12.815/12.816: RADIATIVE TRANSFER PROBLEM SET #1 SOLUTIONS

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1 12.815/12.816: RADIATIVE TRANSFER PROBLEM SET #1 SOLUTIONS TA: NIRAJ INAMDAR 1) Radiation Terminology. We are asked to define a number of standard terms. See also Table 1. Intensity: The amount of energy in a given energy bin centered about a frequency ν is given by de ν dν. Consider a location in space parameterized by the vector r at a time t. Define a differential surface area by the magnitude of its area, da, and the vector normal to its surface, ˆn. If a ray of radiation passes through that surface by making an angle with the normal of θ from within a solid angle of dω, then the specific or monochromatic intensity I ν is defined as de ν dν = I ν r, ˆn, t) cos θ.da.dt.dω.dν; 1) That is, it is defined as the quantity of energy passing through a surface of unit energy, per unit time, per unit solid angle, per unit frequency. It is so- called since it concerns the energy quantity about a single given energy or frequency bin on the spectrum. Note, too, the presence of the cos θ factor, which specifies that we are concerned with the projection of the ray normal to the surface element, since that is the portion that can pass through the surface element. Having defined monochromatic intensity, we can consider the total integrated intensity or just intensity) I as being the monochromatic intensity integrated over all energy or frequency bins I = ν=0 I ν dν 2) Important note. Depending on your field of interest, you will often find that some will variously have a preference for working in units of energy, wavelength or frequency. These quantities can all be considered equivalently; in particular, for massless particles, we make use of the Planck-Einstein relation: E = hν = hc λ When performing integrations, however, care must be taken to account for the Jacobian factor appropriately. Hence, if we are interested in 3) Date: Friday 13 September

2 integrating, say, the Planck function B ν ν, T ) over all frequencies, the correct procedure is to take the integrals in the following way: B ν ν, T )dν = B E E, T ) dν/de.de = B λ λ, T ) dν/dλ.dλ ν E λ where dν/de = 1/h dν/dλ = c/λ 2 Monochromatic intensity: See discussion above. Radiance: This is equivalent to monochromatic intensity; see, e.g. p. 5 of Liou. Flux density: The flux density is a quantity that gives, for a given surface element, the total intensity accessible within the solid angle subtended about the normal to that surface. Hence, we have for monochromatic radiation the monochromatic flux density or monochromatic irradiance) F ν F ν = I ν cos θ.dω 4) Ω Alternatively, it can be thought of as the total energy passing through a unit area, per unit time. Integrating over all wavelengths gives the flux density or irradiance) F : F = F ν dν 5) ν Irradiance: This is equivalent to flux density. Radiant power: The radiant power or total flux) f is defined as the irradiance integrated over all areas of the surface real or imaginary) under consideration: f = F.dA 6) A Emittance: If the flux density or irradiance is from an emitting source, it is called the the emittance. Emittance is the direction-integrated brightness or luminance. Brightness: Equivalent to luminance, brightness refers to the intensity or radiance from an emitting source. Luminance: This is equivalent to brightness. Luminosity: The total flux or radiant power from an emitting source. Total flux: This is equivalent to radiant power. 2

3 TABLE 1. Summary of radiometry terminology. Integrate over Symbol General Equivalent terminology terminology for an emitting body I, I ν Intensity = radiance Brightness = luminance I cos θ.dω F, F ν Flux density = irradiance Emittance F.dA f, W Total flux = radiant power Luminosity L) 3

4 2) Intensity and Flux. a) Isotropic radiation refers to radiation that has the same value regardless of the direction. By definition of specific) flux density F ν, for a given monochromatic intensity I ν, we have F ν = I ν cos θ.dω 7) Ω In a spherical coordinate system, dω = sin θ.dθ.dφ, where θ is the polar angle measured with respect to the vertical) and φ is the azimuthal angle. Substituting this into the above, we get see, e.g., Liou Eq ) F ν = 2π π/2 φ=0 θ=0 I ν θ, φ) cos θ. sin θ.dθ.dφ 8) Where the integration limits are taken over the half-sphere. For isotropic radiation then, I ν θ, φ) I ν, so that F ν = I ν π/2 θ=0 = 2π I ν = πi ν cos θ. sin θ.dθ π/2 θ=0 2π φ=0 d [ sin 2 θ ) /2 ] b) To carry the integration out over the entire solid angle, we must take the polar angle domain as θ [0, π]. Since F ν = 0. π θ=0 d [ sin 2 θ ) /2 ] = 0, The average intensity J ν is defined as J ν = 1 I ν dω 9) 4π Again, in spherical coordinates, J ν = 1 4π = I ν 4π 2π π φ=0 θ=0 2π π φ=0 = I ν 4π 4π = I ν θ=0 I ν θ, φ) sin θ.dθ.dφ dφ sin θ.dθ.dφ for isotropic radiation) c) Consider a narrow cylinder comprising a bundle of rays. Let one end of the cylinder be enclosed by an area da 1, with the other by an area da 2, with each area normal making an angle θ 1 and θ 2 with the ray, 4

5 respectively. Let the ends of the cylinder be a distance D apart. With no sources and sinks, the expression for conservation of energy gives I ν,1 da 1 cos θ 1 dσ 1 dt.dν = I ν,2 cos θ 2 da 2 dσ 2 dt.dν 10) A differential element of solid angle is defined as the projected differential area element on a unit circle. Hence, we have dσ 1 cos θ 2dA 2 and dσ D 2 2 cos θ 1dA 1 D 2 D 2 = cos θ 1dA 1 dσ 2 = cos θ 2dA 2 dσ 1 11) cos θ 1 da 1 dσ 1 = cos θ 2 da 2 dσ 2 Substituting into the energy conservation equation then we get I ν,1 = I ν,2 Based on the above discussion, we see that in general, that if we start with a given solid angle dσ 1, after traversing a distance D, the new solid angle dσ 2 dσ 1 /D 2 if the area under consideration is constant. However, since intensity must remain constant, we have I ν,1 F ν,1 dσ 1 = I ν,2 F ν,2 dσ 2 D 2 F ν,2 dσ 1 F ν,2 F ν,1 /D 2 5

6 3) Blackbody Radiation. a) We have Planck s law per unit frequency given by B ν T ) = 2h ν 3 ) c 2 exp 1 B ν hν 12) If we wish to convert the expression to per unit wavelength, by the chain rule see the note in Problem 1), we have B/ λ = B/ ν dν/dλ. By the Planck-Einstein relation, we have Substituting then gives hν = hc λ dν dλ = c λ 2 13) B λ T ) = 2h c/λ) 3 ) c c 2 exp hc/λ 1 λ 2 hc/λ = 2hc2 1 λ 5 exp ) 1 14) b) Using the relation above, we solve db λ /dλ = 0 for the λ corresponding to the maximum of the Planck function: db λ dλ = d 2hc2 1 ) dλ λ 5 exp hc/λ k B 1 T = 2hc 2 5/λ 6 exp ) [ exp 1 exp = 0 hcλ Some slight rearranging gives Setting we get 5k BT hc/λ exp ξ hc/λ hc/λ hc/λ 5 ξ exp ξ) = 0 ) hc/ λ 2 ) ) ] 2 1 hc/λ ) = 0 This must be solved numerically for ξ; in MATLAB, the following commands may be used to quickly solve the equation: 6

7 >> options = optimset Display, off ); >> xi_val = fsolve@x)-5./x + 1./1-exp-x))),1,options) xi_val = Solving for λ max then we get λ max = hc Using the following physical constants in SI units) gives k B = JK 1 c = ms 1 h = J s 15) λ max = T = b T Since we have used SI units for all fundamental constants, here λ max must be expressed in meters, so that we must have b = K m. c) We wish to integrate 2hc B λ dλ = 0 λ 5 exp dλ ) 1 hc/λ 16) We first make the substitution ξ hc/λ) / ). Differentiating, we get dξ = hc dλ λ 2 dλ = hc Substituting back into Eq. 16) for λ and dλ, we get 2k 4 B T 4 h 3 c 2 0 dξ ξ 2 ξ 3 dξ exp ξ 1 17) The integral can be solved in the following way. First, we recognize that exp ξ 1) 1 can be rewritten as a geometric sum: 1 exp ξ 1 = exp mξ) 18) m=1 7

8 The integral becomes 2k4 B T 4 h 3 c 2 0 exp mξ) ξ 3 dξ 19) m=1 Performing the integration by parts then gives 2k4 B T 4 exp mξ) m 3 ξ 3 + 3m 2 ξ 2 + 6mξ + 6 ) h 3 c 2 m 4 m=1 12k4 B T 4 h 3 c 2 m=1 ξ=0 20) 1 m 4 = 12k4 B T 4 h 3 c 2 ζ4) 21) where the infinite sum evaluates to a Riemann zeta with a particularly nice value: ζ4) = π 4 /90. Hence, 12kB 4 T 4 π 4 h 3 c 2 90 = 2k4 B π5 T 4 15h 3 c 3 π = σ π T 4, where σ 2k4 B π5 15h 3 c 3 d) For monochromatic radiance A at λ, the Planck function can be easily inverted to solve for T : T = hc λk B The physical constants of interest are again [ )] 2hc 2 1 log λ 5 A ) k B = JK 1 c = ms 1 h = J s We have λ = 10 µm = 10 5 m and also A = 9.8 Wm 2 Sr 1 µm 1 = Wm 2 Sr 1 m 1, changing the units for the wavelength bin from µm to m. Substituting into Eq. 22) gives T = K = 26.5 C = 79.7 F. e) It is not realistic to neglect the effect of the atmosphere in part d). It turns out that, centered around 10 µm, the sky emission peaks 1. 1 See, e.g., irwindows.html 8

9 4) LTE. Local thermodynamic equilibrium, or LTE, refers to a condition in which, over the time and distances scales of interest, the system at hand can be taken to be in thermodynamic equilibrium. In particular, it suggests that, in an LTE region, all mechanical pressure), chemical, and thermal gradients are resolved simultaneously to their equilibrium values. An immediate consequence of the latter is that the material volume under consideration may be taken to be at uniform temperature T. In general thermodynamic equilibrium, we have di ν /ds = 0; that is, the radiation field does not change in space. Hence, di ν ds = 0 23) α ν I ν = j ν 24) However, in general thermodynamic equilibrium, I ν = B ν, so that α ν B ν = j ν ; 25) this is Kirchhoff s Law. In LTE, the claim is then that locally, the source function can be replaced with the expression above. radiative transfer equation: Thus, consider the In LTE, we rewrite this as di ν α ν ds = I ν + j ν α ν di ν ds = α νi ν + j ν 26) di ν dτ ν = I ν + B ν, where dτ ν α ν ds and B ν j ν α ν 27) after making the substitution above. In LTE, while T/ τ 0, the fundamental assumption is that T/ τ is small, and in general, all thermodynamic potentials small in the sense L Φ 0 Φ 1 28) where L is the length scale of interest, Φ is a thermal, chemical, or mechanical potential T, µ, and P, respectively) and the subscript 0 denoting some physically reasonable reference value of the potential. 9

10 5) Hydrostatic Equilibrium. a) Consider an infinitesimal cubic volume element with one corner at x, y, z) and with dimensions x, y, and z. Consider only gravity acting in the z-direction the analysis can easily be extended to the other two directions, if gravity should have components along them). The pressure acting on the face at z is given by P z); at z + z, we have P z + z) P z) + dp z 29) dz The force acting on the bottom face due to pressure is thus P z) x y; on the top surface, [ P z) + dp ] dz z x y 30) The net force acting on the element due to pressure is dp x y z 31) dz In equilibrium, this force must be balanced by the weight of the volume element, given by g z ρ x y z. Therefore, dp dz x y z = g zρ x y z 32) dp dz = g zρ 33) b) We still lack a constitutive equation linking ρ and P ; in this case, we may use the ideal gas relation P = ρrt/m for molar mass M to get dp dz = g z RT/M P 34) Assuming g z is approximately constant over the z range of consideration, we may integrate this immediately from z = 0 to z to obtain [ ] z P z) = P 0 exp RT/g z M) 35) where P 0 is the pressure value at z = 0. The scale height H can be seen to be given by RT/g z M). c) At sea level, g z 9.8 ms 2. The molar mass for dry air M Dry air = kg mol 1. Assuming T = 25 C = 298 K and recalling 10

11 that R = JK 1, we get H = RT g z M JK K = 9.8 ms kg mol 1 = 8, 463 m = 8.46 km Mt. Everest has a height of 8, 848 m, roughly one scale height the varition in g z may be considered negligible). Assuming the temperature is the same for both altitudes, the pressure on Everest will be a little more than 1 e-fold less than the sea-level value P 0, about P 0. Since for ideal gases, pressure scales linearly with density, the density will be less by the same amount. d) We have T = 2000 K and M H2 = kg mol 1. For a hot Jupiter, we may suppose that R Hot jupiter O R Jupiter ) and M Hot jupiter O M Jupiter ). Thus the surface gravity will be given by g Hot Jupiter GM Jupiter R 2 Jupiter Therefore = m 3 kg 1 s kg m) 2 = ms 2 H Hot Jupiter JK K ms kg mol 1 = m = 319 km 11

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