# Why is $\mathbf{B}$ a pseudovector?

I got the difference between polar vectors and axial vectors (pseudovectors).

An example of pseudovector is $\mathbf{B}$. But why exactly the magnetic field is a pseudovector and its components parallel to an coordinate axis do no change signs if I reverse that axis?

It is a pseudo-vector because it is the curl of a vector potential, or because its curl is a vector ($$\displaystyle{\vec{J}}$$, $$\displaystyle{\frac{d\vec{E}}{dt}}$$).

Or: consider the Biot-Savart law, which expresses it as an integral over the cross product (pseudo-vector) of 2 vectors.

One can also look at the Lorentz Law:

$$\vec{F}= q\left(\vec{E} + \vec{v}\times \vec{B}\right)$$

$$\vec{B}$$ must be a pseudo-vector for its cross product with a vector to be a vector (force). Further consideration of this law as the only thing that makes sense wrt to covariance in Special Relativity may provide a deeper understanding.

Like many pseudovectors, $$\mathbf{B}$$ is more correctly (or more completely) thought of as a two-form. It's the lower right $$3\times 3$$, skew-symmetric block (the spatial part) of the Faraday tensor, so it actually represents a directed plane. Alternatively, as the exterior derivative of the vector potential one-form, it's a two form. In 3 dimensions, we can equate a directed plane to a vector by taking the latter to be the unit normal to the plane, with appropriate, consistent orientation. The operation that does this "cheat" (which only works in three dimensions) for us and lets us think of planes as defined by vectors is the Hodge dual, which involves multiplication by the Levi-Civita tensor $$\epsilon$$. It's the $$\epsilon$$, and its behavior under a parity transformation, that brings about the "pseudonesss" of $$\mathbf{B}$$.

Written out more fully, the mapping from vector potential to $$\mathbf{B}$$ is $$\mathbf{B}=\mathrm{curl}\mathbf{A} = (\star(\mathrm{d}\mathbf{A}_\flat))^\sharp$$ and is the Gibbs-Heaviside minestrone blending the two operations exterior derivative followed by Hodge dual with raising and lowering of indices thrown in to shift vectors to one forms and back again, showing how $$\mathbf{A}\mapsto\nabla\times\mathbf{A}$$ comes up with a disguised two-form. $$\star$$ is the "pseudo"-culprit here!

If you want a physical visualization why the $$\mathbf{B}$$ vector picks up its anomalous sign flip on an improper isometry, one can't really do better than the diagram below, taken from the Wikipedia Pseudovector Article.

Check the direction of the $$\mathbf{B}$$ field from a current loop when the loop undergoes a reflexion. It's thus easy to see that $$\mathbf{B}$$ maps to a vector field with the opposite direction to the mirror image vector field.

Actually, this little picture is much more general than magnetic fields and current for it is nothing other than a visualization of the exterior derivative of a two form; we can think intuitively of the exterior derivative as the integral over of a differential form over the boundary of an infinitessimal element of space divided by the signed magnitude of the volume form for that space. If this sounds complicated, in 2 dimensions its nothing more than visualizing the curl of a vector field as the circulation around a loop divided by the signed area of the loop - the sign of the latter flips when we make an improper isometry. This is an elegant example in keeping with an approach to exterior calculus that may be helpful:

Peter Schröder and Keenan Krane, "Discrete Differential Geometry: A Quick and Dirty Introduction to the Geometry of Surfaces"

• Is this wikipedia diagram in conflict with the following paragraph from the same article: Consider an electric current loop in the z = 0 plane that inside the loop generates a magnetic field oriented in the z direction. This system is symmetric (invariant) under mirror reflections through this plane, with the magnetic field unchanged by the reflection. But reflecting the magnetic field as a vector through that plane would be expected to reverse it; this expectation is corrected by realizing that the magnetic field is a pseudovector, with the extra sign flip leaving it unchanged.
– JEB
Aug 29, 2017 at 1:44

Generally speaking, any vector you get from doing a cross-product is a pseudovector. This because when defining a coordinate system, you have two options: you can define a right-handed coordinate system, or a left-handed one. In a right-handed system, the index finger points in the $$x$$ direction, the middle finger points in the $$y$$ direction, and the thumb points in the $$z$$ direction.

The same is also true for a left-handed system, but you use your left hand instead. The definition of the cross product is: $$\begin{cases} \mathbf{\hat x} \times \mathbf{\hat y} = \mathbf{\hat z}, \\ \mathbf{\hat y} \times \mathbf{\hat z} = \mathbf{\hat x}, \\ \mathbf{\hat z} \times \mathbf{\hat x} = \mathbf{\hat y}. \end{cases}$$ If you just look at the first line alone, a cross product of something in the $$x$$ direction with something in the $$y$$ direction will always result in something in the $$z$$ direction, regardless of whether our system is right or left-handed. Hence, $$\mathbf{\hat x} \times \mathbf{\hat y}$$ in a right-handed system would point in the opposite direction to $$\mathbf{\hat x} \times \mathbf{\hat y}$$ in a left-handed system! But our choice in system is completely arbituary, and our preference for right-handed systems is only a result of most of us being right-handed.

In physics, some vectors like momentum, acceleration, etc. do not depend on this arbitrary choice, but since the magnetic field is the result of a cross product between the current flow and the position vector, it evidently does. In this sense, it is less "real" than those other vectors, hence the name "pseudovector". We could have certainly used a left-handed system instead, and the magnetic field would reverse in direction, but this doesn't matter; we don't observe magnetic fields, we observe their effects on moving charged particles. And since this effect is also given by a cross product, $$\mathbf F = q \mathbf v \times \mathbf B,$$ as long as we are consistent with our use of the right/left-hand rule, we will get the same direction for the force regardless of our choice is handedness. In that sense force is more "real" of a vector than the magnetic field. (If you've studied any rotational mechanics, other examples of pseudovectors include angular momentum and torque, both of which incidentally result from cross products.)

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.)

NB: Copied in from the older direct duplicate question, which appears to have received less attention.