Timeline for The $2\pi$ in the definition of the Fourier transform
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 31, 2023 at 22:47 | comment | added | Yly | There's an argument in physics that $h$ rather than $\hbar$ is the "right unit" for action, having to do with the proper units of phase space volume in statistical mechanics. The argument is given in Sethna's "Entropy, Order Parameters, and Complexity", ex. 7.3. This is not widely known among physicists, though, who generally prefer to work with $\hbar$, because it eliminates some $2\pi$ factors in various formulas. (Ironically, this is only so because of inconvenient Fourier transform conventions! More $2\pi$'s could be killed by using $h$ and putting $2\pi$ in the FT exponent.) | |
| Mar 24, 2017 at 11:59 | vote | accept | coudy | ||
| Mar 23, 2017 at 14:39 | comment | added | nfdc23 | In terms of Pontryagin duality, for which there is always a "coordinate-free" Plancherel theorem (and Poisson summation formula) using the dual group, this expresses that ${\rm{d}}x$ is the unique self-dual Haar measure under the associated self-duality of $\mathbf{R}$. (Likewise, that Poisson summation $\sum_{n \in \mathbf{Z}} f(n) = \sum_{n \in \mathbf{Z}} \widehat{f}(n)$ holds without extraneous constants for version 2 expresses the good behavior of making $\mathbf{Z}$ be its own annihilator.) | |
| Mar 23, 2017 at 10:39 | comment | added | Adam P. Goucher | That's actually really helpful. I often have difficulty remembering the factors of $2 \pi$ in the convolution identity, and now I no longer need to. | |
| Mar 23, 2017 at 7:49 | history | answered | yuggib | CC BY-SA 3.0 |