🚀 go-pugleaf

 Mark Palenik: 
 >I looked at his site, and found formula 26 on p. 19
 >
 >it is:
 >
 >phi = q/(4*pi*EpsilonNaut)*(e^(-j*k*r1)/r1 - e^(-j*k*r2)/r2)*e^(j*w*t)
 >
 >My guess is, that if he sais curl(grad(phi)) =/= 0, he messed up
 >something with the jacobian.  Of course, you could take
 >curl(grad(phi)) without the jacobian, which would equal zero, but if
 >we want to be physical about it, we'd have to apply it.  My guess is
 >that the math was a bit too tricky for him and he screwed up
 >somewhere.
 
  That usually happens when someone decides to go through a lot of
tedious arithmetic for something that can be solved by inspection.
I can choose cartesian coordinates, and since those are simplest,

  \nabla x (\nabla\phi) = \epsilon_ijk \nabla_i\nabla_j\phi

                        = (\nabla_i\nabla_j - \nabla_j\nabla_i)\phi
           
                        = 0, since partial derivatives commute

  Since it appears that r1 and r2 are variables (and different variables),
if one wished to evaluate that function by the method of tedium, one only
need evaluate one term, say, \exp(jk.r1)/r1, since the other is identical.

 >I'm also confused as to what j and k are.  Are they variables? 
 >Constants?
 
 As near as I can tell, k is the wave vector (i.e., k.r) and j is
is the imaginary unit (engineers tend to use j rather than i as
sqrt(-1) with j = -i).

 >In any event, curl(grad(phi)) does equal zero, no matter how you look
 >at it, if you do the calculations correctly.
 
  Right. Sergey has been claiming the function above is a counter example
to that mathematical identity.