Existence of monochromatic pulses?

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Solution 1

The fundamental reason for this is that a truly monochromatic waveform $$ f(t) = f_0e^{-i\omega t} $$ is active for all real times $t$ ─ it doesn't start, and it doesn't stop ─, and this means that you need an infinite time to produce one, and you need an infinite time to detect it. Since the energy density of the wave is constant, the infinite duration also means that you require infinite energy to set this up.

Now, why do we say that you need infinite time to have a truly monochromatic situation? Let's focus first on the detection side, and suppose that you only have a finite time $T$ to measure the waveform, which is centered at a frequency $\omega_0$ such that $\omega_0 T\gg1$, i.e. $T$ fits many cycles of the central frequency. Now, the real question is: can you distinguish between the frequency $\omega=\omega_0$ and some other frequency $\omega=\omega_0+\delta\omega$ which is close to, but not quite, the central frequency $\omega_0$ which you think you have?

Putting feet to the ground, and supposing that the signals start off in sync, the question ultimately asks how well you can distinguish between $e^{i\omega_0T}$ and $$ e^{i\omega T} = e^{i\omega_0T} e^{i\delta\omega \,T}, $$ where the wave has advanced by a phase $\delta\omega\,T$ over the observation window. Now, here's the problem: what happens if $\delta\omega$ is much smaller than $2\pi/T$? In this case, the two waves at $\omega=\omega_0$ and $\omega=\omega_0+\delta\omega$ will have hardly drifted out of step even over your long observation window, and you will find it hard to distinguish between the two.

Note, moreover, that if you could expand your observation window to a longer time $T_\mathrm{longer} = 2\pi/\delta\omega\gg T$, then your observation window would include times where waves at $\omega=\omega_0$ and $\omega=\omega_0+\delta\omega$ would be $\pi$ out of step, and you would be able to distinguish between them. However, so long as your observation window $T_\mathrm{longer}$ is finite, there will always be detunings $\widetilde{\delta\omega}\ll 2\pi/T_\mathrm{longer}$ that are too close for you to be able to resolve between $\omega=\omega_0$ and $\omega=\omega_0+\widetilde{\delta\omega}$.

It's also worth talking about what happens at the edges of the observation window, as well as the production of the wave. Does your wave have a sharp cutoff, going from a finite amplitude to zero instantly? Then at the boundary it is hardly monochromatic. Instead, you might want it to go smoothly from flat to zero over some transition period $\Delta T$, but then that transition period, during which the amplitude is changing, is going to make it harder for you to distinguish between two waves that are only a small fraction of a radian apart in phase.

So, what does this mean? It tells you that a true monochromatic pulse is impossible to realize in real life, because by "true monochromatic pulse" what we mean is the mathematically idealized model that's been on since forever and will remain turned on until eternity. What you can make is waves that are closer to monochromatic than your experiment can resolve, in which case you can just use the monochromatic approximation without worrying ─ but that doesn't make the wave truly monochromatic.

Finally, it's also important to note that the fact that monochromatic waves are unphysical does not make them less useful. Normally, when we consider monochromatic waves, we are considering the dynamics of some oscillatory quantity $u(t)$ which responds to a linear system. In this case, it is often overwhelmingly simpler to view $u(t)$ as a superposition of plane waves $e^{-i\omega t}$ with some weight $\tilde u(\omega)$, i.e. to deconstruct $u(t)$ as its Fourier transform: $$ u(t) = \int_{-\infty}^\infty \tilde u(\omega) e^{-i\omega t}\mathrm d\omega. $$ If the dynamics is linear, then we can just worry independently about how each monochromatic component will react to the dynamics, without needing to pay attention to the fact that they are unphysical, and then put it together later when we Fourier transform back to the time domain. Whenever we deal with the physics of monochromatic waves (as in, say, the Helmholtz equation, or phasor analysis) that's always the underlying mindset.

Solution 2

This follows from classical Fourier analysis. The frequency spread and time duration of a pulse are related by $$ \Delta \omega \Delta t \approx 2 \pi $$ so to make a truly monochromatic pulse where $\Delta \omega$ is basically $0$ implies this pulse is infinite in duration. Thus, any pulse with a finite duration cannot be truly monochromatic.

Solution 3

To create pulses you start with an pure monochromatic wave and add a pulse shape. If you now apply a Fourier transform to this pulse shaped wave, you will see that your delta spike got broader. So the Fourier transform is your link between the space and time domain. Basically the sharper you want your pulses to be, the more high frequency components you need to archive the desired pulse form.

Here is a visualization of the Fourier transform of a pulsed signal. You can think of the low frequency blue line as your starting monochromatic frequency and the higher order ones are those you need to get your pulse shape

enter image description here

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Mac Sat
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Updated on May 20, 2020

Comments

  • Mac Sat
    Mac Sat over 3 years

    Why there can not be a monochromatic pulse? My physics professor told us that we can't generate a monochromatic light pulse and I was wondering what are the physical limitations causing this.

  • Mac Sat
    Mac Sat over 6 years
    i have this formula in my course but no proof or derivation , am curious to the similarity with the Heisenberg uncertainty principle is there a link ?
  • ZeroTheHero
    ZeroTheHero over 6 years
    @Mac Sat: Yes and no. The Heisenberg uncertainty relations are quantum in nature, and depend on non-commutativity of operators. There is nothing quantum about the Fourier relations, which are entirely classical. The version $\Delta E\Delta t\approx h$ is actually tricky to handle because time is not an observable in quantum mechanics, so the usual interpretation of this inequality as quantum must be done rather carefully. The relations certainly look the same (much to the relief of Heisenberg, apparently) but are not strictly linked.
  • Ruslan
    Ruslan over 6 years
    Well they are linked. Actually you can prove momentum-position uncertainty relation from the fact that momentum-space wavefunction is a Fourier image of the position-space one. As for time-energy — yes, this one is a bit more tricky.
  • Emilio Pisanty
    Emilio Pisanty over 6 years
    "Nearly" and "virtually" monochromatic are not the same as monochromatic.