How does light travel through air molecules?

[BingoTheClowno]

TL;DR Summary

An FAQ on how light behaves

This might be a stupid question but, taking a shower yesterday, I was wondering how is light propagating through the air molecules. For example, take the light coming from a distant star entering our atmosphere. Between my eyes and the star there are billions of air molecules. How do the photons pass through all these molecules and be able to excite my eyes retina. Are the photons exciting each electron of the individual atom and thus propagate from atom to atom, molecule to molecule? How does this exactly work I'd like to know.

 

[mathman]

The air is mostly empty space, so light gets through rather easily, although, as you can see, the sky is light blue - a result of some light being scattered by the air molecules. Without the atmosphere, the sky would be black, except for the sun, moon, and stars.

 

[BingoTheClowno]

But between my eye and the star there must be billions of air molecules in a straight line that the photon must hit. What happens then at the atom level?

 

[ZapperZ]

But between my eye and the star there must be billions of air molecules in a straight line that the photon must hit. What happens then at the atom level?

1. Figure out how many photons are in a "typical light beam". You'd be surprised.

2. Light scatter very weakly with "free" atoms. So even in air, the interaction is very weak.

3. Even when it encounters an atom, an atom is primarily empty space. It can scatter off orbital electrons (VERY weak scattering cross section), or the nucleus (which occupies a VERY small fraction of an atom).

Conclusion: photons don't get scattered that much in air, and even less in the almost-vacuum of space. Even when they get scattered, there's a gazillion more within a typical light beam of sight.

Zz.

 

[CarlB]

Even when they get scattered, there's a gazillion more within a typical light beam of sight.

Back when I was in grad school, the experimentalists told me otherwise. In fact, they said that the human eye was capable of imaging photons on nearly a photon by photon basis. That is, with any given photon, there was a substantial probability that it would result in a nerve impulse and that that nerve impulse would register with the brain. Thus the associated beams, at least for dark adapted viewers, could have relatively small numbers of photons.

For an interesting introduction to photons from the astronomical community, along with instructions on how to measure your pupil diameter with a set of alan wrenches, see:
http://www.bpccs.com/lcas/Articles/your_eyes_key_to_observing.htm

The simple reason that air doesn't absorb standard light much is that air molecules don't absorb it much. The reason that the individual molecules don't divert the beam much (despite having an index of refraction differing very much from that of vacuum) is that individual molecules are very small in comparison to a wavelength of light. If you go to higher frequency radiation, air molecules certainly will interact more strongly, which is why the sky appears blue. (That is, higher frequencies are preferentially scattered, so the scattered light is much more blue than the unscattered light. But even with blue light, air scatters very little, as the sharpness of your shadow indicates.)

Instead, the primary thing in air that scatters light is changes to the local density that are thermodynamically related. This is the source of star twinkle, and is similar in effect to the way that waves of water reflect the light on the bottom of a shallow pool of water.

Of course with anything in physics there are many different ways of expressing the same true fact.

Carl

 

[ZapperZ]

Back when I was in grad school, the experimentalists told me otherwise. In fact, they said that the human eye was capable of imaging photons on nearly a photon by photon basis. That is, with any given photon, there was a substantial probability that it would result in a nerve impulse and that that nerve impulse would register with the brain. Thus the associated beams, at least for dark adapted viewers, could have relatively small numbers of photons.

I'm not sure how this is relevant to the quote from me that you cited. I was saying that there's A LOT of photons from a typical light source. Even being scattered by a billion atoms, there's still MORE of them following behind that you can't really tell that a lot of photons have been scattered already. You need to look very carefully at the night sky to even notice the "twinkling" due to air movement. A billion atoms is nothing when there's a gazillion photon coming at you.

Zz.

 

[Pieter Kuiper]

It is actually not so simple. Air is mostly empty space, but so is iodine vapor, which is rather brown.

In a wave picture, the electric field induces an electrical polarization of the molecule. Unless there is absorption, this polarization is out of phase with the driving field. The oscillationg polarization also causes a new electric field. This is added to the incoming field, so that the field behind the atom is slightly out of phase with the incoming beam.

The phase velocity is now smaller than c, by a factor equal to the index of refraction n.

 

[ZapperZ]

But I'm sure you'll notice that (i) the OP is asking about why light could get transmitted over all that distance and with billions of atoms it its way and (ii) that it is implictly referring only to the visible spectrum. In other words, it wasn't about why was light blocked, or reduced, when it passes through clouds, for instance.

I could, for example, point out that the UV part of the spectrum is mostly missing by the time he/she actually sees the light from that far away, so it is also not true that nothing got absorbed and made it that far.

I've discussed optical transmission, especially through solids, more often than I've cared to count. I think this will be one of our FAQ one of these days when I get find the patience again to write about it. Unless I get some volunteers to write this before then...

Zz.

 

[zoobyshoe]

It is actually not so simple. Air is mostly empty space, but so is iodine vapor, which is rather brown.
In a wave picture, the electric field induces an electrical polarization of the molecule. Unless there is absorption, this polarization is out of phase with the driving field. The oscillationg polarization also causes a new electric field. This is added to the incoming field, so that the field behind the atom is slightly out of phase with the incoming beam.
The phase velocity is now smaller than c, by a factor equal to the index of refraction n.

If a photon hits an individual N2 or O2 molecule and is transmitted through it this way does it always continue on in the exactly the same direction it was going when it hit the molecule?

 

[Pieter Kuiper]

If a photon hits an individual N2 or O2 molecule and is transmitted through it this way does it always continue on in the exactly the same direction it was going when it hit the molecule?

No. You can see this at a surface: some is reflected. In point scattering (of an individual molecule) there will be scattering in all direction. But only in the forward direction, contributions will be in phase.

The description of light as a stream of photons complicates this kind of questions. For a good answer, one needs to describe the scattering of quantum mechanical particles, that also have a phase.

And Zz: I was not criticizing your posting, just replying to the OP.