340 Class Notes, Nov 22

Phy 340 class notes for Fall, 1999

Aug 25	Sep 8	Oct 4	Nov 1	Phy 340 page
Aug 30	Sep 13	Oct 11	Nov 8
	Sep 20	Oct 18	Nov 15
	Sep 27	Oct 25	Nov 22
			Nov 29

Phy 340, Week 14

22 Nov Angular momentum of a rigid, rotating body. Write the integral again in terms of the angular velocity, and notice how the output L depends on the input w. This defines a function. I needed to make some side comments on the meaning of the word "function" because the common notation can get confusing in this context. It's necessary to distinguish between the function, f, and the value of the function, f(x), at the point x. For many cases, this is being unnecessarily fussy, but for what comes next, the distinction is important in understanding the results.

The defining equation here is L = I( w ). The input is a vector, the output is a vector, the function ( I ) is not. It is called a tensor.

Definition: a vector-valued function of a vector variable, satisfying the linearity property,

I( a v₁ + b v₂ ) = a I( v₁ ) + b I( v₂ )

is called a (second rank) tensor.

In order to manipulate tensors, it's useful to write them in terms of components. Just as you can write a vector in terms of components and work with those, you can do the same with tensors. I developed the treatment pretty fully in class, but here I'll write an abbreviated form. Call the basis vectors e_i, where the index i runs over x,y,z (or over 1,2,3 if you prefer).

L = S_i L_i e_i = I( w ) = I( S_i w_i e_i )

The two vectors L and w have the respective components indicated. Now use the linearity property and take the summation sign outside the function

= S_i w_i I( e_i )

Remember now that the output of the tensor (function), I, is a vector. That means you can write it in terms of the same basis. Do so.

I( e_i ) = S_k I_ki e_k

Plug this back into the preceeding equation to get

L = S_k L_k e_k = S_i w_i S_k I_ki e_k

Did you notice the trick I pulled with the index on the L_k? It's a dummy, so you can call it anything you want. I chose to make it match the index on the vector on the right. Now two equal vectors must have the same components -- the coefficients of e_k must match -- so this implies

L_k = S_i I_ki w_i.

You can write this out as a matrix times a column matrix, with the indexes going the the standard order, I_row,column.

Nothing in this development used the scalar product, only that the vectors e_i form a basis. That is, they are linearly independent and span the space.

Apply this now to compute the components of the inertia tensor.

I( e_x ) = I_xxe_x + I_yxe_y + I_zxe_z
= Ú dm r x ( e_x x r ) = Ú dm [ r² e_x - ( r · e_x ) r ]
= Ú dm [ ( y² + z² ) e_x - xy e_y - xz e_z ]

From this, you directly read three of the nine components of I. You then repeat the process for I( e_y ) and I( e_z ) to obtain similar results. for the other six components.

22 Nov Do exam problems 2 and 3 in a couple of different ways. In response to a question, what does the word "vector" mean? I didn't do all the details, but I outlined what the ideas are.

Talk in some detail about the meaning of one of the homework problems, the one with the magnetic field tensor.

Theorem:

w₁ · I( w₂ ) = I( w₁ ) · w₂

This theorem has as a consequence that fact that the components of the inertia tensor form a symmetric matrix in the usual orthonormal basis.

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