Is this an example of Parity violation? I always hear about parity violation in high energy physics, but what about examples in classical physics?
Say we have a wire carrying current in the $+x$ direction, thus generating a magnetic field $\mathbf{B}$ in the (say) $+\phi$ direction.
Now we put a mirror in the $yz$ plane, so that the image is a current flowing in the $-x$ direction with a magnetic field $\mathbf{B}$ in the same $+\phi$ direction.
This of course makes sense because the current $Id\mathbf{x}$ transforms as a vector whereas the magnetic field $\mathbf{B}$ is a pseudovector and does not change sign upon inversion.
But the image in the mirror is not physically possible as the direction of $\mathbf{B}$ is not curling around $I$ following the right hand rule.
Isn't this an example of parity violation? That is the mirrored image could not possibly represent a real physical process and we can therefore tell it apart?
 A: There isn't parity violation in classical physics; it's a phenomenon unique to the weak charged current.
There are axial vectors (a.k.a. psuedovectors) in classical physics, like the magnetic field or the angular momentum vector, which would change sign if computed using a "left-hand rule" instead of the usual right-hand rule. However whenever you compute observables in classical mechanics you always wind up using the right-hand rule an even number of times. For instance you find the direction of a magnetic field by taking a cross product of a current vector $\vec j$ and a displacement $\vec r$,
$$
d\vec B \propto \frac{ dq\,\vec v \times \hat r }{r^2},
$$
where both position and velocity are both ordinary polar vectors which change sign under the parity transformation $(x,y,z)\to(-x,-y,-z)$. The Lorentz force has a second cross product,
$$
\vec F = \vec I \times \vec B.
$$
So if you were to use the left-hand rule throughout your entire calculation, you'd get the same result for the accelerations of your particles, which is what you would actually be able to measure in the lab.
I just recently ran across an excellent essay by Martin Gardner on this subject which predates the discovery of parity nonconservation in the weak interaction; after the fall of parity Gardner expanded the material into a book, which I haven't read but I'm sure is excellent.

tparker asks in a comment,

Could you expand on your comment "whenever you compute observables in classical mechanics you always wind up using the right-hand rule an even number of times"? In QFT, there seem to be pseudoscalar observables - for example the pion. This has nothing to do with the P-violation of the weak interaction. So why are all classical observables even under parity but not quantum observables? 

I wouldn't consider "the pion" an observable.  I'd consider it a particle.  You can observe properties of this particle, such as:


*

*the time before it decays

*its position

*its momentum

*its kinetic energy

*its angular momentum magnitude

*its angular momentum orientation

*its electric charge

*its interaction cross section with some sort of target

*its magnetic dipole moment

*its electric dipole moment


Each of these properties has a definite set of rules for transformation under rotations (scalars unchanged, vectors rotated, tensors more complicated, etc.) and inversions (sign change or not).  However what we physically get out of experiments are sets of scalar observations: in a detector here I count $N$ particles, and in a detector there I count $M$ particles.  About no axis could I rotate a detector to count $-N$ particles, so count rates are scalar.
The pion field is indeed pseudoscalar, but not parity-violating because you don't observe the pion field directly: you observe pions, or their interaction products.  The pion is kind of a challenging case because you have to infer its parity from selection rules.
The classic
argument
goes like this: a sneaky application of the Pauli principle says that a neutron pair or proton pair with angular momentum $J=\hbar$ must have $L=\hbar$ and therefore, since the parity of the orbital angular momentum wavefunctions is $(-1)^L$,  odd parity.
The deuteron has spin and parity $J^P=1^+$, and the pion is spinless.
Since the capture reaction $\pi^- d^+ \to nn$ occurs (that is, the rate isn't zero) when the deuterium "pionic atom" is in its lowest $S$-wave state, it has either (a) different parity for the system before and after the capture occurs, or (b) negative intrinsic parity for the pion.  
It's the second one.  The strong interaction is parity-conserving, but that doesn't preclude the appearance of fields with negative parity, like the pion field or the $L=\text{odd}$ wavefunctions.
In electromagnetism we find ourselves predicting the behavior of scalar quantities like energies and interaction rates by taking scalar products of polar vectors with polar vectors and axial vectors with axial vectors.  For example, two particles with parallel momenta $\vec p_1 \cdot \vec p_2 > 0$ will have generally a larger cross section for interaction than two particles with antiparallel momenta $\vec p_1 \cdot \vec p_2 < 0$, because the former may find themselves in a reference frame where they are spending a lot of time together.  The interaction energy between two magnetic moments is proportional to the scalar product between two axial vectors, usually a spin with a spin.
Not until the weak interaction is involved do we find any interactions where a scalar observable depends on the interaction between a polar vector and an axial vector.  For example, the interaction cross section for neutrinos is proportional to $1-\vec\sigma\cdot\vec p$: neutrinos do not interact with other matter unless their spin $\vec\sigma$ is antiparallel to their momentum.  A particle with a permanent electric dipole moment parallel to its spin would have a (nominally scalar) energy in an electric field proportional to $\vec\sigma\cdot\vec E$ --- a pseudoscalar term.
Nothing prevents us philosophically from writing down a parity-violating interaction term in a model for electromagnetism, just like nothing prevents us from writing down a term where the interaction energy depends on whether or not it's Thursday.  But we don't find any evidence of parity-violating interactions until we start exploring the weak sector.
A: To clarify what's going on here, let's start by considering an example with an electric field. Suppose you have some complicated, asymmetric shape like a guitar, and it has positive charge distributed on it. At some nearby point $\mathbf{r}$, we have an electric field $\mathbf{E}$. Under a full parity inversion $(x,y,z)\rightarrow(-x,-y,-z)$, the entire picture flips, so that not only would we redraw the guitar inverted, but we would also redraw the location of point $\mathbf{r}$ so that it lay on the other side of the guitar. The electric field would also flip, since $\mathbf{E}$ is a vector (not a pseudovector).
Now in your magnetic field example, let's say that $+x$ is to the right, and we're discussing the magnetic field at a point $\mathbf{r}$ above the wire. The field comes out of the page. Under a full parity inversion, the current reverses direction (and every element of current is also moved to a new location, although that effect doesn't really matter, since the current is uniform). The point $\mathbf{r}$ also moves; we would now draw it underneath the wire. Since $\mathbf{B}$ is a pseudovector, it hasn't changed. It's still coming out of the page, and that's consistent with the right-hand rule.
By the way, you also have to be careful when talking about reflections such as $(x,y,z)\rightarrow(-x,y,z)$ that aren't full parity inversions. In such a reflection, you can't just assume that pseudovectors stay the same. A reflection is equivalent to a full parity inversion followed by a 180-degree rotation. For example, a magnetic field in the z direction flips under an x reflection.
