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Mellinger
Lesson5Einstein coefficient & HI
line
Toshihiro HandaDept. of Phys. & Astron., Kagoshima University
Kagoshima Univ./ Ehime Univ.Galactic radio astronomy
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Energy level
▶ Schrödinger equation of an electron
▶ Steady state = energy eigen value■ Parameter separation
▶ Schrödinger equation for steady state
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Energy eigenvalue & bound states
▶ Solution depends on the boundary condition■ Continuous E is possible if E>0.■ Discreet values if E<0 (bound state)
▶ Continuous solution in quantum mechanics!
▶ In many cases, we consider bound states.■ Discreet eigenvalues and eigenfunctions
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Electron in an atom/molecule
▶ Electron in an atom/molecule■ Bound state → discreet energy levels
▶ Energy state
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Energy transition and matter
▶ Electron transition between energy levels■ Emission & absorp of EM wave■ DE=hn
▶ Structure of energy levels=matter identify■ Wavelength of emission & absorption lines■ Matter identification with spectrum
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Einstein coefficient(1)
▶ Emission & absorption: 2 level model■ Transition probability of emission A
Independent of input intensity■ Transition probability of absorption B
Proportional to input intensity
dI = n2 A-n1 B I
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Einstein coefficiant(2)
▶ Steady state dI=0dI = n2 A-n1 B I n2 A = n1 B I I =
▶ Thermal equilibrium b/w matter & radiation■ Therm. eq.→energy is in Boltzmann distribution
= =■ Therm. eq.→I must be blackbody radiation
I = Bn (T) = ■ Impossible to become a blackbody!?
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Stimulated emission(1)
▶ Add another process■ Difficult to see the existence
▶ Introduction of “stimulated” emission■ spontaneous emission A21, absorption prob. B12
■ stimulated emission B21
Emission proportional to input intensity!
dI = n2 A21-n1 B12 I+n2 B21 I
= n2 A21-(n1B12-n2B21) ISeems to reduce the effective absorption coeff.
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Stimulated emission(2)
▶ Steady state dI=0n2 A21=(n1B12-n2B21) I I =
■ Therm. eq.→energy is in Boltzmann distribution
= =, (RH)= ■ Therm. eq.→I must be I = Bn (T) = ■ We got that =1, =
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Relation between Einstein coff.
=1, =
▶ In the case with statistical weight g1, g2
■ Boltzmann dist. =
g1B12=g2B21, A21= B21
▶ A21, B12, B21 are fixed for matter.■ Relation is valid if thermal non-equilibrium■ A21, B12, B21 : Einstein coefficients
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Maser(1)
▶ If stimulated emission truly exist, thendI = n2 A21-(n1B12-n2B21) I
Effective absorption increases. We cannot know it?
▶ What happens, if n2> (B12/B21) n1?■ = . Therefore, it means T<0.■ Negative temperature i.e. inverse population
▶ Stimulated emission can arise!
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Maser(2)
▶ We can make T<0 for 3 level system
pomping
inverse population
maserseed photon
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Maser(3)
▶ MASERMicrowave Amplification by Stimulated Emission of Radiation
■ Developed by Towns in 1954He found ammonia in space, too.
▶ LASER■ Microwave →Light
▶ Characteristics of stimulated photon■ Same freq., phase, polarization as the seed photon
First detected molecule with 3 atoms in space
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Excitation temperature
▶ In general, emission is■ I=(2h n 3/c2) [(n1/n2) -1]
▶ Thermal non-equilib.: n1/n2 is not Boltzmann■ But convenient expressed by “temperature”
▶ Excitation temperature Tex
■ Define as = ■ Tex= -
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Emissivity
▶ Describe emissivity e by Einstein coeff.■ For isotropic radiation…■ Radiation energy dEn for dV dt dW
dE n =h n j(n) n2 A21 dV dt
= j(n) n2 A21 dS dx dt dW■ In the case of radiation only
dIn = = en dx
■ It gives en = j(n) n2 A21
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Absorption coefficient(1)
▶ abs. coeff. k described with Einstein coeff.■ For isotropic absorption…■ Radiation energy dEn into dV dt dW
dE n = -h n j(n) (n1 B12 -n2 B21) In dV dt
= - j(n) (n1 B12 -n2 B21) In dS dx dt dW■ In the case that absorption only
dIn =- =-kn In dx
■ It gives kn = j(n) (n1 B12-n2 B21)
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Absorption coefficient(2)
■ (continued)
kn = j(n) (n1 B12-n2 B21)
= j(n) n1 B12
= j(n) n1 B21
= j(n) n1A21
= j(n) n1A21 [1-]
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Source function▶ Source function Sn=
S n = = =
▶ In therm. eq. Tex=T (temperature in eq.)
▶ LTE: Local Thermal Equilibrium■ Tex’s are the same between all levels.
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Neutral atomic hydrogen
▶ proton + electron■ Proton is a particle with spin 1/2 2 values■ electron is a particle with spin 1/2 2 values
spin of a particle=should be related with mangetizmInteraction between two spins
A10=2.86888×10-15 [s-1], n =1.420405751786[GHz]
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HI emission(1)
▶ A10=2.86888×10-15 [s-1]Enough slow transition to excite under ISM densityShow maser if poping hydrogen maser clock
▶ Absorption coefficientk n = j(n) n0A10
■ g0=1←no degenerate, g1=3←F=+1,0,-1 degen.
■ nH=n0+n1= n0
■ For HI, Tex=Ts (spin temperature)
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HI emission(2)
▶ Approximation as 1≪ =0.07 [K]≪Ts~100 [K]
■ First order approximation1- ≅ nH= n0 =4n0
■ k n = nH A10 j(n)
=2.6×10-15 j(n) [cgs]
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HI emission(3)
▶ The column density can get through■ NH=∫nH dx = ∫ Tstn dn [cgs]
▶ In optically thin case, TB= Ts(1-e-t)= Tstn
■ NH= ∫ TB d n [cgs]
▶ Use Doppler velocity d n = dv,
NH[cm-2]=1.8224×1018 ∫ TB dv [K km s-1]
▶ Caution: this equation is valid only for■ Optically thin
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What is the natural width j(n)?
▶ Duration in quantum transition Dt■ Not 0 ∵ emitted EM wave is not d(t) time
profile■ Not ∞ ∵ finish the transition in finite time span
▶ Gradually increase and gradually decrease▶ Give a width after Fourier transformation
wave packet
wave particle
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report
▶ Attach to your e-mail. Deadline : 13 Nov.■ Submit to [email protected]
▶ Questions
1. Show relations between Einstein coeff.
2. How long the mean time to transit a neutral hydrogen atom?
