# Why is the $B$-Field an axial Vector? [duplicate]

in trying to understand the Wu-Experiment I wonder why the $B$-Field is an axial vector. I know that $\vec{B} = \vec{\nabla} \times \vec{A}$. Under Parity transformation I'd expect $\vec{A} \rightarrow -\vec{A}$, however I don't know whether $\vec{\nabla} \rightarrow -\vec{\nabla}$.

• Short answer: the cross product changes sign. The right hand rule becomes the left hand rule under parity. Nov 17, 2016 at 12:35
• NB This question/answer is older than the duplicate, but for whatever reason the newer version received more attention. Apr 26, 2021 at 6:18

Maybe the best way is to think about $\vec{B}$ in terms of the Biot-Savart law.

Imagine a loop carrying a current $I$ in a plane that is perpendicular to a mirror. The Biot-Savart law says that the B-field at position $\vec{r}$ is given by $$\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi}\, \oint \frac{I\, d\vec{l} \times \vec{r'}}{|\vec{r'}|^2}\ dl,$$ where $\vec{r'} = \vec{r}-\vec{l}$ is the displacement from an element on the loop to where the field is calculated.

This is an axial vector because if we look at this situation in a mirror, the current would appear to flow in the opposite sense, $\vec{l}$ is reversed and the $\vec{B}$ field should actually be in the opposite direction to its mirror image. i.e. An actual mirror image would look like it was obeying a left-hand rule, rather than a right-hand rule.

This is actually exactly the example used on the wikipedia page on pseudovectors, which is another name for an axial vector.

In this example, both $\vec{l}$ and $\vec{r}$ are displacements and are true vectors. Their vector product must be an axial vector.

You are asking about a parity transformation, but as far as I am aware $\vec{B}$ is unchanged by a parity inversion. Axial vectors do not change signs under parity inversions. Angular momentum is another example of an axial vector that does not change under a parity inversion. $\vec{A}$ on the other hand is a true vector and has its sign flipped by a parity inversion. The curl of a true vector is an axial vector and the curl of an axial vector is a true vector. So $\nabla$ is behaving as a true vector in this regard where $\nabla \rightarrow -\nabla$ is odd under a parity inversion (because $\partial/\partial x \rightarrow -\partial/\partial x$ etc.)

• As parity is an inversion and not a mirror image, we'd still have to rotate the right image, correct? If so, this would make sense to me. Nov 17, 2016 at 13:55
• @infinitezero Yes. The current loop is invariant under parity (you move each current element to its diametrical opposite, and you flip the current direction), as is the magnetic field it produces. You can see the right-hand image as a mirror version or as a rotation - those two pathways differ by a parity inversion, and are therefore equivalent. Nov 17, 2016 at 13:58

(Within Netwonian mechanics) You can start with Lorentz Force law $$\vec{F}=q\vec{E}+q\vec{v}\times\vec{B}$$We know that $\vec{F}$ is a physical vector(From Newton's law). We also know that $\vec{v}$ is a physical vector. Therefore $\vec{B}$ must be an axial vector.

• I like this argument. Nov 17, 2016 at 13:58