1
$\begingroup$

Axial vectors like angular momentum $L^k$ are not proper vectors and thus do not follow the vector transformation law: $\bar v^i = \Lambda^i_j v^j$. Instead they are 'pseudovectors' which follow the vector trasnformation law except for a -1 sign under reflections/parity inversions.

These more complicated transformation rules for angular momentum can be made simpler by expressing it as a rank-2 antisymmetric tensor : $L^{ij} = \begin{bmatrix}0&L_z&-L_y\\-L_z&0&L_x\\L_y&-L_x&0\end{bmatrix}$.

In this form the angular momentum vector components under a transformation can be calculated using the usual rank-2 tensor transformation law, namely: $\bar L^{ij} = \Lambda^i_a \Lambda^j_b L^{ab}$.

I was wondering, is a similar procedure true for any pseudotensor. Can we always express it's components as the components of a higher rank new proper tensor?

$\endgroup$
3
  • 5
    $\begingroup$ Hodge dual? $\endgroup$ Commented Jan 2, 2021 at 13:58
  • $\begingroup$ Could you elaborate, does this mean it is true that we can always do this? $\endgroup$
    – Alex Gower
    Commented Jan 2, 2021 at 13:59
  • 1
    $\begingroup$ If you want to understand the object(s) you're working with here you could read about differential forms, angular momentum is an example. Technically the cross product on $\Bbb R^3$ does not produce another vector in $\Bbb R^3$, what you have actually used is the exterior product of two vectors: $\vec x\wedge \vec p=\vec x\otimes \vec p-\vec p\otimes \vec x$. This is an element of the exterior algebra, $\bigwedge^2 \Bbb R^3$. $\endgroup$
    – Charlie
    Commented Jan 2, 2021 at 20:48

1 Answer 1

1
$\begingroup$

I will offer an alternative way of looking at this. You care about transformations, right? More often than not these transformations form groups. For example rotation group is $\text{SO}(3)$, parity and time-reversal group is $Z_2$. Groups have representations over vector spaces, so different vectors simply are from different representation spaces for these groups. There is only one representation of the rotation group $\text{SO}(3)$ for 3d vectors. But you can have two representations of the Parity group - this gives you axial and polar vectors. You also have two representations of the time-reversal group - two more representations. So if you combine time-reversal, parity and rotations, you can have four different kinds of 3d vectors.

So if you want an exhaustive list of possibilities, my hunch is that you will get it if you go after the transformations that you are interested in.

Then, you can find ways to represent your vectors in the terms you like. I would expect that you should always be able to represent your pseudo-vectors in terms of vectors/tensors/forms from the tangent/co-tangent spaces because this is how you nail an abstract vector space to your physical system. But this is a hand-wavy statement - I would be happy to see it proven wrong.


Addendum

  1. I shamelessly conflated irreducible representations with reducible ones. $Z_2$, being an Abelian group only has 1d irreducible representations, but $\mbox{SO}\left(3\right)$ has a single 3d irreducible representation. Combining time-reversal with parity, with rotations gives a group $Z_2\times Z_2\times\mbox{SO}\left(3\right)$. Now that has 4 different irreducible representations over 3d vector space.
  1. One has to distinguish vectors from vectors :-). Representation talk is applicable to any vector space over real numbers. But then there are vectors 'that transform like vectors/tensors', which means vectors from the tangent space of appropriate differentiable manifold (at some specific point). Both kinds are vectors, but the latter 'vectors/tensors' are more constrained. This is why you get vectors that do not transform like 'vectors/tensors'.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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