46.1 Maxwell’s equations

46.1 Maxwell’s equations

We saw in chapter 45 that our quantum theories of free particles could be coupled to a background electromagnetic field by introducing vector potential fields $A_{\mu }=(A_0,\mathbf{A})$ . Electric and magnetic fields are defined in terms of $A_{\mu }$ by the equations

$$\mathbf{E}=-\frac{\partial \mathbf{A}}{\partial t}+\nabla A_0,\mathbf{B}=\nabla \times \mathbf{A}$$

In this chapter we will see how to make the $A_{\mu }$ fields dynamical variables, although restricting to the special case of free electromagnetic fields, without interaction with matter fields. The equations of motion will be:

Definition (Maxwell’s equations in vacuo)

The Maxwell equations for electromagnetic fields in the vacuum are

$$\nabla \cdot \mathbf{B}=0\text{\hspace{1em}(46.1)}$$

$$\nabla \times \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}\text{\hspace{1em}(46.2)}$$

$$\nabla \times \mathbf{B}=\frac{\partial \mathbf{E}}{\partial t}\text{\hspace{1em}(46.3)}$$

and Gauss’s law:

$$\nabla \cdot \mathbf{E}=0\text{\hspace{1em}(46.4)}$$

Digression. In terms of diferential forms these equations can be written very simply as

$$dF=0,d\ast F=0$$

where the first equation is equivalent to $46.1$ and $46.2,$ the second (which uses the Hodge star operator for the Minkowski metric) is equivalent to $46.3$ and $46.4$ . Note that the first equation is automatically satisfied, since by definition $F=dA$ , and the $d$ operator satisfies $d^2=0$

Writing out equation 46.1 in terms of the vector potential gives

$$\nabla \cdot \nabla \times \mathbf{A}=0$$

which is automatically satisfied for any vector field $A$. Similarly, in terms of the vector potential, equation 46.2 is

$$\nabla \times \left(-\frac{\partial \mathbf{A}}{\partial t}+\nabla A_0\right)=-\frac{\partial }{\partial t}(\nabla \times \mathbf{A})$$

which is automatically satisfied since

$$\nabla \times \nabla f=0$$

for any function $f.$

Note that since Maxwell’s equations only depend on $A_{\mu }$ through the gauge invariant fields $B$ and E, if $A_{\mu }=\left(A_0,\mathbf{A}\right)$ is a solution, so is the gauge transform

$$A_{\mu }^{\phi }=\left(A_0+\frac{\partial \phi }{\partial t},\mathbf{A}+\nabla \phi \right)$$

Chapter contents · Next section: 46.2 The Hamiltonian formalism for electromagnetic fields · 中文版本

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