# Behavior of the Electric- and Magnetic-field under time reversal and parity

The behavior of the electric- $\mathbf{E}$ and the magnetic-field $\mathbf{B}$ und time reversal and parity can be calculated in different ways.

My first solution is to study the transformation behavior of the Field-strength-tensor $F^{\mu\nu}$ when acting on it with the Lorentz-transformation for time reversal $$(\mathcal{T}_{\;\;\;\nu}^{\mu})=\text{diag}(-1,1,1,1)$$ and similarly for partiy $$(\mathcal{P}_{\;\;\nu}^{\mu})=\text{diag}(1,-1,-1,-1)$$. The result is that for both time reversal and parity $\mathbf{E}$ and $\mathbf{B}$ behave as: \begin{align}\mathbf{E'} &= -\mathbf{E}\\ \mathbf{B'} &= \mathbf{B} \end{align}

On the other hand, if one follows the argumentation of Jackson and demands the invariance of the e.o.m., mathematically: \begin{align} \mathcal{T}&: \quad \mathbf{x'} = \mathbf{x} \quad \text{and} \quad t' = -t\\ \mathcal{P}&: \quad \mathbf{x'} = -\mathbf{x} \quad \text{and} \quad t' = t \end{align} for the transformations and $$m_0 \gamma' \frac{\mathrm{d}u'^\mu}{\mathrm{d}t'} \stackrel{!}{=} \frac{q}{c} F'^{\mu\nu}u'_\nu$$ for the e.o.m. The equation above implies a different transformation behavior for the field-strength-tensor with the result: \begin{align} \mathcal{T}&: \quad \mathbf{E'} = \mathbf{E} \quad \text{and} \quad \mathbf{B'} = -\mathbf{B}\\ \mathcal{P}&: \quad \mathbf{E'} = -\mathbf{E} \quad \text{and} \quad \mathbf{B'} = \mathbf{B} \end{align}

My question now: How is this issue/ambiguity resolved or what is my misconception.

• While I don't know the answer, note that Jackson's version is the one that makes sense physically; think of a current loop running backwards in time. – Javier Nov 13 '15 at 23:26

The transformation under time reversal of the forms electrondynamics is subtle because the gauge field 1-form $A = A_\mu \mathrm{d}x^\mu$ and the field strength $F = F_{\mu\nu}\mathrm{d}x^\mu\wedge\mathrm{d}x^\nu$ are not the correct physical objects to transform.
This may be seen by observing that the Maxwell equations are $\mathrm{d}F = 0$ and $\mathrm{d}\star F = \star J$, but the former is just a Bianchi identity following from $\mathrm{d}^2 = 0$. The actual equation of motion for the gauge theory is given in terms of the Hodge duals $\star F$ and $\star J$, and it are thus the Hodge duals whose transformation behaviour dictates the transformation behaviour under time reversal.
In the field strength tensor, we have the terms $E_i \mathrm{d}t\wedge\mathrm{d}x^i$ and $B_i \epsilon^{ijk}\mathrm{d}x^j\wedge\mathrm{d}x^k$ and from this one would indeed conclude that it is the electric field that changes sign under time reversal. However, inspecting the Hodge dual that occurs in the equation of motion, we find the opposite behaviour since the star of the terms with $\mathrm{d}t$ contains no $\mathrm{d}t$ terms anymore and vice versa.
The electric field has to be invariant under time reversal. Enter the Lorentz force $\vec F=q\cdot(\vec E + \vec v \times \vec B)$. The force is $ma$, where $a$ is the acceleration, nonrelativistically. Velocity $\vec v$ flips sign under $T$, while the acceleration remains invariant. So, $\vec E$ has to be invariant under $T$. Likewise, because $\vec v$ flips under $T$, so does the $\vec B$ field. These transformation properties of the $\vec E$ and $\vec B$ fields under $T$ are opposite to what happens under parity.