3
$\begingroup$

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?

$\endgroup$
14
  • $\begingroup$ As you note yourself, the cross-product uses the right-hand rule and is therefore not coordinate-independent. $\endgroup$
    – NDewolf
    Nov 26 '20 at 11:38
  • $\begingroup$ How so? You don't need coordinates to define or use the right-hand rule. $\endgroup$ Nov 26 '20 at 11:40
  • $\begingroup$ Using the (arbitrarily chosen) right-hand rule, the cross product of any two vectors can be determined in a purely geometrical way, without referring to any coordinate system. $\endgroup$ Nov 26 '20 at 11:44
  • $\begingroup$ You do not choose a right-hand rule using coordinates, you choose it geometrically. $\endgroup$ Nov 26 '20 at 11:47
  • 1
    $\begingroup$ @JilalJahangir I think they argue, there is no "positive" or "counter-clockwise" notion in 3D space. You use the words, but what to they mean? In usual 3D space, there is indeed no such thing, but you can supply it as additional structure if you wish. $\endgroup$
    – Umaxo
    Nov 26 '20 at 12:01
2
$\begingroup$

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

pseudovector 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.

wheel pseudovector

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

$\endgroup$
2
  • $\begingroup$ Let’s align the $x$-axis along the red vector and the $z$-axis with the rightmost blue vector. Your answer emphasizes the active viewpoint, i.e. a “parity operator” $\hat{P}$ actively transforms the red vector $\mathbf{r}=r\mathbf{i}$ and green vector $\mathbf{g}=g\mathbf{j}$ into their mirrored counterparts $\hat{P}\mathbf{r}=\mathbf{r}$ and $\hat{P}\mathbf{g}=-\mathbf{g}$. What I pointed out in my OP is that the cross product behaves abnormally here because we define $\hat{P}(\mathbf{g}\times\mathbf{r})\equiv\hat{P}\mathbf{g}\times\hat{P}\mathbf{r}$. $\endgroup$ Nov 27 '20 at 8:15
  • $\begingroup$ The abnormal behavior results from choosing to treat the cross product that way. One could equally well treat the blue vector $\mathbf{b}=b\mathbf{k}$ as any other vector. Since $\hat{P}$ is a linear operator we need only know that $\hat{P}\mathbf{k}=\mathbf{k}$. Hence $\hat{P}\mathbf{b}=\hat{P}(b\mathbf{k})=b\mathbf{k}=\mathbf{b}$. $\endgroup$ Nov 27 '20 at 8:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.