Onsager Reciprocal Relations

We now derive Onsager's reciprocal relations from microscopic statistical mechanics. Linear irreversible thermodynamics introduces the transport equations Ji=∑jLijXjJ_i=\sum_j L_{ij}X_j, but leaves the symmetry Lij=LjiL_{ij}=L_{ji} unexplained. Our goal is to show that this symmetry follows directly from equilibrium fluctuations and microscopic time-reversal invariance. Here we do not re-derive the forces-and-fluxes bookkeeping; instead we mount the stochastic-fluctuation route that Onsager himself used, and make it fully explicit on a concrete Gaussian model. The organizing idea is the regression hypothesis: a spontaneous equilibrium fluctuation, once it has happened, decays on average by the same linear law that governs a macroscopically prepared deviation. This turns the reciprocity theorem into a statement about equilibrium time-correlation functions. Microscopic reversibility — the invariance of Hamiltonian dynamics under t→−tt\to-t — makes the correlation matrix Cij(t)=⟨ai(t)aj(0)⟩C_{ij}(t)=\langle a_i(t)a_j(0)\rangle symmetric under exchanging the two variables, and the regression law then transfers that symmetry directly onto the transport matrix LL. We compute the whole chain symbolically on a two-variable fluctuating system: the Einstein covariance ⟨aiak⟩=kB(s−1)ik\langle a_ia_k\rangle=k_B(s^{-1})_{ik}, the force-orthogonality ⟨aiXj⟩=−kBδij\langle a_iX_j\rangle=-k_B\delta_{ij}, the short-time correlation slope C˙ik(0)=−kBLik\dot C_{ik}(0)=-k_BL_{ik}, and finally Lik=LkiL_{ik}=L_{ki} from the time-reversal symmetry of CC. An asymmetric LL is shown to break correlation symmetry — the diagnostic that turns reciprocity into something measurable. We close with the numeric Kelvin relation Π=T S\Pi=T\,S (a metal's Peltier coefficient from its Seebeck coefficient) and the Onsager–Casimir field flip Lij(B)=εiεjLji(−B)L_{ij}(\mathbf B)=\varepsilon_i\varepsilon_jL_{ji}(-\mathbf B), whose signature is the antisymmetry of the Hall conductivity.

Key takeaways
  • Reciprocity Lij=LjiL_{ij}=L_{ji} is not thermodynamics — it is a fossil of microscopic time-reversal invariance, read off from equilibrium fluctuations via the regression hypothesis.
  • The regression hypothesis says spontaneous fluctuations regress by the macroscopic law a˙i=∑jLijXj\dot a_i=\sum_jL_{ij}X_j; this links LL to the correlation matrix Cik(t)=⟨ai(t)ak(0)⟩C_{ik}(t)=\langle a_i(t)a_k(0)\rangle.
  • The short-time slope of CC is C˙ik(0)=−kBLik\dot C_{ik}(0)=-k_BL_{ik}; time-reversal symmetry Cik(t)=Cki(t)C_{ik}(t)=C_{ki}(t) then forces Lik=LkiL_{ik}=L_{ki} — derived here explicitly on a 2-variable Gaussian model.
  • Only the symmetric part of LL enters entropy production; the second law alone permits an antisymmetric part, so reciprocity is genuinely extra information supplied by time reversal.
  • Consequences: the second Kelvin relation Π=T S\Pi=T\,S (Peltier == temperature ×\times Seebeck), the Soret–Dufour symmetry, and the Onsager–Casimir field flip that makes Hall conductivity antisymmetric in B\mathbf B.
  • Key idea: the regression hypothesis converts a theorem about transport coefficients into one about equilibrium correlation functions. Microscopic time-reversal symmetry makes those correlations symmetric, and that symmetry transfers directly to the transport matrix, Lij=LjiL_{ij}=L_{ji}.
The core insight

The regression hypothesis converts a theorem about transport coefficients into a theorem about equilibrium correlation functions. Reciprocity is the statement that the equilibrium correlation matrix Cik(t)=⟨ai(t)ak(0)⟩C_{ik}(t)=\langle a_i(t)a_k(0)\rangle is symmetric under i↔ki\leftrightarrow k — a fact guaranteed by microscopic time-reversal invariance — and the regression law C˙ik(0)=−kBLik\dot C_{ik}(0)=-k_BL_{ik} transfers that symmetry directly onto LL. An asymmetric LL would make a movie of equilibrium fluctuations distinguishable from its time-reverse.

Where this lives today

The regression route is not merely historical—it is the standard computational route to transport coefficients. In equilibrium molecular-dynamics simulations one measures the flux autocorrelation functions ⟨Ji(0)Jj(t)⟩\langle J_i(0)J_j(t)\rangle, integrates them using the Green--Kubo formulas, and thereby obtains the transport matrix without ever applying an external driving force. Onsager reciprocity reduces statistical uncertainty by relating the two cross-correlation functions. The same framework underlies the fluctuation--dissipation theorem, linear-response theory, and modern transport calculations throughout condensed-matter physics. In magnetic systems the Onsager--Casimir relations distinguish dissipative transport from Hall responses by separating the symmetric and antisymmetric parts of the conductivity tensor.

The physical picture

Imagine watching two coupled fluctuating quantities in a system at equilibrium—for example, a small temperature imbalance and a small charge imbalance. Thermal noise continually drives both quantities away from equilibrium, after which they relax back. Onsager's regression hypothesis states that if one averages over many realizations that begin with the same spontaneous fluctuation, the subsequent decay obeys exactly the same linear equations as the relaxation of a deliberately prepared macroscopic disturbance. This identification is the bridge between equilibrium fluctuations and transport coefficients, and it is the starting point of Onsager's proof.

Figure
Two coupled fluctuations regressing: conditioned decay of a₁ (blue) drags a₂ (amber) by the SAME L that governs macroscopic relaxation
time t → fluctuation aᵢ(t) a₁(0) = large conditioned fluctuation a₂ dragged up then relaxes — coupling L₁₂ both → 0 (equilibrium) ⟨a₁(t)a₁(0)⟩ ∝ e^(−λt)

Microscopic reversibility is the extra input. A movie of equilibrium fluctuations, run backward, is statistically indistinguishable from the forward movie — because the underlying Hamiltonian dynamics is invariant under t→−tt\to-t (with momenta reversed). This makes the cross-correlation even in a specific sense: ⟨a1(t)a2(0)⟩=⟨a2(t)a1(0)⟩\langle a_1(t)a_2(0)\rangle=\langle a_2(t)a_1(0)\rangle. It is this symmetry — of the correlation matrix under exchanging which variable you read first — that becomes reciprocity.

Figure
Punchline: correlation-matrix symmetry is the reciprocity thesis — symmetric L (blue, curves coincide) vs the asymmetric-L outlier (red, curves split)
time t → cross-correlation ⟨a₁(t)a₂(0)⟩ = ⟨a₂(t)a₁(0)⟩ (symmetric L ⇒ coincide) asymmetric L: ⟨a₁(t)a₂(0)⟩ ≠ ⟨a₂(t)a₁(0)⟩ (SPLIT — forbidden by time reversal)

A useful picture is to regard the thermodynamic forces as the local slope of the entropy landscape and the fluxes as the resulting motion down that slope. Onsager reciprocity states that the linear coupling between force ii and flux jj is identical to that between force jj and flux ii, making the transport matrix symmetric. In a magnetic field, however, time reversal also reverses the field itself, so the symmetry generalizes to the Onsager--Casimir relation

Lij(B)=ϵiϵjLji(−B),L_{ij}(B)=\epsilon_i\epsilon_jL_{ji}(-B),

allowing antisymmetric transport such as the Hall effect.

The core equations

Near equilibrium the entropy has a local maximum, so its expansion in the macroscopic deviations aia_i begins at quadratic order. Einstein's fluctuation formula therefore predicts a Gaussian equilibrium distribution for the fluctuations. The remainder of the derivation combines these Gaussian statistics with Onsager's regression hypothesis to obtain the reciprocal relations.

S(a)=S0−12∑ijsij aiaj,Xi=∂S∂ai=−∑jsijaj(1)S(a)=S_0-\tfrac12\sum_{ij}s_{ij}\,a_ia_j,\qquad X_i=\frac{\partial S}{\partial a_i}=-\sum_j s_{ij}a_j \qquad (1)

where: aia_i — deviation of macroscopic variable ii from equilibrium; sij=−∂2S/∂ai∂ajs_{ij}=-\partial^2S/\partial a_i\partial a_j — the positive-definite entropy-curvature matrix; XiX_i — the thermodynamic force conjugate to aia_i. The forces are the local entropy gradient: the system is pushed back toward the maximum. The fluctuation statistics follow from P(a)∝eS(a)/kBP(a)\propto e^{S(a)/k_B}, a Gaussian whose covariance and force-orthogonality we will use repeatedly.

⟨aiak⟩=kB (s−1)ik,⟨aiXj⟩=−kB δij(2)\langle a_ia_k\rangle=k_B\,(s^{-1})_{ik},\qquad \langle a_iX_j\rangle=-k_B\,\delta_{ij} \qquad (2)

where: kBk_B — Boltzmann constant; (s−1)ik(s^{-1})_{ik} — inverse entropy-curvature (the covariance up to kBk_B); the second identity says the fluctuation of aia_i is orthogonal to every conjugate force except its own partner. This orthogonality is the algebraic pivot of the whole proof. The regression hypothesis promotes the macroscopic decay law to a statement about conditional averages of spontaneous fluctuations.

ddt⟨ai(t) ak(0)⟩=∑jLij ⟨Xj(t) ak(0)⟩(3)\frac{\mathrm d}{\mathrm dt}\langle a_i(t)\,a_k(0)\rangle=\sum_j L_{ij}\,\langle X_j(t)\,a_k(0)\rangle \qquad (3)

where: LijL_{ij} — the Onsager transport matrix (the same one as the macroscopic law); ⟨ai(t)ak(0)⟩\langle a_i(t)a_k(0)\rangle — the equilibrium two-time correlation. Onsager's central postulate is that the regression of a fluctuation is governed by the macroscopic LL, with no new coefficients. Microscopic reversibility plus stationarity makes the correlation matrix symmetric under exchanging the two variables.

⟨ai(t) ak(0)⟩=⟨ai(0) ak(t)⟩=⟨ak(t) ai(0)⟩(4)\langle a_i(t)\,a_k(0)\rangle=\langle a_i(0)\,a_k(t)\rangle=\langle a_k(t)\,a_i(0)\rangle \qquad (4)

where: the first equality is stationarity (averages depend only on the time difference) combined with time-reversal t→−tt\to-t; the second is the index swap. Reading the correlation "ii first" or "kk first" gives the same number — this is the symmetry that becomes reciprocity. Combining (2), (3) and (4) at short times yields the reciprocal relations, and their field-dependent generalization.

C˙ik(0)=−kB Lik  → (4)   Lik=Lki,Lij(B,ω)=εiεj Lji(−B,−ω)(5)\dot C_{ik}(0)=-k_B\,L_{ik}\;\xrightarrow{\ (4)\ }\;L_{ik}=L_{ki},\qquad L_{ij}(\mathbf B,\boldsymbol\omega)=\varepsilon_i\varepsilon_j\,L_{ji}(-\mathbf B,-\boldsymbol\omega) \qquad (5)

where: εi=±1\varepsilon_i=\pm1 — the time-reversal parity of aia_i; B\mathbf B, ω\boldsymbol\omega — magnetic field and rotation, both odd under time reversal. In zero field with equal-parity variables the last relation collapses to plain symmetry Lij=LjiL_{ij}=L_{ji}. The same correlation symmetry underlies the Green–Kubo formula, which the SM11 companion transport-coefficients-green-kubo uses to compute these coefficients from equilibrium flux autocorrelations.

Lij=1kB∫0∞⟨Ji(0) Jj(t)⟩ dt=Lji(6)L_{ij}=\frac{1}{k_B}\int_0^\infty\langle J_i(0)\,J_j(t)\rangle\,\mathrm dt=L_{ji} \qquad (6)

where: Ji(t)J_i(t) — the microscopic flux; the index symmetry of the integrand (again from time reversal) makes LL symmetric at the microscopic level, the manifestly-computable face of reciprocity.

Derivation AThe regression route to reciprocity on a two-variable model

This derivation makes Onsager's argument fully concrete on a two-variable fluctuating system, so that every step is an explicit matrix identity rather than an appeal to a general theorem. We treat two coupled macroscopic deviations a1,a2a_1,a_2 (for instance a heat imbalance and a charge imbalance) with entropy curvature ss, forces X=−saX=-sa, and a regression law a˙=−Ma\dot a=-Ma where M=LsM=Ls. The goal is to show that the equal-time slope of the correlation matrix is −kBL-k_BL, and that the time-reversal symmetry of that correlation matrix forces LL to be symmetric — recovering, from equilibrium fluctuations alone, the reciprocity result that the SM10 owner states.

Step 1. Write the Gaussian fluctuation statistics from P(a)∝eS/kBP(a)\propto e^{S/k_B} with S=−12aTs aS=-\tfrac12 a^{\mathsf T}s\,a. The covariance is Σ≡⟨a aT⟩=kB s−1\Sigma\equiv\langle a\,a^{\mathsf T}\rangle=k_B\,s^{-1}, so ⟨aiXj⟩=−∑ksjk⟨aiak⟩=−kB( s s−1)ij=−kBδij\langle a_iX_j\rangle=-\sum_k s_{jk}\langle a_ia_k\rangle=-k_B(\,s\,s^{-1})_{ij}=-k_B\delta_{ij}. Why this matters: the orthogonality kills every cross-term except the diagonal, so the short-time slope will pick out exactly LikL_{ik} and nothing else.

Step 2. Model the regression as a linear relaxation a˙=−Ma\dot a=-Ma with M=LsM=Ls (fluxes =L×=L\timesforces, forces =−sa=-sa). Then the two-time correlation matrix is C(t)≡⟨a(t) a(0)T⟩=e−MtΣC(t)\equiv\langle a(t)\,a(0)^{\mathsf T}\rangle=e^{-Mt}\Sigma. Why: conditioning on the state at t=0t=0 and averaging the deterministic regression gives ⟨a(t)∣a(0)⟩=e−Mta(0)\langle a(t)|a(0)\rangle=e^{-Mt}a(0), and multiplying by a(0)Ta(0)^{\mathsf T} and averaging inserts the equal-time covariance Σ\Sigma.

Step 3. Differentiate at t=0t=0: C˙(0)=−MΣ=−(Ls)(kBs−1)=−kB L\dot C(0)=-M\Sigma=-(Ls)(k_Bs^{-1})=-k_B\,L. Component-wise, C˙ik(0)=−kBLik\dot C_{ik}(0)=-k_BL_{ik}. Why: the entropy curvature ss cancels against s−1s^{-1} in the covariance, leaving the transport matrix LL bare — this is the algebraic heart, and it is exactly the force-orthogonality of Step 1 in disguise.

Step 4. Impose microscopic reversibility. Equilibrium correlations are invariant under t→−tt\to-t, and stationarity gives ⟨ai(t)ak(0)⟩=⟨ak(t)ai(0)⟩\langle a_i(t)a_k(0)\rangle=\langle a_k(t)a_i(0)\rangle, i.e. C(t)=C(t)TC(t)=C(t)^{\mathsf T}. Therefore C˙(0)\dot C(0) is symmetric, and by Step 3 so is kBLk_BL. Why: reading the correlation "ii first" versus "kk first" must give the same number, because the reversed movie is equally likely — a purely microscopic fact, not a thermodynamic one.

Step 5. Conclude Lik=LkiL_{ik}=L_{ki}. Equivalently, setting the off-diagonal slopes equal, −kBL12=−kBL21-k_BL_{12}=-k_BL_{21} gives L12=L21L_{12}=L_{21}. The reciprocal relations are the index-symmetry of a correlation matrix. Why: nothing about the second law was used — an antisymmetric LL satisfies S˙≥0\dot S\ge0 perfectly well; time reversal is the ingredient that forbids it. *The five key results of this derivation — the covariance ⟨aiak⟩=kB(s−1)ik\langle a_ia_k\rangle=k_B(s^{-1})_{ik}, the orthogonality ⟨aiXj⟩=−kBδij\langle a_iX_j\rangle=-k_B\delta_{ij}, the slope C˙(0)=−kBL\dot C(0)=-k_BL, correlation-matrix symmetry for symmetric LL (and its breaking for asymmetric LL), and recovery of LL from the numeric correlation slope — all hold exactly.*

Go deeper · Derivation B · 1 minThe Kelvin thermoelectric relation from reciprocity
Derivation BThe Kelvin thermoelectric relation from reciprocity

Here we cash out reciprocity into a single measurable equality — that a material's Peltier coefficient is its temperature times its Seebeck coefficient — which Kelvin conjectured in 1854 on thermodynamic grounds Onsager later showed were incomplete. We couple charge and heat transport in one dimension, in the conjugate representation where the entropy production is S˙=JeXe+JqXq\dot S=J_eX_e+J_qX_q, so that the SM11-derived symmetry Leq=LqeL_{eq}=L_{qe} applies directly. The point is that two independently-measured device coefficients are locked together by a microscopic-reversibility theorem.

Step 1. Write the coupled linear laws Je=LeeXe+LeqXqJ_e=L_{ee}X_e+L_{eq}X_q and Jq=LqeXe+LqqXqJ_q=L_{qe}X_e+L_{qq}X_q, with XeX_e the electrochemical force and Xq=∇(1/T)X_q=\nabla(1/T) the thermal force. Why this pairing: only with the entropy-production conjugate forces does the reciprocity Leq=LqeL_{eq}=L_{qe} hold; using ∇T\nabla T instead of ∇(1/T)\nabla(1/T) without the compensating −1/T2-1/T^2 factor would spuriously break it (see the warning below).

Step 2. The Seebeck coefficient is the open-circuit voltage per unit temperature gradient, S=(∇V/∇T)∣Je=0S=(\nabla V/\nabla T)|_{J_e=0}, obtained by setting Je=0J_e=0 and eliminating XeX_e: it is proportional to Leq/LeeL_{eq}/L_{ee}. Why: at zero current the electrochemical force must exactly cancel the thermally-driven part, and that balance is set by the cross-coefficient over the diagonal one.

Step 3. The Peltier coefficient is the heat carried per unit charge at uniform temperature, Π=(Jq/Je)∣∇T=0\Pi=(J_q/J_e)|_{\nabla T=0}, proportional to Lqe/LeeL_{qe}/L_{ee}. Why: with no temperature gradient, all transport is driven by XeX_e, and the ratio of heat flux to charge flux is fixed by the other cross-coefficient over the same diagonal one.

Step 4. Apply Onsager reciprocity, Leq=LqeL_{eq}=L_{qe}. Since both coefficients are normalized by the same electrical conductivity LeeL_{ee}, the equality of the off-diagonal transport coefficients immediately gives Π=T S\Pi=T\,S, Kelvin's second relation. Why: the factor of TT appears because the thermodynamic force is ∇(1/T)\nabla(1/T), whereas the experimentally quoted Seebeck coefficient is defined using ∇T\nabla T. *For example, with S=−20 μV/KS=-20\,\mu\mathrm{V/K} at T=300 KT=300\,\mathrm{K}, Π=T S=−6.0 mV\Pi=T\,S=-6.0\,\mathrm{mV}, and therefore Π/S=T\Pi/S=T exactly.*

Limiting cases

CaseResultMeaning
Linear near-equilibrium regimeJi=∑jLijXjJ_i=\sum_j L_{ij}X_j, Lij=LjiL_{ij}=L_{ji}Valid for small forces; far from equilibrium the coefficients become force-dependent and reciprocity generally fails.
Magnetic field / rotation (Casimir)Lij(B)=εiεjLji(−B)L_{ij}(\mathbf B)=\varepsilon_i\varepsilon_jL_{ji}(-\mathbf B)Equal-parity variables (εiεj=+1\varepsilon_i\varepsilon_j=+1) give LL symmetric only after flipping B\mathbf B; the antisymmetric-in-B\mathbf B piece is the Hall conductivity.
Second-law bound onlyLsym⪰0L^{\mathrm{sym}}\succeq0, Lii>0L_{ii}>0, LiiLjj≥Lij2L_{ii}L_{jj}\ge L_{ij}^2Positivity of S˙\dot S constrains only the symmetric part; an antisymmetric part is thermodynamically allowed — reciprocity is genuinely extra.
Contragredient variable changeJ′=AJ,  X′=(A−1)TX⇒L′=ALATJ'=AJ,\;X'=(A^{-1})^{\mathsf T}X\Rightarrow L'=ALA^{\mathsf T}Reciprocity survives only transformations that keep S˙=∑JiXi\dot S=\sum J_iX_i invariant; arbitrary redefinitions asymmetrize LL.

Worked examples

Example 1 — Peltier from Seebeck via the Kelvin relation (a real number).

Step 1. Take a metal with Seebeck coefficient S=−20 μV/KS=-20\ \mu\mathrm V/\mathrm K (the sign reflects electron-dominated transport) at T=300 KT=300\ \mathrm K. By reciprocity, its Peltier coefficient is not an independent quantity: Π=T S\Pi=T\,S.

Step 2. Compute Π=(300 K)(−20×10−6 V/K)=−6.0 mV\Pi=(300\ \mathrm K)(-20\times10^{-6}\ \mathrm V/\mathrm K)=\mathbf{-6.0\ mV}. Physically, driving one coulomb of charge through a junction of this metal transports −6 mV×1 C=−6 mJ-6\ \mathrm{mV}\times1\ \mathrm C=-6\ \mathrm{mJ} of heat — the basis of thermoelectric cooling.

Step 3. The reciprocity signature is Π/S=T=300 K\Pi/S=T=\mathbf{300\ K} exactly. Measuring Π\Pi and SS on the same material and forming their ratio yields the absolute temperature — a check that microscopic reversibility is operating, and a self-consistency test used in thermoelectric metrology.

Example 2 — recovering LL from the fluctuation-correlation slope.

Step 1. Take a two-variable system with entropy curvature s=(2.00.50.51.5)s=\begin{pmatrix}2.0&0.5\\0.5&1.5\end{pmatrix} and a symmetric transport matrix L=(0.80.30.30.6)L=\begin{pmatrix}0.8&0.3\\0.3&0.6\end{pmatrix} (in units kB=1k_B=1). The relaxation matrix is M=LsM=Ls and the equilibrium covariance is Σ=s−1\Sigma=s^{-1}.

Step 2. Form the correlation matrix C(t)=e−MtΣC(t)=e^{-Mt}\Sigma and its short-time slope C˙(0)=−MΣ=−L\dot C(0)=-M\Sigma=-L. Numerically differentiating CC at t=10−6t=10^{-6} recovers LL to ∼10−4\sim10^{-4}: Lrec≈(0.8000.3000.3000.600)L_{\mathrm{rec}}\approx\begin{pmatrix}0.800&0.300\\0.300&0.600\end{pmatrix}, matching the input.

Step 3. Check the diagnostic: with the symmetric LL, C(t)C(t) is symmetric at every tt (verified at t=0.05,0.1,0.3,0.7t=0.05,0.1,0.3,0.7). Replace LL by the asymmetric (0.80.50.10.6)\begin{pmatrix}0.8&0.5\\0.1&0.6\end{pmatrix} and C(0.3)C(0.3) is no longer symmetric — the equilibrium fluctuation movie would be distinguishable from its time-reverse, which microscopic reversibility forbids. Reciprocity is exactly the requirement C(t)=C(t)TC(t)=C(t)^{\mathsf T}.

Origins · 1 minregression hypothesis and how this was discovered
RH
regression hypothesis
introduced by Lars Onsager, 1931 · The claim that an equilibrium fluctuation, once spontaneously created, decays on average by the same linear laws that govern macroscopic relaxation. It is the bridge that lets microscopic time-reversal symmetry — a property of correlation functions — descend onto measurable transport coefficients, and is the logical core of Onsager's proof of reciprocity. It was later placed on a rigorous footing by the fluctuation–dissipation theorem and the Green–Kubo formulas.

History
1854
— Kelvin (William Thomson) conjectures the thermoelectric relations Π=TS\Pi=TS and the Thomson relation on thermodynamic grounds, without a complete justification
1905
— Einstein inverts Boltzmann's principle, P∝eS/kBP\propto e^{S/k_B}, giving the Gaussian statistics of equilibrium fluctuations
1931
— Onsager proves the reciprocal relations Lij=LjiL_{ij}=L_{ji} from microscopic reversibility and the regression hypothesis (Phys. Rev. 37, 405 and 38, 2265)
1945
— Casimir generalizes to variables of definite time-reversal parity and to magnetic fields, Lij(B)=εiεjLji(−B)L_{ij}(\mathbf B)=\varepsilon_i\varepsilon_jL_{ji}(-\mathbf B) (Rev. Mod. Phys. 17, 343)
1951
— Callen and Welton prove the fluctuation–dissipation theorem, the quantum-statistical parent of the correlation route
1968
— Onsager receives the Nobel Prize in Chemistry for the reciprocal relations

Going further

References. L. Onsager, 'Reciprocal Relations in Irreversible Processes I & II', Phys. Rev. 37, 405 (1931); 38, 2265 (1931) · H. B. G. Casimir, 'On Onsager's Principle of Microscopic Reversibility', Rev. Mod. Phys. 17, 343 (1945) · S. R. de Groot & P. Mazur, Non-Equilibrium Thermodynamics (Dover, 1984) · Landau & Lifshitz, Statistical Physics Part 1, §120 · Kardar, Statistical Physics of Particles · Reif, Fundamentals of Statistical and Thermal Physics, Ch. 15 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity The key self-consistency checks for this article are: (1) the Einstein covariance ⟨aiak⟩=kB(s−1)ik\langle a_ia_k\rangle=k_B(s^{-1})_{ik} and orthogonality ⟨aiXj⟩=−kBδij\langle a_iX_j\rangle=-k_B\delta_{ij}; (2) the slope C˙(0)=−kBL\dot C(0)=-k_BL and the symmetry L12=L21L_{12}=L_{21}; (3) the Kelvin relation Π=TS\Pi=TS; (4) the Casimir parity relation and Hall antisymmetry.

Builds on: Detailed Balance and Microscopic Reversibility · Fluctuation–Dissipation Theorem · Markov Processes and the Markov Property · Fokker–Planck Equation Leads to: Fluctuation–Dissipation Theorem · Langevin Equation and Stochastic Differential Equations · Molecular Dynamics Simulation

Why is this called the "regression" hypothesis?

"Regression" here means the decay (return) of a fluctuation toward equilibrium, not statistical regression. Onsager's hypothesis is that the average way a spontaneous fluctuation regresses is identical to the way a macroscopically-imposed deviation relaxes. This is far from obvious: a spontaneous fluctuation is selected by the noise, whereas a macroscopic deviation is prepared by an experimenter. The hypothesis — later proved by the fluctuation–dissipation theorem — is what lets you measure transport coefficients from equilibrium noise, and it is the reason the correlation-matrix argument works at all.

How is the SM11 treatment different from the SM10 Onsager article?

The SM10 article onsager-reciprocal-relations owns the phenomenology: how to choose conjugate flux–force pairs, the bilinear entropy production, and the statement Lij=LjiL_{ij}=L_{ji} with its Kelvin and Soret–Dufour consequences. This SM11 article does not repeat that bookkeeping; it executes the stochastic derivation explicitly on a two-variable Gaussian fluctuation model — computing the covariance, the force-orthogonality, the correlation-matrix slope −kBL-k_BL, and the symmetry that follows from time reversal — and frames reciprocity as a diagnostic on equilibrium correlation functions. It is the same result seen from the fluctuation side, which is the side that connects to Green–Kubo and molecular dynamics.

Why does only the symmetric part of L enter entropy production?

Entropy production is the quadratic form S˙=∑ijXiLijXj=XTLX\dot S=\sum_{ij}X_iL_{ij}X_j=X^{\mathsf T}LX. Any antisymmetric matrix LaL^{\mathrm a} satisfies XTLaX=0X^{\mathsf T}L^{\mathrm a}X=0 identically, because the quadratic form only sees the symmetric part 12(L+LT)\tfrac12(L+L^{\mathsf T}). So the second law (S˙≥0\dot S\ge0) constrains LsymL^{\mathrm{sym}} to be positive-semidefinite but says nothing about LaL^{\mathrm a}. Reciprocity — which sets La=0L^{\mathrm a}=0 in zero field — is therefore not a thermodynamic consequence; it is the additional content that time-reversal invariance supplies.

What exactly breaks reciprocity in a magnetic field?

The Lorentz force q v×Bq\,\mathbf v\times\mathbf B depends on velocity, and velocity is odd under time reversal while B\mathbf B is treated as an external parameter. Reversing time therefore does not leave the dynamics invariant unless you also reverse B\mathbf B. The corrected statement is the Onsager–Casimir relation Lij(B)=εiεjLji(−B)L_{ij}(\mathbf B)=\varepsilon_i\varepsilon_jL_{ji}(-\mathbf B). For the conductivity tensor this means the symmetric part is even in B\mathbf B (ordinary dissipative conduction) while the antisymmetric part is odd — precisely the Hall conductivity, which changes sign when the field reverses.

Is the linear regime essential, or can reciprocity extend far from equilibrium?

The plain reciprocity Lij=LjiL_{ij}=L_{ji} is a linear-response statement: it assumes fluxes are linear in forces, i.e. small forces. Far from equilibrium the coefficients become force-dependent and the symmetry generally fails. There are important nonlinear generalizations — for example, fluctuation theorems (see the SM11 articles on fluctuation-theorems and stochastic-thermodynamics) impose exact symmetries on the full distribution of entropy production arbitrarily far from equilibrium, and reciprocity re-emerges as their linear-response limit. But you cannot simply extrapolate Lij=LjiL_{ij}=L_{ji} into the strongly driven regime.

Does the derivation require the variables to be even under time reversal?

Yes — the plain symmetry Lij=LjiL_{ij}=L_{ji} assumes both aia_i and aja_j are even under time reversal (e.g. energy density, particle density). For variables of definite but possibly odd parity εi=±1\varepsilon_i=\pm1 (such as a current or a velocity), the correct statement carries the parity factors: Lij=εiεjLjiL_{ij}=\varepsilon_i\varepsilon_jL_{ji} at zero field. Mixed-parity couplings (one even, one odd variable) are antisymmetric. This is the same bookkeeping that, combined with the field flip, gives the full Onsager–Casimir relation, and it is why the Hall coefficient — coupling two velocity-like components — comes out antisymmetric.

How is reciprocity actually used in a simulation?

In equilibrium molecular dynamics you record the fluxes (stress tensor for viscosity, heat current for conductivity, particle current for diffusion) along a long trajectory and form the autocorrelation matrix ⟨Ji(0)Jj(t)⟩\langle J_i(0)J_j(t)\rangle. The Green–Kubo formula Lij=1kB∫0∞⟨Ji(0)Jj(t)⟩ dtL_{ij}=\frac{1}{k_B}\int_0^\infty\langle J_i(0)J_j(t)\rangle\,\mathrm dt then gives the transport matrix. Onsager symmetry lets you replace the two noisy cross-correlations ⟨JiJj⟩\langle J_iJ_j\rangle and ⟨JjJi⟩\langle J_jJ_i\rangle by their average, halving the variance, and it provides a consistency check: a computed asymmetric LL signals insufficient sampling or a broken time-reversal symmetry in the model.

Before this

Where this leads