# What is Fourier transform of space variable? on the similar grounds what is the Laplace transform of the same?

I understand that the transform of time domain is frequency domain and the transformation of time to frequency domain is done by Fourier/Laplace transforms. I am confused about the transformation of space variables. What is the domain to which we transform in to? I read somewhere that Fourier transform of space variables is momentum and couldn't understand much. Also, if there be any, what is the difference between both Fourier and Laplace transforms?

• You can think of it as a transformation into complex frequency domain. The pure modes are $e^{st}=e^{\Re s t}\cos(\Im s t)+ie^{\Re s t}\sin(\Im s t)$ for $t \ge 0$. So it can include damped modes and instable exponential increasing modes. That's one way to look at it. – DisintegratingByParts Jul 26 '15 at 18:58

The Fourier Transform of a spatial variable is no different mathematically from a Fourier Transform of a temporal variable. The mathematics is agnostic to parameter interpretation.

For the Fourier Transform pair for the time-frequency domain are often written

$$F(\omega) = \mathscr{F}(f)(\omega) = \int_{-\infty}^{\infty} f(t) e^{i \omega t} \, dt$$

$$f(t) = \mathscr{F}^{-1}(F)(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega) e^{-i \omega t} \, d\omega$$

while the analogous notation for the spatial-spatial frequency domain are often written

$$F(k) = \mathscr{F}(f)(k) = \int_{-\infty}^{\infty} f(x) e^{i kx} \, dx$$

$$f(x) = \mathscr{F}^{-1}(F)(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} F(k) e^{-i kx} \, dk$$

Certainly, the only difference between these pairs is symbolic.

However, in physics and engineering, one assigns units to these symbols. For the time-frequency transform pair, units of time and inverse time are assigned to the canonical parameters $$t$$ and $$\omega$$, respectively, and hence the reason we have a time-domain-frequency domain pair. For example, units could be in seconds and inverse seconds (i.e. radians/second).

When we move to the spatial Fourier Transform, the canonical units $$x$$ and $$k$$ are, for example, meters and inverse meters. The interpretation of inverse meters is that of a "wave number," and represents a spatial frequency for a traveling (or standing) wave.

The interpretation of the spatial Fourier Transform yielding momentum originates in quantum mechanics for which we have the relationship $$p=k\hbar$$, where $$\hbar$$ is the Dirac constant or reduced Planck's constant. Then, letting $$k=p/\hbar$$, we have

$$F(p) = \mathscr{F}(f)(p) = \int_{-\infty}^{\infty} f(x) e^{i px/\hbar } \, dx$$

$$f(x) = \mathscr{F}^{-1}(F)(x) = \frac{1}{2\pi\hbar }\int_{-\infty}^{\infty} F(p) e^{-i px/\hbar } \, dp$$

where $$F(p)$$ is called the momentum representation of $$f(x)$$

• Use \hbar for $\hbar$. I was in the process of writing an answer quite similar to this but gave up on it since it felt more like a physics question to me after I delved into the QM aspect of the duality between momentum space and physical space. – Cameron Williams Jul 26 '15 at 19:12
• @CameronWilliams Thanks! I'll edit. – Mark Viola Jul 26 '15 at 19:12
• Thanks man now I have a better understanding now. I would just like to ask that on the similar grounds what will be the interpretation of Laplace transform? As the Laplacian variable is s = (sigma) + i(omega)..In frequency domain "sigma" signifies the transient damping of signal so what is the interpretation in spatial frequency and momentum domain? Does it similarly signify damping? – AVyas Jul 27 '15 at 6:31
• @AVyas You're welcome. My pleasure. And you're correct. Whether Laplace or Fourier, the waves can be evanescent. – Mark Viola Jul 27 '15 at 12:27