Be a Continuously Differentiable Path With Length L Let F Be a Vector Field on R3

Vector Field Theory

In Table of Integrals, Series, and Products (Seventh Edition), 2007

10.71–10.72 Vector integral theorems

10.711

Gauss's divergence theorem. Let V be a volume bounded by a simple closed surface S and let f be a continuously differentiable vector field defined in V and on S. Then, if d S is the outward drawn vector element of area,

KE 39 ${\int }_S\text{f}\cdot d\text{S}={\int }_V\text{div\hspace{0.17em}f}dV$

10.712

Green's theorems. Let Φ and Ψ be scalar fields which, together with ∇² Φ and ∇² Ψ, are defined both in a volume V and on its surface S, which we assume to be simple and closed. Then, if d/dn denotes differentiation along the outward drawn normal to S, we have

10.713

Green's first theorem

KE 212 ${\int }_S\Phi \frac{\partial \Psi }{\partial n}dS={\int }_V\left(\Phi {\nabla }^2\Psi +\text{grad}\Phi \cdot \text{grad}\Psi \right)dV$

10.714

Green's second theorem

KE 215 ${\int }_S\left(\Phi \frac{\partial \Psi }{\partial n}-\Psi \frac{\partial \Phi }{\partial n}\right)dS={\int }_V\left(\Phi {\nabla }^2\Psi -\Psi {\nabla }^2\Phi \right)dV$

10.715

Special cases

1.

${\int }_S\left(\Phi \text{grad}\Phi \right)\cdot d\text{S}={\int }_V\left(\Phi {\nabla }^2\Phi +{\left(\text{grad}\Phi \right)}^2\right)dV$

2.

MV 81 ${\int }_S\frac{\partial \Phi }{\partial n}dS={\int }_V{\nabla }^2\Phi dV$

10.716

Green's reciprocal theorem. If Φ and Ψ are harmonic, so that ∇² Φ = ∇² Ψ = 0, then

3.

MM 105 ${\int }_S\Phi \frac{\partial \Psi }{\partial n}dS={\int }_S\Psi \frac{\partial \Phi }{\partial n}dS$

10.717

Green's representation theorem. If Φ and ∇² Φ are defined within a volume V bounded by a simple closed surface S, and P is an interior point of V, then in three dimensions

4.

KE 219 $\Phi \left(P\right)=-\frac{1}{4\pi }{\int }_V\frac{1}{r}{\nabla }^2\Phi dV+\frac{1}{4\pi }{\int }_S\frac{1}{r}\frac{\partial \Phi }{\partial n}dS-\frac{1}{4\pi }{\int }_S\Phi \frac{\partial }{\partial n}\left(\frac{1}{r}\right)dS$

5.

$\Phi \left(P\right)=\frac{1}{4\pi }{\int }_S\frac{1}{r}\frac{\partial \Phi }{\partial n}dS-\frac{1}{4\pi }{\int }_S\Phi \frac{\partial }{\partial n}\left(\frac{1}{r}\right)dS$

In the case of two dimensions, result (4) takes the form

6.

MM 116 $\begin{array}{l}\Phi \left(p\right)=\frac{1}{2\pi }{\int }_S{\nabla }^2\Phi \left(q\right)ln\left|p-q\right|dS\\ +\frac{1}{2\pi }{\int }_C\Phi \left(q\right)\frac{\partial }{\partial n_q}ln\left|p-q\right|dq-\frac{1}{2\pi }\int ln\left|p-q\right|\frac{\partial }{\partial n_q}\Phi \left(q\right)dq\end{array}$

where C is the boundary of the planar region S, and result (5) takes the form

7.

VL 280 $\Phi \left(p\right)=\frac{1}{2\pi }{\int }_C\Phi \left(q\right)\frac{\partial }{\partial n_q}ln\left|p-q\right|dq-\frac{1}{2\pi }{\int }_{C}ln\left|p-q\right|\frac{\partial }{\partial n_q}\Phi \left(q\right)dq$

10.718

Green's representation theorem in Rⁿ. If Φ is twice differentiable within a region Ω in Rⁿ bounded by the surface Σ with outward drawn unit normal n, then for p ∉ ∑ and n > 3

$\Phi \left(p\right)=\frac{-1}{\left(n-2\right){\sigma }_n}{\int }_{\Omega }\frac{{\nabla }^2\Phi \left(q\right)}{{\left|p-q\right|}^{n-2}}d{\Omega }_q+\frac{1}{\left(n-2\right){\sigma }_n}{\int }_{\sum }\left(\frac{1}{{\left|p-q\right|}^{n-2}}\frac{\partial \Phi \left(q\right)}{\partial n_q}-\Phi \left(q\right)\frac{\partial }{\partial n_q}\frac{1}{{\left|p-q\right|}^{n-2}}\right)d{\sum }_q,$

where

VL 279 ${\sigma }_n=\frac{2{\pi }^{n/2}}{\Gamma \left(n/2\right)}$

is the area of the unit sphere in Rⁿ .

10.719

Green's theorem of the arithmetic mean. If Φ is harmonic in a sphere, then the value of Φat the center of the sphere is the arithmetic mean of its value on the surface.

KE 223

10.720

Poisson's integral in three dimensions. If Φ is harmonic in the interior of a spherical volume V of radius R and is continuous on the surface of the sphere on which, in terms of the spherical polar coordinates (r, θ, ϕ), it satisfies the boundary condition Φ(R, θ, ϕ) = f (θ, ϕ), then

$\Phi \left(r,\theta ,\varphi \right)=\frac{R\left(R^2-r^2\right)}{4\pi }{\int }_0^{\pi }{\int }_{-\pi }^{\pi }\frac{f\left({\theta }^{\prime },{\varphi }^{\prime }\right)sin{\theta }^{\prime }d{\theta }^{\prime }d{\varphi }^{\prime }}{{\left(r^2+R^2-2rRcos\gamma \right)}^{3/2}},$

where

KE 241 $cos\gamma =cos\theta cos{\theta }^{\prime }+sin\theta sin{\theta }^{\prime }cos\left(\varphi -{\varphi }^{\prime }\right)\cdot$

10.721

Poisson's integral in two dimensions. If Φis harmonic in the interior of a circular disk S of radius R and is continuous on the boundary of the disk on which, in terms of the polar coordinates (r, θ), it satisfies the boundary condition Φ(R, θ) = f (θ), then

$\Phi \left(r,\theta \right)=\frac{(R^2-r^2)}{2\pi }{\int }_{-\pi }^{\pi }\frac{f\left(\varphi \right)d\varphi }{r^2+R^2-2rRcos\left(\theta -\varphi \right)}\cdot$

10.722

Stokes' theorem. Let a simple closed curve C be spanned by a surface S. Define the positive normal n to S, and the positive sense of description of the curve C with line element d r, such that the positive sense of the contour C is clockwise when we look through the surface S in the direction of the normal. Then, if f is continuously differentiable vector field defined on S and C with vector element S = n dS,

MM 143 ${\oint }_C\text{f}\cdot \text{d}\text{r}={\int }_S\text{curl\hspace{0.17em}f}\cdot d\text{S},$

where the line integral around C is taken in the positive sense.

10.723

Planar case of Stokes' theorem. If a region R in the (x, y)-plane is bounded by a simple closed curve C, and f ₁(x, y), f ₂(x, y) are any two functions having continuous first derivatives in R and on C, then

MM 143 ${\oint }_C\left(f_{1}dx+f_{2}dy\right)=\int {\int }_R\left(\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right)dxdy,$

where the line integral is taken in the counterclockwise sense.

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Deterministic Partial Differential Equations

Jinqiao Duan , Wei WANG , in Effective Dynamics of Stochastic Partial Differential Equations, 2014

2.3 Integral Equalities

In this and the next two sections we recall some equalities and inequalities useful for estimating solutions of spdes as well as pdes.

Let us review some integral identities. For more details, see [1, Sec. 7.3], [17, Ch. 7] or [121, Appendix C].

Green's theorem in ${\mathbb{R}}^2$ : Normal form

$\underset{C}{\oint }\mathbf{v}·\mathbf{n}\mathit{dc}={\int \int }_D\nabla ·\mathbf{v}\mathit{dA}\text{,}$

where $\mathbf{v}$ is a continuously differentiable vector field, $C$ is a piecewise smooth closed curve that encloses a bounded region $D$ in ${\mathbb{R}}^2$ , and $\mathbf{n}$ is the unit outward normal vector to $C$ . The curve $C$ is positively oriented (i.e., if you walk along $C$ in the positive orientation, the region $D$ is to your left).

Green's theorem in ${\mathbb{R}}^2$ : Tangential form

$\underset{C}{\oint }\mathbf{v}·\mathbf{T}\mathit{dc}=\underset{C}{\oint }\mathit{Mdx}+\mathit{Ndy}={\int \int }_D\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)\mathit{dx}\mathit{dy}\text{,}$

where $\mathbf{v}=M(x\text{,}y)\mathit{dx}+N(x\text{,}y)\mathit{dy}$ is a continuously differentiable vector field, $C$ is a piecewise smooth closed curve that encloses a bounded region $D$ in ${\mathbb{R}}^2$ , and $\mathbf{T}$ is the unit tangential vector to $C$ . The curve $C$ is positively oriented (i.e., if you walk along $C$ in the positive orientation, the region $D$ is to your left).

Divergence theorem in ${\mathbb{R}}^k$

${\int }_S\mathbf{v}·\mathbf{n}\mathit{ds}={\int }_D\nabla ·\mathbf{v}\mathit{dA}\text{,}$

where $\mathbf{v}$ is a continuously differentiable vector field, $S$ is a closed surface that encloses a bounded region $D$ in ${\mathbb{R}}^k$ , and $\mathbf{n}$ is the outward unit normal vector to $S$ . In particular, taking $\mathbf{v}$ as a vector function with one component equal to a scalar function $u$ and the rest of components zero, we have

${\int }_D{u}_{x_i}\mathit{dx}={\int }_{\partial D}{\mathit{un}}_i\mathit{ds}\text{,}i=1\text{,}\dots \text{,}k\text{,}$

where $\mathbf{n}=(n_1\text{,}\cdots \text{,}{n}_k)$ is the outward unit normal vector to the boundary $\partial D$ of the domain $D$ in ${\mathbb{R}}^k$ .

Applying this theorem to $\mathit{uv}$ and ${\mathit{uv}}_{x_i}$ , we get the following two integration by parts formulas.

Integration by parts formula in ${\mathbb{R}}^k$

${\int }_D{u}_{x_i}v\mathit{dx}=-{\int }_D{\mathit{uv}}_{x_i}\mathit{dx}+{\int }_S{\mathit{uvn}}_i\mathit{ds}\text{,}i=1\text{,}\dots \text{,}k\text{,}$

and

${\int }_{D}u\Delta v\mathit{dx}=-{\int }_D\nabla u·\nabla v\mathit{dx}+{\int }_{\partial D}u\frac{\partial v}{\partial \mathbf{n}}\mathit{ds}\text{,}$

where $\mathbf{n}=(n_1\text{,}\dots \text{,}{n}_k)$ is the outward unit normal vector to the boundary $\partial D$ of the domain $D$ in ${\mathbb{R}}^k$ .

Stokes's theorem in ${\mathbb{R}}^3$

$\underset{C}{\oint }\mathbf{v}·d\mathbf{r}={\int \int }_S(\nabla \times \mathbf{v})·\mathbf{n}d\sigma \text{,}$

where $\mathbf{v}$ is a continuously differentiable vector field and $C$ is the boundary of an oriented surface $S$ in ${\mathbb{R}}^3$ . The direction of $C$ is taken as counterclockwise with respect to the surface $S$ 's unit normal vector $\mathbf{n}$ . Stokes's theorem relates the surface integral of the curl of a vector field over a surface to the line integral of the vector field over its boundary.

Green's identities in ${\mathbb{R}}^k$

$\begin{array}{cc}\hfill {\int }_D\Delta \mathit{udx}=& {\int }_{\partial D}\frac{\partial u}{\partial \mathbf{n}}\mathit{ds}\text{,}\hfill \\ \hfill {\int }_D\nabla u·\nabla \mathit{vdx}=& -{\int }_{D}u\Delta \mathit{vdx}+{\int }_{\partial D}u\frac{\partial v}{\partial \mathbf{n}}\mathit{ds}\text{,}\hfill \\ \mathrm{and}\hfill \\ \hfill {\int }_D(u\Delta v-v\Delta u)\mathit{dx}=& {\int }_{\partial D}\left(u\frac{\partial v}{\partial \mathbf{n}}-v\frac{\partial u}{\partial \mathbf{n}}\right)\mathit{ds}\text{,}\hfill \end{array}$

where $\mathbf{n}$ is the unit outward normal vector to the boundary $\partial D$ of the domain $D$ in ${\mathbb{R}}^k$ .

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Local derivative with new parameter

Abdon Atangana , in Derivative with a New Parameter, 2016

2.4 Definition of partial derivative with new parameter

In this section, we present some useful definition of partial β-derivatives.

Definition 2.4.1

Let f be a function of two variables x and y, then, the β-derivative of f respect to x is defined as follows:

(2.69) $\begin{array}{l}\hfill {}_0^A{D}_x^{\beta }(f(x,y))=\underset{\epsilon \to 0}{lim}\frac{f\left(x+\epsilon {\left(x+\frac{1}{\mathrm{\Gamma }(\beta )}\right)}^{1-\beta },y\right)-f(x,y)}{\epsilon }.\end{array}$

Definition 2.4.2

Let x, y be a system of Cartesian coordinates in two-dimensional Euclidean space, and let i, j be the corresponding basis of unit vectors. The β -divergence of a continuously differentiable vector field F = Ui + V j is equal to the scalar-valued function:

(2.70) $\begin{array}{l}\hfill div\beta \mathbf{F}={}_0^A{\nabla }^{\beta }\cdot \mathbf{F}={}_0^A{D}_x^{\beta }(U)+{}_0^A{D}_y^{\beta }(U).\end{array}$

Although expressed in terms of coordinates, the result is invariant under orthogonal transformations, as the physical interpretation suggests. The mixed (β,α)-divergence of F is defined as:

(2.71) $\begin{array}{l}\hfill div\beta ,\alpha \mathbf{F}={}_0^A{\nabla }^{\beta ,\alpha }\cdot \mathbf{F}={}_0^A{D}_x^{\beta }(U)+{}_0^A{D}_y^{\alpha }(U).\end{array}$

Definition 2.4.3

Let x, y be a system of Cartesian coordinates in two-dimensional Euclidean space, and let i, j be the corresponding basis of unit vectors. The β-gradient of a continuously differentiable function f is equal to the vector field:

(2.72) $\begin{array}{l}\hfill grad\beta f={}_0^A{\nabla }^{\beta }f={}_0^A{D}_x^{\beta }(f(x,t))i+{}_0^A{D}_y^{\beta }(f(x,y))j.\end{array}$

The mixed (β,α)-grad of f is defined as:

(2.73) $\begin{array}{l}\hfill grad\beta ,\alpha f={}_0^A{\nabla }^{\beta ,\alpha }f={}_0^A{D}_x^{\beta }(f)i+{}_0^A{D}_y^{\alpha }(f)j.\end{array}$

Definition 2.4.4

The β-Laplace operator in two dimensions of a function f is given by

(2.74) $\begin{array}{l}\hfill {}_0^A{\mathrm{\Delta }}^{\beta }f=\frac{{\partial }^{2\beta }f(x,y)}{\partial x^{2\beta }}+\frac{{\partial }^{2\beta }f(x,y)}{\partial y^{2\beta }},\end{array}$

where x and y are the standard Cartesian coordinates of the xy-plane. The mixed (β,α)-Laplace transform method is defined as:

(2.75) $\begin{array}{l}\hfill {}_0^A{\mathrm{\Delta }}^{\beta }f=\frac{{\partial }^{2\beta }f(x,y)}{\partial x^{2\beta }}+\frac{{\partial }^{2\alpha }f(x,y)}{\partial y^{2\alpha }}.\end{array}$

Definition 2.4.5

In Cartesian coordinates, the β-curl of a continuously vector field F is, for F, composed of [Fx,Fy,Fz]:

$\left|\begin{array}{lll}\hfill \mathbf{i}\hfill & \hfill \mathbf{j}\hfill & \hfill \mathbf{k}\hfill \\ \hfill \frac{{\partial }^{\beta }}{\partial x^{\beta }}\hfill & \hfill \frac{{\partial }^{\beta }}{\partial y^{\beta }}\hfill & \hfill \frac{{\partial }^{\beta }}{\partial z^{\beta }}\hfill \\ \hfill F_x\hfill & \hfill F_y\hfill & \hfill F_z\hfill \end{array}\right|,$

where i, j, and k are the unit vectors for the x-, y-, and z-axes, respectively. This expands as follows:

(2.76) $\begin{array}{l}\hfill \left(\frac{{\partial }^{\beta }{F}_z}{\partial y^{\beta }}-\frac{{\partial }^{\beta }{F}_y}{\partial z^{\beta }}\right)\mathbf{i}+\left(\frac{{\partial }^{\beta }{F}_x}{{\partial }^{\beta }z}-\frac{{\partial }^{\beta }{F}_z}{{\partial }^{\beta }x}\right)\mathbf{j}+\left(\frac{{\partial }^{\beta }{F}_y}{\partial x^{\beta }}-\frac{{\partial }^{\beta }{F}_x}{\partial y^{\beta }}\right)\mathbf{k}.\end{array}$

Although expressed in terms of coordinates, the result is invariant under proper rotations of the coordinate axes but the result inverts under reflection.

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Modern Map Methods in Particle Beam Physics

Martin Berz , in Advances in Imaging and Electron Physics, 1999

1.5.2 Scalar and Vector Potentials

In this section, we introduce a new set of scalar and vector fields called potentials that allow the determination of the electric and magnetic fields by differentiation and that allow a simplification of many electrodynamics problems. We begin with some definitions.

We call ψ a scalar potential for the n-dimensional vector field $\overrightarrow{F}$ if $\overrightarrow{F}=\overrightarrow{\nabla }\psi$ .

We call $\overrightarrow{A}$ a vector potential for the three-dimensional vector field $\overrightarrow{F}$ if $\overrightarrow{F}=\overrightarrow{\nabla }\times \overrightarrow{A}$ .

The question now arises under what condition a given vector field $\overrightarrow{F}$ on Rⁿ has a scalar potential or a vector potential. We first address the case of the scalar potential and answer this question for the general n-dimensional case. For the purposes of electrodynamics, the case n = 3 is usually sufficient, but for many questions connected to Lagrangian and Hamiltonian dynamics, the general case is needed.

In the following, we study the existence and uniqueness of scalar potentials. Let $\overrightarrow{F}$ be a continuously differentiable vector field on Rⁿ .

1.

If $\overrightarrow{F}=\overrightarrow{\nabla }\psi$ , then

(1.300) $\partial F_i/\partial x_j=\partial F_j/\partial x_i\text{for}\text{all}i,j=1,\dots ,n.$

2.

If $\partial F_i/\partial x_j=\partial F_j/\partial x_i$ for all i, j = l,…, n, then there exists a scalar potential ψ such that

(1.301) $\overrightarrow{F}=\overrightarrow{\nabla }\psi .$

3.

A scalar potential ψ for the vector $\overrightarrow{F}$ is uniquely specified up to a constant.

Note the difference of (1) and (2), which are actually the converses of each other. We also note that in the three-dimensional case, the condition $\partial F_i/\partial x_j=\partial F_j/\partial x_i$ is equivalent to the more readily recognized condition $\overrightarrow{\nabla }\times \overrightarrow{F}=\overrightarrow{0}$ .

For the proof, we proceed as follows.

1.

If there exists a twice continuously differentiable ψ such that $\overrightarrow{F}=\overrightarrow{\nabla }\psi$ , then $F_i=\partial \psi /\partial x_i$ and $F_j=\partial \psi /\partial x_j$ , and thus

(1.302) $\frac{\partial F_i}{\partial x_j}=\frac{{\partial }^2\psi }{\partial x_j\partial x_i}=\frac{{\partial }^2\psi }{\partial x_i\partial x_j}=\frac{\partial F_j}{\partial x_i}\cdot$

2.

Now assume that $\overrightarrow{F}$ is continuously differentiable and

(1.303) $\frac{\partial F_i}{\partial x_j}=\frac{\partial F_j}{\partial x_i}\text{for}\text{all}i,j=1,\dots ,n.$

Define ψ as a path integral of $\overrightarrow{F}$ from (0,…,0) to (x₁, x₂,…x_n ) along a path that first is parallel to the x ₁ axis from (0,0,…,0) to (x ₁,0,…,0), then parallel to the x ₂ axis from (x₁,0,…,0) to ( $x_1,x_{2}0,\dots ,0$ ), and so on, and finally parallel to the x_n axis from ( $x_1,x_2,\dots ,x_{n-1},0$ ) to ( $x_1,x_2,\dots ,x_{n-1},x_n$ ). Then we have

(1.304) $\begin{array}{ll}\hfill \psi & ={\int }_{\left(0,\dots ,0\right)}^{\left(x_1,\dots ,x_n\right)}\overrightarrow{F}\cdot d\overrightarrow{r}\hfill \\ \hfill & ={\int }_0^{x_1}{F}_1\left(x_1,0,\dots 0\right)d{x}_1+{\int }_0^{x_2}{F}_2\left(x_1,x_2,0,\dots ,0\right)d{x}_2+\cdots \hfill \\ \hfill & +{\int }_0^{x_n}{F}_n\left(x_1,x_2,\dots ,x_n\right)d{x}_n.\hfill \end{array}$

In the following, we show that ψ indeed satisfies the requirement $\overrightarrow{F}=\overrightarrow{\nabla }\psi$ . We first differentiate with respect to x ₁ and obtain

$\begin{array}{ll}\hfill \frac{\partial \psi }{\partial x_1}& =F_1\left(x_1,0,\dots ,0\right)+{\int }_0^{x_2}\frac{\partial F_2\left(x_1,x_2,0,\dots ,0\right)}{\partial x_1}d{x}_2+\cdots \hfill \\ \hfill & +{\int }_0^{x_n}\frac{\partial F_n\left(x_1,x_2,\dots ,x_n\right)}{\partial x_1}d{x}_n\hfill \\ \hfill & =F_1\left(x_1,0,0,\dots 0\right)+{\int }_0^{x_2}\frac{\partial F_1\left(x_1,x_2,0,\dots ,0\right)}{\partial x_2}d{x}_2+\cdots \hfill \\ \hfill & +{\int }_0^{x_n}\frac{\partial F_1\left(x_1,x_2,\dots ,x_n\right)}{\partial x_n}d{x}_n\hfill \end{array}$

$\begin{array}{ll}\hfill & =F_1\left(x_1,0,0,\dots ,0\right)\hfill \\ \hfill & +\left[F_1\left(x_1,x_2,0,\dots ,0\right)-F_1\left(x_1,0,0,\dots ,0\right)\right]+\cdots \hfill \\ \hfill & +\left[F_1\left(x_1,x_2,\dots ,x_{n-1,}{x}_n\right)-F_1\left(x_1,x_2,\dots ,x_{n-1},0\right)\right]\hfill \\ \hfill & =F_1\left(x_1,x_2,\dots ,x_n\right).\hfill \end{array}$

where Eq. (1.303) was used for moving from the first to the second line. Similarly, we have

$\begin{array}{ll}\hfill \frac{\partial \psi }{\partial x_2}& =F_2\left(x_1,x_2,0,\dots ,0\right)+{\int }_0^{x_3}\frac{\partial F_3\left(x_1,x_2{x}_3,0,\dots ,0\right)}{\partial x_2}d{x}_3+\cdots \hfill \\ \hfill & +{\int }_0^{x_n}\frac{\partial F_n\left(x_1,x_2,\dots ,x_n\right)}{\partial x_2}d{x}_n\hfill \\ \hfill & =F_2\left(x_1,x_2,0,\dots ,0\right)+{\int }_0^{x_3}\frac{\partial F_2\left(x_1,x_2{x}_3,0,\dots ,0\right)}{\partial x_3}d{x}_3+\cdots \hfill \\ \hfill & +{\int }_0^{x_n}\frac{\partial F_2\left(x_1,x_2,\dots ,x_n\right)}{\partial x_n}d{x}_n\hfill \\ \hfill & =F_2\left(x_1,x_2,\dots ,x_n\right).\hfill \end{array}$

We continue in the same way for $\partial \psi /\partial x_3$ and $\partial \psi /\partial x_4$ , and ultimately we derive the following for $\partial \psi /\partial x_n$ :

(1.305) $\frac{\partial \psi }{\partial x_n}=F_n\left(x_1,x_2,\dots ,x_n\right),$

which shows that indeed the scalar field ψ defined by Eq. (1.304) satisfies $\overrightarrow{F}=\overrightarrow{\nabla }\psi$ .

3.

Assume that there exist two scalars ψ₁ and ψ₂ which satisfy

$\begin{array}{l}\overrightarrow{F}=\overrightarrow{\nabla }{\psi }_1\\ \overrightarrow{F}=\overrightarrow{\nabla }{\psi }_2.\end{array}$

Then $\overrightarrow{\nabla }\left({\psi }_1-{\psi }_2\right)=\overrightarrow{0}$ , which can only be the case if

${\psi }_1-{\psi }_2=\text{constant}\text{.}$

Therefore, a scalar potential ψ is specified up to a constant, which concludes the proof.

We note that in the three-dimensional case, the peculiar-looking integration path in the definition of the potential ψ in Eq. (1.304) can be replaced by any other path connecting the origin and (x, y, z), because $\overrightarrow{\nabla }\times \overrightarrow{F}=\overrightarrow{0}$ ensures path independence of the integral according to the Stokes theorem (Eq. 1.298). Furthermore, the choice of the origin as the starting point is by no means mandatory; indeed, any other starting point ${\overrightarrow{x}}_0$ leads to the addition of a constant to the potential, because due to path independence the integral from ${\overrightarrow{x}}_0$ to $\overrightarrow{x}$ can be replaced by one from ${\overrightarrow{x}}_0$ to $\overrightarrow{0}$ and one from $\overrightarrow{0}$ to $\overrightarrow{\mathrm{x}}$ , and the first one always produces a constant contribution.

Next we study the existence and uniqueness of vector potentials. Let $\overrightarrow{\mathrm{F}}$ be a continuously differentiable vector field.

1.

If $\overrightarrow{F}=\nabla \times \overrightarrow{A}$ , then

(1.306) $\overrightarrow{\nabla }\cdot \overrightarrow{F}=0.$

2.

If $\overrightarrow{\nabla }\cdot \overrightarrow{F}=0$ , there exists a vector potential $\overrightarrow{A}$ such that

(1.307) $\overrightarrow{F}=\overrightarrow{\nabla }\times \overrightarrow{A}.$

3.

A vector potential $\overrightarrow{A}$ for the vector $\overrightarrow{F}$ is uniquely determined up to a gradient of a scalar.

For the proof, we proceed as follows.

1.

If there exists $\overrightarrow{A}$ such that $\overrightarrow{F}=\overrightarrow{\nabla }\times \overrightarrow{A}$ , then

$\begin{array}{ll}\hfill \overrightarrow{\nabla }\cdot \overrightarrow{F}& =\overrightarrow{\nabla }\cdot \left(\overrightarrow{\nabla }\times \overrightarrow{A}\right)\hfill \\ \hfill & =0,\hfill \end{array}$

as direct evaluation of the components reveals.

2.

Since $\overrightarrow{\nabla }\cdot \overrightarrow{F}=0$ ,

(1.308) $\frac{\partial F_x}{\partial x}+\frac{\partial F_y}{\partial y}+\frac{\partial F_z}{\partial z}=0.$

Considering

$\begin{array}{l}{\left(\overrightarrow{\nabla }\times \overrightarrow{A}\right)}_x=\frac{\partial A_z}{\partial y}-\frac{\partial A_y}{\partial z}\\ {\left(\overrightarrow{\nabla }\times \overrightarrow{A}\right)}_y=\frac{\partial A_x}{\partial z}-\frac{\partial A_z}{\partial x}\\ {\left(\overrightarrow{\nabla }\times \overrightarrow{A}\right)}_z=\frac{\partial A_y}{\partial x}-\frac{\partial A_x}{\partial y},\end{array}$

define $\overrightarrow{A}$ as

(1.309) $\begin{array}{l}{A}_x={\int }_0^z{F}_y\left(x,y,z\right)dz\\ A_y=-{\int }_0^z{F}_x\left(x,y,z\right)+{\int }_0^x{F}_z\left(x,y,0\right)dx\\ A_z=0.\end{array}$

Then $\overrightarrow{A}$ indeed satisfies the requirement $\overrightarrow{F}=\overrightarrow{\nabla }\times \overrightarrow{A}$ , as shown by computing the curl:

$\begin{array}{ll}\hfill {\left(\overrightarrow{\nabla }\times \overrightarrow{A}\right)}_x& =F_x\left(x,y,z\right)\hfill \\ \hfill {\left(\overrightarrow{\nabla }\times \overrightarrow{A}\right)}_y& =F_y\left(x,y,z\right)\hfill \\ \hfill {\left(\overrightarrow{\nabla }\times \overrightarrow{A}\right)}_z& =-{\int }_0^z\frac{\partial F_x\left(x,y,z\right)}{\partial x}dz+F_z\left(x,y,0\right)-{\int }_0^z\frac{\partial F_y\left(x,y,z\right)}{\partial y}dz\hfill \\ \hfill & =-{\int }_0^z\left[\frac{\partial F_x\left(x,y,z\right)}{\partial x}+\frac{\partial F_y\left(x,y,z\right)}{\partial y}\right]dz+F_z\left(x,y,0\right)\hfill \\ \hfill & ={\int }_0^z\frac{\partial F_z\left(x,y,z\right)}{\partial z}dz+F_z\left(x,y,0\right)\hfill \\ \hfill & =F_z\left(x,y,z\right)-F_z\left(x,y,0\right)+F_z\left(x,y,0\right)\hfill \\ \hfill & =F_z\left(x,y,z\right),\hfill \end{array}$

where Eq. (1.308) was used in moving from the second to the third line of the z component. Indeed, there exists a vector $\overrightarrow{A}$ which satisfies $\overrightarrow{F}=\overrightarrow{\nabla }\times \overrightarrow{A}$ .

3.

Assume that there exist two vectors ${\overrightarrow{A}}_1$ and ${\overrightarrow{A}}_2$ which satisfy

$\begin{array}{l}\overrightarrow{F}=\overrightarrow{\nabla }\times {\overrightarrow{A}}_1\\ \overrightarrow{F}=\overrightarrow{\nabla }\times {\overrightarrow{A}}_2,\end{array}$

then

$\overrightarrow{\nabla }\times \left({\overrightarrow{A}}_1-{\overrightarrow{A}}_2\right)=\overrightarrow{0}.$

Thus, according to the previous theorems about the existence of scalar potentials, there exists ξ which is unique up to a constant such that

${\overrightarrow{A}}_1-{\overrightarrow{A}}_2=\overrightarrow{\nabla }\xi .$

So, the scalar potential $\overrightarrow{A}$ is unique up to the additional $\overrightarrow{\nabla }\xi$ , which concludes the proof.

A closer study of the proof reveals that instead of having $A_z=0$ , by symmetry it is possible to construct vector potentials that satisfy either $A_x=0$ or $A_y=0$ .

We now address the question of nonuniqueness of scalar and vector potentials in more detail. As was shown, different choices of the integration constant in the case of the scalar potential and specific choices of ξ in the case of the vector potential all lead to valid potentials. These transformations between potentials are simple examples of so-called gauge transformations which lead to the same physical system via different potentials. The various possible choices that yield the same field are called different gauges. We will discuss the matter in detail after developing a more complete understanding of the electrodynamics expressed in terms of potentials.

It is illuminating to note that the different gauges based on the non-uniqueness of the potentials, which at first may appear somewhat disturbing, can be accounted for very naturally with the concept of an equivalence relation. An equivalence relation is a relationship "˜" between two elements of a given set that for all elements of the set satisfies

(1.310) $\begin{array}{l}a~a,\\ a\sim b\Rightarrow b\sim a,\text{and}\\ a\sim b,b\sim c\Rightarrow a\sim c.\end{array}$

For the case of scalar potentials, the relationship ˜ _s on the set of scalar functions is that $V_1{~}_s{V}_2$ if $V_1-V_2$ is a constant. For the case of vector potentials, the relationship ˜_ν on the set of vector functions is that ${\overrightarrow{A}}_1{~}_{\upsilon }{\overrightarrow{A}}_2$ if and only if ${\overrightarrow{A}}_1-{\overrightarrow{A}}_2$ can be written as the gradient of a scalar. In terms of gauges, in both cases potentials are related if one can be obtained from the other via a gauge transformation. One can quite readily verify that both these relations satisfy the condition in Eq. (1.310).

For a fixed element a, all elements related to it constitute the equivalence class of a, denoted by [a]; therefore,

(1.311) $\left[a\right]=\left\{x|x\sim a\right\}.$

Apparently, the equivalence classes [] _s of scalar potentials and []_υ of vector potentials describe the collection of all functions ψ, $\overrightarrow{A}$ satisfying $\overrightarrow{F}=\overrightarrow{\nabla }\psi$ and $\overrightarrow{F}=\overrightarrow{\nabla }\times \overrightarrow{A}$ , respectively, which are those that can be obtained through valid gauge transformations from one another. The concept of classes allows the description of scalar and vector potentials in a very natural way; indeed, we can say that the scalar potential ψ to a vector function $\overrightarrow{F}$ is really an equivalence class under the relationship ∼ _s of all those functions whose divergence is $\overrightarrow{F}$ , and the vector potential $\overrightarrow{A}$ to $\overrightarrow{F}$ is the equivalence class under ˜_ν of all those whose curl is $\overrightarrow{F}$ . Within this framework, both scalar and vector potentials are unique, while the underlying freedom of gauge is maintained and accounted for.

The theorems on the existence of scalar and vector potentials allow us to simplify the equations of electrodynamics and make them more compact. Suppose we are given $\overrightarrow{J}$ and ρ, and the task is to describe the system in its entirety. Utilizing Maxwell's equations, we can find $\overrightarrow{E}$ and $\overrightarrow{B}$ of the system, and for the description of the electrodynamic phenomena we need a total of six components.

By using scalar and vector potentials, we can decrease the number of variables in the electromagnetic system. The Maxwell equation $\overrightarrow{\nabla }\cdot \overrightarrow{B}=0$ implies that there exists $\overrightarrow{A}$ such that $\overrightarrow{B}=\overrightarrow{\nabla }\times \overrightarrow{A}$ . On the other hand, Faraday's law $\overrightarrow{\nabla }\times \overrightarrow{E}+\partial \overrightarrow{B}/\partial t=\overrightarrow{0}$ can then be written as

(1.312) $\overrightarrow{\nabla }\times \left(-\overrightarrow{E}-\frac{\partial \overrightarrow{A}}{\partial t}\right)=\overrightarrow{0}.$

This ensures that there is a scalar potential Φ for $-\overrightarrow{E}-\partial \overrightarrow{A}/\partial t$ that satisfies

(1.313) $-\overrightarrow{E}-\frac{\partial \overrightarrow{A}}{\partial t}=\overrightarrow{\nabla }\Phi .$

Thus, the two homogeneous Maxwell's equations are satisfied automatically by setting

(1.314) $\overrightarrow{B}=\overrightarrow{\nabla }\times \overrightarrow{A}\text{and}\overrightarrow{E}=-\overrightarrow{\nabla }\Phi -\frac{\partial \overrightarrow{A}}{\partial t}.$

The other two inhomogeneous Maxwell equations determine the space-time behavior of $\overrightarrow{A}$ and Φ. Using the constitutive relations, $\overrightarrow{D}=\varepsilon \overrightarrow{E}$ and $\overrightarrow{B}=\mu \overrightarrow{H}$ , we express these two Maxwell equations in terms of $\overrightarrow{A}$ and Φ. The Ampère–Maxwell law takes the form

(1.315) $\begin{array}{l}\overrightarrow{\nabla }\times \left(\frac{1}{\mu }\overrightarrow{\nabla }\times \overrightarrow{A}\right)=\overrightarrow{J}+\frac{\partial }{\partial t}\left[\varepsilon \left(-\overrightarrow{\nabla }\Phi -\frac{\partial \overrightarrow{A}}{\partial t}\right)\right]\\ {\overrightarrow{\nabla }}^2\overrightarrow{A}-\varepsilon \mu \frac{{\partial }^2\overrightarrow{A}}{\partial t^2}=\overrightarrow{\nabla }\left(\overrightarrow{\nabla }\cdot \overrightarrow{A}+\varepsilon \mu \frac{\partial \Phi }{\partial t}\right)-\mu \overrightarrow{J}.\end{array}$

On the other hand, Coulomb's law appears as

(1.316) $\begin{array}{ll}\hfill \overrightarrow{\nabla }\cdot \left[\varepsilon \left(-\overrightarrow{\nabla }\Phi -\frac{\partial \overrightarrow{A}}{\partial t}\right)\right]& =\rho \hfill \\ \hfill {\overrightarrow{\nabla }}^2\Phi +\frac{\partial }{\partial t}\left(\overrightarrow{\nabla }\cdot \overrightarrow{A}\right)& =-\frac{\rho }{\varepsilon }.\hfill \end{array}$

Altogether, we have obtained a coupled set of two equations. To summarize, the following set of equations are equivalent to the set of Maxwell's equations:

(1.317) $\overrightarrow{B}=\overrightarrow{\nabla }\times \overrightarrow{A}$

(1.318) $\overrightarrow{E}=\overrightarrow{\nabla }\Phi -\frac{\partial \overrightarrow{A}}{\partial t}$

(1.319) ${\overrightarrow{\nabla }}^2\overrightarrow{A}-\varepsilon \mu \frac{{\partial }^2\overrightarrow{A}}{\partial t^2}=\overrightarrow{\nabla }\left(\overrightarrow{\nabla }\cdot \overrightarrow{A}+\varepsilon \mu \frac{\partial \Phi }{\partial t}\right)-\mu \overrightarrow{J}$

(1.320) ${\overrightarrow{\nabla }}^2\Phi +\frac{\partial }{\partial t}=\left(\overrightarrow{\nabla }\cdot \overrightarrow{A}\right)=-\frac{\rho }{\varepsilon }.$

The fields $\overrightarrow{B}$ and $\overrightarrow{E}$ can be determined by the solutions $\overrightarrow{A}$ and Φ of Eqs. (1.319) and (1.320). However, as seen in the theorems in the beginning of this section, the potentials have the freedom of gauge in that $\overrightarrow{A}$ is unique up to the gradient ${\overrightarrow{\nabla }}_u$ of a scalar field u, and Φ is unique up to a constant. For the coupled situation, it is possible to formulate a general gauge transformation that simultaneously affects $\overrightarrow{A}$ and Φ without influence on the fields. Indeed, if u is an arbitrary smooth scalar field depending on space and time, then the transformation

(1.321) $\overrightarrow{A}\to {\overrightarrow{A}}^{\prime }=\overrightarrow{A}+\overrightarrow{\nabla }u$

(1.322) $\Phi \to {\Phi }^{\prime }=\Phi -\frac{\partial u}{\partial t}$

does not affect the fields $\overrightarrow{E}$ and $\overrightarrow{B}$ and the two inhomogeneous equations. Equation (1.317) for the magnetic field has the form

$\begin{array}{ll}\hfill {\overrightarrow{B}}^{\prime }& =\overrightarrow{\nabla }\times {\overrightarrow{A}}^{\prime }=\overrightarrow{\nabla }\times \left(\overrightarrow{A}+\overrightarrow{\nabla }u\right)\hfill \\ \hfill & =\overrightarrow{\nabla }\times \overrightarrow{A}+\overrightarrow{\nabla }\times \overrightarrow{\nabla }u=\overrightarrow{B},\hfill \end{array}$

whereas the corresponding one (Eq. 1.318) for the electric field has the form

$\begin{array}{ll}\hfill {\overrightarrow{E}}^{\prime }& =-\overrightarrow{\nabla }{\Phi }^{\prime }-\frac{\partial {\overrightarrow{A}}^{\prime }}{\partial t}=-\overrightarrow{\nabla }\left(\Phi -\frac{\partial u}{\partial t}\right)-\frac{\partial }{\partial t}\left(\overrightarrow{A}+\overrightarrow{\nabla }u\right)\hfill \\ \hfill & =-\overrightarrow{\nabla }\Phi -\frac{\partial \overrightarrow{A}}{\partial t}+\overrightarrow{\nabla }\frac{\partial u}{\partial t}-\frac{\partial \overrightarrow{\nabla }u}{\partial t}=\overrightarrow{E}.\hfill \end{array}$

Furthermore, Eq. (1.319) transforms as

$\begin{array}{ll}\hfill & {\overrightarrow{\nabla }}^2{\overrightarrow{A}}^{\prime }-\varepsilon \mu \frac{{\partial }^2{\overrightarrow{A}}^{\prime }}{\partial t^2}-\overrightarrow{\nabla }\left(\overrightarrow{\nabla }\cdot {\overrightarrow{A}}^{\prime }+\varepsilon \mu \frac{\partial {\Phi }^{\prime }}{\partial t}\right)+\mu \overrightarrow{J}\hfill \\ \hfill & ={\overrightarrow{\nabla }}^2\overrightarrow{A}+{\overrightarrow{\nabla }}^2\left(\overrightarrow{\nabla }u\right)-\varepsilon \mu \frac{{\partial }^2\overrightarrow{A}}{\partial t^2}-\varepsilon \mu \frac{{\partial }^2\overrightarrow{\nabla }u}{\partial t^2}\hfill \\ \hfill & -\overrightarrow{\nabla }\left(\overrightarrow{\nabla }\cdot \overrightarrow{A}+{\overrightarrow{\nabla }}^{2}u+\varepsilon \mu \frac{\partial \Phi }{\partial t}-\varepsilon \mu \frac{{\partial }^{2}u}{\partial t^2}\right)+\mu \overrightarrow{J}\hfill \\ \hfill & ={\overrightarrow{\nabla }}^2\overrightarrow{A}-\varepsilon \mu \frac{{\partial }^2\overrightarrow{A}}{\partial t^2}-\overrightarrow{\nabla }\left(\overrightarrow{\nabla }\cdot \overrightarrow{A}+\varepsilon \mu \frac{\partial \Phi }{\partial t}\right)+\mu \overrightarrow{J},\hfill \end{array}$

and Eq. (1.320) reads

$\begin{array}{ll}\hfill & {\overrightarrow{\nabla }}^2{\Phi }^{\prime }+\frac{\partial }{\partial t}\left(\overrightarrow{\nabla }\cdot {\overrightarrow{A}}^{\prime }\right)+\frac{\rho }{\varepsilon }\hfill \\ \hfill & ={\overrightarrow{\nabla }}^2\Phi -{\overrightarrow{\nabla }}^2\frac{\partial u}{\partial t}+\frac{\partial }{\partial t}\left(\overrightarrow{\nabla }\cdot \overrightarrow{A}\right)+\frac{\partial }{\partial t}{\overrightarrow{\nabla }}^{2}u+\frac{\rho }{\varepsilon }\hfill \\ \hfill & ={\overrightarrow{\nabla }}^2\Phi +\frac{\partial }{\partial t}\left(\overrightarrow{\nabla }\cdot \overrightarrow{A}\right)+\frac{\rho }{\varepsilon },\hfill \end{array}$

and altogether the situation is the same as before.

In the gauge transformation (Eqs. 1.321 and 1.322), the freedom of choosing u is left for the convenience depending on the problem. When $\overrightarrow{A}$ and Φ satisfy the so-called Lorentz condition,

(1.323) $\overrightarrow{\nabla }\cdot \overrightarrow{A}+\varepsilon \mu \frac{\partial \Phi }{\partial t}=0,$

then the gauge is called the Lorentz gauge. Suppose there are solutions ${\overrightarrow{A}}_0$ and Φ₀ to Eqs. (1.319) and (1.320), then

(1.324) ${\overrightarrow{A}}_L=A_0+\overrightarrow{\nabla }{u}_L$

(1.325) ${\Phi }_L={\Phi }_0-\frac{\partial u_L}{\partial t}$

are also solutions as shown previously. Now,

$\begin{array}{ll}\hfill \overrightarrow{\nabla }\cdot {\overrightarrow{A}}_L+\varepsilon \mu \frac{\partial {\Phi }_L}{\partial t}& =\overrightarrow{\nabla }\cdot \left({\overrightarrow{A}}_0+\overrightarrow{\nabla }{u}_L\right)+\varepsilon \mu \frac{\partial }{\partial t}\left({\Phi }_0-\frac{\partial u_L}{\partial t}\right)\hfill \\ \hfill & =\overrightarrow{\nabla }\cdot {\overrightarrow{A}}_0+\varepsilon \mu \frac{\partial {\Phi }_0}{\partial t}+{\overrightarrow{\nabla }}^2{u}_L-\varepsilon \mu \frac{{\partial }^2{u}_L}{\partial t^2}.\hfill \end{array}$

By choosing u_L as a solution of

(1.326) ${\overrightarrow{\nabla }}^2{u}_L-\varepsilon \mu \frac{{\partial }^2{u}_L}{\partial t^2}=-\left(\overrightarrow{\nabla }\cdot {\overrightarrow{A}}_0+\varepsilon \mu \frac{\partial {\Phi }_0}{\partial t}\right),$

obviously the Lorentz gauge condition

(1.327) $\overrightarrow{\nabla }\cdot {\overrightarrow{A}}_L+\varepsilon \mu \frac{\partial {\Phi }_L}{\partial t}=0$

is satisfied. Then the two inhomogeneous equations are decoupled, and we obtain two symmetric inhomogeneous wave equations

(1.328) ${\overrightarrow{\nabla }}^2{\overrightarrow{A}}_L-\varepsilon \mu \frac{{\partial }^2{\overrightarrow{A}}_L}{\partial t^2}=-\mu \overrightarrow{J}$

(1.329) ${\overrightarrow{\nabla }}^2{\varphi }_L-\varepsilon \mu \frac{{\partial }^2{\overrightarrow{\Omega }}_L}{\partial t^2}=-\frac{\rho }{\epsilon }.$

Another useful gauge is the Coulomb gauge, which satisfies the Coulomb condition

(1.330) $\overrightarrow{\nabla }\cdot \overrightarrow{A}=0.$

Supposing ${\overrightarrow{A}}_0$ and Φ₀ are solutions to Eqs. (1.319) and (1.320), then

(1.331) ${\overrightarrow{A}}_C={\overrightarrow{A}}_0+\overrightarrow{\nabla }{u}_C$

(1.332) ${\Phi }_C={\Phi }_0-\frac{\partial u_C}{\partial t}$

are also solutions. Now observe

$\begin{array}{ll}\hfill \overrightarrow{\nabla }\cdot {\overrightarrow{A}}_C& =\overrightarrow{\nabla }\cdot \left({\overrightarrow{A}}_0+\overrightarrow{\nabla }{u}_C\right)\hfill \\ \hfill & =\overrightarrow{\nabla }\cdot {\overrightarrow{A}}_0+{\overrightarrow{\nabla }}^{2}uc.\hfill \end{array}$

By choosing u_c as a solution of

(1.333) ${\overrightarrow{\nabla }}^2{u}_C=-\overrightarrow{\nabla }\cdot {\overrightarrow{A}}_0,$

we obtain that

(1.334) ${\overrightarrow{\nabla }}^2\cdot {\overrightarrow{A}}_C=0$

holds. Then the two inhomogeneous equations read

(1.335) ${\overrightarrow{\nabla }}^2\cdot {\overrightarrow{A}}_C-\varepsilon \mu \frac{{\partial }^2{\overrightarrow{A}}_C}{\partial t^2}=\varepsilon \mu \frac{\partial \overrightarrow{\nabla }{\Phi }_C}{\partial t}-\mu \overrightarrow{J}$

(1.336) ${\overrightarrow{\nabla }}^2{\Phi }_C=-\frac{\rho }{\varepsilon },$

and while there is no symmetry, it is convenient that the scalar potential Φ _c is the "instantaneous" Coulomb potential due to the time-dependent charge density $\rho \left(\overrightarrow{x},t\right)$ .

Sometimes a gauge where A_z , the z component of $\overrightarrow{A}$ , is set to 0, is useful. In the proof of the existence of vector potentials we saw that it is possible to choose $\overrightarrow{A}$ in such a way; we now assume A does not satisfy this condition outright, and we attempt to bring it to this form. The gauge condition in this case is

(1.337) $A_z=0.$

Supposing ${\overrightarrow{A}}_0$ and Φ₀ are solutions of Eqs. (1.319) and (1.320), then

(1.338) ${\overrightarrow{A}}_S={\overrightarrow{A}}_0+\overrightarrow{\nabla }{u}_S$

(1.339) ${\Phi }_S={\Phi }_0-\frac{\partial u_S}{\partial t}$

are also solutions. In particular, we have

$A_{Sz}=A_{0z}+\frac{\partial u_S}{\partial z},$

and by choosing u_s as a solution of

(1.340) $\frac{\partial u_S}{\partial z}=-A_{0z},$

we obtain that, as needed,

(1.341) $A_{Sz}=0.$

In a similar way we can of course also construct vector potentials in which the x or y component vanishes.

We conclude the discussion of potentials with the special case of time-independent free space, in which the whole argument becomes very simple. With the constitutive relations $\overrightarrow{D}={\varepsilon }_0\overrightarrow{E}$ and $\overrightarrow{B}={\mu }_0\overrightarrow{H}$ , Maxwell's equations are

(1.342) $\overrightarrow{\nabla }\cdot \overrightarrow{E}=0,\overrightarrow{\nabla }\times \overrightarrow{E}=\overrightarrow{0}$

(1.343) $\overrightarrow{\nabla }\cdot \overrightarrow{B}=0,\overrightarrow{\nabla }\times \overrightarrow{B}=\overrightarrow{0}.$

From the curl equations we infer that there are scalar potentials Φ and Φ _M for $\overrightarrow{E}$ and $\overrightarrow{B}$ , respectively, such that we have

(1.344) $\overrightarrow{E}=-\overrightarrow{\nabla }\Phi$

(1.345) $\overrightarrow{B}=-\overrightarrow{\nabla }{\Phi }_M.$

Applying the divergence equations leads to Laplace equations for the electric and magnetic scalar potential:

(1.346) ${\overrightarrow{\nabla }}^2\Phi =0$

(1.347) ${\overrightarrow{\nabla }}^2{\Phi }_M=0.$

Read full chapter

URL:

https://www.sciencedirect.com/science/article/pii/S1076567008702271

Vector Calculus

James R. Kirkwood , in Mathematical Physics with Partial Differential Equations, 2013

Stokes' Theorem

Stokes' Theorem is a generalization of Green's Theorem to ℝ³. In Stokes' Theorem we relate an integral over a surface to a line integral over the boundary of the surface. We assume that the surface is two-sided that consists of a finite number of pieces, each of which has a normal vector at each point. This allows us to consider surfaces such as cubes that do not have normal vectors at the edges. We assume an orientation of the surface, which means we agree on the direction of the unit normal vector. This is necessary because there will be two unit normal vectors at each point p on the surface, $\hat{n}(p)$ and $-\hat{n}(p)$ . This induces an orientation for the boundary of the surface.

Theorem (Stokes' Theorem):

Let S be an oriented surface, and let ∂S denote the oriented boundary of S. If ${\displaystyle \stackrel{\rightharpoonup }{F}}$ is a continuously differentiable vector field on S, then

${\iint }_S(\nabla \times {\displaystyle \stackrel{\rightharpoonup }{F}})dS={\int }_{\partial S}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds.$

We give a sketch of the central idea in the proof of Stokes' Theorem, which is simply Green's Theorem. We begin by dividing the surface S into small squares, and focus on one square, which we call S _ij (see Figure 2-3-8).

Figure 2-3-8.

Let ${\displaystyle \stackrel{\rightharpoonup }{F}}=F_x\hat{i}+F_y\hat{j}+F_z\hat{k}$ be a vector field and $\hat{s}=x\hat{i}+y\hat{j}+z\hat{k}\text{,}$ so that $ds=dx\hat{i}+dy\hat{j}+dz\hat{k}.$ We compute

${\oint }_{\partial S_{ij}}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds$

by which we mean the line integral around the boundary of S _ij . We choose the axes so that S _ij is perpendicular to the z-axis. Then

${\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds=F_{x}dx+F_{y}dy+F_{z}dz.$

Consider S _ij as shown in Figure 2-3-9. We have

${\int }_{C_1}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\int }_{x=x_1}^{x_1+\mathrm{\Delta }x}{F}_x(x\text{,}{y}_1)dx$

Figure 2-3-9.

and

${\int }_{C_3}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\int }_{x=x_1+\mathrm{\Delta }x}^{x_1}{F}_x(x\text{,}{y}_1+\mathrm{\Delta }y)dx=-{\int }_{x=x_1}^{x_1+\mathrm{\Delta }x}{F}_x(x\text{,}{y}_1+{\mathrm{\Delta }}_y)dx$

so

${\int }_{C_1}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds+{\int }_{C_3}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\int }_{x=x_1}^{x_1+\mathrm{\Delta }x}{F}_x(x\text{,}{y}_1)dx-{\int }_{x=x_1}^{x_1+\mathrm{\Delta }x}{F}_x(x\text{,}{y}_1+\mathrm{\Delta }y)dx={\int }_{x=x_1}^{x_1+\mathrm{\Delta }x}[F_x(x\text{,}{y}_1)-F_x(x\text{,}{y}_1+\mathrm{\Delta }y)]dx=-{\int }_{x=x_1}^{x_1+\mathrm{\Delta }x}[F_x(x\text{,}{y}_1+\mathrm{\Delta }y)-F_x(x\text{,}{y}_1)]dx.$

Also,

${\int }_{C_2}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\int }_{y=y_1}^{y_1+\mathrm{\Delta }y}{F}_y(x_1+\mathrm{\Delta }x\text{,}y)dy$

and

${\int }_{C_4}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\int }_{y=y_1+\mathrm{\Delta }y}^{y_1}{F}_y(x_1\text{,}y)dy=-{\int }_{y=y_1}^{y_1+\mathrm{\Delta }y}{F}_y(x_1\text{,}y)dy\text{,}$

so

${\int }_{C_2}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds+{\int }_{C_4}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\int }_{y=y_1}^{y_1+\mathrm{\Delta }y}[F_y(x_1+\mathrm{\Delta }x\text{,}y)-F_y(x_1\text{,}y)]dy.$

By Green's Theorem we have

${\oint }_{\partial S_{ij}}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\int }_{C_1}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds+{\int }_{C_3}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds+{\int }_{C_2}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds+{\int }_{C_4}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\oint }_{\partial S_{ij}}{F}_{x}dx+F_{y}dy={\iint }_{S_{ij}}\left(\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\right)dxdy.$

Recall that

$(\nabla \times {\displaystyle \stackrel{\rightharpoonup }{F}})\cdot \hat{k}=\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}\text{,}$

so that in this special case

${\oint }_{\partial S_{ij}}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\iint }_S(\nabla \times {\displaystyle \stackrel{\rightharpoonup }{F}})\cdot \hat{k}dS.$

In the general case, the square will not be in the xy-plane, but if we replace $\hat{k}$ by the unit normal vector $\hat{n}$ , we get

${\oint }_{\partial S_{ij}}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\iint }_S(\nabla \times {\displaystyle \stackrel{\rightharpoonup }{F}})\cdot \hat{n}dS={\iint }_S(\nabla \times {\displaystyle \stackrel{\rightharpoonup }{F}})dS.$

For a second way to get the same result, note that

${\int }_{C_2}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds+{\int }_{C_4}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\int }_{y=y_1}^{y_1+\mathrm{\Delta }y}[F_y(x_1+\mathrm{\Delta }x\text{,}y)-F_y(x_1\text{,}y)]dy\approx \frac{\partial F_y}{\partial x}\mathrm{\Delta }x\mathrm{\Delta }y$

and

${\int }_{C_1}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds+{\int }_{C_3}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds=-{\int }_{x=x_1}^{x_1+\mathrm{\Delta }x}[F_x(x\text{,}{y}_1+\mathrm{\Delta }y)-F_x(x\text{,}{y}_1)]dx\approx -\frac{\partial F_x}{\partial y}\mathrm{\Delta }x\mathrm{\Delta }y.$

Now we put the squares together. Figure 2-3-10 describes the idea.

Figure 2-3-10.

For adjoining squares, the contributions at the common boundary cancel one another because they are in opposite directions. Thus, when we sum over all squares the only pieces that do not cancel are those on the boundary of S. Thus, we get

${\iint }_S(\nabla \times {\displaystyle \stackrel{\rightharpoonup }{F}})dS={\int }_{\partial S}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds.$

Example:

We verify Stokes' Theorem in the case ${\displaystyle \stackrel{\rightharpoonup }{F}}=3y\hat{i}+4x\hat{j}+(2x+2z)\hat{k}$ , where S is the upper half of the hemisphere x ²+y ²+z ²=1.

Now ∂S is x ²+y ²=1, so we let

$\sigma (t)=(\cos t\text{,}\sin t\text{,}0)$

and then on ∂S,

$\sigma ΄(t)=(-\sin t\text{,}\cos t\text{,}0)\text{,}$

${\displaystyle \stackrel{\rightharpoonup }{F}}(\sigma (t))=(3\sin t\text{,}4\cos t\text{,}2\cos t)\text{,}$

${\displaystyle \stackrel{\rightharpoonup }{F}}(\sigma (t))\cdot \sigma ΄(t)=-3{\sin }^{2}t+4{\cos }^{2}t.$

We then have

${\int }_{\partial S}{\displaystyle \stackrel{\rightharpoonup }{F}}\cdot ds={\int }_0^{2\pi }(-3{\sin }^{2}t+4{\cos }^{2}t)dt=\pi .$

We next compute

${\iint }_S(\nabla \times {\displaystyle \stackrel{\rightharpoonup }{F}})dS.$

We have

$\nabla \times {\displaystyle \stackrel{\rightharpoonup }{F}}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ 3y& 4x& 2x+2z\end{array}\right|=-2\hat{j}+\hat{k}.$

We parameterize S with spherical coordinates

$x=\sin \theta \cos \phi \text{,}y=\sin \theta \sin \phi \text{,}z=\cos \theta \text{,}$

so that

$T_{\theta }=\cos \theta \cos \phi \hat{i}+\cos \theta \sin \phi \hat{j}-\sin \theta \hat{k}\text{,}$

$T_{\phi }=-\sin \theta \sin \phi \hat{i}+\sin \theta \cos \phi \hat{j}\text{,}$

and

$T_{\theta }\times T_{\phi }={\sin }^2\theta \cos \phi \hat{i}+{\sin }^2\theta \sin \phi \hat{j}+\sin \theta \cos \theta \hat{k}$

is normal to the surface. Then

$(\nabla \times \overrightarrow{F})\cdot \hat{n}=(\nabla \times \overrightarrow{F})\cdot (T_{\theta }\times T_{\phi })=-2{\sin }^2\theta \sin \phi +\sin \theta \cos \theta$

and

${\iint }_S(\nabla \times {\displaystyle \stackrel{\rightharpoonup }{F}})\cdot \hat{n}dS={\int }_{\theta =0}^{\pi /2}\left({\int }_{\phi =0}^{2\pi }(-2{\sin }^2\theta \sin \phi +\sin \theta \cos \theta )d\phi \right)d\theta =\pi .$

Note that we could have computed the normal vector using T _φ×T _θ instead of T _θ ×T _φ, in which case we would have

${\iint }_S(\nabla \times {\displaystyle \stackrel{\rightharpoonup }{F}})\cdot \hat{n}dS=-\pi .$

To make the signs of the integrals agree in Stokes' Theorem, we must follow the "right-hand rule" that determines the outward pointing normal for a surface that is not closed according to the direction of the line integral of the boundary. If the signs are opposite, you should have taken the cross product in the reverse order.

The following theorem gives results that will be important in the solution of partial differential equations.

Theorem (Green's Identities):

a.

(Green's First Identity): If f and g are scalar functions with continuous partial derivatives in a region R, and if V is a region within R with surface S, then

${\iiint }_{\mathit{V}}(f{\nabla }^{2}g+\nabla f\cdot \nabla g)dV={\iint }_{S}f\frac{\partial g}{\partial n}dS\text{,}$

where $\frac{\partial g}{\partial n}=\nabla g\cdot \hat{n}\text{,}$ the directional derivative of g in the direction that is outward normal to the surface S.

b.

(Green's Second Identity): If f and g are scalar functions with continuous partial derivatives in a region R, and if V is a region within R with surface S, then

${\iiint }_{\mathit{V}}(f{\nabla }^{2}g-g{\nabla }^{2}f)dV={\iint }_S\left(f\frac{\partial g}{\partial n}-g\frac{\partial f}{\partial n}\right)dS.$

We leave the proof of the theorem as exercises 13 and 14.

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