# Emission radiance in participating media

1. Mar 4, 2015

### Hypatio

This link discusses radiation in participating media. Eq. 9.13 gives a prediction of the changing power of a ray along the path length as:

$I_\eta(x)=I_\eta(0)\exp(-\tau_\eta)+I_{b\eta}[1-\exp(-\tau_\eta)]$

where $\tau_\eta$ is the absorption coefficient times the length.

So, the first term gives energy of the incident ray $I_\eta$ lost to absorption as it propogates, and the second term gives the energy gained from emission.

With this in mind, I want to know how to figure out what the dimensionless coefficient of emission is. This way I can just multiply $I_{b\eta}$ by that coefficient and know the emission radiance everywhere for that temperature. Is this coefficient the value of $(1-\exp(-\tau_\eta))$ at one meter of propogation? after one centimeter? Is it related to the slope of this curve near x=0? How do I find it?

Also, the above equation is derived from Eq. 9.11 in that reference, where we are told that emission is the absorption coefficient times the radiance of a blackbody ($\kappa_\eta I_{b\eta}$). I assume that this is not the same as the dimensional absorption coefficient, but I don't know how to get the dimensionless one (that has a range of 0 to 1), from the dimensional one.

Last edited: Mar 4, 2015
2. Mar 4, 2015

### optophotophys

The provided link is not sufficient for understanding the exact situation. After some search, I found a textbook which looks like discussing similar topic. So, my thought is based on the chap. 10.5.2 in http://www.thermalfluidscentral.org/e-books/book-viewer.php?b=37&s=11

From the product of the dimensionless coefficient of emission$(1-\exp(-\tau_\eta))$ and $I_{b\eta}$, you can only know the emission radiance propagating into a certain direction originated from the concerning volume (see Fig. 10.27 in the textbook). If you want to know the total emission radiance density at some location, this is a different matter. You need to know the shape of the material etc... and need to integrate the whole emission radiance propagating in various directions.

I think $\kappa_\eta$ is just the dimensional absorption coefficient. What made you think that it should be dimensionless?

3. Mar 5, 2015

### Hypatio

Thanks for the reference, it's really good. What is discussed in 10.5.2 is very close to what I am trying to figure. To clarify, I don't think I need to know the total emission radiance density after radiative transport. I need to know the total emission radiance at the point where radiation originates. It appears that I need to solve something like Eq. 10.120 in the text.

I want to know the energy of all photons instantaneously emitted inside an arbitrary volume having some absorption coefficient and temperature. I don't care about what happens to the photons once they are emitted (whether it is absorbed, scattered, transmitted, etc.), that is secondary, I just want to know how much energy per volume.

But, if it is true that this energy is equal to $\kappa_\eta I_{\eta b}(T)dV$, then I don't understand the phenomenology, because my value ought to be linearly dependent on the volume as it is an extensive property. Moreover, the result seems wrong if the absorption coefficient is >1 m^-1, since this means that radiation is greater than blackbody.

Edit: Rereading the text, I don't think it is discussing what I want. I don't want energy conservation of a propogating ray. I want the net emission within a volume, which is linearly extensive.

Last edited: Mar 5, 2015
4. Mar 5, 2015

### optophotophys

$\kappa_\eta I_{\eta b}dV$ is proportional to the volume. You have the same $\kappa_\eta$ irrespective of the concerning volume size. $\kappa_\eta$ is the absorption coefficient per unit length. So $\kappa_\eta I_{\eta b}dV$ is an extensive quantity.

5. Mar 6, 2015

### Hypatio

What happens when the absorption coefficient is greater than 1 unit? I must be missing something in the dimensional analysis because it seems that if $\kappa_\eta$=50 m^-1, the intensity will be 50 times greater than a blackbody, which is impossible.

Can you show me where I am failing here?

6. Mar 8, 2015

### optophotophys

eq (10.119) looks like just an approximation (and contains some error). From eq (10.118), we can have a thermal radiation intensity in a given direction from an small element of medium $\delta s$ as
$$\delta I=(1-\exp(-\kappa \delta s)) I_b$$.
If the element size is enough small or $\kappa$ is enough small, we have
$$\delta I=-\kappa \delta s I_b$$.

7. Mar 8, 2015

### Hypatio

Actually, I've recognized that the problem is that I am looking at $dI_{\lambda b}$, and not $I_{\lambda b}$. $dI$ can, of course, be infinite, because it's just describing the change in intensity over the ray path.

My problem is that I want to know $I_{\lambda,emitted}$, not $dI_{\lambda}$. That is: the radiation emitted from each point in the volume. So It seems that Eq. 10.118 is more appropriate, or do I need to integrate Eq. 10.119?

But then it seems like everything radiates like a blackbody, because (1-exp(-KS)) always goes to 1 as S goes to zero...

I really appreciate the help thusfar.

8. Mar 9, 2015

### optophotophys

Eq. 10.118 is the one you need.

No. Everything behaves as a blackbody, when the size is infinitely large $S\rightarrow\infty$. It is intuitively reasonable in this simplified situation. An infinitely large object absorb all the incident energy (in this textbook, surface reflection is neglected). This means that the object is a blackbody.