The Casimir force between conducting plates

SI units throughout; the metric signature would be (+, −, −, −), though no four-vector is needed here. Put perfectly conducting plates of area A = L² at z = 0 and z = d, with L ≫ d and periodic conditions in x and y. At each plate the transverse field of topic 11.03, quoted, must have zero tangential E and zero normal B, so its z-dependence is sin(nπz/d) or cos(nπz/d) with k_z = nπ/d and ω = c√(k∥² + n²π²/d²): derived. For n ≥ 1 two polarisations survive, TE with E parallel to the plates and TM with E in the plane of k and the normal. For n = 0 only TM does, a field E_z uniform across the gap; a TE field uniform in z would be tangential at the plates and must vanish. Each family holds A d²k∥/(2π)² modes. For d = 1.00 µm the n = 1 family starts at πc/d = 9.42 × 10¹⁴ s⁻¹, a free-space wavelength of 2d = 2.00 µm, yet the n = 0 family carries every longer wavelength too.

Each mode is an oscillator with ground-state energy ½ħω, quoted from Unit 9. Summing over the modes above, with s = √(k∥² + n²π²/d²) as the variable, the gap holds (ħc/4π)Σ′ₙ F(nπ/d) per unit area, where F(κ) = 2∫ s² f(s/K) ds from κ to ∞ and Σ′ gives n = 0 half weight for its single polarisation: derived. The comparison is free space in the same volume A d, (ħc/4π)(d/π)∫ F(κ) dκ, and the reason is mechanical. Move a plate out by δd and the gap gains A δd of volume while the vacuum outside loses exactly that much, so the only energy that can change is the gap's minus free space's. Casimir's 1948 set-up, a plate inside a large conducting box, makes the same subtraction explicit. The cutoff f must be the same function on both sides, with f(0) = 1 and f → 0 well above K. With f = e^(−λs) the free-space term is 3ħcd/(π²λ⁴) per unit area, 1.54 mJ m⁻² for λ = 50 nm and d = 1.00 µm, millions of times the answer we are after.

Write G(n) = F(nπ/d). The difference is a sum minus the integral of the same smooth function, and the Euler-Maclaurin formula, quoted, gives it as −G′(0)/12 + G‴(0)/720 − …, with the ½ on n = 0 exactly the trapezoid weight it needs. Differentiate F: F′(κ) = −2κ² f(κ/K), so F′(0) = F″(0) = 0 and F‴(0) = −4f(0) = −4, while F⁽⁵⁾(0) carries a factor 1/K². Then G‴(0) = (π/d)³F‴(0) = −4π³/d³ and ΔE/A = (ħc/4π)(−4π³/(720d³)) = −π²ħc/(720d³): derived. Only f(0) = 1 entered, which is why every smooth cutoff gives the same limit; that the remainder vanishes for each such f is stated, not proved here. A sharp cutoff fails that condition. With a step at |k| = K, the same difference swings from +480 to −302 times the Casimir value as Kd/π runs from 4 to 4.5, and its swings grow as K²: no limit exists.

Differentiate: P = −∂(ΔE/A)/∂d = −π²ħc/(240d⁴), derived, the minus sign meaning attraction. With π²ħc/240 = 1.300 × 10⁻²⁷ J m, P is 1.30 mPa at 1.00 µm, 13.0 Pa at 100 nm and 1.3 × 10⁵ Pa, about an atmosphere, at 10 nm, where the ideal model has already failed. Dimensions: J m ÷ m⁴ = J m⁻³ = Pa. Limits: as ħ → 0 it vanishes, as it must for a classical field at zero temperature, which has no zero-point motion; as d → ∞ it dies as d⁻⁴, faster than the d⁻² pressure of any residual voltage. Measured, quoted: Lamoreaux (1997), torsion pendulum, sphere and plate, 0.6 to 6 µm, agreement to about 5%; Mohideen and Roy (1998), atomic-force microscope, 0.1 to 0.9 µm, about 1%; Bressi and co-workers (2002), parallel plates themselves, 0.5 to 3 µm, the coefficient to 15%. Sphere-plate data are compared through F = 2πR ΔE/A, the medium example.

The cutoff is not an embarrassment; it is the metal. A real conductor reflects well only below its plasma frequency ωₚ and turns transparent above it, so modes much shorter than c/ωₚ never feel the plates. For gold ħωₚ ≈ 9.0 eV and c/ωₚ ≈ 22 nm. The ideal result is the limit d ≫ c/ωₚ, where the modes that decide the force, with wavelengths near d, are reflected almost perfectly. The figure shows what happens away from it: two smooth cutoffs that agree as λ/d → 0 split in opposite directions as λ grows, so the departure belongs to the metal, not the vacuum. Lifshitz's theory (1956), quoted, computes it from the metal's dielectric function; for a plasma-model metal P ≈ P₀[1 − 16δ₀/(3d) + …] with δ₀ = c/ωₚ, about 10% below ideal for gold at 1 µm. Also quoted: at gaps of a few tens of nanometres the force crosses over to the unretarded van der Waals form ∝ d⁻³, and at room temperature thermal photons become comparable near 3 µm, where whether to treat the metal with dissipation or without is still argued between experiments.

The derivation used only how Σ½ħω changes with d, never Σ½ħω itself, so it says nothing about the absolute zero-point energy that gravity would feel. Nor does the force alone prove the zero-point energy real. Lifshitz's theory, and later Schwinger's source theory, obtain the same pressure as the retarded van der Waals attraction between fluctuating charges and currents in the two plates, never mentioning the field's zero-point energy, a point Jaffe (2005) made explicit; it is named here, not derived. The coupling gives it away. A real metal has ωₚ² = ne²/(ε₀m), which vanishes with e, and then so does the force; the ideal result, free of e, is the limit in which the plates respond infinitely strongly at every relevant frequency. The two accounts are one physics in two bookkeepings: modes of the field with boundary conditions, or correlated fluctuations of the charges that impose them. What the measurements do establish is that the quantised field gets the force between neutral bodies right where retardation matters.