drift, and retain the uncertainty of every applied correction.

systematic-error cancellation by reversal separates a wanted effect from an offset with a different symmetry. For example, a resistance measurement with thermoelectric offset $V_0$ gives $V_{+}=RI+V_0$ and $V_{-}=-RI+V_0$. The difference givescancelling a stable offset, while $(V_{+}+V_{-})/2$ diagnoses it. Alternate the current direction quickly enough that the offset is approximately unchanged. This does not cancel a current-scale calibration error or resistance changes from Joule heating; vary the current magnitude and monitor temperature to test those effects. Control measurements must target the actual proposed bias.

Randomizing the order of settings prevents slow drift from consistently imitating a physical trend. Interleaving a reference measurement helps track that drift. Reversing geometry, exchanging matched detectors and changing alignment can expose effects tied to one part of the apparatus. Repeating the measurement with an independent method is particularly valuable when its dominant biases differ; agreement between two instruments sharing the same calibration does not test a common calibration error.

Apply justified corrections, propagate their uncertainties and include correlations when the same calibration enters several observations. A systematic component shared by all observations does not shrink as $1/\sqrt{N}$. A measurement uncertainty budget should distinguish the physical origin of each effect from the method used to evaluate its uncertainty: statistical Type A and other Type B evaluations do not simply mean random and systematic, respectively.^[ref]

In a metal, a partially filled energy band or overlapping energy bands supplies electrons with nearby unoccupied states. At thermal equilibrium, oppositely directed group velocities are equally populated and the net electric current vanishes. An applied electric field distorts the occupation distribution, producing a small mean drift velocity. A filled energy band alone has no such low-field intraband response because its velocity contributions cancel over the entire Brillouin zone.

In the Drude model, let $n$ be the mobile electron number density, let $e>0$ be the elementary charge, and let $\tau$ be the momentum relaxation time. The mean motion obeysThe steady state is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m$. The current density is opposite to the electron drift velocity:This gives microscopic Ohm's law, with electrical conductivity $\sigma$ in $\mathrm{S}{\mathrm{m}}^{-1}$ and electrical resistivity ${\rho }_{\mathrm{el}}$ in $\Omega \mathrm{m}$. For a uniform wire of length $\ell$ and cross-sectional area $A$, its electrical resistance is $R={\rho }_{\mathrm{el}}\ell /A$. The dissipated power per volume is $\mathbf{J}\cdot \mathbf{E}=\sigma E^2$, the Joule heating supplied by the external source.

The free electron model with Fermi-Dirac statistics improves the microscopic picture. The Pauli exclusion principle fills states up to the Fermi energy at low temperature. For a spherical, parabolic energy band with two spin states,Only states near the Fermi surface change their occupation in a weak-field response; the large equilibrium Fermi velocity is not the small drift velocity. The same $n{e}^2\tau /m$ formula results for this isotropic parabolic model when $\tau$ is evaluated for transport near the Fermi surface. In a crystal, the energy band group velocity, effective mass and possibly direction-dependent scattering replace the simplest bare-particle picture. The mean free path is approximately ${\ell }_{\mathrm{mfp}}=v_F\tau$.

A perfectly periodic static lattice is already included in Bloch states and does not by itself give a finite electrical resistivity. Momentum relaxation comes from defects, impurities, boundaries, and electron-phonon scattering. Independent scattering mechanisms approximately add their rates, giving Matthiessen's rule,The approximately temperature-independent ${\rho }_0$ is the residual resistivity. In ordinary simple metals, the phonon contribution is approximately proportional to temperature at high temperature; a three-dimensional acoustic-phonon transport model gives a $T^5$ contribution well below the relevant Debye temperature. Other mechanisms can dominate in particular materials or regimes, so neither power law is universal.

The classical Drude model incorrectly treats the equilibrium carrier statistics. Fermi-Dirac statistics explains the small electronic heat capacity, while electrons carrying both charge and heat give the approximate Wiedemann-Franz law ${\kappa }_e/(\sigma T)={\pi }^2{k}_B^2/(3e^2)$ for degenerate carriers with appropriate nearly elastic scattering. Both the simple electrical conductivity formula and this thermal relation depend on the assumptions about energy bands and scattering, rather than being exact for all metals.