Review Mode Density and Planck Dist - Optical Spectroscopy | PHYS 552, Study notes of Optics

Download Review Mode Density and Planck Dist - Optical Spectroscopy | PHYS 552 and more Study notes Optics in PDF only on Docsity! 1 ReviewModeDensityAndPlanckDist Some important concepts that were discussed last time: The density of field modes (# of modes) in the cavity per unit volume. ( ) 2 2 3c dd ω ωρ ω ω π = This is the number of modes between ω and dω . It is derived from the boundary conditions of 2 2 2 1E E c t 2 ∂∇ = ∂ solutions. This is a purely classical result from imposing boundary conditions on classical Maxwell Equations in theory. Then we look at the possible energy values of the field that is restricted to exist in these modes. The mode numbers, ( )x y zv , v , v have integer values. These mode designations specify the spatial properties of the allowed electric field through the allowed wave numbers. ( )2k π λ= xk / Lxvπ= yk / Lyvπ= zk / Lzvπ= ( ) ( ) ( ) ( ) ( )x x x y zE r, E cos k x sin k y sin k zt t= i i , etc. The energy of the field in the cavity is: ( ) ( )( )2 -1 20 01 E r, B r,2 cavity Energy dV t tε μ= +∫ Next, we solved the modes for E but we can get B from: ( ) ( )E r, - B r,t t t ∂ ∇× = ∂ Maxwell Eq. The Maxwell wave equations give as plane wave solutions: ( ) ( ) 2 2 2 E r, - E r,t tt ω∂ = ∂ This will be true for every separate mode. Any field can be expressed as a sum over the mode solutions. Now we bring in quantum restrictions on the energy. The above equation is the differential equation for a harmonic oscillator. The quantum solution for harmonic oscillator energy is: k,n k 1E n 2 k ω⎛ ⎞= +⎜ ⎟ ⎝ ⎠ 2 This is for the kth mode. So, this kth mode has a fundamental frequency, ( )kck c / Lk kvω π= = , and the energy for this mode can only exist in steps of kω . But there is room for populating this mode with any number of kω energy lumps (photons). Note the zero ( )kn 0= value. It is always there. Let’s get the ( )Eρ we need for Fermi’s Golden Rule. . . Note that if we want to have the total number of states between and dω ω ω+ we would have to multiply by the volume, V. ( ) ( )2 2 3V V / cd dρ ω ω ω ω π= This is over all angles (i.e. all directions of k ) and includes 2 polarizations. If we look at those directions of k within a solid angle dΩ , we have to multiply by d / 4πΩ . This would give for the total number of modes between and dω ω ω+ in dΩ : ( ) ( )2 3 3d 1V V / c4 4d dρ ω ω ω ω ππ Ω⎡ ⎤ ⎛ ⎞= ⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠ This can now be used to calculate the total energy in the cavity between E and E + dE that are in modes, which correspond to the solid angle dΩ to be: Remember dE dω= ( ) ( )d dV V E 4 4 dEdρ ω ω ρ π π Ω Ω⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ( ) 2,d 3 3 V 1N 4 cE dEd dω πΩ ⎡ ⎤= Ω⎢ ⎥⎣ ⎦ So, the density of the number of states in the cube with the energy between E and dE and with k’ in the direction dΩ is: ( ) ( ) 3 2 ,d 3 N L 12 E E 2 c Ed d d ω ρ π Ω ⎛ ⎞= Ω ≡⎜ ⎟ ⎝ ⎠ The factor 2 is from two orthogonal polarizations This is the density of the number of energy states between E and dE. ***Note: we do not yet know the energy in these states because we do not know the number of energy quanta in these states. This is given by (for the mode k): k,n k 1E n 2 k ω⎛ ⎞= +⎜ ⎟ ⎝ ⎠