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I have read in this answer that

"to represent angular momentum as a vector you need to use a right hand rule. This is annoying, because physics at ordinary scales is reflection invariant but people do this to avoid tensors, because they have more intuition for vectors."

In which way does the right hand rule achieve this mapping to vectors?

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  • $\begingroup$ Related: physics.stackexchange.com/q/9864/2451 , physics.stackexchange.com/q/69345/2451 $\endgroup$ – Qmechanic♦ Sep 8 '18 at 15:19
  • $\begingroup$ Note that a second rank antisymmetric tensor can be represented as something that behaves as a vector under rotation but is invariant under inversion. Such an object it is not actually a vector but an axial vector or pseudovector. Examples are the magnetic field and angular momentum. $\endgroup$ – my2cts Sep 8 '18 at 19:46
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    $\begingroup$ I would say it is done to avoid bivectors (which, as is common for elements of an associative algebra, can be encoded as matrices and thus tensors). Bivectors work in any number of dimensions and don't have handedness concerns. In 3 dimensions, we can define the cross product as $u\times v=(u\wedge v)I^{-1}$ where $I$ is the pseudoscalar. Handedness comes in in the choice of unit pseudoscalar that $I$ refers to. Using this the angular momentum bivector is $L=r\wedge p$. $\endgroup$ – Derek Elkins Sep 9 '18 at 18:46
  • $\begingroup$ @DerekElkins You can extend this in an answer $\endgroup$ – veronika Sep 10 '18 at 20:40
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Angular momentum is most naturally a skew two-index tensor $L^{ab} = x^ap^b-x^bp^a$. Indeed, in higher dimensions, the two-index language is essential. In 3-d the $SO(3)$ invariance of the Levi-Civita symbol $\epsilon^{abc}$ gives us the option of converting $L^{ab}$ to index object to a one-index (cartesian vector) object $L^a= \epsilon^{abc} L^{bc}$ - but we have to make a choice of sign when we do this. The right-hand rule is just a convention, as are right-handed co-ordinate systems. You can do otherwise (and at least one classic book on mechanics uses a left-handed coordinate system) but then you will have difficulty communicating with other people...

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    $\begingroup$ Dear Mike i am sure that you wrote something nice but I understand nothing except the last two sentences $\endgroup$ – veronika Sep 9 '18 at 7:22
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    $\begingroup$ @veronika Sorry! I assumed from you mentioning "tensor" that you were familiar with the idea. My answer was essentially saying "yes" The RH rule is a trick to avoid tensors --- and my formulae were attempt to show how this is done. Unfortunately, you need to learn something about tenors to understand what I said though. $\endgroup$ – mike stone Sep 9 '18 at 13:35
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Not quite an answer to the question asked, but a clarification of one of the ideas you used to build the question.

As I have noted elsewhere right- (or left-)hand rules tend to appear in pairs when computing physical behaviors, so the resulting physics is reflection invariant.

Yes, you use a right hand rule to compute the angular momentum of a spinning wheel (a quantity, not a behavior), but you use another right-hand rule to find the direction of a gyroscope's precession (a behavior) from that angular momentum.

Likewise you use a right-hand rule to find the direction of a magnetic filed created by a solenoid (a quantity) but another right-hand rule to find the direction of a ion's curvature in that field (a behavior) from the field.

In both cases the behavior is unchanged if you use left-hand rules instead of right-hand rules. (Though the convention for direction of angular momentum or magnetic field is reversed.)

In other words, the presence of handed rules in physics doesn't automatically make the behaviors of systems described parity odd (or worse fail to have a well defined parity).

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    $\begingroup$ Thank you. Could you explain what distinguishes a measured quantity from a behavior? $\endgroup$ – veronika Sep 9 '18 at 7:29
  • $\begingroup$ Mmmm. That's a very good question, and I'm glad you asked it. Alas, I don't have a completely satisfying answer. Let's say that the direction of these ``quantities'' is a matter of definition because their affects on the movement of matter and energy lie at right angles to the direction of the quantity itself, which introduces an ambiguity that we have to settle by convention. $\endgroup$ – dmckee♦ Sep 9 '18 at 16:51

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