# Rigorous-ish mathematical meaning of having two total differentials in the denominator

When I'm doing, say, $\pi^0 \rightarrow \gamma \gamma$ decay, and I want to find the angular distribution for the number of photons with energy $E$, $N(E, \Omega)$. To get the probability I then need to find

$$\int d\Omega \int dE \frac{dN(\Omega, E)}{dE d\Omega}$$

What exactly is $\frac{dN(\Omega, E)}{dE d\Omega}$? Is it shorthand for $\frac{\partial^2 N(\Omega, E)}{\partial E \partial\Omega}$?

Moreover, if we may treat $dN$ as a total differential does it follow that

$$\frac{dN(\Omega, E)}{dE d\Omega} = \frac{1}{dE d\Omega}\left(\frac{\partial N}{\partial E}dE + \frac{\partial N}{\partial \Omega}d\Omega \right) ?$$

and if so, how do I understand, say, $\frac{dE}{dE d\Omega}$ ?

• You may regard it as the measure derivative (I think the correct name would be "Radon-Nikodym derivative") of the probability measure "dN" with respect to the standard Lebesgue measure $\text d ^3 \mathbf p$, times a Jacobian factor to take into account the smooth change of variables $\mathbf p \to (E,\Omega)$. – pppqqq Aug 21 '18 at 20:36
• @pppqqq That looks like it should be an answer – David Z Aug 21 '18 at 21:18
• If you wish to interpret the derivative as an actual derivative, you can think of the function $N$ as “the total number with energy at most $E$ and angle at most $\Omega$”. However, this interpretation doesn’t really matter because you always end up working with just differentials anyway. – knzhou Aug 21 '18 at 21:20
• @knzhou That also looks like it should be an answer - at least it seems more like an answer than a comment. – David Z Aug 21 '18 at 21:35
• It should be $d^2 N(\Omega, E)$ instead of $d N(\Omega, E)$. – Danijel Aug 22 '18 at 7:26

These kinds of differentials and derivatives are common in physics. I already answered this for the moment of inertia here, so let's start with that because it's simpler. The moment of inertia is $$I = \int r^2 \, dm$$ but $$m$$ is not, in any obvious sense, a coordinate, like $$x$$ would be in a $$dx$$ integral. The idea is that we're summing over all the masses by integrating over $$dm$$, but there's no canonical function $$m$$ whose value ranges from zero to the total mass $$M$$. However, we can introduce such a parametrization by using another coordinate system, such as cylindrical coordinates, $$I = \int r^2 \frac{dm}{dr} dr.$$ Here we compute the moment of inertia by integrating over cylinders of radius $$r$$. We think of $$dm/dr$$ as the amount of mass in a cylindrical shell per unit thickness. We don't speak of it as really being a derivative, because $$m$$ is not naturally a function. Nonetheless, you can interpret $$dm/dr$$ as an actual derivative by saying that $$m(r)$$ is the amount of mass within radius $$r$$.
The same thing is going on here. We really want to compute $$N = \int dN$$ but we parametrize this integral with energy and angle, for $$N = \int d\Omega \, dE \, \frac{d^2 N}{d\Omega \, dE}.$$ Just as for $$dm/dr$$, the derivative really means "the contribution to $$N$$ in a small region of $$E$$ and $$\Omega$$, per unit $$E$$ and $$\Omega$$". It is not really treated like a function; mathematically $$dN$$ is properly just a measure. You can turn this into a "real" derivative by defining $$N(E, \Omega)$$ to mean the total contribution to $$N$$ counting only particles of energy less than $$E$$ and angle less than $$\Omega$$, but since you never need or care about the function $$N(E, \Omega)$$, this isn't typically done. Still, of course, it can be.
In measure theory, there is a theorem (named after Radon-Nikodym), stating that when a measure $\mu$ is absolutely continuous with respect to one another $\nu$ (meaning that a zero $\nu$-measure set is also a zero $\mu$-measure set), then there exists an $L^1(\nu$) function $\rho$ such that $$\text d \mu =\rho\,\text d \nu\qquad\qquad(1).$$ See the link above for a more precise statement. (Actually, the theorem is an iff since, conversely, a measure $\text d \mu$ in the form (1) would be absolutely continuous with respect to $\text d \nu$)
The function $\rho$ is usually denoted $$\rho =\frac{\text d \mu}{\text d \nu}.$$ It is a "derivative" only in a measure theoretic sense.
Having said that, let me mention that in physics we often do not pay too much attention to the hypothesis of the Radon-Nikodym theorem, in particular to absolute continuity, as long as a formal expression in the form (1) exists. For instance, the mass distribution of point-mass at the origin of space $\mathbb R ^3$ would be $$\rho = \delta ^3(\mathbf x),$$ and you can write the mass-measure as $$\text d m = \rho \text d^3 \mathbf x = \delta ^3 (\mathbf x ) \text d ^3 \mathbf x.\qquad \qquad (1')$$ Of course this does not mean that $\text d m$ is absolutely continuous, but formal manipulations using the RHS of (1') are still possible, if one restricts himself to well behaved integrands.