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In physics, whenever Fourier analysis is utilised to analyse a problem the term "Fourier mode" is often used, e.g. "a given function can be represented in terms of its Fourier modes". My question is, what exactly is meant by the term "Fourier mode"?

Is it in reference to a given wave oscillating at a fixed frequency? And then a given function is built up from an (infinite) superposition of these waves, each oscillating at a fixed frequency (ranging over a continuous frequency spectrum)?!

user35305
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    This has been asked before. Could you explain in what way the question and answers linked there do not address what you want to know? Also, the second paragraph of the question here is one very long and confusing sentence. You might want to edit it to be more readable. – DanielSank Aug 27 '16 at 22:12
  • @DanielSank I have updated the question to hopefully make it more comprehensible. I have read the answers in your link, but I'm still unsure of their relation and meaning in Fourier analysis?! – user35305 Aug 27 '16 at 22:42
  • You say that in the context of Fourier analysis in physics term "frequency mode" is used. I can try to guess when the author means but an example would make it more obvious. I'll write an answer meanwhile and can revise based on your updates. – DanielSank Aug 27 '16 at 22:47
  • "In particular, what is the relation to the Fourier modes of a given function?" I don't know what that is meant to ask. "Is it in reference to a given time-independent wave oscillating at a fixed frequency?" That's a contradiction in terms. A thing that is time-independent is not oscillating. – DanielSank Aug 27 '16 at 22:53
  • @DanielSank Good point. I didn't think the wording through correctly there, I meant oscillating at a time independent frequency. I have further updated the question. – user35305 Aug 27 '16 at 23:16

1 Answers1

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A mode in physics is, generally speaking, the spatial part of a waveform. A Fourier mode, more specifically, is a wave that oscillates sinusoidally in space. Thus, when we write the Fourier transform of a wave function $f(\mathbf r,t)$ as $$ f(\mathbf r,t) = \int\tilde f(\mathbf k,t)e^{i\mathbf k\cdot \mathbf r} \: \mathrm d\mathbf k, $$ the quote you give,

a given function can be represented terms of its Fourier modes,

means exactly that: we have a bunch of Fourier modes, which are the functions $ f(\mathbf k,t)e^{i\mathbf k\cdot \mathbf r}$, with $\mathbf k$ a static parameter, and then we add them all up (via integration over this parameter) to give us back our $f(\mathbf r,t)$.

The term 'mode' isn't used by itself very much, but it does appear with other qualifiers such as the normal modes of a system, which specifically requires a regular sinusoidal oscillation at each and every point of the system. Similarly to Fourier modes, these normal modes can then be used to represent any arbitrary wave as a sum of normal modes.

It's also important to note that 'mode' in general can apply to both continuous and discrete systems, as does 'normal mode'. You can in principle define Fourier modes for an infinite or cyclical chain of discrete coupled masses, characterized by a sinusoidal dependence on the (discrete) spatial index, but this is rarely used. In all of these cases, though, what sets the Fourier modes apart from other modes of oscillation is the sinusoidal spatial dependence.

Emilio Pisanty
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  • You've got the Fourier transform over spatial modes, but OP is asking about frequency. To link these, we need a wave equation. – DanielSank Aug 28 '16 at 00:40
  • @Daniel I've yet to see a paper using the term Fourier mode for a temporal oscillation, or indeed other than in the sense I described. If there are other usages I'd be interested, but to be honest the OP is much too muddled to draw distinctions that fine. – Emilio Pisanty Aug 28 '16 at 00:47
  • I'm not really sure what you mean. A "mode" almost always has spatial and a temporal aspects and a wave equation usually links them. Also, note that I can expand a function in a Fourier series (or transform) regardless of whether it's a function of $x$ or $t$. – DanielSank Aug 28 '16 at 01:25
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    @EmilioPisanty Ah ok, so at a given instant in time each term $f(\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{r}}$ is a waveform oscillating sinusoidally at some fixed spatial frequency $\mathbf{k}$ - a Fourier mode of the function $f(\mathbf{r},t)$ at a given instant in time, t. Each Fourier mode then represents a constituent of the function $f(\mathbf{r},t)$, in the sense that one can synthesise the function in terms of an infinite set of sinusoidal waveforms by integrating over the continuous range of spatial frequencies?! ... – user35305 Aug 28 '16 at 09:12
  • @EmilioPisanty ... Is they idea then that $f(\mathbf{r},t)$ satisfies some sort of wave equation and consequently the Fourier modes satisfy a corresponding wave equation, which dictates how they must evolve in time such that, at each instant $t$, the function $f(\mathbf{r},t)$ can be represented in terms of these Fourier modes as $f(\mathbf{r},t)=\int\tilde{f}(\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{r}}d\mathbf{k}$? – user35305 Aug 28 '16 at 09:20
  • @user35305 On your first comment, yeah, pretty much that. On your second content, almost. The Fourier transform as I wrote it applies to all functions, which means it's also not as useful. If f obeys a wave equation, however, the idea of the mode decomposition is to turn that PDE wave equation (hard) into a bunch of uncoupled ODEs (easy). We sometimes still call these "the wave equation" because they're it's heirs, so to speak, but they're just plain ODEs. – Emilio Pisanty Aug 28 '16 at 12:15
  • @EmilioPisanty Ok, so would you say that I've understood the intuitive meaning of Fourier modes (from what I wrote in my first comment? How does one interpret the time dependence in $\tilde{f}(\mathbf{k},t)$ in general? Is the point that the waveform $f(\mathbf{r},t)$ will be oscillating spatially, but in general, this oscillation will change in time, and so this is captured in the Fourier modes by allowing the amplitude of each Fourier mode to vary in time (such that its spatial oscillation remains time independent, by the amplitude of the oscillation at each spatial point varies in time)? – user35305 Aug 28 '16 at 12:52
  • @user35305 The time dependence on $\tilde f$ is chosen such that each $\tilde f(\mathbf k,t)e^{i\mathbf k\cdot\mathbf r}$ is a solution of the wave equation by itself. (Then, by linearity, the full superposition $f(\mathbf r,t)$ is also a solution.) The rest of your comments are pretty muddled, though. – Emilio Pisanty Aug 28 '16 at 13:23
  • @EmilioPisanty I was trying to parse your comment on the fact that any function can be represented in terms of Fourier modes and doesn't necessarily obey a wave equation?! – user35305 Aug 28 '16 at 14:16
  • I thought that the point of Fourier transforms was that a given function (satisfying mild conditions) can be represented in terms of an infinite number of sinusoidally oscillating functions ranging over a continuous set of frequencies?! – user35305 Aug 28 '16 at 14:30
  • @user35305 Yes. Every (suitably regular) function can be expanded as a Fourier transform, and this happens regardless of whether it obeys a wave equation or not. If it doesn't obey a wave equation (or if, for example, it obeys a nonlinear equation), the Fourier decomposition will generally exist but it will not be particularly helpful: the real trick is turning a PDE, or a set of coupled ODEs, into a set of uncoupled ODEs, and if that doesn't happen then the trick isn't useful. – Emilio Pisanty Aug 28 '16 at 15:25
  • Past this point, however, this much back and forth is not the way to go - you really just need to bury your nose in a book and think things through. – Emilio Pisanty Aug 28 '16 at 15:25
  • @EmilioPisanty Ok fair enough. Thanks very much for your help. Just one last thing, is it correct to say that each Fourier mode is a waveform, spatially oscillating sinusoidally at some fixed spatial frequency $\mathbf{k}$, and that the original function is then synthesised from an infinite set of such Fourier modes, each one oscillating a different frequency (of course, one integrates over the continuous range of frequencies)?! – user35305 Aug 28 '16 at 16:31
  • @user35305 Yes, that is correct, as already explained in the answer and the above comments. – Emilio Pisanty Aug 28 '16 at 16:44