[BUG] curl question

[Matthias Kawski]

Kimberly,

here is an idea for a fun analysis -- which explains the very
delicate balance which makes the paddle wheel "irrotational"
in the "perfect case" of the exponent being -1.

The curl of a vector field in cylindrical coordinates

F(r,theta,z) = Fr rhat + Ftheta thetahat + Fz zhat

is

(curl F) = (1/r d/dtheta Fz - d/dz Ftheta) rhat
         + (d/dz Fr - d/dr Fz) thetahat
         + 1/r ( d/dr(r*Ftheta) - d/dtheta Fr) zhat

In our case the field is parallel to the xy-plane, i.e. Fz=0,
and it is independent of z and theta ('rotaionally invariant).
In the special case of

  F = r^alpha thetahat

the curl is easily computed to be

(curl r^alpha thetahat) = (1/r) d/dr ( r^(alpha+1))
                        = (alpha+1)r^\alpha zhat

In particular, for the magnetic field alpha=-1 and the curl
vanishes away from zero. But for alpha larger or smaller than
-1 one obtains "positive" or "negative" curl, i.e. the paddle-
wheel may rotate either way about its own axis as it floats
around the origin.

In the vector field anal;yzer simply add an exponent like
^(3/2) or ^(3/4) or ^(1/2) into the formula, e.g.

-y/(x^2+y^2)^(3/2)*(1-box(16*(x^2+y^2)))
            ^^^^^

and the animnations for slows (set to show "curl", top buttons)
will nicely animate the rotations in either direction.
Indeed going even closer, to say exponents 0.9 and 1.1 etc, one
obtaisn vector fields that are virtually indistinguishable to
the eye, but therotating boxes nicely show the imperfections.
{{This is completely analogous to the con/divergence of the
integrals of 1/x^p for p<1, p=1, p>1 which look same, but
analysis shows the difference)).


Extension:
I used to bring sliced wine-corks (half colored) and a round
tub of water to my classes. We brought the water into a swirling
motion, and then checked what the corks did: The result was neat,
near the outside they spun backwards due boundary layer effects,
near the center was a deep funnel and the corks spun forward.
Here the idea is to replace the velocity field by more general
dependence on r, i.e. the field if  F = f(r) thetahat. Basically
the same formula as above shows that for general monotone f(r)
one always will find a distance at which the corks to not spin!

But the real explanation is via line integrals:
Draw regions in the plane that are bounded by arcs of circles and
rays from the origin (on the FLOW panel, use the Flow-Equipot
option which will draw such contours as long as they are small.
Check the "show circ" box!
Again for expoenents such as 1/2, 0.9, 1, 1.1, 3/2 study the
line integrals around the contours -- it is a very subtile
cancellation of the work along the innner and outer arcs
that makes the integrals vanish.
Clearly the lengths of these racs are proportional to r. Hence
they will cancel if and only if the field strength is exactly
inversely proportional to r. In all other cases the curl is either
positive or negative, and the "tug" at inside or at outside is
bigger for the rotating solid.

The vanishing of the integral in the magnetic field is one of
the big conservation laws -- which I consider on a similar
level as conmservation of energy, mass, momentum.

Enjoy!

Matthias
**********************************************************
Matthias Kawski                http://math.asu.edu/~kawski
Dept. of Mathematics and Statistics         kawski at asu.edu
Arizona State University            office: (480) 965 3376
Tempe, Arizona 85287-1804           home:   (480) 893 0107
**********************************************************

On Fri, 30 Jul 2004, Kimberly J. Burch wrote:

> Dear Tevian,
>
> I was hoping you could clear something up.  We are trying to look at the
> curl of a vector field using the paddle wheel argument that we learned at
> the workshop.  However, the following vector field is giving us trouble...
>
> F(x,y)= y/(x^2+y^2) i - x/(x^2+y^2) j
>
> If you plot the vector field in Maple you see the vectors are longer near
> the origin (and decrease in length as you move away from the origin).  If
> you place a paddle wheel in the flow it would seem that since the arrows
> are longer at the bottom side of the paddle wheel that the paddle wheel
> should spin counter-clockwise.  But, curl=0 by computation!
>
> One thought we had was that the paddle wheel does turn counter-clockwise
> but the clockwise flow of the water negates this motion so that the paddle
> wheel does not spin.  I am disappointed if this is the case. If we want
> to teach the students a model of how to intuitively understand the curl,
> this example seems to break the model.
>
> Sorry to bother you with this.  If you are too busy please feel free to
> ignore this.  Between 5 or so professors here you would think we could
> come up with a good explanation and yet we all think it should spin!
>
> -Kimberly
>
> -------------------------------------------------------------------------------
> Dr. Kimberly Jordan Burch, Assistant Professor
> Department of Mathematical Sciences
> 202 Richardson Hall, Montclair State University
> Montclair, NJ 07043
> Phone: (973) 655-5184
> http://www.csam.montclair.edu/~burch
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