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In statistical mechanics (in particular I'm interested in applying the Landau Kinetic Equation to a hard spheres gas), it is often useful to get the gradient or the Fourier transform of the interaction potential.

However, the interaction potential for hard spheres is $\infty$ for distances smaller than $2r$ and $0$ elsewhere.

How can I take the Gradient or the Fourier transform of such a potential if it isn't even a distribution/generalized function like the Dirac delta function?

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  • $\begingroup$ It’s proportional to a spherical Bessel function of the first kind. You can literally just google “fourier transform of hard sphere potential”. The trick, by the way, for dealing with the infinite potential inside of r is to change variables from U(r) to exp(-U(r)/kT)-1, which is finite for all r. $\endgroup$ Commented Nov 27 at 17:11
  • $\begingroup$ Why is it infinite for less than 2r? Why not for less than r? If it were couldn’t (wouldn’t) you just change scales? $\endgroup$ Commented Nov 28 at 0:11
  • $\begingroup$ Also, Fourier transforms are not limited to functions, distributions, generalized functions. Operators defined by boundary conditions can also have Fourier transforms. $\endgroup$ Commented Nov 28 at 0:14
  • $\begingroup$ @kangermu thank you for your halp! It is 2r because the centers can't be closer than the sum of their radio r+r. But it doesn't matter up to a change of variables, as you said. I'm aware it works for distributions, but I had only seen Fourier transform of functions with a finite or countably infinite number of points where it equaled infinite. The hard sphere potential is definitely weird for having a continuum of inifity there. $\endgroup$ Commented 2 days ago

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