Is a layer of gas with sufficiently large optical thickness really radiating as a black body?

Question

Can a parcel of gas with large value of optical thickness really radiate like a black body? I have in mind a simple (most likely oversimplified) model which yields

$$I_\nu = I_\nu(0) e^{-\tau\nu} + I_\nu^B [1-e^{-\tau_\nu}]$$

It says, that radiation of a layer of optical thickness $\tau$ radiates based on

• the radiation entering on one side, attenuated by absorption
• the intrinsic radiation, also attenuated by absorption

I have seen this equation in several books. But I do not understand the quantity $J_\nu$ in the equation

$$d I\lambda = -k_\lambda \rho I_\lambda ds + j_\lambda \rho ds$$

which is replaced after introducing the source function $J_\lambda=j_\lambda/k_\lambda$ by

$$\frac{d I\lambda(s)}{k_\lambda \rho ds} = -I_\lambda(s) + J_\lambda(s)$$

In my textbook, a few sentences after having introduced this, it is said, that

$$J_\lambda(s) = B_\lambda^B(T)$$

where $B_\lambda^B(T)$ is the Planck-Function. I cannot follow this argument: It would mean, that an absorbing gas with sufficiently big optical thickness radiates to the outside according to Planck's law - but by what argument this assumption is justified?

See here the original text:

Answer 1

The expression $J_{\nu} = B_\nu$ (where $J_\nu$ is the source function) is true for a source of thermal radiation that is in local thermodynamic equilibrium. i.e. Where the population of various energy levels and the speeds of particles are determined by a single temperature. This is just Kirchoff's law of thermal radiation.

If that is the case, then you can see from the equation of radiative transfer (the first equation you have written), that when $\tau_\nu$, the optical depth, becomes large, then the specific intensity, $I_\nu \rightarrow B_\nu$.

So it is indeed true that if you have a sufficiently thick slab (and by this we mean optically thick, with large $\tau_\nu$), then the specific intensity from the slab will approach the Planck function value. Of course for the emergent spectrum to resemble the Planck function then this must be true at all frequencies. The emergent spectrum will in fact be the Planck function at the temperature at which the optical depth to the surface is $\sim 2/3$ and the radiation can escape.

Note that the emission of a blackbody spectrum is not a property of the gas/material alone, it is a combination of thermal emission (the $J_\nu = B_\nu$ part) and the geometry (size and density) of the system

Edit: Incidentally, the notation you are using is very confusing. Usually $S_\nu$ is used for the source function and $J_\nu$ is the mean specific intensity. The confusing thing is that for a blackbody, $S_\nu = B_\nu = I_\nu = J_\nu$ !