Radiation Transport in a Gas

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1 Radiation Transport in a Gas By analogy to a particle gas, define a photon distribution function by, f ν ν, Ω; r, t)dvdωd r = Number of photons of a frequency in ν, ν + dν), in a volume at rd r), with direction in Ω,dΩ) ) A related quantity, which is actually the one used most often, is the Spectral Radiation Intensity, defined as, I ν = hνf ν c W/m 2 /sterad/hz) or J/m 2 /sterad) 2) so I ν = Energy crossing unit area per unit time, normal to Ω within a solid angle. Another related quantity is the Energy Density: u ν = f ν hvdω J/m /Hz) ) which can also be written as, u ν = Iν dω 4) c Finally, the Energy Flux Vector or radiant heat flux) is defined as, S ν = f ν hνcωdω = I ν ΩdΩ 5) with units of W/m 2 /Hz. For isotropic radiation, S ν = 0, even though u ν may not be zero or I ν ). For the definition 5), we count the individual energy fluxes of all photons, of any direction, then we add them vectorially. Equilibrium Distribution This was derived from Bose-Einstein statistics, and it constitutes an example of an isotropic radiation field that can be obtained, say, inside a blacked enclosure at constant temperature. The various Equilibruim or Black Body) quantities are 2ν 2 f ν = c hν u ν = 4πhνf ν = 8π c 2hν I ν = cf ν hν = c 2 6) e hν kt 7) e hν kt B kt ν 8) e hν

2 Notice u ν or I ν are maximum when, λt = 2900μm) K Wein s Law) 9) The total energy density in the radiation field is, u = u ν dν = 0 8π 5 5 k 4 T 4 0) hc) and the one-sided total energy flux is, uc q = = 2π5 k 4 T 4 = σt 4 σ = W/m 2 /K 4 ) ) 4 5 h c 2 But, of course, the net two-sided) energy flux is zero. The relevance of q is that it is the power flux emitted from the surface of a black body, where only one side exists. The Radiation Transport Equation This is the analogue of the Boltzmann equation for photons. There are no forces on a photon, so all we have is, ) I ν t + c I Iν ν = 2) t collisions notice the equation could be in terms of f nu, but I ν is customary). Also, c = cω. Comparing the terms on the left, Iν/ t) Δx c I ν) << usually. So, ignore photon accumulation : cδt ) cω Iν I ν = ) t collisions or, in terms of distance s along the beam ds = cdt), Ω I ν = ) Iν t collisions 4) Structure of the Collision Operator Photons can be absorbed, scattered, or emitted along their paths. Absorption + Scattering is also called Extinction. We ignore for now scattering. Define the Spectral Absorption Coefficient k ν, such that k ν I ν = Energy absorbed in the unit spectral range about ν, per unit length along the Ω-directed path, per steradian k ν has dimensions of inverse length. In fact, k ν = λ ν is the Photon Mean Free Path. Including scattering k ν Extinction coefficient. Define also the Spectral Emission per unit length and per unit frequency, into unit solid angle about Ω J ν ). 2

3 With these definitions, 4) becomes Ω I ν = J ν k ν I ν 5) We must relate k ν and J ν to atomic properties of the gas through which the photons propagate. Let N n = Number of atoms in n th excitation level. Let α nm = Cross section for a photon-induced n m transition. where the frequency ν is such that hν = ɛ m ɛ n. Because of uncertainty and various Broadening mechanisms see later), the energy level difference ɛ m ɛ n is spread over some finite although often narrow) part of the spectrum, rather than it being sharply defined, and so several transitions may contribute to a particular frequency ν: k ν = knm = N n α nm with ν = ɛ m ɛ n )/h) 6) n<m,all m The emission can be spontaneous independent of the radiation field), or it can be induced or stimulated proportional to J v, and in the same direction). We introduce a spontaneous emission coefficient, β mn, and an induced emission coefficient, γ mn, and write, J ν k ν I ν = N n α nm I ν + β mn N m + γ mn N m I ν ) 7) This would add up to zero for any m, n) if the radiation were Black Body radiation equilibrium): β mn B mn N m γ I ν ) equil. = B ν = = mn α α nm N n γ mn N nm N n 8) m γ mn N m Comparing this to B ν = 2hν, we conclude that, c 2 e hν/kt β mn γ mn = 2hν c 2 ; α nm γ mn N n N m ) equil. 9a;9b) Now, in equilibrium we already know that, Nn ) N m equil. = hν e + kt 20) g m and so 9b) simplifies to, α nm γ mn = g m 2) Equations 9a) and 2) are Einstein s relationships, and, as usual, allow us to express all three kinetic coefficients α, β, γ) in terms of only one of them. This is sometimes chosen to

4 be absorption α), other times spontaneous emission β). Substituting Einstein s relations into 7), and choosing to express everything in terms of absorption, we obtain, Ω I ν = [ Nm 2hν α nm N n N n g m c 2 + I ν ) I ν ] 22) Particular Cases a) For low enough frequencies ν c 2 I ν 2h ) /), or, at a given frequency, for strong enough radiation, we can neglect the ν term arising from spontaneous emission), and retain only stimulated emission: ) Nm /g m Ω I ν = Nn α nm I ν 2) N n / and if upper levels are more populated in proportion to their degeneracy) than lower levels, the intensity increases along the beam path. This is amplification, or LASER action. Notice that UV lasers are more difficult that MW MASER) or visible laster higher ν requires higher I ν ), and X-ray lasers are only possible in the enormous radiation field of a nuclear fireball. Notice also that Nm/gm would be e hν/kt, if the levels were equilibrated at some N n/ population temperature T. An inversion is necessary that will over-populate the upper levels; this is sometimes described as a negative temperature. b) Gas energy levels populated according to some T, any frequency. This implies a partial equilibrium situation, in which the atomic/molecular levels are equilibrated among themselves, but the radiation field is out of equilibrium. This situation is very common, and is called Local Thermodynamic Equilibrium LTE). Of course, no lading is possible at LTE. In this case, as noted, N m = e hν kt, and so N n g m ) [ ) ] 2hν Ω I ν = Nn α nm + I c 2 ν e hν kt Iν }{{}}{{} hν k ν e kt I ν e kt hν ) Ω I ν 2hν c 2 [ = k ν e hν 2hν /c 2 kt ) hν I e + kt v The first term in brackets is B ν, the Black Body radiation intensity. Also, define a modified absorption coefficient absorption minus stimulated emission) by, ] k ν hν = k ν e kt ) 24) and we obtain the LTE Radiation Transport Equation: Ω I ν = k ν B ν I ν ) 25) 4

5 Integrated relationships for a spectral For a the absorption of the emission are concentrated, and the total spontaneous emission is an atomic property. The decay rate from m to n is A mn ; it can be distributed differently in ν, depending on T, collisions, etc., but the total is constant. Emission is isotropic, so per atom, Amn per sterad. The net energy emission rate per unit volume, per 4π sterad, for the whole, is then Nm Amn hν. Comparing to the expression Ω I 4π ν ) sp.em. = m,n<m β mnn m, which is per unit frequency, we see that, A mn β mn dν = hν 26) 4π and then, from the Einstein relation between β mn and α nm, the integral over the of the absorption cross-section is, α nm dν = c2 g m 2hν βmndν = A 8π mn g m c 2 ν 2 27) and also, the integrated cross-section for stimulated emission is, γ mn dν = g m Amn c 2 αnmdν = 28) 8π ν 2 Solution along a ray Say x is the distance along which we want to calculate the evolution of I ν, and we know the state of the gas along xk ν x),b ν x)). The equation of propagation is, di ν = k ν B ν I ν ) 29) dx and using variation of parameters, the solution is, [ I ν x) = e 0 I + k ε 0 ν0) ν )B ν ε)e x x ε k ν ε)dε kν ε )dε 0 ] dε 0) The first factor is an attenuation factor. Inside the bracket, I ν 0) is the intensity of the ray incident on the medium. The integral inside the brackets is the cumulative new radiation that has accumulated along the ray s path in 0 x), at the rate k ν B ν per unit length, except that the portions emitted upstream are attenuated only from their emission ε) to the observation point x). If the gas is uniform constant k ν,b ν ), equation 0) reduces to, I ν x) = B ν +[I ν 0) B ν ]e k ν x ) 5

6 Again, I ν 0)e k ν x is the remaining part of the original radiation, and B ν e k ν x )isthe accumulated gas emission. Radiation Signature of a gas If I ν 0) = 0, the intensity at x = L is purely emitted radiation, given by k I ν L) = Bν e ν L ) 2) The quantity k ν L is called the optical thickness of the gas. Notice that this includes not only the actual thickness L, but also the absorption coefficient k ν, which can depend very strongly on frequency ν. Specifically, over some parts of the spectrum k ν L, and the gas is said to be optically thin there. Under these conditions, 2) can be approximated e ɛ ɛ) as, I ν L) B ν k ν L ) This means we see a radiative spectrum that reflects proportionately the gas absorptivity k ν which can now be interpreted as emissivity ), with a slowly changing black body envelope, B ν : On the other hand, in the vicinity of a strong absorption, we may have, for the same L, k ν L and the gas is optically thick there. In that case, Equation 2) gives, I ν L) B ν 4) and we see, over a limited part of the spectrum, a black body radiator. Any structure k ν may have in this spectral region is lost due to the black-body saturation. Application-Line Reversal Pyrometer The temperature of a flame or a moderate temperature plasma) can be measured to within ±20K or so with the following arrangement: A calibrated tungsten ribbon, treated to emit as a black body, is placed behind the flame, and heated by an electrical current. The calibration must indicate the ribbon s temperature, T b, for a chosen current I. The ribbon is observed through the flame, using a dispersive element spectroscope), to generate an observable spectrum. In this situation, Equation ) applies, with I v 0) = B v T b ). We observe IvL) = B v T )+[B v T b ) B v T )]e 5) As we vary T b with I), when T b = T we observe I v L) =B v T b ), i.e., literally as if the flame were not there. When T b <T, flame emission dominates, and in regions of the spectrum where k v L>> thick), we see I ν L) B ν T ). Is is customary to see the flame with some common salt, so that the yellow sodium doublet is definitively thick. As a consequence, when T b <T, the strong yellow is seen clearly against a darker background B ν T b )+B ν T )k ν L in the thin regions. Conversely, when T b >T, the yellow is actually dark B ν T )) against the thing background B ν T b )). The point of inversion is pretty sharply defined, and T b = T is easily found. k v L 6

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