0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Ah, I misread that, sorry. In so far, all I wrote is non-sense of course. $\endgroup$ – Sebastian Riese Dec 15 '19 at 18:19
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What makes you think that I am confusing parity with mirror reflection? I am thinking exclusively of parity. $\endgroup$ – mithusengupta123 Dec 15 '19 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 '19 at 15:00
  • $\begingroup$ I have a different question. $\endgroup$ – mithusengupta123 Dec 15 '19 at 15:05
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ But $\epsilon_{ijk}^\prime= -\epsilon_{ijk}$. Do you agree? $\endgroup$ – mithusengupta123 Dec 15 '19 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 $\endgroup$ – mithusengupta123 Dec 15 '19 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 '19 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 '19 at 17:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.