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The electric and magnetic field components $E_i$ and $B_i$, defined from the electromagnetic field tensor $F_{\mu\nu}$ are $$E_i=cF_{0i},~~B_i=-\frac{1}{2}\epsilon_{ijk}F^{jk}.$$ Since $\epsilon_{ijk}$ is an invariant pseudotensor, it transforms as $\epsilon_{ijk}\xrightarrow{{\rm parity}}-\epsilon_{ijk}$. It can also be quickly checked that $F^{jk}\xrightarrow{{\rm parity}} F^{jk}$.

But this leads to a wrong conclusion that $B_i\xrightarrow{{\rm parity}}-B_i$. This means $\vec{B}$ is not an axial vector which is definitely false. What is the mistake?

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  • $\begingroup$ Ah, I misread that, sorry. In so far, all I wrote is non-sense of course. $\endgroup$
    – Sebastian Riese
    Dec 15, 2019 at 18:19

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Under parity, position changes as ${\bf r}\to -{\bf r}$, the electric field transforms as ${\bf E}\to -{\bf E}$ and the magnetic field tranforms as ${\bf B}\to {\bf B}$. Do not confuse parity with reflection in a mirror where $(x,y,z)\to (-x,y,z)$. Under mirror reflection ${\bf E}$ stays the same while ${\bf B}$ changes sign.

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  • $\begingroup$ What makes you think that I am confusing parity with mirror reflection? I am thinking exclusively of parity. $\endgroup$
    – Solidification
    Dec 15, 2019 at 14:45
  • $\begingroup$ I thought you might be confused because you thought that ${\bf B}$ should change sign under parity, although it does not in fact do so. Although formally paritiy is ${\bf r}\to -{\bf r}$, I like to explain the physics of parity non-conservation in weak interactions via mirror reflection because most people, physicists included, have more experience with mirrors than with space inversion. $\endgroup$
    – mike stone
    Dec 15, 2019 at 15:00
  • $\begingroup$ I have a different question. $\endgroup$
    – Solidification
    Dec 15, 2019 at 15:05
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The mistake is the transformation behaviour of $\epsilon_{ijk}$ you assume. Let's look at the transformation of the angular momentum $$ \vec L = \vec r \times \vec p ,$$ which in components gives $$ L_i = \epsilon_{ijk} r_j p_k . $$

Under parity we have $\vec L' = \vec L$, while $\vec r' = -\vec r$ and $\vec p' = -\vec p$ (those are proper vectors), this means, that in components we can identify: $$ L_i' = \epsilon_{ijk}' r_j' p_k' = \epsilon_{ijk}' (-r_j) (-p_k) = L_i .$$ Which tells us, that $\epsilon_{ijk}$ has to stay invariant under these transformation. You already have the relevant point: it is pseudo-tensor (density) of third order, and a pseudo-tensor of arbitrary order transform as $T' =T$ under parity.

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  • $\begingroup$ But $\epsilon_{ijk}^\prime= -\epsilon_{ijk}$. Do you agree? $\endgroup$
    – Solidification
    Dec 15, 2019 at 20:10
  • $\begingroup$ "However, the Levi-Civita symbol is a pseudotensor because under an orthogonal transformation of Jacobian determinant −1, for example, a reflection in an odd number of dimensions, it should acquire a minus sign if it were a tensor. As it does not change at all, the Levi-Civita symbol is, by definition, a pseudotensor." en.wikipedia.org/wiki/Levi-Civita_symbol#Properties $\endgroup$
    – Solidification
    Dec 15, 2019 at 20:12
  • $\begingroup$ Sorry for continuing to be confusing. The transformation behaviour I gave is for tensors, while pseudo-tensors are, by definition, invariant under parity. I'll fix the paragraph. $\endgroup$
    – Sebastian Riese
    Dec 16, 2019 at 17:53
  • $\begingroup$ (That's, btw, why it is usually referred to as the "Levi-Civita symbol", because it is not a tensor, but just some quantity defined to be invariant under arbitrary orthogonal coordinate trasformations). $\endgroup$
    – Sebastian Riese
    Dec 16, 2019 at 17:54

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