Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This CW question is aimed at developing some intuition (grokking) about a certain formula of Fourier analysis. Any kind of explanation (physical, geometrical, analytical ...) is welcome.


If we have a function $$\begin{array}{cc}\phi\colon \mathbb{R}\times\mathbb{R}^n\to \mathbb{C},&\phi=\phi(t, x),\end{array}$$ we can take the space-time Fourier transform $$\widetilde{\phi}(\tau, \xi)=\int_{\mathbb{R}\times\mathbb{R}^n}\phi(t,x) e^{-i(t\tau+x\cdot\xi)}\, dt dx$$ and the spatial Fourier transform (on the time slice $t=0$) $$\widehat{f}(0, \xi)=\int_{\mathbb{R}^n} \phi(0, x)e^{-i x \cdot \xi}\, dx.$$

From the (space-time) Fourier inversion formula it follows that $$\tag{1}\widehat{f}(0, \xi)=\int_{\mathbb{R}}\widetilde{\phi}(\tau,\xi)\, d\tau.$$

Can you give some explanation of formula (1) that allows us to grok it?

share|improve this question

3 Answers 3

up vote 1 down vote accepted

Perhaps the following makes it more grokkable:

If $\phi$ allows separating the space and time parts, $\phi(t,x) = a(t)\cdot b(x)$, then

$$\tilde{\phi}(\tau,\,\xi) = \widehat{a}(\tau)\cdot\widehat{b}(\xi)$$

and $(1)$ is simply the Fourier inversion applied to the time part. The span of separated functions is dense, hence $(1)$ is valid for all $\phi$ by continuity.

share|improve this answer
    
This one gives some insight. Very nice. –  Giuseppe Negro Jul 24 '13 at 17:18

I think you may be off by a factor of $2 \pi$, but here's how I see it:

$$\int_{\mathbb{R}} d\tau \, \widetilde{\phi}(\tau, \xi) = \int_{\mathbb{R}} d\tau \,\int_{\mathbb{R}\times\mathbb{R}^n} dt \, dx \,\phi(t,x) e^{-i(t\tau+x\cdot\xi)} $$

Changing the order of integration:

$$\int_{\mathbb{R}} d\tau \, \widetilde{\phi}(\tau, \xi) =\int_{\mathbb{R}\times\mathbb{R}^n} dt \, dx \,\phi(t,x) e^{-i x\cdot\xi} \,\int_{\mathbb{R}} d\tau \, e^{-i t\tau} $$

The innermost integral is equal to $2 \pi \delta(t)$. Using the sifting property of the delta function, we get

$$\int_{\mathbb{R}} d\tau \, \widetilde{\phi}(\tau, \xi) = 2 \pi \int_{\mathbb{R}^n} dx \,\phi(0,x)\, e^{-i x\cdot\xi} = 2 \pi \widehat{f}(0,\xi)$$

share|improve this answer
    
I should have mentioned that I am "choosing units so that $2\pi=1$". :-) –  Giuseppe Negro Jul 24 '13 at 16:52
1  
@GiuseppeNegro: ah, so $\pi=1/2$ and indeed we may square the circle after all. Brilliant! –  Ron Gordon Jul 24 '13 at 16:54

Here is a measure-theoretic interpretation. Let $$\mu(d\tau, d\xi)=\widetilde{\phi}(\tau, \xi)d\tau d\xi$$ be the absolutely continuous measure on $\mathbb{R}\times\mathbb{R}^n$ associated to the space-time Fourier transform. Its marginal distributions are then $$\tag{2}\mu_\xi(d\xi)=\left(\int_{\mathbb{R}}\widetilde{\phi}(\tau, \xi)\, d\tau\right)d\xi,$$ $$\tag{3} \mu_\tau(d\tau)=\left(\int_{\mathbb{R}^n}\widetilde{\phi}(\tau, \xi)\, d\xi\right)d\tau.$$ So the formula (1) in Question above tells us that, when viewed as an absolutely continuous measure, the spatial Fourier transform is a marginal distribution of the space-time Fourier transform.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.