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?

  • $\begingroup$ 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. $\endgroup$ Commented Jul 26, 2015 at 18:58

1 Answer 1


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 $f(x)$

  • $\begingroup$ 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. $\endgroup$ Commented Jul 26, 2015 at 19:12
  • $\begingroup$ @CameronWilliams Thanks! I'll edit. $\endgroup$
    – Mark Viola
    Commented Jul 26, 2015 at 19:12
  • $\begingroup$ 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? $\endgroup$ Commented Jul 27, 2015 at 6:31
  • $\begingroup$ @AVyas You're welcome. My pleasure. And you're correct. Whether Laplace or Fourier, the waves can be evanescent. $\endgroup$
    – Mark Viola
    Commented Jul 27, 2015 at 12:27

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