Maxwell equations and symmetry

Question

Do the full inhomogeneous Maxwell equations obey parity (P) and time reversal (T) symmetry separately or only the full CPT symmetry?

I believe the homogeneous Maxwell equations obey parity and time reversal symmetry separately - is that right? The homogeneous Maxwell equations reduce to a wave equation in which space and time appear as second order derivatives.

Here's my effort at answering my first question.

In the first inhomogeneous equations one has:

$$\mathbf{\nabla \cdot E} = \frac{\rho}{\epsilon_0}$$

If I change the sign of x,y,z, in $\mathbf{\nabla . E}$ then I must change the sign of $\rho$ to balance it.

In the third equation one has:

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

As the sign of x,y,z has changed then the sign of t must change to balance it.

Thus we must have CPT symmetry. This can be checked in the last equation:

$$\mathbf{\nabla \times B} = \mu_0 \mathbf{j} + \frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}$$

The sign of x,y,z in $\mathbf{\nabla \times B}$ has changed. This is balanced by a sign change of t in $\frac{1}{c^2}\frac{\partial \mathbf{E}}{\partial t}$. The sign of $\mathbf{j}$ has changed because the charge has changed sign.

Thus the inhomogeneous Maxwell equations obey CPT symmetry rather than C,P or T alone.

Is this reasoning right?

Answer 1

If you define $C:q\rightarrow -q$, $P:(x,y,z)\rightarrow (-x,-y,-z)$ and $T:t\rightarrow -t$, then all the Maxwell eaquations are invariant under $C$, $P$, $T$ or any combination of them. To see this you just have to notice how the transformations act on sources and coordinates.

The charge conjugation acts non trivially as \begin{align} C\rho&=-\rho,\\ Cj_i&=-j_i,\\ CE_i&=-E_i,\\ CB_i&=-B_i, \end{align} with everything else unchanged. Notice the charge conjugation I defined is not exactly the charge conjugation defined in Quantum Field Theory. You can call it a classical counterpart.

Under parity you just have to be careful when dealing with vectors (such as $\vec E$) and pseudovector (such as $\vec B$). Vectors change components under spatial reflection while pseudovectors do not. The objects that change sign under $P$ are \begin{align} Pj_i&=-j_i,\\ PE_i&=-E_i,\\ P\frac{\partial}{\partial x_i}&=-\frac{\partial}{\partial x_i}. \end{align}

Under time reversal, the velocity gets a minus sign, therefore $\vec B$ changes while $\vec E$ does not. So the non trivial actions of $T$ are \begin{align} Tj_i&=-j_i,\\ TB_i&=-B_i,\\ T\frac{\partial}{\partial t}&=-\frac{\partial}{\partial t}. \end{align}