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I am learning radiometry and one of the equation is radiance which is given as the radiant flux per unit projected area per unit solid angle. In equation:

$$L = {d^2\Phi \over {cos(\theta)dAd\omega}} (eq. 1)$$

Now further in the book I read they use intensity which is the angular density of radiant flux:

$$I = {d\Phi \over d\omega} (eq. 2)$$

And they explain that because of the cosine law I is attenuated by $cos(\theta)$ the angle of incidence between the surface normal and the incident light direction (or view direction). So far so good.

My problem is that they substitute $Icos(\theta)d\omega$ to the numerator in equation 1 which gives something like:

$${Icos(\theta)d\omega \over {cos(\theta)dAd\omega}} \rightarrow {I\over{dA}}$$

All that seems logical to me but the question is: in equation 1 the numerator is $d^2\Phi$. So is it legal to replace it with just $Icos(\theta)d\omega$. What does the exponent 2 means (after d and before phi) mathematically in that case? How should I read it and interpret it?

Thank you so much for your "smart" help.

For reference: www.astrowww.phys.uvic.ca/~tatum/stellatm/atm1.pdf (p12)

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  • $\begingroup$ In this context $d^2\Phi$ in the numerator means you are taking not just one but two partial derivatives of $\Phi$ (in particular, one with respect to $A$ and the other with respect to $\omega$). So $\frac{\partial^2 f}{\partial x\partial y}$ is shorthand for $\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)$ essentially, or equivalently for $\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x}\right)$ (differentiation with respect to two variables in different orders yields the same result with only weak regularity hypotheses required). $\endgroup$ – anon Jul 14 '13 at 7:46
  • $\begingroup$ It is not clear to me why $d^2\Phi$ can be replaced by $I\cos(\theta)d\omega$, though. You shouldn't be able to end up with a quantity divided by a differential (e.g. $I/dA$). Is there some sort of text or worksheet you're referring to? If possible can you provide a reference, something so others can see what's going on more clearly? (It's possible I am just not familiar with physics, ultimately.) $\endgroup$ – anon Jul 14 '13 at 7:53
  • $\begingroup$ Anon, you can have a look at this reference astrowww.phys.uvic.ca/~tatum/stellatm/atm1.pdf page 12, however in this PDF the author doesn't give a formal definition of radiance, so it's a bit hard to follow. The author speaks of both intensity and radiance. Thank you. $\endgroup$ – Marc Ourens Jul 14 '13 at 8:19
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    $\begingroup$ It just seems as a case of sloppy notation to me. He should have written $\delta I/\delta A$, because the intensity through an infinitesimal surface element should also be infinitesimal. He in fact does so in equation 1.14.1 further on page 15. $\endgroup$ – Raskolnikov Jul 14 '13 at 8:46
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The answer to your questions is that in physics this type of repalcement is possible. So asking math people why may cause discussion. Hygenically as a amthematician you can not straight do it but in physics this type of operation with operators is correct and one time under the constarints brought in that field of work consensus: $d^2\Phi$ acts as an operator and cn be substituted in this case. It pre-supposes a periodic behaviour of the solution of the system that allows straight to apply the mathematical ugly but physically elegant: $d^2\Phi=Icos(\theta)d\omega$ (Oups!).

So the mathematical background of calculation is correct but the heuristic straight forward substitution is not mathematically nice.


Let me try to put the equations in infitessimal notation (just try to recall what I remind from radiation theory), there are three of them:

$$\delta I=L\,cos\theta\,\delta A \quad (1)$$ $$\delta \Phi=I\,\delta \omega \quad (2)$$ $$\delta^2 \Phi=L\,\delta A \, (\delta\omega\, cos\theta) \quad (3)$$

With the differential intensity $\delta I$ (of the point source in a given direction on $\delta A$).

It should be possible now to re-construct all your equations by the above eauation.

Your equation $1$ follows from here equaetion $3$. Your equation $2$ from here equation $2$ and with equation $1$ and $3$ we obtain $$\delta^2 \Phi=L\,\delta A \, (\delta\omega\, cos\theta)=\delta I\; \delta\omega\quad (4)$$ Now introduce the condition $$\delta I_{(\theta)}=I_{(n)} \,cos(\theta) \quad (5)$$ And this is physics, if I recall correctly this can be argumented by Lambert's cosine law.

Now you should have all arguments toegether.

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  • $\begingroup$ Thanks a lot, not re-assuring but that could do it. I have to say I didn't find the reasoning in the PDF particularly clear, so I don't find it logical, I just accepted it, but it seems odd to me, and I am unsatisfied with it. Thanks again. $\endgroup$ – Marc Ourens Jul 14 '13 at 8:57
  • $\begingroup$ See extended. Hope helps now. $\endgroup$ – al-Hwarizmi Jul 14 '13 at 11:05
  • $\begingroup$ This is (more than) very nice. Thank you so much! Great help. $\endgroup$ – Marc Ourens Jul 14 '13 at 12:22
  • $\begingroup$ Sorry al-Hwarizmi could you please confirm the following reasoning is good: you get equation 1 by dividing the equation for intensity by $d\omega$ on both side. Then equation 4 is straightforward. What you then suggest is to replace the $d^2\Phi$ in the equation for radiance with equation 4 which leads to L (radiance on the left side) is equal to ${{LdAd\omega cos(\theta)} \over {cos(\theta)dAd\omega}}$. This L = L. Is that the right way of proving it? Thanks again. $\endgroup$ – Marc Ourens Jul 14 '13 at 12:54
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    $\begingroup$ So you say don't think in terms of substituting $LdAd\omega cos(\theta)$ with $d^2\Phi$? When you say two different I appear in equation 4 (second equation 4, there's 2 equations 4 in the post), you are talking about how the differential intensity relates to intensity along the normal direction ($I_{n}$)? One is the differential intensity (left inside) and the other just the "normal" intensity. However don't you have to show the equality holds in order to prove that radiance doesn't change with direction? And sub eq. 4bis into the right inside of eq.4? Sorry just try to be super clear;-) $\endgroup$ – Marc Ourens Jul 14 '13 at 13:29

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