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Fourier transform of $1/ sqrt{m^2+p_1^2+p_2^2+p_3^2}$

Mathematics Asked by Sebastien B on November 26, 2021

Let $m>0$ and consider the function $f:mathbb R^3tomathbb C$
defined through
$$ f(p_1,p_2,p_3) = frac{1}{sqrt{m^2+p_1^2+p_2^2+p_3^2}}.$$

I would like to compute the Fourier transform of $f$.

This particular function is of interest as one which naturally appears in some problems of special relativity.


What I already know:

  • Although $f$ is neither integrable nor square integrable, the Fourier transform of $f$ is well defined as the Fourier transform of a tempered distribution.
  • Using symbolic calculus software, I expect that
    $int_{-infty}^{+infty}f(p_1,p_2,p_3) e^{-i x_1p_1} dp_1 = 2K_0(x_1 sqrt{m^2+p_2^2+p_3^2})$, where $K_0$ is a modified Bessel function of the second kind.

Questions:

  1. Is the Fourier transform of $f$ explicitly computable?
  2. If it is, how could I compute it?

One Answer

$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},} newcommand{braces}[1]{leftlbrace,{#1},rightrbrace} newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack} newcommand{dd}{mathrm{d}} newcommand{ds}[1]{displaystyle{#1}} newcommand{expo}[1]{,mathrm{e}^{#1},} newcommand{ic}{mathrm{i}} newcommand{mc}[1]{mathcal{#1}} newcommand{mrm}[1]{mathrm{#1}} newcommand{pars}[1]{left(,{#1},right)} newcommand{partiald}[3][]{frac{partial^{#1} #2}{partial #3^{#1}}} newcommand{root}[2][]{,sqrt[#1]{,{#2},},} newcommand{totald}[3][]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}} newcommand{verts}[1]{leftvert,{#1},rightvert}$ $ds{underline{underline{mbox{With} vec{R} equiv verts{m}vec{r}}}}$: begin{align} &bbox[10px,#ffe]{iiint_{largemathbb{R}^{3}}{expo{-icvec{p}cdotvec{r}} over root{m^{2} + p^{2}}},dd^{3}vec{p}} = m^{2}iiint_{largemathbb{R}^{3}}{expo{-icvec{p}cdotvec{R}} over sqrt{p^{2} + 1}},dd^{3}vec{p} \ = & m^{2}int_{0}^{infty}{1 over root{p^{2} + 1}} overbrace{pars{int_{Omega_{Largevec{p}}}expo{-icvec{p}cdotvec{R}},{ddOmega_{vec{p}} over 4pi}}} ^{ds{sinpars{pR} over pR}} 4pi p^{2},dd p \[5mm] = & {4pi m^{2} over R}int_{0}^{infty}{psinpars{pR} over root{p^{2} + 1}},dd p \[5mm] = & -,{4pi m^{2} over R}, partiald{}{R}int_{0}^{infty}{cospars{pR} over root{p^{2} + 1}},dd p \[5mm] = & -,{4pi m^{2} over R}, partiald{mrm{K}_{0}pars{R}}{R} end{align}

$ds{mrm{K}_{0}}$ is a Modified Bessel Function. See A & S $ds{bfcolor{black}{9.6.21}}$.

begin{align} &bbox[10px,#ffe]{iiint_{largemathbb{R}^{3}}{expo{-icvec{p}cdotvec{r}} over root{m^{2} + p^{2}}},dd^{3}vec{p}} = -4pi m^{2},{mrm{K}_{1}pars{R} over R} \[5mm] = & bbox[10px,#ffd,border:1px solid navy]{-4pi verts{m},{mrm{K}_{1}pars{verts{m}r} over r}} \ & end{align} See A & S $ds{bfcolor{black}{9.6.28}}$.

Answered by Felix Marin on November 26, 2021

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