[Eternal Student (OP)]

Hi.

Does anyone know of an experiment that shows the oscillation in either the electric or magnetic field (or both) when a light ray travels through some place?

    For a radio wave, we can just put a metal rod (an antennae) in the path of the wave and show that an electrical signal is generated in the antenna.   We can argue that this is the result of forces from the electric field,   F = Ee and also the magnetic field (which is moving relative to the static charges in the rod, giving us some v>0 in the usual formula)  F = e (v x B) ).  There is plenty of equipment, like a good multimeter, that will directly show we get an electrical signal when it's at these sorts of radio frequencies.    Can we do anything like this with e-m radiation in the visible spectrum?   Is there an antenna that picks up light (rather than reflects it) and equipment fast enough to respond and show we have a rapidly oscillating electrical signal?

    It doesn't have to be an antenna.  Is there perhaps some experiment that has been done with a line of charges and then a laser beam was fired along the path of that line of charges?  The laser being used to ensure you get a nice ray of light which is all in phase.  (Hopefully causing some rapid oscillation in those charges that we could measure).   
 
    Anyway, is there any experiment that you know of that directly (or as directly as possible) shows the oscillation in the electric (or magnetic) field for e-m radiation when the frequency is up in the visible band (or higher)?
Is there some ultra-fine probes that we can attach to a multi-meter and just stick them into space and directly measure the oscillating E field as the light ray passes through?

Best Wishes.

[evan_au]

I understand that you can show polarisation of a microwave beam by building a wall with row of parallel wires.
- When the wall is oriented in one direction, it "short-circuits" the E-field and the microwaves are blocked
- Rotate the wall by 90°, and the microwaves get through, as the wires don't short-circuit the magnetic field.

A similar arrangement (on a smaller scale) is used to produce polarised lenses for visible light - there is a grid of long, parallel crystals in the lense.

[Eternal Student (OP)]

Hi.

  Thanks for your reply, @evan_au .   I do quite like the idea of polarising lenses and I am busy thinking about it.   It might be enough for my purposes and I'm very grateful for your time.

   I'd rather have a more direct demonstration that there are oscillating E and B fields in light,  for example, if there was some electrical device you could connect to the crystalline structures in a polarising filter and show that there is an electrical current generated.   Otherwise, you just have to accept that polarising lenses do work the way we think and the evidence for that is not at all convincing for the typical person.

     Rather than simply blocking light of one polarisation, a polarising lens does sometimes "appear" (I'm not saying they do, only that it may appear) to actively interact with passing light and turn it.
     You've probably heard of the experiment that is done to show quantum mechanical effects with polarised light.
Two polarising lenses  turned at 90 degrees will block all light passing through.  However, if you insert a third lens in between that is offset by only 45 degrees,  then you will get some light transmitted through the end of the optical system.   This is weird because if the polarising filter is just blocking some light then every filter you add should only remove more light, yet inserting the third filter has somehow boosted the amount of light you get at the end.
     See, for example, this reference:  https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Tutorials_(Rioux)/Quantum_Fundamentals/08%3A_Quantum_Principles_Illuminated_with_Polarized_Light
     
or this video, if you prefer a more dynamic demonstration:   Approx. 1 min 30 secs.  "Three polarizing filters: a simple demo of a creepy quantum effect", from YouTube.   

     You can explain this with QM but a stubborn few will stick with classical physics and suggest that polarising filters just don't work as we thought.  If you reduce it down to a simple level,  we can fire just two photons (with the same polarisation) at a polarising filter that is off-set by 45 degrees and the filter doesn't seem to be just blocking something.  You don't observe something consistent with just stopping (blocking) 0, 1 or 2 of the photons.  Most of the time you observe one photon thrown out at the other side and it has been turned so that it is unlike either of the incoming photons.

    Anyway, I'm not trying to be awkward.   I've just read or heard something presented on a lecture course I was trying to follow and it seems to be wrong, or at least mis-leading.   I'm trying to reconcile my existing understanding with what was said.  This post is already too long, so I'm not even going to try and discuss the lecture right now.   (Besides which, my own difficulties in understanding something aren't all that interesting to others, it's my problem not yours / theirs).

   I've always been prepared to accept the oscillatory nature of the E field and B field in light as "a given".  It follows from the theory and can be quite directly observed in radio waves.   However, I just wondered if it could be stated in a more concrete way:  If there is some very direct demonstration of the oscillating E and B fields in visible light.

Best Wishes.

[evan_au]

If I ask the same questions in a different order, I get a different answer, which is a little disturbing to me

In my primitive understanding, passing light through a polarising filter is like a test or a measurement of the polarization.
- If a photon passes through the filter, it has "passed" the test, and you have measured its polarisation as being in-line with the filter.
- If you put another polarising filter at 90°, everything passing the first filter will fail the second filter.
- However, if you put a 45° filter in between two 90° polarising filters, you are conducting another test/measurement. The outcome of this measurement is that cos²(45°) = 50% cos(45°) = 71% of photons passed the second test, and 50% of that (ie one quarter) 71% of that (ie half) now get through the third filter (if they were perfect polarisers)

In a quantum world, taking a measurement changes the thing you are measuring. So it is no longer light polarised vertically, any photons passing the 45° filter are now polarised at 45°. And a fraction of these will pass the third filter.

Corrected, as advised below by Eternal Student...

[paul cotter]

Hi, Eternal Student. As you stated, it's easy to demonstrate the electric field of a radio wave. With light however, we are dealing with a very short wavelength and this makes it very difficult to demonstrate either the discrete electric or magnetic component in a direct way. With regard to polarising filters I can't help wondering is there a rotation taking place? Many compounds with chiral asymmetry that exist in right and left handed configurations will rotate the plane of polarisation if in a transparent medium( it's many years since I studied this stuff-BC would be the authority on this ).

[Colin2B]

With regard to polarising filters I can't help wondering is there a rotation taking place?

You are still looking at this through classical eyes. The problem is that in subsequent measurements you are changing the bases (similar to coordinates) that you are using to take the measurements.

[Eternal Student (OP)]

Hi.

   Yes that's the right idea @evan_au .   Except that you need to take squares, so the probability of a photon passing the filter is 0.5 not 0.71...    Under classical field theory (so that is E and B fields) the intensity of light received is proportional to the square of the amplitude of the wave that is transmitted.   You resolve the incoming wave into components and get the fraction Cos 45°  for the amplitude that is transmitted exactly as you stated.  That means  Cos² 45°  =  ½  is the fraction in Intensity.   Then you can jump back to the particle and photon interpretation where  intensity  is just proportional to the number of photons received (per unit time and per unit area of the receiver).  Hence, you get the probability ½ for each photon.

Many compounds with chiral asymmetry that exist in right and left handed configurations will rotate the plane of polarisation...

   I completely agree.  It's not clear that polarising lenses based on crystalline macromolecules really are just blockers rather than rotators of polarised light.   (I also agree with the other statement but that's why I'm asking if anyone knows of a suitable experiment).

Best Wishes.

LATE EDITING:   Overlap with Colin2B who just posted.  At a glance, looks OK and this post doesn't need adjustment.

[evan_au]

I can't help wondering is there a (polarisation) rotation taking place?

This is exactly how Liquid Crystal Displays work (or LCD displays, as they are tautologically identified).
- The liquid crystal material is sandwiched between two polarizing filters
- The liquid crystal rotates the polarisation of light under the control of an electric field. This allows light to pass (or not pass), depending on whether a voltage is applied to transparent electrodes.

https://en.wikipedia.org/wiki/Liquid-crystal_display

[alancalverd]

The transfer of energy from light to any nonmagnetic molecule can only arise from interactions with the E field, so just measure the heating effect of photons on an absorber. Graphite is a good one.

[Eternal Student (OP)]

Hi.

   I like the idea @alancalverd  and thank you for your time.
   
    You do seem to be establishing that light carries energy.   I'm not sure if it tells me much about light being an oscillation in the E field.   If you fired lumps of blu-tack  (that's sticky clay stuff - but other brands are available)  at a cup of tea so that they just stick to the cup,  then you are transferring kinetic energy to the cup.   That's got to give you a warm cup of tea eventually.   I think that's how they do it at our local Cafe, it takes years and the cup is sticky.   I'm fairly sure they don't use oscillating E field blu-tack, just the regular stuff.

   More seriously,  I think your suggestion solves half the problem and I'm continuing to think about it, thank you.

Best Wishes.

[Colin2B]

.Yes that's the right idea @evan_au .   Except that you need to take squares, so the probability of a photon passing the filter is 0.5 not 0.71...    Under classical field theory (so that is E and B fields) the intensity of light received is proportional to the square of the amplitude of the wave that is transmitted.   You resolve the incoming wave into components and get the fraction Cos 45°  for the amplitude that is transmitted exactly as you stated.  That means  Cos² 45°  =  ½  is the fraction in Intensity.   Then you can jump back to the particle and photon interpretation where  intensity  is just proportional to the number of photons received (per unit time and per unit area of the receiver).  Hence, you get the probability ½ for each photon.

Yes, you’ve hit on the main reason why a lot of people get confused with the polarisation experiments.

Sorry, I didn’t have time to reply properly before, just unpacking having returned from a 2 week trip to be alongside a relative in hospital. Not a lot of time available.

As you are aware, polarising plastics use long thin chains of molecules such that the electrons will oscillate along the chain and absorb the photons, whereas perpendicular to the chain there there is no absorption and the light passes through. The experiment you are looking for is a tricky one as you are trying to measure specific motion of electrons in the material and to do that means disturbing the very electrons you are looking to measure.
This experiment is takes a different approach and might be similar to what you are looking for https://www.sciencedaily.com/releases/2018/04/180410100941.htm

[Eternal Student (OP)]

Hi.

  That does look like the sort of thing I was after.   Thank you very much.

Best Wishes to you and your relative.

[alancalverd]

It is obvious that light carries energy* - you can measure it with a bolometer, or infer it from photochemistry. The question is how can it transfer that energy to an absorber?

Shine white light through an organic filter - say a theatrical gel. Some frequencies have been absorbed (and any theatre lighting technician will tell you how much because the filters get hot!) but the filter pigment has negligible magnetic susceptibility, so the electric field of the incident light must be interacting with the molecules in a frequency-specific manner. 

All this presumes, of course, that you assume light to be an electromagnetic wave in the first place. So far, nobody has produced a better model of refraction, diffraction and interference, and the application of Maxwell's equations does indeed give us an exact value for its velocity derived entirely from static measurements. 

*There is often a perceptual problem when dealing with high frequency EM radiations. Thanks to transient photoelectrochemistry, our bodies are exquisitely sensitive to the intensity of visible radiation, in the 2 - 5 eV range, but the rest of our senses are very dull in comparison, so we don't associate physiological effects with low incident powers outside that range, hence it is easy to forget that visible light transfers energy.   

[Colin2B]

I like the idea @alancalverd  and thank you for your time.
   
    You do seem to be establishing that light carries energy.   I'm not sure if it tells me much about light being an oscillation in the E field.   If you fired lumps of blu-tack  (that's sticky clay stuff - but other brands are available)  at a cup of tea so that they just stick to the cup,  then you are transferring kinetic energy to the cup.   That's got to give you a warm cup of tea eventually.   I think that's how they do it at our local Cafe, it takes years and the cup is sticky.   I'm fairly sure they don't use oscillating E field blu-tack, just the regular stuff.

Worth having a deep think about Alan’s suggestion. Light doesn’t have mass, so that’s not the source of any energy transfer. When looking at electrons you would expect the charge to be affected by an electric field, which would transfer energy and is certainly the origin of the photoelectric effect. Unless you can think of a better transfer mechanism.
I did wonder about experiments using the optical Mossbaur effect with polarised light. There seem to be some experiments out there, but you would have to go through them to decide whether they answer your question.

Note, just got a warning of another post while I’ve been writing. Similar thoughts from Alan

[yor_on]

There are indirect proofs for light behaving as EM, oscillating, if that is what you're thinking of ES?

This one?

" Another idea I had how to show that light is an electromagnetic wave would be to use a stronger (and well cooled) antenna (a stick in which electrons can be accelerated) than Hertz. The electromagnetic waves in the antenna have the frequency in which the electrons move up and down inside. So if one would increase the frequency more and more (by applying strong alternating voltage with ever increasing frequency), one should eventually see these radiations in form of light (as manifestation of the osciallations of electrons which are electromagnetic waves).  "

https://physics.stackexchange.com/questions/145330/what-is-the-experimental-evidence-that-light-is-an-electromagnetic-wave

It's a interesting question, what made you want to ask there?

[Bored chemist]

Does this count?
https://skullsinthestars.com/2008/05/04/classic-science-paper-otto-wieners-experiment-1890/

[paul cotter]

yor-on, it is not possible to run a signal frequency up to that of visible light. Low microwave frequencies are the highest achievable with current electronics, well short of visible light. By the way, I thought you only did catastrophe, now you are digressing into physics-a good move imho.

[Eternal Student (OP)]

Hi.

  Sory, I haven't been keeping much of an eye on this thread any longer.  I'll catch up and follow the links people had suggested.

It's a interesting question, what made you want to ask there?

    I didn't ask on Physics stack exchange, if that's what you mean.  That was someone else.
    I asked here for a reason that was releated to something I read or heard from an online lecture course recently.  I can try and summarise that but it will take me a day or so to dig up all the details again.   I'd like to do that because otherwise I have a tendancy to just "ramble".

Best Wishes.

[Bored chemist]

it is not possible to run a signal frequency up to that of visible light.

That's a matter of definition
https://en.wikipedia.org/wiki/Frequency_comb

[yor_on]

Please do so ES. And define your thoughts on it. I'm interested.