3 Casimir effect c. All vacuum energies considered so far are infinite. In reality, the metal wal...
3 Casimir effect c. All vacuum energies considered so far are infinite. In reality, the metal walls are conduc tors only at finite frequencies, and thus do not impose boundary conditions at infinite frqu cies. (The cutoff is given roughly by the plasma frequencywp Thus, we need (and should) not consider infinite frequencies in our equations. To remove We will derive the Casimir effect in thee d sions, making use of the Euler-Maclaurin formula imen- them, we introduce a cutoff function n=0 1 F(0)30 x 4 1F" (0) Show that the energy difference Eq. (2) becomes where θη-1 for n > 0, θ0 1/2, and θη-0 for n <0. (You don't need to prove this formula.) Let us consider a square box with conducting walls of length L. Let E be the vacuum energy inside the box. We now insert a conducting plate ΔE-he L &aT)-Ï0(k,)dk, where dialing the volume parallel to oneofthe walls, 瞧)-「dkAV(:) ATA , dividing the volume into two. Denote ER and e vacuum energy in the two volumes uum energy due to insertion d. Show that The change in vac of the wall is evidently g(k)-RF(n) 4R3 where F(n) a. Show that » Use polar coordinates in the k k, plane and define 2k2+ k2 . substitute n and α, where k, nπ/17, K- (IT / R, and . substitute 722 + α2 e. With that, the expression for the energy The function n is needed because there are two polarization states for n>1.) The wavenumbers ky, k^ can take any values, a good approximation if L is large. Since R iss however, only wavenumbers knT/R are allowed in a-direction difference is 4R3 0 b. Since L - R and L are both large, we can find simplified expressions for the vacuum ener- gies in the larger volumes by assuming that now any values are allowed even for k, replacing the summation by an integration. Show that Use theEuler-Maclaurin formula to calculate ΔΕ and calculate the Casimir force F Note that we do not need to know the exact form of the cutoff function f when calculating the above result for the measurable force. Ap- parently our ignorance of the behavior of the sys- affect our abil- ity to make physical predictions. This will be 3 L2 (L-R) tem at high frequencies does not x / dkdkydk-\/k2+*2+k2. (3)instrumental in renormalization in field theory.
3 Casimir effect c. All vacuum energies considered so far are infinite. In reality, the metal walls are conduc tors only at finite frequencies, and thus do not impose boundary conditions at infinite frqu cies. (The cutoff is given roughly by the plasma frequencywp Thus, we need (and should) not consider infinite frequencies in our equations. To remove We will derive the Casimir effect in thee d sions, making use of the Euler-Maclaurin formula imen- them, we introduce a cutoff function n=0 1 F(0)30 x 4 1F" (0) Show that the energy difference Eq. (2) becomes where θη-1 for n > 0, θ0 1/2, and θη-0 for n 1.) The wavenumbers ky, k^ can take any values, a good approximation if L is large. Since R iss however, only wavenumbers knT/R are allowed in a-direction difference is 4R3 0 b. Since L - R and L are both large, we can find simplified expressions for the vacuum ener- gies in the larger volumes by assuming that now any values are allowed even for k, replacing the summation by an integration. Show that Use theEuler-Maclaurin formula to calculate ΔΕ and calculate the Casimir force F Note that we do not need to know the exact form of the cutoff function f when calculating the above result for the measurable force. Ap- parently our ignorance of the behavior of the sys- affect our abil- ity to make physical predictions. This will be 3 L2 (L-R) tem at high frequencies does not x / dkdkydk-\/k2+*2+k2. (3)instrumental in renormalization in field theory.