Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am still unsure if Fourier Transform has any fundamental significance in Physics. Is it anything more than a calculation tool? For example sometimes people Fourier transform an equation to solve it relatively easily and then transform back. So from this point of view this is not a fundamental topic. Can anybody please clear my doubt: Would our knowledge in physics be reduced if Fourier transform were never discovered?

Finally, is it correct to say that Fourier series is more fundamental than Fourier transform? or are these equivalent?

share|cite|improve this question

closed as not constructive by dmckee Jan 2 '13 at 16:12

As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, visit the help center for guidance.If this question can be reworded to fit the rules in the help center, please edit the question.

These thought experiments are really hard to follow. Fourier transform/series emerge whenever one works with functions in such a way that the sines/cosines have an easy enough interpretation. It's not possible for things like the Fourier transform "not to be ever discovered". It could have been written/discovered later but ultimately people would be forced to write things that look easier with the Fourier transform and it would be discovered at least as convenience. All the identities that FT obeys would be known at some moment, anyway. – Luboš Motl Jan 2 '13 at 10:40
Also, the relationship between F. series and F. transform goes both ways. Fourier series are relevant for periodic functions, so they may be obtained from Fourier transform as a special case - the transform is a combination of delta-functions. On the other hand, Fourier transform may be obtained as a limit of Fourier series where the periodicity is sent to infinity. It's almost the same maths and the relationship between the two methods is both-sided and kind of a trivial bookeeping device. – Luboš Motl Jan 2 '13 at 10:41
The question seems to call for either the construction of a Big List (tm) or for an extended discussion, neither of which is well suited to the Q&A format. – dmckee Jan 2 '13 at 16:13
I think this question is poorly worded, but mathematically any Schwartz class of functions can be decomposed into Fourier components, fundamentally (it is an automorphism on that space), and I think most physicists mostly model physical systems as a Schwartz space. I suppose a good question would be, how do Banach spaces fit into physics? Can physical systems really have non-compact topologies? Of course Banach spaces make it easier to manipulate some of the objects (and spaces of those objects) that we use to operate on physical space, but are any physical spaces strictly Banach? – daaxix Jan 2 '13 at 17:48

The Fourier transform is at the heart of the uncertainty principle, and so is deeply fundamental to physics. As the components of momentum for a particle broaden, the sharpness (location) of the particle via a Fourier transform increases. Conversely, as the location of a particle becomes highly uncertain, its momentum state becomes more precise. See the Feynman Lectures, Volume III for more information on this remarkable relationship.

share|cite|improve this answer
This rests on the understanding of the FT as a decomposition into plane waves, rather than sinusoids, and the resulting parallel between the two. I've always considered that key to seeing the FT as something really fundamental. – Colin K Jan 2 '13 at 8:42
Dear Colin, decomposition into complex exponentials and decomposition into sines and cosines are equivalent and related by a completely trivial two-to-two relabeling of the amplitudes, via $\exp(ix)=\cos x + i\sin x$. Complex numbers are of course deep but once one understands basic things such as exponentials of complex numbers, the relationship between the two expressions is a trivial technicality and the complex exponentials are obviously the "more natural and easier" way to decompose functions according to those who have enough math background to work with complex numbers. – Luboš Motl Jan 2 '13 at 10:37
Yes. That's what I said. – Colin K Jan 2 '13 at 11:47

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