Show description

Physics Formulary by ir. A. 10 Spin For the spin operators are defined by their commutation relations: [S x , Sy ] = i¯ hSz . Because the spin operators do not act in the physical space (x, y, z) the uniqueness of the wavefunction is not a criterium here: also half odd-integer values are allowed for the spin. Because [L, S] = 0 spin and angular momentum operators do not ¯ σ, with have a common set of eigenfunctions. The spin operators are given by S = 12 h σx = 0 1 1 0 , σy = 0 −i i 0 1 0 0 −1 , σz = The eigenstates of Sz are called spinors: χ = α+ χ+ + α− χ− , where χ+ = (1, 0) represents the state with spin up (Sz = 12 h ¯ ) and χ− = (0, 1) represents the state with spin down (Sz = − 12 h ¯ ).

If Pr 1: δ/δT ≈ 3 Pr. 8 Heat conductance For non-stationairy heat conductance in one dimension without flow holds: κ ∂2T ∂T = +Φ ∂t c ∂x2 where Φ is a source term. If Φ = 0 the solutions for harmonic oscillations at x = 0 are: T − T∞ x x = exp − cos ωt − Tmax − T∞ D D Physics Formulary by ir. A. Wevers 44 with D = 2κ/ω c. At x = πD the temperature variation is in anti-phase with the surface. The onedimensional solution at Φ = 0 is 1 x2 T (x, t) = √ exp − 4at 2 πat This is mathematical equivalent to the diffusion problem: ∂n = D∇2 n + P − A ∂t where P is the production of and A the discharge of particles.

5 Waveguides and resonating cavities The boundary conditions for a perfect conductor can be derived from the Maxwell equations. If n is a unit vector ⊥ the surface, pointed from 1 to 2, and K is a surface current density, than holds: n · ( D2 − D1 ) = σ n · ( B2 − B1 ) = 0 n × ( E2 − E1 ) = 0 n × ( H2 − H1 ) = K In a waveguide holds because of the cylindrical symmetry: E(x, t) = E(x, y)ei(kz−ωt) and B(x, t) = B(x, y)ei(kz−ωt) . From this one can now deduce that, if Bz and Ez are not ≡ 0: ∂Ez i ∂Bz − εµω k εµω 2 − k 2 ∂x ∂y ∂Bz i ∂Ez + εµω Ex = k εµω 2 − k 2 ∂x ∂y Bx = ∂Ez i ∂Bz + εµω k εµω 2 − k 2 ∂y ∂x ∂Bz i ∂Ez − εµω Ey = k εµω 2 − k 2 ∂y ∂x By = Now one can distinguish between three cases: 1.