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

Question

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.

Answer 1

The transformation under time reversal of the forms in electrodynamics 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.

This highlights a general and important fact: The Hodge star does not commute with coordinate transformations that change the handedness of the underlying coordinate system, since its definition crucially relies on the ordering and handedness of the vectors in the system. Therefore, as soon as we consider transformations whose determinant is negative (since that is the abstract sign of changing handedness), care must be taken for all geometric objects whether the correct physical interpretation is to have the transformation act on them or their duals.

Answer 2

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.

Answer 3

In order to answer your first question, one may observe that the scalar potential actually is invariant under T (why would such a static potential, generated by a point charge, otherwise flip under T; it does not), while the vector potential A flips sign under T. This is a variance with the application of your "Lorentz T transformation" to what would otherwise be the four-vector potential (Phi, A). So, in order to apply T as a matrix multiplication to your F tensor, you transform one Lorentz index with Tmunu, the other with Pmunu, giving you consistency with the arguments presented to you in your second line of reasoning.

Answer 4

For the parity of B, the arguments can be confusing because there are at least two ways of thinking about the question of mirror symmetry and parity. I like the following.

Suppose that you observe an experiment in a mirror, without knowing that is the case. The mirror experiment is called "the magnetic force on a moving charge". The velocity is parallel to the mirror and so unchanged from what it actually was. The force is into the mirror, and so you see it flipped from what it was. You are asked to deduce the direction if the magnetic field from this data, and so you conclude that the field points opposite to what it actually did. In other words, V,B,F form a triple. You assume it is right-handed, VXB=F, and so if F flipped but V did not, then B flipped.

On the other hand (haha) you could watch a film of a "mirror physicist" perform the experiment. Being mirrored, they would use a LEFT hand rule and so they would conclude that B is the same as it was in the actual lab. Using a left hand rule is equivalent to reversing the numbering sense of all three spatial axes, so that -i"X"-j=-k. In a parity transformation a right handed set of axes becomes a left handed set, V becomes -V and F becomes -F; but also "right" becomes "left" and so "X" picks up a "-" sign. Thus "VXB=F" in mirror-world is -V(-X)-B=-F which is just VXB=F in the actual lab.