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The classical Heisenberg model assigns a three-component unit vector to each site and couples pairs by their dot product. Positive $J$ favors parallel alignment. With no applied magnetic field or anisotropy, simultaneous rotations of all vectors preserve the energy, so an ordered state chooses a direction and exhibits spontaneous symmetry breaking. The vector order parameter is the spatially averaged orientation. A quartic Landau free energy describes its mean-field onset near a continuous transition; fluctuations can change the critical exponents.^[ref]

For the flux $W_{ij}=-M{\partial }_i{\mu }_j+N_{ij}$ of a vector order parameter, microscopic reversibility requires independent Gaussian white noise components with covarianceThe current changes sign under time reversal. The resulting path-probability ratio contributes $-(k_{B}T{)}^{-1}\int W_{ij}{\partial }_i{\mu }_j$ to its logarithm; integration by parts converts this to the free-energy loss from the conserved dynamics.

Write ${\mu }_j=\delta F/\delta p_j$ and $\int ={\int }_{t_1}^{t_2}dt\int d\mathbf{r}$. Eliminating the noise gives $f=\dot{p}+\Gamma \mu$. The Onsager--Machlup path probability is thereforeThese are conditional path densities given the corresponding starting configuration. For a time-even vector order parameter, reversal reads the configurations backward, changes $\dot{p}$ to $-\dot{p}$, and keeps $\mu$ unchanged on each corresponding configuration.

The two Gaussian white noise measures have the same normalization. With a common midpoint discretization, the noise-to-path Jacobian also agrees under reversal by time-reversal invariance of a path Jacobian. The common factor $\mathcal{K}[p]$ may include that path-dependent Jacobian; it need not be a universal constant. This makes the shared-factor assertion precise while leaving the requested ratio unaffected.

For the isotropic zero-field classical Heisenberg model of a ferromagnet, the analytic vector Landau expansion uses the Cartesian components of the vector order parameter near $\mathbf{m}=0$. rotational symmetry makes its scalar terms functions of $\mathbf{m}\cdot \mathbf{m}$. Global spin inversion symmetry also requires $f(\mathbf{m})=f(-\mathbf{m})$. Through fourth order this givesA positive quartic coefficient stabilizes the unconstrained truncated free energy. The disordered phase has $b>0$ and its minimum at $m=0$; the aligned phase has $b<0$ and a minimum at nonzero $m$. At the continuous transition $b=0$. The attractive alignment interaction supplies an energetic tendency toward negative quadratic curvature; orientational entropy favors the disordered phase at high temperature, so the sign of $b$ is a free-energy question rather than a consequence of the pair energy alone.

The analyticity assumption matters: the variable printed as $m$ is a nonnegative magnitude, so reversing $\mathbf{m}$ does not send that scalar to $-m$. Terms proportional to $|\mathbf{m}|$ or $|\mathbf{m}{|}^3$ would be rotationally invariant but nonanalytic at the vector origin. Standard analytic Landau theory excludes them. Without that assumption, the stated pair symmetry alone would not exclude odd powers of the magnitude.^[ref]

Differentiation gives $f^{\prime }(m)=2m(b+2d{m}^2)$. For $b>0$ the origin is the global minimum; for $b<0$, $m=\sqrt{-b/(2d)}$ has $f^{\prime \prime }=-4b>0$ and $f=-b^2/(4d)<f(0)$. Therefore, close enough to the transition for the quartic expansion to apply,The order parameter approaches zero continuously with mean-field critical exponent $\beta =1/2$: the transition is continuous. In the aligned phase a particular direction is chosen. The vector-spin rotation symmetry is spontaneously broken from $SO(3)$ to $SO(2)$, or from $O(3)$ to $O(2)$ including reflections. The manifold of vector order parameter directions is a sphere, so this is not just a two-state sign choice.^[ref]

The plot uses illustrative $d=1/2$ and temperature measured in units of $T_c$. A unit-spin average must satisfy $m\le 1$; if the quartic model is used literally far below the transition, its constrained minimum is $\min \{1,\sqrt{(T_c-T)/(2d{T}_c)}\}$. In practice the near-transition expansion is not a controlled prediction of deep low-temperature saturation. Also, a finite zero-field statistical ensemble that averages all directions has $⟨\mathbf{m}⟩=0$ even below $T_c$; the curve describes the magnitude in a selected ordered state, obtained by appropriate boundary conditions or the thermodynamic limit before removing a small magnetic field.