Categories: Electromagnetism, Physics.

Maxwell’s equations

In physics, Maxwell’s equations govern all macroscopic electromagnetism, and notably lead to the electromagnetic wave equation, which describes the existence of light.

Gauss’ law

Gauss’ law states that the electric flux $$\Phi_E$$ through a closed surface $$S(V)$$ is equal to the total charge $$Q$$ contained in the enclosed volume $$V$$, divided by the vacuum permittivity $$\varepsilon_0$$:

\begin{aligned} \Phi_E = \oint_{S(V)} \vb{E} \cdot \dd{\vb{A}} = \frac{1}{\varepsilon_0} \int_{V} \rho \dd{V} = \frac{Q}{\varepsilon_0} \end{aligned}

Where $$\vb{E}$$ is the electric field, and $$\rho$$ is the charge density in $$V$$. Gauss’ law is usually more useful when written in its vector form, which can be found by applying the divergence theorem to the surface integral above. It states that the divergence of $$\vb{E}$$ is proportional to $$\rho$$:

\begin{aligned} \boxed{ \nabla \cdot \vb{E} = \frac{\rho}{\varepsilon_0} } \end{aligned}

This law can just as well be expressed for the displacement field $$\vb{D}$$ and polarization density $$\vb{P}$$. We insert $$\vb{E} = (\vb{D} - \vb{P}) / \varepsilon_0$$ into Gauss’ law for $$\vb{E}$$, multiplied by $$\varepsilon_0$$:

\begin{aligned} \rho = \nabla \cdot \big( \vb{D} - \vb{P} \big) = \nabla \cdot \vb{D} - \nabla \cdot \vb{P} \end{aligned}

To proceed, we split the net charge density $$\rho$$ into a “free” part $$\rho_\mathrm{free}$$ and a “bound” part $$\rho_\mathrm{bound}$$, respectively corresponding to $$\vb{D}$$ and $$\vb{P}$$, such that $$\rho = \rho_\mathrm{free} + \rho_\mathrm{bound}$$. This yields:

\begin{aligned} \boxed{ \nabla \cdot \vb{D} = \rho_{\mathrm{free}} } \qquad \quad \boxed{ \nabla \cdot \vb{P} = - \rho_{\mathrm{bound}} } \end{aligned}

By integrating over an arbitrary volume $$V$$ we can get integral forms of these equations:

\begin{aligned} \Phi_D &= \oint_{S(V)} \vb{D} \cdot \dd{\vb{A}} = \int_{V} \rho_{\mathrm{free}} \dd{V} = Q_{\mathrm{free}} \\ \Phi_P &= \oint_{S(V)} \vb{P} \cdot \dd{\vb{A}} = - \int_{V} \rho_{\mathrm{bound}} \dd{V} = - Q_{\mathrm{bound}} \end{aligned}

Gauss’ law for magnetism

Gauss’ law for magnetism states that magnetic flux $$\Phi_B$$ through a closed surface $$S(V)$$ is zero. In other words, all magnetic field lines entering the volume $$V$$ must leave it too:

\begin{aligned} \Phi_B = \oint_{S(V)} \vb{B} \cdot \dd{\vb{A}} = 0 \end{aligned}

Where $$\vb{B}$$ is the magnetic field. Thanks to the divergence theorem, this can equivalently be stated in vector form as follows:

\begin{aligned} \boxed{ \nabla \cdot \vb{B} = 0 } \end{aligned}

A consequence of this law is the fact that magnetic monopoles cannot exist, i.e. there is no such thing as “magnetic charge”, in contrast to electric charge.

Faraday’s law of induction states that a magnetic field $$\vb{B}$$ that changes with time will induce an electric field $$E$$. Specifically, the change in magnetic flux through a non-closed surface $$S$$ creates an electromotive force around the contour $$C(S)$$. This is written as:

\begin{aligned} \oint_{C(S)} \vb{E} \cdot \dd{\vb{l}} = - \dv{t} \int_{S} \vb{B} \cdot \dd{\vb{A}} \end{aligned}

By using Stokes’ theorem on the contour integral, the vector form of this law is found to be:

\begin{aligned} \boxed{ \nabla \times \vb{E} = - \pdv{\vb{B}}{t} } \end{aligned}

Ampère’s circuital law

Ampère’s circuital law, with Maxwell’s correction, states that a magnetic field $$\vb{B}$$ can be induced along a contour $$C(S)$$ by two things: a current density $$\vb{J}$$ through the enclosed surface $$S$$, and a change of the electric field flux $$\Phi_E$$ through $$S$$:

\begin{aligned} \oint_{C(S)} \vb{B} \cdot d\vb{l} = \mu_0 \Big( \int_S \vb{J} \cdot d\vb{A} + \varepsilon_0 \dv{t} \int_S \vb{E} \cdot d\vb{A} \Big) \end{aligned} \begin{aligned} \boxed{ \nabla \times \vb{B} = \mu_0 \Big( \vb{J} + \varepsilon_0 \pdv{\vb{E}}{t} \Big) } \end{aligned}

Where $$\mu_0$$ is the vacuum permeability. This relation also exists for the “bound” fields $$\vb{H}$$ and $$\vb{D}$$, and for $$\vb{M}$$ and $$\vb{P}$$. We insert $$\vb{B} = \mu_0 (\vb{H} + \vb{M})$$ and $$\vb{E} = (\vb{D} - \vb{P})/\varepsilon_0$$ into Ampère’s law, after dividing it by $$\mu_0$$ for simplicity:

\begin{aligned} \nabla \cross \big( \vb{H} + \vb{M} \big) &= \vb{J} + \pdv{t} \big( \vb{D} - \vb{P} \big) \end{aligned}

To proceed, we split the net current density $$\vb{J}$$ into a “free” part $$\vb{J}_\mathrm{free}$$ and a “bound” part $$\vb{J}_\mathrm{bound}$$, such that $$\vb{J} = \vb{J}_\mathrm{free} + \vb{J}_\mathrm{bound}$$. This leads us to:

\begin{aligned} \boxed{ \nabla \times \vb{H} = \vb{J}_{\mathrm{free}} + \pdv{\vb{D}}{t} } \qquad \quad \boxed{ \nabla \times \vb{M} = \vb{J}_{\mathrm{bound}} - \pdv{\vb{P}}{t} } \end{aligned}

By integrating over an arbitrary surface $$S$$ we can get integral forms of these equations:

\begin{aligned} \oint_{C(S)} \vb{H} \cdot d\vb{l} &= \int_S \vb{J}_{\mathrm{free}} \cdot \dd{\vb{A}} + \dv{t} \int_S \vb{D} \cdot \dd{\vb{A}} \\ \oint_{C(S)} \vb{M} \cdot d\vb{l} &= \int_S \vb{J}_{\mathrm{bound}} \cdot \dd{\vb{A}} - \dv{t} \int_S \vb{P} \cdot \dd{\vb{A}} \end{aligned}

Note that $$\vb{J}_\mathrm{bound}$$ can be split into the magnetization current density $$\vb{J}_M = \nabla \cross \vb{M}$$ and the polarization current density $$\vb{J}_P = \pdv*{\vb{P}}{t}$$:

\begin{aligned} \vb{J}_\mathrm{bound} = \vb{J}_M + \vb{J}_P = \nabla \cross \vb{M} + \pdv{\vb{P}}{t} \end{aligned}

Redundancy of Gauss’ laws

In fact, both of Gauss’ laws are redundant, because they are already implied by Faraday’s and Ampère’s laws. Suppose we take the divergence of Faraday’s law:

\begin{aligned} 0 = \nabla \cdot \nabla \cross \vb{E} = - \nabla \cdot \pdv{\vb{B}}{t} = - \pdv{t} (\nabla \cdot \vb{B}) \end{aligned}

Since the divergence of a curl is always zero, the right-hand side must vanish. We know that $$\vb{B}$$ can vary in time, so our only option to satisfy this is to demand that $$\nabla \cdot \vb{B} = 0$$. We thus arrive arrive at Gauss’ law for magnetism from Faraday’s law.

The same technique works for Ampère’s law. Taking its divergence gives us:

\begin{aligned} 0 = \frac{1}{\mu_0} \nabla \cdot \nabla \cross \vb{B} = \nabla \cdot \vb{J} + \varepsilon_0 \pdv{t} (\nabla \cdot \vb{E}) \end{aligned}

We integrate this over an arbitrary volume $$V$$, and apply the divergence theorem:

\begin{aligned} 0 &= \int_V \nabla \cdot \vb{J} \dd{V} + \pdv{t} \int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V} \\ &= \oint_S \vb{J} \cdot \dd{S} + \pdv{t} \int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V} \end{aligned}

The first integral represents the current (charge flux) through the surface of $$V$$. Electric charge is not created or destroyed, so the second integral must be the total charge in $$V$$:

\begin{aligned} Q = \int_V \varepsilon_0 \nabla \cdot \vb{E} \dd{V} \quad \implies \quad \nabla \cdot \vb{E} = \frac{\rho}{\varepsilon_0} \end{aligned}

And we thus arrive at Gauss’ law from Ampère’s law and charge conservation.

© Marcus R.A. Newman, a.k.a. "Prefetch". Available under CC BY-SA 4.0.