Maxwell's Equations

Step 1: Gauss's Law — from Coulomb's law and superposition

$$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\mathbf{r}')(\mathbf{r}-\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|^3}d\tau'$$

Taking the divergence and using $\nabla \cdot \left(\frac{\hat{\mathbf{r}}}{r^2}\right) = 4\pi\delta^3(\mathbf{r})$:

$$\nabla \cdot \mathbf{E} = \frac{1}{\epsilon_0}\int \rho(\mathbf{r}')\delta^3(\mathbf{r}-\mathbf{r}')d\tau' = \frac{\rho}{\epsilon_0}$$

Step 2: No Magnetic Monopoles — from the Biot-Savart law

$$\mathbf{B} = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J} \times \hat{\boldsymbol{\mathscr{r}}}}{\mathscr{r}^2}d\tau' = \nabla \times \mathbf{A}$$

Since B is the curl of A, its divergence vanishes identically:

$$\nabla \cdot \mathbf{B} = \nabla \cdot (\nabla \times \mathbf{A}) = 0$$

Step 3: Faraday's Law — from experimental observation (see Chapter 8)

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$

Derived from $\oint \mathbf{E} \cdot d\boldsymbol{\ell} = -d\Phi_B/dt$ via Stokes' theorem.

Step 4: Ampere-Maxwell Law — Ampere's law plus displacement current

$$\nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$$

The displacement current term is required for mathematical consistency (see next derivation).

Step 5: Summary — the four equations in vacuum

$$\boxed{\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon_0}, \quad \nabla \cdot \mathbf{B} = 0, \quad \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}, \quad \nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}}$$

Step 1: Take the divergence of Ampere's law (pre-Maxwell)

$$\nabla \cdot (\nabla \times \mathbf{B}) = \mu_0 \nabla \cdot \mathbf{J}$$

Step 2: The left side is identically zero (divergence of a curl)

$$0 = \mu_0 \nabla \cdot \mathbf{J}$$

This implies $\nabla \cdot \mathbf{J} = 0$, but the continuity equation requires $\nabla \cdot \mathbf{J} = -\partial\rho/\partial t$. Contradiction!

Step 3: Use Gauss's law to express ∂ρ/∂t in terms of E

$$\frac{\partial\rho}{\partial t} = \epsilon_0 \frac{\partial}{\partial t}(\nabla \cdot \mathbf{E}) = \epsilon_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t}$$

Step 4: The continuity equation becomes

$$\nabla \cdot \mathbf{J} + \epsilon_0 \nabla \cdot \frac{\partial \mathbf{E}}{\partial t} = 0 \implies \nabla \cdot \left(\mathbf{J} + \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\right) = 0$$

Step 5: Replace J with J + J_d in Ampere's law to restore consistency

$$\nabla \times \mathbf{B} = \mu_0\left(\mathbf{J} + \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\right)$$

Step 6: Verify — take the divergence of the corrected equation

$$\nabla \cdot (\nabla \times \mathbf{B}) = \mu_0\nabla \cdot \mathbf{J} + \mu_0\epsilon_0\nabla \cdot \frac{\partial \mathbf{E}}{\partial t} = \mu_0\left(-\frac{\partial\rho}{\partial t} + \frac{\partial\rho}{\partial t}\right) = 0 \;\checkmark$$

Step 7: Define the displacement current density

$$\boxed{\mathbf{J}_d = \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}}$$

This is not a real current of moving charges — it is a time-varying electric field that acts as a source of magnetic field.

Step 1: Take the divergence of the Ampere-Maxwell law

$$\nabla \cdot (\nabla \times \mathbf{B}) = \mu_0 \nabla \cdot \mathbf{J} + \mu_0\epsilon_0 \frac{\partial}{\partial t}(\nabla \cdot \mathbf{E})$$

Step 2: The left side vanishes identically

$$0 = \mu_0 \nabla \cdot \mathbf{J} + \mu_0\epsilon_0 \frac{\partial}{\partial t}(\nabla \cdot \mathbf{E})$$

Step 3: Apply Gauss's law: ∇·E = ρ/ε₀

$$0 = \mu_0 \nabla \cdot \mathbf{J} + \mu_0\epsilon_0 \frac{\partial}{\partial t}\left(\frac{\rho}{\epsilon_0}\right)$$

Step 4: Simplify (the ε₀ cancels)

$$0 = \mu_0\left(\nabla \cdot \mathbf{J} + \frac{\partial\rho}{\partial t}\right)$$

Step 5: Since μ₀ ≠ 0, we obtain the continuity equation

$$\boxed{\frac{\partial\rho}{\partial t} + \nabla \cdot \mathbf{J} = 0}$$

Charge conservation is not an independent assumption — it is built into Maxwell's equations automatically.

Step 1: Decompose charge and current into free and bound contributions

$$\rho = \rho_f + \rho_b, \quad \mathbf{J} = \mathbf{J}_f + \mathbf{J}_b + \mathbf{J}_p$$

where $\rho_b = -\nabla \cdot \mathbf{P}$, $\mathbf{J}_b = \nabla \times \mathbf{M}$, $\mathbf{J}_p = \partial\mathbf{P}/\partial t$

Step 2: Substitute into Gauss's law

$$\nabla \cdot \mathbf{E} = \frac{1}{\epsilon_0}(\rho_f - \nabla \cdot \mathbf{P}) \implies \nabla \cdot (\epsilon_0\mathbf{E} + \mathbf{P}) = \rho_f$$

Step 3: Define the electric displacement field D

$$\mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P} \implies \nabla \cdot \mathbf{D} = \rho_f$$

Step 4: Substitute bound current into the Ampere-Maxwell law

$$\nabla \times \mathbf{B} = \mu_0\left(\mathbf{J}_f + \nabla \times \mathbf{M} + \frac{\partial\mathbf{P}}{\partial t}\right) + \mu_0\epsilon_0\frac{\partial\mathbf{E}}{\partial t}$$

Step 5: Rearrange, grouping ∂D/∂t = ε₀∂E/∂t + ∂P/∂t

$$\nabla \times \left(\frac{\mathbf{B}}{\mu_0} - \mathbf{M}\right) = \mathbf{J}_f + \frac{\partial\mathbf{D}}{\partial t}$$

Step 6: Define the auxiliary magnetic field H

$$\mathbf{H} = \frac{\mathbf{B}}{\mu_0} - \mathbf{M} \implies \nabla \times \mathbf{H} = \mathbf{J}_f + \frac{\partial\mathbf{D}}{\partial t}$$

Step 7: The complete Maxwell equations in matter

$$\boxed{\nabla \cdot \mathbf{D} = \rho_f, \quad \nabla \cdot \mathbf{B} = 0, \quad \nabla \times \mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t}, \quad \nabla \times \mathbf{H} = \mathbf{J}_f + \frac{\partial\mathbf{D}}{\partial t}}$$

Only free charges and currents appear as sources — bound sources are absorbed into D and H.

Step 1: Write down Faraday's law and the Ampere-Maxwell law in vacuum

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}, \qquad \nabla \times \mathbf{B} = \mu_0\epsilon_0\frac{\partial \mathbf{E}}{\partial t}$$

Step 2: Take the curl of Faraday's law

$$\nabla \times (\nabla \times \mathbf{E}) = -\frac{\partial}{\partial t}(\nabla \times \mathbf{B})$$

Step 3: Substitute the Ampere-Maxwell law for ∇×B on the right

$$\nabla \times (\nabla \times \mathbf{E}) = -\mu_0\epsilon_0\frac{\partial^2 \mathbf{E}}{\partial t^2}$$

Step 4: Apply the vector identity ∇×(∇×E) = ∇(∇·E) - ∇²E

$$\nabla(\nabla \cdot \mathbf{E}) - \nabla^2\mathbf{E} = -\mu_0\epsilon_0\frac{\partial^2 \mathbf{E}}{\partial t^2}$$

Step 5: In vacuum, ∇·E = 0 (Gauss's law with ρ = 0), so the first term vanishes

$$-\nabla^2\mathbf{E} = -\mu_0\epsilon_0\frac{\partial^2 \mathbf{E}}{\partial t^2}$$

Step 6: The electromagnetic wave equation for E

$$\boxed{\nabla^2\mathbf{E} = \mu_0\epsilon_0\frac{\partial^2 \mathbf{E}}{\partial t^2}}$$

Step 7: Repeat the procedure (take curl of Ampere-Maxwell, substitute Faraday) to get the identical equation for B

$$\nabla^2\mathbf{B} = \mu_0\epsilon_0\frac{\partial^2 \mathbf{B}}{\partial t^2}$$

Both E and B satisfy the same wave equation, propagating at speed $v = 1/\sqrt{\mu_0\epsilon_0}$.

Step 1: The general wave equation has the form

$$\nabla^2 f = \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2}$$

where $v$ is the propagation speed.

Step 2: Comparing with the EM wave equation, identify the speed

$$\frac{1}{v^2} = \mu_0\epsilon_0 \implies v = \frac{1}{\sqrt{\mu_0\epsilon_0}}$$

Step 3: Substitute the SI values of the fundamental constants

$$\mu_0 = 4\pi \times 10^{-7}\;\text{T·m/A}, \qquad \epsilon_0 = 8.854 \times 10^{-12}\;\text{C}^2/(\text{N·m}^2)$$

Step 4: Compute the product μ₀ε₀

$$\mu_0\epsilon_0 = (4\pi \times 10^{-7})(8.854 \times 10^{-12}) = 1.113 \times 10^{-17}\;\text{s}^2/\text{m}^2$$

Step 5: Take the square root of the reciprocal

$$v = \frac{1}{\sqrt{1.113 \times 10^{-17}}} = \frac{1}{1.054 \times 10^{-8.5}} \approx 2.998 \times 10^8\;\text{m/s}$$

Step 6: This equals the measured speed of light!

$$\boxed{c = \frac{1}{\sqrt{\mu_0\epsilon_0}} = 2.998 \times 10^8\;\text{m/s}}$$

Maxwell concluded: "light is an electromagnetic disturbance propagated through the field according to electromagnetic laws." This was the great unification of optics and electromagnetism.