Pseudo-gauge ambiguity

The definition of conserved current is not unique. One example is the magnetization current and dipole charge density. Let Jμ = (ρ, J) represent the conserved conductive electric current. For a polarizable and magnetizable material, the total charge density and electric current are given by $\tilde{\rho }=\rho +\nabla \cdot P$ and $\tilde{J}=J+\nabla \times M$, respectively, where P is the electric dipole density, and M is the magnetization density. In the covariant form, we have: ${\tilde{J}}^{\mu }=J^{\mu }+{\partial }_{\nu }{\mathcal{M}}^{\mu \nu }\text{with}{\mathcal{M}}^{\mu \nu }=-{\mathcal{M}}^{\nu \mu }\mathrm{.}$ (46) Obviously, the total current ${\tilde{J}}^{\mu }$ is conserved if the conduction current Jμ is conserved, and the total electric charge remains unchanged provided the surface dipole density vanishes. The transformation of a conserved current that preserves both the original conservation law and total conserved charge is called a pseudo-gauge transformation. The example above demonstrates that the total current and conduction current differ by a pseudo-gauge transformation (with the magnetization ${\mathcal{M}}^{\mu \nu }$ serving as the pseudo-gauge field). This example also highlights that a pseudo-gauge transformation is not a true gauge transformation because it alters the physical content of the transformed current. Further insight into the pseudo-gauge transformation can be obtained by examining Maxwell’s equations: ${\partial }_{\mu }{F}^{\mu \nu }={\tilde{J}}^{\nu }\mathrm{.}$ (47) One could subtract $-{\partial }_{\rho }{\mathcal{M}}^{\nu \rho }$ from both sides and find ${\partial }_{\mu }{H}^{\mu \nu }=J^{\nu }\mathrm{,}$ (48) where the new field-strength tensor is defined as $H^{\mu \nu }\equiv F^{\mu \nu }+{\mathcal{M}}^{\mu \nu }$. This demonstrates that without imposing additional constraints, the two sets of fields, $\left(F^{\mu \nu }\mathrm{,}{\tilde{J}}^{\mu }\right)$ and $\left(H^{\mu \nu }\mathrm{,}{J}^{\mu }\right)$, describe the same physical laws, and one can freely choose which set to use. (If further constraints are imposed, such as the Bianchi equation ${\partial }_{[\rho }{F}_{\mu \nu ]}=0$, which is not preserved under a general pseudo-gauge transformation, then only certain pseudo-gauges that respect the Bianchi equation are permitted.)

Similarly, let us consider angular momentum conservation (note the analogy with Eq. (48), where ${\text{Σ}}^{\mu \nu \rho }$ and Θ^ρv - Θ^vρ are analogous to $H^{\mu \nu }$ and Jμ in Eq. (48)): ${\partial }_{\mu }{\text{Σ}}^{\mu \nu \rho }={\text{Θ}}^{\rho \nu }-{\text{Θ}}^{\nu \rho }\mathrm{,}$ (49) which is preserved under the transformation ${\text{Σ}}^{\mu \rho \sigma }\to {\widetilde{\text{Σ}}}^{\mu \rho \sigma }\equiv {\text{Σ}}^{\mu \rho \sigma }-{\text{Φ}}^{\mu \rho \sigma }\mathrm{,}$ (50) ${\text{Θ}}^{\mu \nu }\to {\widetilde{\text{Θ}}}^{\mu \nu }\equiv {\text{Θ}}^{\mu \nu }+\frac{1}{2}{\partial }_{\lambda }{\text{Φ}}^{\lambda \mu \nu }\mathrm{,}$ (51) with ${\text{Φ}}^{\lambda \mu \nu }=-{\text{Φ}}^{\lambda \nu \mu }$ denotes an arbitrary local field. However, this transformation violates the conservation law of the energy-momentum tensor. It can be modified as following: ${\text{Σ}}^{\mu \rho \sigma }\to {\widetilde{\text{Σ}}}^{\mu \rho \sigma }\equiv {\text{Σ}}^{\mu \rho \sigma }-{\text{Φ}}^{\mu \rho \sigma }\mathrm{,}$ (52) ${\text{Θ}}^{\mu \nu }\to {\widetilde{\text{Θ}}}^{\mu \nu }\equiv {\text{Θ}}^{\mu \nu }+\frac{1}{2}{\partial }_{\lambda }\left({\text{Φ}}^{\lambda \mu \nu }-{\text{Φ}}^{\mu \lambda \nu }-{\text{Φ}}^{\nu \lambda \mu }\right)\mathrm{,}$ (53) which preserves both Eq. (49) and Eq. (1). Given a spacelike hypersurface Ξ, the total energy-momentum and total angular momentum across Ξ are $P^{\nu }={\displaystyle \int }d{\text{Ξ}}_{\mu }{\text{Θ}}^{\mu \nu }\mathrm{,}$ (54) $\begin{array}{l}{M}^{\rho \sigma }={\displaystyle \int }d{\text{Ξ}}_{\mu }{M}^{\mu \rho \sigma }\\ ={\displaystyle \int }d{\text{Ξ}}_{\mu }\left(x^{\rho }{\text{Θ}}^{\mu \sigma }-x^{\sigma }{\text{Θ}}^{\mu \rho }+{\text{Σ}}^{\mu \rho \sigma }\right)\mathrm{.}\end{array}$ (55) One can check that Pμ and $M^{\rho \sigma }$ are invariant under pseudo-gauge transformation (52) and (53) if the pseudo-gauge field ${\text{Φ}}^{\mu \rho \sigma }$ vanishes at the boundary of Ξ⁴.

One consequence of the pseudo-gauge transformation is the freedom to choose the symmetry properties of the spin tensor. To illustrate this, we consider an example in which we aim to transform the general spin tensor ${\text{Σ}}^{\mu \rho \sigma }=-{\text{Σ}}^{\mu \sigma \rho }$ into a completely antisymmetric form. We can choose ${\text{Φ}}^{\mu \rho \sigma }={\text{Σ}}^{\left(\mu \rho \right)\sigma }-\frac{1}{2}{\text{Σ}}^{\sigma \mu \rho }$. After applying the pseudo-gauge transformation, we obtain ${\text{Σ}}^{\mu \rho \sigma }\to {\widetilde{\text{Σ}}}^{\mu \rho \sigma }=\frac{1}{2}\left({\text{Σ}}^{\mu \rho \sigma }-{\text{Σ}}^{\rho \mu \sigma }+{\text{Σ}}^{\sigma \mu \rho }\right)\mathrm{,}$ (56) ${\text{Θ}}^{\mu \nu }\to {\widetilde{\text{Θ}}}^{\mu \nu }={\text{Θ}}^{\mu \nu }+\frac{1}{4}{\partial }_{\lambda }\left(3{\text{Σ}}^{\nu \mu \lambda }+{\text{Σ}}^{\mu \nu \lambda }-{\text{Σ}}^{\lambda \nu \mu }\right)\mathrm{.}$ (57) Note that the obtained ${\widetilde{\text{Σ}}}^{\mu \rho \sigma }$ is totally antisymmetric; therefore, it can be parameterized as ${\widetilde{\text{Σ}}}^{\mu \rho \sigma }=-{ϵ}^{\mu \rho \sigma \nu }{\tilde{S}}_{\nu }\mathrm{,}$ (58) where ${\tilde{S}}^{\mu }$ denotes the corresponding spin (pseudo)vector. Thus, the spin density tensor is thus ${\tilde{S}}^{\mu \nu }=-{ϵ}^{\mu \nu \rho \sigma }{u}_{\rho }{\tilde{S}}_{\sigma }$. The main difference between this spin density tensor and that used in Sect. 3 is that ${\tilde{S}}^{\mu \nu }$ contains three degrees of freedom corresponding to the three spatial spin vectors, whereas $S^{\mu \nu }$ has six degrees of freedom, with three for spatial spin and three for boost. Thus, in some cases, it is more convenient to use ${\widetilde{\text{Σ}}}^{\mu \rho \sigma }$ to construct the spin hydrodynamics. By following a procedure similar to that adopted in Sect. 3, we can derive constitutive relations in this context. In doing so, we decompose ${\tilde{S}}^{\mu }$ into ${\tilde{S}}^{\mu }={\sigma }^{\mu }+n_5{u}^{\mu }$, where ${\sigma }^{\mu }$ represents the spatial spin with the condition $\sigma \cdot u=0$, and n₅ is a pseudoscalar field (hence subscript 5). We also decompose ${\widetilde{\text{Θ}}}^{\mu \nu }$ into: ${\widetilde{\text{Θ}}}^{\mu \nu }=\epsilon u^{\mu }{u}^{\nu }-P{\text{Δ}}^{\mu \nu }+{\widetilde{\text{Θ}}}_{s\left(1\right)}^{\mu \nu }+{\tilde{q}}^{\mu }{u}^{\nu }-{\tilde{q}}^{\nu }{u}^{\mu }+{\tilde{\varphi }}^{\mu \nu }\mathrm{,}$ (59) where, as in Sect. 3, we assumed the Landau-Lifshitz frame ${\widetilde{\text{Θ}}}_s^{\mu \nu }{u}_{\nu }=\epsilon u^{\mu }\mathrm{,}$ (60) such that ${\widetilde{\text{Θ}}}_{s\left(1\right)}^{\mu \nu }$ is purely transverse to uμ. It is important to note that although we use the same symbols $\epsilon$, P, and uμ as in Sect. 3, their actual values may differ because the energy-momentum tensors and spin tensors in these two cases are different (but connected by the pseudo-gauge transformations (56) and (57)). We adopt a power counting scheme similar to the one we chose in Sect. 3, $\epsilon \mathrm{,}P\mathrm{,}T\mathrm{,}{u}^{\mu }\sim O(1)\mathrm{,}$ (61) ${\tilde{S}}^{\mu }\mathrm{,}{\tilde{q}}^{\mu }\mathrm{,}{\tilde{\varphi }}^{\mu \nu }\sim O(\partial )\mathrm{.}$ (62) Using the same form for the entropy current and the first law for local thermodynamics presented in Sect. 3, one can then find the divergence of the entropy current to be $T{\partial }_{\mu }{s}^{\mu }={\widetilde{\text{Θ}}}_{s\left(1\right)}^{\mu \nu }{\partial }_{(\mu }{u}_{\nu )}+{\widetilde{\text{Θ}}}_a^{\mu \nu }\left({\tilde{\mu }}_{\mu \nu }+T{\partial }_{[\mu }{\beta }_{\nu ]}\right)+O\left({\partial }^3\right)\mathrm{.}$ (63) We note that in deriving this result, we have utilized the fact that contracting the equation of motion (49) with $u^{\rho }$ reveals that ${\tilde{q}}^{\mu }$ is not an independent current, but is determined by ${\tilde{S}}^{\mu }$ through the following relation: ${\tilde{q}}^{\mu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{u}_{\nu }{\nabla }_{\rho }{\tilde{S}}_{\sigma }\mathrm{.}$ (64) This is because when the spin tensor is completely antisymmetric, the components responsible for the boost are gauged away, meaning that the corresponding torque for the boost in the antisymmetric part of the energy-momentum tensor cannot be an independent current either. Owing to this relationship, we can show that $n_5=\tilde{S}\cdot u$ is actually an O(∂³) quantity (and thus does not appear on the right-hand side of Eq. (63)). In fact, through direct calculation, one can find that the higher order terms that are neglected in Eq. (63) contains only one term: $\propto n_5$, $\frac{1}{2}{n}_5{ϵ}^{\mu \nu \rho \sigma }{u}_{\sigma }\left[{\partial }_{\mu }\left(\beta {\tilde{\mu }}_{\nu \rho }\right)+{\nabla }_{\mu }{u}_{\rho }\left({\partial }_{\nu }\beta +D{\beta }_{\nu }\right)\right]\mathrm{,}$ (65) which infers that $n_5\propto {ϵ}^{\mu \nu \rho \sigma }{u}_{\sigma }\left[{\partial }_{\mu }\left(\beta {\tilde{\mu }}_{\nu \rho }\right)+{\nabla }_{\mu }{u}_{\rho }\left({\partial }_{\nu }\beta +D{\beta }_{\nu }\right)\right]\sim O\left({\partial }^3\right)$ and thus can be neglected [62]. Therefore, from Eq. (63), we derive the constitutive relations for spin hydrodynamics with a completely antisymmetric spin tensor as follows [62]: ${\widetilde{\text{Θ}}}_{s\left(1\right)}^{\mu \nu }=\zeta \theta {\text{Δ}}^{\mu \nu }+2\eta {\sigma }^{\mu \nu }\mathrm{,}$ (66) ${\widetilde{\text{Θ}}}_{a\left(1\right)}^{\mu \nu }={\tilde{\varphi }}_{\left(1\right)}^{\mu \nu }={\eta }_s{\text{Δ}}^{\mu \rho }{\text{Δ}}^{\nu \sigma }\left({\mu }_{\rho \sigma }-T{\varpi }_{\rho \sigma }\right)\mathrm{.}$ (67) Although these relations take the same form as those obtained in Sect. 3, it is important to note that they apply specifically to the pseudo-gauge of a completely antisymmetric spin tensor. These relations are particularly convenient for describing the evolution of spatial spin degrees of freedom.

Thus, choosing different forms for the spin tensor (loosely referred to as different pseudo-gauges) leads to different forms for the constitutive relations. In an extreme case, one might even select ${\text{Φ}}^{\mu \rho \sigma }={\text{Σ}}^{\mu \rho \sigma }$, which completely eliminates the spin tensor and renders the energy-momentum tensor totally symmetric (this choice is commonly referred to as the Belinfante gauge [111-113]). While this may seem to eliminate all information about spin in hydrodynamics, the energy density, viscous tensors, and heat current remain influenced by spin, meaning that the dynamics of spin are still embedded within these quantities. For discussions on the transformation from canonical to Belinfante gauges, see Refs. [69, 71, 75, 99, 114, 115]. In addition, other pseudo-gauges have been employed and discussed in the context of spin hydrodynamics [85, 86, 88, 116-119].

Spin hydrodynamics for strong vorticity

The power counting scheme employed in the previous discussions is motivated by the observation that at global equilibrium, the spin potential μμν is determined by the thermal vorticity ${\varpi }_{\mu \nu }=\left({\partial }_{\nu }{\beta }_{\mu }-{\partial }_{\mu }{\beta }_{\nu }\right)/2$, which is naturally assumed to be an O(∂) quantity. However, this assumption may not hold true because the global equilibrium allows for arbitrarily large rotations (vorticity). When the vorticity is large, the assignment ${\varpi }_{\mu \nu }\sim O(\partial )$ becomes inadequate; instead, it is more appropriate to consider that ${\varpi }_{\mu \nu }\sim O(1)$. We explore this situation in this subsection, following closely the discussions in Ref. [72]. Before going into the details, it is useful to compare spin hydrodynamics with magnetohydrodynamics (MHD), in which the magnetic field is treated as an O(1) quantity (See Ref. [120] for a review of relativistic MHD).

The MHD describes the coupled evolution of fluid energy-momentum (or temperature and velocity) and the electromagnetic field in the low-energy and long-wavelength regimes. The fundamental equations consist of the conservation laws for the energy-momentum tensor and Maxwell’s equations. Owing to the screening effect, the electric fields within the fluid are gapped and parametrically small compared to the magnetic field. This renders the electric field inactive in the low-energy, long-wavelength regime. In contrast, there is no screening of the magnetic field, allowing it to exhibit its own dynamics even in this regime. Consequently, the magnetic field can be large and is treated as an O(1) quantity, despite the fact that $B=\nabla \times A$ involves one spatial gradient. The presence of an O(1) magnetic field breaks the SO(3) symmetry in the constitutive relations for Θ^μv, introducing anisotropy even in ideal hydrodynamics. Specifically, we can define a normalized vector $b^{\mu }=B^{\mu }/B$, where $B=\sqrt{-B^{\mu }{B}_{\mu }}$, satisfying b²=-1 and $b\cdot u=0$, as an additional building block for hydrodynamic constitutive relations. For example, for a partiy-even and charge neutral fluid, the energy-momentum tensor can be decomposed into ${\text{Θ}}^{\mu \nu }=\epsilon u^{\mu }{u}^{\nu }-P_{\perp }{\text{Ξ}}^{\mu \nu }+P_{\parallel }{b}^{\mu }{b}^{\nu }+{\text{Θ}}_{\left(1\right)}^{\mu \nu }\mathrm{,}$ (68) where ${\text{Ξ}}^{\mu \nu }={\text{Δ}}^{\mu \nu }+b^{\nu }{b}^{\nu }$ is a projector transverse to both uμ and Bμ. The terms $P_{\perp }$ and $P_{\parallel }$ represent the pressures in directions transverse and parallel to the magnetic field, respectively. Note that when we allow an environmental parity violation (e.g., when there is a density imbalance between the right- and left-hand particles in the fluid) and a finite charge density, the term $u^{(\mu }{b}^{\nu )}$ can appear in the zeroth order. The term ${\text{Θ}}_{\left(1\right)}^{\mu \nu }$ (which is assumed to be symmetric because the spin degree of freedom is typically disregarded in MHD) denotes a collection of terms that are at least of order O(∂) in the gradient expansion and consistent with the Onsager relations. For a parity-even fluid, all such terms are expressed as ${\text{Θ}}_{\left(1\right)}^{\mu \nu }={\displaystyle {\sum }_{i=1}^7{\lambda }_i}{\eta }_i^{\mu \nu \rho \sigma }{\nabla }_{\rho }{u}_{\sigma }$, where ${\lambda }_i$ is the corresponding transport coefficient [121-23]: ${\eta }_1^{\mu \nu \rho \sigma }=b^{\mu }{b}^{\nu }{b}^{\rho }{b}^{\sigma }\mathrm{,}$ (69a) ${\eta }_2^{\mu \nu \rho \sigma }={\text{Ξ}}^{\mu \nu }{\text{Ξ}}^{\rho \sigma }\mathrm{,}$ (69b) ${\eta }_3^{\mu \nu \rho \sigma }=-{\text{Ξ}}^{\mu \nu }{b}^{\rho }{b}^{\sigma }-{\text{Ξ}}^{\rho \sigma }{b}^{\mu }{b}^{\nu }\mathrm{,}$ (69c) ${\eta }_4^{\mu \nu \rho \sigma }=-2\left[b^{(\mu }{\text{Ξ}}^{\nu )\rho }{b}^{\sigma }+b^{(\mu }{\text{Ξ}}^{\nu )\sigma }{b}^{\rho }\right]\mathrm{,}$ (69d) ${\eta }_5^{\mu \nu \rho \sigma }=2{\text{Ξ}}^{\rho (\mu }{\text{Ξ}}^{\nu )\sigma }-{\text{Ξ}}^{\mu \nu }{\text{Ξ}}^{\rho \sigma }\mathrm{,}$ (69e) ${\eta }_6^{\mu \nu \rho \sigma }=-b^{(\mu }{b}^{\nu )\rho }{b}^{\sigma }-b^{(\mu }{b}^{\nu )\sigma }{b}^{\rho }\mathrm{,}$ (69f) ${\eta }_7^{\mu \nu \rho \sigma }={\text{Ξ}}^{\rho (\mu }{b}^{\nu )\sigma }+{\text{Ξ}}^{\sigma (\mu }{b}^{\nu )\rho }\mathrm{,}$ (69g) where $b^{\mu \nu }={ϵ}^{\mu \nu \rho \sigma }{u}_{\rho }{b}_{\sigma }$ is a cross projector that appears only when charge-conjugation symmetry is violated (e.g., when a net charge density is presented).

Similar to the discussions above regarding MHD, we can consider a scenario for spin hydrodynamics where the vorticity is treated as zeroth-order in gradients, while the gradients of other thermodynamic quantities are treated as first-order. In line with the MHD, this framework has been referred to as gyrohydrodynamics in Ref. [72]. To simplify the notation, we reuse Bμ to denote the unit vector along the vorticity, $b^{\mu }={\varpi }^{\mu }/\sqrt{-{\varpi }_{\mu }{\varpi }^{\mu }}={\omega }^{\mu }/\sqrt{-{\omega }_{\mu }{\omega }^{\mu }}\mathrm{,}$ (70) with ${\varpi }^{\mu }=-{ϵ}^{\mu \nu \rho \sigma }{u}_{\nu }{\partial }_{\rho }{\beta }_{\sigma }/2=\beta {\omega }^{\mu }$ the thermal vorticity vector. We chose the pesudo-gauge such that the spin tensor is totally antisymmetric. Using $u^{\mu }\mathrm{,}{b}^{\mu }$ as well as $g^{\mu \nu }\mathrm{,}{ϵ}^{\mu \nu \rho \sigma }$ as building blocks, we can decompose Θ^μv and ${\text{Σ}}^{\mu \nu \rho }$ into the following irreducible forms: $\begin{array}{c}{\text{Θ}}^{\mu \nu }=\epsilon u^{\mu }{u}^{\nu }-P_{\perp }{\text{Ξ}}^{\mu \nu }+P_{\parallel }{b}^{\mu }{b}^{\nu }+P_{\times }{b}^{\mu \nu }\\ +q^{\mu }{u}^{\nu }-u^{\mu }{q}^{\nu }+{\text{Θ}}_{s\left(1\right)}^{\mu \nu }+{\varphi }^{\mu \nu }\mathrm{,}\end{array}$ (71) ${\text{Σ}}^{\mu \nu \rho }=-{ϵ}^{\mu \nu \rho \lambda }{S}_{\lambda }=-{ϵ}^{\mu \nu \rho \lambda }\left(n_5{u}_{\lambda }-S_{\parallel }{b}_{\lambda }+S_{\perp \lambda }\right)\mathrm{,}$ (72) where $P_{\perp \mathrm{,}\parallel \times }$ represent pressures (which will be counted as O(1) quantities in gradient expansion) in different directions, whose physical meaning will become clear shortly. The quantity $S_{\parallel }=b\cdot S$ denotes the spin component in the direction of the vorticity, while $S_{\perp }^{\mu }={\text{Ξ}}^{\mu \nu }{S}_{\nu }$ denotes the spin component transverse to the vorticity. As before, we chose the Landau-Lifshitz frame, with qμ, ${\text{Θ}}_{s\left(1\right)}^{\mu \nu }$, ${\varphi }^{\mu \nu }$, and $S_{\perp }^{\mu }$ transverse to uμ. Note again that, with this fully antisymmetric choice of spin tensor, the qμ vector is no longer independent, but is determined by Sμ through Eq. (64).

The power counting scheme is such that $S_{\parallel }$ is counted as order one, whereas ${\varphi }^{\mu \nu }=-{\varphi }^{\nu \mu }$, $S_{\perp }^{\mu }$, n₅, and qμ (see Eq. (64)) are counted as at least O(∂). Additionally, we will count Sμ as $O(\hslash )$ (since spin is totally quantum in nature) in comparison to other thermodynamic quantities, which can appear even at the classical level and are therefore assigned $O\left({\hslash }^0\right)$. This allows for a double expansion in both $\partial$ and ℏ. For the entropy current, we can write $s^{\mu }=s{u}^{\mu }+s_{\left(1\right)}^{\mu }$ and use Eq. (33). It is straightforward to derive the divergence of the entropy current, and after some calculations, it was found that up to $O\left(\hslash {\partial }^2\mathrm{,}{\partial }^3\right)$ [72]: $\begin{array}{c}{\partial }_{\mu }{s}^{\mu }=\left[s-\beta \left(\epsilon +P_{\perp }\right)\right]\theta -\left(P_{\parallel }-P_{\perp }-{\mu }_{\parallel }{S}_{\parallel }\right)b^{\mu }{b}^{\nu }{\partial }_{\mu }{\beta }_{\nu }\\ +P_{\times }{b}^{\mu \nu }\left({\partial }_{\mu }{\beta }_{\nu }+\beta {\mu }_{\mu \nu }\right)+{\text{Θ}}_{s\left(1\right)}^{\mu \nu }{\partial }_{(\mu }{\beta }_{\nu )}+{\varphi }^{\mu \nu }({\partial }_{[\mu }{\beta }_{\nu ]}\\ +\beta {\mu }_{\mu \nu })+{\partial }_{\mu }\left(s_{\left(1\right)}^{\mu }-\beta {\mu }^{\mu }{n}_5\right)+O\left(\hslash {\partial }^2\mathrm{,}{\partial }^3\right)\mathrm{.}\end{array}$ (73) The first line provides the zero-order contribution to the entropy production, which is expected to vanish so that they represent non-dissipative contributions. This gives $\epsilon +P_{\perp }=Ts\mathrm{,}{P}_{\parallel }=P_{\perp }+{\mu }_{\parallel }{S}_{\parallel }\mathrm{,}{P}_{\times }=0.$ (74) The first relation is the Gibbs-Duhem relation, indicating that $P_{\perp }$ can be interpreted as the thermodynamic pressure. The second relation shows that the pressure along the vorticity direction differs from the thermodynamic pressure by an amount due to spin polarization ${\mu }_{\parallel }{S}_{\parallel }$. This term is similar to the MB term in the magnetohydrodynamic constitutive relation. The third relation shows that there is no spin torque at the leading order.

At O(∂) order, the requirement of a semi-positive entropy production gives that ${\text{Θ}}_{s\left(1\right)}^{\mu \nu }=T{\eta }^{\mu \nu \rho \sigma }{\partial }_{(\rho }{\beta }_{\sigma )}+T{\xi }^{\mu \nu \rho \sigma }\left({\partial }_{[\rho }{\beta }_{\sigma ]}+\beta {\mu }_{\rho \sigma }\right)\mathrm{,}$ (75) ${\varphi }^{\mu \nu }=T{\gamma }^{\mu \nu \rho \sigma }\left({\partial }_{[\rho }{\beta }_{\sigma ]}+\beta {\mu }_{\rho \sigma }\right)+T{\xi }^{\text{'}\mu \nu \rho \sigma }{\partial }_{(\rho }{\beta }_{\sigma )}\mathrm{,}$ (76) $s_{\left(1\right)}^{\mu }=\beta {\mu }^{\mu }{n}_5\mathrm{,}$ (77) where ${\eta }^{\mu \nu \rho \sigma }$ and ${\gamma }^{\mu \nu \rho \sigma }$ are the usual and rotational viscous tensors representing the response of the symmetric and antisymmetric parts of the energy-momentum tenor to fluid shear and expansion, and the difference between vorticity and spin potential, respectively, and ${\xi }^{\mu \nu \rho \sigma }$ and ${{\xi }^{\prime }}^{\mu \nu \rho \sigma }$ are two cross viscous tensors. Note that the cross viscous tensors are not independent of each other but inter-related according to Onsager’s reciprocal principle, ${\xi }^{\text{'}\mu \nu \rho \sigma }\left(b\right)={\xi }^{\rho }\left(-b\right)$. By decomposing these tensors into irreducible structures, one obtains a number of new transport coefficients (viscosities) that characterize the response of the fluid to gradients of fluid velocity and spin potential [72]: $\begin{array}{c}{\eta }^{\mu \nu \rho \sigma }={\zeta }_{\perp }{\text{Ξ}}^{\mu \nu }{\text{Ξ}}^{\rho \sigma }+{\zeta }_{\parallel }{b}^{\mu }{b}^{\nu }{b}^{\rho }{b}^{\sigma }+{\zeta }_{\times }\left(b^{\mu }{b}^{\nu }{\text{Ξ}}^{\rho \sigma }+{\text{Ξ}}^{\mu \nu }{b}^{\rho }{b}^{\sigma }\right)\\ +{\eta }_{\perp }\left({\text{Ξ}}^{\mu \rho }{\text{Ξ}}^{\nu \sigma }+{\text{Ξ}}^{\mu \sigma }{\text{Ξ}}^{\nu \rho }-{\text{Ξ}}^{\mu \nu }{\text{Ξ}}^{\rho \sigma }\right)\\ +2{\eta }_{\parallel }\left(b^{\mu }{\text{Ξ}}^{\nu (\rho }{b}^{\sigma )}+b^{\nu }{\text{Ξ}}^{\mu (\rho }{b}^{\sigma )}\right)\\ +2{\eta }_{H_{\perp }}\left({\text{Ξ}}^{\mu (\rho }{b}^{\sigma )\nu }+{\text{Ξ}}^{\nu (\rho }{b}^{\sigma )\mu }\right)\\ +2{\eta }_{H_{\parallel }}\left(b^{\mu }{b}^{\nu (\rho }{b}^{\sigma )}+b^{\nu }{b}^{\mu (\rho }{b}^{\sigma )}\right)\mathrm{,}\end{array}$ (78) $\begin{array}{c}{\gamma }^{\mu \nu \rho \sigma }={\gamma }_{\perp }\left({\text{Ξ}}^{\mu \rho }{\text{Ξ}}^{\nu \sigma }-{\text{Ξ}}^{\mu \sigma }{\text{Ξ}}^{\nu \rho }\right)\\ +2{\gamma }_{\parallel }\left(b^{\mu }{\text{Ξ}}^{\nu [\rho }{b}^{\sigma ]}-b^{\nu }{\text{Ξ}}^{\mu [\rho }{b}^{\sigma ]}\right)\\ +2{\gamma }_H\left(b^{\mu }{b}^{\nu [\rho }{b}^{\sigma ]}-b^{\nu }{b}^{\mu [\rho }{b}^{\sigma ]}\right)\mathrm{,}\end{array}$ (79) $\begin{array}{c}{\xi }^{\mu \nu \rho \sigma }=2{\xi }_{\parallel }\left(b^{\mu }{\text{Ξ}}^{\nu [\rho }{b}^{\sigma ]}+b^{\nu }{\text{Ξ}}^{\mu [\rho }{b}^{\sigma ]}\right)\\ +{\zeta }_{H_{\perp }}{\text{Ξ}}^{\mu \nu }{b}^{\rho \sigma }+{\zeta }_{H\parallel }{b}^{\mu }{b}^{\nu }{b}^{\rho \sigma }\\ +2{\xi }_H\left(b^{\mu }{b}^{\nu [\rho }{b}^{\sigma ]}+b^{\nu }{b}^{\mu [\rho }{b}^{\sigma ]}\right)\mathrm{,}\end{array}$ (80) where the η’s, $\zeta$’s, $\gamma$’s, and Ξ’s are transport coefficients. Especially, those with subscript “H” are Hall-type transport coefficients which do not contribute to the entropy production and thus their sign are not constrained by the second law of local thermodynamics. One may wonder why the term $\propto b^{\mu \nu }{b}^{\rho \sigma }$ (such term would contribute to an $O(\partial )$ analog of $P_{\times }$ term in Eq. (71)) does not appear in ${\gamma }^{\mu \nu \rho \sigma }$. This is because it is not independent of the other terms in ${\gamma }^{\mu \nu \rho \sigma }$ [121]. Note that the expression for ${\xi }^{\mu \nu \rho \sigma }$ is different from that in Ref. [72] but equivalently yields the same constitutive relations once substituted into Eq. (75).

A spin Cooper-Frye formula

To apply spin hydrodynamics to specific physical systems, we need to know the appropriate observables for the detection of spin degrees of freedom in the fluid. In principle, the presence of the spin degree of freedom in the fluid should modify the usual hydrodynamic quantities such as the energy density and fluid velocity, but when the spin density is not large (nevertheless it is always suppressed by ℏ comparing to the traditional hydrodynamic quantities), such modification is small. In heavy ion collisions, the natural observable is the spin polarization of hadrons, including spin-1/2 hyperons and spin-1 vector mesons. Hyperons are of special interest because they primarily decay via weak interactions such that the momentum of one of the daughter particles tends to align with the spin direction of the hyperon. To obtain the spin polarization observables of a hadron from spin hydrodynamics, machinery is required to convert the outcomes of spin hydrodynamics, such as fluid velocity, temperature, and spin potential, to measurable hadronic observables.

In the application of traditional hydrodynamics to heavy-ion collisions, the hadron momentum spectra are typically obtained using the Cooper-Frye formula: $E_p\frac{d{N}_i}{d{p}^3}={\displaystyle {\int }_{\text{Ξ}}}d{\text{Ξ}}_{\mu }\left(x\right)p^{\mu }{f}_i\left(x\mathrm{,}p\right)\mathrm{,}$ (81) where the integral is over the freeze-out hypersurface (where particlization occurs) Ξ, and fi(x,p) is the distribution function of species i of the hadrons in the fluid. Any possible degeneracy of the hadrons should be accounted for in fi. For example, when dissipative effects are neglected, the distribution function fi is typically taken as the Fermi-Dirac or Bose-Einstein functions $f_{F\mathrm{,}B}\left(p\cdot \beta -{\mu }_i\right)$ with ${\mu }_i$ is the chemical potential. The above Cooper-Frye formula has been widely used in hydrodynamic simulations in heavy-ion collisions and has proven to be very successful. Therefore, to extend traditional hydrodynamics to spin hydrodynamics, we also need to generalize the above Cooper-Frye formula to a spin Cooper-Frye formula.

Let us consider a system in which thermal equilibrium is reached locally but not necessarily globally. The density operator $\widehat{\rho }$ for the description of such an ensemble is obtained by maximizing the entropy functional under the constraints of the given energy-momentum and angular momentum (or spin) densities: $\begin{array}{c}S\left[\widehat{\rho }\right]=-\text{Tr}\left(\widehat{\rho }\ln \widehat{\rho }\right)+\lambda \left(\text{Tr}\widehat{\rho }-1\right)-{\displaystyle {\int }_{\text{Ξ}}}d{\text{Ξ}}_{\mu }\left[\text{Tr}\left(\widehat{\rho }{\widehat{\text{Θ}}}^{\mu \nu }\right)-{\text{Θ}}^{\mu \nu }\right]{\beta }_{\nu }\\ +\frac{1}{2}{\displaystyle {\int }_{\text{Ξ}}}d{\text{Ξ}}_{\mu }\left[\text{Tr}\left(\widehat{\rho }{\widehat{\text{Σ}}}^{\mu \nu \rho }\right)-{\text{Σ}}^{\mu \nu \rho }\right]{\mu }_{\nu \rho }\mathrm{,}\end{array}$ (82) where ${\text{Θ}}^{\mu \nu }\left(x\right)$ and ${\text{Σ}}^{\mu \nu \rho }\left(x\right)$ are the actual local energy-momentum tensor and spin tensor, respectively, and ${\beta }_{\nu }\left(x\right)$ and ${\mu }_{\nu \rho }\left(x\right)$ are the corresponding Lagrange multipliers. The Lagrange multiplier $\lambda$ is introduced to normalize $\widehat{\rho }$ and is related to the partition function Z as $\exp (1-\lambda )=Z$. The resultant density operator is the local-equilibrium density operator [124-127]: ${\widehat{\rho }}_{\text{LE}}=\frac{1}{Z_{\text{LE}}}\exp \left\{-{\displaystyle {\int }_{\text{Ξ}}}d{\text{Ξ}}_{\mu }\left(x\right)\left[{\widehat{\text{Θ}}}^{\mu \nu }\left(x\right){\beta }_{\nu }\left(x\right)-\frac{1}{2}{\widehat{\text{Σ}}}^{\mu \rho \sigma }\left(x\right){\mu }_{\rho \sigma }\left(x\right)\right]\right\}\mathrm{,}$ (83) where $Z_{\text{LE}}$ denote the local-equilibrium partition function. Now, we see that ${\widehat{\rho }}_{\text{LE}}$ is determined by the local thermodynamic quantities βμ and ${\mu }_{\rho \sigma }$. If we calculate the spin density ${\sum }^{\mu \rho \sigma }\left(x\right)$ using this density operator, we obtain a relationship between ${\sum }^{\mu \rho \sigma }\left(x\right)$ and the local thermodynamic quantities (and possibly their derivatives). However, this is not particularly useful in the context of heavy-ion collisions because what is measured is the spin density in momentum space, rather than the coordinate space. To express such a relation for the spin density in momentum space or phase space, the most natural approach is to use the Wigner function.

To illustrate how this can be achieved, we consider a Dirac fermion system as an example. The Wigner operator is defined as follows: $\widehat{W}\left(x\mathrm{,}p\right)={\displaystyle \int }{d}^{4}s{e}^{-ip\cdot s}\overline{\widehat{\psi }}\left(x+\frac{s}{2}\right)\otimes \widehat{\psi }\left(x-\frac{s}{2}\right)\mathrm{,}$ (84) where ${\left[\overline{\widehat{\psi }}\otimes \widehat{\psi }\right]}_{ab}\equiv {\overline{\widehat{\psi }}}_b{\widehat{\psi }}_a$ with a, b spinor indices. We choose the canonical pseudo-gauge, in which the energy-momentum tensor operator and spin tensor operator are given by ${\widehat{\text{Θ}}}^{\mu \nu }=\overline{\widehat{\psi }}i{\gamma }^{\mu }{\partial }^{\nu }\widehat{\psi }-{\eta }^{\mu \nu }\widehat{\mathcal{L}}\mathrm{,}$ (85) ${\widehat{\text{Σ}}}^{\mu \nu \rho }=\frac{1}{4}\overline{\widehat{\psi }}\left\{{\gamma }^{\mu }\mathrm{,}{\sigma }^{\rho \sigma }\right\}\widehat{\psi }=-\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }\overline{\widehat{\psi }}{\gamma }_{\sigma }{\gamma }_5\widehat{\psi }\mathrm{.}$ (86) where $\widehat{\mathcal{L}}$ is the Lagrangian (in the following, we consider free fermions, so that $\mathcal{L}=\overline{\psi }\left(i{\gamma }^{\mu }{\partial }_{\mu }-m\right)\psi$ is in quadratic form, and the second term in ${\widehat{\text{Θ}}}^{\mu \nu }$ vanishes when using the equation of motion of the field operator) and ${\sigma }^{\rho \sigma }=i\left[{\gamma }^{\rho }\mathrm{,}{\gamma }^{\sigma }\right]/2$. Note that the second equation indicates that the spin vector ${\widehat{S}}_{\sigma }=(1/2)\overline{\widehat{\psi }}{\gamma }_{\sigma }{\gamma }_5\widehat{\psi }$ is half the axial current. Both ${\widehat{\text{Θ}}}^{\mu \nu }\left(x\right)$ and ${\widehat{\text{Σ}}}^{\mu \nu \rho }\left(x\right)$ are local Heisenberg operators. We can extend them into operators in the phase space by using the Wigner transformation, for example, $\begin{array}{c}{\widehat{\text{Σ}}}^{\mu \nu \rho }\left(x\mathrm{,}p\right)=-\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{\displaystyle \int }{d}^{4}s{e}^{-ip\cdot s}\overline{\widehat{\psi }}\left(x+\frac{s}{2}\right){\gamma }_{\sigma }{\gamma }_5\widehat{\psi }\left(x-\frac{s}{2}\right)\\ =-\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{\text{Tr}}_{\text{D}}\left[{\gamma }_{\sigma }{\gamma }_5\widehat{W}\left(x\mathrm{,}p\right)\right]\mathrm{,}\end{array}$ (87) where ${\text{Tr}}_{\text{D}}$ is the trace over the Dirac space. It is easy to see that ${\displaystyle \int }{d}^{4}p/{\left(2\pi \right)}^4{\widehat{\text{Σ}}}^{\mu \nu \rho }\left(x\mathrm{,}p\right)={\widehat{\text{Σ}}}^{\mu \nu \rho }\left(x\right)$. The integration of ${\widehat{\text{Σ}}}^{\mu \nu \rho }\left(x\mathrm{,}p\right)$ over a certain spacelike hypersurface gives us the spin tensor in momentum space (whose exact meaning will be clarified later), whose ensemble average under ${\widehat{\rho }}_{\text{LE}}$ is exactly the quantity that we are looking for. Therefore, we must calculate the Wigner function under local equilibrium: $W\left(x\mathrm{,}p\right)=\langle \widehat{W}\left(x\mathrm{,}p\right)\rangle =\text{T}r\left[{\widehat{\rho }}_{\text{LE}}\widehat{W}\left(x\mathrm{,}p\right)\right]\mathrm{,}$ (88) where Tr denotes the trace over a complete set of microstates in the system. To proceed, the local-equilibrium density operator can be rewritten as ${\widehat{\rho }}_{\text{LE}}=\exp \left(\widehat{A}+\widehat{B}\right)/Z_{\text{LE}}$ with the abbreviations $\widehat{A}=-{\widehat{P}}^{\mu }{\beta }_{\mu }\left(x\right)\mathrm{,}$ (89) $\widehat{B}=-{\displaystyle \int }d{\text{Ξ}}_{\mu }\left(y\right)\left[{\widehat{\text{Θ}}}^{\mu \nu }\left(y\right)\text{Δ}{\beta }_{\nu }\left(y\right)-\frac{1}{2}{\widehat{\text{Σ}}}^{\mu \rho \sigma }\left(y\right){\mu }_{\rho \sigma }\left(y\right)\right]\mathrm{,}$ (90) where ${\widehat{P}}^{\mu }={\displaystyle \int }d{\text{Ξ}}_{\nu }\left(y\right){\widehat{\text{Θ}}}^{\nu \mu }\left(y\right)$, $\text{Δ}{\beta }_{\mu }\left(y\right)={\beta }_{\mu }\left(y\right)-{\beta }_{\mu }\left(x\right)$. The purpose of rewriting ${\widehat{\rho }}_{\text{LE}}$ in this form is that, the correlation length between the spin tensor and the energy-momentum tensor is typically small. Within this correlation length, we can assume that local thermodynamic quantities, such as βμ, vary only slightly. Given that ${\mu }_{\rho \sigma }$ is also small at the hypersurface Ξ (which is a reasonable assumption for heavy-ion collisions, although it may not hold for a strongly polarized medium), we assign $\text{Δ}{\beta }_{\nu }\sim {\mu }_{\rho \sigma }\sim O(\partial )$, therefore, $\widehat{A}\sim O(1)$, $\widehat{B}\sim O(\partial )$. Using this power-counting scheme, we can expand the right-hand side of Eq. (88) order by order in $\partial$ by applying the identity $e^{\widehat{A}+\widehat{B}}=e^{\widehat{A}}+e^{\widehat{A}}{\displaystyle {\int }_0^1}d\lambda e^{-\lambda \widehat{A}}\widehat{B}{e}^{\lambda \widehat{A}}+\cdot \cdot \cdot$, and obtain $W(x\mathrm{,}p)=W_0\left(x\mathrm{,}p\right)+W_1\left(x\mathrm{,}p\right)+\cdots \mathrm{,}$ (91) where $W_0\left(x\mathrm{,}p\right)={\langle \widehat{W}\left(x\mathrm{,}p\right)\rangle }_0\equiv \frac{1}{Z_0}\text{Tr}\left(e^{\widehat{A}}\widehat{W}\left(x\mathrm{,}p\right)\right)\mathrm{,}$ (92) $W_1\left(x\mathrm{,}p\right)\equiv {\langle \widehat{W}\left(x\mathrm{,}p\right)\rangle }_{\left(\text{Θ}\right)}+{\langle \widehat{W}\left(x\mathrm{,}p\right)\rangle }_{\left(\text{Σ}\right)}\mathrm{,}$ (93) with $\begin{array}{ll}{\langle \widehat{W}\left(x\mathrm{,}p\right)\rangle }_{\left(\text{Θ}\right)}\hfill & \equiv -{\displaystyle {\int }_0^1}d\lambda {\displaystyle \int }d{\text{Ξ}}_{\rho }\left(y\right)\text{Δ}{\beta }_{\nu }\left(y\right){\langle {\widehat{\text{Θ}}}^{\rho \nu }\left(y-i\lambda \beta (x)\right)\widehat{W}(x\mathrm{,}p)\rangle }_{\mathrm{0,}c}\mathrm{,}\hfill \\ {\langle \widehat{W}\left(x\mathrm{,}p\right)\rangle }_{\left(\text{Σ}\right)}\hfill & \equiv \frac{1}{2}{\displaystyle {\int }_0^1}d\lambda {\displaystyle \int }d{\text{Ξ}}_{\nu }\left(y\right){\mu }_{\rho \sigma }\left(y\right){\langle {\widehat{\text{Σ}}}^{\nu \rho \sigma }\left(y-i\lambda \beta (x)\right)\widehat{W}(x\mathrm{,}p)\rangle }_{\mathrm{0,}c}\mathrm{,}\hfill \end{array}$ (94) and $Z_0=\text{Tr}{e}^{\widehat{A}}$. Here, ${\langle \cdots \rangle }_{\mathrm{0,}c}$ means the connected part of the correlation. The calculation then will depend on the shape of the hypersurface Ξ. For illustration, we consider Ξ to be the 3-space at some time t so that its normal direction is ${\widehat{t}}^{\mu }=(\mathrm{1,}0)$. The calculation is then straightforward using the free field operator $\begin{array}{ll}\widehat{\psi }(x)\hfill & ={\displaystyle \sum _{\sigma =1}^2}\frac{1}{{\left(2\pi \right)}^{3/2}}{\displaystyle \int }\frac{d^{3}k}{2E_k}\left[u_{\sigma }\left(k\right)e^{-ik\cdot x}{\widehat{a}}_{\sigma }\left(k\right)+v_{\sigma }\left(k\right)e^{ik\cdot x}{\widehat{b}}_{\sigma }^{†}\left(k\right)\right]\mathrm{,}\hfill \end{array}$ (95) where $E_k=\sqrt{k^2+m^2}$ and ${\widehat{a}}_{\sigma }\left(k\right)\mathrm{,}{\widehat{b}}_{\sigma }\left(k\right)$ are annihilation operators for particles and antiparticles satisfying the anti-commutation relation $\left\{{\widehat{a}}_{\sigma }\left(k\right)\mathrm{,}{\widehat{a}}_{{\sigma }^{\prime }}^{†}\left(q\right)\right\}=\left\{{\widehat{b}}_{\sigma }\left(k\right)\mathrm{,}{\widehat{b}}_{{\sigma }^{\prime }}^{†}\left(q\right)\right\}=2E_k{\delta }_{\sigma {\sigma }^{\prime }}{\delta }^3\left(k-q\right)$ and the relation ${\langle {\widehat{a}}_{\sigma }^{†}\left(k\right){\widehat{a}}_{{\sigma }^{\prime }}\left(q\right)\rangle }_0={\langle {\widehat{b}}_{\sigma }^{†}\left(k\right){\widehat{b}}_{{\sigma }^{\prime }}\left(q\right)\rangle }_0=2E_k{\delta }_{\sigma {\sigma }^{\prime }}{\delta }^3\left(k-q\right)n_F\left(k\cdot \beta \right)$. In the following, we consider only the particle branch; the antiparticle branch is completely similar. The zeroth-order Wigner function can be easily obtained: $W_0\left(x\mathrm{,}p\right)=2\pi (\cancel{p}+m)\theta \left(p_0\right)\delta \left(p^2-m^2\right)n_F\left(p\cdot \beta \right)$, which is spin independent: ${\text{Tr}}_D\left[{\gamma }^{\mu }{\gamma }_5{W}_0\left(x\mathrm{,}p\right)\right]=0$.