Behavior of electric and magnetic fields under time reversal and parity

Question

The behavior of the electric field $\mathbf{E}$ and the magnetic field $\mathbf{B}$ under 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 equation of motion (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.

Answer 5

I think ACuriousMind's answer is unfortunately misleading. It has some truth to it, but it is not really the correct answer from a group theoretic standpoint. In particular it is true and important to know that the Hodge star operator anti-commutes with parity inversion. But as Jagerber48 points out, the Hodge star argument cannot ever simultaneously account for the properties of both $\vec{E}$ and $\vec{B}$ under the action of $\mathcal{P}$ and $\mathcal{T}$.

The problem here is that when we begin to care about reflections, we are not merely looking at representations of the proper orthochronus Lorentz group $SO_+^{\uparrow}(1,3)$ as we are so often used to, but of the full Lorentz group $O(1,3)$. The entire Lorentz group has four distinct connected components of group elements, which can be reasonably named the identity component, the parity inverted component, the time reversed component, and the time-parity reversed component. $$ \mathcal{I}, \>\>\mathcal{P},\>\> \mathcal{T},\>\> \mathcal{PT}.$$ While these names are physically meaningful, the fact that all these operators are involutions, and that $\mathcal{P}$ and $\mathcal{T}$ are reflecting exactly opposite compact subspaces means we can equally understand these labels as $$\mathcal{I},\>\> \mathcal{P},\>\> -\mathcal{P},\>\> -\mathcal{I}$$

The elements $\mathcal{I}$ and $-\mathcal{I} = \mathcal{PT}$, being proportional to the identity, will commute with every element of the Lorentz group; the set of those group elements with this property is called the center of the Lorentz group, often denoted $\mathcal{Z}\left[O(1,3)\right] = Z_2$. I.e. the center here is a discrete subgroup of two elements.

It seems obvious to state that the two elements of the center act differently when applied to $4$-vectors. A more surprising fact is that acting upon $F^{\mu\nu}$ with the $\mathcal{PT}$ from our vector representation, is the same as doing nothing at all! (Try it yourself)

This is because the second tensor power of the vector representation of $O(1,3)$ does not inherit a faithful representation of the full Lorentz group, but instead forms a representation of $PO(1,3)$, known as the projective orthogonal group (this is exactly analogous to how a spinor is nontrivially affected under a rotation by $2\pi$, yet vectors are left unaffected, $O$ doubly covers $PO$ in even dimensions). The Projective orthogonal groups are precisely defined as the 'centerless' versions of the orthgonal groups. What this means abstractly is that in $PO(1,3)$ any Lorentz transforms which are related by application of an element of the center, are identified. So out of the four connected components of $O(1,3)$, both $\mathcal{I}$ and $\mathcal{PT}$ become $\mathcal{I}$ in $PO(1,3)$, and both $\mathcal{P}$ and $\mathcal{T} (= -\mathcal{P})$ become $\mathcal{P}$ in $PO(1,3)$. This is precisely what your argument shows, and this is how we know we are in this representation of the Projective Orthogonal group: if an object transforms under $PO(1,3)$, there are only two kinds of Lorentz transforms for it, not four.

But this is bad, because we know $\vec{E}$ and $\vec{B}$ fields have distinct time reversal properties. The reason for the 'error' in our ability to perform this operation is in the assumption that when our representation of the proper orthochronus Lorentz group (the component connected to the identity) is faithful, that so to must be our representation of $O(1,3)$.

There are two ways out of this problem: the more cumbersome one, is the option of realizing the adjoint representation (where $F$ lives) as a subgroup of the vector representation of $O(3,3) \supset O(1,3)$. This explicitly $6$ dimensional representation of the Lorentz group has a non-trivial center, as we desire.

The second option, is to perform a discrete central extension on our 'accidentally' projective Lorentz transforms. I.e. for every Lorentz transform we can apply to $F$: $$\Lambda^T F \Lambda$$ we introduce the possibility of being transformed instead by $$-\left(\Lambda^T F \Lambda\right).$$ Thus when acting on rank two tensors, transforms from $\mathcal{I}$ and $\mathcal{P}$ are given as usual, and the transforms from $\mathcal{T}$ and $\mathcal{PT}$ acquire an additional minus sign. With this upgraded transformation property the second tensor power becomes a faithful representation of the full Lorentz group! Where before we found we could only end up transforming $F$ by elements of the two components $\mathcal{I}$ and $\mathcal{P}$, by the inclusion of the minus sign we have found our way back to Lorentz transforms in the other two discrete connected components $\mathcal{T}=-\mathcal{P}$ and $\mathcal{PT} = -\mathcal{I}$

Thus the answer to your question is that abstractly, for any particular real-representation $r$ we know that, $$\mathcal{T}_r=-\mathcal{P}_r.$$ However under the tensor product of vector representations we have seen $$\mathcal{P}_v \otimes \mathcal{P}_v = \mathcal{P}_F$$ but of course then $$\mathcal{T}_v \otimes \mathcal{T}_v = -\mathcal{P}_v \otimes -\mathcal{P}_v = \mathcal{P}_F$$ and so we find we pass to a representation of the Projective orthogonal group. Then, through a central extension of tagging $\mathcal{T}$ and $\mathcal{PT}$ transformations with an overall negative sign, we find ourselves once again in a faithful representation of the full Lorentz group $O(1,3)$ and regain the ability to apply the time reversal operator:

$$\mathcal{T}_F = -\mathcal{P}_F$$ which when acting on $F$ appropriately reverses the magnetic field and leaves invariant the electric field, as expected.