Copasetic Flow

Summary :  This one took awhile.  I got busy in the lab  These notes start with rotation matrix properties and the transpose products of matrices.  special relativity via tensors also begins.  Specifically, the Lorentz transformation tensor components are reviewed and the number of independent parameters are counted. Did a few concrete checks that a matrix times its transpose is symmetric.  Sure enough, it is. $\begin{pmatrix} a & b \\ c & d \\ \end{pmatrix}\begin{pmatrix} a & c \\ b & d \\ \end{pmatrix} = \begin{pmatrix} a^2 + b^2 & ca + bd \\ ca+bd & c^2+d^2 \\ \end{pmatrix}$ $\begin{pmatrix}a & b & c\\ d & e & f \\ g & h & i \\ \end{pmatrix}\begin{pmatrix} a & d & g\\ b & e & h \\ c & f & i \\ \end{pmatrix} = \begin{pmatrix} a^2 + b^2 + c^2 & ad + be + cf & ag + bh  + ci\\ ad + be + cf & d^2 + e^2 + f^2 & gd + eh + fi \\ ga + hb + ic ...

Summary :  Still More Tensor Identities Three more identities with gradients, divergences, Laplacians, and cross products.  Later today, the fun stuff, special relativity begins! $\vec{A} \cdot \left(\vec{B} \times \vec{C}\right) = \vec{B} \cdot \left(\vec{C} \times \vec{A}\right)$ $= A_i \epsilon_{ijk} B_j C_k$ As long as we only cycle the indices in the Levi-Civita symbol, we won't cause a sign change, so the above is also equal to $= A_k \epsilon_{ijk} B_i C_j$ Which we can commute to get $= B_i \epsilon_{ijk} C_j A_k = \vec{B} \cdot \left(\vec{C} \times \vec{A}\right)$ Done! $\vec{\nabla} \cdot \left(\vec{\nabla} \times \vec{A} \right) = 0$ $=\partial_i \epsilon_{ijk} \partial_j A_k$ $= 0$ If $i$ and $j$ are equal, then the Levi-Civita evaluates to zero.  If they are not equal, then swapping the two indices produces the same mixed partial derivative result, but with a negative sign inserted by swapping indices in the Levi-Civita symbol.  ...

Summary:  Having worked through the examples that looked the most difficult, today's notes contain examples that are pick-up work from the easy problems.  These are simple-ish tensor index identities, including the divergence of the position vector, the cross product of the position vector, the Laplacian of one over the displacement squared, and the curl of a gradient. $\nabla \cdot \vec{r} = 3$ $= \dfrac{\partial}{\partial x_i} r_i$ Keep in mind that $r_1 = x$, $r_2 = y$, and $r_3 = z$.  Using the rules of partial differentiation, when the partial operates on the variable it is with respect to it will return 1, and when it operates on any other variable, it will return 0.  The results sum to 3. $\vec{\nabla} \times \vec{r} = 0$ $=\epsilon_{ijk} \partial_j r_k$ $= 0$ For the $\epsilon{ijk}$ to evaluate to a non-zero result, $j$ and $k$ have to not be equal.  However, as discussed above, if $J \ne k$, then the partial derivative evaluates to zero. ...

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...

What's Gone Before and What Will Ensue :  Yesterday, he first step of an exercise regarding the rotation of cross products was worked out.  Today, the identity proved yesterday will be used to show that when the two vectors in a cross product are rotated by the same rotation matrix, the resulting vector of the cross product is rotated by the same rotation matrix.  In the end, yet another property will be proven using tensor index notation. The identity from yesterday is: $\epsilon_{ijk}W_{iq}W_{jl}W_{km} = det\left(W\right)\epsilon_{qlm}$ We'll also need the definition of the cross product in index notation $\vec{A} \times \vec{B} = \epsilon_{ijk}A_jB_k$ and the rotation matrix transpose identity $M^TM = 1$ also known as $M_{iq}M_{in} = \delta_{qn}$ We want to prove that if $\vec{A}$ and $\vec{B}$ are both rotated by the same rotation matrix, $M_{in}$, then so is the result of the cross product, $\vec{V}$ First, rotate the two input vectors $\vec{A^{\prime}...