Timeline for What is the 3-dimensional Fourier transform of $1/k^4$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 22, 2022 at 19:11 | comment | added | newtothis | @CarloBeenakker, yes. I considered that as the imaginary part of the $e^{ikr}/(k(k+ia)(k-ia))$ but I still didn't get result. I was however, able to derive it by using a regularisation method mentioned by another commenter $e^{ikr}/(k^2 + a^2)^2$. Thanks! | |
| Nov 22, 2022 at 10:55 | comment | added | Carlo Beenakker | @newtothis -- you need the integral $$\int_0^\infty \frac{\sin kr}{k(k^2+\epsilon^2)}\,dk=\pi\frac{1-e^{-\epsilon r}}{2\epsilon^2},$$ which you then expand in powers of $\epsilon$ to get the result in the answer. | |
| Nov 22, 2022 at 8:47 | comment | added | newtothis | Could you provide some steps on how you are solving the regluarised integral? I seem to eventually get an integrand like $e^{ikr}/[k(k + ia)(k-ia)]$. But this doesn't give me the result you have. | |
| Mar 23, 2022 at 4:17 | comment | added | HoangNguyen | Carlo, I do find your comment on the force rather than the potential being the "real" thing that is measurable insightful... Indeed, if this question were raised to a mathematician, the divergence of U would overwhelm the physics behind it. Thank you again. | |
| Mar 21, 2022 at 15:00 | comment | added | HoangNguyen | That is it. Thank you so much, Carlo. Following your cue, we may also consider 1/(k^2+epsilon^2)^2 then the result is pi^2*exp(-epsilon * r)/epsilon. Upon Taylor expanding exp(-epsilon * r), the term that is independent of epsilon is indeed -pi^2*r (consistent with yours). On the other hand, the divergent term is pi^2/epsilon and thus dependent on "regularization" scheme, which is expected. Thanks again! [Before posting my question, I had tried both ways, but I was too dumbstruck with the epsilon in the denominator that I failed to realize the Taylor expansion. Life is good.] | |
| Mar 21, 2022 at 7:26 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |