# I Fourier transform of the components of a vector

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1. Dec 16, 2016

### redtree

Given the Fourier conjugates $\vec{r}$ and $\vec{k}$ where $\vec{r} = [r_1,r_2,r_3]$ and $\vec{k} = [k_1,k_2,k_3]$ , are $r_1$ /$k_1$, $r_2$/$k_2$, $r_3$/$k_3$ also Fourier conjugates, such that:
$\begin{split} f(\vec{r})&=[f_1(r_1),f_2(r_2),f_3(r_3)] \\ F[f(\vec{r})]&=F[f_1(r_1),f_2(r_2),f_3(r_3)] \\ F[f_1(r_1),f_2(r_2),f_3(r_3)]&=[\hat{f}_1(r_1),\hat{f}_2(r_2),\hat{f}_3(r_3)] \\ [\hat{f}_1(r_1),\hat{f}_2(r_2),\hat{f}_3(r_3)]&=\hat{f}(\vec{k}) \end{split}$

2. Dec 16, 2016

I believe the answer is yes, that e.g. $k_x$ and $x$, etc. are Fourier conjugates, by themselves, just as $\omega$ and $t$ are Fourier conjugates. In the multi-dimensional Fourier transform, the 3 coordinates are treated independently, and the time/frequency is also treated independently for a 4-dimensional F.T. $\\$ editing... One additional comment I could add is that for the Fourier transform theory and equations such as the convolution equation to work in F.T. space, there must be a linear response (editing) of the output vector field to the input vector field. e.g. in linear circuit theory, if you write $\tilde{V}_{out}(\omega)=\tilde{m}(\omega) \tilde{V}_{in}(\omega)$ you must have $V_{out}(t)=\int m(t-t')V_{in}(t') \, dt'$. The last integral equation is indicative of a linear temporal response.