Maxwell's equations: from vector algebra to geometric algebra. Maxwell representations, part I.

Transcript

this is the first video in a video series on maxwell's equations and its different representations we're going to start with maxwell's equations in matter i'm going to assume linear isotropic media that's a specific relationship between d and e and b and h where e and mu are constants and from that we'll arrive at the geometric algebra form for an r3 basis where we have a e1 e2 e3 which are all euclidean vectors all squaring two plus one we'll see that the maxwell's equations can be represented as a hybrid differential operator that includes space and time derivatives a field which incorporates both electric and magnetic fields we call this the faraday and a multi-vector current charge density current density hybrid that includes the charge density and the vector current density i've been lazy with my arrows here so this is intended to be an arrow the specific form of the faraday is electric field plus i times c b it will then go into the geometric algebra space time algebra representation where we take the direct basis which has basis elements gamma zero gamma one gamma two gamma three and in this basis gamma zero square is two plus one but gamma one gamma two gamma three all square to minus one we're not going to use the matrix any of the possible matrix representations of these but we will you just assume that they obey those square value relationships mentioned and we'll see that the the form of maxwell's equation also singular as it was in in the r3 geometric algebra case has this form grad f equals j where this is a non-arrow grad symbol that incorporates both the space and time derivative terms in a different way than this subtly different and also a four vector uh current density which is a is a constant times this multi-vector current relationship from there we're going to derive two other representations we'll derive the tensor relationships and we will derive the differential forms uh relationship i think for this video we'll probably just get here to the euclidean space basis uh geometric algebra representation let's start and see see how far we get i've written out maxwell's equations in their uh in matter formulation and what we're going to do is we're going to normalize this representation somewhat so look in particular we have here we have a spatial derivative operator that has dimensions one over length and here we have a time derivative operator that has dimensions one over time so this means that our fields must have different units that's undesirable we're also going to plug in our values for h and d in terms of b and and e and we are going to see how we can make this uh a little bit tidier to start with even before we get to geometric algebra so reminder that we are going to substitute d equals epsilon e and e equals h i guess it's h over mu or h equals mu b and we're also going to use the speed of light in the medium that's the one over the square root of mu epsilon and we're going to write eta as the square root of mu over epsilon so with these transformations let's tidy things up so we're going to start here and move this side over so if to convert the one over time so that it has the same units as the gradient we'll add in one over c and then we'll scale our field by pi c so now we have dimensions of gradient equals dimensions of one over c d dt so these dimensions equal one over length and we have now dimensions of the electric field equals the dimensions of b times c i don't really care exactly what units those are we're working in si units but uh it's really not quite relevant this stage so we're going to scale our all of our magnetic fields by c we need to substitute in h equals mu b so right here b mu mu there we go now we want to scale this all by c and d is e times epsilon i'll just tidy this up all right now epsilon c mu is epsilon mu times c which is epsilon u times one over square root of epsilon mu which is square root of epsilon u which is one over c so what do you know everything just happens to work out perfectly so that we get one over c d dt d d d and then down and there was a mistake here that should have just been grabbed on d equals rho it's only now that we switch d equals e epsilon that we say rho over epsilon and any other so let's look at it we have electric field electric field electric field all the ds are gone bc bc bc all of our derivatives with respect to time are now one over c d dt here and one of c d dt here and we have all of our grad dots and grand crosses so there's no geometric algebra here this is just a way to type to put things in uniform uh uniform units and the only left thing left to do here is let's look at c u so c u is u over square root of mu epsilon which is square root of mu over epsilon which is our eta so that's eta all right here's maxwell's equations still in vector form but ready to put into a simpler representation i'm just going to regroup all of these so we have grad cross e plus minus one over c d dt let's add a j plus one over c d e d t and grad dot b c zero having made this transformation with this transformation still just maxwell's equations in vector form but we can now bring geometric algebra into the mix so what we're going to use is the fundamental identity of geometric algebra which says that product of two vectors is the dot product plus their wedge product and in particular we're going to write the dual form of the wedge product so that is a dot b plus i times a cross b and here i is b1 e2 e3 and we have i squared is minus one that i won't verify here today but uh it's easy to do so in particular we can go and substitute a equals grad into this so that we have grad b say is grab dot b plus i grad cross b and we see the perfect opportunity to apply that here in maxwell's equations because we have we have grad e and i grad cross e and we have grad dot bc and we have a grad cross bc so let us go and combine these pairwise equations into two gradient equations operating each on a electric or magnetic field so the first one for the electric field we have grad e is grad dot e which is r over epsilon plus i times minus one over c d bc dt now i have graph b c is energy is i times a to j plus four over c d e d t we're almost there so the only thing that we have here that is a little bit non-uniform is we have grad e and we have a derivative of e on both sides but here we have grad b and here on this side we have a i times bc we can move the i into this term let's do that so we have ibc in this equation and here we don't have an idc so we can multiply through by i to get i squared on the on the right hand side so that we have grad and i commute because i commutes with uh any grade multi any multi vectors in r3 so we have grad ivc and over here we have just minus one now let's just let's rewrite this a little one more time although we could skip steps but we have let's not all right so here's the first stage of simplifying our maxwell's equations not really well it's arguable whether it's simplifying but it is at least merging maxwell's equations we've merged maxwell's equations successively into into two equations and we see that we have the same fields involved in both and what we can actually just do is just add these two equations up so we're left now with grad plus one over c d dt times e plus i see b over epsilon minus f of j so just to verify our our terms here we had to grab e we have grad icb there's those two terms and we have one over d one over c d dt of icb and uh one over c d dt of e and lo and behold we now have a multi-vector equation for maxwell's maxwell's equation in multi-vector form we'll just tie this up one last little bit by writing f equals e plus y c b and we'll write j as rho over epsilon minus eta vector j so in this electromagnetic field what we're calling the faraday we have vector plus a tri-vector times a vector which gives us a by vector so this is a vector by vector multi-vector and our charge and current density hybrid has got the charge density and the current density as a vector and i wrote the arrow here even though i've been lazy with arrows everywhere else just to distinguish the two so now that we have a multi-vector j we distinguish it from our vector j and if we do that now our maxwell's equation is the spatial derivatives plus the time derivatives of f equals j one equation there we go now it's worthwhile to just reverse the logic here and make sure that we got everything correct so let's let's go back and do that and so if we go and verify so we want to verify that grad plus one over c dt of e plus icb is rho over epsilon minus energy alright so let's expand out the left-hand side so we have grad e plus grad icb is 1 over c d dt of e is 1 over c dt ycd and write this out a little bit uh so grab dot e let's grab wedge e plus i c grad dot p plus i c grad wedge b plus one over c d dt e and finally plus one over c dt of icb so we have here a scalar bivector here we have a scalar times trivector so that is a tri-vector this has grade three and we have a five vector times a tri-vector so that is actually a vector this one is grade one grade one and finally we have a bivector or we have a vector times the tri-vector so that is a bi-vector that is grade two so we now go and group all of our terms we should have one equation for every grade so we write out our scalar term we have grad dot e because the scalar part of rho epsilon minus n and j which is rho epsilon over epsilon and any other scalar terms no there's the zero at grade one we have y c times grad wedge b so that is i squared times c times grad cross b and we have plus one over c d e d t this equals minus energy grade two terms we have grad wedge e which is i times grad cross e 3 grade 1 grade 1 grade 2 plus 1 over c dt y c b zero finally our grade three term is just y c grad dot b equals zero all we have to do is rescale these in some cases so first case we have grad dot e this is rule over epsilon that's just gauss's law and now here i squared is minus one let's just bring these this to the other side as we normally do so we'll have and expand out that so grad cross b is minus it's bringing the dt here to the other side we have minus a minus factor but now we want another minus factor so it's a plus and we want one over c squared minus eta over c j and in this equation we will just cancel our eyes so we're left with grad cross let's see e cos minus d dt b and finally canceling ic we have ground dot b equals zero i'm just checking if i've got all my signs right and do not there should be a plus here let's see what went wrong oh okay because i divided by minus one and i didn't up the sign so that is plus now we have um originally we had a term epsilon mu u dt but that is one over c squared d dt okay that checks out and eta over c is the square root of u epsilon times the square root of mu over epsilon which is mu all right so this this guy here is just u j plus atom u d e d t which is correct all right there we go so we've gone backwards and forwards from the vector form of maxwell's equations to the geometric algebra form of microsoft's equations back to the vector form all we had to do was uh apply grade selection operations to the entire multi-vector maxwell's equation and then we get out the four traditional vector form of maxwell's equations at least for linear istropic media