Copasetic Flow

 I've run into an interesting math problem. Project TouCans occasionally makes dx contacts. That's not a problem, that's actually really awesome! However, KO6BTY and I would like to be able to map the F2 skips our radio signal makes on its way to the receiving station. We have plenty of data about the F2 layer of the ionosphere that is captured by ionosondes around the world. The issue is how do we determine which set of ionosonde data to use? Our first guess is that we should use the ionosonde closest to the path of the signal at any point. OK... Now, how do we determine that? First, we'll calculate the shortest distance from each ionosonde to the path of the QSO between ham radio stations. Then, we can sort the distances so that we can only include the ionosondes that are closest to the path. Armed with that data, we can perhpas used the distances to calculate a weighted,  estimated height of the F2 layer at any point along the signals path as it skips from ...

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:  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}...