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, $\frac{dE}{dE d\Omega}$ ?

  • 1
    $\begingroup$ 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 $\mathbf p \to (E,\Omega)$. $\endgroup$
    – pppqqq
    Aug 21, 2018 at 20:36
  • $\begingroup$ @pppqqq That looks like it should be an answer $\endgroup$
    – David Z
    Aug 21, 2018 at 21:18
  • $\begingroup$ 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. $\endgroup$
    – knzhou
    Aug 21, 2018 at 21:20
  • $\begingroup$ @knzhou That also looks like it should be an answer - at least it seems more like an answer than a comment. $\endgroup$
    – David Z
    Aug 21, 2018 at 21:35
  • $\begingroup$ It should be $d^2 N(\Omega, E)$ instead of $d N(\Omega, E)$. $\endgroup$
    – Danijel
    Aug 22, 2018 at 7:26

2 Answers 2


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.

Mass distributions, differential cross-sections, probability densities etc. in physics can often be regarded as measure-theoretic derivatives, with respect to some fixed measure (usually the Lebesgue measure) on the relevant space.

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.


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