0
$\begingroup$

Please "explain" angular momentum.... BUT!!

I am able to justify how the cross product between the postition from an axis and the velocity of a particle, provides "angular velocity." Such an operation maintains "information" about the instantaneous axis of rotation, the magnitudes of the perpedicular disance from the axis and the velocity and mass.

I also know that we can obtain the same result by taking the position vector, converting its components into a skew symmetric matrix and multiplying that matrix by the linear momentum vector.

Good.

What I am UNABLE to do (and the reason I solicit guidance) is to explain how a skew symmetric matrix of position components times the momentum vector gives the angular momentum.

In other words, it seems I am still tied to the cross product.

How can I "explain" the meaning of the angular momentum WITHOUT resorting to the cross product and going DIRECTLY to the skew symmetric form?

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
-1
$\begingroup$

A matrix is best understood intuitively by how it is acting on vectors. In this case the position matrix is constructed such that it acts exactly the same way a crossproduct with the position vector would act. So the intuition is exactly the same and you cannot hope to get a different intuition from the matrix than from the cross product with the corresponding vector. It doesn't mean that you resort to the cross product intuition, it means that you resort to the intuition which is the same for both mathematical descriptions.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Then could you say it for me? For the cross product, I can state: a x b = |a| |b| sin (angle) n-perp. And I can intuit the distance from the axis using the sin of the angle and the velocity perpendicular to that distance: the angular momentum. The geometric form of the cross product contains the information for the ang.mom, when the two vectors are distance and velocity. I can SEE that. I can SAY that. How do you say it for the matrix times the vector? But I cannot say it for the matrix. Can $\endgroup$
    – JT Coriolis
    Commented May 9, 2017 at 14:53
  • $\begingroup$ You shouldn't consider the vector notation. Consider the index notation and write the components, then you will see what the matrix actually does. And, not surprisingly, what it does is exactly the same as what the vector product with $x$ does. So, there is no new intuition behind the matrix vector product... $\endgroup$
    – Photon
    Commented May 10, 2017 at 8:27
  • 1
    $\begingroup$ Oh. I just did it. There it was sitting in my face and I could not see. It is so obvious. Thank you for your patience and your reticence to do it for me. Nice. $\endgroup$
    – JT Coriolis
    Commented May 10, 2017 at 8:33

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