Maxwell equations and symmetry 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?
 A: 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}
