# Why is this not a violation of parity invarance for EM

I read that Wu's experiment illustrates that parity violation is possible for weak processes. In that experiment, when Co-60 undergoes beta decay, the emitted electrons come out opposite to the direction of nuclear spin. The mirror image process does not occur.

I also read that parity is conserved for EM and strong processes. But consider the following from EM. Say I have a positively-charged particle moving to the right, and the magnetic field is into the page. Then force on the particle is up (i.e. the particle is accelerated upwards initially). And here, the mirror image also does not occur! The mirror image (the mirror I am thinking of lies in the plane that cuts your head between the eyes) would be moving left, magnetic field into page, acceleration still pwards - but EM would predict that the acceleration is supposed to be downwards!!

Why isn't this an example of parity violation?

• Oct 13, 2018 at 15:55

The magnetic field is a "pseudovector" (more properly, a 2-form), as opposed to the electric field, which is a vector (or a 1-form). That is, under parity, $\mathbf B$ is left unchanged. You can see this from the Lorentz force, $$\mathbf F = q(\mathbf E + \mathbf v\times \mathbf B)$$ where since force is a vector, $\mathbf E$ must also be a vector. Since $\mathbf v \times \mathbf B$ is a product, if $\mathbf B$ were a vector, this term would not transform properly under parity. Thus $\mathbf B$ does not change when we perform a parity transformation.
However I think the more correct way to see this is from the relativistic formulation of electrodynamics. Introduce the electromagnetic field tensor $$F_{\mu\nu} =\begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix}$$ and note that the purely spatial part $F_{ij}$, $1 \le i, j \le 3$ is equivalent to the magnetic field. Since there are two indices, the components are invariant under parity transformations. The electric field is given by $E_i = F_{0i}$, so it changes sign under parity, it is a vector.
The more sophisticated yet way to see this decomposition is that if there is timelike 1-form $dt$ we can decompose the field strength 2-form as $$F = E\wedge dt + B.$$ We see that $E$ is a 1-form (equivalent to a vector after raising the index), but $B$ is a 2-form (often called pseudovector, because not enough people know about the wonders of differential forms).
Now your transformation is not quite $P : (x,y,z) \mapsto (-x,-y,-z).$ It is $RP : (x,y,z) \mapsto (-x,y,z)$, or $P$, then a rotation by $\pi$ in the $yz$-plane. Here the $x$-axis is along $\mathbf v$ and the $z$-axis along $\mathbf B$. Since $\mathbf v$ is perpendicular to the plane of the rotation, it is just affected by the reflection, and is to the left. $\mathbf B$, lying in the plane of rotation, is rotated half a revolution, and is now out of the page, so the force is upward. The force being a vector, this is precisely what we expect. I hope this answers your question.