Copasetic Flow

Summary:  The one that took four days.  A detector that worked finally arrived for the experiment, so work on EM II has been somewhat slower.  Also, the example here uses a lot of material from prior examples and requires being on your toes.  This example is all about showing that a rather abstruse looking rotation matrix is in fact a rotation matrix.  It involves recognizing dot and cross products when they're written in tensor index notation and having rock solid index skills.  At the end of the day though, it's pretty cool, but it still seems like there should be an even simpler way to do this than the one shown here. The game is to show that the following is a rotation matrix in that when multiplied by its transpose, the result is the identity matrix: $M_{ij} = \delta_{ij}cos \alpha + n_i n_j \left(1 - cos \alpha\right) + \epsilon_{ijk}n_k sin \alpha$ Keep in mind that $n_i$ is defined to be a unit vector.  The transpose relation that we're su...

Summary of what's gone on before.  Got through the index notation for gradients and whatnot.  I was left a little bit baffled by the notation for the orthogonal transpose identity.  Consequently, I'm digging back into it. In this set of notes, the transpose, orthogonal identity, $M_{ki}M_{kj} = \delta_{ij}$, is first hammered out.  It then becomes obvious what's going on.  The dummy summing of the two row indices gave us the equivalent of a matrix multiply where it's row times row instead of row times column.  The rows of a matrix that should have been transposed however are the same as the columns of one that wasn't.  In other words by forcing a different type of matrix multiply, they teased out the transpose for free. Here's the hammering through bit. Let's take as an example, the simple rotation matrix about the z axis.  Keep in mind that it has already been explained above why this will work for any orthogonal matrices and that this is j...