A.1. Field theory


# Chapter A. Appendices

## A.1. Field theory

### Green, Gauss, Stokes formulae

Let $D$ be a bounded domain in $\mathbb{R}^2$ and $L=\partial D$ be its boundary. Then $$-\int_{L} \mathbf{A}\cdot \mathbf{n} \,ds= \iint_D (\nabla \cdot \mathbf{A})\,dS \label{eq-A.1.1}$$ where the left-hand side expression is a linear integral, the right-hand side expression is an area integral and $\mathbf{n}$ is a unit inner normal to $L$. This is Green formula.

Let $V$ be a bounded domain in $\mathbb{R}^3$ and $\Sigma=\partial V$ be its boundary. Then $$-\iint_{\Sigma} \mathbf{A}\cdot \mathbf{n} \,dS= \iiint_D (\nabla \cdot \mathbf{A})\,dV \label{eq-A.1.2}$$ where the left-hand side expression is a surface integral, the right-hand side expression is a volume integral and $\mathbf{n}$ is a unit inner normal to $\Sigma$. This is Gauss formula.

Remark 1.

1. Here sign "$-$"appears because $\mathbf{n}$ is a unit inner normal.
2. Gauss formula holds in any dimension. It also holds in any "straight" coordinate system.
3. In the curvilinear coordinate system $(u_1,\ldots, u_n)$ in also holds but divergence must be calculated as $$\nabla \cdot \mathbf{A}= \sum_k J^{-1}\partial_{u_k} (A^k J)= \sum_k \partial_{u_k}A^k + \sum_k (\partial_{u_k} \ln J) \, A^k \label{eq-A.1.3}$$ where $Jdu_1\ldots du_n$ is a volume form in these coordinates. F.e. in the spherical coordinates if $\mathbf{A}$ is radial, $\mathbf{A}= A_r \mathbf{r}/r$ (with $r=|\mathbf{r}|$ we have $\partial_{r} \ln J)= (n-1)r^{-1}$ and therefore $\nabla\cdot A_r \mathbf{r}/r= \partial_r A_r + (n-1)r^{-1}A_r$.

Let $D$ be a bounded domain in $\mathbb{R}^2$ and $L=\partial D$ be its boundary, counter–clockwise oriented (if $L$ has several components then inner components should be clockwise oriented). Then $$\oint_{L} \mathbf{A}\cdot \, d\, \mathbf{r}= \iint_D (\nabla \times \mathbf{A})\cdot\mathbf{n}\,dS \label{eq-A.1.4}$$ where the left-hand side expression is a line integral, the right-hand side expression is an area integral and $\mathbf{n}=\mathbf{k}$. This is Green formula again.

Let $\Sigma$ be a bounded piece of the surface in $\mathbb{R}^3$ and $L=\partial \Sigma$ be its boundary. Then $$\oint_{L} \mathbf{A}\cdot \, d\, \mathbf{l}= \iint_\Sigma (\nabla \times \mathbf{A})\cdot\mathbf{n}\,dS \label{eq-A.1.5}$$ where the left-hand side expression is a line integral, the right-hand side expression is a surface integral and $\mathbf{n}$ is a unit normal to $\Sigma$; orientation of $L$ should match to direction of $\mathbf{n}$. This is Stokes formula.

</spanRemark 2.

1. We can describe orientation in the Green formula as "the pair $\{d\mathbf{r}, \mathbf{n}\}$ has a right-hand orientation"
2. We can describe orientation in the Stokes formula as "the triple $\{d\mathbf{r}, \boldsymbol{\nu}, \mathbf{n}\}$ has a right-hand orientation" where $\boldsymbol{\nu}$ is a normal to $L$ which is tangent to $\Sigma$ and directed inside of $\Sigma$.
3. Stokes formula holds in any dimension of the surface $\Sigma$ but then it should be formulated in terms of differential forms $$\int _\Sigma d\omega = \int _{\partial\Sigma}\omega\tag{Stokes formula}$$ which is the material of Analysis II class (aka Calculus II Pro).

### Properties of $\nabla$

#### Definitions

Definition 1.

1. Operator $\nabla$ is defined as $$\nabla = \mathbf{i} \partial_x + \mathbf{j} \partial_y+ \mathbf{k} \partial_z. \label{eq-A.1.6}$$
2. It could be applied to a scalar function resulting in its gradient ($\operatorname{grad}\phi$) \begin{equation*} \nabla \phi = \mathbf{i} \partial_x\phi + \mathbf{j} \partial_y\phi+ \mathbf{k} \partial_z\phi \end{equation*}
3. and to vector function $\mathbf{A}=A_x\mathbf{i}+A_y\mathbf{j}+A_z\mathbf{k}$ resulting in its divergence ($\operatorname{div}\mathbf{A}$) \begin{equation*} \nabla \cdot \mathbf{A} = \partial_xA_x + \partial_y A_y+ \partial_zA_z \end{equation*}
4. and also in its curl ($\operatorname{curl}\mathbf{A}$) or rotor ($\operatorname{rot}\mathbf{A}$), depending on the mathematical tradition: \begin{equation*} \nabla \times \mathbf{A} = \left| \begin{matrix} \mathbf{i} & \mathbf{j} &\mathbf{k} \\ \partial_x & \partial_y & \partial_z\\ A_x & A_y &A_z\end{matrix}\right| \end{equation*} which is equal to \begin{equation*} (\partial_y A_z-\partial_z A_y)\mathbf{i}+ (\partial_z A_x-\partial_x A_z)\mathbf{j}+ (\partial_x A_y-\partial_y A_x)\mathbf{k}. \end{equation*}

#### Double application

Definition 2. $$\Delta= \nabla^2 = \nabla\cdot \nabla= \partial_x^2 + \partial_y^2+ \partial_z^2. \label{eq-A.1.7}$$ is Laplace operator or simply Laplacian.

Four formulae to remember: \begin{gather} \nabla (\nabla \phi)= \Delta \phi,\label{eq-A.1.8}\\[3pt] \nabla \times (\nabla \phi)= 0,\label{eq-A.1.9}\\[3pt] \nabla \cdot (\nabla \times \mathbf{A})= 0,\label{eq-A.1.10}\\[3pt] \nabla \times (\nabla \times \mathbf{A})= -\Delta \mathbf{A} + \nabla (\nabla \cdot \mathbf{A}) \label{eq-A.1.11} \end{gather} where all but the last one are obvious and the last one follows from $$\mathbf{a}\times (\mathbf{a} \times \mathbf{b})= - \mathbf{a}^2 \mathbf{b}+ (\mathbf{a}\cdot \mathbf{b}) \mathbf{a} \label{eq-A.1.12}$$ which is the special case of $$\mathbf{a}\times (\mathbf{b} \times \mathbf{c})= \mathbf{b}(\mathbf{a}\cdot\mathbf{c})- \mathbf{c}(\mathbf{a}\cdot\mathbf{b}). \label{eq-A.1.13}$$

#### Application to the product

Recall Leibniz rule how to apply the first derivative to the product which can be symbolically written as \begin{equation*} \partial (uv)= (\partial_u + \partial_v)(uv)= \partial_u (uv)+\partial_v (uv)= v\partial_u (u) +u\partial_v (v)=v\partial u +u\partial v \end{equation*} where subscripts "$u$" or "$v$" mean that it should be applied to $u$ or $v$ only.

Since $\nabla$ is a linear combination of the first derivatives, it inherits the same rule. Three formulae are easy \begin{gather} \nabla ( \phi\psi)= \phi\nabla \psi +\psi \nabla \phi,\label{eq-A.1.14}\\[3pt] \nabla \cdot ( \phi\mathbf{A})= \phi\nabla \cdot \mathbf{A} +\nabla \phi\cdot \mathbf{A} , \label{eq-A.1.15}\\[3pt] \nabla \times ( \phi\mathbf{A})= \phi\nabla \times \mathbf{A} +\nabla \phi\times \mathbf{A} , \label{eq-A.1.16}\end{gather} and the fourth follows from the Leibniz rule and (\ref{eq-A.1.13}) $$\nabla \times ( \mathbf{A}\times \mathbf{B})= (\mathbf{B}\cdot\nabla)A-\mathbf{B}(\nabla\cdot \mathbf{A}) - (\mathbf{A}\cdot\nabla)B+\mathbf{A}(\nabla\cdot \mathbf{B}). \label{eq-A.1.17}$$