Onsager Reciprocal Relations

We take from the SM10 article onsager-reciprocal-relations the phenomenology of linear irreversible thermodynamics — fluxes $J_i=\sum_j L_{ij}X_j$ driven by conjugate thermodynamic forces $X_j$ , the bilinear entropy production $\dot S=\sum_i J_iX_i\ge0$ , the reciprocity result $L_{ij}=L_{ji}$ , and its Casimir extension in a magnetic field. That article establishes the result and its thermoelectric payoff. 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 $t\to-t$ — makes the correlation matrix $C_{ij}(t)=\langle a_i(t)a_j(0)\rangle$ symmetric under exchanging the two variables, and the regression law then transfers that symmetry directly onto the transport matrix $L$ . We compute the whole chain symbolically on a two-variable fluctuating system: the Einstein covariance $\langle a_ia_k\rangle=k_B(s^{-1})_{ik}$ , the force-orthogonality $\langle a_iX_j\rangle=-k_B\delta_{ij}$ , the short-time correlation slope $\dot C_{ik}(0)=-k_BL_{ik}$ , and finally $L_{ik}=L_{ki}$ from the time-reversal symmetry of $C$ . An asymmetric $L$ is shown to break correlation symmetry — the diagnostic that turns reciprocity into something measurable. We close with the numeric Kelvin relation $\Pi=T\,S$ (a metal's Peltier coefficient from its Seebeck coefficient) and the Onsager–Casimir field flip $L_{ij}(\mathbf B)=\varepsilon_i\varepsilon_jL_{ji}(-\mathbf B)$ , whose signature is the antisymmetry of the Hall conductivity.

⚡ Key Takeaways

Reciprocity $L_{ij}=L_{ji}$ 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 $\dot a_i=\sum_jL_{ij}X_j$ ; this links $L$ to the correlation matrix $C_{ik}(t)=\langle a_i(t)a_k(0)\rangle$ .
The short-time slope of $C$ is $\dot C_{ik}(0)=-k_BL_{ik}$ ; time-reversal symmetry $C_{ik}(t)=C_{ki}(t)$ then forces $L_{ik}=L_{ki}$ — derived here explicitly on a 2-variable Gaussian model.
Only the symmetric part of $L$ 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 $\Pi=T\,S$ (Peltier $=$ temperature $\times$ Seebeck), the Soret–Dufour symmetry, and the Onsager–Casimir field flip that makes Hall conductivity antisymmetric in $\mathbf B$ .
Key idea: The regression hypothesis converts a theorem about transport coefficients into a theorem about equilibrium correlation functions; microscopic reversibility makes those correlations symmetric under exchanging the two variables, and that symmetry descends intact onto the transport matrix — $L_{ij}=L_{ji}$ .

💡 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 $C_{ik}(t)=\langle a_i(t)a_k(0)\rangle$ is symmetric under $i\leftrightarrow k$ — a fact guaranteed by microscopic time-reversal invariance — and the regression law $\dot C_{ik}(0)=-k_BL_{ik}$ transfers that symmetry directly onto $L$ . An asymmetric $L$ would make a movie of equilibrium fluctuations distinguishable from its time-reverse.

The physical picture

Imagine watching two coupled fluctuating quantities in a system at equilibrium — say a small temperature imbalance $a_1$ between two halves of a bar and a small charge imbalance $a_2$ . Both wander spontaneously, driven by thermal noise, and both slowly relax back. The regression hypothesis is a claim about the average of that relaxation: if you condition on a large fluctuation of $a_1$ and average over the ensemble, the subsequent decay of $a_1$ and $a_2$ obeys the same linear laws you would measure by deliberately imposing a temperature difference and watching the currents. Onsager elevated this from a plausible statement to the engine of a proof.

Figure

Two coupled fluctuations regressing: conditioned decay of a₁ (blue) drags a₂ (amber) by the SAME L that governs macroscopic relaxation

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 $t\to-t$ (with momenta reversed). This makes the cross-correlation even in a specific sense: $\langle a_1(t)a_2(0)\rangle=\langle a_2(t)a_1(0)\rangle$ . 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)

A helpful mnemonic: forces are how far up the entropy hill you are pushed, fluxes are how fast you slide down. Reciprocity says the coupling of slide-direction $i$ to push-direction $j$ equals the coupling of $j$ to $i$ — the friction-like matrix is symmetric, exactly as a Hessian is. In a magnetic field the sliding becomes chiral (the Lorentz force curves orbits), reversing time no longer restores the dynamics unless you also flip $\mathbf B$ , and the symmetry weakens to the Casimir form that ties $L(\mathbf B)$ to $L(-\mathbf B)$ and generates the antisymmetric Hall piece.

The core equations

Near equilibrium the entropy is a maximum, so its expansion in macroscopic deviations $a_i$ is quadratic and the probability of a fluctuation is set by Einstein's inversion of Boltzmann's principle. We start there, then feed the resulting Gaussian statistics into the regression law.

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

\langle a_ia_k\rangle=k_B\,(s^{-1})_{ik},\qquad \langle a_iX_j\rangle=-k_B\,\delta_{ij} \qquad (2)

where: $k_B$ — Boltzmann constant; $(s^{-1})_{ik}$ — inverse entropy-curvature (the covariance up to $k_B$ ); the second identity says the fluctuation of $a_i$ 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.

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

\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 $t\to-t$ ; the second is the index swap. Reading the correlation " $i$ first" or " $k$ 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.

\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: $\varepsilon_i=\pm1$ — the time-reversal parity of $a_i$ ; $\mathbf B$ , $\boldsymbol\omega$ — magnetic field and rotation, both odd under time reversal. In zero field with equal-parity variables the last relation collapses to plain symmetry $L_{ij}=L_{ji}$ . 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.

L_{ij}=\frac{1}{k_B}\int_0^\infty\langle J_i(0)\,J_j(t)\rangle\,\mathrm dt=L_{ji} \qquad (6)

where: $J_i(t)$ — the microscopic flux; the index symmetry of the integrand (again from time reversal) makes $L$ 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 $a_1,a_2$ (for instance a heat imbalance and a charge imbalance) with entropy curvature $s$ , forces $X=-sa$ , and a regression law $\dot a=-Ma$ where $M=Ls$ . The goal is to show that the equal-time slope of the correlation matrix is $-k_BL$ , and that the time-reversal symmetry of that correlation matrix forces $L$ to be symmetric — recovering, from equilibrium fluctuations alone, the reciprocity result that the SM10 owner states.

Step 1. Write the Gaussian fluctuation statistics from $P(a)\propto e^{S/k_B}$ with $S=-\tfrac12 a^{\mathsf T}s\,a$ . The covariance is $\Sigma\equiv\langle a\,a^{\mathsf T}\rangle=k_B\,s^{-1}$ , so $\langle a_iX_j\rangle=-\sum_k s_{jk}\langle a_ia_k\rangle=-k_B(\,s\,s^{-1})_{ij}=-k_B\delta_{ij}$ . Why this matters: the orthogonality kills every cross-term except the diagonal, so the short-time slope will pick out exactly $L_{ik}$ and nothing else.

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

Step 3. Differentiate at $t=0$ : $\dot C(0)=-M\Sigma=-(Ls)(k_Bs^{-1})=-k_B\,L$ . Component-wise, $\dot C_{ik}(0)=-k_BL_{ik}$ . Why: the entropy curvature $s$ cancels against $s^{-1}$ in the covariance, leaving the transport matrix $L$ 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 $t\to-t$ , and stationarity gives $\langle a_i(t)a_k(0)\rangle=\langle a_k(t)a_i(0)\rangle$ , i.e. $C(t)=C(t)^{\mathsf T}$ . Therefore $\dot C(0)$ is symmetric, and by Step 3 so is $k_BL$ . Why: reading the correlation " $i$ first" versus " $k$ first" must give the same number, because the reversed movie is equally likely — a purely microscopic fact, not a thermodynamic one.

Step 5. Conclude $L_{ik}=L_{ki}$ . Equivalently, setting the off-diagonal slopes equal, $-k_BL_{12}=-k_BL_{21}$ gives $L_{12}=L_{21}$ . The reciprocal relations are the index-symmetry of a correlation matrix. Why: nothing about the second law was used — an antisymmetric $L$ satisfies $\dot S\ge0$ perfectly well; time reversal is the ingredient that forbids it. *Verified symbolically and numerically in verify_aq-sm-onsager-reciprocal-relations-sm11.py: the covariance $\langle a_ia_k\rangle=k_B(s^{-1})_{ik}$ , the orthogonality $\langle a_iX_j\rangle=-k_B\delta_{ij}$ , the slope $\dot C(0)=-k_BL$ , correlation-matrix symmetry for symmetric $L$ (and its breaking for asymmetric $L$ ), and recovery of $L$ from the numeric correlation slope.*

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 $\dot S=J_eX_e+J_qX_q$ , so that the SM11-derived symmetry $L_{eq}=L_{qe}$ 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 $J_e=L_{ee}X_e+L_{eq}X_q$ and $J_q=L_{qe}X_e+L_{qq}X_q$ , with $X_e$ the electrochemical force and $X_q=\nabla(1/T)$ the thermal force. Why this pairing: only with the entropy-production conjugate forces does the reciprocity $L_{eq}=L_{qe}$ hold; using $\nabla T$ instead of $\nabla(1/T)$ without the compensating $-1/T^2$ factor would spuriously break it (see the warning below).

Step 2. The Seebeck coefficient is the open-circuit voltage per unit temperature gradient, $S=(\nabla V/\nabla T)|_{J_e=0}$ , obtained by setting $J_e=0$ and eliminating $X_e$ : it is proportional to $L_{eq}/L_{ee}$ . 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, $\Pi=(J_q/J_e)|_{\nabla T=0}$ , proportional to $L_{qe}/L_{ee}$ . Why: with no temperature gradient, all transport is driven by $X_e$ , and the ratio of heat flux to charge flux is fixed by the other cross-coefficient over the same diagonal one.

Step 4. Impose $L_{eq}=L_{qe}$ . The two ratios share the denominator $L_{ee}$ , and the reciprocity of the numerators gives $\Pi=T\,S$ — the second Kelvin relation, now a theorem. Why: the factor of $T$ arises because the thermal force is $\nabla(1/T)=-\nabla T/T^2$ , converting the $1/T$ -representation coefficient into the laboratory Seebeck coefficient. *Verified in verify_aq-sm-onsager-reciprocal-relations-sm11.py: with a metal Seebeck $S=-20\ \mu\mathrm V/\mathrm K$ at $T=300\ \mathrm K$ the Kelvin relation predicts $\Pi=T\,S=-6.0\ \mathrm{mV}$ , and $\Pi/S=T$ exactly — the testable reciprocity signature.*

Limiting cases

Case	Result	Meaning
Linear near-equilibrium regime	$J_i=\sum_j L_{ij}X_j$ , $L_{ij}=L_{ji}$	Valid for small forces; far from equilibrium the coefficients become force-dependent and reciprocity generally fails.
Magnetic field / rotation (Casimir)	$L_{ij}(\mathbf B)=\varepsilon_i\varepsilon_jL_{ji}(-\mathbf B)$	Equal-parity variables ( $\varepsilon_i\varepsilon_j=+1$ ) give $L$ symmetric only after flipping $\mathbf B$ ; the antisymmetric-in- $\mathbf B$ piece is the Hall conductivity.
Second-law bound only	$L^{\mathrm{sym}}\succeq0$ , $L_{ii}>0$ , $L_{ii}L_{jj}\ge L_{ij}^2$	Positivity of $\dot S$ constrains only the symmetric part; an antisymmetric part is thermodynamically allowed — reciprocity is genuinely extra.
Contragredient variable change	$J'=AJ,\;X'=(A^{-1})^{\mathsf T}X\Rightarrow L'=ALA^{\mathsf T}$	Reciprocity survives only transformations that keep $\dot S=\sum J_iX_i$ invariant; arbitrary redefinitions asymmetrize $L$ .

⚠️ The symmetry lives in the conjugate representation, not any representation

$L_{ij}=L_{ji}$ holds only for the flux–force pairs defined by the entropy production $\dot S=\sum_iJ_iX_i$ . A nonsingular linear redefinition of variables preserves the symmetry only under contragredient (inverse-transpose) transformations. The classic error is to take $\nabla T$ rather than $\nabla(1/T)=-\nabla T/T^2$ as the heat force without the compensating factor: the coefficient matrix then looks asymmetric even though the physics is unchanged. Symmetry is a property of the pairing, not of the raw currents.

Worked examples

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

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

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

Step 3. The reciprocity signature is $\Pi/S=T=\mathbf{300\ K}$ exactly. Measuring $\Pi$ and $S$ 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 $L$ from the fluctuation-correlation slope.

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

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

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

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

\Pi=TS

and the Thomson relation on thermodynamic grounds, without a complete justification

1905

— Einstein inverts Boltzmann's principle,

P\propto e^{S/k_B}

, giving the Gaussian statistics of equilibrium fluctuations

1931

— Onsager proves the reciprocal relations

L_{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,

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

Where this lives today

The regression route is not a historical curiosity: it is exactly how transport coefficients are extracted in modern molecular-dynamics simulation. One runs an equilibrium trajectory, measures the flux autocorrelation matrix $\langle J_i(0)J_j(t)\rangle$ , and integrates it — the Green–Kubo estimator — with the Onsager symmetry used to average the two cross-correlations and cut statistical noise. The same correlation-matrix symmetry underlies the fluctuation–dissipation theorem and, through it, linear-response theory across condensed-matter physics. In spintronics and quantum-Hall systems the Casimir field-flip is the workhorse identity that separates the dissipative (symmetric) conductivity from the non-dissipative Hall (antisymmetric) part, and the quantized Hall plateaus are protected precisely by this antisymmetry.

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

Provenance. The regression-hypothesis / correlation-matrix derivation follows Onsager (1931), Landau–Lifshitz §120, and de Groot & Mazur Ch. IV; the Casimir field flip is from Casimir (1945). The SM10 companion onsager-reciprocal-relations owns the forces-and-fluxes phenomenology and the primary statement $L_{ij}=L_{ji}$ ; this article recalls that result and supplies the explicit two-variable Gaussian derivation. A verifier should recompute: (1) the Einstein covariance $\langle a_ia_k\rangle=k_B(s^{-1})_{ik}$ and orthogonality $\langle a_iX_j\rangle=-k_B\delta_{ij}$ ; (2) the slope $\dot C(0)=-k_BL$ and the symmetry $L_{12}=L_{21}$ ; (3) the Kelvin relation $\Pi=TS$ ; (4) the Casimir parity relation and Hall antisymmetry. All checks are in verify_aq-sm-onsager-reciprocal-relations-sm11.py.

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 $L_{ij}=L_{ji}$ 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 $-k_BL$ , 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 $\dot S=\sum_{ij}X_iL_{ij}X_j=X^{\mathsf T}LX$ . Any antisymmetric matrix $L^{\mathrm a}$ satisfies $X^{\mathsf T}L^{\mathrm a}X=0$ identically, because the quadratic form only sees the symmetric part $\tfrac12(L+L^{\mathsf T})$ . So the second law ( $\dot S\ge0$ ) constrains $L^{\mathrm{sym}}$ to be positive-semidefinite but says nothing about $L^{\mathrm a}$ . Reciprocity — which sets $L^{\mathrm a}=0$ 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 $q\,\mathbf v\times\mathbf B$ depends on velocity, and velocity is odd under time reversal while $\mathbf B$ is treated as an external parameter. Reversing time therefore does not leave the dynamics invariant unless you also reverse $\mathbf B$ . The corrected statement is the Onsager–Casimir relation $L_{ij}(\mathbf B)=\varepsilon_i\varepsilon_jL_{ji}(-\mathbf B)$ . For the conductivity tensor this means the symmetric part is even in $\mathbf B$ (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 $L_{ij}=L_{ji}$ 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 $L_{ij}=L_{ji}$ into the strongly driven regime.

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

Yes — the plain symmetry $L_{ij}=L_{ji}$ assumes both $a_i$ and $a_j$ are even under time reversal (e.g. energy density, particle density). For variables of definite but possibly odd parity $\varepsilon_i=\pm1$ (such as a current or a velocity), the correct statement carries the parity factors: $L_{ij}=\varepsilon_i\varepsilon_jL_{ji}$ 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 $\langle J_i(0)J_j(t)\rangle$ . The Green–Kubo formula $L_{ij}=\frac{1}{k_B}\int_0^\infty\langle J_i(0)J_j(t)\rangle\,\mathrm dt$ then gives the transport matrix. Onsager symmetry lets you replace the two noisy cross-correlations $\langle J_iJ_j\rangle$ and $\langle J_jJ_i\rangle$ by their average, halving the variance, and it provides a consistency check: a computed asymmetric $L$ signals insufficient sampling or a broken time-reversal symmetry in the model.