Transformation Properties of the Cross Product

Question

In Goldstein's Classical Mechanics, the cross product $\mathbf{V}^{*}$ of two vectors $\mathbf{D}$ and $\mathbf{F}$, with components $D_i$ and $F_i$ with respect to some Cartesian reference frame, respectively, is defined as a vector with components $$V_i^{*}=D_jF_k-D_kF_j,\qquad\text{$i,j,k$ in cyclic order.}$$ This definition implies that the direction of $\mathbf{V}^{*}$ depends on the handedness of the reference frame used, however, whilst in most elementary treatments of vector algebra, the cross product is defined as a vector with a well-defined direction independent of any reference frame, to be determined via the right-hand rule.

Indeed, when $\mathbf{D}=D\mathbf{i}$ and $\mathbf{F}=F\mathbf{j}$ with respect to a right-handed Cartesian reference frame, we have according to the above $$V_3^{*}=DF,\qquad V_1^{*}=V_2^{*}=0\qquad\implies\qquad\mathbf{V}^{*}=DF\mathbf{k}.$$ Now, the basis vectors $\mathbf{i}'=-\mathbf{i}$, $\mathbf{j}'=-\mathbf{j}$, $\mathbf{k}'=-\mathbf{k}$ and associated coordinate axes form a left-handed Cartesian reference frame. With respect to this reference frame, we have $\mathbf{D}=-D\mathbf{i}'$ and $\mathbf{F}=-F\mathbf{j}'$. Hence we now have $$V_3^{*}=(-D)(-F)=DF,\qquad V_1^{*}=V_2^{*}=0\qquad\implies\qquad\mathbf{V}^{*}=DF\mathbf{k}'=(-DF)\mathbf{k}.$$

I don't really understand the purpose of defining the cross product in a coordinate-dependent way which clearly yields different results depending on the handedness of the frame used. I understand that many physics texts wish to distinguish between two kinds of vectors (polar and axial), but in Goldstein's book this distinction merely seems to stem from trading the traditional, coordinate-independent geometrical definition for the above coordinate-dependent one.

The viewpoint of the above discussion is a passive one, i.e. the coordinate axes are inverted. Some texts prefer the active viewpoint, where instead of transitioning from one reference frame to another, points, vectors and physical objects are rotated with respect to one fixed reference frame. In this case one inverts all vectors: $\mathbf{D}'=-\mathbf{D}$ for all $\mathbf{D}$. If $\mathbf{V}^{*}=\mathbf{D}\times\mathbf{F}$, one would argue that $(\mathbf{V}^{*})'=\mathbf{D}'\times\mathbf{F}'=(-\mathbf{D})\times(-\mathbf{F})=\mathbf{D}\times\mathbf{F}=\mathbf{V}^{*}$. But why wouldn't $(\mathbf{V}^*)'=-\mathbf{V}^{*}$, i.e. why does one compute the cross product of the inverted vectors instead of inverting the cross product itself?

Answer 1

The cross product is what's called a pseudovector. It transforms normally under rotations but it gains an additional minus sign under reflections.

Look at the following picture of a reflection

Let's define the blue vector as the cross product of green $\times$ red. After a reflection the blue vector, still being defined as green $\times$ red, points in the opposite direction. It is not a proper vector because if you mirror a normal vector along the y axis$^\dagger$ only the y component gets flipped. A parity transformation is just a composition of three reflections and what I just said holds for any odd number of reflections. An even number of reflections can be written as a rotation so you wouldn't notice a difference there.

The coordinate way of writing the cross product is perfectly consistent with the geometric definition.

Angular momentum is another example of a pseudovector and I found this picture (from the wikipedia page on pseudovectors) very illuminating. It shows the angular momentum of the wheels and notably it points in the same direction before and after the reflection.

$\dagger$ With y axis I mean the left-right direction so along the green vector.