Reading 2/4/2008: 2.4-2.6 (Continued)

Two points.

1. A lot of people have asked why there exists a set of coordinates wherein the stress tensor is diagonal. This is actually just a fact from linear algebra. Remember that S_ij can be thought of as a matrix S. Since the stress tensor is symmetric, the matrix is symmetric in the sense that S^T = S. This is just because S_ij is just the component in the ith row and jth column. Check it if you don't believe it.

Now from linear algebra, there's a theorem that says every symmetric matrix admits an orthogonal diagonalization. That is, there exists a an orthogonal matrix P and a diagonal matrix D such that

S = P^-1DP

Note that the diagonal elements of D are just the eigenvalues of S. Also, P is invertible because every orthogonal matrix is (det P = +-1).

Now why does this answer the question? Because if we let the columns of P become our new basis for R^2, then S "looks like" D. More precisely, the linear transformation S has representation D with respect to the basis formed by the columns of P. Thus those columns represent the coordinate system in which S is diagonal.

2. With regard to my little tangent before, maybe to help clarify a bit.

SO(3) is the set of rotations in three space. Its elements are just 3x3 matrices that are not only orthogonal (as all rotations are) but actually have determinant +1 to rule out reflections. It's the Special Orthogonal Group. We call it a group because it's actually a more sophisticated mathematical object, but that's not important. The key point is that multiplication and addition are defined on this set by matrix multiplication and division. It might help to think of matrices as vectors in R⁹.

What's interesting is that SO(3) is actually a Lie group. This basically means that it has the operation of matrix addition and that the group "looks like some version of Rⁿ". This is obvious to us, because we can think of a matrix easily as just a vector in R⁹. But this gives it lots of special properties, because it means everything is sufficiently nice.

We can define a "path" in SO(3) to be a set of matrices parameterized by a real variable t. Say &gamma(t) might be a bunch of matrices that vary smoothly at t varies from 0 to 1. What do we mean by vary smoothly? Well that's simple - each entry in the matrix is just a real function of t, and all of those functions are smooth.

So now we have this structure on our set of rotations. How do we think about an infinitesimal rotation? Well, it should be close to the identity - because no rotation at all is just multiplying by the identity matrix. And it should have some notion of "direction", because if we integrated a bunch of infinitesimal rotations (in some sense), we ought to be able to get to a real rotation.

What we do is we let &gamma be an arbitrary path starting at the identity (so &gamma(0) = I) and defined from t = 0 to some later value of t (it doesn't matter). We can then define an infinitesimal rotation by taking the derivative of this path at 0. Why does this make sense? Well intuitively, say &gamma(1) is the rotation we want to move "towards", then we should be able to get there by integrating &gamma'(t) from 0 to 1 - integrating the infinitesimal rotations. This is just the fundamental theorem of calculus in disguise.

A lie algebra is just the set of all objects of the form &gamma'(0) where &gamma is, again, a path starting at the identity matrix and going somewhere else. The calculation I made in my previous post was investigating what the lie algebra of SO(3) (denoted by so(3), to be confusing!) actually looks like. All I did was let &gamma be an arbitrary path, I then wrote &gamma in terms of an arbitrary rotation matrix (every matrix in SO(3) can be written in terms of three angles, which we can then think of as functions of t), and then differentiated with respect to t and set t=0. That's where the matrix graphic in my post comes from. Since &gamma was arbitrary, d&theta/dt, d&phi/dt, and d&psi/dt are all arbitrary as well. That's why we calculate that so(3) is just the set of antisymmetric matrices.

Hopefully that makes slightly more clear what I was talking about.