Is there a physical method to prove for example when the zeta regularization of a series

$$ 1+2^{k}+3^{k}+............= \zeta (-k) $$

gives the correct result: Casimir effect, vacuum energy and when does it 'fail'?

For example we plug the zeta regulator inside the renormalization group equations and this group tells us if zeta regularization fails or give exact results.


I walk through a physical argument similar to what Trimok has shown in my blog article Ramanujan and the Casimir Effect. You basically assume that each available classical e-m mode is filled to exactly the level of one-half of a "quantum", and put a small box inside one twice as big. The gist of the argument goes like this (quoting from my article):

We simplify things a little by putting everything in a one-dimensional box. So the energy modes are 1,2,3, etc. We can choose our units so that the pressure is numerically equal to the energy.

Now make the box twice small. The energy modes are 2,4,6...etc. But the PRESSURE is energy per unit volume (length, since it's one-dimensional)...so the pressures are (get this:) 4,8,12... What's the difference in pressure between the big box and the small box?

1 + 2 + 3 .... - (4 + 8 + 12...)

which is...1 - 2 + 3 - 4 + 5 .... = 0.25!

To get the actual Casimir effect, you then put the big box in yet a bigger box. You get a pressure difference equal to one-quarter of what you got in the first set of boxes. Continuing this process gives you a chain of boxes from which you can calculate the pressure relative to infinity.

  • $\begingroup$ and why simply keep always :) the FINITE part and throw the INFINITIES away ?? in this case zeta regularizatio would give the correct answer $\endgroup$ – Jose Javier Garcia Aug 3 '13 at 20:07
  • $\begingroup$ Thanks, Jose. I think this is the first time I've been picked as the winning answer. I don't know what zeta regularization is but I like to create a chain of physical logic to justify the math. That's the only way I have confidence that the math is giving me the right answer. $\endgroup$ – Marty Green Aug 3 '13 at 20:33

With casimir effect, you could modify the though experiment so as to avoid infinities (while using renormalization).

For instance, instead of considering $2$ (conducting) plates $D_1,D_2$ with distance $d$ apart, you can consider 4 plates $D_0,D_1,D_2,D_3$, with distance $(D_0,D_1) = (D_2,D_3) = L $. After having renormalised modes, for instance, in one dimension, you will have, for energy between two plates due to internal modes :

$$E(\epsilon, l) = \frac{1}{2} \sum_{n=1}^{+ \infty} \frac{n \pi}{l}~e ^{- \large \epsilon ~(\large \frac{n \pi}{l})}$$

The force between the 2 plates $D_1,D_2$, due to internal modes, is $f(\epsilon,l) = - \frac{\partial E(\epsilon,l)}{\partial l}$

To calculate the total force between the 2 plates $D_1,D_2$, you have to substract the force due to the internal modes $D_0,D_1$ or $D_2,D_3$, so the total energy is :

$$f_{CASIMIR}(d, L) = lim_{\epsilon \rightarrow 0} ~~(f(\epsilon, d) - f(\epsilon, L))$$

The term $f(\epsilon, d) - f(\epsilon, L)$ does not contain infinite terms in $\frac{1}{\epsilon}, \frac{1}{\epsilon^2}, etc...$, so there is no problem.

No, we are going to send the plates $D_0,D_3$ at infinity, this is the limit $lim_{L \rightarrow + \infty}$ . This means simply that these plates , in some sense, don't exist any more.

So, finally :

$$f_{CASIMIR}(d) = lim_{L \rightarrow + \infty} ~~ lim_{\epsilon \rightarrow 0}~~ (f(\epsilon, d) - f(\epsilon, L))$$


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