Bivectorial Nonequilibrium Thermodynamics: Cycle Affinity, Vorticity Potential, and Onsager’s Principle

Abstract

We generalize an idea in the works of Landauer and Bennett on computations, and Hill’s in chemical kinetics, to emphasize the importance of kinetic cycles in mesoscopic nonequilibrium thermodynamics (NET). For continuous stochastic systems, a NET in phase space is formulated in terms of cycle affinity $\nabla \wedge \big (\mathbf{D}^{-1}\mathbf{b}\big )$ and vorticity potential $\mathbf{A}(\mathbf{x})$ of the stationary flux $\mathbf{J}^{*}=\nabla \times \mathbf{A}$. Each bivectorial cycle couples two transport processes represented by vectors and gives rise to Onsager’s notion of reciprocality; the scalar product of the two bivectors $\mathbf{A}\cdot \nabla \wedge \big (\mathbf{D}^{-1}\mathbf{b}\big )$ is the rate of local entropy production in the nonequilibrium steady state. An Onsager operator that relates vorticity to cycle affinity is introduced.

Appendix

Here we summarize the mathematics used to derive the results in the present work. In the main text, we used the notion of multivariable calculus and the notion of wedge product without the introduction of differential form for simplicity. However, the concept of differential form and the associated exterior calculus are needed to derive the formula of generalized curl and cross product for dimensions higher than 3 [1, 6]. We shall introduce and use the differential form calculus here. Throughout the text, we will use Cartesian coordinate to describe the entire Euclidean $\mathbb {R}^n$.

1.1 A: Differential Form and Integration

In vector calculus, the infinitesimal work done by a force $\varvec{f}$ from time t to $t+\delta t$ on a path $\mathbf{y}(t)$ is given by

$$\begin{aligned} \delta \mathcal {W}=\varvec{f}(\mathbf{y}(t)) \cdot \delta \mathbf{y}(t)=\sum _{j=1}^n f_j(\mathbf{y}(t)) \delta y_j(t+\delta t) \end{aligned}$$

(A 1)

where $\delta \mathbf{y}(t)$ denotes the infinitesimal vector $\mathbf{y}(t+\delta t)-\mathbf{y}(t)$. Usually, to emphasize the infinitesimal limit $\delta t \rightarrow 0$, we replace $\delta$ with $\mathrm{d}$, leading to a notation $\varvec{f}(\mathbf{y}(t)) \cdot \mathrm{d}\mathbf{y}(t)$. However, in the mathematics of differential form, the operater $\mathrm{d}$ is generalized and understood differently. In the main text, we used $\mathrm{d}$ as the standard infinitesimal difference operator in calculus. Here, we shall use $\delta$ as the infinitesimal operator and $\mathrm{d}$ as the exterior derivative of differential form, as we will introduce below.

We first introduce the concept of 1-forms, which are linear functions that maps a vector to a real number. Notice that the infinitesimal work in Eq. (A 1) actually takes the (tangent) vector $\delta \mathbf{y}(t)$ at a point $\mathbf{y}(t)$ and return us a number. The infinitesimal work is thus generally a differential 1-form at a given point, say $\varvec{\xi }$, associated with a force vector $\varvec{f}=(f_{1},f_{2},\ldots ,f_{n})$,

$$\begin{aligned} \omega _{\varvec{\xi }}(\mathbf{u})=\sum _{j=1}^{n}f_{j} (\varvec{\xi }) \mathrm{d}x_{j}(\mathbf{u}). \end{aligned}$$

(A 2)

It takes an infinitesimal vector $\mathbf{u}$ as an input and gives us the infinitesimal work generated when going from $\varvec{\xi }$ to $\varvec{\xi }+\mathbf{u}$. The basis of a 1-form is $\{\mathrm{d}x_{j}\}$, which are themselves 1-forms¹: $\mathrm{d}x_i$ takes a vector $\mathbf{u}$ and gives us its ith component,

$$\begin{aligned} \mathrm{d}x_i \left( \sum _j u_j \mathbf {e}_j\right) =\mathbf {e}_i\cdot \left( \sum _j u_j \mathbf {e}_j\right) =u_i \end{aligned}$$

(A 3)

where $\mathbf {e}_i$ is the unit vector in the ith direction. To match up Eq. (A 1) with Eq. (A 2), simply take $\mathbf{u}:=\delta \mathbf{y}(t)$ and $\varvec{\xi }=\mathbf{y}(t)$. The relation $\mathrm{d}x_i (\square ) = \mathbf {e}_i\cdot \square$ is what allows us to write the differential form in a vectorized expression in the main text,

$$\begin{aligned} \omega _{\varvec{\xi }}(\mathbf{u})&=\sum _i f_i(\varvec{\xi })\mathrm{d}x_i \left( \sum _j u_j \mathbf {e}_j\right) \end{aligned}$$

(A 4a)

$$\begin{aligned}&=\sum _i f_i(\varvec{\xi })\mathbf {e}_i\cdot \left( \sum _j u_j \mathbf {e}_j\right) =\varvec{f}(\varvec{\xi }) \cdot \mathbf{u}. \end{aligned}$$

(A 4b)

A differential form is what we can integrate over a manifold. The integral of the work over a path $\Gamma =\{\mathbf{y}(s), 0\le s \le t\}$ is then

$$\begin{aligned} \int _{\Gamma } \omega = \int _{\Gamma } \sum _{j=1}^n f_j \mathrm{d}x_j = \int _{0}^t \sum _{j=1}^{n}f_{j} (\varvec{\mathbf{y}}(s)) \mathrm{d}x_{j}(\delta \mathbf{y}(s)) \end{aligned}$$

(A 5)

where inputs of the differential form are suppressed concisely before the parameterization in the last step. To carry out the computation, one would proceed with $\mathrm{d}x_{j}(\delta \mathbf{y}(s))=\delta \mathbf{y}_j(s)=\mathbf{y}_j'(s)\mathrm{d}s$, which makes Eq. (A 5) an usual one dimensional integration w.r.t. s. This identification of differential form turns out to be significant for the general integral of m-form on a general manifold and the generalization of Stoke’s theorem.

1.2 B: Bivector and 2-Form

Stokes theorem in $\mathbb {R}^3$ tells us that a line integral of a vector field on a closed loop is equal to the surface integral of the vector field’s curl. The intuition behind is that the curl of the vector field gives the vorticity of the vector field of an infinitesimal plane object. When integrating all the infinitesimal planes that tile the surface, neighboring circulation cancels and all the vorticity of the infinitesimal plane combines to give the vorticity on the big loop on the boundary. This intuition is still valid in $\mathbb {R}^n$. To see that, we shall first introduce how the “planary object” is represented in general $\mathbb {R}^n$: it is given by the notion of a (simple) bivector.

Two parameters are needed to parametrize a surface, and a surface can be cut into infinitesimal two dimensional parallelograms with edges given by the two infinitesimal tangent vectors of a point. Moreover, circulation of a vector field over an infinitesimal plane can have two orientations. Putting these together, we use the notion $\mathbf{u}\wedge \mathbf{v}$ to represent an oriented parallelogram object spanned by two vectors $\mathbf{u}$ and $\mathbf{v}$, thus the name bivector. The orientation of the object is reflected by the anti-symmetry of the wedge product $\wedge$, $\mathbf{u}\wedge \mathbf{v}=-\mathbf{v}\wedge \mathbf{u}$. The wedge product $\wedge$ is a linear operation satisfying $(c_1 \mathbf{u}_1+c_2 \mathbf{u}_2)\wedge \mathbf{v}= c_1 \mathbf{u}_1\wedge \mathbf{v}+ c_2 \mathbf{u}_2 \wedge \mathbf{v}$ where $c_1,c_2 \in \mathbb {R}$. With these, we can get the component form of the bivector

$$\begin{aligned} \mathbf{u}\wedge \mathbf{v}= \sum _{1\le i<j \le n} (u_i v_j - u_j v_i) \mathbf {e}_i \wedge \mathbf {e}_j \end{aligned}$$

(A 6)

with basis $\mathbf {e}_i \wedge \mathbf {e}_j$. Importantly, one can show that the area of the parallelogram, denoted as $\Vert \mathbf{u}\wedge \mathbf{v}\Vert$ is given by

$$\begin{aligned} \Vert \mathbf{u}\wedge \mathbf{v}\Vert ^2 = \sum _{1\le i<j \le n} (u_i v_j - u_j v_i)^2. \end{aligned}$$

(A 7)

An inner product between two bivectors $\mathbf {A}=\sum _{i<j} A_{ij} \mathbf {e}_i \wedge \mathbf {e}_j$ and $\mathbf {B}=\sum _{i<j} B_{ij} \mathbf {e}_i \wedge \mathbf {e}_j$ is thus naturally defined as

$$\begin{aligned} \mathbf {A}\cdot \mathbf {B} = \sum _{1\le i<j \le n} A_{ij} B_{ij} \end{aligned}$$

(A 8)

with $\left( \mathbf {e}_i \wedge \mathbf {e}_j\right) \cdot \left( \mathbf {e}_k \wedge \mathbf {e}_l\right) =\delta _{ik} \delta _{jl}$ for $i<j$ and $k<l$.

We note that, in general, bivectors are objects that can be expressed as $\mathbf {A}=\sum _{i<j} A_{ij} \mathbf {e}_i \wedge \mathbf {e}_j$. Not all bivector can be expressed as the wedge product of two vectors. Such bivectors are called simple bivectors, and only simple bivectors can have the geometrical meaning as a parallelogram spanned by two vectors: a general bivector can be the sum of many simple bivectors, superposition of many parallelogram. We also note that due to the anti-symmetry of the wedge product, the component $A_{ij}$ of a bivector $\mathbf{A}$ can be represented by an anti-symmetric matrix. Then, the inner product between two bivectors, as shown in Eq. (A 8), is the half of the Frobenius product of their anti-symmetric components.

Now, similar to a 1-form taking a vector to a real number, a 2-form takes a bivector to a real number.The basis of a 2-form is given by $\{\mathrm{d}x_i \wedge \mathrm{d}x_j\}$ for $1\le i <j \le n$. Specifically, for $1\le i<j \le n$,

$$\begin{aligned} \mathrm{d}x_i \wedge \mathrm{d}x_j \left( \sum _{k<l} A_{kl} \mathbf {e}_k \wedge \mathbf {e}_l \right) = \left( \mathbf {e}_i \wedge \mathbf {e}_j \right) \cdot \left( \sum _{k<l} A_{kl} \mathbf {e}_k \wedge \mathbf {e}_l \right) = A_{ij}. \end{aligned}$$

(A 9)

Again, the relation between $\mathrm{d}x_i \wedge \mathrm{d}x_j \left( \mathbf {A} \right) = \left( \mathbf {e}_i \wedge \mathbf {e}_j \right) \cdot \mathbf {A}$ is what allow us to rewrite a differential 2-form in a vectorized form in the main text,

$$\begin{aligned} \omega (\mathbf {A})&=\sum _{i<j}B_{ij}\mathrm{d}x_i \wedge \mathrm{d}x_j \left( \mathbf {A} \right) = \sum _{i<j}B_{ij}\left( \mathbf {e}_i \wedge \mathbf {e}_j \right) \cdot \mathbf {A} \end{aligned}$$

(A 10a)

$$\begin{aligned}&= \sum _{i<j} B_{ij} A_{ij}=\mathbf {B}\cdot \mathbf {A}. \end{aligned}$$

(A 10b)

When integration the differential 2-form over a surface, the $\mathbf {A}$ as the input of the 2-form here would be the infinitesimal bivector given by the two infinitesimal tangent vectors at a point, representing the infinitesimal tangent parallelogram at the point.

1.3 C: Exterior Derivative and the Curl of a Vector Field

The concept of curl in vector calculus is useful because of the Stokes theorem in $\mathbb {R}^3$. We shall thus use the generalized version of the Stokes theorem, the Stokes–Cartan theorem, to motivate the notion of exterior derivative and get the general definition of the curl of a vector field in $\mathbb {R}^n$.

The Stokes–Cartan theorem states that the integral of a differential form $\omega$ over the boundary of some oriented manifold $\varOmega$ is equal to the integral of its exterior derivative $\mathrm{d}\omega$ over the whole of $\varOmega$:

$$\begin{aligned} \oint _{\partial \varOmega }\omega =\int _{\varOmega }\mathrm{d}\omega . \end{aligned}$$

(A 11)

In a sense, the exterior derivative $\mathrm{d}$ is defined so that Eq. (A 11) holds for a general manifold. We shall thus understand it with Stokes–Cartan theorem: the exterior derivative of a differential form $\omega$ can be interpreted, geometrically, as the integral over the boundary of an infinitesimal parallelepiped $h\varOmega$,

$$\begin{aligned} \mathrm{d}\omega = \lim _{h\rightarrow 0} \frac{1}{\Vert h \varOmega \Vert } \int _{\partial (h \varOmega )} \omega \end{aligned}$$

(A 12)

where $\Vert h \varOmega \Vert$ denotes the volume of $h\varOmega$. With this, one sees that the twice exterior derivative of any differential form has to be zero, $\mathrm{d}\mathrm{d}\omega =0$. This is by applying the Stokes–Cartan theorem twice for a form that is itself the exterior derivative of another form $\varphi =\mathrm{d}\omega$ (such form is called exact). For an arbitrary compact region $\varOmega$, we have

$$\begin{aligned} \int _{\varOmega }\mathrm{d}\mathrm{d}\omega =\int _{\varOmega }\mathrm{d}\varphi = \int _{\partial \varOmega }\varphi = \int _{\partial \varOmega }\mathrm{d}\omega = \int _{\partial \partial \varOmega } \omega . \end{aligned}$$

(A 13)

Since $\partial \partial \varOmega =\emptyset$ and $\varOmega$ is arbitrary, we have $\mathrm{d}\mathrm{d}\omega =0$. A form with zero exterior derivative is said to be closed. Therefore, every exact form is closed.

The exterior derivative of a k-form is a $(k+1)$-form. The exterior derivative of a general form $\alpha \wedge \beta$ obeys the product rule: Suppose $\alpha$ is a p-form, then

$$\begin{aligned} \mathrm{d}\left( \alpha \wedge \beta \right) =\mathrm{d}\alpha \wedge \beta +\left( -1\right) ^p \left( \alpha \wedge \mathrm{d}\beta \right) . \end{aligned}$$

(A 14)

Applying this and $\mathrm{d}\mathrm{d}\omega =0$, one can get the exterior derivative of the 1-form in Eq. (A 2),

$$\begin{aligned} \mathrm{d}\left( \sum _{j=1}^{n}f_{j}\mathrm{d}x_{j}\right)&= \sum _{j=1}^n \left( \mathrm{d}f_j\right) \wedge \mathrm{d}x_j = \sum _{1\le i,j\le n} \left( \frac{\partial f_{j}}{\partial x_i} \mathrm{d}x_i\right) \wedge \mathrm{d}x_{j} \end{aligned}$$

(A 15a)

$$\begin{aligned}&= \sum _{1\le i<j\le n} \left( \frac{\partial f_{j}}{\partial x_i} - \frac{\partial f_{i}}{\partial x_j} \right) \mathrm{d}x_i \wedge \mathrm{d}x_{j} \end{aligned}$$

(A 15b)

where $\mathrm{d}x_i \wedge \mathrm{d}x_j=-\mathrm{d}x_j \wedge \mathrm{d}x_i$ is used. By Stokes–Cartan theorem, we the have

$$\begin{aligned} \oint _{\partial \varOmega } \sum _{j=1}^{n}f_{j}\mathrm{d}x_{j} =\int _{\varOmega }\sum _{1\le i<j\le n} \left( \frac{\partial f_{i}}{\partial x_j} - \frac{\partial f_{i}}{\partial x_j} \right) \mathrm{d}x_i \wedge \mathrm{d}x_{j}. \end{aligned}$$

(A 16)

As introduced in Sec. B, we can rewrite Eq. (A 16) in the vectorized form,

$$\begin{aligned} \oint _{\partial \varOmega } \varvec{f}\cdot \mathrm{d}\mathbf{x}=\int _{\varOmega } \nabla \wedge \varvec{f}\cdot \mathrm{d}\varvec{\sigma } \end{aligned}$$

(A 17)

where inner product between bivectors was introduced in Eq. (A 8). Hence, $\nabla \wedge \varvec{f}$ as a bivector is the curl of the vector field $\varvec{f}$.

Since $\mathrm{d}\mathrm{d}\omega =0$, a gradient vector field is always curl-free, i.e. $\nabla \wedge \nabla U=0$ where U is a scalar potential. For the converse, we apply Poicaré lemma, which states that every closed form is exact (locally) on a contractible domain. Since we are concern with processes on the entire Euclidean manifold $\mathbb {R}^n$, which is contractible, we can conclude that curl-free vector field is globally a gradient field.

1.4 D: Bivector Potential of a Divergence-Free Vector Field

Using exterior derivatives and differential forms, the integral of a vector field $\mathbf{F}(\mathbf{x})$ over an $(n-1)$-dimensional closed surface $\Sigma$ as flux is

$$\begin{aligned} \oint _{\Sigma } \sum _{k=1}^{n}F_{k}(\mathbf{x})\mathrm{d}\sigma _{k}&=\int _{\varOmega } \mathrm{d}\left( \sum _{k=1}^{n}F_{k}(\mathbf{x})\mathrm{d}\sigma _{k}\right) \end{aligned}$$

(A 18a)

$$\begin{aligned}&=\int _{\varOmega }\sum _{k=1}^{n}\left[ \sum _{j=1}^{n}\left( \partial _{j}F_{k}\right) \mathrm{d}x_{j}\right] \wedge \mathrm{d}\sigma _{k} \end{aligned}$$

(A 18b)

$$\begin{aligned}&=\int _{\varOmega }\sum _{k,j=1}^{n}\left( \partial _{j}F_{k}\right) \mathrm{d}x_{j}\wedge \mathrm{d}\sigma _{k} \end{aligned}$$

(A 18c)

$$\begin{aligned}&=\int _{\varOmega }\sum _{j=1}^{n}\left( \partial _{j}F_{j}\right) \mathrm{d}V, \end{aligned}$$

(A 18d)

where $\varOmega$ is the n-volume contained by the closed $(n-1)$-surface $\Sigma$, and

$$\begin{aligned} \mathrm{d}\sigma _{k}=(-1)^{k-1} \mathrm{d}x_{1}\wedge \mathrm{d}x_{2}\wedge \cdots \wedge \mathrm{d}x_{k-1}\wedge \mathrm{d}x_{k+1}\wedge \cdots \wedge \mathrm{d}x_{n} \end{aligned}$$

(A 19)

with $\mathrm{d}x_{k}$ missing. The $(-1)^{k-1}$ factor is to ensure $\mathrm{d}x_{j}\wedge \mathrm{d}\sigma _{k}=\delta _{jk}\mathrm{d}V$ where

$$\begin{aligned} \mathrm{d}V=\mathrm{d}x_1 \wedge \mathrm{d}x_2 \wedge \cdots \wedge \mathrm{d}x_n \end{aligned}$$

(A 20)

is the infinitesimal n dimensional volume element. The $(n-1)$-form $\mathrm{d}\sigma _i$ is the Hodge dual of the 1-form $\mathrm{d}x_i$, often denoted as $\mathrm{d}\sigma _i = \star \mathrm{d}x_i$.

Now if a vector field $\mathbf{F}(\mathbf{x})$ is divergence free, i.e. $\sum _{j=1}^{n}\left( \partial _{j}F_{j}\right) =0$, then the $(n-1)$-form has a zero exterior derivative,

$$\begin{aligned} \mathrm{d}\left( \sum _{k=1}^{n}F_{k}(\mathbf{x})\mathrm{d}\sigma _{k}\right) =0. \end{aligned}$$

(A 21)

Poicaré lemma then guarantees that on a contractible domain,

$$\begin{aligned} \sum _{k=1}^{n}F_{k}(\mathbf{x})\mathrm{d}\sigma _{k}=\mathrm{d}\omega , \end{aligned}$$

(A 22)

where $\omega$ is expected to be a $(n-2)$-form with the general expression

$$\begin{aligned} \omega =\sum _{1 \le i<j \le n}u_{ij}(\mathbf{x})\mathrm{d}\eta _{ij}, \end{aligned}$$

(A 23)

in which

$$\begin{aligned} \mathrm{d}\eta _{ij}=(-1)^{i-1+j-2}\mathrm{d}x_{1}\cdots \mathrm{d}x_{i-1}\wedge \mathrm{d}x_{i+1}\wedge \cdots \mathrm{d}x_{j-1}\wedge \mathrm{d}x_{j+1}\cdots \mathrm{d}x_{n} \end{aligned}$$

(A 24)

with $\mathrm{d}x_{i}$ and $\mathrm{d}x_{j}$ missing. The $(-1)^{i-1+j-2}$ factor is to ensure $\left( \mathrm{d}x_i \wedge \mathrm{d}x_j\right) \wedge \mathrm{d}\eta _{ij}=\mathrm{d}V$ so that $\mathrm{d}\eta _{ij} = \star \left( \mathrm{d}x_i \wedge \mathrm{d}x_j \right)$. Then,

$$\begin{aligned} \mathrm{d}\omega&= \sum _{1 \le i<j \le n}\sum _{k=1}^{n}\left( \partial _{k}u_{ij}\right) \mathrm{d}x_{k}\wedge \mathrm{d}\eta _{ij} \end{aligned}$$

(A 25a)

$$\begin{aligned}&=\ \sum _{1 \le i<j \le n}\left\{ \left( \partial _{i}u_{ij}\right) \mathrm{d}x_{i}\wedge \mathrm{d}\eta _{ij}+\left( \partial _{j}u_{ij}\right) \mathrm{d}x_{j}\wedge \mathrm{d}\eta _{ij}\right\} \end{aligned}$$

(A 25b)

$$\begin{aligned}&=\ \sum _{1 \le i<j \le n} (-1)\left( \partial _{i}u_{ij}\right) \mathrm{d}\sigma _{j}+\sum _{1 \le i<j \le n}\left( \partial _{j}u_{ij}\right) \mathrm{d}\sigma _{i} \end{aligned}$$

(A 25c)

$$\begin{aligned}&=\ \sum _{1 \le j<i \le n} (-1)\left( \partial _{j}u_{ji}\right) \mathrm{d}\sigma _{i}+\sum _{1 \le i<j \le n}\left( \partial _{j}u_{ij}\right) \mathrm{d}\sigma _{i} \end{aligned}$$

(A 25d)

$$\begin{aligned}&=\ \sum _{i=1}^{n}\sum _{j=1}^{n}\left( \partial _{j}A_{ij}\right) \mathrm{d}\sigma _{i}. \end{aligned}$$

(A 25e)

In the last step we have introduced $A_{ij}(\mathbf{x})=u_{ij}(\mathbf{x})$ for $i<j$, $A_{ij}(\mathbf{x})= -A_{ji}(\mathbf{x})$ for $i>j$ and $A_{ii}(\mathbf{x})=0$. It is easy to verify that

$$\begin{aligned} F_{i}(\mathbf{x})=\sum _{j=1}^{n}\left( \partial _{j}A_{ij}(\mathbf{x})\right) \end{aligned}$$

(A 26)

is a divergence free vector field:

$$\begin{aligned} \nabla \cdot \mathbf{F}(\mathbf{x})=\sum _{i=1}^{n}\partial _{i}F_{i}(\mathbf{x})=\sum _{i,j=1}^{n}\left( \partial _{i}\partial _{j}A_{ij}(\mathbf{x})\right) =0. \end{aligned}$$

(A 27)

The vector potential $\mathbf{A}(\mathbf{x})$ of a divergence-free field is a bivector with anti-symmetric matrix components. Since we consider diffusion on the whole $\mathbb {R}^n$, which is contractible. Eq. (A 22) and so Eq. (A 26) are thus globally valid.

In $\mathbb {R}^3$, a divergence free vector field has a vector potential through the same curl differential operator as the one we used to compute the vorticity of a vector field. For general $\mathbb {R}^n$, this is no longer true. To distinguish the two in general, we have used $\nabla \wedge$ as the curl operator that maps a vector field to its vorticity bivector, with $\wedge$ reminding us the result is a bivector. Here, we use $\nabla \times$ to denote the differential operator that links a divergence-free vector field to its bivector potential. Eq. (A 26) is then expressed as

$$\begin{aligned} \mathbf {F}(\mathbf{x})=\nabla \times \mathbf {A}(\mathbf{x}). \end{aligned}$$

(A 28)

The close relation between them can be seen by integration by part: For a divergence-free vector field $\mathbf {F}=\nabla \times \mathbf {A}$, we have

$$\begin{aligned}&\int _{\mathbb {R}^{n}}\left( \nabla \times \mathbf {A}\right) \cdot \mathbf{u}\ \mathrm{d}V \ =\int _{\mathbb {R}^{n}}\sum _{i,j}\partial _{j}A_{ij}u_{i}\mathrm{d}V \end{aligned}$$

(A 29a)

$$\begin{aligned}= & {} \int _{\mathbb {R}^{n}}\sum _{i<j}A_{ij}\left( \partial _{i}u_{j}-\partial _{j}u_{i}\right) \mathrm{d}V=\int _{\mathbb {R}^{n}} \mathbf {A}\cdot \left( \nabla \wedge \mathbf {u}\right) \mathrm{d}V. \end{aligned}$$

(A 29b)