Maxwell’s Equations

Electrostatics

$\nabla \cdot \vec{E} = \frac{ρ}{ε_{0}}$
$\nabla \times \vec{E} = 0$
$\nabla^{2} Φ = - \frac{ρ}{ε_{0}}$

Notes

$\nabla \times \vec{E} (\vec{x}) = \nabla \times [\frac{1}{4 π ε_{0}} \int ρ (\vec{x}) \frac{\vec{x} - {\vec{x}}^{'}}{| \vec{x} - \vec{x^{'}} |^{3}} d^{3} x^{'}] = \frac{1}{4 π ε_{0}} \int ρ (\vec{x}) \nabla \times \frac{\vec{x} - {\vec{x}}^{'}}{| \vec{x} - \vec{x^{'}} |^{3}} d^{3} x^{'} = \dots = 0$

$Φ (\vec{x}) = \frac{1}{4 π ε_{0}} \int \frac{ρ (\vec{x})}{| \vec{x} - {\vec{x}}^{'} |} d^{3} x^{'}$

$\nabla \cdot (\nabla Φ) = - \frac{ρ}{ε_{0}}$ This gives us the Poisson equation: $\nabla^{2} Φ = - \frac{ρ}{ε_{0}}$

If we are dealing with a space with no charges, $ρ = 0$ then we get Laplace’s equation: $\nabla^{2} Φ = 0$ .

Green’s Theorem to solve Boundary Value Problem

Say we have Dirchlet (the potential on the surface, $Φ$ on $S$ ) or the Neumann condition (the derivative on the surface, $\frac{\partial Φ}{\partial n} = \nabla Φ \cdot \hat{n}$ ) then you can uniquely determine the potential inside the region.

Let $Φ, ψ$ be scalar field and let $\vec{A} = ϕ \nabla ψ$ be a vector field. Then, $\int \nabla \cdot (ϕ \nabla ψ) d V = \oint_{S} ϕ (\nabla ψ) \hat{n} d a$ . By the chain rule, we have $\int (\nabla ϕ \nabla ψ + ϕ \nabla^{2} ψ) d V = \oint_{S} ϕ \frac{\partial ψ}{\partial n} d a$ .

Let $\vec{B} = ψ \nabla ϕ$ be another vector field. Then, $\int (\nabla ψ \nabla ϕ + ψ \nabla^{2} ϕ) d V = \oint_{S} ψ \frac{\partial ϕ}{\partial n} d a$ .

Subtracting these two expressions, we have $\int_{V} (ψ \nabla^{2} ϕ - ϕ \nabla^{2} ψ) d V = \oint_{S} (ϕ \frac{\partial Ψ}{\partial n} - ψ \frac{\partial ϕ}{\partial n}) d a$ .

Let $ψ = \frac{1}{| \vec{x} - {\vec{x}}^{'} |} = \frac{1}{R}$ and $ϕ = Φ$ .

$\nabla^{2} \frac{1}{r} = - 4 π δ^{3} (\vec{x} - {\vec{x}}^{'})$

Integrating over $x^{'}$ , $\int_{V} (\frac{1}{R} \frac{ρ}{ε_{0}} - Φ ({\vec{x}}^{'}) (- 4 π δ^{3} (R))) d V = \int_{V} (\frac{1}{R} \frac{ρ}{ε_{0}}) d V + 4 π Φ (\vec{x}) = \int_{S} (Φ ({\vec{x}}^{'}) \frac{\partial}{\partial n^{'}} \frac{1}{R} - \frac{1}{R} \frac{\partial Φ}{\partial n^{'}}) d a^{'}$ . Thus, $Φ (\vec{x}) = \frac{1}{4 π ε_{0}} \int_{V} \frac{ρ ({\vec{x}}^{'})}{| \vec{x} - {\vec{x}}^{'} |} d^{3} x^{'} + \int_{S} \dots d a^{'}$ Note: We integrate over the enclosed volume and we must know both BC in this formulation.

Note: $\frac{\partial ψ}{\partial n} = \nabla ψ \cdot \hat{n}$ .

Uniqueness Theorem

$- \nabla^{2} Φ = - \frac{ρ}{ε_{0}}$ and either of the Dirchlet or Neumann boundary conditions are satisfied, then $Φ$ is unique.
…

Proof of 1. Assume we have $Φ_{1}, Φ_{2}$ which satisfy the equation and the BC. Then, let $U = Φ_{2} - Φ_{1}$ . So, $\nabla^{2} U = 0$ , satisfying Laplace’s equation and at the surface since both boundaries are the same, then either $U = 0$ on the surface or the derivative is zero on the surface. So, $\int_{V} (U \nabla^{2} + \nabla U \cdot \nabla U) d V = \int_{S} U (\frac{\partial U}{\partial n}) d a = 0$ . So, $\int | \nabla U |^{2} d V = 0$ . Then, $\nabla U = 0$ implies that $U (x) = C$ . If we have Dirchlet, $C = 0$ since the boundaries must be equal otherwise it could be some arbitrary $C$ . QED.

Green Functions

G(\vec{x},\vec{x}’) = G(\vec{x}’,\vec{x})

$\nabla^{2} G (\vec{x}, {\vec{x}}^{'}) = - 4 π δ (\vec{x} - {\vec{x}}^{'})$
$G (\vec{x}, \vec{x}) = \frac{1}{| \vec{x} - {\vec{x}}^{'} |} + F (\vec{x}, {\vec{x}}^{'})$ such that $\nabla^{' 2} F = 0$ inside $V$ . I.e. homogeneous in $V$ .

Types:

Dirchlet: $G_{D} = 0$ on surface
Neumann: $G_{N} (\vec{x}, {\vec{x}}^{'})$ $\frac{\partial G_{N}}{\partial n} = - \frac{4 π}{A_{S}}$ .

$\nabla^{' 2} G_{N} = - 4 π δ (\vec{x} - {\vec{x}}^{'}), \int_{V} \nabla^{'} (\nabla G_{N}) d V = - 4 π = \int_{S} (\nabla^{'} G_{N}) \hat{n} d S$ . So, $\frac{\partial G_{N}}{\partial n} = C$ such that $C$ times the area of the surface is equal to $- 4 π$ .

$Φ (\vec{x}) = \frac{1}{4 π ε_{0}} \int ρ ({\vec{x}}^{'}) G (\vec{x}, {\vec{x}}^{'}) d^{3} x^{'} + \frac{1}{4 π} \int_{S} [G (\vec{x}, {\vec{x}}^{'}) \frac{\partial Φ}{\partial n} - Φ ({\vec{x}}^{'}) \frac{\partial G}{\partial n^{'}}]$

If we have Dirchlet BC. Then note that $G_{D} = 0$ implies that the term with the Neumann boundary condition vanishes and so we can solve for the Potential exactly without knowledge of the Neumann BC. I.e. $Φ (\vec{x}) = \frac{1}{4 π ε_{0}} \int_{V} ρ ({\vec{x}}^{'}) G_{D} d V - \frac{1}{4 π} \int_{S} Φ ({\vec{x}}^{'}) \frac{\partial G_{D}}{\partial n^{'}} d a^{'}$ .

Similarly, if we have Neumann BC, then $\partial_{n} G_{N} = \frac{- 4 π}{A_{S}}$ implies …

Energy

$W = \sum_{i = 1}^{N} W_{i} = \frac{1}{4 π ε_{0}} \sum_{i = 1}^{N} \sum_{j = 1}^{i - 1} \frac{q_{i} q_{j}}{x_{i j}}$

Let us have $n$ point charges scattered around the space. Then, we can express the work to arrange this, i.e. the potential energy of the system, is $W = \frac{1}{2} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} W_{i j} = \frac{1}{8 π ε_{0}} \sum_{i = 1}^{N} \sum_{j \neq i}^{N} \frac{q_{i} q_{j}}{| {\vec{x}}_{i j} |}$

For a continuous charge distribution, we then get, $W = \frac{1}{2} \frac{1}{4 π ε_{0}} \iint \frac{ρ (\vec{x}) ρ ({\vec{x}}^{'})}{| \vec{x} - {\vec{x}}^{'} |} d^{3} x d^{3} x^{'} = \frac{1}{2} \int \frac{1}{4 π ε_{0}} ρ (\vec{x}) \frac{ρ ({\vec{x}}^{'})}{| \vec{x} - {\vec{x}}^{'} |} d^{3} x d^{3} x^{'} = \frac{1}{2} \int ρ (\vec{x}) Φ (\vec{x}) d V = - \frac{ε_{0}}{2} \int Φ \nabla^{2} Φ d V$

$\nabla \cdot (Φ \nabla Φ) = Φ \nabla^{2} Φ + \nabla Φ \cdot \nabla Φ = Φ \nabla^{2} Φ + | \vec{E} |^{2}$ Using a vector identity, $W = - \frac{ε_{0}}{2} \int Φ \nabla^{2} Φ d V = \frac{ε_{0}}{2} \int | \vec{E} |^{2} d V - \frac{ε_{0}}{2} \int (Φ \nabla Φ) \cdot \hat{n} d a = \frac{ε_{0}}{2} \int | \vec{E} |^{2} d V \Rightarrow w = \frac{1}{2} ε_{0} | \vec{E} |^{2}$ . The second term vanishes since at infinity, the distribution appears as a point charge which has a decaying field at infinity. (The second term does not vanish if the region integrated is finite or close to the distribution).