What Is a Pseudovector?
Introduction
Orientation of a Coordinate System
Suppose there are two Cartesian coordinate systems $S = \left\{ \mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3} \right\}$ and $S^{\prime} = \left\{ \mathbf{e}_{1}^{\prime}, \mathbf{e}_{2}^{\prime}, \mathbf{e}_{3}^{\prime} \right\}$ in $3$-dimensional space. For instance, let $S = \left\{ \hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}} \right\}$ and $S^{\prime} = \left\{ \hat{\mathbf{x}}, 2\hat{\mathbf{y}}, 3\hat{\mathbf{z}} \right\}$. Suppose the coordinates of some vector $\mathbf{r}$ with respect to $S$ are as follows.
$$ \mathbf{r}_{S} = 1 \cdot \hat{\mathbf{x}} + 2 \cdot \hat{\mathbf{y}} + 3 \cdot \hat{\mathbf{z}} = (1,2,3)_{S} $$
Since a vector itself does not change no matter which coordinate system it is referred to, its coordinates with respect to $S^{\prime}$ must be as below.
$$ \mathbf{r}_{S^{\prime}} = 1 \cdot \hat{\mathbf{x}} + 1 \cdot 2\hat{\mathbf{y}} + 1 \cdot 3\hat{\mathbf{z}} = (1,1,1)_{S^{\prime}} $$
Here, the transformation between the two sets of coordinates is carried out by the following matrix $Q$.
$$ Q = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} $$
$$ \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}_{S^{\prime}} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & 0 & 1/3 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}_{S} $$
In this way, even the same vector has different coordinates depending on which coordinate system is chosen. On the other hand, if we choose yet another coordinate system $S^{\prime \prime} = \left\{ \hat{\mathbf{x}}, \hat{\mathbf{y}}, -\hat{\mathbf{z}} \right\}$, the coordinates of the same vector $\mathbf{r}$ are as follows.
$$ \mathbf{r}_{S^{\prime \prime}} = 1 \cdot \hat{\mathbf{x}} + 2 \cdot \hat{\mathbf{y}} + (-3) \cdot (- \hat{\mathbf{z}}) = (1, 2, -3)_{S^{\prime \prime}} $$
The coordinates changed both in $S \to S^{\prime}$ and in $S \to S^{\prime \prime}$, but the two changes are different in nature. $S^{\prime}$ only stretches the scales of the axes, so the directions the axes point to remain the same, whereas $S^{\prime \prime}$ reverses the very direction in which the $z$-axis points. What measures this difference is the orientation of a coordinate system.
Consider the matrix whose columns are the basis vectors of the new coordinate system written in the coordinates of the original one. For the two coordinate systems above, these are respectively as follows.
$$ P_{S \to S^{\prime}} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} , \qquad P_{S \to S^{\prime \prime}} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix} $$
Whether two coordinate systems have the same orientation or not is determined by the sign of the determinant of this matrix. If it is positive, they have the same orientation; if negative, opposite orientations.
$$ \det P_{S \to S^{\prime}} = 6 > 0 , \qquad \det P_{S \to S^{\prime \prime}} = -1 < 0 $$
A coordinate system with the same orientation as the standard basis $\left\{ \hat{\mathbf{x}}, \hat{\mathbf{y}}, \hat{\mathbf{z}} \right\}$ of $\mathbb{R}^{3}$ is called right-handed, and one with the opposite orientation is called left-handed. The $S^{\prime}$ above is a right-handed coordinate system, and $S^{\prime \prime}$ is a left-handed coordinate system.
What Happens When the Orientation Is Reversed
Above, we placed no restrictions on the coordinate systems, but from now on, let us consider only orthonormal coordinate systems, whose basis vectors are mutually perpendicular and all of length $1$. Then the coordinate transformation matrix becomes an orthogonal matrix $Q$, and $\det Q = \pm 1$.
Let $S$ be a right-handed coordinate system, and let $S^{\prime}$ again denote the coordinate system obtained by flipping only the $z$-axis. (This is different from the $S^{\prime}$ above.)
$$ \mathbf{e}_{1}^{\prime} = \mathbf{e}_{1}, \qquad \mathbf{e}_{2}^{\prime} = \mathbf{e}_{2}, \qquad \mathbf{e}_{3}^{\prime} = - \mathbf{e}_{3} $$
Then $Q = \diag(1, 1, -1)$ and $\det Q = -1$, so $S^{\prime}$ is a left-handed coordinate system.
Now place two arrows in space, one in the direction of $\mathbf{e}_{1}$ and the other in the direction of $\mathbf{e}_{2}$. Both lie in the $xy$-plane, so their components do not change even when the $z$-axis is flipped.
| Components in $S$ | Components in $S^{\prime}$ | |
|---|---|---|
| First arrow | $(1, 0, 0)$ | $(1, 0, 0)$ |
| Second arrow | $(0, 1, 0)$ | $(0, 1, 0)$ |
Applying the component formula for the cross product as it is in each coordinate system gives $(0, 0, 1)$ in both. However, $(0, 0, 1)$ in $S$ refers to an arrow in the direction of $\mathbf{e}_{3}$, while $(0, 0, 1)$ in $S^{\prime}$ refers to an arrow in the direction of $\mathbf{e}_{3}^{\prime} = -\mathbf{e}_{3}$. The same calculation was performed, yet in space they point in exactly opposite directions.
Let us look again at the components alone. If the result of the cross product transformed in the same way as position, its components in $S^{\prime}$ should have been the following.
$$ Q \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} $$
But what we actually obtained is $(0, 0, 1)$. The sign is off by one flip, and the size of that discrepancy is exactly $\det Q = -1$.
$$ \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = (\det Q) Q \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} $$
The cause is clear. The component formula for the cross product was built on the premise of a right-handed coordinate system, so using the same formula in a left-handed coordinate system flips the orientation convention one more time. Hence quantities made by cross products inevitably follow a different transformation law from that of position or velocity in the first place. The definition below separates these two and gives them names.
Definition
As in the introduction, consider two orthonormal coordinate systems $S$, $S^{\prime}$ and the orthogonal matrix $Q \in \operatorname{O}(3)$ between them. Suppose some physical quantity is measured in each of the two coordinate systems, yielding components $\mathbf{v}$ and $\mathbf{v}^{\prime}$. If the following holds no matter how the two coordinate systems are chosen, this physical quantity is called a polar vector.
$$ \mathbf{v}^{\prime} = Q \mathbf{v} $$
If a physical quantity measured in the same way, yielding components $\mathbf{a}$ and $\mathbf{a}^{\prime}$, satisfies the following, it is called a pseudovector or an axial vector.
$$ \mathbf{a}^{\prime} = (\det Q) Q \mathbf{a} $$
Explanation
The determinant of an orthogonal matrix is $\det Q = \pm 1$, and among them those with $\det Q = 1$ are rotations, that is, elements of the special orthogonal group $\operatorname{SO}(3)$. For rotations, the following holds.
$$ R \in \operatorname{SO}(3) \implies (\det R) R = R $$
Therefore, polar vectors and pseudovectors cannot be distinguished by rotations. The difference shows up in the rest of the orthogonal group $\operatorname{O}(3)$, namely in transformations with $\det Q = -1$ such as reflections and spatial inversion, which reverse the orientation of space. The coordinate transformation of point reflection through the origin $Q = -I$ satisfies $\det(-I) = (-1)^{3} = -1$ in $3$ dimensions, so we obtain the following.
$$ \begin{align*} \mathbf{v}^{\prime} &= (-I) \mathbf{v} = -\mathbf{v} \\ \mathbf{a}^{\prime} &= \left( \det (-I) \right)(-I) \mathbf{a} = (-1)(- \mathbf{a}) = \mathbf{a} \end{align*} $$
Under the coordinate transformation of point reflection through the origin, the intuitive vector we have in mind (= polar vector) reverses its direction, but a pseudovector does not. The same goes for reflections. Considering the reflection $Q = \diag(1, 1, -1)$ across the $xy$-plane, we have $\det Q = -1$, so the following holds.
$$ \begin{align*} \mathbf{v}^{\prime} &= Q \mathbf{v} = (v_{1}, v_{2}, -v_{3}) \\ \mathbf{a}^{\prime} &= (\det Q) Q \mathbf{a} = -(a_{1}, a_{2}, -a_{3}) = (-a_{1}, -a_{2}, a_{3}) \end{align*} $$
A polar vector keeps the components within the mirror plane and flips only the sign of the $z$-component perpendicular to the mirror, whereas a pseudovector does exactly the opposite, keeping the perpendicular component and flipping the two in-plane components. That the cross product of two polar vectors actually transforms in this way can be confirmed in the properties below.
The Meaning of "Pseudo"
Although it is called a pseudovector, from the viewpoint of linear algebra it is undeniably a vector. Polar vectors and pseudovectors are distinguished by the difference in how coordinate transformations apply to them, so both are 3-dimensional vectors in $\mathbb{R}^{3}$. In other words, what "pseudo" actually modifies is not the vector itself but the rule by which the vector follows coordinate transformations.
Example
| List | |
|---|---|
| Polar vectors | position $\mathbf{r}$, velocity $\mathbf{v}$, momentum $\mathbf{p}$, force $\mathbf{F}$, electric field $\mathbf{E}$ |
| Pseudovectors | angular velocity $\boldsymbol{\omega}$, angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$, torque $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$, magnetic field $\mathbf{B}$ |
That angular momentum and torque are pseudovectors follows immediately from $(1)$, since they are cross products of polar vectors. For the magnetic field, this can be seen from the Lorentz force law $\mathbf{F} = q \mathbf{v} \times \mathbf{B}$. Since the force on the left-hand side and the velocity are both polar vectors, $\mathbf{B}$ must be a pseudovector for the equation to hold. In this way, the distinction of pseudovectors can also serve as a tool for checking whether a newly formulated equation makes sense.
A quantity that transforms like a scalar under rotations but changes sign under spatial inversion is called a pseudoscalar, and the scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ of three polar vectors is a typical example.
Difficult Definition
The definition above requires introducing coordinate systems first and describing in words how the components are obtained. Borrowing the language of representations, it can be defined without mentioning coordinate systems. For the orthogonal group $\operatorname{O}(3)$, consider the following two maps.
$$ \rho^{+}(Q) := Q, \qquad \rho^{-}(Q) := (\det Q) Q $$
Since the determinant is a homomorphism, both are representations of $\operatorname{O}(3)$.
$$ \rho^{-}(Q_{1}Q_{2}) = \det(Q_{1}Q_{2}) Q_{1}Q_{2} = \left[ (\det Q_{1}) Q_{1} \right] \left[ (\det Q_{2}) Q_{2} \right] = \rho^{-}(Q_{1}) \rho^{-}(Q_{2}) $$
An element of $\mathbb{R}^{3}$ equipped with $\rho^{+}$ is called a polar vector, and an element of $\mathbb{R}^{3}$ equipped with $\rho^{-}$ is called a pseudovector. Set up this way, the definition immediately makes clear that whether a triple is a pseudovector is not a property of the triple itself but a matter of how the orthogonal group acts on the space in which it lies. The two representations coincide when restricted to $\operatorname{SO}(3)$, but as irreducible representations of $\operatorname{O}(3)$ they are not isomorphic to each other, and this is why the two cannot be distinguished by rotations alone.
Properties
The cross product of two polar vectors is a pseudovector. For $Q \in \operatorname{O}(3)$ and $\mathbf{u}, \mathbf{v} \in \mathbb{R}^{3}$, the following holds.
$$ (Q \mathbf{u}) \times (Q \mathbf{v}) = (\det Q) Q (\mathbf{u} \times \mathbf{v}) \tag{1} $$
If $\mathbf{u}$ and $\mathbf{v}$ are polar vectors, the left-hand side is the cross product of the two transformed vectors, so $(1)$ says that $\mathbf{u} \times \mathbf{v}$ follows exactly the transformation law of a pseudovector.
Proof
For any $\mathbf{w} \in \mathbb{R}^{3}$, since the scalar triple product is the determinant of the matrix whose rows are the three vectors, we have the following.
$$ \begin{align*} \left[ (Q\mathbf{u}) \times (Q\mathbf{v}) \right] \cdot (Q \mathbf{w}) &= \det \begin{bmatrix} (Q\mathbf{u})^{\mathsf{T}} \\ (Q\mathbf{v})^{\mathsf{T}} \\ (Q\mathbf{w})^{\mathsf{T}} \end{bmatrix} = \det \left( \begin{bmatrix} \mathbf{u}^{\mathsf{T}} \\ \mathbf{v}^{\mathsf{T}} \\ \mathbf{w}^{\mathsf{T}} \end{bmatrix} Q^{\mathsf{T}} \right) \\ &= (\det Q) \det \begin{bmatrix} \mathbf{u}^{\mathsf{T}} \\ \mathbf{v}^{\mathsf{T}} \\ \mathbf{w}^{\mathsf{T}} \end{bmatrix} = (\det Q) \left[ (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} \right] \end{align*} $$
Meanwhile, since an orthogonal matrix preserves the inner product, the following holds.
$$ \left[ (\det Q) Q (\mathbf{u} \times \mathbf{v}) \right] \cdot (Q \mathbf{w}) = (\det Q) \left[ (\mathbf{u} \times \mathbf{v}) \cdot \mathbf{w} \right] $$
Since $Q$ is invertible, as $\mathbf{w}$ ranges over all of $\mathbb{R}^{3}$, $Q\mathbf{w}$ also ranges over all of $\mathbb{R}^{3}$. The two vectors on both sides have the same inner product with every vector, so the two vectors are equal.
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