[continued from previous message]   
      
   perpendicular to the surface if the integral of the magnetic field is a   
   integer multiple of 2pi.   
      
   This result too may be gussied up in fancy mathematical language.  (And   
   it's not just jargon, but crucial in understanding this problem more   
   deeply.)  The wavefunction of our charged particle on the sphere, or any   
   other surface, is not really a function, but a section of a complex line   
   bundle over the surface.  Giving such a line bundle a connection,   
   calculating the curvature, integrating the curvature over the surface,   
   and dividing by 2 pi, we must get an integer!  (Again, I'm assuming the   
   surface is compact and oriented: e.g. a sphere, torus, or donut with   
   more holes...)  This integer is called the first Chern number c_1 of the   
   line bundle.  A deeper theorem says that given a surface we can cook up   
   a line bundle with any desired c_1, and another theorem says   
   that line bundles over surfaces are completely classified by their first   
   Chern number.   
      
   Okay, now -- did you notice the following physical problem with what I   
   did?  Say we have a sphere in R^3, and a magnetic field perpendicular to   
   the sphere, such that the integral over the sphere is 2 pi n.  (As we   
   saw above n has to be an integer.)  Unless n=0, this means that   
   there is a net magnetic flux flowing in or out of the sphere -- which   
   contradicts the fact that div B = 0!  I.e., Gauss' theorem says that   
   the integral of the normal component of B over the sphere is 0, since   
   div B = 0.   
      
   Well, this didn't bother Dirac!  In fact, he came up with all this stuff   
   when he was studying magnetic monopoles!  He considered the possibility that   
   div B was nonzero in the vicinity of some monopole, but considered a big   
   sphere around the monopole, assumed div B = 0 in the vicinity of this big   
   sphere, and used quantum mechanics to show (as above) that total   
   magnetic charge inside the sphere must be an integer multiple of 2 pi!   
   I've dropped all the units, but if you throw hbar and the electron   
   charge back in correctly, you get a "quantum of magnetic charge".   
      
   There's another way around the problem, too, which mathematicians can   
   tolerate, if not physicists: just say "I'm considering a particle in a   
   magnetic field on some ABSTRACT S^2, not one sitting inside R^3, so I   
   don't need to worry about the ball "inside" my sphere."   
      
   Or, if you like, you can say, "How do you know there's not a WORMHOLE   
   inside my sphere, so that the B field can be pouring in from somewhere   
   else?"  In other words, "My sphere is a nontrivial 2-cycle in a   
   3-manifold, so it is possible for a closed 2-form (the magnetic field)   
   to have nonzero integral over it."   This is the basis for a suggestion   
   by Wheeler, that charged particles are really the mouths of wormholes,   
   and that actually the divergence of E and B are zero everywhere.   
      
   As you can see, weaseling out of this problem can take many interesting   
   forms!  That is, perhaps, the essence of mathematical physics.  :-)   
      
      
   Leatherworking   
      
   Okay, this is a ``just-for-fun'' posting on braids.   
   I had a mind-bending exploration of topology the   
   other night with a friend that I'd like to recommend to all of you.   
   Take a long thing piece of paper in it and cut two parallel long slits   
   in it like so:   
      
      _________________________________________________________________   
     |  _____________________________________________________________  |   
     |  _____________________________________________________________  |   
     |_________________________________________________________________|   
      
      
   This represents the trivial braid on three strands.  You can actually   
   get nontrivial braids by ``cheating'' and grabbing the end marked $A$   
   (below) and sliding it through one of the slits --- say the top one,   
   around the position marked $X$ (the exact position doesn't matter, of   
   course), and then pulling it back to about its original position:   
      
      _________________________________________________________________   
     |  ____________________________________________X________________  |   
    A|  _____________________________________________________________  |   
     |_________________________________________________________________|   
      
   If you do this in the simplest possible way, you will get a braid in   
   which two of the strands cross around each other twice, while the third   
   strand is not tangled with the other two --- but all the strands have a   
   360 degree twist in them now!!   
      
   (So really we are working here not with braids but ``framed braids,'' in   
   which each strand has a certain twist to it.  Framed braids also form a   
   group which has the ordinary braid group as a quotient.)   
      
   Okay, here's your puzzle.  Get your piece of paper to look like this as   
   a braid --- with no strand having any twist in it:   
      
    ____   ___   ___   ______   
        \ /   \ /   \ /   
         \     \     \   
    ____/ \   / \   / \   ___   
           \ /   \ /   \ /   
            /     /     /   
    _______/ \___/ \___/ \___   
      
   This sort of braid, where top and bottom strand take turns going   
   over the middle strand, is the typical braid found in hairdos.  Here   
   however the exact number of crossings counts.  Note two neat things about   
   this braid.  First, each strand winds up in its original position (top to   
   top, middle to middle, bottom to bottom) - i.e. its image in the symmetric   
   group is the identity.  Second, if we get rid of any one strand the   
   remaining two are unlinke (i.e. form a trivial braid on two strands).  Thus   
   it's a braid analog to the ``Borromean rings'' (three linked circles no   
   pair of which are linked).   
      
   Anyway, getting your piece of paper to look like this without any   
   cutting and pasting is a topological trick well-known to   
   leather-workers, who can make seamless leather braids this way.  My   
   friend and I were unable to make this braid except using the following   
   trick.  Grab the strands near the left (as in the first picture) and   
   braid them to look like the desired braid, ignoring the fact that near   
   the right things are getting all screwed up.  Now look at what you have   
   at the right --- the inverse braid of the one you want (no surprise,   
   since the whole braid is still the identity braid)!  While preserving the left   
   half, which is the way you want it, now use ``cheating'' moves on the   
   right half (i.e., grab the right end and slip it through the slits) to   
   kill off the unwanted junk (the inverse braid of the one you want).  You   
   can do it with three, or perhaps even just two, ``cheating moves'' --- if   
   you're clever!  You are now left with the desired braid as in the third   
   picture!   
      
   Now there has got to be a more straightforward way of doing this!   One   
   should simply be able to create the desired braid by 2 or 3 cheating   
   moves.  Unfortunately my friend and I never succeeded.  It's sort of   
   like we knew how to differentiate but not how to integrate.  But we   
   learned some interesting topology in the process --- and that's what   
   counts!  So I strongly recommend that everyone make a 2-slitted strip of   
   paper (leather would be better) and see what kinds of framed braids they   
   can make.  There is clearly an interesting sort of group lurking here:   
   the subgroup of framed braids that can be generated by ``cheating moves''.   
   I am sure that topologists have figured this stuff out already, but it's   
   more fun to mess with it yourself in this case.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)