Why can't objects be seen if the wavelength is too long?

• I
• yosimba2000
In summary: This appears over and over in all of physics as the frequency /bandwidth limit. Most famously perhaps as Heisenberg uncertainty, but it is really a general information... limitation.
yosimba2000
Sorry, this has been asked a few times, but nothing's getting through to me. I really need to satisfy this curiosity :(

I've read a lot of "water wave" analogies, but they don't make sense.
A water wave has physical peaks and valleys, and I can see how when the peak travels and hits a ship, the water wave "dies". I can see how when a water wave encounters a small object, that small object has basically no effect on the propagation of the water wave.

But if we look at a single lightwave, the typical illustration of it is like a water wave, with peaks and valleys. But the peaks and valleys of the E/M wave are not physical. They are are the graphical representation of the strength of the EM vector; it's not like the lightwave itself is moving up and down (because it travels in a straight line). It's only the direction and strength of the EM vectors along that line that point up and down.

So how does the frequency at which the E/M vectors point up and down determine how well we can see an object?

You can also visualize what I'm imagining like this: We have an Atom at Position X. We shoot at the Atom polarized light (E-vector points up and down). If we just observe Position X, at any moment, we see an Atom, and we see an E-vector arrow pointing up or down at varying strengths. How does the speed at which the arrow changes direction at Position X affect how we visualize the atom? It's not like the lightwave skips Position X, or goes under or over it, and therefore the Atom can't "respond" to it... the Atom should always be feeling the effects of the lightwave. But how does that translate to how well the Atom can be seen?

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The Bill
Do you mean see with the human eye?

PeroK said:
Do you mean see with the human eye?
Well in general, the relationship of visualizing/detecting small things with too long a wavelength. Like trying to see an atom with visible light.

yosimba2000 said:
Well in general, the relationship of visualizing/detecting small things with too long a wavelength. Like trying to see an atom with visible light.
There's an explanation here, for example:

https://www.physicscentral.com/explore/action/atom.cfm

Technically, the interaction between light and an individual atom is explained by quantum mechanics.

The Bill
I have read that article yesterday, but it still uses a water/physical peaks and valleys analogy. Is there really no satisfying classical explanation?

yosimba2000 said:
Is there really no satisfying classical explanation?
For why light does or does not interact with an individual atom? Not really. You need quantum mechanics for that.

You can explain why an EM wave in kilometre wavelengths doesn’t interact with an antenna a meter long. You can see the interaction with the antenna as the EM wave driving the electrons in the antenna to move, which makes them emit an EM wave themselves. The resulting wave is the sum of the two, and there is interference which either reduces the energy of the wave compared to the undisturbed wave (absorbtion) or changes the direction of flow of the energy (diffraction/reflection) or both. A one meter antenna simply isn't big enough to generate any significant emission when driven by the kilometre-scale wave because the incident wave's field is pretty much constant on the meter scale, so it makes no significant difference to the wave, neither absorbing nor reflecting it.

Atoms do act in some senses as tiny dipole antennas, but you do need to worry about quantum effects. I don't know enough about QED to really comment.

The Bill, vanhees71 and PeroK
yosimba2000 said:
I have read that article yesterday, but it still uses a water/physical peaks and valleys analogy

So you don't want an anology and you don't want the mathematics - what exactly do you want?

sophiecentaur, vanhees71 and phinds
Some (non-quantum) mathematics:

The intensity of a light of wavelength λ scattering off a sphere of radius r at a distance x is proportional to $(r/x)^2(r/\lambda)^4$. The first term is just the inverse square law. The second term is about (1/5000)4 and the first term is driven by how close you can get - 1 centimeter perhaps? So you're down about 23 orders of magnitude from the initial intensity.

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vanhees71 and Ibix
It is the same reason that old people, who lose the high frequencies in their hearing, cannot distinguish fricatives.
It is the same reason that a small aperture lens will lose acuity and we build huge telescopes
It is same reason why a basson and a Tuba are bigger than a piccolo

This appears over and over in all of physics as the frequency /bandwidth limit. Most famously perhaps as Heisenberg uncertainty, but it is really a general information result.

vanhees71
hutchphd said:
old people,
Hey!

sophiecentaur
I can't hear you...

vanhees71, Vanadium 50 and Melbourne Guy
yosimba2000 said:
Is there really no satisfying classical explanation?
EM theory is totally classical and it provides an excellent explanation. It may not all be 'intuitive' but it's up to the student to sort that out. It involves non-trivial maths.

vanhees71
Ibix said:
You can explain why an EM wave in kilometre wavelengths doesn’t interact with an antenna a meter long.
You really ought to replace "doesn't interact" with "may interact weakly". But even that is putting it too strongly. A simple ferrite rod antenna, perhaps 10cm long with the appropriate coil wrapped round it, will easily receive 1500m radio broadcasts.
Not all antennae are lengths of wire.

nasu, Ibix and vanhees71
sophiecentaur said:
You really ought to replace "doesn't interact" with "may interact weakly".
I think this is the key. The things that scatter very strongly are things that are in some way at resonance. Tuned antennas. Atomic energy transitions. Molecular Dye transitions. Organ pipes. These are the objects we see and hear: the rest get lost .
The appropriate language for this discussion is the mathematics of Fourier and linear response theories.

sophiecentaur and Ibix
Resonance is all about matching with space and I see it as a crossover between classical and QT because very small atoms radiate with relatively long wavelengths at times. Plus the ‘Q’ factor tends to be very high with narrow spectral lines as with highly tuned small antennae.
But the thread is not to do with absorption and emission but with scattering - not a narrow band effect at all.

Melbourne Guy said:
This is great, I am really liking this one. Guess I should have started with diffraction first :)

Melbourne Guy, hutchphd, berkeman and 1 other person
sophiecentaur said:
Resonance is all about matching with space and I see it as a crossover between classical and QT because very small atoms radiate with relatively long wavelengths at times. Plus the ‘Q’ factor tends to be very high with narrow spectral lines as with highly tuned small antennae.
But the thread is not to do with absorption and emission but with scattering - not a narrow band effect at all.
One must be careful! You can apparently get dispersion theory at the level of linear-response theory from a purely classical simple model of a dielectric. You just assume that the response of the charged particles to small perturbations due to the external incoming em. waves is that of harmonic oscillators, and choose the oscillator frequencies of this model from experiment. This gives you a quite accurate description.

You can also use quantum mechanics in semiclassical approximation, i.e., treating the matter quantum-mechanically (non-relativistic QM will do for usual matter) and the em. field as classical and treat the perturbation due to the external em. wave in first-order time-dependent perturbation theory. This leads after all to the same result, but the oscillator frequencies turn out to be given by the differences between the energy levels of the bound state describing the medium. The typical energy levels in atomic physics are such that the corresponding differences, defining the resonance frequencies for the incoming em. waves, lead to wave lengths much larger than the extension of the atom (e.g., visible light between 400 and 800 nm vs. the Bohr radius of a hydrogen atom of around 0.05 nm). That's why the electric-dipole approximation underlying this quantum-mechanical treatment of dispersion theory works so well.

PeroK
vanhees71 said:
You can also use quantum mechanics in semiclassical approximation, i.e., treating the matter quantum-mechanically
The QM approach must be difficult to cope with because the atomic energy levels are only changed by one photon on one atom and the resulting scatter has to be due to interactions with a lot of separate atoms. It's as if the interaction is taking place way out on the skirts of any 'resonance' which could cause a change corresponding to the photon energy. How does that work - keeping the QM explanation?
vanhees71 said:
defining the resonance frequencies for the incoming em. waves, lead to wave lengths much larger than the extension of the atom (e.g., visible light between 400 and 800 nm vs. the Bohr radius of a hydrogen atom of around 0.05 nm).
I don't see that the displacement / distortion of an orbital would have to relate to the wavelength. The displacement of charges in a metal with an induced RF frequency is only a tiny (seriously, really tiny) fraction of the wavelength, whether at resonance or not.

We can explain reflection of EM waves by small objects by using classical physics. For instance, a metal rod subjected to an EM wave will experience a potential between its ends and a current will flow in it. The electrons in the rod are being subjected to acceleration. This then causes the rod to re-radiate. We can view the action as a receiving antenna and a transmitting antenna, and the action is the same as in radar.
If the object is resonant at the frequency of the EM radiation, by virtue of being, say, half a wavelength long, then if we take electrical measurements of it we notice that it exhibits a resistance. This resistance is called the radiation resistance, Rr, and is an indication that the rod is coupled to free space, and can either absorb energy from a passing wave, or if excited by a generator can radiate an EM wave. If the rod is shorter than a half wavelength, we find that its coupling to free space is reduced, and this is seen as a drastic reduction in the radiation resistance. The reason for this action is that the electrons in a shorter rod are subject to less acceleration, and hence radiate less. When we reduce the length of a rod the radiation resistance falls approximately with the square of the length, so there is a rapid reduction.
In addition to the reduction in radiation resistance with length, we also find that the short rod has capacitive reactance in series with the resistance. This increases rapidly as the length is reduced, and it means that for a given potential applied between the ends of the rod the current will be less.
The electric field of the incoming EM wave creates a potential difference between the ends of the rod or object. This then creates a current in the rod. For a very short rod, the reactance is by far the greater part of the impedance of the rod, and so it dictates the current. From Ohms Law, I = V/X. This current also flows through the radiation resistance and causes the rod to re-radiate a power of I^2 Rr.
Notice that as the rod is shortened below its resonant length, its reactance increases rapidly and its radiation resistance reduces rapidly, creating a large decrease in the energy absorbed from a wave. We can also say that, assuming no losses in the object, all the absorbed energy is re-radiated.
To place some numbers on the effect, if the length of the rod is L and incoming field strength is E, then the voltage across the rod is EL. The reactance, X, of a rod is proportional to 1/L. And the radiation resistance is proportional to L^2.
Using Ohm's Law, the current, I, in the rod is then V/X = EL/(1/L) = EL^2
The received power is then I^2 Rr = (EL^2)^2 L^2 = E^2 L^6
Therefore the absorbed power, for a very short rod or object, depends on the length to the power 6.

PeroK

1. Why can't objects be seen if the wavelength is too long?

Objects can't be seen if the wavelength is too long because the human eye is only able to detect a limited range of wavelengths, known as the visible spectrum. Wavelengths that are too long or too short for the human eye to detect are not perceived as visible light, and therefore cannot be seen.

2. What is the relationship between wavelength and visibility?

The relationship between wavelength and visibility is that shorter wavelengths correspond to higher frequencies and are perceived as colors such as blue and violet, while longer wavelengths correspond to lower frequencies and are perceived as colors such as red and orange. Wavelengths outside of the visible spectrum, such as infrared and ultraviolet, cannot be seen by the human eye.

3. How does the size of an object affect its visibility based on wavelength?

The size of an object does not affect its visibility based on wavelength, as long as the object is within the visible spectrum. However, if the object is emitting or reflecting wavelengths outside of the visible spectrum, its size can affect its visibility. For example, a small object emitting infrared light would not be visible to the human eye, but a larger object emitting the same infrared light may be visible.

4. Can objects with longer wavelengths be detected by other means?

Yes, objects with longer wavelengths can be detected by other means such as infrared cameras, which are able to detect and display infrared light that is not visible to the human eye. Other animals, such as snakes and bees, are also able to see longer wavelengths and may be able to detect objects that are invisible to humans.

5. How does the atmosphere affect visibility based on wavelength?

The Earth's atmosphere is able to filter out certain wavelengths of light, making them invisible to the human eye. For example, most ultraviolet light is absorbed by the Earth's atmosphere, making it invisible to us. Additionally, atmospheric conditions such as fog or pollution can scatter or block certain wavelengths of light, affecting visibility. This is why objects may appear less visible in foggy or hazy conditions.

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