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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 $\phi$

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    $\begingroup$ So you're reading a chapter about distribution theory and you got this far while still being unclear about the role of test functions? To be honest, it sounds like you should do a whole lot of in-depth revision of the earlier parts of that chapter. $\endgroup$ – Emilio Pisanty Jun 4 at 19:28
  • $\begingroup$ Yes, I knkw what a test function is! Anyhow, I'll make myself clearer: when he makes the example of mass distribution, he says thag evaluating the total mass of a distribution $\rho$ means evaluating $\langle \rho | 1 \rangle $. He also makes the example of the moment of inertia of a body with said mass distribution: $ I=\langle \rho | r^2 \rangle $. As for the dipole, he doesn't specify what that $ \phi $ is. And yes, his conclusion is that a dipole should be identified with $ - \delta'(0) $, but I can't really figure out why he comes to this conclusion (heuristically) $\endgroup$ – Operatore_Nabla Jun 5 at 11:04
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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.

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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.

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  • $\begingroup$ Yes I know what test functions are, read my comment above, I think I made a bit clearer what I'm having trouble with. Note that the "intuitive" side of the reasoning is unclear to me, rather than the rigorous one. $\endgroup$ – Operatore_Nabla Jun 5 at 11:09

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