Pseudoscalars, Polar Vectors, and Axial Vectors

Source: Copilot exchange

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What is a pseudoscalar?

A pseudoscalar is a quantity that behaves like an ordinary scalar under rotations, but changes sign under a parity transformation (mirror reflection or spatial inversion).

More formally:

A scalar $s$ is unchanged by parity:

$$s(\mathbf{x}) \rightarrow s(-\mathbf{x}) = s$$

A pseudoscalar $p$ picks up a minus sign:

$$p(\mathbf{x}) \rightarrow p(-\mathbf{x}) = -p$$

Why is it called “pseudo”?

Because it looks like a scalar with respect to rotations. Rotating the coordinate system does not affect its value. The difference only appears when you perform an orientation-reversing transformation such as a mirror reflection.

Example

Take two polar vectors $\mathbf{a}$ and $\mathbf{b}$.

Their cross product

$$\mathbf{a}\times\mathbf{b}$$

is an axial vector (pseudovector).

Now dot that with another polar vector $\mathbf{c}$:

$$(\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}.$$

This is a scalar number, but under parity:

• $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ each change sign.
• $\mathbf{a}\times\mathbf{b}$ does not change sign because it is axial.
• Therefore the whole quantity changes sign.

Hence

$$(\mathbf{a}\times\mathbf{b})\cdot\mathbf{c}$$

is a pseudoscalar.

Geometrically it represents an oriented volume (the scalar triple product).

Connection to the Helmholtz decomposition of an axial field

An axial vector field $A_i$ can be Helmholtz-decomposed as

$$A_i = (\nabla\times V)_i + \partial_i\phi_A .$$

Here:

• $\nabla\times V$ is an axial vector if $V$ is polar.
• $\partial_i\phi_A$ must also be axial.

Since $\nabla$ is a polar vector operator, the quantity whose gradient produces an axial vector cannot be an ordinary scalar. It must be a pseudoscalar:

$$\phi_A \xrightarrow{P} -\phi_A.$$

Then

$$\nabla \phi_A$$

transforms as an axial vector.

So the longitudinal part of an axial vector field is indeed encoded in a pseudoscalar potential.

This is why, in theories where the fundamental sources provide no pseudoscalar degree of freedom, one often finds that only the curl-type (transverse) axial component can be sourced independently, while the longitudinal axial component is absent or constrained away.

In short

Quantity	Under parity
Scalar $s$	$s \to s$
Polar vector $\mathbf{v}$	$\mathbf{v}\to-\mathbf{v}$
Axial vector $\mathbf{A}$	$\mathbf{A}\to\mathbf{A}$
Pseudoscalar $p$	$p\to-p$

and

$$\boxed{\text{gradient of a pseudoscalar} = \text{axial vector}}$$

which is the key point in the quoted argument.

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What is a polar vector?

A polar vector is the ordinary kind of vector you first learn in physics.

Examples:

• Position: $\mathbf{r}$
• Velocity: $\mathbf{v}$
• Force: $\mathbf{F}$
• Electric field: $\mathbf{E}$
• Gradient: $\nabla\phi$

The defining property is how it behaves under parity (spatial inversion):

$$(x,y,z)\rightarrow(-x,-y,-z).$$

A polar vector changes sign:

$$\mathbf{v}\rightarrow-\mathbf{v}.$$

For example, position transforms as

$$\mathbf{r}\rightarrow-\mathbf{r}.$$

because every coordinate flips.

Contrast with an axial vector

An axial vector (or pseudovector) does not change sign under parity:

$$\mathbf{A}\rightarrow\mathbf{A}.$$

Examples:

• Angular momentum $\mathbf{L}=\mathbf{r}\times\mathbf{p}$
• Torque $\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}$
• Magnetic field $\mathbf{B}$

Why?

Because they are built from a cross product of two polar vectors. Under parity:

$$\mathbf{r}\rightarrow-\mathbf{r},\qquad \mathbf{p}\rightarrow-\mathbf{p},$$

so

$$\mathbf{L} = \mathbf{r}\times\mathbf{p} \rightarrow (-\mathbf{r})\times(-\mathbf{p}) = \mathbf{r}\times\mathbf{p}.$$

The two minus signs cancel.

The connection to pseudoscalars

Notice:

• $\nabla$ is a polar vector operator.
• If $\phi$ is an ordinary scalar, then $\nabla\phi$ is a polar vector.
• If $\phi_A$ is a pseudoscalar, then $\nabla\phi_A$ is an axial vector.

This is because under parity:

$$\nabla\rightarrow-\nabla, \qquad \phi_A\rightarrow-\phi_A,$$

and the two minus signs cancel:

$$\nabla\phi_A \rightarrow \nabla\phi_A.$$

So in the quoted argument, the longitudinal part of an axial field

$$A_i = (\nabla\times V)_i + \partial_i\phi_A$$

requires $\phi_A$ to be a pseudoscalar. That is exactly why the author says the independent longitudinal content of $A_i$ is encoded in a pseudoscalar potential.

A useful memory aid

Object	Parity transformation
Scalar $s$	$s\to s$
Pseudoscalar $p$	$p\to -p$
Polar vector $\mathbf{v}$	$\mathbf{v}\to-\mathbf{v}$
Axial vector $\mathbf{A}$	$\mathbf{A}\to\mathbf{A}$

So polar vectors flip under a mirror inversion; axial vectors do not.