Photons passing through a stormy sea...

[gfellow]

I have looked through the search to see if there was a satisfactory answer, but had no luck, so let quickly mention that I have always stood in awe at the collective knowledge that springs from this forum. It cannot be unlike tuning in to Douglas Adam's "Deep Thought".
Having hopefully stroked the collective ego, I will now ask my question:

Supposedly, once emitted, a photon is unaffected by electromagnetic influences and will not veer from its path*. My question is, if polarized laser light passes close to an electric discharge on its way to a given target, will the disturbance in any way influence the path or quality of the photon?

*Here, I am omitting the 'bending' of light, as in streaming electrons in a fluctuating magnetic field emitting photons (as in a TV tube) and also turning a blind eye to gravitational considerations (which are minute unless we are talking solar masses, black holes etc.)

The answer is of some import to me, as I would like to inaugurate a conversation with an interferometry expert in the not-too-far-future, and do not want to come off as a complete boob. Consequently any help you can give me on this question would be greatly appreciated.

[ShinAce]

My question is, if polarized laser light passes close to an electric discharge on its way to a given target, will the disturbance in any way influence the path or quality of the photon?

Not that I know of. It wouldn't matter if the light is coherent, polarized, or what have you. Photons do not carry charge, so they won't be affected by electricity.

Make note that an electron that is stationary is commonly called a charge, while an electron in motion would be a current. An 'electric discharge' sounds like a current, but you haven't specified if you're talking about lightning, or the charged plates of a capacitor.

[WayneFrancis]

Note that a CRT doesn't bend light. As you point out it bends the path of the electrons. That stream in it self isn't emitting photons you see that comes from the electrons hitting a coating of phosphor.

So as I understand it the magnetic fields doesn't change the path of the photon at all.

[a1call]

In absence of any matter, I agree that light will not be affected by an electromagnetic field.
However, an electromagnetic field can indirectly affect light by effecting the matter which the light passes through.
A common example would be in an LCD. Here a liquid crystal such as Benzate is sandwiched between two sheets of glass. if an electromagnetic field is applied to that apparatus, the light will be circularly polarized. The effect of this is that if a horizontally polarised light passes through it will change its polarization to vertical.
ETA: The above is what I remember from my studies more than 30 years ago so corrections are welcome.
Now that I think about it, I think the apparatus is circularly polarized in absence of an electromagnetic field and in presence of the field the light will exit with the same polarization it enters. It will exit 90 degrees rotated in absence of an EM field.

[ShinAce]

In absence of any matter

Yeah, I neglected that part.

[onomatomanic]

As the off-site thread linked in the OP already pointed out, the view that light remains entirely unaffected by the presence of EM fields, in the absence of matter, is strictly true only in a classical treatment.

In a quantum-mechanical treatment, there is a mechanism which allows for interactions, which is or is at least similar to Renormalization: Photons occasionally "decay" into a particle-antiparticle pair, which then mutually annihilate to (re)produce a photon. These particles are charged, and, fleeting though they be, potentially provide a way for electrically neutral particles to couple to an EM field.

I wasn't able to find an accessible discussion of this specific case, but according to my limited understanding of papers such as these ones, the net effect remains identically zero to first order even when this is taken into account, because of certain symmetrical properties of photons. "First order" meaning scenarios like the simple pair-creation/annihilation described above, as opposed to "higher-order" cases in which either more than one photon enters or exits the Renormalization process, or in which the short-lived particles in turn decay into yet other particles before eventually recombining into the exiting photon (cf the right-hand side of Figure 1 from the article linked above). And for some of those higher-order cases, the net effect does turn out to be non-zero.

That being said, such higher-order cases happen only rarely. What that means in practical terms is that for such effects to become relevant (i.e. observable), you need either very strong fields, such that the induced asymmetry becomes significant, or you need to wait a long time (let the photons travel across a long distance), such that a significant number of higher-order cases can occur, or both. Alternatively, one of the publications also discussed the field of "tuned" EM fields, meaning fields which have just the right strength and/or structure to preferentially elicit a particular higher-order case. This latter version is essentially similar to the classical phenomenon of resonance in e.g. acoustics.

From what I gathered, when one or more of these conditions are met, the effect of the EM field in a vacuum would be much like that of a refractive and dispersive medium: If a perfectly collimated beam goes in, what comes out will be a weaker, less collimated beam going in a somewhat different direction, dependent on the polarization of the EM field, with the remainder scattered all over the place.

[gfellow]

Everyone, thank you so much for your input, you folks never disappoint; this helps a lot!
(ShinAce, for clarification I was referring to an electrical discharge, as in lightning.)

A quick addendum, if I may.
Would it be theoretically possible to have a laser emit photons with a unique signature - polarized photons came readily to mind, but it could also be a specific wavelength, or a combination of both - and have a reasonable chance that some of these can pass in proximity to the discharge, continue towards a sensor target, and still be distinguished from the static of the discharge it passes?

[cjameshuff]

Note that if there is a discharge, the light is not passing through a vacuum. Photons can certainly interact with hot ionized gas or free electrons.

Would it be theoretically possible to have a laser emit photons with a unique signature - polarized photons came readily to mind, but it could also be a specific wavelength, or a combination of both - and have a reasonable chance that some of these can pass in proximity to the discharge, continue towards a sensor target, and still be distinguished from the static of the discharge it passes?

Of course. If they aren't passing through the discharge or from a source on the near side, they aren't in line of sight of it. If they are passing through it (requiring the discharge to be transparent to the wavelengths involved), a glow discharge is largely ionized gas, a high intensity electric arc produces mostly black body radiation. A decently monochromatic laser could easily put a lot more power in a particular narrow wavelength range than the discharge.

[chornedsnorkack]

Since electron and positron can annihilate into 2 photons, 2 photons of suitable energy, direction and spin must be able to produce electron-positron pair.

Are 2 photons capable of undergoing a scattering (outcome 2 photons, not any other particles)?

[korjik]

Everyone, thank you so much for your input, you folks never disappoint; this helps a lot!
(ShinAce, for clarification I was referring to an electrical discharge, as in lightning.)

A quick addendum, if I may.
Would it be theoretically possible to have a laser emit photons with a unique signature - polarized photons came readily to mind, but it could also be a specific wavelength, or a combination of both - and have a reasonable chance that some of these can pass in proximity to the discharge, continue towards a sensor target, and still be distinguished from the static of the discharge it passes?

Shooting a laser through the disturbance then measuing its frequency with a spectrometer would do that.

[onomatomanic]

Are 2 photons capable of undergoing a scattering (outcome 2 photons, not any other particles)?

I don't know. Fundamentally, the Feynman diagrams looks sound to me, though I think the simplest case is a third-order one rather than second-order one which you may or may not be describing: Both photons pair-produce, virtual charged particles scatter via photon exchange, both pairs annihilate to reproduce two photons.

That being said, the articles I linked above specifically mentioned that such corrections vanish, at any order, unless there is an external field which introduces an additional asymmetry, as in the OP's scenario.

I suspect that what reconciles those two aspects is gauge invariance, which doesn't allow the photon to do some things that other neutrally charged particles are allowed to do. And that's as far as I'm going to stick out my intellectual neck.

[gfellow]

Forum members, thank you so much for your input. I go into the World a wiser man.

[trinitree88]

Forum members, thank you so much for your input. I go into the World a wiser man.

gfellow. Add a bit more wisdom. While two photons can produce a pair, it only requires sufficient energy from one. If it is of ~ 1.022 Mev, it will annihilate and the pair appears....electron /positron. More typically the photon has more energy than that and some is left over for kinematics of the pair, or a now-redshifted photon leaving the scene. It's one of the ways high energy photons dissipate energy. The other two are Compton scattering....transferring energy to charged particles........and ionization, separating neutral atoms into cations and their electrically matched electrons.
Also, when the electron positron pair annihilates, it is most commonly into two gamma photons @ ~511 kev (.511 Mev each)..~ 180 degrees ....to conserve energy and momentum. But, it is also seen as three, co-planar photons, dividing the 1.022 Mev in three. the relative proportion of each is a branching ratio. pete


SEE:http://en.wikipedia.org/wiki/Annihilation

notice in the cosmic ray spallation track electron /positron pairs near the bottom (they leave out the successive gammas, Compton scattering, and ionizations...all ultimately leading to heat, which is why you have germanium calorimeters in particle physics detectors.)
SEE:http://en.wikipedia.org/wiki/Cosmic_ray

SEE; http://www.sciencedirect.com/science...21961470900297

[chornedsnorkack]

gfellow. Add a bit more wisdom. While two photons can produce a pair, it only requires sufficient energy from one.

A single photon has no energy whatsoever.

Also, when the electron positron pair annihilates, it is most commonly into two gamma photons @ ~511 kev (.511 Mev each)..~ 180 degrees ....to conserve energy and momentum. But, it is also seen as three, co-planar photons, dividing the 1.022 Mev in three. the relative proportion of each is a branching ratio. pete

No.

Electron and positron with opposite spins can only annihilate to even number of photons, 2, 4, 6 or more. Electron and positron with parallel spins can only ever annihilate to odd number of photons, 3, 5, 7 or more.

[ctcoker]

A single photon has no energy whatsoever.

What are you talking about? We know MeV and higher energy gamma photons exist, and even radio photons have energy and momentum. It is true that single photons in a vacuum cannot spontaneously pair produce, but that's because they can't simultaneously conserve energy and momentum when doing so (they need to be close to some other massive particle to dump some of the momentum into).

[cjameshuff]

Photon energy is dependent on reference frame. It's not that a single photon has zero energy, it's that you need another particle to provide a reference frame before you can even define a photon's energy.

[trinitree88]

What are you talking about? We know MeV and higher energy gamma photons exist, and even radio photons have energy and momentum. It is true that single photons in a vacuum cannot spontaneously pair produce, but that's because they can't simultaneously conserve energy and momentum when doing so (they need to be close to some other massive particle to dump some of the momentum into).

ctcoker. Yep. E=hv applies to individual photons, or to any of the families of neutrinos and antineutrinos as individuals, too.SEE: http://www.obrien-dull.com/profpages/primer.html

P.S. I spent a bit of time building detectors @ MIT....with a lot of very bright people who knew their stuff.Learned lots of cool stuff.


pete

[George]

The Zeeman Effect demonstrates magnetic field influence.

[cjameshuff]

The Zeeman Effect demonstrates magnetic field influence.

It demonstrates an influence of magnetic fields on emission of photons, not on the photons themselves.

[George]

Yep. Thanks for the distinction.