What does $\mathrm d^2 x$ exactly mean? 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)
 A: 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.
