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Radiation From a Plane-Parallel Atmosphere

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Radiation From a Plane-Parallel Atmosphere **

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Problem: Assuming a plane-parallel atmosphere, (1) show that emission lines will appear if "j[lambda]>>1" and there is no radiation entering the atmosphere from outside. (2) Further, show that, if radiation enters from below, the relation between specific intensity and source function will determine whether emission or absorption lines are seen.

 

Hints:

 

The emergent intensity at the top of the atmosphere is the general solution to the transfer equation.

Calculate the integral in the source function.

If Sλ = Bλ for LTE, then τ >> 1. Calculate the source function for τ >> 1 and for τ << 1.

Find an expression for I[lambda]*0 in terms of j[lambda] and judge whether there will be emission or absorption when "j[lambda]>>1."

 

 

Useful Equations:

 

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NULL=j[lambda]/kappa[lambda]

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Solution (1):

 

Calculating the integral from the equation for radiative transfer:

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-S[lambda]*(int(exp(-tau[lambda]), tau[lambda] = tau[lambda, 0] .. 0))

-S[lambda]*(exp(-tau[lambda, 0])-1)

(1)

Therefore,

 

I[lambda](0) = I[lambda, 0]*exp(-tau[lambda, 0])-S[lambda]*(exp(-tau[lambda, 0])-1)

I[lambda](0) = I[lambda, 0]*exp(-tau[lambda, 0])-S[lambda]*(exp(-tau[lambda, 0])-1)

(2)

 

If Iλ,0 = 0 (no radiation is entering from outside), then, in the case where τ >> 1 (Sλ = Bλ for LTE), the exponent is nearly zero, and

 

I[lambda] = S[lambda] and S[lambda] = B[lambda]

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The emergent radiation will appear as blackbody radiation.

 

In the case where τ << 1, then e x 1-τ, so

 

`&sime;`(I[lambda](0), -S[lambda]*(1-tau[lambda, 0]-1))

`&sime;`(I[lambda](0), S[lambda]*tau[lambda, 0])

(3)

 

For a slab of thickness l,

 

tau[lambda, 0] = kappa[lambda]*l*rho

tau[lambda, 0] = kappa[lambda]*l*rho

(4)

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and Sλ defined as

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S[lambda]=j[lambda]/kappa[lambda]

j[lambda]/kappa[lambda]

(5)

Therefore,

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`&sime;`(`#msub(mn("I"),mi("&lambda;",fontstyle = "normal"))`(0), (j[lambda]/kappa[lambda]*kappa[lambda])*l*rho)

`&sime;`(`#msub(mn("I"),mi("&lambda;",fontstyle = "normal"))`(0), j[lambda]*l*rho)

(6)

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Therefore, emission lines will appear for j[lambda]>>1.

 

``

2*Solution:

 

I[lambda](0) = I[lambda, 0]*(-l*rho*kappa[lambda]+1)+S[lambda]*kappa[lambda]*l*rho

I[lambda](0) = I[lambda, 0]*(-l*rho*kappa[lambda]+1)+S[lambda]*kappa[lambda]*l*rho

(7)

``

I[lambda]*0 = -l*rho*I[lambda, 0]*kappa[lambda]+l*rho*S[lambda]*kappa[lambda]+I[lambda, 0]

 

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I[lambda]*0 = I[lambda, 0]-kappa[lambda]*l*rho(I[lambda, 0]-S[lambda])

0 = I[lambda, 0]-kappa[lambda]*l*rho(I[lambda, 0]-S[lambda])

(8)

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If  Iλ,0 > Sλ, then absorption lines will appear on the spectrum. This situation obtains for the Sun's photosphere, for example.

 

If  Iλ,0 < Sλ, then emission lines will appear on the spectrum. This situation obtains for the Sun's chromosphere, for example.

 

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