Mathematical rigorous definition for an electrical dipole

Question

I've been reading Laurent Schwartz's Mathematics for the physical sciences, and in the chapter on distributions he makes many cool examples of ways to define in a mathematical rigorous way certain entities given for granted in physics (e.g. a distribution of mass that may be discrete using the Dirac's delta). What is unclear to me is the definition of electrical dipole of moment +1 in one dimension as a distribution. The way he goes is the following: a dipole is the "limit" of thd system $T_\epsilon$ of two charges $\frac{1}{\epsilon},-\frac{1}{\epsilon} $ respectively at the positions $0,\epsilon$ for $\epsilon \to 0$. He then says that this corresponds to the distribution \begin{equation} \langle T_\epsilon | \phi \rangle = \frac{1}{\epsilon} \phi(\epsilon) - \frac{1}{\epsilon} \phi(0) = \phi'(0). \end{equation} Would there be someone patient enough as to make it a bit clearer for me? I understand the reasoning of the dipole as a limit, but I can't understand what he means by the undefined function $\phi$

Answer 1

The quantity $\phi$ is his smooth test function. The dipole distribution itself is $T=-\delta'(x)$ with $$ \langle T|\phi\rangle\equiv -\int \delta'(x) \phi(x) \,dx= \int \delta(x) \phi'(x)\,dx= \phi'(x). $$ The integration by parts is really the definition of the derivative of $\delta(x)$ rather than a real integration by parts.

Does not Schwartz describe test functions in his book? After all he invented them and whole idea of a rigorous approach to Dirac's delta.

Answer 2

I prefer to think of dipoles as irreducible representations of the SO(3) group. Now you could have such representations over different fields, e.g. it could be electromagnetic field, it could be current density, it could be gravitational potential etc.

For me, the most familiar one is the dipoles that occur in electromagnetism. So let us think about the dipole that is the component of some charge density $\rho\left(\mathbf{r}\right)$. Assuming you cannot actually go an poke that charge density, the only way you can observe it, is through its potential:

$\phi\left(\mathbf{r}\right)=\int d^3 r' G\left(\mathbf{r}-\mathbf{r'}\right)\rho\left(\mathbf{r}\right)$

Where $G\left(\mathbf{r}-\mathbf{r'}\right)$ is the relevant Greens function. That greens function is well-behaved as long as the observer does not go into the region actually occupied by charge.

As a result all the necessary information about the charge density can be expressed as:

$\rho\left(\mathbf{r}\right)=\rho_0\delta\left(\mathbf{r}\right)+\rho_1^\mu\partial_\mu \delta \left(\mathbf{r}\right)+\rho_2^{\mu\nu}\partial_{\mu\nu}\delta\left(\mathbf{r}\right)...$

So that: $\phi\left(\mathbf{r}\right)=\rho_0G\left(\mathbf{r}\right)+\rho_1^\mu\partial_\mu G\left(\mathbf{r}\right)+\rho_2^{\mu\nu}\partial_{\mu\nu}G\left(\mathbf{r}\right)+\dots$

Basically you do Taylor expansion on the Greens function in the region where charge density is not zero (and then integrate the charge density, e.g. $\rho_0=\int d^3 r' \rho\left(\mathbf{r'}\right)$), but you package it as 'Taylor expansion' of the delta function.

So then the question comes about what to do with the tensors $\rho_{0,1,2,\dots}$. If you choose to decompose them into the irreducible representations of SO(3) group (representation follows from the derivatives), then $\rho_0$ will be a monopole (trivial), $\rho_1$ will be the dipole (trivial), $\rho_2$ will have a quadrupole component and some other stuff which may or may not be discarded (depends on treatment).

Another way to approach this is to think about charges. The charge density of a point charge is $\rho=q\delta\left(\mathbf{r}\right)$. Now consider the charge density due to two opposite charges at $\mathbf{r}$, separated by vector $\mathbf{\epsilon\mathbf{\hat{p}}}$:

$\rho=+q\delta\left(\mathbf{r}-\frac{\epsilon}{2}\mathbf{\hat{p}}\right)+(-q)\delta\left(\mathbf{r}-\left(-\frac{\epsilon}{2}\mathbf{\hat{p}}\right)\right)=-q\epsilon\left(\frac{\delta\left(\mathbf{r}+\frac{\epsilon}{2}\mathbf{\hat{p}}\right)-\delta\left(\mathbf{r}-\frac{\epsilon}{2}\mathbf{\hat{p}}\right)}{\epsilon}\right)=-\mathbf{p}.\boldsymbol{\nabla}\delta\left(\mathbf{r}\right)$

In the process we took the limit $\epsilon\to0$ and $\epsilon q\to p$ and defined $p\mathbf{\hat{p}}=\mathbf{p}$, which we call the dipole moment. Clearly, limit of a delta function works only in the generalized sense.

Answer 3

I think it can be intuitively understood as follows. The Dirac delta (for the Riemann-Stieltjes integral) is the point mass (for Lebesgue integration) and there is a Schwartz distribution version of that (as a linear functional on a space of test functions). The reason the linear functional approach is useful is that operations (for example differentiation) can be transferred from the space where they makes sense (smooth test functions) to the dual (where they don't). The claim is that the derivative of a point mass is the dipole. The idea is to look at an approximate identity, Gaussians converging to a point mass. Here we recall that this is an "identity" for the operation of a moving average (convolution). Now the derivatives of the approximate identity should converge to the derivative of the point mass. The derivative of the Gaussian (an even function) is an odd function, sort of like two Gaussians next to each other, one up and one down: a wave. Indeed, convolution of some smooth function with this will give an approximate derivative (it's approximately the usual formula). Hope that helps. I remember getting this insight someplace back in grad school- and it convinced me that Schwartz' ideas are super cool, and they deserve much more attention. For one more example, white noise
should be the derivative of Brownian motion, and can be made rigorously as a distribution-valued process. I saw this approach taken in a book by maybe Gelfand-Shilov? Now this is a stationary process, hence measure-preserving for the shift flow. One can show it is measure theoretically isomorphic to an infinite entropy Bernoulli flow. The proof is that it is isomorphic to the increment flow on Brownian paths, \tau_s:B(t) mapsto B(t+s)-B(s). Then one can directly prove Ornstein's vwB property. Now this has an interesting interpretation: it is the limit of random coin tosses (a Bernoulli shift) as the time between tosses goes to 0. This makes some rigorous sense of the idea of choosing every other point of the real numbers.....an uncountable coin toss! "Take every other point...."