Transcript METO 621
METO 621 Lesson 9 Solution for Zero Scattering • If there is no scattering, e.g. in the thermal infrared, then the equation becomes dI I B (T) d S • This equation can be easily integrated using an integrating factor e dI d e I e (I e ) Be d d Solution for Zero Scattering Solution for Zero Scattering • Consider a straight path between point P1 and P2 . The optical path from P1 to an intermediate point P is given by P P P1 P1 0 0 (P1,P) ds ds ds (P) (P1) • Integrating along the path from P1 to P2 (P2 ) d t (P2 ) (P1 ) dt (Ie ) I (P ) e I (P ) e 2 1 dt (P1 ) (P2 ) dt B(t)e t (P1 ) Solution for Zero Scattering • Dividing through by e(P2) we get I (P2 ) I (P1)e (P2 ) (P1 ) (P2 ) dt B(t)e t (P2 ) (P1 ) I (P1)e (P1 ,P2 ) (P2 ) t(P,P2 ) dt B (t)e (P1 ) • But what does this equation tell us about the physics of the problem? Physical Description Solution for Zero Scattering • Break up the path from P1 to P2 into small elements ds with optical depths d • When sn is zero then en is equal to 1 • Hence dB is the blackbody emission from the element ds • The intensity at P1 is I[t(P1)] • This intensity will be absorbed as it moves from P1 to P2 , and the intensity at P2 will be I[t(P1)]exp[-(t(P2)-t(P1)] Solution for Zero Scattering • Now consider each small element P with a D, with an optical depth • Emission from each element is BD • The amount of this radiation that reaches P2 is BDt exp(P,P2)] where is the optical depth between P and P2 • Hence the total amount of radiation reaching P2 from all elements is n 0 D B ( )e ( P , P2 ) ( P2 ) B (t )e ( P1 ) dt t ( P , P2 ) Isothermal Medium – Arbitrary Geometry Redefine the origin such that (P1 0, then ( P2 I (P2 I (P1 e ( P2 B ( ( P2 t dte 0 I 0e ( P2 B 1 e ( P2 If the medium is optically thin, i.e. τ(P2) <<1 then the second term becomes B τ(P2). If there is no absorption or scattering then τ=0 and the intensity in any direction is a constant, i.e. I[τ(P2)]=I[τ(P1)] Isothermal Medium – Arbitrary Geometry If we consider the case when τ>>1 then the total intensity is equal to B(T). In this case the medium acts like a blackbody in all frequencies, i.e. is in a state of thermodynamic equilibrium. If ones looks toward the horizon then in a homogeneous atmosphere the atmosphere has a constant temperature. Hence the observed intensity is also blackbody Zero Scattering in Slab Geometry • Most common geometry in the theory of radiative transfer is a plane-parallel medium or a slab • The vertical optical path (optical depth) is given the symbol as distinct from the slant optical path s • Using z as altitude (z) = s |cosq|s m • The optical depth is measured along the vertical downward direction, i.e. from the ‘top’ of the medium Half-range Intensities Half-Range Quantities in Slab Geometry • The half-range intensities are defined by: I ( ,q,) I ( ,q /2, ) I ( ,q,) I ( ,q /2, ) • Note that the negative direction is for the downward flux, Half-Range Quantities in Slab Geometry • The radiative flux is also defined in terms of half-range quantities. ˆ) F d cosq I ( 2 /2 0 0 d 2 /2 0 0 d dq sin q cosq I ( ,q, ) d mm I ( , m, ) 2 /2 0 0 ˆ ) d F d cosq I ( 2 /2 0 0 d d mm I ( , m, ) dq sin q cosq I ( ,q, ) Half Range Quantities • In the limit of no scattering the radiative transfer equations for the half-range intensities become dI ( , m, ) m I ( , m, ) B( ) d dI ( , m, ) m I ( , m, ) B( ) d Formal Solution in Slab Geometry • Choose the integrating factor e /m,,for the first equation, then d / m dI 1 / m B ( ) / m I e I e e ( d m d m This represents a downward beam so we integrate from the “top” of the atmosphere (=0) to the bottom (*. • Slab geometry * ( d '/ m */ m d ' I e I ( *, m , ) e I (0, m , ) 0 d ' * d ' ' / m e B ( ' ) 0 m or I ( *, m , ) I (0, m , )e * /m * 0 d ' m B ( ' )e ( * ') / m Slab Geometry • For an interior point, <* , we integrate from 0 to . The solution is easily found by replacing * with I ( , m, ) I (0, m, )e / m 0 d ' m B ( ')e ( ' )/ m