I define a vector as any object $(a_i,a_j, a_k)$ such that it transforms the same way as the coordinates themselves. That is if $x'_i = R_{ij}x_j$, then $a'_i = R_{ij}a_j$ (using the Einstein convention of summing over repeated indices). Please correct me if this is not a proper definition.

I now want to take the cross product of two vectors, $\vec{c} = \vec{a}\times \vec{b}$ and prove (by this definition) that the output is also a vector. Using the Levi-Civita symbol, we have

$c'_i = R_{ij}c_j = det(R) R_{ij}\epsilon_{jlm}a_l b_m$

But this should somehow be the same as the cross product of the transformed $\vec{a}$ and $\vec{b}$. That is $c'_i = \epsilon_{ijk}a'_jb'_k = \epsilon_{ijk}R_{jl} a_l R_{km}b_m$

How do I reconcile these two? I know I should get $\rm{det(R)}$ in the second expression somehow but I can't see how to express it in this notation.

  • 1
    $\begingroup$ Actually, from the point of view of transformation properties, the cross product is not really a vector, but rather a “pseudovector”: en.wikipedia.org/wiki/Pseudovector. Or better, if you want to avoid that mess: use the bivector $a \wedge b$ instead. $\endgroup$ – Hans Lundmark Apr 24 '18 at 7:51
  • $\begingroup$ Doesn't the det(R) take care of the pseudovector-ness of the output? In your link, it also says that $Rv_1 \times Rv_2 = det(R) R(v_1 \times v_2)$. I just can't see how to prove that statement using index notation. $\endgroup$ – user1936752 Apr 24 '18 at 8:18
  • $\begingroup$ Well, if you include the determinant, then yes. But you need to assume that $R$ is an orthogonal matrix too, maybe that's what's causing you trouble? (And then $\det R= \pm 1$.) $\endgroup$ – Hans Lundmark Apr 24 '18 at 8:33
  • $\begingroup$ I see your point. I've edited the question now. The trouble I'm having is to express det(R) in index notation and show the equivalence. $\endgroup$ – user1936752 Apr 24 '18 at 9:23

Your Answer

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

Browse other questions tagged or ask your own question.