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It is indeed compactly-supported Poincare duality via currents, for which the standard reference is de Rham's Differentiable Manifolds. For example, the Dirac delta function is the dual of a point! Anyway, to get to a quick understanding, see the beginning of Section 7.3 of Nicolaescu's notes http://www3.nd.edu/~lnicolae/Lectures.pdf

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Hints: Note that the derivative of the sign function $${\rm sgn}^{\prime}(z)~=~2\delta(z) \tag{A}$$ is twice the Dirac delta distribution. This fact seems to be at the heart of OP's question. Repeated differentiations of the Mestel disk potential $$\Phi~:=~ v_0^2 \ln(r+|z|), \qquad r~:=~\sqrt{R^2+z^2}, \tag{B}$$ leads to $$\frac{\partial \Phi}{\partial ... 8 The delta function \delta(x) has unit area, but the function \delta(2x) is "half as wide" and thus has half as much area; thus you can pick up extra factors from 'how fast' you cross the peak of the delta function. The general identity is$$\delta(f(x)) = \sum \frac{\delta(x-x_i)}{\big| df/dx|_{x=x_i} \big|}$$where the x_i are the roots of f. In ... 2 Both states \Psi_{k,\sigma} and \Psi_{k',\sigma'} are meant to be states of the same particle species i.e. they have the same values of the squared mass k^2. The inner product of one-particle states from different species s is zero which one might indicate by additional s,s' labels and a Kronecker symbol \delta_{s,s'}. Weinberg claims about the ... 1 Conventionally in QFT, particles and antiparticles are defined with positive energy k^0\geq 0 only. (Recall that would-be negative energy states are reinterpreted as matter/antimatter of the opposite kind in order to make the vacuum stable.) 14 This is notation from Distribution Theory in Functional Analysis. The theory of distributions is meant to make things like the Dirac Delta rigorous. In this context, just to give you one overview, a distribution is a functional on the space of test functions. We define the space of test functions over \mathbb{R} as \mathcal{D}(\mathbb{R}) being the ... 10 This is not a peculiar physicist notation oddly enough. The notation allows one to interpret 1/x as a distribution (which makes sense since it's being added to the delta distribution on the right hand side of the equation). For a suitable test function \varphi, one defines this distribution as$$ \mathrm{pv}(1/x)(\varphi) = \lim_{\epsilon\to ...

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