# Integrating over a charge distribution without using Radon-Nikodym

Goodmorning, I'm studying for a basic Electrostatic course and I have a doubt about how to justify in terms of measure theory the physicists' writing

$$dq=\rho \ d\tau$$ or $$\rho = \frac{dq}{d\tau}$$,

where $$\tau$$ is the $$\mathbb{R}^3$$ Lebesgue's measure (wich they call "infinitesimal element of volume") $$dq$$ is the "infinitesimal element of charge" and $$\rho$$ is the volumetric charge density.

For now I see them this way:

I can define a measure $$q$$ over $$\mathbb{R}^3$$ in this way: $$q(V)=\int_V \rho \ d\tau$$. Now, this measure is absolutely continuous with respect to Lebesgue's measure and $$\rho$$ is the Radon-Nikodym derivative $$\rho=\frac{dq}{d\tau}$$.

Moreover, from the properties of Radon-Nikodym derivative it follows that $$\int_V f \ dq=\int_V f\ \frac{dq}{d\tau} d\tau=\int_V f\rho \ d\tau$$ and maybe in this sense, as a notation, I can say that $$dq=\rho \ d\tau$$.

Now I come to my doubts:

1. Can I justify this writing without using Radon-Nikodym but just in terms of change of variables (Area formula)?

2. If I can, wich are the parametrization functions involved? Does the domains of integration have to change?

Physics is not mathematics. No-one solving an electrostatic problem cares as to whether the charge distribution is absolutley continuous or not. Certainly the textbooks will always asume that $$dq=\rho d[{\rm Vol}]$$. The exception is the case of surface charges when $$dq= \sigma d[{\rm Area}]$$ or line charges when $$dq= \lambda d[\rm length]$$.
If the distributions are singular (as withe surface charge) we apply distribution theory. The test functions used in distribution theory are smooth so we can stick to plain old Riemann integrals, or at worst Riemann-Stieltjes. No need for Lebesgue. Just use calculus 101 and, if necessary (for polar coordinates say) a Jacobean, to write the integrals as $${\mathbb R}^3$$ triple integrals.
• @Fillipo You can write the surface charge as a distribution- valued integral $\rho({\bf r}) = \int \sigma(\xi,\eta) \delta^3({\bf r}- {\bf r'}(\xi,\eta))\sqrt{g} d\xi d\eta$ where $\xi, \eta$ parameterise the surface and $\sqrt{g} d\xi d\eta$ is the element of area on the surface. Jan 3, 2023 at 14:46
• One is the charge in some region (i.e the number of electrons multiplied by $1.6\times 10^{-19}$ Coulombs) and the other is Euclidean volume of the region. Is there anything more than that? It does not seem mysterious to me. Jan 3, 2023 at 17:49