# Wrong transformation of magnetic field components (defined from $F^{\mu\nu}$) under parity

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?

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

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.

• What makes you think that I am confusing parity with mirror reflection? I am thinking exclusively of parity. – mithusengupta123 Dec 15 '19 at 14:45
• 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. – mike stone Dec 15 '19 at 15:00
• I have a different question. – mithusengupta123 Dec 15 '19 at 15:05

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.

• But $\epsilon_{ijk}^\prime= -\epsilon_{ijk}$. Do you agree? – mithusengupta123 Dec 15 '19 at 20:10
• "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 – mithusengupta123 Dec 15 '19 at 20:12
• 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. – Sebastian Riese Dec 16 '19 at 17:53
• (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). – Sebastian Riese Dec 16 '19 at 17:54