Why do cross products not change their sign under an inversion of coordinates?

Question

I always understood that the direction of a cross product was determined by the right-hand rule regardless of the handedness of the coordinate system we're working in. The only way a cross product wouldn't change its sign under a coordinate inversion is if we were to assume a left-hand rule for computing cross products in a left-handed coordinate system. But if we were to do this then we would have to specifically mention that the right-hand rule only holds true in a right-handed coordinate system (which almost no one ever does). Also, this means that the directions of quantities like angular momentum depend upon the coordinate system being used which just seems wrong.

Also also, if we were to use a left-hand rule for left-handed coordinate systems, $\hat{i} \times \hat{j} = \hat{k}$ even in a left-handed coordinate system which is incorrect.

This question was prompted by Problem 1.10 from Griffiths and its solution.

Answer 1

Griffiths has made a mistake. He's asked about a weird situation, when he meant to ask about a simple situation. Let's understand Griffith's weird situation, and then let's understand the simple situation.

To start, if you use the algebraic formula for cross product,

\begin{equation} \vec a\times\vec b = (a_yb_z-a_zb_y,a_zb_x-a_xb_z,a_xb_y-a_yb_x) \end{equation}

then you've already decided that your ____-hand rule matches the handedness of your coordinate system. If you work in a right-handed coordinate system, this formula reproduces the right-hand rule. If you work in a left-handed coordinate system, this formula reproduces the left-hand rule. If you start in an original right-handed coordinate system, then using this formula to define $\times_\mathrm{orig}$ gives a right-handed rule, while using this formula in the inverted coordinate system to define $\times_\mathrm{inv}$ gives a left-handed rule.

Griffiths is making the following statement: If you use the algebraic formula for the cross product in both the original and inverted basis, then the components of $\vec a\times \vec b$ are the same in the inverted basis as they were in the original basis. In other words, $(\vec a\times_\mathrm{orig}\vec b)_{x}=(\vec a\times_{inv}\vec b)_{x'}$ and similarly for $y,z$. Of course, since $x'=-x$, that means that the vector they represent are negatives of each other, $(\vec a\times_\mathrm{orig}\vec b)=-(\vec a\times_{inv}\vec b)$.

To sum up, Griffiths has proved that, in terms of vectors, if you take a right-handed cross product in the original coordinate system, it is negative the left-handed cross product in the inverted coordinate system. People usually don't notice this result, or particularly care. People usually care about a completely different result: If you take a right-handed cross product $\vec a\times\vec b$, and a right-handed cross product $(-\vec a)\times(-\vec b)$, they both give the same result. This you can probably prove yourself without too much headache, and is clearly what Griffiths was trying to get at based on the solution.

Answer 2

Also, this means that the directions of quantities like angular momentum depend upon the coordinate system being used which just seems wrong.

Seems wrong, but isn't. The "direction" of angular momentum only matters insofar as it's perpendicular to the plane of rotation. Whether you define clockwise to be "out of the plane" vs. "into the plane" is entirely convention, but of course you need to stick with that convention once you choose it. It's the same as us choosing to call "positive charge" positive - we could have just as easily said protons are negatively charged and electrons are positive.

Answer 3

Griffiths problem 1.10(c) seems to be about a physical transformation (reflection),* while your question seems to be about a change of coordinates. Those are different things. If you reflect a right hand, it becomes a left hand. If you change from a right-handed coordinate system to a left-handed coordinate system, your right hand remains a right hand. Nothing has happened to it; you're just describing it in a different way.

In the problem from Griffiths, you start with some vectors, say $(1,0,0)$ and $(0,1,0)$ (whose cross product is $(0,0,1)$), and you negate the components of those vectors, without changing the coordinate system. They become $(-1,0,0)$ and $(0,-1,0)$, whose cross product is $(0,0,1)$, not, as you might expect, $(0,0,-1)$.

If you don't do anything to the vectors, but simply write their components in a different coordinate system, say $\hat X=-\hat x,\hat Y=-\hat y,\hat Z=-\hat z$, then they are $(X,Y,Z)=(-1,0,0)$ and $(X,Y,Z)=(0,-1,0)$, and their cross product is $(X,Y,Z)=(0,0,-1)$. The right-hand rule in these coordinates looks different, but it isn't actually different: it's still a right-hand rule.

If $(1,0,0)$ and $(0,1,0)$ are ordinary vectors, then their cross product $(0,0,1)$ can be treated as a pseudovector, which gets an extra factor of $-1$ on reflection, therefore becoming $(0,0,1)$, which is the correct cross product of the reflected vectors. A pseudovector does not get an extra factor of $-1$ in a coordinate change. The pseudovector $(x,y,z)=(0,0,1)$ is the pseudovector $(X,Y,Z)=(0,0,-1)$, which is again the correct cross product – as it must be, since you are just describing the same situation in a different way.

The direction of angular momenta does not depend on the coordinate system being used.

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* Actually, it's not clear: problems 1.10(c) and (d), which just say "inversion", seem to be about physical reflection, but (a) and (b), which talk about "translation of coordinates" and "inversion of coordinates", muddy the issue. For the purposes of this answer I'm ignoring (a) and (b). See also Jahan Claes's answer.

Answer 4

There are two ways to define the cross product. The one feels more physical is the "right hand rule" based definition. The more rigorous one is based on it being distributive with the products for the basis vectors defined explicitly, i.e.:

$$ \hat{x} \times \hat{y} = \hat{z} $$

and the others. Under the second definition, it is not the cross product that is right handed, but rather the coordinate system. In this definition, the cross product is really following the sign of the permutation of the basis vectors. So, as in this problem, when you invert the coordinate system, you end up with the same component for the product, but it is the negative of the vector in absolute physical space.

If, instead, you want the cross product to always refer to the right handed vector, then under an inversion the cross product equation would need to be redefined to its own negative. It is certainly true that if you invert the coordinate system, the "right hand rule" product of two vectors will have the negative coordinates that it has in the old coordinate system.

You can see that my explanation agrees with the solution in Griffiths because $\vec{A} \times \vec{B}$ is simply changed to $(-\vec{A}) \times (-\vec{B})$ for the same $\times$. Since the operator is the same, then it is still true in the inverted system that

$$ \hat{x} \times \hat{y} = \hat{z} $$

as before. But the inverted system is left handed, so this means that the cross product in the inverted system is left handed.

In effect, what has happened was both input vectors' components are negated, which causes a cancellation, and the operator itself changes handedness. These three effects leave a net negative from the perspective of physical space, causing the cross product vector to negate in physical space. It does not change the components of the result though, when you write the components in the transformed coordinate system. This is because negating both inputs cancelled to have no effect, and the operation is the same.

Good question though, and I have spent a lot of time baking my head over these questions. The issue is, from circumstance to circumstance and book to book, people do not really mean the same thing and use terms the same way.