Copasetic Flow

Google Chrome hasn't been letting me into the lgdc.uml.edu database, (maybe their certificate expired?) There's a way to tell 'curl' to get the information anyway. Up until a few days ago, I could retrieve data from the Pt. Arguello ionosonde using Now, I just get where the advanced button does not let me into the site anyway. So, I put curl to work with: curl --insecure "https://lgdc.uml.edu/common/DIDBGetValues?ursiCode=PA836&charName=MUFD,hmF2,hmF1&fromDate=2023.03.22&toDate=2023.03.23 and immediately got back the data I was looking for:

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.  ...

Now available as a Kindle ebook for 99 cents ! Get a spiffy ebook, and fund more physics The following is a pretty lengthy procedure, but converting the divergence, (nabla, del) operator between coordinate systems comes up pretty often. While there are tables for converting between common coordinate systems , there seem to be fewer explanations of the procedure for deriving the conversion, so here goes! What do we actually want? To convert the Cartesian nabla to the nabla for another coordinate system, say… cylindrical coordinates. What we’ll need: 1. The Cartesian Nabla: 2. A set of equations relating the Cartesian coordinates to cylindrical coordinates: 3. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system: How to do it: Use the chain rule for differentiation to convert the derivatives with respect to the Cartesian variables to derivatives with respect to the cylindrical variables. The chain ...