# Maxwell’s Equations

## Electrostatics

• $$\nabla\cdot\vec{E}=\frac{\rho}{\varepsilon_0}$$
• $$\nabla\times\vec{E}=0$$
• $$\nabla^2\Phi=-\frac{\rho}{\varepsilon_0}$$

### Notes

$$\nabla\times\vec{E}(\vec{x}) = \nabla\times\left[\frac{1}{4\pi\varepsilon_0}\int\rho(\vec{x})\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x'}|^3}\:d^3x'\right] = \frac{1}{4\pi\varepsilon_0}\int\rho(\vec{x})\nabla\times\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x'}|^3}\:d^3x' = \cdots = 0$$

$$\Phi(\vec{x}) = \frac{1}{4\pi\varepsilon_0}\int\frac{\rho(\vec{x})}{|\vec{x}-\vec{x}'|}\:d^3x'$$

$$\nabla\cdot(\nabla\Phi) = -\frac{\rho}{\varepsilon_0}$$ This gives us the Poisson equation: $$\nabla^2\Phi=-\frac{\rho}{\varepsilon_0}$$

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

#### Green’s Theorem to solve Boundary Value Problem

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

Let $$\Phi,\psi$$ be scalar field and let $$\vec{A}=\phi\nabla\psi$$ be a vector field. Then, $$\int\nabla\cdot(\phi\nabla\psi)dV=\oint_S\phi(\nabla\psi)\hat{n}da$$. By the chain rule, we have $$\int(\nabla\phi\nabla\psi+\phi\nabla^2\psi)dV = \oint_S\phi\frac{\partial\psi}{\partial n}da$$.

Let $$\vec{B}=\psi\nabla\phi$$ be another vector field. Then, $$\int(\nabla\psi\nabla\phi+\psi\nabla^2\phi)dV = \oint_S\psi\frac{\partial\phi}{\partial n}da$$.

Subtracting these two expressions, we have $$\int_V(\psi\nabla^2\phi-\phi\nabla^2\psi)dV = \oint_S\left(\phi\frac{\partial\Psi}{\partial n}-\psi\frac{\partial\phi}{\partial n}\right)da$$.

Let $$\psi=\frac{1}{|\vec{x}-\vec{x}'|} = \frac{1}{R}$$ and $$\phi=\Phi$$.

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

Integrating over $$x'$$, $$\int_V\left(\frac{1}{R}\frac{\rho}{\varepsilon_0}-\Phi(\vec{x}')(-4\pi\delta^3(R))\right)\:dV = \int_V\left(\frac{1}{R}\frac{\rho}{\varepsilon_0}\right)\:dV +4\pi\Phi(\vec{x}) = \int_S\left(\Phi(\vec{x}')\frac{\partial}{\partial n'}\frac{1}{R}-\frac{1}{R}\frac{\partial\Phi}{\partial n'}\right)\:da'$$. Thus, $$\Phi(\vec{x}) = \frac{1}{4\pi\varepsilon_0}\int_V\frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}\:d^3x' + \int_S\cdots da'$$ Note: We integrate over the enclosed volume and we must know both BC in this formulation.

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

#### Uniqueness Theorem

1. $$-\nabla^2\Phi=-\frac{\rho}{\varepsilon_0}$$ and either of the Dirchlet or Neumann boundary conditions are satisfied, then $$\Phi$$ is unique.

Proof of 1. Assume we have $$\Phi_1,\Phi_2$$ which satisfy the equation and the BC. Then, let $$U=\Phi_2-\Phi_1$$. So, $$\nabla^2U=0$$, 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, $$\int_V(U\nabla^2+\nabla U\cdot\nabla U)dV = \int_S U\left(\frac{\partial U}{\partial n}\right)d a = 0$$. So, $$\int|\nabla U|^2 dV=0$$. 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})

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

Types:

1. Dirchlet: $$G_D=0$$ on surface
2. Neumann: $$G_N(\vec{x},\vec{x}')$$ $$\frac{\partial G_N}{\partial n}=-\frac{4\pi}{A_S}$$.

$$\nabla'^2G_N=-4\pi\delta(\vec{x}-\vec{x}'), \int_V\nabla'(\nabla G_N)dV = -4\pi = \int_S(\nabla' G_N)\hat{n}dS$$. So, $$\frac{\partial G_N}{\partial n}=C$$ such that $$C$$ times the area of the surface is equal to $$-4\pi$$.

$$\Phi(\vec{x})=\frac{1}{4\pi\varepsilon_0}\int\rho(\vec{x}')G(\vec{x},\vec{x}')d^3x' + \frac{1}{4\pi}\int_S\left[G(\vec{x},\vec{x}')\frac{\partial\Phi}{\partial n}-\Phi(\vec{x}')\frac{\partial G}{\partial n'}\right]$$

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. $$\Phi(\vec{x})=\frac{1}{4\pi\varepsilon_0}\int_V\rho(\vec{x}')G_D\:dV-\frac{1}{4\pi}\int_S\Phi(\vec{x}')\frac{\partial G_D}{\partial n'}\:da'$$.

Similarly, if we have Neumann BC, then $$\partial_n G_N=\frac{-4\pi}{A_S}$$ implies …

## Energy

$$W = \sum_{i=1}^{N}W_i = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^N\sum_{j=1}^{i-1}\frac{q_iq_j}{x_{ij}}$$

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 $$W=\frac{1}{2}\sum_{i=1}^N\sum_{j\neq i}^NW_{ij}=\frac{1}{8\pi\varepsilon_0}\sum_{i=1}^N\sum_{j\neq i}^N\frac{q_iq_j}{|\vec{x}_{ij}|}$$

For a continuous charge distribution, we then get, $$W = \frac{1}{2}\frac{1}{4\pi\varepsilon_0}\iint\frac{\rho(\vec{x})\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}\:d^3xd^3x' = \frac{1}{2}\int\frac{1}{4\pi\varepsilon_0}\rho(\vec{x})\frac{\rho(\vec{x}')}{|\vec{x}-\vec{x}'|}\:d^3xd^3x' = \frac{1}{2}\int\rho(\vec{x})\Phi(\vec{x})\:dV = -\frac{\varepsilon_0}{2}\int\Phi\nabla^2\Phi\:dV$$

$$\nabla\cdot(\Phi\nabla\Phi)=\Phi\nabla^2\Phi+\nabla\Phi\cdot\nabla\Phi=\Phi\nabla^2\Phi+|\vec{E}|^2$$ Using a vector identity, $$W = -\frac{\varepsilon_0}{2}\int\Phi\nabla^2\Phi\:dV = \frac{\varepsilon_0}{2}\int|\vec{E}|^2\:dV-\frac{\varepsilon_0}{2}\int(\Phi\nabla\Phi)\cdot\hat{n}da = \frac{\varepsilon_0}{2}\int|\vec{E}|^2\:dV\Rightarrow w=\frac{1}{2}\varepsilon_0|\vec{E}|^2$$. 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).

Created: 2023-06-25 Sun 02:28

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