Fourier transform of the components of a vector

[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{equation}
\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}
\end{equation}##

 

[Charles Link]

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{equation}
\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}
\end{equation}##

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.