## Sunday, September 12, 2004

On Sep 12, 2004, at 6:19 PM, main_engineering wrote:

Jack: Now you can do the problem for two concentric spheres and see if it is
topologically equivalent to the cylinder problem with the movable
piston inside.

Thank you! I'll work on it.

Actually I did it in last message - same story.

Milonni uses a rectangular cavity in chapter 2, sec. 2.7 to derive the
Casimir force. It is much easier than your cylinder with a piston. Where is
the error in his calculation?

His calculation is overly-complicated with lots of fudge factors. I am leaving for a week and will look at it in detail when I get back. Oh! Aha! I see! He asks the wrong question. Milonni does not include the modes outside the plates. Rather he uses

U(x) - U(infinity) where x = separation between the plates.

This is the conceptual error since what he should do is compare

-dU(x)/dx with -dU(L - x)/dx at x.

That is, the correct thing to calculate is not

U(x) - U(infinity)

But is

U(x) + U(L - x)

With the total ZPF force at x from both sides

-(d/dx)[U(x) + U(L - x)] = 0

"The Question is: What is The Question?" John A. Wheeler

The exact FREE PHOTON FIELD calculation with no real on-mass-shell charges, i.e. interaction j.A = 0, has

U(x) linear in x and U(L - x) linear in L - x with the SAME coefficients!

I now use 3 plates one at 0, one at x and one at L, and look at the ZPF forces on the plate at x from the virtual photons between 0 and x (inside region) and between x and L (outside region).

I compute U(x) and U(L - x) exactly from a finite series and I find that

F(from between the plates on one plate) = dU(x)/dx = -dU(L - x)/dx = - F(from outside the plates on that same one plate).

Therefore, the net ZPF force (not to be misidentified with the Casimir force) on the given plate from FINITE NUMBERS of both the inside and outside free virtual photon discrete cavity modes adds to zero exactly.

Also by exact counting of FINITE number of modes

U(x) = A(hc/a^2)x

U(L - x) = A(hc/a^2)(L - x)

Where a is the short-wave cutoff and L is the long-wave cut off.

Since the total ZPF force adds to zero exactly on each plate, it does not matter that the separate contributions to the ZPF force go to infinity as the short-wave cutoff goes to zero. The long-wave cutoff L is irrelevant to the ZPF force on the plate at x.

The VdW Casimir force BETWEEN TWO NEUTRAL MOLECULES is an entirely different problem. What you have are sheets of these molecules making up the plates and you need to integrate over the plates to wind up with the Casimir force

Note that Milonni computes

Total Casimir Force = - (pi^2hc/240x^4)(AREA OF PLATES)

In contrast A. Zee computes for this same problem

pihc/24x^2 with no area dependence!

Therefore, the two mainstream pundits do not even agree with each other!

Probably Milonni's is closer to experiment - but the techniques used in both cases are wrong.

The only right way to do it is to explicitly compute the VdW force using the actual electron-photon interaction. You cannot do it correctly using only free photon quantum field theory without any j.A coupling - that is my main point.

My claim is that when you do use ONLY free photon quantum field theory without any explicit j.A coupling then the net ZPF force is exactly ZERO and it is not the VdW Casimir force that has a qualitatively different origin.

He too sums the modes within a cavity of volume V = d*L^2, where L^2 is the
area of "plates" at the ends of the cavity and d is their separation, to
find the energy. He simply takes the difference in the energies contained in
the cavity, between when d is large and when d is small and then takes the
gradient wrt d. Pressure exerted from outside is NOT even included in this
problem. It deals solely with the allowed frequency modes contained in the
cavity as a function of d.

If I use his calculation and consider a cube (square cylinder) of length L
with a movable piston inside, then I get the energies,

U(d) = -((pi^2*hc)/(720*d^3))*L^2 (Milonni 2.108)

U(L-d) = -((pi^2*hc)/(720*(L-d)^3))*L^2

The force per unit area on the piston does not cancel out!

F(d) = (pi^2*hc)/(240*d^4)

F(L-d) = -(pi^2*hc)/(240*(L-d)^4)

His method is independent of the high freq. cutoff where yours is not.

The only difference between this and your original piston in a cylinder is
that the cross section of the cylinder is square instead of round. The force
is due to a negative energy density inside (fewer allowed modes inside), not
a positive pressure from outside.

main_engineering