Fourier transform fallacy? (Optics)

[Jufa]

Here it goes. I have been taught that a finite pulse of light does not have a single frequency. By finite pulse I was given an example of a source of light that has been emitted during a finite amount of time and, consequently, covers a finite region of space. Then I was taught that you can actually distinguish these frequencies by making this pulse encounter a diffracting medium such as a diffraction grating or a prism. At that point I said: "Okay, I don't find this intuitive to happen, but I have to take it because it does happen." It is the explanation I was given later what made me think more about this subject and I ended up encountering some problems in it that, hopefully, someone will fix for me because I already assume that there must be something that I am misunderstanding.
The common answer is that you can decompose this finite signal into an infinite set of infinite harmonic waves with a defined frequency and you then find the actual spectrum of the signal, the same you would find experimentally.
I will now point out what I find that is not consistent about this explanation:

1- One of the conclusions you can reach following this reasoning is that measuring the frequency spectrum of a finite pulse of light actually means finding its Fourier decomposition not within the space covered by the wave, but within all the space (from minus infinite to plus infinite). I find this correspondence between measurement and decomposition in a set of space where the wave does not actually "live" quite counterintuitive.
By "Measuring the spectrum of frequencies of a pulse of light (that must include, at least, one period) " I understand "Measuring the frequency of oscillation of the electric field defined all along the distance covered by the pulse in a certain moment of time." Suppose you create an electromagnetic wave by making an electron oscillate with a certain frequency within an antenna. The wave you get oscillates with the same frequency as the electron does. And if you perform the sort of measurement I described you should get a single frequency (Dirac's Delta). No matter if the antenna has been working for a minute or for decades. I'm not 100% sure of the latter mentioned and it might be the main point of the question. Maybe you cannot know the frequency of the electron oscillation with arbitrary precision (due to quantum phenomena) but, if so, how can the accuracy of this frequency depend on the time the antenna has been working?

2- Let's assume there is not such a quantum reason (I have not found it anywhere). Then what we have is two ways of measuring that provide different results: The diffraction will tell us that the pulse is made of a spectrum of frequencies and the simple measurement of the field oscillation experiment will tell us that the pulse has a single frequency. One would say: "When performing the experiment I described you get a single frequency because the other frequencies are "hided" due to destructive interferences. So the most general experiment you can perform in order to measure all the frequencies of a pulse (even the hided ones) is the one where, for instance, the pulse passes through a diffraction prism. That way waves with different frequency will separate and will not interfere destructively anymore. Then you compute the frequencies of the waves that come out of the prism." And, as far as I know, that is what does happen experimentally. But this suggests me two inconsistencies:
-How do we measure the frequencies that come out of the prism? These waves are finite as well and that means that they must "hide" some sort of spectrum. So the measurement would never end.
-This is the inconsistency that worries me the most: Fourier Theorem and Superposition Theorem all together tell us that the electric field defined by the finite pulse and the infinite sum of infinite harmonic waves are equivalent. That would predict the "prism experiment" already mentioned but it also predicts that you could perform this experiment anywhere, since the extent of the harmonic functions is infinite. That leads to an information transfer faster than c (light speed) which makes no sense. Obviously, the latter does not happen experimentally, which makes me think that there's something about Superposition Theorem or Fourier's Theorem (or even both) that I misunderstand.
Thank you for reading.

 

[Charles Link]

The frequency spread/bandwidth of the sinusoidal pulse is approximately $\Delta f \approx \frac{1}{T}$ where $T$ is the duration of the pulse. A good spectrum analyzer should be able to show this kind of result for sinusoidal pulses where the sinusoidal frequency is in the r-f region or lower. $\\$ In the optical regime, a prism spectrometer is not going to have the necessary resolving power to measure any kind of bandwidth/line width, because the $\Delta \lambda_{minimum}$ for the instrument will be much larger than the line width of the pulsed optical signal. With a very high resolution diffraction grating spectrometer, you might be able to observe/measure the line widths $\Delta \lambda$ of very short laser pulses.

 

[ZapperZ]

Summary: Inconsistency regarding to Fourier decomposition of finite pulses. Someone help me.

Here it goes. I have been taught that a finite pulse of light does not have a single frequency. By finite pulse I was given an example of a source of light that has been emitted during a finite amount of time and, consequently, covers a finite region of space. Then I was taught that you can actually distinguish these frequencies by making this pulse encounter a diffracting medium such as a diffraction grating or a prism. At that point I said: "Okay, I don't find this intuitive to happen, but I have to take it because it does happen." It is the explanation I was given later what made me think more about this subject and I ended up encountering some problems in it that, hopefully, someone will fix for me because I already assume that there must be something that I am misunderstanding.
The common answer is that you can decompose this finite signal into an infinite set of infinite harmonic waves with a defined frequency and you then find the actual spectrum of the signal, the same you would find experimentally.
I will now point out what I find that is not consistent about this explanation:

1- One of the conclusions you can reach following this reasoning is that measuring the frequency spectrum of a finite pulse of light actually means finding its Fourier decomposition not within the space covered by the wave, but within all the space (from minus infinite to plus infinite). I find this correspondence between measurement and decomposition in a set of space where the wave does not actually "live" quite counterintuitive.
By "Measuring the spectrum of frequencies of a pulse of light (that must include, at least, one period) " I understand "Measuring the frequency of oscillation of the electric field defined all along the distance covered by the pulse in a certain moment of time." Suppose you create an electromagnetic wave by making an electron oscillate with a certain frequency within an antenna. The wave you get oscillates with the same frequency as the electron does. And if you perform the sort of measurement I described you should get a single frequency (Dirac's Delta). No matter if the antenna has been working for a minute or for decades. I'm not 100% sure of the latter mentioned and it might be the main point of the question. Maybe you cannot know the frequency of the electron oscillation with arbitrary precision (due to quantum phenomena) but, if so, how can the accuracy of this frequency depend on the time the antenna has been working?

2- Let's assume there is not such a quantum reason (I have not found it anywhere). Then what we have is two ways of measuring that provide different results: The diffraction will tell us that the pulse is made of a spectrum of frequencies and the simple measurement of the field oscillation experiment will tell us that the pulse has a single frequency. One would say: "When performing the experiment I described you get a single frequency because the other frequencies are "hided" due to destructive interferences. So the most general experiment you can perform in order to measure all the frequencies of a pulse (even the hided ones) is the one where, for instance, the pulse passes through a diffraction prism. That way waves with different frequency will separate and will not interfere destructively anymore. Then you compute the frequencies of the waves that come out of the prism." And, as far as I know, that is what does happen experimentally. But this suggests me two inconsistencies:
-How do we measure the frequencies that come out of the prism? These waves are finite as well and that means that they must "hide" some sort of spectrum. So the measurement would never end.
-This is the inconsistency that worries me the most: Fourier Theorem and Superposition Theorem all together tell us that the electric field defined by the finite pulse and the infinite sum of infinite harmonic waves are equivalent. That would predict the "prism experiment" already mentioned but it also predicts that you could perform this experiment anywhere, since the extent of the harmonic functions is infinite. That leads to an information transfer faster than c (light speed) which makes no sense. Obviously, the latter does not happen experimentally, which makes me think that there's something about Superposition Theorem or Fourier's Theorem (or even both) that I misunderstand.
Thank you for reading.

When you solve problems like this, you assume that the "boundary conditions" do not have a significant effect on what you are doing. In your case, you assume that the wave is in a steady state, and that when it started and when it will end are far enough away in time that they are not affecting your measurement.

This is not unique. In solid state physics, we usually consider the crystal structure to be infinite in all directions such that the boundary of the material does not play a role when we are "far" enough away from it. We then work out solution and see if they match our observations. If they do, then our assumptions are valid for our purposes.

What you need to figure out is if the sum of all the Fourier components actually will produce a finite pules of the shape that you want. If it does, then ignoring the temporal boundary conditions is valid.

Zz.

 

[hutchphd]

A little bit of knowledge is a dangerous thing. Many of the questions you ask (and there are a lot of them!...that's good) require careful consideration of the sizes of effects and differentiation between mathematical constructs and actual real world systems. It is not my intent to comprehensively answer your query but to point out a few things for your consideration. I'm sure others will help.

The formal question of Fourier synthesis does assume the the artificial construct of the infinite (time, frequency, space, wavenumber) within which you can arbitrarily accurately represent any particular function. In practice any laboratory experiment is finite in space and time. But these numbers are always huge often 10¹⁰ compared to the measurements in question. So it is a very very useful approximation for the real world description.

As you correctly surmised, your assumption about the "pure" emission from an oscillating electron is subject to the same limitations in energy precision as is the concurrent photon. If the electron state has a finite lifetime, then there is an uncertainty in energy and it all gives the same result .

The issues of superluminal influence that concern you are not typically addressed because they do not materially affect the outcome in most circumstances. There are places where they need to be considered explicitly but for mostly the effect of using "instant" time is not important, and it makes life much simpler.

If you have further specific questions they will be addressed, and Google can help, too.

.

 

[Nugatory]

Suppose you create an electromagnetic wave by making an electron oscillate with a certain frequency within an antenna. The wave you get oscillates with the same frequency as the electron does. And if you perform the sort of measurement I described you should get a single frequency (Dirac's Delta). No matter if the antenna has been working for a minute or for decades.

The oscillating electron is measuring the time-varying electromagnetic field at only a single point, and many different waveforms will produce the same oscillation at a single point. To see what the Fourier transform is telling us is there we need to look at the behavior of multiple electrons in multiple locations; that will be different with different waveforms and pulse widths.

 

[Dale]

Summary: Inconsistency regarding to Fourier decomposition of finite pulses. Someone help me.

The wave you get oscillates with the same frequency as the electron does. And if you perform the sort of measurement I described you should get a single frequency (Dirac's Delta).

If the electron has been oscillating for a finite time then the electron’s frequency band is infinite also.

 

[Jufa]

A little bit of knowledge is a dangerous thing. Many of the questions you ask (and there are a lot of them!...that's good) require careful consideration of the sizes of effects and differentiation between mathematical constructs and actual real world systems. It is not my intent to comprehensively answer your query but to point out a few things for your consideration. I'm sure others will help.

The formal question of Fourier synthesis does assume the the artificial construct of the infinite (time, frequency, space, wavenumber) within which you can arbitrarily accurately represent any particular function. In practice any laboratory experiment is finite in space and time. But these numbers are always huge often 10¹⁰ compared to the measurements in question. So it is a very very useful approximation for the real world description.

As you correctly surmised, your assumption about the "pure" emission from an oscillating electron is subject to the same limitations in energy precision as is the concurrent photon. If the electron state has a finite lifetime, then there is an uncertainty in energy and it all gives the same result .

The issues of superluminal influence that concern you are not typically addressed because they do not materially affect the outcome in most circumstances. There are places where they need to be considered explicitly but for mostly the effect of using "instant" time is not important, and it makes life much simpler.

If you have further specific questions they will be addressed, and Google can help, too.

.

Thanks for your time. As to other replies, I appreciate your explanation, which helped me to overcome some of my doubts. After carefully reading your reply I think I can restate what actually "bothers" me now (I'm affraid it will take more than a few lines):
It seems to me that if you "ask" a pulse of light for its time duration (trivially related to the distance covered by the pulse) and for its frequency it will not answer both questions with arbitrary precision. In fact, the accuracy of both questions satisfy this sort of relation $\Delta f \Delta t \simeq 1$. Within the quantum physics paradigm, this would not bother me (just because I have already been training myself to not being bothered by quantum phenomena that actually bothers me). It would be a sort of Heisenberg relation. I could understand it analogously. What bothers me now is that I have been taught this in classical optics and as far as I know the sort of experiments I describe was performed before we had any quantum view of the world. It seems to me that everyone I ask about this topic assumes that asking for the frequency of a finite pulse (let's assign this pulse a certain state) will only provide you answers that are not actually the frequency of this state but the frequency of other states (infinite pulses) that properly summed result in the state of the finite pulse. I find this assumption quite "quantum". It seems to me that the measurement of the frequency of a pulse is equivalent to a sort of operator the eigenstates of which are the infinite harmonic functions and that this operator does not commutate with another sort of operator that tells you the distance covered by the pulse.

Classically I could understand it by the following reasoning: <<If I want to accurately measure the "pure" frequency of a pulse I have to perform the measure several times in order to get a lower uncertainty. Then if the pulse I am working with only contains a finite number of periods I will have some sort of uncertainty in the frequency.>>

I do understand this and it does lead as well to the same result $\Delta f \Delta t \simeq 1$ but I don't see its correspondence with Fourier Synthesis. I had only seen this correspondence between probabilities and actual intensity in Quantum Physics, when you assume that light is not emitted continuosly.

 

[hutchphd]

I do understand this and it does lead as well to the same result ΔfΔt≃1ΔfΔt≃1 \Delta f \Delta t \simeq 1 but I don't see its correspondence with Fourier Synthesis.

I am unsure what your confusion is. The Fourier transform of a Gaussian bump in time gives a Gaussian bump in frequency obeying the uncertainty relation. Any other shaped bump in time will produce a larger corresponding"uncertainty" relation. Many information theoretical results (like required radio bandwidth) rely upon these relationships. These results depend not at all upon Quantum Mechanics per se.

 

[Jufa]

Okay. I guess there is something basic I'm missing. I'm sorry, I guess this should be labelled as "undergrad" but I have already questioned some teachers at my university teachers and they did not come up with a clear answer. I will think it deeper. Thank you for your helping!

 

[hutchphd]

These are not foolish questions, and so don't hesitate to ask them. But I don't quite understand your exact confusion.

 

[ZapperZ]

Okay. I guess there is something basic I'm missing. I'm sorry, I guess this should be labelled as "undergrad" but I have already questioned some teachers at my university teachers and they did not come up with a clear answer. I will think it deeper. Thank you for your helping!

If this is what you asked them, then it is not that they can't come up with a clear answer, it is because this isn't a clear question!

The ability to answer something depends on the clarity and completeness of the question. We can't investigate something when we don't know exactly what it is that we're asking or searching for.

The mathematics of HUP isn't new. I've said that many times already on here, and that the correspondence between HUP and Fourier transform is in the math. I'm not sure why this is puzzling or a "fallacy".

Zz.

 

[Charles Link]

It might be worth mentioning to the OP that sources that are nearly monochromatic or a single frequency are never precisely such. There are different degrees of being nearly monochromatic, and the coherence length is a measure of this. The coherence length can be measured with a Michelson interferometer or similar techniques. The source being tested will have the fringe patterns start to fade when the path difference from the two arms (the two paths from each of the mirrors) is of the same magnitude as the coherence length. See https://en.wikipedia.org/wiki/Coherence_length

 

[Jufa]

If this is what you asked them, then it is not that they can't come up with a clear answer, it is because this isn't a clear question!

The ability to answer something depends on the clarity and completeness of the question. We can't investigate something when we don't know exactly what it is that we're asking or searching for.

The mathematics of HUP isn't new. I've said that many times already on here, and that the correspondence between HUP and Fourier transform is in the math. I'm not sure why this is puzzling or a "fallacy".

Zz.

Sure ZapperZ, you are right. I think that for me a clear answer would be a reformulation of my question in a way that it is clear and then answer it. I know it is too much and the job of reformulation is actually mine jajaja. Anyway, I truly appreciate you sharing your time and knowledge. This sort of feedback helps me a lot because it makes me realize that I'm losing myself somewhere while reasoning. I'll try to rethink it and in case I come up with some specific doubts I will let you know. Apologise me for the title of the thread, I guess it is not quite appropriate.

 

[gmax137]

Maybe this is off topic (if so, I apologize) but this reminds me of a long-ago physics class where we came to a similar conclusion, in the context of musical instruments; with the bottom line being tuba notes must be held longer than, say, piccolo notes. If you try to play a tuba too fast all you get is mush. This appeared on the chalkboard as ΔfΔt≃1.

 

[Charles Link]

One other input for the OP besides my post 12. It may interest the OP that any time you modulate a single frequency, either by pulsing it, or even by having amplitude fluctuations, (such as AM radio), you introduce a finite bandwidth $\Delta f$ centered around the fundamental frequency $f_o$ to what would otherwise be a very narrow bandwidth. The spectral shape around that $\Delta f$ can be Gaussian in shape, or it can consist of side bands or other fine structure. $\\$ Parseval's theorem is also quite important for relating Fourier amplitudes to power spectral densities. See: https://en.wikipedia.org/wiki/Parseval's_theorem $\\$It may also interest you that a diffraction grating spectrometer with photodiodes as receivers will give an intensity (power density) spectrum, rather than a Fourier decomposition of the received signal.

 

[Dale]

It would be a sort of Heisenberg relation. I could understand it analogously. What bothers me now is that I have been taught this in classical optics and as far as I know the sort of experiments I describe was performed before we had any quantum view of the world.

Actually, the Heisenberg relation is derived from the Fourier transform.

I find this assumption quite "quantum".

This is actually a little backwards. The Heisenberg uncertainty principle is based on the classical Fourier transform. The only thing that is quantum about it is that the scale is based on Planck's constant. Since you here have $\Delta f \Delta t \approx 1$ and there is no $\hbar$ involved it is classical.

 

[Jufa]

Thank you. I did not notice that. This makes much more sense.

 

[Jufa]

Thanks everyone. I think there is something basic that I'm missing so I will try to get a wider knowledge regarding to Fourier Synthesis and Spectroscpy. In fact I would like to understand why does a diffraction grating perform the Fourier transform of a signal instead of Fourier Series. I don't understand why the wave is not decomposed within its actual domain of action where, as a premise, it is periodic, but within all the space.
My intuitive reasoning path fails dramatically: I imagine that a harmonic infinite wave, when passing through a diffraction grating, gives a single frequency all time you watch at the screen and that a finite harmonic frequency should give the same but only during the time that it is actually passing through the grating which, in this case would be finite.

 

[sophiecentaur]

does a diffraction grating perform the Fourier transform of a signal instead of Fourier Series.

Those are not alternative options. The transform is a process and the Series is the result of a Transform, A grating will perform a transform from the time domain variation of the incoming ( very wide bandwidth) wave to a spatial domain distribution (angles) of the emerging spread beam. Truncating the time that the incident wave is applied will spread / blur / distort the angle distribution of the emerging waves because the transform assumes an infinitely long input time or (for a discrete transform) an infinitely repeating 'signal' (tape loop equivalent). The 'series' consists of harmonics of the truncation function and not of the original function. The basics are the same as for audio spectrum analysis and the comb of products that are seen in that case are harmonics of the period of the string of audio samples. The corresponding optical spectrometer process is (for me at least) a bit less straightforward but Fourier is Fourier.

You may need to read your textbook passages several times in order to get this stuff straight and view this (reversible) process forwards and backwards to see what is actually going on. Fourier analysis never truly sticks to the ideal -∞/+∞ limits of the Fourier Integral so it will always produce artefacts which may or may not be significant.

 

[Charles Link]

A prism spectrometer works by refraction with dispersion, where the index of refraction $n=n(\lambda)$. Thereby, after sending collimated light=(parallel rays) through the faces of prism, different wavelengths emerge at slightly different angles. Light at a given incident angle comes to focus at a given location in the focal plane of a lens=thereby giving the spectrum as function of position. (The parallel rays of the light incident on the prism are created by focusing the light onto the entrance slit of the spectrometer, which is located at the focal point of another lens to create a collimated beam=parallel rays). Similar optics are employed in a diffraction grating spectrometer, where a diffraction grating replaces the prism as the dispersive element. $\\$ When you study the derivation of the formula for the far field intensity pattern (which is the same as the pattern in the focal plane of the focused parallel rays that emerge from the grating) for light of wavelength $\lambda$ from a diffraction grating $I(\theta)=I_o \frac{\sin^2(N \phi/2)}{\sin^2(\phi/2)}$ where $\phi=\frac{2 \pi d \sin(\theta)}{\lambda}$, you will see the dispersion that results, where the energy of wavelength $\lambda$ emerges at angle $\theta$ that depends upon the wavelength. $\\$ (Note: The Fourier transform type mathematics that arises in computations with the diffraction grating, (to compute $I(\theta)$ from the incident monochromatic beam on the grating), is distinct from the computation of the Fourier transform $\tilde{E}(\omega)$ from $E(t)$). $\\$ For r-f (radio frequencies) we often have $\tilde{V}(\omega)$ computed from $V(t)$. Most often for optical frequencies, we simply have intensity $I=I(\lambda)$. The intensity pattern $I$ is proportional to $|\tilde{E}(\omega)|^2$, with frequency $\omega$ ( or wavelength $\lambda$ ) being a function of position in the focal plane where the spectrum is observed. Oftentimes, the spectral intensity $I(\lambda)$ is what gets measured=typically with a spectrometer. (At optical frequencies, we normally are not given an $E(t)$ to process, i.e. we do not compute $\tilde{E}(\omega)$). $\\$ (Note: I edited this a couple of times. Hopefully it now gives some helpful info).

 

[Jufa]

Those are not alternative options. The transform is a process and the Series is the result of a Transform, A grating will perform a transform from the time domain variation of the incoming ( very wide bandwidth) wave to a spatial domain distribution (angles) of the emerging spread beam. Truncating the time that the incident wave is applied will spread / blur / distort the angle distribution of the emerging waves because the transform assumes an infinitely long input time or (for a discrete transform) an infinitely repeating 'signal' (tape loop equivalent). The 'series' consists of harmonics of the truncation function and not of the original function. The basics are the same as for audio spectrum analysis and the comb of products that are seen in that case are harmonics of the period of the string of audio samples. The corresponding optical spectrometer process is (for me at least) a bit less straightforward but Fourier is Fourier.

You may need to read your textbook passages several times in order to get this stuff straight and view this (reversible) process forwards and backwards to see what is actually going on. Fourier analysis never truly sticks to the ideal -∞/+∞ limits of the Fourier Integral so it will always produce artefacts which may or may not be significant.

I think I'm getting closer to let you know what exactly "bothers" me so that someone can let me know exactly what I am misunderstanding. In the last post, I said a few things that are actually wrong as you pointed out (by FOURIER SERIES instead of Fourier Transform I meant FOURIER TRANSFORM WITHIN A FINITE DOMAIN, as the pulse is already periodic, which leads to a signal made out of Series that, in this case, would be a truncated series of only one term, since the signal is already harmonic) I don't see why we decompose our wave into infinite waves. What bothers me is not that they are infinite. If they were finite but larger in time (and consequently in space) than our finite pulse it would bother me as well. What does bother me is that they, added together, do not describe the same system as the finite pulse, because, as I pointed out in my first post, these waves could be separated from each other at any point of the space and the time one from each other (using spectrometry) leading either to a superluminal paradox or a causality paradox. So I don't see why this decomposition can really have a physical meaning. It is true that this decomposition does contain the same energy as the finite pulse and it does describe the same electromagnetic field propagation through all space but this does not mean that they surely describe the same system (I think that I have just clearly shown that they do not).

Let me insist in what does concern me. I think that every person that replies assumes one thing and is this thing what actually bothers me. This assumption is that THE FREQUENCY SPECTRUM MEASURED ASSUMES A DECOMPOSITION OF THE PULSE INTO INFINITE HARMONIC WAVES.
I say: "Okay, the frequency of the pulse is WELL-DEFINED, even though we cannot measure it accurately because we only have one pulse and a single measure would have large uncertainty, it MUST BE A SINGLE ONE. In all the derivations I find of the frequency bandwidth of finite pulses they all start assuming that we are dealing with a full periodic signal that includes a whole number of periods so I think it is fair to assume that this frequency does exist. Now let's picture the following experiment: you make an infinite harmonic wave with the same frequency as the one mentioned above go through a prism. And we look at its spectrum which, as far as I know, should remain at the screen as long as the wave is still going through the prism. Let's say we just look at the spectrum during a finite time. Let's say this amount of time is the same amount of time that would take to the finite pulse to go through the same prism. Now I guess that during this amount of time, as at any other moment we should see a single angle of deviation at the screen with a very tiny bandwidth and this bandwidth would be just a consequence of the non-ideal resolution of our spectrometer.
Now, during this finite time what would have happened if instead of the harmonic infinite wave we had put our finite pulse? Well, it seems that what happens is that instead of a tiny width we get a sort of a continuous spectrum. Both results are quite different. In fact, they are so different that we should be able to distinguish both scenarios just by looking to the screen. But aren't they the same scenarios? During this time, the light that has passed through the spectrometer is STRICTLY THE SAME, if by light we refer to the electromagnetic field perturbation that propagates through space at a velocity c (which I think it is a shared premise). How can the outcome of the grating be affected by whatever is passing through it later to it or whatever has already passed through it before? In other words. How can the grating "know" if the pulse of light that is going through it is a finite pulse or is just a part of an infinite pulse.
When someone says: "As the time is finite you will get a wider bandwidth of frequency because there is uncertainty" I don't understand it. I mean, maybe the frequency you get by measuring twice the same pulse is slightly different because of experimental error, but every measurement will give you only ONE SINGLE frequency. If you then represent all these frequencies in a sort of a histogram you will get the bandwidth but not in a single measurement.
I hope that this post highlights my misunderstanding of the whole process so that someone can tell me where am I losing myself.
Thank you for reading.

 

[Charles Link]

Let's consider normal incidence of a beam of monochromatic light onto a grating, and the maximum that occurs at $\theta$ in the far field at $m=1$, where $(1)(\lambda)=d \sin(\theta)$. There are phase differences, across the grating, basically $\phi(x)=\frac{2 \pi x \sin(\theta)}{\lambda}$ between the different portions of the beam that constructively interfere. (Basically, for the $n$th slit, the phase is $\phi_n=n 2 \pi$). $\\$ If the source is coherent over the time corresponding to this variation in $\phi$, then the interference will be very complete, and the diffraction peak very narrow (in $\Delta \theta$). (All $N$ slits constructively interfere). $\\$ If alternatively, the source has a short coherence time,(so that portions of the beam have slight additional phase variations), the result will be a wider (in $\Delta \theta$) diffraction maximum, which corresponds to a wider observed $\Delta \lambda$. $\\$ e.g. with a very short pulse of frequency $f=c/\lambda$, only a portion of the $N$ slits on the grating will contribute to the constructive interference at angle $\theta$ at any instant. Fewer lines on the grating results in a wider (in $\Delta \theta$) observed diffraction maximum. The frequency may be precisely $f$, but the short pulse duration will inevitably result in a measured $\Delta \lambda$ that is larger, because of the short pulse duration.

 

[hutchphd]

I hope that this post highlights my misunderstanding of the whole process so that someone can tell me where am I losing myself.

I believe that your intuition for these processes is failing you because you are not appreciating the relative sizes of these effects Put some actual values into your thought experiments and work it through. Even a short light pulse has thousands of wavelengths

 

[sophiecentaur]

In all the derivations I find of the frequency bandwidth of finite pulses they all start assuming that we are dealing with a full periodic signal that includes a whole number of periods so I think it is fair to assume that this frequency does exist.

The Fourier Transform works over all space and all time. Whenever you use it in a practical case, you introduce errors. The fact that we use Fourier analysis so often shows that the errors are very often 'acceptable' - just the same as the effects of noise are always there but not necessarily drastic. Having a repeated pulse improves both the accuracy of the FT and the uncertainty due to noise. Those two effects may give useful results even if it makes you feel uneasy. We're talking real world.

 

[sophiecentaur]

How can the grating "know" if the pulse of light that is going through it is a finite pulse or is just a part of an infinite pulse.

If you remember that all this works for all waves, as well as em, so speed is not an issue. The result, measured at a place on the screen, is due to the interference between waves that are arriving at anyone time. The interference pattern will only form is the coherence is adequate - i.e. when the path difference is less than the coherence length. Fine on boresight, even for very broadband signals but off beam you need better and better coherence length (that's narrower and narrower bandwidth) to establish a regular pattern.
As I mentioned further up the thread, the FT works when there's any sort of periodic variation - in space or in time. A picture can be analysed in terms of basic spatial frequencies and transmitted in a form that allows compression (as with jpegs) - with 'acceptable' degradation of course.

 

[marmoset]

On the mathematical side it may be worth mentioning there are formalizations of our intuitive notion of frequency/wavelength which do allow them to be localized (see e.g. the spectrogram https://en.wikipedia.org/wiki/Spectrogram and time frequency analysis) and which allow you to meaningfully talk about for example a time-varying frequency, like the sound of a siren. In fact they give us a spectrum as a function of time, a 2D object, rather than the usual 1D Fourier transform.

I mention this since arguably the response of a detector is better modeled by one of these 'local Fourier transforms' (time-frequency distributions) than by the usual Fourier transform. To give an example related to the OP: If you have two pulses of different frequency separated by a very large time, and you operate your detector for a time interval which only contains one pulse, you will not detect the frequency of the second pulse at all - yet both frequencies are in the ordinary Fourier transform! In the spectrogram one frequency appears at one time, the other another; the time axis in the spectrogram arguably provides a good model for the 'window' in which we actually measure the frequency.

An even simpler case is if you mistime when you start running your detector - if you only record data before a pulse has arrived, you will not record anything at all!

 

[Dale]

I don't see why we decompose our wave into infinite waves.

We do it because the infinite waves are mathematically easy. The Fourier transform is what is known as a decomposition into a basis. You can, in principle, use all sorts of different functions (many of them not infinite) as your basis functions. Cos and sin functions are not magic, they are just an easy and convenient basis to use, but a cos or sin function is infinite. So when you use this easy basis you are using infinite basis functions.

What does bother me is that they, added together, do not describe the same system as the finite pulse, because, as I pointed out in my first post, these waves could be separated from each other at any point of the space and the time one from each other (using spectrometry) leading either to a superluminal paradox or a causality paradox.

This is not correct, I am not certain what makes you believe this. If you add your infinite Fourier terms then you really do wind up back with your original finite pulse. Why do you think that they do not? Also, I am not sure why you think that there is any causality paradox involved. Can you explain the thought process that leads you to either of these statements?

This assumption is that THE FREQUENCY SPECTRUM MEASURED ASSUMES A DECOMPOSITION OF THE PULSE INTO INFINITE HARMONIC WAVES.

I would say that this is less of an assumption and more of a definition. What do we mean when we say the words "frequency spectrum"? By those words we mean precisely the decomposition of an arbitrary function into a sin and cos basis, which is a basis of infinite functions. We can and do use other basis functions for decomposition, and arrive at a transformed version of our signal, we just do not call that decomposition the "frequency spectrum". That term is specifically reserved for the cos and sin basis.

I say: "Okay, the frequency of the pulse is WELL-DEFINED, even though we cannot measure it accurately because we only have one pulse and a single measure would have large uncertainty, it MUST BE A SINGLE ONE.

Why? Why must a finite pulse correspond to any single cos or sin function?

Let's say we just look at the spectrum during a finite time.

By looking at the spectrum for a finite time you are introducing a truncation that causes a broadening into a continuous spectrum. This is unrelated to the resolution of the spectrometer, even a spectrometer with higher frequency resolution will not be able to resolve more detail in the signal.

Let's say this amount of time is the same amount of time that would take to the finite pulse to go through the same prism. Now I guess that during this amount of time, as at any other moment we should see a single angle of deviation at the screen with a very tiny bandwidth and this bandwidth would be just a consequence of the non-ideal resolution of our spectrometer.
Now, during this finite time what would have happened if instead of the harmonic infinite wave we had put our finite pulse? Well, it seems that what happens is that instead of a tiny width we get a sort of a continuous spectrum. Both results are quite different.

You would indeed get a continuous spectrum, which is identical to the continuous spectrum that you obtain by looking at the infinite spectrum for a finite time. The results are not different, they are identical. It doesn't matter if you have a finite signal that you look at for an infinite duration or an infinite signal that you look at for the same finite duration. The result is the same.

 

[Charles Link]

To respond to the 3rd item mentioned by @Dale above, (for the OP), one of the primary uses for Fourier analysis is to determine how much energy is contained in a given $\Delta \omega$ at frequency $\omega$. This is very well quantified by Parseval's theorem that I mentioned in post 15 above. It really doesn't matter for most purposes what the duration of the sinusoid at that frequency ultimately is. The computation using Parseval's theorem will give the energy observed during the sample period. $\\$ And to add to @marmoset 's description above, (post 26=with the sound of a siren), a Fourier or spectral analysis, if the sample frame is very long, will not tell everything about the source. It can contain a lot of info, but it takes a real time type of sampling, like he described above, in order to observe things like audio signals (AM modulation) that may be an important feature of what otherwise might simply be seen as a fairly steady signal at the carrier frequency.

 

[Jufa]

I'm sorry if the explanation of my doubt was sort of misleading guys. All I did not understand was why it was needed to perform the Fourier Transform along an infinite space if the pulse just took a finite time interval t_o (to me it seemed an arbitrary decision doing it within this infinite interval). I knew that this matched with experience and with some sort of physical intuition, but had no clue of the meaning of doing it. Now I have understood that this is because we are always within the approximation that t_0<<<T being T the minimum of time exposure our light detector has. Therefore it is fair to assume that what our detector does is to perform a decomposition of our pulse into longer ones. That is, our "infinit" is nothing else than T/t_0.
Indeed, the time interval in which we compute the Fourier transform, i.e. -infinity < t < + infinity stands for
0 < t <T/t_0. So we are actually decomposing our pulse into pulses of the length of the time that it takes for our detector since it starts receiving light stimulation until it stops being aware of it (which is always a longer time than the time it takes the pulse to actually pass through the detector).
One can actually perform the Fourier transform of not one pulse but plenty of them with an arbitrary phase difference between (which is what actually happens when dealing with decoherent light beams, more than a single pulse) and making use of the fact the expected amplitude of the sum of n random phases goes like sqrt(n), one can see that the Fourier Transform of a big number n of pulses that our detector processes at a time is (using randomness of initial phase) sqrt(n) times the transform for a single short pulse.

 

[Jufa]

I just hope this thread was useful for anyone else than me. And hopefully, I can receive some feedback. To me, it seemed kind of delusive the way in which this phenomenon was treated at my University, here in Barcelona. Even when consulting directly to some teachers no one told me explicitly the reason why we decomposed pulses into infinite ones. It seems, someone please correct me if I'm wrong, that all it takes to obtain a signal with a single frequency component is to have a long enough pulse so that the time that it takes to our detector to stop detecting when there's no pulse anymore, is neglectable with the whole time of detection. Surely this only happens with t_0-->infinity but I think it would be interesting to give students some notions of what this infinite quantity refers to.

Thanks for your time :).