The problem I stumbled upon is the same as in this question.

Namely, I tried to find transformations of $\vec{E}$ and $\vec{B}$ under parity and time reversal by transforming the field tensor $F^{\mu\nu}$. These are Lorentz transformations given by $T=\mathrm{diag}[-1,1,1,1]$ and $P=\mathrm{diag}[1,-1,-1,-1]$. After applying the transformation rule of the tensor: $$F^{'\mu\nu}=\Lambda^\mu_{\ \ \rho}\Lambda^\nu_{\ \ \sigma} F^{\rho\sigma}$$ I get the same result for both parity and time reversal: $$\vec{E'} = - \vec{E}$$ $$\vec{B'} = \vec{B}$$

I am not satisfied by the answer given in the linked post because first I don't understand the language of differential forms and second I would like to make things clear by getting answer to a few simple questions.

  • Am I being lied to when I'm said that the $F^{\mu\nu}$ tensor transforms under L.t. the way I described above? If not my guess is then that even though the tensor does transform like described, the interpretation of its components does not stay the same as before the reversal. And thus I'm led to a false conclusion in my analysis because $F^{0i} = -\frac1c E_i$ but $F^{'0i} \not = -\frac1c E_i$. Is this reasoning correct?

  • Is it still "legal" to prove the invariability of Maxwell's equations under these transformations by simply showing that $\partial'_\mu F^{'\mu\nu}=\mu_0 J'^\nu$ and $\epsilon^{'\mu\nu\rho\sigma} \partial'_\nu F'_{\rho\sigma} = 0$ still hold? I would assume it is enough if the above reasoning is sound. I.e. if tensors really transform as I think they do, then the change of meaning of the components wouldn't mind, because here we are dealing with the relations among whole tensors, not just some of their components.

  • $\begingroup$ 'Am I being lied to ' This is not a scientific statement. Consider a reformulation. $\endgroup$
    – my2cts
    Apr 19, 2022 at 20:06

2 Answers 2


As Ulrich Jentschura pointed out, if you try to apply this logic to the 4-potential, you already get the wrong answer. Applying your $T$ matrix to the 4-potential will flip the sign of the scalar potential, but this isn't physically correct; the scalar potential is T-even.

So what's wrong with your approach, exactly?

Well, you're assuming that $F$ transforms a certain way, and then you're noticing that it doesn't give you the correct answer. So the only possibility is that the assumption was wrong.

In general, when a physical theory has P and T symmetry, then each quantity in that theory may be either even or odd under each operation. Your idea that all tensors transform under P and T according to your $P$ and $T$ matrices is equivalent to the following hypothesis:

  • Hypothesis: Assuming that symmetry under the full Lorentz group holds, a component of a four-tensor is P-odd iff it has an odd number of spatial indices. A component of a four-tensor is T-odd iff it has an odd number of temporal indices.

The problem is that this hypothesis isn't true. Actually, for the electromagnetic field tensor $F$, it's true for P, but it's false for T.

However, a weaker statement is true:

  • Definition: A polar tensor is a tensor that satisfies the Hypothesis with respect to P: that is, its components with an odd number of spatial indices are P-odd while the others are P-even.

  • Definition: A pseudotensor is the opposite of a polar tensor: its components with an odd number of spatial indices are P-even while the others are P-odd.

  • Theorem: Assuming that symmetry under the full Lorentz group holds, every four-tensor is either a polar tensor or a pseudotensor. Furthermore, the analogous statement holds with respect to T: every four-tensor either satisfies the Hypothesis with respect to T, or the opposite. I will refer to these as "T-polar" and "T-pseudo" since I'm not aware of any better terminology.

  • Example: There is no tensor of order 2 that can have 16 nonzero components that are all T-even. The reason is that such objects would lead to contradictions when contracted against well-behaved four-vectors like $u$ (i.e., ones that do satisfy the theorem): the resulting scalar would be neither T-even nor T-odd; its magnitude would change under T. This contradicts the assumption of T-symmetry.

How do we know whether $F$ is a polar tensor or a pseudotensor? Well, basically, just use the theorem. $F_{01}$ for example is equal to $\partial_0 A_1 - \partial_1 A_0$. Under P, the signs of $A_1$ and $\partial_1$ change, while those of $\partial_0$ and $A_0$ do not, so the result is that $F_{01}$ does change sign. It has an odd number of spatial components. Therefore, $F$ is a polar tensor. Under T, the signs of $\partial_0$ and $A_1$ change, while those of $\partial_1$ and $A_0$ do not, so $F_{01}$ does not change sign. This tells us that $F$ is a T-pseudotensor.

Here, we relied on the assumption that $A$ is a polar vector and a T-pseudovector. How do we know that's true? It's because $A$ is proportional to $J$, and $J$ is a polar vector and a T-pseudovector. How do we know that's true? It follows from the definition of $J$ and the fact that electric charge is invariant under both P and T. How do we know that's true? It's true by definition. We define P and T such that they don't change the electric charge of any particle.

In general, to determine whether a quantity is polar or pseudo, and to determine whether it's T-polar or T-pseudo, if it's a fundamental quantity, we choose an appropriate definition, and otherwise, it follows from the relationship between the quantity of interest and the fundamental quantities, as determined by the laws of nature.

  • $\begingroup$ Just to make sure I understood, it turns out that when textbooks speak about tensor transformations under Lorentz tr. what they really mean is under proper orthochronous Lorentz transformations. In this special case I can apply the transformation rule from my post without problems. If I include $P$ transf. I get tensors and pseudotensor which I knew of before, they transform the same way up to $det(\Lambda)$ jumping out in front of pseudotensors. But if I add $T$, then I get to define "T-pseudo" tensors and the problem here is that I wasn't aware that $F$ was a T-pseudotensor $\endgroup$
    – lojle
    Apr 19, 2022 at 20:44
  • $\begingroup$ @lojle Yes, in many cases "Lorentz transformation" means "proper orthochronous Lorentz transformation", and in any case, the tensor quantities that appear in "Lorentz covariant" equations might be polar tensors or pseudotensors; we use the same notation for both because they transform the same way under the proper orthochronous Lorentz group. $\endgroup$
    – Brian Bi
    Apr 19, 2022 at 20:49
  • $\begingroup$ Sorry to bother you but could you just check my if these transf. rules are okay. For P-tensor, T-pseudotensor $A$: $$A'^\mu=\Lambda^0_0 \Lambda^\mu_{\: \nu}A^\nu$$ And for P-pseudotensor, T-tensor $B$: $$B'^\mu=\Lambda^0_0 det(\Lambda) \Lambda^\mu_{\: \nu}B^\nu$$ $\endgroup$
    – lojle
    Apr 19, 2022 at 21:11
  • $\begingroup$ @lojle I'd have to think about that carefully. It should really be posted as a separate question. $\endgroup$
    – Brian Bi
    Apr 19, 2022 at 21:56

I subscribe to this Wikipedia article. The transformation laws that you find are correct. Indeed $\vec E$ and $\vec B$ are not really 3D vectors, as they are components of $F^{\mu\nu}$ and antisymmetric Lorentz tensor of rank 2. $A^0,\vec A$ is a four vector. $A^0$ changes sign under time reversal, while $\vec A$ changes sign under parity reversal. I don't see how the covariant formulation of electrodynamics can work without these facts.

However, another Wikipedia article makes quite opposite claims.

The difference is that in the second approach time reversal is not considered to imply charge inversion.

  • $\begingroup$ @my2acts E and B are 3D vectors else how can you cross product B with u to find the Lorentz force for a charged particle inside a magnetic field? $\endgroup$
    – Miss Mulan
    Apr 19, 2022 at 21:30

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