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The chosen steady-state model isHere $U$ is the terminal voltage drop along positive electric current through the module. Its Seebeck effect term opposes the cooling supply. The cold Peltier effect term removes heat; Joule heating and returning heat conduction reduce the available cooling. At zero electric current, $Q_c=Q_h=-K\Delta T$: the passive leak warms c and removes heat from h, consistent with the stated signs. Subtracting the two face equations cancels heat conduction and gives the electrical power, an independent conservation of energy check.

The factor one half is an ideal-model result, not a universal rule for every contact. To see its origin, let $x$ run from cold to hot along a uniform equivalent leg of length $L$, cross-sectional area $A$, electrical resistivity $\rho$ and thermal conductivity $\kappa$. With constant coefficients, insulated sides and no Thomson effect, steady conservation of energy givesThe Fourier's law heat current in the positive $x$ direction is $-\kappa A{T}^{\prime }$. At $x=0$ it equals $-K\Delta T-I^{2}R/2$, and at $x=L$ it equals $-K\Delta T+I^{2}R/2$, using $K=\kappa A/L$ and $R=\rho L/A$. Adding the local transported heat current $\alpha TI$ gives the two face equations. Summing equivalent legs preserves this accounting. Nonuniform material properties, side losses or asymmetric contact dissipation require a revised model.

For an original toy module, stipulate $\alpha =0.040V/K$, $R=2.0\Omega$, $K=0.40W/K$, $T_c=280\mathrm{K}$ and $T_h=300\mathrm{K}$. At $I=3.0\mathrm{A}$,The coefficient of performance is defined here only for refrigeration with $Q_c>0$ and $P_{\mathrm{in}}>0$. The hot-side assembly must remove $37.0\mathrm{W}$, not just the extracted $16.6\mathrm{W}$. These are stipulated calculations, not measurements or a commercial module specification.

At the same fixed face temperatures but $I=-3.0\mathrm{A}$, the same boundary definitions give $Q_c=-50.6\mathrm{W}$, $Q_h=-35.0\mathrm{W}$, $U=-5.2\mathrm{V}$ and $P_{\mathrm{in}}=15.6\mathrm{W}$. The formerly cold face is now heated; the formerly hot reservoir supplies $35.0\mathrm{W}$. Still $Q_h-Q_c=15.6\mathrm{W}$. Reversal changes the linear terms, while $I^{2}R$ and the leak at fixed temperatures remain unchanged. Holding the old labels fixed makes this check meaningful.

The corresponding steady entropy production of the module is the outgoing reservoir entropy minus the incoming reservoir entropy:The Peltier effect terms cancel because they carry the same entropy $\alpha I$ at each end. This cancellation explains the reversible coupling; it does not make finite-resistance operation reversible. In the toy example the two positive contributions are approximately $0.06214$ and $0.001905W/K$, giving ${\dot{S}}_{\mathrm{gen}}=0.06405W/K$ for either current direction. In cooling operation, combining this inequality with the energy balance gives the Carnot refrigerator limit $\mathrm{COP}\le T_c/(T_h-T_c)$, equal to $14$ for these reservoir temperatures. A coefficient of performance above one would describe transported heat per electrical work, not the creation of energy.

Thermoelectric cooling uses the Peltier effect to remove heat from a load using electrical work. A typical thermoelectric module connects many pairs of different semiconductors electrically in a series circuit and thermally in parallel between two faces. For an idealized pair with $S_p>S_n$, positive conventional current crosses n to p at the cold bridge and p to n at the hot return. With common bridge material and isothermal contacts, the intervening metal contributions telescope: the net pair coefficient is $S_p-S_n$. For $N$ identical pairs define the module Seebeck coefficient $\alpha =N(S_p-S_n)>0$.

Choose fixed face temperatures $T_h\ge T_c>0$, $\Delta T=T_h-T_c$, module electrical resistance $R>0$ and open-circuit thermal conductance $K>0$. All three module parameters $\alpha ,R,K$ are constants over the chosen operating range. Positive $I$ is the orientation absorbing Peltier effect heat at c. Define $Q_c$ as net heat extracted from the cold reservoir per unit time, $Q_h$ as net heat delivered to the hot reservoir per unit time, and $P_{\mathrm{in}}$ as electrical power entering the module. Despite the conventional letters $Q_c,Q_h$, both denote power in watts, not accumulated energy. Ferrotec gives the corresponding single-stage equations and distinguishes module performance from bypass leaks.^[ref]

A thermometer measures temperature through a physical response whose relation to temperature is known. Examples include the length change from thermal expansion, a change in electrical resistance, and the pressure of a gas thermometer. If the response function $X=f(T)$ is an injective function over the working range, its measured value determines $T=f^{-1}(X)$ through the inverse function. Thermometer calibration establishes that relation for the particular instrument and measurement conditions. Two matching readings are meaningful only after calibration to a common scale and with the probes sampling comparable equilibrium states. ^[ref]

For a contact thermometer, the sensing part must reach thermal equilibrium with the region being measured. That region may differ from the rest of the object: an immersed probe near a heater can read a local temperature while the surrounding water remains cooler. A stable reading can also be a steady state set by competing heat transfer paths to the sample and the room. Good contact, adequate immersion, mixing where appropriate, and checks for spatial differences help establish which temperature the reading represents. These are measurement controls inferred from the thermal equilibrium requirement, not a guarantee that every flat trace is equilibrium.

A thermometer also changes the sample it touches. Consider an original toy calculation with an initially uniform sample at $T_o=350\mathrm{K}$, a probe at $T_p=300\mathrm{K}$, and constant whole-body heat capacities $C_o=100\mathrm{J}{\mathrm{K}}^{-1}$ and $C_p=1\mathrm{J}{\mathrm{K}}^{-1}$. Suppose together they form an isolated system, with no work or phase transition and negligible contact energy. The first law of thermodynamics and thermal equilibrium giveThe final thermometer reading underestimates the sample's original temperature by about $0.495\mathrm{K}$, although it correctly measures their new common temperature. The disturbance becomes small when the participating probe heat capacity is small compared with the sample's. A large controlled heat bath is a different limit: it supplies or removes heat while approximately maintaining its own temperature.

Waiting time is a separate issue from disturbance. Under a toy Newton's law of cooling model for a uniform probe in a maintained heat bath, a constant contact conductance $G$ givesThe time constant $\tau$ controls this exponential decay. For stipulated values $T_b=350\mathrm{K}$, $T_p(0)=300\mathrm{K}$ and $\tau =5\mathrm{s}$, the probe reads about $331.61\mathrm{K}$ after one time constant and $347.51\mathrm{K}$ after three. These are calculated values, not an instrument specification or measured data. Real probes may have multiple relaxation times and additional heat transfer paths; choose an equilibration criterion from the required accuracy, rather than assuming that a fixed waiting time works universally.

A useful original model treats the thermopile absorber as one approximately uniform temperature $T_h(t)$ with heat capacity $C_{\mathrm{th}}$, connected to a fixed reference temperature $T_c$ by thermal conductance $G_{\mathrm{th}}$. Both parameters are positive constants here. Let $P_{\mathrm{abs}}(t)$ be an added absorbed power above the initial background. Linearize the outgoing heat transfer about that background. With $\Delta T=T_h-T_c$ measured from an initial state having $\Delta T=0$, conservation of energy gives the first-order linear differential equationThe left side is the rate of stored energy; the two terms on the right are added power and outgoing heat transfer. The effective thermal conductance can include several paths, rather than just heat conduction along the wires. The model assumes a stable reference, negligible electrical loading, a common absorber temperature, and changes small enough for constant parameters.

For a step from zero to constant $P_0$ at $t=0$, subtract the steady solution $\Delta T_{\infty }=P_0/G_{\mathrm{th}}$. The difference obeys exponential decay, so the initial condition fixesThe time constant has units $(\mathrm{J}{\mathrm{K}}^{-1})/(\mathrm{W}{\mathrm{K}}^{-1})=\mathrm{s}$. With the same constructed $N=20$ and $\alpha =40\mu \mathrm{V}{\mathrm{K}}^{-1}$, choose $C_{\mathrm{th}}=0.02\mathrm{J}{\mathrm{K}}^{-1}$, $G_{\mathrm{th}}=0.01\mathrm{W}{\mathrm{K}}^{-1}$ and $P_0=0.03\mathrm{W}$. ThenAt one time constant, the response is $1-e^{-1}\approx 0.632$ of its final value: $V_{\mathrm{oc}}(2\mathrm{s})\approx 1.52\mathrm{mV}$. Reaching $95\%$ requires $t=-\tau \ln (0.05)\approx 5.99\mathrm{s}$. A voltmeter read too soon therefore underestimates the steady signal even when its electrical calibration is perfect.

The steady ratio $V_{\infty }/P_0=N\alpha /G_{\mathrm{th}}$ is $0.080\mathrm{V}{\mathrm{W}}^{-1}$ in this constructed model. Halving thermal conductance doubles this ratio but also doubles the time constant if heat capacity stays fixed. Reducing heat capacity alone speeds the response without changing its steady value. Increasing $N$ gives a larger signal only at fixed individual temperature differences: adding legs can also change thermal conductance, heat capacity and electrical resistance. The algebra identifies coupled design choices, rather than predicting a particular manufactured detector.

A finite electrical resistance produces Thermal electrical noise. For nearly uniform absolute temperature $T$ and effective measurement signal bandwidth $B$, its root mean square open-circuit voltage isHere $k_{\mathrm{B}}$ is the Boltzmann constant. Thermal electrical noise in the sensing element is an important contribution, and the signal bandwidth enters its voltage variance.^[ref] If identical pairs have fixed electrical resistance $r$ and fixed $\Delta T$, then $R_{\mathrm{pile}}=Nr$: the signal grows as $N$ while this noise grows as $\sqrt{N}$. The resulting signal-to-noise ratio grows only as $\sqrt{N}$ under those assumptions. Reference temperature drift, environmental electromagnetic radiation, and meter noise remain separate limits; reducing signal bandwidth trades fast information for less integrated electrical noise.

With ${ϵ}^{01}=+1$, the Noether density for the angular shift gains $\theta {\partial }_x\cos \varphi /(4\pi )$. Its integral gives the displayed topological shift. Mechanical charge $Q_0$ determines the classical rotating-kink energy, while canonical symmetry charge $Q_{\theta }$ is conjugate to the periodic angle and is integral after quantization. Reversing the epsilon convention reverses the theta sign.

Locally this density is a total derivative, so it changes neither the bulk field equations nor the energy. Its contribution to the angular canonical momentum is a spatial boundary derivative. With ${ϵ}^{01}=+1$, it produces the theta-angle shift of a sigma-model kink charge.

A vitamin also called vitamin B1. Its active coenzyme form supports reactions in carbohydrate and amino acid metabolism; deficiency therefore affects energy-demanding tissues rather than representing a simple shortage of dietary energy.

At fixed gradient and magnet length, the signed thin-lens focal length is proportional to particle momentum. For total energy error $\delta =(E-E_0)/E_0$, differentiating $p^2{c}^2=E^2-m^2{c}^4$ gives $d\ln p/d\ln E=1/{\beta }^2$. The ultrarelativistic result is $f=f_0(1+\delta )$ to first order. If energy means kinetic energy instead, the coefficient is $(K+m{c}^2)/(K+2m{c}^2)$.

Let a strong shock wave of radius $R$ and speed $V$ sweep a uniform mass density ${\rho }_0$ into a shell of mass $4\pi {\rho }_0{R}^3/3$ and speed $2V/(\gamma +1)$. If the cavity pressure is $\alpha$ times the immediate post-shock pressure, shell momentum balance gives $R\dot{V}=3(\alpha -1)V^2$. Constant total energy gives $R^3{V}^2=\mathrm{constant}$ and hence $\alpha =1/2$. Including shell internal and kinetic energy, and treating the cavity volume as $4\pi R^3/3$ to thin-shell accuracy, yields $R=\xi (E/{\rho }_0{)}^{1/5}{t}^{2/5}$ with ${\xi }^5=75(\gamma -1)(\gamma +1{)}^2/[16\pi (5\gamma -3)]$. This coefficient is approximate; resolving the shell's finite volume changes it.

Combine the two massless neutrinos in ${\tau }^{-}\to {\ell }^{-}{\bar{\nu }}_{\ell }{\nu }_{\tau }$ into $q$. In the parent rest frame, four-momentum conservation givesThe lower bound follows from the nonnegative Minkowski norm of the sum of two future-directed null four-vectors; the upper bound follows from $E_{\ell }^{\ast }\ge m_{\ell }$. Substituting gives the displayed energy bounds. They are kinematic: they do not specify the weak interaction decay amplitude or the probability density function inside the interval.

If the charged daughter is also massless, put $x=2E_{\ell }^{\ast }/m_{\tau }\in [0,1]$. A parent with Lorentz factor $\gamma$ and speed $\beta$ in natural units producesHere ${\theta }^{\ast }$ is the daughter angle relative to the boost direction in the parent rest frame. Two opposite parent momenta therefore need not produce two opposite visible daughter momenta: the invisible system can recoil differently in each particle decay.

For the uniform noninteracting electron model with two spin states and kinetic energy of nonrelativistic particles $E=p^2/(2m_e)$, count all available one-particle states up to an energy $E\ge 0$:This extensive density of states has units of inverse energy; $g(E)/V$ is the density of states per unit volume. Availability and occupation are different: $g(E)$ is positive above $E_F$, but at zero temperature those states are empty. The occupied energy distribution is $g(E)f(E)$, with the Fermi-Dirac distribution reduced to $f(E)=1$ below $E_F$ and $f(E)=0$ above it. The convention at the exact boundary does not affect a continuum integral.

With the dimensionless quantity $x=E/E_F$, the occupied count per $dx$, divided by the total particle number, is $3\sqrt{x}/2$ for $0<x<1$. Its area is one. Weighting this by $E$ gives the internal energy and its mean per particle:Thus the highest occupied energy and the mean energy differ. For the Fermi energy of a model electron gas, $U/N\simeq 1.0155\mathrm{eV}$. At fixed $N$, $E_F\propto V^{-2/3}$, so the derivative $P=-(\partial U/\partial V{)}_N$ gives the electron degeneracy pressure $P=2n{E}_F/5$. The same state-counting geometry therefore predicts a nonzero pressure at zero temperature.

For isotropic direct-gap parabolic bands, $m_r^{-1}=m_e^{\ast -1}+m_h^{\ast -1}$ and transition energy is $E_g+{\hslash }^2{k}^2/(2m_r)$. Differentiating the two-spin state count $k^3/(3{\pi }^2)$ per volume gives the displayed density for $E>E_g$, and zero below. Density per angular frequency is $\hslash g_j(\hslash \omega )$, not the same dimensioned quantity.

For a quantum rotor with angle period $2\pi$, let $u_m=e^{im\varphi }/\sqrt{2\pi }$ and prepare $\psi (\varphi ,0)=\sqrt{4/(3\pi )}{\cos }^2\varphi$. Its only nonzero angular-momentum amplitudes are $c_0=\sqrt{2/3}$ and $c_{\pm 2}=1/\sqrt{6}$. Their energies are $0$ and $2{\hslash }^2/I$, respectively, soThe state revives after $T=\pi I/\hslash$ and its density at $T/2$ is the initial density rotated by $\pi /2$. At $T/4$ there are four equal angular maxima. This is interference between a stationary component and a degenerate pair of counter-rotating components; it is not a uniformly rotating rigid profile. Angular-momentum and energy probabilities stay constant.

In a minimal hole description of a metal–ligand–metal path, two holes initially occupy separate metal orbitals and the ligand has no hole. Creating a virtual ligand hole costs a charge-transfer energy $\Delta$. Two hops generate an effective metal–metal hopping of order $t_{pd}^2/\Delta$; two more hops through metal double occupancy generate the antiferromagnetic scale in the title. Ligand double occupancy adds a channel with denominator $2\Delta +U_p$. Orbital geometry and intra-atomic exchange determine when the antiferromagnetic channel dominates.

In an orthonormal basis of localized atomic orbitals, a ring of three equivalent sites with real hopping $t$ has Hamiltonian $H=(E_0-t)I+tJ$, where $J$ is the three-by-three all-ones matrix. The normalized symmetric state $(|1⟩+|2⟩+|3⟩)/\sqrt{3}$ has energy $E_0+2t$. Its two-dimensional orthogonal complement has energy $E_0-t$. Thus the symmetric state is the unique spatial ground state for $t<0$, with an energy difference $3|t|$ to the excited eigenstates. Equivalently, the three discrete Bloch states have energies $E_0+2t\cos (2\pi l/3)$, $l=0,\pm 1$. Nonorthogonal atomic orbitals instead require a generalized eigenvalue problem.

For an original classical calculation of the return step, neglect the residual ion's Coulomb force after release. Take a spatially uniform driving electric field $E(t)=E_0\cos \omega t$, an electron of electric charge $-e$ and mass $m$, and zero initial position and velocity at $t_i$. With $\alpha =\omega t_i$ and $\beta =\omega t$, integrating Newton's second law givesA return requires $\beta >\alpha$ and $x(t)=0$. Define the cycle-averaged free-electron quiver kinetic energySearching the first-return solutions for release phases $0<\alpha <\pi /2$ gives a maximum near $\alpha =0.3134$ and $\beta =4.3990$, with $K_{\mathrm{return}}\simeq 3.1731U_p$. For a chosen ionization energy $I_p$, recombination into the initial state can therefore emit a photon with energy up to approximately $I_p+3.17U_p$ in this model. This is an upper return-energy scale, not a prediction of the emitted yield. It omits release probabilities, quantum interference, propagation, phase matching and depletion of the neutral gas.

The calculation explains why a long driving wavelength can increase the available return energy: at fixed $E_0$, $U_p$ grows as ${\omega }^{-2}$. It does not say that conversion efficiency improves. The electron must return with a probability amplitude for emission that survives the microscopic dynamics and adds coherently across the gas.

Consider an original hypothetical atom with a lower angular momentum multiplet $F=1$, $m=-1,0,+1$, and one opposite-parity excited level $F^{\prime }=0,m^{\prime }=0$. Stipulate that all spontaneous emission returns to these three lower quantum states, with no other decay channel. This is an isolated four-state model, not the actual sodium or rubidium D-line structure. In particular, the physical rubidium cell above contains multiple hyperfine structure levels.

The electric-dipole selection rule permits $\Delta F=-1$ here; ${\sigma }^{+}$ absorption with $\Delta m=+1$ excites only $m=-1$. Decay from $F^{\prime }=0$ has three equal integrated branching probabilities $b_{-}=b_0=b_{+}=1/3$: rotational symmetry gives equal squared Clebsch-Gordan coefficients. The emitted photons carry the three permitted angular momentum projections; equal integrated branching does not imply equal intensity in every observation direction. Thus $m=0$ and $m=+1$ are both dark. The chosen beam cannot prepare a unique pure $m=+1$ state.

Let $R\ge 0$ be the absorption rate of an atom in $m=-1$, and let $\Gamma$ be the excited-state spontaneous emission rate, both in inverse seconds. Assume $R,\gamma \ll \Gamma$, negligible stimulated emission, and no maintained quantum coherence between sublevels. Eliminating the short-lived excited population gives normalized lower populations $p_{-}+p_0+p_{+}=1$ to this approximation. At each scattering event the atom leaves $m=-1$ with probability $2/3$; the remaining $1/3$ simply returns it there.

Model depolarizing collisions and other population relaxation by resets to the unpolarized distribution at rate $\gamma \ge 0$. The rate equations areThis relaxation prescription assumes that Zeeman splitting is small compared with thermal energy, so the unpolarized distribution approximates thermal equilibrium. It is not an energy-independent relaxation law at arbitrarily low temperature.

Every transfer has a nonnegative rate: optical transfers $-\to 0$ and $-\to +$ each have rate $R/3$, and relaxation transfers between each distinct pair have rate $\gamma /3$. These define a continuous-time Markov chain. The three derivatives sum to zero, and at $p_i=0$ the corresponding derivative cannot be negative. Hence the model conserves population and preserves nonnegative populations. Optical cycles that return to $m=-1$ contribute to scattering even though they cause no net population transfer.

Independent stochastic, approximately fixed-noise and fractional-response contributions add in variance. The relative resolution is $A/\sqrt{E}\oplus B/E\oplus C$. Signal-counting and sampling fluctuations enter $A$; fixed equivalent energy noise enters $B$; residual response nonuniformity enters $C$. The powers approximate a fitted energy range rather than describing every leakage or Pile-up effect exactly.

For $f(y)=y-y^2+\gamma y^3/3$ with $0<\gamma <1$, the fold points of a critical manifold are $y_{\pm }=(1\pm \sqrt{1-\gamma })/\gamma$. When $\delta \ll ϵ\ll 1$, the fastest variable satisfies $z\simeq f(y)$ even during jumps of the intermediate variable. At fixed $x$, the outer branches of $x=f(y)$ attract the fast $y$ dynamics because $f^{\prime }(y)>0$; its middle branch repels. If $y_{-}<\alpha <y_{+}$, the slow equation $\dot{x}=\alpha -y$ drives the low branch to its fold point of a critical manifold and the high branch back to the other fold point of a critical manifold. Fast jumps at almost fixed $x$ close a relaxation cycle. This explains why a reduction justified on the slow arcs must also be checked on the jump time scale. For $\gamma >1$, the cubic is strictly increasing; the reduced energy $V=[x-f(\alpha ){]}^2/2+ϵ(y-\alpha {)}^2/2$ has $\dot{V}=-(y-\alpha )[f(y)-f(\alpha )]\le 0$, precluding this oscillation mechanism.

Tidal capture converts sufficient relative orbital energy into internal tidal heating to leave two initially unbound galaxies on a bound orbit. For identical spherical galaxies of mass $m_g$, a distant rapid flyby of impact parameter $p$ and relative speed $v$ gives the approximate capture condition $pv<[(32/3)G^2{m}_g^2⟨r^2⟩{]}^{1/4}$. This requires the impulse approximation and weak gravitational focusing; it is not a general criterion for subsequent coalescence.