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Recently I had to deal with Fourier transformations and delta functions, and I was wondering how about that. I know, that its trivial to show in cartesian coordinates, but i couldn't do it in spherical coordinates. Somehow one should be able to reduce it to something which only depends on |k| (which i could do) and show that it is a delta function (which i couldn't).

curious one
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  • I don't understand the question. Do you know the expression for the delta-function you use in spherical coordinates or do you want to derive this? I suspect you are looking for a derivation of the result $\delta^3({\bf x}) \equiv \delta(x)\delta(y)\delta(z) = \frac{\delta(r)}{2\pi r^2}$ where $r=|{\bf x}|$ and where the latter function is the delta function in spherical coordinates. – Winther Aug 06 '15 at 19:10
  • The result mentioned above is shown here for polar coordinates. The approach used to get the result is the same for spherical coordinates, i.e. you need to calculate the Jacobian of the coordinate transformation. – Winther Aug 06 '15 at 19:12
  • Depends, what you mean with 'the expression for the delta-function you use in spherical coordinates'. I already know that this is a delta function since I know that this is $\delta(x)\delta(y)\delta(z)= \frac{\delta(\textbf{r})}{2 \pi r^2}$. But lets assume, that I don't know, that this Integral is a delta function, and I only know how to integrate in spherical coordinates.

    I take coordinates where x is along the z-axis. So I have to do the integration over $\phi$ (which is trivial), $\theta$ (which should give me $2 \frac{sin(kx)}{kx}$, but maybe this is wrong) and $k$.

     – curious one Aug 06 '15 at 20:31
  • You might try to first derive something like $\delta({\bf x}) \propto \int_0^\infty {\rm d}k,k^2 \frac{\sin(kx)}{kx} = \frac{1}{x}\frac{d}{dx}\int_0^\infty {\rm d}k,(e^{ikx}+e^{-ikx})$. For a spherically symmetric function $f({\bf x}) = f(x)$ then try integrating $\int\delta({\bf x})f(x) {\rm d}^3x \propto \int_0^\infty{\rm d x} xf(x)\frac{d}{dx}\int_0^\infty {\rm d}k (e^{ikx}+e^{-ikx})$ by parts to 'see this'. If this calculation makes sense is another question. – Winther Aug 06 '15 at 21:10
  • Shouldn't it be $\frac{f(x)}{x}$ in the last integral? Nevertheless, this is exactly the point, where i got stuck. Let's say, the integral $\int_0^\infty \mathrm{d}k ~cos(kx)$ does vanish (in some sense), then we are still left with the equation $ \lim_{x \to \infty} \frac{f(x)}{x} \int_0^\infty \mathrm{d}k~ cos(kx)=f(0)$ but this doesn't make sense, if we already assumed that the integral over k vanishes for all x – curious one Aug 06 '15 at 22:13
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    When considering the 3D Dirac delta distribution $\delta^3({\bf x})$ using coordinates, one should use a coordinate system that is defined in an open neighborhood of the support ${{\bf 0}}$ of the distribution. Spherical coordinates fail in that department. – Qmechanic Aug 07 '15 at 12:21
  • Do I understand you right, that the problem is, that for k=0 the coordinates are not well defined? But why does then the formula $\delta(x) \delta(y) \delta(z) =\frac{ \delta(r)}{2 \pi r^2}$ exist? – curious one Aug 08 '15 at 09:10

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There is no need to start and work with any kind of co-ordinates in here, it can be done only using Delta & Fourier transform properties, as follows:

First, take a look at the fourier transform of the delta: $$\mathcal{F}[\delta^3(\textbf{x})] = \int \mathrm{d}^3 \textbf{x} \delta^3(\textbf{x}) {e^{-i \textbf{kx}}} = 1$$ Now use the fact that the inverse fourier transform of fourier transform of some function (assume you can apply the transform) is the function+use the result above and the definition of inverse transform to obtain: $$\delta^3(\textbf{x})=\mathcal{F^{-1}}[\mathcal{F[\delta^3(\textbf{x})]}]=\mathcal{F^{-1}}[1]= \int \mathrm{d}^3 \textbf{k} \frac {e^{i \textbf{k x}}} {(2 \pi)^3} $$

Nitay
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