Casimir effect for metal plates

Casimir (1948) :

There is an attraction between two metal

plates which is due to the perturbation of

the photon vacuum produced by metal

plates . <br />The force per area is :     f = - π^2/240c/a^4

Introduction

Assume there are two parallel metal plates (in the x-y plane) separated by a distance a in z-direction. In between the plates the photon momentum is quantized in the perpendicular direction. Energy and number of photon modes  between the plates differ  from their vacuum values

Sin[k_z a] = 0 ; <br />k_z = (π n)/a ; <br />n = ? 1, ? 2 ... (1)
RowBox[{RowBox[{RowBox[{RowBox[{Cell[], ω}], , =, , (k^2 + ((π n)/a)^2)^(1/2)}], ;}],    , }] (2)
k = (k_x, k_y) - parallel momentum ; (2)

Energy and force

(dE/dV) _vac = 2∫^2k/(2π)^2∫k_t/(2π) 1/2ω = ∫^2k/(2π)^2∫k_t/(2π) ω ; (3)

2comes from two polirizations of photons . 1/2ω is the contribution of each photon mode ... 1/2) . Deriving (dE/dV) _vacconventional running waves boundary conditions are assumed .

(dE/dV) _met =    1/(2a) Underoverscript[∑, n = -∞, arg3] ∫ ^2k/(2π)^2 ω ; (4)

Underoverscript[∑, n = -∞, arg3] is the summation over the discrete perpendicular momentum . For

running waves the summation reads  1/aUnderoverscript[∑, n = -∞, arg3] ... eeded . Each standing wave corresponds to two running waves, thus giving  a factor of

two . This fact needs to be compensated for by the explicit

factor 1/2 in front of the summation, resulting in  1/(2a) Underoverscript[∑, n = -∞, arg3] . 

Renormalization

(dE/dV) _Ren = (dE/dV) _met - (dE/dV) _vac =   ∫ ^2k/(2π)^2 (1/ ... (π n)/a)^2)^(1/2) - ∫_ (-∞)^∞k_t/(2π) (k^2 + k_t^2)^(1/2)) ; (5)
Scale   k_t, k_tν,   k_t = νπ/a (6)
(dE/dV) _Ren = 1/(2a) ∫ ^2k/(2π)^2 (Underoverscript[∑, n = -∞ ... n)/a)^2)^(1/2) - ∫_ (-∞)^∞ν (k^2 + ((π ν)/a)^2)^(1/2)) (6)
Simplify notation, identifying in the integral ν = n (7)
(dE/dV) _Ren = 1/(2a) ∫ ^2k/(2π)^2 (Underoverscript[∑, n = -∞ ... ^2 (Underoverscript[∑, n = -∞, arg3] -∫_ (-∞)^∞n) ω ; (7)
ω = (k^2 + ((π n)/a)^2)^(1/2) (8)
Energy per area A is (9)
E ≡dE/dA = adE/dV = 1/2∫ ^2k/(2π)^2 (Underoverscript[∑, n = -∞, arg3] -∫_ (-∞)^∞n) ω ; (9)
Force per are (10)
f = -∂E /∂a = 1/2∫ ^2k/(2π)^2 (Underoverscript[∑, n = - ... ∞)^∞n) (π n)^2/a^3 1/ω       (10)
= 1/(8a^3) ∫ ^2k (Underoverscript[∑, n = -∞, arg3] -∫_ (-∞)^∞n)   n^2/ω ; (10)

Scaling   f  =-Const/a^4

^2k = 2π k k = π   ( k^2) ;  (11)
f = π/(8 a^3) Underoverscript[∫, 0, arg3]    (k^2) ( Underovers ... ipt[∑, n = 1, arg3] -Underoverscript[∫, 0, arg3] n)   n^2/ω ; (12)
k^2 = (π/a)^2 ξ ;     ω = (k^2 + ((π n)/a)^2)^(1/2) = π/a (ξ + n^2)^(1/2) ; (13)
f = π/(4a^3) (π/a)^(2 - 1) Underoverscript[∫, 0, arg3]   & ... = 1, arg3] -Underoverscript[∫, 0, arg3] n)   n^2/(ξ + n^2)^(1/2) ; (14)
f = -Const/a^4 ; Const = π^2/4Underoverscript[∫, 0, arg3]    ... arg3] n - Underoverscript[∑, n = 1, arg3] )   n^2/(ξ + n^2)^(1/2) ; (15)

Calculation of Const

Cutoff with the help of Gamma function

Γ (n) = Underoverscript[∫, 0, arg3] τ^(n - 1) Exp[- τ ] τ ;  (16)
τ = z t ;  (16)
Γ (n) = z^nUnderoverscript[∫, 0, arg3] t^(n - 1) Exp[- z t ] t ;  (16)
1/z^n = 1/(Γ (n)) Underoverscript[∫, 0, arg3] t^(n - 1) Exp[- z t ] t ;  (16)
1/z^(1/2) = 1/π^(1/2) Underoverscript[∫, 0, arg3] Exp[- z t ] t/t^(1/2) ; ... sp;       (  Γ (1/2) = π^(1/2)   ) ; (16)
1/(ξ + n^2)^(1/2) = 1/π^(1/2) Underoverscript[∫, 0, arg3] Exp[- (ξ + n^2) t ] t/t^(1/2) ; (17)
Const = π^2/4Underoverscript[∫, 0, arg3]   ξ ( Underoversc ... arg3] n - Underoverscript[∑, n = 1, arg3] )   n^2/(ξ + n^2)^(1/2) ; (18)
Underoverscript[∫, 0, arg3]   ξ  Exp[- ξ t] = 1/t ; (19)
Const = π^2/41/π^(1/2)    Underoverscript[∫, 0, arg3]  ... ] n - Underoverscript[∑, n = 1, arg3] )   n^2  Exp[- n^2 t ] ; (20)
Const = π^(3/2)/4 Underoverscript[∫, 0, arg3] t/t^(3/2)   f[t] ; (21)
f[t] = (Underoverscript[∫, 0, arg3] n -   Underoverscript[∑, n = 1, arg3] )   n^2  Exp[- n^2 t ] ; (21)

Numerical calculations

Const = π^(3/2)/4 Underoverscript[∫, 0, arg3] f[t]/t^(3/2) t ; (22)
f[t] = ( Underoverscript[∫, 0, arg3] n - Underoverscript[∑, n = 1, arg3] )   n^2  Exp[- n^2 t ] ;  (22)
Underoverscript[∫, 0, arg3] n  n^2  Exp[- n^2 t ] = -ͦ ... , arg3] n Exp[- n^2 t ] = -∂/∂t1/2π/t^(1/2) = π^(1/2)/(4t^(3/2)) ; (22)
f[t_] := π^(1/2)/(4t^(3/2)) - Underoverscript[∑, n = 1, arg3]   n^2  Exp[- n^2 t ] ; (23)
3/2 Plot[ ... t, .3, 10}, FrameTrue, FrameLabel {t, f[t]/ t }, DefaultFontTimes[20]] (23)

[Graphics:HTMLFiles/Casimir-NoColor_48.gif]

Figure 1

⁃Graphics⁃

Const = π^(3/2)/4Underoverscript[∫, 0.05, arg3] f[t]/t^(3/2) t//N = 0.0411233 ; (24)
Analytical calculation (presented below) gives (25)
Const = π^2/240 (25)

(240/π^2) Const

1.

Analytical calculation

Const = π^(3/2)/4Underoverscript[∫, 0, arg3] t/t^(3/2)    (Unde ... -   Underoverscript[∑, n = 1, arg3] )   n^2  Exp[- n^2 t ] ; (26)
Const = π^(3/2)/4Underoverscript[∫, 0, arg3] t/t^(3/2)    (-d/d ... 1/2) -   Underoverscript[∑, n = -∞, arg3]    Exp[- n^2 t ]) (26)

Poisson summation formula

Let (27)
Overscript[f, ~][k] = Underoverscript[∫, -∞, arg3] x f[x] Exp[ k x] (27)
Then the Poisson summation formula is valid (28)
Underoverscript[∑, n = -∞, arg3] f[n] = Underoverscript[∑, m = -∞, arg3] Overscript[f, ~][2π m] (28)
Example (29)
f[x] = Exp[-t x^2]  (29)
Overscript[f, ~][k] = Underoverscript[∫, -∞, arg3] x f[x] Exp[-t x^2 +  k x] = π/t^(1/2) Exp[-k^2/(4t)]  (29)
Underoverscript[∑, n = -∞, arg3] Exp[-t n^2] = π/t^(1/2) Underoverscript[∑, m = -∞, arg3] Exp[-(π^2m^2)/t] (29)
Using the Poisson summation one gets (30)
Const = π^(3/2)/8Underoverscript[∫, 0, arg3] t/t^(3/2)    (-d/d ... 721;, m = 1, arg3] 1/(π^2m^2)^2 = 3/(8π^2) Underoverscript[∑, m = 1, arg3] 1/m^4 (30)
Const = 3/(8π^2) Underoverscript[∑, m = 1, arg3] 1/m^4 ; (31)
Underoverscript[∑, m = 1, arg3] 1/m^4 = ζ[4] = π^4/90 ; (32)

Zeta[4]

π^4/90

Const = 3/(8π^2) π^4/90 = π^2/240 ; (33)
Const = π^2/240 (34)
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