The Cross Product and Angular Momentum

In summary, the problem is that the cross product is not associative, so it is difficult to go from one term in angular momentum to another without getting lost in non-associativity. A way to avoid this problem is to use the determinant form of the angular momentum.
  • #1
JTC
100
6
Hello

I need help to explain the affect of the cross product without the its current symbolism, but for angular momentum.

I can explain angular momentum in terms of the cross product of 3D space formulated like this:
|r| |v| * sin(angler.v) e-perp to r and v Eq.1
(I can explain this to students: length of the "position" and "perpedicular velocity" vector, etc.)

THEN, in Euclidean Space, this cross product is equivalent to:
(ry*vz-rz*vy) e1 + (ry*vz-rz*vy) e2 +(ry*vz-rz*vy) e3 Eq. 2
And I can work out how E1 becomes E2 in Euclidean space.

And I can teach the students how to compute the angular momentum of a particle moving in a plane, but about an axis perpendicular to the plane.

HOWEVER:

I also know that there is an easier way to construct the cross product that does not carry the baggage of non-associativity

I take the first vector, r, and create a skew symmetric matrix out of its components.
And multiply that matrix by the column components of v. Let me call this Eq. 3

Good:

Now the question:

How can I go from a PHYSICAL description of the angular momentum, BYPASS the toxic cross product and go right to the skew symmetric form of computing this affect?

In other words, I am only able to go from E1 to E2 to E3
But I prefer to go from E1 direct to E3
In fact, I would really like to go direct and only to E3.

In other words, how can I explain what this skew symmetric form of the first vector does to the second vector that is the EQUIVALENT of talking about lengths and perpedicularity, but IN THE CONTEXT OF CLASSICAL ANGULAR MOMENTUM?

Maybe this is a silly question.
 
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  • #2
JTC said:
HelloTHEN, in Euclidean Space, this cross product is equivalent to:
(ry*vz-rz*vy) e1 + (ry*vz-rz*vy) e2 +(ry*vz-rz*vy) e3 Eq. 2
And I can work out how E1 becomes E2 in Euclidean space.

Maybe this is a silly question.
how does E1 become E2 if angular momentum is a vector quantity and E1 is a scalar?
 
  • #3
Ben Wilson said:
how does E1 become E2 if angular momentum is a vector quantity and E1 is a scalar?

Sorry... a typo in E1... I forgot to list the vector nature.
I Just fixed it.
A detail, yes, but an important one: thanks
 
  • #4
JTC said:
Sorry... a typo in E1... I forgot to list the vector nature.
how would you do that?
 
  • #5
Is the issue really to avoid "non-associativity" (related to BAC-CAB and the Jacobi identity)
or is it to avoid "non-commutativity"?

Is there a problem with just using the determinant formulation (rather than the cross-product notation)?
Might it seem contrived to represent a position vector by an antisymmetric matrix?

From a purist's viewpoint, it might be good to avoid the cross-product altogether and focus on the bivector associated with r and v [with its area and handedness], rather than a vector perpendicular to it.
 
  • #6
robphy said:
Is the issue really to avoid "non-associativity" (related to BAC-CAB and the Jacobi identity)
or is it to avoid "non-commutativity"?

Is there a problem with just using the determinant formulation (rather than the cross-product notation)?
Might it seem contrived to represent a position vector by an antisymmetric matrix?

From a purist's viewpoint, it might be good to avoid the cross-product altogether and focus on the bivector associated with r and v [with its area and handedness], rather than a vector perpendicular to it.
No, the associativity, commutativity is not the issue.

Yes, I WOULD like to avoid the cross product altogether, but I do not how to discuss angular momentum without it.

With the cross product, I can say: "this term, r x v, is a measure of the distance of the particle from the rotational axis, and the sin(Angle) enters and so does the magnitude and the cross product presvers the plane."

But I have NO IDEA how to discuss angular momenum without the cross product. YOU talk about the determinant formulatoin. Can you elaborate?
How would such a formulatoin (going right to the skew matrix) explain the "group" of terms in angular momentum AND their plane of action?
 
  • #7
Why do you want to avoid the cross product? Is the intention to confuse your students? To avoid useful math almost always leads to more confusion than it helps. I always call it the "didactical deformation of physics", and I fight it whenever I can!:mad:
 
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1. What is the cross product?

The cross product is a mathematical operation that takes two vectors as inputs and produces a third vector that is perpendicular to both of the input vectors. It is denoted by a "x" symbol and is also known as the vector product.

2. How is the cross product calculated?

The cross product is calculated using the following formula: A x B = (AyBz - AzBy, AzBx - AxBz, AxBy - AyBx). This means that the resulting vector will have components in the x, y, and z directions.

3. What is the physical significance of the cross product?

The cross product has many physical applications, including calculating torque and angular momentum. It is also used in electromagnetism to describe the relationship between electric and magnetic fields.

4. How is the cross product related to angular momentum?

The cross product is directly related to angular momentum, as it is used to calculate the magnitude and direction of angular momentum. The direction of the resulting vector from the cross product determines the direction of the angular momentum vector.

5. Can the cross product be used with non-orthogonal vectors?

No, the cross product is only defined for orthogonal (perpendicular) vectors. If the input vectors are not orthogonal, the cross product will be zero, indicating that there is no perpendicular vector between them.

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