# Is a magnet and bouncy balls a case of parity violation?

This may well be a very odd question; however, I'm currently studying parity violation and it came to mind that, if a Cobalt-60 atom decaying by the weak force and emitting more electrons opposite the magnetic field direction is a parity violation (since in its mirror image, more electrons are emitted in the direction of the magnetic field), why could I not simply take a macroscopic magnet which I have designed to emit particles (e.g. bouncy balls) in one direction and not the other. Then, in a mirror, the magnetic field direction, since it is an axial vector, would not change, but the direction of particle emission would and so the mirror image would not be symmetrical and there would be a parity violation. Could someone tell me if and where I'm wrong?

Here is another take. It does not matter if you can do it macroscopically. Sure you can build a bar magnet that only ejects particles from one side and not the other. But under a mirror that would not look suspicious as it is a macroscopical object, you could easily create another macroscopical magnet that ejects particles from the opposite end.

But when it comes down to elementary particles and your particle decays in such a way that it emits electrons in the direction of spin, then you should be able to find particles that emit electrons in the opposite direction. The mirror image of the phenomena should be equally valid. Yet this is not what happens.

• Thanks Mauricio, I understand the gist of your answer but I'm still not fully persuaded- surely, even if you can construct the mirror image of a microscopic bar magnet emitting particles, this mirror image is still not symmetrical since the direction of particle emission has changed but the magnetic field direction, being an axial vector, has not. Thus, given that the mirror image is not symmetrical, surely by the definition of parity, it is a violation regardless of whether or not this mirror image could be constructed in the non-mirrored world? Sep 5 at 21:30
• The mirror image does not need to be symmetrical, think of a car rolling next to a mirror, the angular momentum of the wheels is not inverted in the mirror image but nobody is claiming that it implies some parity violation. If we replace the mirror with another car it is very similar to the previous mirror image and physics remains valid. But with the Co60, you cannot replace the mirror with another nucleus that reproduces the previous mirror image. Sep 5 at 22:03
• This macroscopic object has an electron ejector at $\vec r$, so the electron momentum is proportional to it: $\vec p =\alpha \vec r$, under parity inversion, that reads $-\vec p =-\alpha \vec r$, which is the same equation, aka parity symmetry. If you can build something such that $\vec p = \alpha \vec B$, that device violates parity.
– JEB
Sep 6 at 0:36
• @JEB sure but if you try that macroscopically $\alpha$ would most probably be a quantity that flips signs under parity. Sep 6 at 21:48

Yes: you can do this, an no, it will not violate parity. Parity violation in Wu's famous experiment is not really about the magnetic field, it's about the nuclear spin, $${\vec J}$$. The magnetic field ($$\vec B$$) is there so one can align a macroscopic number of 60-Co spins.

With that, the probability, $$p(\theta)$$, of decaying into an direction $$\theta$$ (as defined by the momentum direction $$\hat k$$) can be written in terms of the available physical quantities:

$$\frac{dp}{d\theta} = a + b(\hat k \cdot \hat J)+ c||\hat k \times \hat J||^2$$

Here $$a,b,c$$ are unknown numbers to be measured by experiment.

Under a parity inversion, $$\hat k \cdot \hat J$$ changes sign, while the $$\hat k \times \hat J$$ does not, so:

$$\hat P\{\frac{dp}{d\theta}\} = a - b(\hat k \cdot \hat J)+ c||\hat k \times \hat J||^2 \ne \frac{dp}{d\theta}$$

Where the inequality holds of $$b\ne 0$$. In other words, any alignment of the decay electron with nuclear spin violates parity.

Note that, in terms of $$\theta$$: $$\frac{dp}{d\theta} = a + b\cos{\theta}+ c\sin^2{\theta}$$

so $$a, b, c$$ are monopole, dipole, and quadrupole terms, respectively, so even (odd) order terms conserve (violate) parity.

For your contraption, aligning $$\vec p_{\rm ball}$$ with $$\vec B$$, you are going to have to introduce other axial vectors into the problem (say, something that measures $$\vec B$$), and given Maxwell's equations and Newtons's Laws: it will not violate parity, whatever it is.