View source for MATLAB/Divergent series investigations/Casimir effect

{{Original research}}

==Introduction==
While the template labels this as ''Original research'', it is '''amateur''' original research.  The greatest theoretical physicists on this planet have thought deeply about the '''[[w:Casimir effect|Casimir effect]]''' for many decades.  Our goal is to make it understandable to more people.  It is ''Research''  because the outcome is unknown. The method proposed here for calculating the Casimir effect may prove entirely wrong. 

*This investigation was inspired by this ''Numberphile''  Youtube video: '''[http://www.youtube.com/watch?v=w-I6XTVZXww ASTOUNDING: 1+2+3+...]'''

But, if this method is correct, '''you''' can be the first to do it.  All you need is a simple computer program that can add perhaps 100,000 terms of a '''[[w:divergent series]]'''.  The code follows that already posted on Wikipedia's '''[[MATLAB/Divergent series investigations]]'''.  In that code, a trick was employed to convert the obviously infinite (divergent) series,

 1 + 2 + 3 + 4 + ...

into a series that converges to a number very close to -1/12:

 1.0 + 2.0 + 2.9 + 3.8 + bigger numbers +...+ smaller numbers +...+ negative numbers +...+ very small numbers

Only this time, we need the triple sum to include all  the waves that can exist inside a rectangular box.  Each wave has positive and well-known "ground state" energy of <math>\frac 1 2\hbar\omega</math>.  Somehow these energies need to add up to a negative number (...because an infinite number of waves are present ''! ?'' ).

==From the wikipedia article==
The Wikipedia article '''[[w:Casimir effect]]''' shows how the vacuum energy of a box is infinite, and proportional to the following divergent series, ''S'':

{{cot|Algebraic steps}}
Following Wikipedia's '''[[w:Casimir_effect#Derivation_of_Casimir_effect_assuming_zeta-regularization|Derivation of Casimir effect assuming zeta-regularization]]''' we consider a box of area A and thickness a<<A. The energy density per unit area, ''<E>/A'', is given by the following expression in the limit  <math>s\to 0</math>:

:<math>\frac{\langle E(s) \rangle}{A} = 
\frac{\hbar c^{1-s}}{4\pi^2} \sum_n \int_0^\infty 2\pi qdq
\left \vert q^2 + \frac{\pi^2 n^2}{a^2} \right\vert^{(1-s)/2},</math>

So as to not get our '''[http://www.youtube.com/watch?v=w-I6XTVZXww "knickers in a twist"]''' with [[w:zeta function regularization]] and the [[w:Riemann zeta function ]], we set s=0 exactly

:<math>\frac{4\pi^2\langle E \rangle}{ 2\pi\hbar c A} = 
\sum_n \int_0^\infty qdq
\sqrt{ q^2 + \frac{\pi^2 n^2}{a^2} }\,</math>

We may further refine the integral with the change of variable to <math>x</math> where,

: <math>q = \frac \pi a x</math> 

:<math>\frac{4\pi^2\langle E \rangle}{ 2\pi\hbar c A} = 
\sum_n \int_0^\infty  \left(\frac \pi a x\right)d \left(\frac \pi a x\right)
\sqrt{  \left(\frac \pi a x\right)^2 + \frac{\pi^2 n^2}{a^2} }
</math> = <math>\left(\frac \pi a \right)^3
\sum_n \int_0^\infty xdx
\sqrt{ x^2 + n^2}</math>
{{cob}}

:<math>\frac{4\pi^2}{ 2\pi\hbar c A}\left(\frac a \pi\right)^3 \langle E \rangle= 
\sum_n \int_0^\infty xdx
\sqrt{ x^2 + n^2}</math>

Hence we seek

:<math>S=\sum_n \int_0^\infty  qdq
\sqrt{ q^2 + x^2 },</math>

As was done to '''[[MATLAB/Divergent_series_investigations|1+2+3+4+...]]''', we "tame" this series with a damping factor, and later take <math>\epsilon \to 0</math>  and <math>\delta\to 0</math> along a path that yields a finite answer for the real part of the series.

==matlab code==
{{cot|see matlab code for these calculations (casimirSumIntegral.m)}}
<pre>


</pre>
{{cob}}

==Tame the infinity==
In a '''[MATLAB/Divergent series investigations|previous exercise]''' the [[w:divergent series]] [[w:1+2+3+...|1+2+3+...]] was "tamed" by introducing a damping term into an integral approximation to the series.  Then, when the tam

:<math>\tilde S=\int_0^N dn\int_0^Xxdx \ e^{(-\epsilon+i\delta)\sqrt{ x^2 + n^2}}\;
\sqrt{ x^2 + n^2}</math>

Setting <math>\Delta=-\epsilon+i\delta</math>, we have,

:<math>\tilde S=\int_0^{N} dn\int_0^Xxdx \ e^{\Delta\sqrt{x^2 + n^2}}\;
\sqrt{ x^2 + n^2} </math>

which is easily converted into a single integral by a sort of "spherical coordinates" with ''{{nowrap begin}}r<sup>2</sup> = x<sup>2</sup> + n<sup>2</sup>{{nowrap end}}'', ''{{nowrap begin}}dxdn=rdrd&theta;{{nowrap end}}'', and ''{{nowrap begin}}x=rcos&theta;{{nowrap end}}''.  After integrating over ''&theta;'' from 0 to ''&pi;/2'', we have:

:<math>\tilde S=\int_0^{\infty} r^3 \ e^{\Delta r}\;dr = \frac{6}{\Delta^4}</math>

The real part vanishes if <math>\Delta^4</math> is imaginary.  Denoting the real and imaginary parts of <math>\Delta = \Delta ' + i\Delta ''</math>, we have,for example 

:<math> \Delta ' = -cos\frac{\pi}{8} \ \mathrm{and} \ \Delta ''=sin\frac{\pi}{8}</math>

==On the physical nature of these equations==

Even though we don't have an answer, our equations are making sense, especially if we focus attention on the real parts of <math>S</math> and <math>\tilde S</math>:

:<math>S(s)=\sum_{n=1}^\infty \int_0^\infty xdx \ 
cos \left(s\Delta ' \sqrt{x^2+n^2}  \right)
\;
e^{-s\Delta ''\sqrt{x^2+n^2} }
\sqrt{ x^2 + n^2}</math>


:<math>\tilde S(s)=\int_0^\infty dn\int_0^\infty xdx \ 
cos \left(s\Delta ' \sqrt{x^2+n^2}  \right)
\;
e^{-s\Delta ''\sqrt{x^2+n^2} }
\sqrt{ x^2 + n^2}</math>

Even though <math>\tilde S(s)=0</math> for all <math>s</math>, the expression for <math>\tilde S</math> contains large terms.  The expression for <math>S</math> also contains large terms.  Moreover, in the limit <math>s\to 0</math>, the contributions to both <math>S</math> and <math>\tilde S</math> come from large values of <math>n</math> and <math>x</math>.  Large values of <math>n</math> and <math>x</math> correspond to high frequency (short wavelength).  Yet, the Casimir effect is dominated by the lowest frequencies (longest wavelengths) that the box can contain , as can be seen by the following manipulation of these expressions:

==Highlighting the importance of low-order modes==
&nbsp; Since <math>\tilde S(s)=0</math>, we have <math>S(s)=S(s)-\tilde S(s) =</math>

<math>\sum_{n=1}^\infty \int_0^\infty xdx \ 
cos \left(s\Delta ' \sqrt{x^2+n^2}  \right)
\;
e^{-s\Delta ''\sqrt{x^2+n^2} }
\sqrt{ x^2 + n^2} </math> <math> - </math>  <math>
\int_0^\infty dn\int_0^\infty xdx \ 
cos \left(s\Delta ' \sqrt{x^2+n^2}  \right)
\;
e^{-s\Delta ''\sqrt{x^2+n^2} }
\sqrt{ x^2 + n^2}</math>

The two terms in this expression are identical, except that one contains an integral over n, while the other is a sum.  We can emphasize this by defining an operator that acts on any function f:

:<math>\left\{  \sum_{n=1}^\infty \ - \ \int_0^\infty dn\right\}f(n)\equiv \sum_{n=1}^\infty f(n) -  \int_0^\infty f(n)dn</math>

In our case,

:<math>f(n) = \int_0^\infty xdx \ 
cos \left(s\Delta ' \sqrt{x^2+n^2}  \right)
\;
e^{-s\Delta ''\sqrt{x^2+n^2} }
\sqrt{ x^2 + n^2}</math>

If it can be established that this operator tends to suppress high order modes, the following scenario explains the Casimir effect: Empty space is occupied by an infinite number of ground state energy levels, and the shorter wavelengths contain most of the energy.  A small metal box also contains an infinite amount of energy, with the shorter wavelengths containing most of the energy.  The energy levels in the box are quantized (discrete), but the spectra of the box and empty space differ only at the lower frequencies.  Hence, while the higher energy modes contain most of the energy, the difference in energy density between as small box and empty space is dominated by the lowest energy modes.

Unfortunately, f(x) is large and unpredictable.  I need to reverse the operations?