# What's the meaning of partial derivative for radiance?

$$L\equiv\frac {\partial^2 \Phi}{\partial A\,\partial\omega\,\cos\theta}$$

where:

$\Phi$ is the radiant flux

$\omega$ is the solid angle

$A\cos\theta$ is the projected surface

Why are partial derivatives used and not full derivatives as in:

$$L\equiv\frac {d^2 \Phi}{dA\,d\omega\,\cos\theta}$$ even though sometimes this formula is also (wrongly?) used ?

I am no mathematician, but $\partial$ and $d$ are not the same and they shouldn't be interchangeable.

The radiance definition is just one example, but I noticed that most physics definitions use partials and not full derivatives. So, if you know why partials are used in other examples, it would maybe help me figure it out for the radiance, which I am particularly interested in.

• I forgot: Is the usual symbol for the flux $\Phi$ or $\phi$? You seem to use them interchangeably. Also, a solid angle $\omega$ rather than $\Omega$ seems pleasantly unique and unconventional. – pyramids Mar 24 '15 at 0:32
• sorry for the small letter for the flux, I meant $\Phi$ – Andrei Mar 24 '15 at 0:37
• One of the most important papers about radiometry definitions is Geometrical considerations and nomenclature for Reflectance - Nicodemus et. al (1977), which uses $\omega$ for solid angle. Also The SI system and SI units for Radiometry and photometry by Palmer and James M. uses the same notation, so I think $\omega$ is the standard conventional notation for solid angle in radiometry. Maybe in other fields there are other notations, but I am trying to get radiometry straight for now. – Andrei Mar 24 '15 at 0:45
• Then apologies for confusing you with my outside influences! – pyramids Mar 24 '15 at 0:47
• No worries.. I am glad that, at least this, was not confusing for me, even though I have to admit that it was a little bit confusing the first time I read the wikipedia definition, which uses $\Omega$, as you suggested. – Andrei Mar 24 '15 at 1:27

General case is:

$$L\equiv\frac {d^2 \Phi}{dA\,d\omega\,\cos\theta}$$

You must use this (in actual fact) difficult formula when:

$\omega = f(A)$ or $A = f(\omega)$ or
$\omega = f(t)$ and $A = f(t)$ (it can be temperature for example) and so on...

Particular case is:

$$L\equiv\frac {\partial^2 \Phi}{\partial A\,\partial\omega\,\cos\theta}$$ You can use it when $A$ and $\omega$ are not connected.
Usually it is true (I always use it).

### Remember, it's always you to decide which pill to eat. • after a few days of reading a lot of other stuff, I am finally able to understand what you said, but now I think there's a problem with your answer. Since Radiance is defined (hence the triple equal in my question) as being the partial derivative of radiant flux, only wrt $A$ and $\omega$ (so basically x,y,z if you care to use a 3D orthogonal system) then I think you should never use the full derivative because you'll get something else and not the radiance. – Andrei Mar 26 '15 at 14:23
• I understand that in some situations maybe the radiant flux depends on temperature and in a strange way maybe also the solid angle depends on time, but that is of no interest for radiance which is defined only if anything else, but space($A$ and $\omega$), to be constant and that's the whole point of defining it with partial derivatives. What do you think, am I right? – Andrei Mar 26 '15 at 14:26
• First of all in my sources there is not "≡" (identity) but "=" (equality) is used, and I think "≡" (identity) is a mistake: when dA = 0 L ≢ ... cause of division by zero. – Oleksii Zhyglov Mar 30 '15 at 5:53

This is one of the rare cases where the full and the partial derivative are (at least typically) the same. There is a difference if the way you tilt your projection surface influences the solid angle into which you radiate. Whoever uses the full derivative must implicitly assume that such a curious dependency cannot happen (what does the lamp care how you look at it?).

Derivatives in these kinds of definitions of measurement quantities are almost always meant to normalize, in a way that partial derivatives do most naturally. The reason is that full derivatives can only lead to nice, reproducible normalizations if they are determined by a law of nature (think Newton's law, which has a full derivative). Where you have design freedom in how to arrange dependencies, the full derivative would depend on your choices and hence not make for good universal definitions.

• Thank you for your answer! Unfortunately, most likely, I am not advanced enough to understand what do you mean by things like "normalize in a way that partial derivatives do most naturally" or "design freedom in how to arrange dependencies". I get a hint of what you try to say, but I feel like I still don't understand the concept since I wouldn't be able to explain why partial derivatives are used in the definition of radiant flux for example, which is $\Phi\equiv\frac{\partial Q}{\partial t}$. Is this telling me that there might be an energy function that depends on time $Q(t)$? – Andrei Mar 24 '15 at 1:24
• Then it is a bad or poorly communicated answer! However, you are entirely correct in surmising that $Q=Q(t)$ denotes an energy. It is the total energy radiated. Its derivative wrt time is the (instantenous) power radiated. – pyramids Mar 24 '15 at 1:39