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?

  • 1
    $\begingroup$ Where did you get stuck trying to prove this? $\endgroup$
    – WillO
    Feb 8, 2014 at 14:54
  • $\begingroup$ See my addition to post $\endgroup$ Feb 8, 2014 at 15:20
  • $\begingroup$ The $C$ symmetry is not defined in a classical (non quantum) context. Therefore it is pointless to ask if Maxwell's equations are invariant under an unexisting symmetry. You may want to change your question, and ask yourself is the QED lagrangian is invariant under CPT. And the answer is yes. $\endgroup$ Feb 8, 2014 at 15:41
  • 1
    $\begingroup$ Don't forget about vectors versus pseudovectors, when you're doing the P transformation. $\endgroup$ Feb 8, 2014 at 18:08
  • $\begingroup$ Is the QED lagrangian invariant under C,P and T separately? $\endgroup$ Feb 8, 2014 at 19:00

1 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}

  • $\begingroup$ Do you have any text in which the CPT symmetry appears for the Maxwell's equations? $\endgroup$ Apr 7, 2019 at 19:20

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