Copasetic Flow

Summary of what's gone on before.  In the last two days, unbeknownst to me at the time, we first showed that the Lorentz/Fitzgerald contraction only happens in the direction of the velocity of a frame.  Coupling this with rotational symmetry, yesterday, it was shown how the vector version of the Lorentz transform could be derived.  Yesterday's treatment needs to be repeated from end to start to truly  count as a derivation though, I suppose. Picking up where I left off yesterday, the second half of the vector Lorentz transform, the expression for time in the primed frame can be written as $t^{\prime} = \gamma\left(t - \vec{v}\cdot\vec{r}\right)$ Once again, this shows that only the component of the position vector parallel to the velocity gets to play.  However, here you should notice that the velocity vector is not normalized, not divided by it's magnitude.  In other words, while only the component of the position vector parallel to the velocity gets...

I'm deeper into the planning for this summer's experimental search for H-Rays [1].  I've abandoned my previous superconducting magnet designs in favor of a much simpler pre-existing yoke magnet that's sitting out in the hallway. The pole pieces are in the center of the picture and are retractable using the knobs on the edges of the magnet.  The resulting gap between the pole pieces is where the Dewar will sit,see the picture below: The Dewar sitting between the pole pieces has created a new theoretical issue.  It's only theoretical for the moment though.  Once we have numbers to go with the theory, we'll find out if our Dewar will wind up looking like this[3][4]: As  +Peter Terren  can tell you, rapidly changing magnetic fields like we hope to generate for quenching our superconducting sample will cause Lenz's law eddy currents that create opposing magnetic fields and resulting forces applied to the surfaces that contain the conductors. ...

In studying form my EM midterm, I came across a practice problem involving finding the potential along the z axis due to a charged disc centered on the same axis.  After thinking about the problem a bit, I turned to what's becoming one of my favorite online references, Dr. J.B. Tatum's text on electricity and magnetism [2].  Sure enough, there was a solution that could be adapted to the task at hand.  It involved several 'clever' trig substitutions one of which were not immediately clear to me, so I've expanded upon it here.  After the clever trig trick, read on to find out more interesting stuff about Dr. Tatum, an emeritus professor at the University of Victoria [1]. The Basic Problem The practice problem mentioned above is described by the following diagram from Dr. Tatum's text.  The first and handiest innovation in Dr. Tatum's treatment is to parameterize the problem using the angle marked as theta and the limit of that angle labeled as alpha. T...

In electromagnetism, we came across the following formula describing a potential on a plane (picture 1) In our problem, we were given a plane that looked like the following.  The potential inside the small square is V and the potential everywhere else is on the plane is zero (picture 2) We were to estimate why the potential on the plane didn't go to zero everywhere when the z in the pre-integral numerator  seemed to indicate that it did. Very roughly speaking, the points within the denominator blow up to infinity and conspire to cancel the numerators 0.  There's a more elegant and rigorous way, (courtesy of my professor), to show that everything is all right though.  Here's a sketch of the technique. What we want to do is show that when the denominator goes to infinity while the numerator is zero, things cancel nicely and we still get a finite answer.  When the denominator goes to infinity, the following conditions apply (picture 3) We want...

Just a quick note on the more generic and far lest cited sibling of the Pythagorean theorem, the law of cosines.  If you're like me, you were taught this formula sometime between the eighth and the eleventh grades and promptly forgot it after your exam.  It's a formula that relates the length of one side of a triangle  (any side), to its opposite angle and the lengths of the other two sides.  The formula and its associated diagram, ( courtesy of Wikipedia ), are shown below: If you keep it in mind, you'll start to see it show up all over the place, honest!  For example it turned up in today's electromagnetism homework.  We were tasked on determining the electrostatic potential at any point in a plane due to a ring of charge.  Some of us put a significant amount of effort into determining a general formula for the distance from a point in the plane to any point on the circle.  An industrious eight grader however, could ...

There are an increasing number of apparent correspondences between EM this semester and our section on complex analysis in math methods last semester.  These are just notes on a few of them. Uniqueness of the Electrostatic Potential Solution and Liouville's theorem After stating that we would be solving Poisson's equation to determine electrostatic potentials, our professor then launched into a proof that the solutions, once found, would be unique.  We first defined a potential psi equal to the difference between non-unique solutions, (assuming for the moment in our proof by contradiction that there could be more than one unique solution).  We placed psi back into Poisson's equation and ran through the following steps: Ultimately we wound up proving that at best psi is a constant, but that it must be zero everywhere on the surface that defines the Dirichlet boundary conditions that the two 'non-unique' solutions both satisfy, so it's constant value ...

Today's post is a little plain Jane, (as far as I know), in that it's just a review of our classes' derivation of the boundary conditions for the electrical displacement field at an interface between two materials.  Here's a question.  Does anyone know of anything, cool or clever to take away from the following derivation?  At the moment, it seems necessary, but not tantalizing. First, we'll need the Divergence Theorem, (picture 1), stating that the volume integral of the divergence of a vector field is equal to the surface integral of the same field with respect to the normal of the surface that bounds the volume. Given the above tools, and starting with a boundary between two materials, and the typical pillbox, (picture 2), where delta A is the area of the top and bottom of the box, n is the unit vector normal to the top of the pillbox and the pillbox straddles the boundary between the two materials.  In the end we'll let the sides of the pill...

Reviewing for the GRE physics subject test, I've frequently come across questions like: Given a distribution of discrete charges, (as shown in the diagram), determine the electric field at a given point r, (also shown in the diagram). Formula for the Electric Field from a Distribution of Charges The distribution of charges I've seen in sample questions have, so far, involved only finite numbers of point charges, so the electric field equation can be written as a sum: where is the distance from a charge contributing to the sum to the location where the electric field is being measured, q is the value of the particular charge, and u is the unit vector pointing from the contributing charge to the field measurement point. This can be separated into x and y components: where theta is the angle between the particular charge and the point where the field is being measured with respect to the x axis. Using these formulas, it's just a matter of summing up the contribution of each ...

This installment in the ‘It’s Obvious. Not!’ series relates to the second edition of the book “div grad curl and all that” by H.M. Schey, published by W. W. Norton. Near the end of the example I referenced here, the author of “div grad curl and all that” states that the following integral is ‘trivial’ and results in an answer of 1/6 pi, (specifically, this falls on page 26 of the second edition). As far as I can tell, the solution is more tedious than it is trivial. I’m hoping there really is a trivial solution. If you know it, please add it to the comments below. I’m posting two versions of the ‘tedious’ solution here. The integral in question: The author suggests switching to polar coordinates before solving the integral using the following substitutions: The substitution that’s not mentioned is: So, now to solve the ‘trivial’ integral, first use the substitutions mentioned above: Factoring out the -r squared term in square root: Using the trigonometry...