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We always say that when a given light wave interacts with atoms bound in a molecule, only waves with wavelength close to the inter-atomic-spacing are able to probe the system. In other context (macroscopic oscillations in a system), one also talks about the wavelength of some fluctuation in a system being larger than the system size, in which case such fluctuations are omitted/ignored.

Questions:

  • What is it that links the wavelength of a wave to its interaction with a system? Be it acoustic waves or EM. Physical intuition would be greatly appreciated, but please don't hesitate showing the math behind it as well, if you see it fit!

  • How does one go about quantifying such problems? i.e. if I have $\lambda_1$ slightly larger than system size $d$, or less larger, how do I conclude whether to consider such oscillations in the system or not?

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  • $\begingroup$ This is not correct (and never was), neither classically nor quantum mechanically. See en.wikipedia.org/wiki/Superresolution for an explanation of techniques that overcome the naive limits in imaging. Other famous examples of small systems affecting large wavelengths are the scattering phenomena, the most well known of which is the blue sky due to Rayleigh scattering. $\endgroup$ – CuriousOne Dec 23 '14 at 13:02
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    $\begingroup$ @CuriousOne What do you mean it is not correct? what's not correct? so any light wavelength will interact with the atomic-binding? Maybe you meant that it's just a bad approximation...but this "approximation" is exactly what I'm trying to understand with my question. $\endgroup$ – user929304 Dec 23 '14 at 13:05
  • $\begingroup$ What's not correct is the assumption that one can make a firm rule to correlate system size and the wavelength it interacts with. $\endgroup$ – CuriousOne Dec 23 '14 at 13:14
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In principle, a wave of any size will interact with a system of any size. The question should therefore be posed differently: how is the interaction of the two affected by their relative size?

Let's take the simple example of scatter. You are familiar (whether you know it or not) with Rayleigh scatter - it's an elastic light scattering phenomenon that makes the sky blue. The Rayleigh scatter cross section (effective probability of interaction) is given by

$$\sigma = \frac{2\pi^4}{3}\frac{d^6}{\lambda^4}\left(\frac{n^2-1}{n^2+2}\right)^2$$

In this equation, $d$ is the diameter of the particle, $n$ is its refractive index, and $\lambda$ is the wavelength of the scattered light. This expression applies when $d<<\lambda$ - typically 1/10th or smaller. So right there we have an interaction that occurs with a wave that's much bigger than the "system" (the particle, in this case).

As wavelengths become shorter, the scattering mechanism is better described as Mie scatter (wavelength on the same order as the particle) - this is characterized by resonances, meaning that some sizes of particles will scatter better than others, but it's not a monotonic relationship (like for Rayleigh scatter).

At even shorter wavelengths, light (or other waves, e.g. acoustic waves) start to behave more "normally" - that is the regime you usually think about when you are talking about direct visualization, optical microscopy, ultrasound imaging etc.

But just because the interactions of long waves with small objects don't easily make pretty images doesn't mean they are not happening - the physics may be a bit harder, and the interaction more statistical and less deterministic, but nonetheless - they do interact.

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  • $\begingroup$ Phonons answer is not "wrong" - he tackled the question a bit differently than I did. You should vote for whichever answer best helped you - or neither, if they did not. $\endgroup$ – Floris Dec 30 '14 at 12:11
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I'm just going to add something orthogonal to Floris' correct answer, because the question is posed in a very general manner, which allows for relatively diverse yet correct kinds of answers.


Here's a crude-intuitive way of seeing it:

A wavelength is just the spatial period of a wave, be it of mechanical or electromagnetic nature. Meaning the distance needed for the wave to repeat its shape in space, of course there are numerous other ways of saying the same thing, e.g. you can udnerstand the wavelength of a wave as the distance it travels starting from one peak to a consequent one.

Now in order to understand what one means by system size in these scenarios, take the example of a rope stretched and held by its ends by two people. Let's say the rope, i.e. our system, has a length $d.$


Link between wavelength and system size with a simple example: I'm sure that you've seen before that when one of the ends' is moved up&down a transverse wave is induced in the system, i.e. the up/down vibration propagates along the rope till the other end and is then reflected backward, ignore losses, this is also called a standing wave! Note it's called transverse because the direction of wave propagation and oscillations are orthogonal to each other. Now clearly if you repeat the initial perturbation with different distances, you induce waves of different wavelength in your the rope. If e.g. you have $\lambda=d/10$, then you will observe 10 complete repetitions of the wave (more simply 10 complete vibrations) until it reaches the other end. Furthermore if choose $\lambda=2d$ you will not even observe one complete period of your wave, but only half of it. The picture below should further clarify this:

enter image description here

Now if you choose very large wavelengths, e.g. $\lambda=100d$, then you'd need a rope 100 times longer to notice the wave(i.e. to see one complete oscillation), as with your current rope of length $d$, you will only see $1/100$ of the oscillation, in simplest words: you don't even feel the wave propagating along the rope, but it is propagating through your system. On the other hand if you consider very small wavelengths, e.g. $\lambda=d/100$, you'd be able to fit 100 complete oscillations in a distance $d$, again this means that the oscillations would have to be so small and complete cycles happening so frequently that you may not even see them. Another visualisation for different $\lambda$'s:

enter image description here

Long story short, if you work around wavelengths $\lambda$ around the same dimensions as your system size $d$, you are more likely to observe/notice them, because then you're sure you'd fit at least one complete cycle of your wave into your system.

Now in more sophisticated examples, only sometimes one uses the same line of thought when considering fluctuations in a system. For example it works when one talks about interfacial fluctuations in a liquid-liquid system, but it fails for the case of inter-atomic distances and light scattering (because this is related to angular resolution of your system, i.e. aperture size (here atomic distance) vs wavelength, which decide whether the light can be diffracted or not, to better understand this case, read about electron microscopes and angular resolution).

Hope this gives you a better intuition! For nicer visualizations, see here!

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What I know is that for polarizing an atom, i.e. creating a dipole, we consider the wave-length of the e.m. field greater by a couple of orders of magnitude than the dimensions of the atom. Does it mean that we test the atom?

On the other hand, if we study the structure of a crystal, we send on it e.m. waves of $\lambda$ bigger than the distance between the atoms in the lattice, s.t. the wave see the lattice as planes on which it gets reflected.

Some more data: for studying the structure of atoms, the photon wavelengths are bigger than the atom dimensions, e.g. the Lyman-$\alpha$ line is of $10^{-7}$cm, while the Bohr radius is $10^{-8}$cm. Gamma rays that can be emitted by nuclei have wavelength of the order of 5 x $10^{-10}$ cm while the nuclear radii are cca. $10^{-12}$ cm.

Thus, at least these tests, use some greater wavelengths than the system dimensions.

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