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mathematics-physics-wiki/docs/en/physics/mathematical-physics/vector-analysis/vector-operators.md
2024-01-25 18:06:06 +01:00

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Vector operators

Properties of the gradient, divergence and curl

Proposition: let a,b \in \mathbb{R}, f,g: \mathbb{R}^3 \to \mathbb{R} be differentiable scalar fields and \mathbf{u}, \mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3 be differentiable vector fields. Then we have the following identities:

Linearity:

\begin{align*} \nabla (af + bg) &= a \nabla f + b \nabla g, \ \nabla \cdot (a\mathbf{u} + b \mathbf{v}) &= a (\nabla \cdot \mathbf{u}) + b (\nabla \cdot \mathbf{v}), \ \nabla \times (a\mathbf{u} + b \mathbf{v}) &= a (\nabla \times \mathbf{u}) + b (\nabla \times\mathbf{v}). \end{align*}

Multiplication rules:

\begin{align*} \nabla (fg) &= f \nabla g+ g \nabla f, \ \nabla \cdot (f \mathbf{u}) &= f (\nabla \cdot \mathbf{u}) + \langle \nabla f, \mathbf{u} \rangle, \ \nabla \cdot (\mathbf{u} \times \mathbf{v}) &= \langle \nabla \times \mathbf{u}, \mathbf{v} \rangle - \langle \mathbf{u}, \nabla \times \mathbf{v} \rangle, \ \nabla \times (f\mathbf{u}) &= f (\nabla \times \mathbf{u}) + \nabla f \times \mathbf{u}. \end{align*}

??? note "Proof:"

Will be added later.

The laplacian

Definition: the laplacian of a differentiable scalar field f: \mathbb{R}^3 \to \mathbb{R} is defined as

\nabla^2 f(\mathbf{x}) := \nabla \cdot \nabla f(\mathbf{x}),

for all \mathbf{x} \in \mathbb{R}^3.

The notation may be unorthodox for some. An alternative notatation for the laplacian is \Delta f, though generally deprecated.

We can also rewrite the laplacian for curvilinear coordinate systems as has been done below.

Theorem: the laplacian of a differentiable scalar field f: \mathbb{R}^3 \to \mathbb{R} for a curvilinear coordinate system is given by

\nabla^2 f(\mathbf{x}) = \frac{1}{g(\mathbf{x})} \partial_i \Big(\sqrt{g(\mathbf{x})} g^{ij}(\mathbf{x}) \partial_j f(\mathbf{x}) \Big),

for all \mathbf{x} \in \mathbb{R}^3.

??? note "Proof:"

Will be added later.

The laplacian for a ortho-curvilinear coordinate system may also be derived and can be found below.

Corollary: the laplacian of a differentiable scalar field f: \mathbb{R}^3 \to \mathbb{R} for a ortho-curvilinear coordinate system is given by

\nabla^2 f(\mathbf{x}) = \frac{1}{h_1 h_2 h_3} \bigg(\partial_1 \Big(\frac{h_2 h_3}{h_1} \partial_1 f(\mathbf{x}) \Big) + \partial_2 \Big(\frac{h_1 h_3}{h_2} \partial_2 f(\mathbf{x}) \Big) + \partial_3 \Big(\frac{h_1 h_2}{h_3} \partial_3 f(\mathbf{x}) \Big) \bigg),

for all \mathbf{x} \in \mathbb{R}^3.

??? note "Proof:"

Will be added later.

Please note that the scaling factors may also depend on the position \mathbf{x} \in \mathbb{R}^3 depending on the coordinate system.

Proposition: the laplacian of a differentiable vector field \mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3 is given by

\nabla^2 \mathbf{v}(\mathbf{x}) = \nabla \big(\nabla \cdot \mathbf{v}(\mathbf{x})\big) - \nabla \times \big(\nabla \times \mathbf{v}(\mathbf{x})\big),

for all \mathbf{x} \in \mathbb{R}^3.

??? note "Proof:"

Will be added much later.

Potentials

Definition: a vector field \mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3 is irrotational or curl free if

\nabla \times \mathbf{v}(\mathbf{x}) = \mathbf{0},

for all \mathbf{x} \in \mathbb{R}^3.

If \mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3 is the gradient of some scalar field \Phi: \mathbb{R}^3 \to \mathbb{R} it is irrotational since

\nabla \times\big (\nabla \Phi(\mathbf{x})\big) = \mathbf{0},

for all \mathbf{x} \in \mathbb{R}^3.

Proposition: an irrotational vector field \mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3 has a scalar potential \Phi: \mathbb{R}^3 \to \mathbb{R} such that

\mathbf{v}(\mathbf{x}) = \nabla \Phi(\mathbf{x}),

for all \mathbf{x} \in \mathbb{R}^3.

??? note "Proof:"

Will be added later.

In physics the scalar potential is generally given by the negative of the gradient, both are correct but one is more stupid than the other.

Definition: a vector field \mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3 is solenoidal or divergence-free if

\nabla \cdot \mathbf{v}(\mathbf{x}) = 0,

for all \mathbf{x} \in \mathbb{R}^3.

If \mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3 is the curl of some vector field \mathbf{u}: \mathbb{R}^3 \to \mathbb{R}^3 it is solenoidal since

\nabla \cdot \big(\nabla \times \mathbf{u}(\mathbf{x}) \big) = 0,

for all \mathbf{x} \in \mathbb{R}^3.

Proposition: a solenoidal vector field \mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3 has a vector potential \mathbf{u}: \mathbb{R}^3 \to \mathbb{R}^3 such that

\mathbf{v}(\mathbf{x}) = \nabla \times \mathbf{u}(\mathbf{x}),

for all \mathbf{x} \in \mathbb{R}^3.

??? note "Proof:"

Will be added later.

The theorem below is the Helmholtz decomposition theorem and states that every vector field can be written in terms of two potentials.

Theorem: every vector field \mathbf{v}: \mathbb{R}^3 \to \mathbb{R}^3 can be written in terms of a scalar \Phi: \mathbb{R}^3 \to \mathbb{R} and a vector \mathbf{u}: \mathbb{R}^3 \to \mathbb{R}^3 potential as

\mathbf{v}(\mathbf{x}) = \nabla \Phi(\mathbf{x}) + \nabla \times \mathbf{u}(\mathbf{x}),

for all \mathbf{x} \in \mathbb{R}^3.

??? note "Proof:"

Will be added later.

It then follows that the scalar and vector potentials can be determined for a volume V \subset \mathbb{R}^3 with a boundary surface A \subset \mathbb{R}^3 that encloses the domain V.

Corollary: the scalar \Phi: \mathbb{R}^3 \to \mathbb{R} and vector \mathbf{u}: \mathbb{R}^3 \to \mathbb{R}^3 potentials for a volume V \subset \mathbb{R}^3 with a boundary surface A \subset \mathbb{R}^3 that encloses the domain V are given by

\begin{align*} \Phi(\mathbf{x}) &= \frac{1}{4\pi} \int_V \frac{\nabla \cdot \mathbf{v}(\mathbf{r})}{|\mathbf{x} - \mathbf{r}|}dV - \frac{1}{4\pi} \oint_A \bigg\langle \frac{1}{|\mathbf{x} - \mathbf{r}|} \mathbf{v}(\mathbf{r}), d\mathbf{A} \bigg\rangle, \ \ \mathbf{u}(\mathbf{x}) &= \frac{1}{4\pi} \int_V \frac{\nabla \times \mathbf{v}(\mathbf{r})}{|\mathbf{x} - \mathbf{r}|}dV - \frac{1}{4\pi} \oint_A \frac{1}{|\mathbf{x} - \mathbf{r}|} \mathbf{v}(\mathbf{r}) \times d\mathbf{A}, \end{align*}

for all \mathbf{x} \in \mathbb{R}^3.

??? note "Proof:"

Will be added later.