by PenultimateManStanding » Wed 08 Aug 2007, 20:11:57
$this->bbcode_second_pass_quote('EnergyUnlimited', '
')I really cannot work out what would be actually "focused" or "defocused" by those "lenses".
Casimir vacuum?
It cannot be true.
What do you mean it cannot be true:
In the original calculation done by Casimir, he considered the space between a pair of conducting metal plates a distance a apart. In this case, the standing waves are particularly easy to calculate, since the transverse component of the electric field and the normal component of the magnetic field must vanish on the surface of a conductor. Assuming the parallel plates lie in the x-y plane, the standing waves are
\psi_n(x,y,z,t) = e^{-i\omega_nt} e^{ik_xx+ik_yy} \sin \left( k_n z \right)
where ψ stands for the electric component of the electromagnetic field, and, for brevity, the polarization and the magnetic components are ignored here. Here, kx and ky are the wave vectors in directions parallel to the plates, and
k_n = \frac{n\pi}{a}
is the wave-vector perpendicular to the plates. Here, n is an integer, resulting from the requirement that ψ vanish on the metal plates. The energy of this wave is
\omega_n = c \sqrt{{k_x}^2 + {k_y}^2 + \frac{n^2\pi^2}{a^2}}
where c is the speed of light. The vacuum energy is then the sum over all possible excitation modes
\langle E \rangle = \frac{\hbar}{2} \cdot 2 \int \frac{dk_x dk_y}{(2\pi)^2} \sum_{n=1}^\infty A\omega_n
where A is the area of the metal plates, and a factor of 2 is introduced for the two possible polarizations of the wave. This expression is clearly infinite, and to proceed with the calculation, it is convenient to introduce a regulator (discussed in greater detail below). The regulator will serve to make the expression finite, and in the end will be removed. The zeta-regulated version of the energy per unit-area of the plate is
\frac{\langle E(s) \rangle}{A} = \hbar \int \frac{dk_x dk_y}{(2\pi)^2} \sum_{n=1}^\infty \omega_n \vert \omega_n\vert^{-s}
In the end, the limit s\to 0 is to be taken. Here s is just a complex number, not to be confused with the shape discussed previously. This integral/sum is finite for s real and larger than 3. The sum has a pole at s=3, but may be analytically continued to s=0, where the expression is finite. Expanding this, one gets
\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}
where polar coordinates q^2 = k_x^2+k_y^2 were introduced to turn the double integral into a single integral. The q in front in the Jacobian, and the 2π comes from the angular integration. The integral is easily performed, resulting in
\frac{\langle E(s) \rangle}{A} = -\frac {\hbar c^{1-s} \pi^{2-s}}{2a^{3-s}} \frac{1}{3-s} \sum_n \vert n\vert ^{3-s}
The sum may be understood to be the Riemann zeta function, and so one has
\frac{\langle E \rangle}{A} = \lim_{s\to 0} \frac{\langle E(s) \rangle}{A} = -\frac {\hbar c \pi^{2}}{6a^{3}} \zeta (-3)
But ζ( − 3) = 1 / 120 and so one obtains
\frac{\langle E \rangle}{A} = \frac {-\hbar c \pi^{2}}{3 \cdot 240 a^{3}}
The Casimir force per unit area Fc / A for idealized, perfectly conducting plates with vacuum between them is
{F_c \over A} = - \frac{d}{da} \frac{\langle E \rangle}{A} = -\frac {\hbar c \pi^2} {240 a^4}
It's much easier to see at the wiki article on Casimir. Technology will save us. That isn't a magnet anyway. Maybe this is what the UFO thing is about, huh?