In short

Three related effects arise whenever electricity and heat meet in a conductor.

Seebeck effect. A junction between two different metals develops an EMF that depends on the temperature difference between that junction and a reference junction at a different temperature. For small temperature differences,

\boxed{\;\varepsilon_{S} \;=\; S_{AB}\,(T_{h} - T_{c})\;}

where S_{AB} is the Seebeck coefficient (or thermoelectric power) of the pair of metals A and B, in volts per kelvin. A copper–constantan thermocouple has S_{AB}\approx 41\ \muV/K at room temperature; a chromel–alumel (Type K) thermocouple, \approx 41\ \muV/K too; platinum–platinum–rhodium (Type R), \approx 10\ \muV/K.

Peltier effect. When a current I crosses a junction between two different metals, heat is absorbed or released at the junction at a rate

\dot{Q}_\text{P} \;=\; \Pi_{AB}\,I

where \Pi_{AB} is the Peltier coefficient in volts (joules per coulomb of charge passing). The sign depends on the direction of current flow — the same junction that heats up when I flows one way cools down when I reverses.

Thomson effect. Along a single conductor carrying a current with a temperature gradient along its length, heat is absorbed or released at a rate per unit volume

\dot{q}_\text{Th} \;=\; -\sigma_{T}\,I\,\frac{dT}{dx}

where \sigma_{T} is the Thomson coefficient of the material.

Kelvin relations — the three coefficients are not independent. Thermodynamics links them:

\Pi_{AB} \;=\; S_{AB}\,T,\qquad \sigma_{A} - \sigma_{B} \;=\; T\,\frac{dS_{AB}}{dT}.

The Peltier coefficient is just the Seebeck coefficient multiplied by absolute temperature; the Thomson coefficient is fixed once you know how S_{AB} varies with T. One physical phenomenon, three measurable faces.

Uses. Thermocouples are the workhorse temperature sensor in industry: they measure the exhaust of a CRYOGENIC UPPER STAGE engine at ISRO's Satish Dhawan Space Centre (4 K to 2000 K, no other sensor covers that range), the core coolant in a NPCIL PHWR reactor at Kaiga (up to 573 K pressurised heavy water), and the flame temperature in a LPG burner in a laboratory. Peltier modules (TEC coolers) are used in laser-diode stabilisers, laboratory temperature baths, and portable picnic coolers that plug into a car's 12 V socket.

Place the junction of two dissimilar wires — copper and iron, say — into the flame of a Bunsen burner in a school physics laboratory. Twist together the other two ends (the "reference junction") and dip them in a beaker of ice water. A sensitive millivoltmeter across any break in the loop reads a few millivolts. Hold a candle flame under the hot junction and the reading climbs. Plunge the hot junction back into cold water and the reading falls. Nothing chemical is happening. Nothing mechanical is moving. The cell is not a cell; there is no electrolyte, no battery, no generator spinning. Yet the voltmeter reads a real voltage, and if you close the circuit, a real current flows — a current powered entirely by the temperature difference between the two junctions.

This is thermoelectricity, discovered by Thomas Seebeck in 1821 when he noticed that a compass needle deflected near a closed loop of bismuth and copper with one junction warmed by his hand. The same effect, ruggedised and calibrated, is the thermocouple — the single most common temperature sensor in industry. Walk through the Satish Dhawan Space Centre in Sriharikota during a GSLV Mk III countdown and you will find hundreds of thermocouples threaded into the cryogenic upper stage, reading the liquid-hydrogen tank at 20 K and the combustion-chamber wall at 3000 K simultaneously. Walk through the reactor hall at the Kaiga nuclear plant in Karnataka and you will find thermocouples monitoring heavy-water coolant at 573 K, fed into the reactor's safety interlocks. These instruments work because two equations — the Seebeck EMF and its cousin the Peltier heat — connect temperature to voltage through a material's atomic structure.

This chapter builds the three thermoelectric effects from the starting point of free electrons in a metal feeling a temperature-dependent chemical potential, derives the Kelvin relations that link them thermodynamically, and shows you how to read a type K thermocouple's output to get a temperature in degrees Celsius to within a quarter of a degree.

Why a junction generates an EMF — the intuitive picture

In a metal, the free (conduction) electrons are not perfectly stationary even in the absence of an external field. They are constantly jostled by thermal agitation. At a higher temperature, the electron gas has higher average kinetic energy; at a lower temperature, lower kinetic energy. So far, this is just thermodynamics applied to free electrons.

Now put two different metals, A and B, in contact at a single point — a junction. The conduction electrons on both sides can tunnel across, settling into whichever metal offers a lower Fermi energy. Different metals have different Fermi energies because their lattices have different atomic cores, different electron densities, and different binding strengths. So at the instant of contact, electrons rush from the metal with higher Fermi energy (say, A) into the metal with lower Fermi energy (B). This transfer charges up metal B slightly negatively and leaves A slightly positive, setting up an electric field at the junction that opposes further flow. Equilibrium is reached when the electrostatic potential step across the junction exactly balances the Fermi-level difference:

e\,V_\text{contact} \;=\; E_\text{F,A} - E_\text{F,B}. \tag{1}

Why: the Fermi level is the electrochemical potential of the electron gas — its "pressure" for electrons. Electrons flow from higher Fermi level to lower, just as water flows from higher pressure to lower. The flow stops when the electrostatic potential step V_\text{contact} has lifted the energy of electrons in B by eV_\text{contact}, bringing the two Fermi levels into alignment. This is the Volta contact potential, and it exists at every dissimilar-metal junction, even in isolation.

So far, this gives a fixed voltage step at each junction that does not depend on temperature — if you simply touched a copper wire to an iron wire at room temperature, there would be a contact potential of a few tenths of a volt sitting across the interface. But you cannot measure it directly, because in any closed measuring loop you must go through the opposite junction too, and the two opposite steps cancel.

The crucial point — temperature-dependent Fermi levels

The Fermi level of a metal shifts with temperature. For most metals it shifts very slightly (a few kT, milli-electron-volts), but crucially, it shifts differently for different metals. Copper's Fermi level moves one way with temperature; constantan's (a Cu-Ni alloy) moves another way, at a different rate.

Now take the thermocouple loop: copper wire runs from terminal 1, joins constantan at the hot junction (at temperature T_h), constantan runs to the cold junction (at temperature T_c), joins another copper wire there, and that copper runs back to terminal 2.

Going around the loop, the electrostatic potential steps are:

Sum these:

\varepsilon_\text{loop} \;=\; \frac{1}{e}\Bigl[\bigl(E_{F,\text{Cu}}(T_h) - E_{F,\text{Con}}(T_h)\bigr) - \bigl(E_{F,\text{Cu}}(T_c) - E_{F,\text{Con}}(T_c)\bigr)\Bigr]. \tag{2}

Why: the loop integral of the electrostatic field is zero around any closed circuit in electrostatics — that is Kirchhoff's voltage law. But the loop integral of the thermoelectric field (the extra field created by Fermi-level differences) is not zero when the two junctions are at different temperatures. That non-zero circulation is the Seebeck EMF.

If both junctions are at the same temperature, the two brackets are identical and cancel — no net EMF. But if T_h \neq T_c, the brackets differ, and a real EMF appears across the open ends of the loop.

Linearising for small \Delta T

For small temperature differences, expand each Fermi-level difference linearly in temperature:

E_{F,\text{Cu}}(T) - E_{F,\text{Con}}(T) \;\approx\; E_{F,\text{Cu}}(T_0) - E_{F,\text{Con}}(T_0) \;+\; (T - T_0)\,\kappa_{AB},

where \kappa_{AB} is the rate of change of the Fermi-level difference with temperature. Substituting into (2) and noting that the constant parts cancel:

\varepsilon_\text{loop} \;=\; \frac{\kappa_{AB}}{e}\,(T_h - T_c).

Define the Seebeck coefficient S_{AB} \equiv \kappa_{AB}/e:

\boxed{\;\varepsilon_{S} \;=\; S_{AB}\,(T_h - T_c)\;} \tag{3}

This is the Seebeck law. The Seebeck coefficient has units of volts per kelvin. It is a material-pair property (different for copper-constantan, Cu-Fe, chromel-alumel, etc.), and in typical metal pairs its magnitude is in the tens of microvolts per kelvin. A 100 K difference with a 40 µV/K thermocouple gives 4 mV — a perfectly measurable voltage with a good digital multimeter.

Explore how the thermocouple voltage scales

The figure below is an interactive plot of the Seebeck voltage for a Type K (chromel–alumel) thermocouple as a function of the hot-junction temperature, with the cold junction fixed at the standard reference of 0°C (273.15 K). The Seebeck coefficient for Type K is approximately constant at 41 µV/K over most of the useful range.

Interactive: Type K thermocouple voltage versus hot-junction temperature A linear curve of Seebeck voltage in millivolts as a function of hot junction temperature in Celsius, with the cold junction at zero Celsius. Slope is 41 microvolts per kelvin. hot junction T_h (°C) Seebeck voltage ε (mV) 0 250 500 750 1000 1250 0 10 20 30 40 50 ε = 41 µV/K × (T_h − T_c) drag the red point along T_h
Drag the red point along the temperature axis. A Type K thermocouple reading 10 mV corresponds to a hot-junction temperature of about 244°C (with the cold junction in an ice bath). A reading of 40 mV puts the hot junction near 975°C — well above where an NTC thermistor would have melted. This wide range is why thermocouples dominate industrial temperature measurement.

The Law of Intermediate Metals — why you can solder on copper lead wires

A practical worry: the hot end of a thermocouple connects to a meter at room temperature, which is on a bench somewhere else. You are forced to join the thermocouple metals to ordinary copper wires to reach the meter. Do those extra copper junctions add extra EMFs and mess up the reading?

They do not, as long as both ends of any added copper stretch are at the same temperature. This is the law of intermediate metals: if you insert a third metal C into a thermocouple circuit between two existing junctions, and both new C-A and C-B junctions are at the same temperature, the thermocouple's net EMF is unchanged.

The proof is immediate from equation (2): inserting a symmetric piece of metal C adds equal-and-opposite Fermi-level steps that cancel. In practice this means the two screw terminals on the back of a digital thermometer — where you attach the thermocouple leads — must be at one common temperature, but you are free to use ordinary copper wiring everywhere else.

The Peltier effect — run it backwards

Seebeck's effect converts a temperature difference into a voltage. The reverse — driving a current through a junction to create a temperature difference — is called the Peltier effect, discovered by Jean Peltier in 1834.

Force a current I through a copper-bismuth junction. At one of the two junctions in the loop, the junction absorbs heat from its surroundings and cools slightly. At the other junction, heat is released and the junction warms. Reverse the current and the hot and cold junctions swap.

The rate of heat absorbed (or released) at a single junction is

\boxed{\;\dot{Q}_\text{P} \;=\; \Pi_{AB}\,I\;} \tag{4}

where \Pi_{AB} is the Peltier coefficient of the A–B junction, with units of volts (joules per coulomb). The sign convention: \dot{Q}_\text{P} is positive when heat flows into the junction from the surroundings as a positive current flows from A to B.

Why the Peltier coefficient equals S_{AB}\,T

The Seebeck and Peltier effects are the two faces of one physical process — electrons carrying entropy across a junction. When a current I crosses a junction, each electron carries not only its charge -e but also an entropy per electron that differs between the two metals. That entropy difference times the temperature is heat, and heat must be absorbed or released to conserve energy.

Thermodynamically (the argument is due to Lord Kelvin):

Step 1. Consider a tiny thermocouple loop with junctions at T and T + dT.

Step 2. Run a tiny current I around it, reversibly, for a short time \Delta t, so a charge q = I\Delta t circulates.

Step 3. The electrical work done per cycle is the loop EMF times the charge:

W \;=\; \varepsilon_{S}\,q \;=\; S_{AB}\,dT\,q.

Step 4. The heat absorbed at the hot junction is \dot{Q}_\text{P}\,\Delta t = \Pi_{AB}(T + dT)\,q. The heat released at the cold junction is \Pi_{AB}(T)\,q. The net heat absorbed is

\Delta Q \;\approx\; \frac{d\Pi_{AB}}{dT}\,dT\,q.

Step 5. For a reversible cycle, entropy change is zero:

\frac{Q_h}{T_h} + \frac{-Q_c}{T_c} \;=\; 0 \quad\Longrightarrow\quad \frac{\Pi_{AB}(T+dT)}{T+dT} \;=\; \frac{\Pi_{AB}(T)}{T}.

Rearranging, \Pi_{AB}/T = constant, so \Pi_{AB}(T) = S_{AB}(T)\cdot T when the reversible heat is matched to the reversible work. The full relation (allowing S_{AB} to depend on T) is:

\boxed{\;\Pi_{AB} \;=\; S_{AB}\,T\;} \tag{5}

Why: equation (5) is the first Kelvin relation. It says the Peltier coefficient is completely determined by the Seebeck coefficient — you do not need to measure it separately. Given that S_{AB} for copper-constantan is about 41 µV/K at T = 300 K, the Peltier coefficient is \Pi_{AB} = 41\times 10^{-6}\times 300 \approx 0.012 V — about 12 millivolts of heat-per-coulomb carried across the junction.

Applications of the Peltier effect

A commercial thermoelectric cooler (TEC) — the flat ceramic plate module you can buy at an electronics shop in Lamington Road, Mumbai — is a stack of many bismuth-telluride semiconductor junctions in series electrically, but in parallel thermally. Drive a few amps through the module and one face gets cold while the other gets hot. Typical performance: a 40 mm square module dissipating 30 W can cool one face to 15°C below the hot-face temperature.

Uses in India:

The Thomson effect — heat along a single wire

Within a single wire of uniform material carrying a current I with a temperature gradient dT/dx along its length, heat is absorbed or released in the bulk, not just at any junction. This is the Thomson effect, predicted by Lord Kelvin in 1854 and verified experimentally by 1867.

The volumetric heat rate is

\boxed{\;\dot{q}_\text{Th} \;=\; -\sigma_{T}\,I\,\frac{dT}{dx}\;} \tag{6}

where \sigma_{T} is the Thomson coefficient of the material (units: V/K). The sign depends on the sign conventions for I and dT/dx, and whether the carriers are electrons or holes.

The second Kelvin relation

A second thermodynamic argument (the same reversible-cycle logic, applied to a finite stretch of wire with a temperature gradient) gives the relation between the Thomson coefficients of the two metals in the thermocouple and the Seebeck coefficient:

\boxed{\;\sigma_{A} - \sigma_{B} \;=\; T\,\frac{dS_{AB}}{dT}\;} \tag{7}

Why: equation (7) is the second Kelvin relation. It says the difference of Thomson coefficients equals T times the temperature derivative of the Seebeck coefficient. For pairs where S_{AB} is very nearly constant with T (like Cu-constantan between room temperature and 200°C), dS_{AB}/dT \approx 0, so \sigma_{\text{Cu}} \approx \sigma_{\text{Constantan}} — nearly zero Thomson effect. For pairs where S_{AB} varies strongly (iron, for example), the Thomson effect is significant.

The point of the two Kelvin relations is that the three thermoelectric effects — Seebeck, Peltier, Thomson — are one phenomenon wearing three faces. If you measure the Seebeck coefficient S_{AB}(T) over a temperature range, you have determined everything: Peltier at any temperature is S_{AB}(T)\cdot T, and the Thomson coefficient difference is T\,dS_{AB}/dT. No independent measurement is needed.

Static diagram — the three effects on one loop

A thermocouple loop showing Seebeck, Peltier, and Thomson effectsA closed loop made of two wires, metal A on top and metal B on bottom, joined at two junctions. The left junction is hot at T_h, the right is cold at T_c. Current I flows around the loop. Arrows mark Seebeck EMF, Peltier heat at the junctions, and Thomson heat along the wires. metal A metal B T_h hot T_c cold I Π · I absorbed Π · I released Thomson heat along A Thomson heat along B ε_S = S_AB (T_h − T_c)
A thermocouple loop showing all three thermoelectric effects simultaneously. The Seebeck EMF $\varepsilon_{S}$ drives a current $I$ around the loop. At the hot junction, heat flows *into* the junction from the surroundings at rate $\Pi_{AB} I$ (Peltier absorption). At the cold junction, $\Pi_{AB} I$ worth of heat is *released* to the surroundings. Along each wire, the Thomson heat $-\sigma_{T} I \frac{dT}{dx}$ is absorbed or released in the bulk.

Worked examples

Example 1: A Type K thermocouple in a Kaiga reactor coolant loop

A Type K chromel-alumel thermocouple is inserted into the heavy-water coolant loop of a PHWR reactor at the NPCIL Kaiga plant. The cold junction is maintained at a reference ice-point (0°C). A digital voltmeter across the open ends reads V = 11.82 mV. Given that Type K has Seebeck coefficient S_{AB} = 41\ \muV/K, compute the coolant temperature in °C.

Type K thermocouple in reactor coolantA thermocouple loop with hot junction in reactor coolant pipe at temperature T, cold junction in an ice bath at 0°C, voltmeter across open ends reading 11.82 millivolts. coolant, T hot junction ice, 0°C cold junction V 11.82 mV chromel alumel chromel
A Type K thermocouple in a typical industrial configuration. The chromel leg (red) runs from the meter to the hot junction and back as a separate wire; the alumel leg (dark) bridges the two junctions. The voltmeter reads the Seebeck EMF across the open ends.

Step 1. Identify knowns. V = 11.82 mV = 11.82\times 10^{-3} V; S_{AB} = 41\times 10^{-6} V/K; T_{c} = 0°C = 273.15 K.

Why: a Type K thermocouple's sensitivity is given; the cold junction sits in an ice-water bath at 0°C by convention; the meter reads 11.82 mV. Three numbers are enough.

Step 2. Apply the Seebeck law (3) and solve for the temperature difference.

T_{h} - T_{c} \;=\; \frac{\varepsilon_{S}}{S_{AB}} \;=\; \frac{11.82\times 10^{-3}\ \text{V}}{41\times 10^{-6}\ \text{V/K}} \;=\; 288.3\ \text{K}.

Why: the ratio of the measured EMF to the sensitivity gives the temperature difference in kelvins directly. A kelvin step equals a Celsius step (they differ only in zero-point), so 288.3 K difference is also 288.3°C difference.

Step 3. Convert to the coolant temperature.

T_{h} \;=\; T_{c} + 288.3\ \text{K} \;=\; 0\text{°C} + 288.3\ \text{K} \;=\; 288.3\text{°C}.

Why: with the cold junction at 0°C, the hot junction is 288.3°C above zero — well inside the expected range for PHWR primary-coolant heavy water (~290°C at 10 MPa).

Step 4. Cross-check with the absolute scale.

T_{h} \;=\; 288.3 + 273.15 \;=\; 561.5\ \text{K}.

That matches the design coolant temperature at the Kaiga Unit-3 reactor (about 565 K at full power — a perfect match given typical 4 K calibration tolerance).

Result: The coolant temperature is T_{h} = 288.3°C, or 561.5 K.

What this shows: A 12 mV signal on a voltmeter, combined with a single material constant, locates the temperature deep inside the reactor's coolant pipe to within a kelvin — without any moving part, any chemical, or any contact between the electronics and the hot fluid. The thermocouple is intrinsically safe: its only connection to the reactor is through the sheathed end of the wire, and a melted thermocouple fails open (no reading), which the safety interlocks treat as a fault. That is why thermocouples remain the industry-standard temperature sensor half a century after the reactor was designed.

Example 2: A Peltier cooler to ice a bottle of water

A bismuth-telluride Peltier module with coefficient \Pi_{AB} = 0.20 V is run at I = 6.0 A in a car's 12 V socket. How much heat is pumped from the cold face in 10 minutes, and (ignoring conduction losses) how long would it take to chill a 500 mL bottle of water from 30°C to 5°C?

A TEC module pumping heatA rectangular Peltier module with cold face on top and hot face on bottom. An arrow labelled Q dot in points into the cold face from a water bottle, an arrow labelled Q dot out points out of the hot face into a fan. A 12 volt DC supply drives 6 amps through the module. water 30 → 5°C Q_in cold face hot face heat sink + fan 12 V I = 6 A Π_AB = 0.20 V Q̇_cold = Π · I = 1.2 W (heat pumped from water)
A 12 V, 6 A Peltier cooler. Heat $\dot{Q}_\text{in}$ is pumped from the water into the cold face; heat $\dot{Q}_\text{out}$ is dumped at the hot face plus the electrical power dissipated in the module ($VI$ = 72 W total). Only the Peltier component actively pumps; Joule heating inside the module must be carried away by the fan.

Step 1. Apply the Peltier formula (4) to compute the cooling rate at the cold face.

\dot{Q}_\text{P} \;=\; \Pi_{AB}\,I \;=\; 0.20\times 6.0 \;=\; 1.20 \text{ W}.

Why: this is the rate at which the Peltier effect alone pumps heat from the cold face to the hot face. It is an active heat transfer — heat moves against the natural temperature gradient, paid for by the electrical work.

Step 2. Heat pumped in 10 minutes.

Q \;=\; \dot{Q}_\text{P}\,t \;=\; 1.20\times 600 \;=\; 720 \text{ J}.

Why: t = 10 min = 600 s. The module pumps 720 J of heat from cold side to hot side in that interval. Meanwhile, the module also draws V I = 12 \times 6 = 72 W of electrical power, almost all of which ends up as heat at the hot face (Joule heating I^{2}R inside the module plus the pumped heat). The efficiency of Peltier cooling is modest — here, COP = 1.2/72 ≈ 0.017, versus COP of 3–4 for a household compressor fridge.

Step 3. Heat needed to cool 500 mL of water from 30°C to 5°C.

Water: m = 0.5 kg, specific heat c = 4186 J/(kg·K), \Delta T = 25 K.

Q_\text{needed} \;=\; m\,c\,\Delta T \;=\; 0.5\times 4186\times 25 \;=\; 5.23\times 10^{4} \text{ J} \;=\; 52.3 \text{ kJ}.

Step 4. Time required.

t \;=\; \frac{Q_\text{needed}}{\dot{Q}_\text{P}} \;=\; \frac{5.23\times 10^{4}}{1.20} \;=\; 4.36\times 10^{4} \text{ s} \approx 12.1 \text{ hours}.

Why: at 1.2 W of active pumping, chilling 0.5 L takes over twelve hours — which is why a 12 V Peltier car-cooler is marketed as "keeps cold things cold, won't freeze warm drinks". To actually make ice, you need dozens of modules in series or a much larger current, both of which become impractical for a car socket.

Result: The Peltier pumps 720 J in 10 minutes. Chilling 500 mL of water by 25°C takes about 12 hours of continuous operation with this module — assuming zero conduction leakage, which in reality roughly doubles the time.

What this shows: The Peltier effect is thermodynamically legitimate — it genuinely pumps heat from cold to hot — but it is not competitive with vapour-compression refrigeration for large heat loads. Where it shines is in small, silent, solid-state, fine-temperature-control applications: a laser diode, a CCD astronomical camera, a laboratory sample holder. The Kelvin relation \Pi = ST tells you that at room temperature with a decent Seebeck coefficient of a few hundred µV/K (semiconductors, not metals), \Pi can reach 0.1–0.3 V, which is what this example uses.

Common confusions

If you came here to understand what a thermocouple does and why a Peltier cooler works, you have it. What follows is the full Onsager-Kelvin treatment that derives the Kelvin relations from the deeper principle of microscopic reversibility, and a brief note on the modern thermoelectric materials programme.

The Onsager derivation — microscopic reversibility forces \Pi = ST

Lord Kelvin's 1854 derivation of equation (5) was a beautiful piece of reasoning, but he had to assume that the Seebeck and Peltier effects were reversible (that the only irreversible process in the circuit was the Joule heating I^{2}R, which could be separated from the thermoelectric contributions). A century later, Lars Onsager in 1931 proved the reversibility assumption from a much deeper principle — the symmetry of microscopic equations of motion under time reversal — and recovered the Kelvin relations as a rigorous consequence. Onsager got the 1968 Nobel Prize in Chemistry for this work.

The full argument works in the framework of irreversible thermodynamics. In a conductor with a current density \vec{j} and a heat flow \vec{q}, both driven by an electric field \vec{E} and a temperature gradient \nabla T, the linear phenomenological relations are:

\vec{j} \;=\; L_{11}\vec{E} + L_{12}\left(-\frac{\nabla T}{T}\right), \qquad \vec{q} \;=\; L_{21}\vec{E} + L_{22}\left(-\frac{\nabla T}{T}\right).

The Onsager reciprocity theorem states L_{12} = L_{21} — the off-diagonal coefficients are equal. In the thermoelectric context, L_{12}/L_{11} is the Peltier coefficient and L_{22}/(TL_{21}) is the Seebeck coefficient (times T). Substituting:

\Pi \;=\; \frac{L_{12}}{L_{11}} \;=\; \frac{L_{21}}{L_{11}} \;=\; ST,

which is equation (5). The Kelvin relation is not an accident of thermodynamic bookkeeping; it is a consequence of the time-reversal symmetry of Schrödinger's equation itself.

The figure of merit ZT and modern materials

For a good thermoelectric material, you want a large Seebeck coefficient S, high electrical conductivity \sigma, and low thermal conductivity \kappa. The dimensionless figure of merit is

ZT \;=\; \frac{S^{2}\sigma T}{\kappa}.

At ZT = 1, a TEC can cool by about 60 K; at ZT = 2, about 100 K. Bismuth telluride (Bi_{2}Te_{3}) has ZT \approx 1 at room temperature — that is the commodity TEC material.

India has a small but active thermoelectric-materials research programme. Groups at IIT Kharagpur, IISc Bangalore, and ARCI Hyderabad are developing nanostructured half-Heusler alloys, skutterudites, and \text{SnSe}-based materials, aiming at ZT > 2 for applications ranging from waste-heat recovery on industrial exhausts (LNG plants at Dahej, Gujarat) to body-temperature wearables that could power a wristwatch from the wearer's skin heat.

Seebeck coefficient sign and carrier type

Metals with positive Seebeck coefficients (iron, copper, gold) are called "p-type" in a loose sense — their Fermi surface geometry makes the thermoelectric response positive. Metals with negative Seebeck (bismuth, constantan — a Cu-Ni alloy specifically tuned to be negative) are "n-type". A standard thermocouple pairs a positive-S metal with a negative-S metal to maximise S_{AB} = S_{A} - S_{B}.

The same sign rule holds for semiconductors, which is why a Peltier module has alternating p-type and n-type bismuth-telluride legs wired in series electrically and in parallel thermally. Electrons carry heat from cold to hot in the n-leg; holes carry heat from cold to hot in the p-leg; both contributions add at the ceramic faces instead of cancelling out.

The Nernst-Ettingshausen effect — thermoelectricity in a magnetic field

Add a magnetic field perpendicular to the current and temperature gradient, and two more cross-effects appear: the Nernst effect (a temperature gradient \nabla T and field B produce an EMF in the third direction) and the Ettingshausen effect (a current I and field B produce a temperature gradient in the third direction). These are the thermoelectric cousins of the Hall effect. They are small in ordinary metals but large in topological materials and in graphene, and are active research areas at TIFR and IISc.

Where this leads next