(E B) Rate of Absorption and Stimulated Emission. π 2 E 0 ( ) 2. δ(ω k. p. 59. The rate of absorption induced by the field is. w k

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Claude Melton
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1 p. 59 Rate of Absorption and Stimulated Emission The rate of absorption indued by the field is π w k ( ω) ω E 0 ( ) k ˆ µ δω ( k ω) The rate is learly dependent on the strength of the field. The variable that you an most easily measure is the intensity I (energy flux through a unit area), whih is the time-averaged value of the Poynting vetor, S S 4π (E B) I S 4π E 8π E 0 Another representation of the amplitude of the field is the energy density I 1 U 8π E 0 (for a monohromati field) Using this we an write w k 4π U ( ω ) k ˆ µ δ(ω k ω) or for an isotropi field where E 0 xˆ 1 E 0 ŷ E 0 ẑ 3 E 0 4π w k U ( ω ) µ 3 k δ(ω k ω) or more ommonly w k B k U (ω k ) 4π B k 3 µ k Einstein B oeffiient (this is sometimes written as B k (π 3 ) µ k when the energy density is in ν).

2 p. 60 U an also be written in a quantum form, by writing it in terms of the number of photons N Nω E 0 8π ω3 U N π 3 B is independent of the properties of the field. It an be related to the absorption ross-setion, σ A. total energy absorbed / unit time σ A total inident intensity (energy / unit time / area ) ω w k ω Bk U (ω k ) I U (ω k ) ω σ A B k More generally you may have a frequeny dependent absorption oeffiient σ ω B k ( ) B k g ( ) where g(ω) is a lineshape funtion. A ( ) ω ω The golden rule rate for absorption also gives the same rate for stimulated emission. We find for two levels m and n : w w B U(ω ) B U(ω ) sine U(ω ) U(ω ) B B The absorption probability per unit time equals the stimulated emission probability per unit time. Also, the ross-setion for absorption is equal to an equivalent ross-setion for stimulated emission, (σ A ) (σ SE ).

3 p. 61 Now let s alulate the hange in the intensity of inident light, due to absorption/stimulated emission passing through sample (length L). di N n σ A Idx + N m σ SE Idx m n di ( I N N n m )σ a dx N n ; N m populations NN n N m : pop. differene I I 0 e Nσ a L for high freq. N N n N N L e σ a N:m 3 σ : m n L :m or I 10 C L I 0 C : mol / L : / moleule 303 N n σ A

4 p. 6 SPONTANEOUS EMISSION What doesn t ome naturally out of semi-lassial treatments is spontaneous emission transitions when the field isn t present. To treat it properly requires a quantum mehanial treatment of the field, where energy is onserved, suh that annihilation of a quantum leads to reation of a photon with the same energy. We need to treat the partiles and photons both as quantized objets. You an dedue the rates for spontaneous emission from statistial arguments (Einstein). For a sample with a large number of moleules, we will onsider transitions between two states m and n with E m > E n. E m m E n W W n The Boltzmann distribution gives us the number of moleules in eah state. N m / N n e ω / kt For the system to be at equilibrium, the time-averaged transitions up W must equal those down W. In the presene of a field, we would want to write for an ensemble? N m B U(ω ) N n B U(ω ) but learly this an t hold for finite temperature, where N m < N n, so there must be another type of emission independent of the field. So we write W W Nm ( A + B U (ω )) N n B U (ω ) If we substitute the Boltzmann equation into this and use B B, we an solve for A :

5 p. 63 A B U (ω )(e ω /kt 1 ) For the energy density we will use Plank s blakbody radiation distribution: ω 3 1 ω π 3 e ω /kt 1 U ( ) U ω U ω is the energy density per photon of frequeny ω. N ω is the mean number of photons at a frequeny ω. N ω The total rate of emission from the exited state is ω 3 A B Einstein A oeffiient π 3 ω 3 w B U (ω ) + A using U (ω ) N π C 3 ω 3 B π 3 ( N +1) Notie, even when the field vanishes (N 0), we still have emission. Remember, for the semilassial treatment, the total rate of stimulated emission was ω 3 w B N π 3 ( ) If we use the statistial analysis to alulate rates of absorption we have ω 3 w BN π 3 The A oeffiient gives the rate of emission in the absene of a field, and thus is the inverse of the radiative lifetime: 1 τ rad A

6 p. 64 Relaxation Leads to Line-broadening What happens to the probability of absorption if an exited state deays exponentially? k First-order result: V sinωt k relaxes exponentially... for instane by oupling to ontinuum P k exp[ Γt] b k i t dτ kv t t 0 or i t b k eiω kt Vk () t If we add relaxation to desription of b k : i b () Γ k e iω kt Vk t t b k (We write this in analogy to oupling to ontinuum n where Γ w nk.) Now we have t b k i e iω k t sinωtv k Γ () b k t E 0 ω k e i(ω k +ω) e i(ω k ω)t t iω b k () The solution to the differential equation µ k Γ + i t y ay be α is

7 p. 65 yt () Ae at + beiαt a + iα () Ae Γt / + E 0 ω k µ k b k t iω Let s look at absorption only long time limit: () E 0ω k µ k b k t ω The probability of transition: e i (ω k ω )t ω k ω iγ / e i (ω k +ω)t P k b k E 0 µ k 1 4 (ω k ω) +Γ /4 e i (ω k ω)t Γ / + i(ω k + ω) Γ / + i(ω k ω) Lorentzian lineshape: Pk Γ ω kl ω The linewidth is related to the system rather than how we introdued the perturbation. Linewidth related to relaxation dynamis.

Line Radiative Transfer

Line Radiative Transfer http://www.v.nrao.edu/ourse/astr534/ineradxfer.html ine Radiative Transfer Einstein Coeffiients We used armor's equation to estimate the spontaneous emission oeffiients A U for À reombination lines. A