Monday Math 128

We have previously discussed the Fourier transform (here and here, especially). In this post, we noted that (using the symmetric angular convention) the space transform for an n dimensional space is
$\tilde{f}(\mathbf{k})=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}f(\mathbf{x})e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,d^nx$
and the inverse is
$f(\mathbf{x})=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}\tilde{f}(\mathbf{k})e^{\imath\mathbf{k}\cdot\mathbf{x}}\,d^nk$ .
We can also do the same for a vector field:
$\mathbf{\tilde{v}}(\mathbf{k})=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}\mathbf{v}(\mathbf{x})e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,d^nx$
and
$\mathbf{v}(\mathbf{x})=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}\mathbf{\tilde{v}}(\mathbf{k})e^{\imath\mathbf{k}\cdot\mathbf{x}}\,d^nk$ .
We note from vector calculus then that for a vector field, the components of the transform are the transforms of the components:
$\begin{eqnarray}\mathcal{F}[\mathbf{v}(\mathbf{x})](\mathbf{k})&=&\mathcal{F}[v_x\mathbf{\hat{x}}+v_y\mathbf{\hat{y}}+v_z\mathbf{\hat{z}}](\mathbf{k})\\&=&\mathcal{F}[v_x](\mathbf{k})\mathbf{\hat{x}}+\mathcal{F}[v_y](\mathbf{k})\mathbf{\hat{y}}+\mathcal{F}[v_z](\mathbf{k})\mathbf{\hat{z}}\end{eqnarray}$ .

We also used integration by parts here to show that for a one-dimensional function f(x), with $\lim_{x\to\pm\infty}f(x)=0$ , the derivative has Fourier transform:
$\mathcal{F}[f'(x)](k)=\imath{k}\mathcal{F}[f(x)](k)$ .
Similarly, we can use vector integration by parts for our multi-dimensional transforms. Working in three dimensions from here:
First, one form of the divergence theorem states:
$\iiint_{V}\mathbf{\nabla}f\,dV=\iint_{S}f\,d\mathbf{A}$ ,
where S is the boundary of the volume V, with outward normal.
Letting f=φψ, and using the gradient product rule $\mathbf{\nabla}(\phi\psi)=(\mathbf{\nabla}\phi)\psi+\phi(\mathbf{\nabla}\psi)$ ,
$\iiint_{V}(\mathbf{\nabla}\phi)\psi\,dV+\iiint_{V}\phi(\mathbf{\nabla}\psi)\,dV=\iint_{S}\phi\psi\,d\mathbf{A}$
$\iiint_{V}(\mathbf{\nabla}\phi)\psi\,dV=\iint_{S}\phi\psi\,d\mathbf{A}-\iiint_{V}\phi(\mathbf{\nabla}\psi)\,dV$ .
Letting $\psi(\mathbf{x})=e^{-\imath\mathbf{k}\cdot\mathbf{x}}$ , we see
$\iiint_{V}(\mathbf{\nabla}\phi)e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,d^3x=\iint_{S}{\phi}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,d\mathbf{A}-\iiint_{V}\phi\mathbf{\nabla}\left(e^{-\imath\mathbf{k}\cdot\mathbf{x}}\right)\,d^3x$ .
and since $\mathbf{\nabla}e^{-\imath\mathbf{k}\cdot\mathbf{x}}=-\imath\mathbf{k}e^{-\imath\mathbf{k}\cdot\mathbf{x}}$ , we have
$\iiint_{V}(\mathbf{\nabla}\phi)e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,d^3x=\iint_{S}{\phi}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,d\mathbf{A}+\imath\mathbf{k}\iiint_{V}\phi{e}^{-\imath\mathbf{k}\cdot\mathbf{x}}\,d^3x$ .
Now, if as $r=\left|\mathbf{x}\right|\to\infty$ , φ(x) goes to zero faster than $\frac1r$ , then, as we expand the volume V to cover all space, the surface integral will go to zero, and we obtain
$\iiint(\mathbf{\nabla}\phi)e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,d^3x=\imath\mathbf{k}\iiint\phi\mathbf{\nabla}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,d^3x$ ,
which means
$\mathcal{F}\left[\mathbf{\nabla}\psi(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\mathcal{F}\left[\psi(\mathbf{x})\right](\mathbf{k})$ ,
in analogy to our one-dimensional rule.

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Tags: Divergence Theorem, Fourier Transform, Math, Monday Math, Wavenumber

This entry was posted on July 26, 2010 at 12:14 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

One Response to “Monday Math 128”

Monday Math 129 « Twisted One 151's Weblog Says:
August 2, 2010 at 12:16 am | Reply
[...] Math 129 By twistedone151 Continuing from last week, we now consider the Fourier transform of the divergence of a vector field in three dimensions. [...]

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Monday Math 128

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