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If I have a scalar field \begin{align} f: \mathbb{R}^3 &\rightarrow \mathbb{C}\\ (x, y, z) &\mapsto f(x, y, z) \end{align}

We can define an operator $P$ that takes a function like $f$ and gives a new function $Pf$. It is clear that for scalar fields like $f$ we have

$$ Pf(x, y, z) = f(-x, -y, -z) $$

But how is $P$ defined for vector fields? Suppose we have

\begin{align} \textbf{F}: \mathbb{R}^3 &\rightarrow \mathbb{R}^3\\ (x, y, z) &\mapsto F_x(x, y, z)\hat{\textbf{x}} + F_y(x, y, z)\hat{\textbf{y}} + F_z(x, y, z)\hat{\textbf{z}} \end{align}

Where each of $F_x, F_y$, and $F_z$ are scalar fields How is $P\mathbf{F}$ defined in this case?

$$ P\mathbf{F}(x, y, z) = \mathbf{F}(-x, -y, -z) = F_x(-x, -y, -z)\hat{\mathbb{x}} + F_y(-x, -y, -z)\hat{\mathbb{y}} + F_z(-x, -y, -z)\hat{\mathbb{z}} $$

Or do the basis vectors get transformed as well?

$$ P\mathbf{F}(x, y, z) = -\mathbf{F}(-x, -y, -z) = -F_x(-x, -y, -z)\hat{\mathbb{x}} - F_y(-x, -y, -z)\hat{\mathbb{y}} - F_z(-x, -y, -z)\hat{\mathbb{z}} $$

Or maybe just the vectors change and not the components:

$$ P\mathbf{F}(x, y, z) = -\mathbf{F}(x, y, z) = -F_x(x, y, z)\hat{\mathbb{x}} - F_y(x, y, z)\hat{\mathbb{y}} - F_z(x, y, z)\hat{\mathbb{z}} $$

I feel like the answer is going to depend on whether $\textbf{F}$ is a "axial" or "polar" vector like $\textbf{B}$ or $\textbf{E}$. But then I feel like the statement "$\textbf{B}$ is an axial vector" is more of a statement about how we choose to define some parity operator rather than something intrinsic about the magnetic field.

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  • $\begingroup$ The first one (I think) $\endgroup$ Commented Apr 1, 2022 at 0:28

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In a naive sense, if we think of parity inversion as similar to "reflecting an image through mirrors" then when we flip the coordinates we should expect the coordinates flip. So I would say

$$ P\mathbf{F}(x, y, z) = - \mathbf{F}(-x, -y, -z) $$

Suppose we have a car with positive $x$ position and moving with velocity $\textbf{v}$ to the $+x$ direction. If we flip this scene then now the car will have negative $x$ position and be moving in the $-x$ direction. This is the second option in the original question.

However, in physics terminology, some vectors are pseudovectors. An example would be angular momentum

$$ \textbf{L} = \textbf{r}\times\textbf{p} $$

If we act the parity operator $P$ on $\textbf{r}$ and $\textbf{p}$ then both vectors flip signs, so their cross product must maintain the same sign.

The frustration, expressed in the original question, is that $\textbf{L}$ is an element of the same vector space as $\textbf{r}$ and $\textbf{p}$ so it makes no sense that the action of the parity operator on $\textbf{L}$ would differ from its action on $\textbf{r}$ and $\textbf{p}$

That is, there seems to be an arbitrary choice made that

$$ P\textbf{L} = (P\textbf{r}) \times (P\textbf{p}) $$

This, of course, disagrees with the definition of the parity operator I gave at the top of this answer, hence the apparent paradox.

When things are approached in this manner there is no "principled" way to determine whether any given vector is a (polar) vector or a pseudovector. To determine how the parity operator acts on any vector you must ADDITIONALLY be aware of the constitutive relations for the original vector (e.g. you must know that $\textbf{L}=\textbf{r}\times \textbf{p}$ and that $\textbf{r}$ and $\textbf{p}$ are polar vectors).

One mathematically rigorous way to resolve this is to introduce $\mathbf{L}$ as the exterior product of $\mathbf{r}$ and $\mathbf{p}$:

$$ \mathbf{L} = \mathbf{r} \wedge \mathbf{p} $$

Now $\mathbf{L}$ is an object in a different vector space than $\mathbf{r}$ and $\mathbf{p}$ so it is not absurd for the parity operator to have a different action on it. Here $\mathbf{L}$ is a bivector or 2-blade, but there are also ways were these objects can instead be expressed as differential forms.

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