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The rabbi's copout



Once again David Bowman has stirred my thoughts. I'm working on a
long answer to John's and David's comments, and while doing so I
suddenly understood (I think) the physical basis for the surface
at infinity term. Please let me try my physical argument out on
you. Since it is less threatening to do so, I will work entirely
in the electrostatic case. As David points out, there are no real
formal differences. My approach will be more elementary than
David's, but I will let him judge whether I have caught the idea
or not.

Consider an infinite simple cubic array of lattice constant a of
identical particles, each of which carries charge Q. For
purposes of discussion I will consider one of these charges to be
located at the origin (0,0,0) of cartesian coordinates (x,y,z).
The charges are located at (la,ma,na), where l, m, and n span all
the integers of both signs between plus and minus infinity.

I will now show that two good physical arguments can appear to be
in conflict, and the resolution of the conflict is surface at
infinity terms!

First, consider a cubical surface centered on the origin which
has edges of length a parallel to the axes. This surface bisects
the space between nearest neighbors and has a name in solid state
theory which I forget for the moment. Let us use Gauss's law to
calculate the electric field flux outward through this surface.
Easy! the net flux must be

Q
Phi = -----
E eps0

By symmetry we can infer the normal component of the electric
field averaged over each square face:

_ Q
E = -----------
^ 2
6 eps0 a

Right? Right!

Gee, maybe we can see this another way, using symmetry and
Coulomb's law instead. Consider the horizontal surface z = a/2.
One of our cube faces from the previous problem lies in this
surface. Let's find the normal component of the electric field
on that cube face, a daunting problem with an infinite number of
Coulomb terms. But symmetry comes to our rescue! For each charge
at (la,ma,na) where z > a/2 there is a symmetrically disposed
charge at (la, ma,(-n+1)a) which has a normal electric field
component that exactly cancels the normal field component due to
the first charge. It is clear that the normal component of the
electric field is zero everywhere on this surface.

E = 0
^

The six faces of the cube are indistinguishable, so to find the
total flux out of the cube we must multiply zero by six. Hooray,
we did it another way!

Right? (Erk!)

We have a contradiction. Which, if either, of these answers is
correct? Like Sholem Aleichem's wise rabbi, I'll have to say
that *both* of them are right, because the *problem* is wrong.
Specifically, the hypothesis of the possibility of the existence
of an infinite simple cubic lattice is wrong for analysis at
this degree of precision. We should expect to obtain conflicting
answers if the model does not contain the physical boundaries of
the system.

Let's consider a less infinite problem (and one which is chosen
to give the correct answer). Let's discard all charges more than
one million lattice constants from the xy plane. This still
"pretty infinite" at the locus of our pet charge at the origin,
and symmetrical. If we consider the surface at z = a/2 which
corresponds to our cube face, however, the environment is stll
pretty infinite, but it is no longer symmetrical. In addition to
the charges that gave us zero contributions to the normal field
there is one planar array of charges in excess located at the
points (la, ma, - 1000000 a). The surface charge density on this
plane averages Q/a^2, and at the distance of the origin it will
look pretty much like a uniformly charged sheet with that surface
charge density. The resulting electric field at z = a/2 will be

Q
E = -----------
^ 2
2 eps0 a

By symmetry I see the electric field will be of equal magnitude
and pointed downward on the surface at z = - a/2. The symmetry
in the other four faces of the cubical Gaussian surface is
complete, so the normal component of the electric field vanishes
on those surfaces. (Please don't ask me to chop my model in the
x and y directions too. No new physics will be learned if I do.)
With complete information we can test the gaussian compliance of
our Coulomb calculation and, voila, it works!

When one passes to the Palmerian limit (1,000,000 => infinity)
in this problem nothing changes in the analysis. That, I suspect,
is what is meant by the surface at infinity term - physically.

There is a calculation I do when teaching my astrophysics course
in which the gravitational collapse time for a uniformly dense
sphere of dust is calculated. One starts with a hypothetically
infinite, homogeneous, isotropic dust universe. Nothing happens
because the gravitational field at each dust particle has
indeterminate direction. One imageines a small modification to
this system. A tiny gap in the form of a spherical shell is
introduced, hypothetically, breaking the former homogeneity. Now
the spherical portion inside the gap collapses under its own
self gravity, oblivious of the sperically symmetric universe
around it because there's no field inside a cavity in the center
of a spherically symmetric mass distribution. This makes it
apparent that very small fluctuations in gravitating systems can
produce condensations. The system is highly unstable.
(Incidentally, in my version of this collapse the rest of the
universe collapses outward. There is no pretense that this is a
real situation.)

Leigh