[alancalverd]

Because as you stated, at the closest point of approach, 10 coulomb of charge was moving past the test point at 0.1 m/s   

[Petrochemicals]

I don't know whether you have the capabilities, or more to the point whether it would work but you could try slowing light down.

[Eternal Student (OP)]

Hi.

   I'm sure I'll miss some posts under the number that have appeared.   Sorry, post again if it seems important (maybe leave it for a day or so).

   @hamdani yusuf   posted some stuff about vibrations in molecules caused by IR radiation,   way back in post #59.
I like it and I'm still considering it.   As you mentioned it really only works for IR radiation not visible light but the frequencies are close and it's hardly worth quibbling over.
    Technically it still doesn't really measure the oscillations in the E field directly, it's just that it should be there otherwise it's hard to explain what is happening.   Just to be clear:  Yes, you'd have a hard time explaining why IR radiation of certain frequencies is absorbed by CO₂  without using an oscillation in the E field as a model for the IR radiation - but it could be done.   E.g. The explanation does assume that a simple ball-and-spring model was sufficient and this could be challenged.  The molecule is better modelled as a Quantum Mechanical structure and there was some sort of "photon and atoms" interaction happening.   Additionally I'm not sure how easily you can assert that the molecules were vibrating in some mechanical sense, that's not something you can directly observe.   You could measure the temperature of the CO₂ and claim that is a measure of these vibrations but that's still a very indirect measure of vibration since temperature is a very complicated thing, all sorts of modes of supporting energy contribute to temperature  (I know you've had a thread discussing temperature elsewhere on this forum).
   None the less, I like it.  It goes high up on the list of experiments that suggest light (or IR radiation in this case) has an oscillating E field.

 Best Wishes.

[Eternal Student (OP)]

Hi.

I don't know whether you have the capabilities, or more to the point whether it would work but you could try slowing light down.

   Yes we could,  we could put the light in a dense medium.   However, this affects the wavelength and speed only.  The frequency of oscillations would be unchanged.
   It's the frequency, or "the speed of oscillation" if you prefer, that makes it so hard to directly measure the oscillation in the E field.   We don't seem to have a lot of equipment that can respond to such rapid fluctuations in the E field.

    However we could try something similar.  In fact, you've made a first class suggestion when I think about.   We don't need to change the speed, we can just change the frequency.
   Emit some visible light,  then get in space rocket and move fast.    The light should have a relativistic Doppler shift and then all we have to observe is the oscillation in the E field of some radio waves (which we can more or less do to everyones satisfaction, just stick a pole in the air to act as an radio antenna and directly measure an oscillating current flow).

Best Wishes.

[Eternal Student (OP)]

Hi again.

   Probably one of the last replies for a while, I think I've gone through all the other posts.

Posts #21, 22, 23  and some related posts discussing  an atom placed in the centre of some lasers.
   

I would have THOUGHT no, to your question about lasers of different frequencies

   I would have thought that a while ago.  Now I'm not so sure.

..So the answer is yes, if the combined energies of the beams hit the sweet spot then the electron will transition..

    That does seem like Colin2B was fairly certain.   I didn't know such experiments had already been done.

It's certainly not something that gets a mention in conventional textbooks discussing the photo-electric effect.  If you add an asterisked comment like this, then the entire message or implication of the result is eroded:
"  ... You can't eject an electron by using light of another frequency, or just by increasing the intensity,  there is an absolute minimum frequency of light that must be used**.
 
**  Except when you use a few frequencies all at the same time."

- - - - - - - - - -

   Anyway, at about this time I'm thinking that the oscillations in the E field of visible light should be observable.  At least as much as we can directly observe the oscillations in radio frequency e-m radiation.   There might not be good enough equipment available at the moment but it's at least theoretically an observable and will be directly measured when equipment catches up and improves.
   I might not be making it clear what I mean by "theoretically observable", a few people will be thinking - either you have measured it or you haven't.   I'll try to re-phrase it this way:   There is a quantum mechanical wave function describing visible light (a collection of photons and not just one).  However, that wave function is certainly NOT the function (like E=E₀ Cos ωt ) that describes oscillation in the E field or B field (there shouldn't be a lot of dispute about that).  The oscillation in the E field IS (underline IS ) an observable,  the wave function is not.    This is the sense in which I would think that the oscillations in the E field are an observable.   There is a suitable operator acting on the wave function such that the value of the E field along the path of the light (at a given time) could be obtained as the eigenvalues of that operator.
   Compare that with, for example, post #27 from @alancalverd   where it was implied that we only model light as an oscillation in the E field, we can't detect the oscillations.  I'm thinking that we probably can detect the oscillations in the E field directly.   They are "there to be seen" just as much as anything is there to be seen and measured, it may be an observable from a fundamentally quantum mechanical system but that's a separate issue, they are none-the-less observable.

Best Wishes.

[Bored chemist]

it means the numbers you provided which must be imaginary numbers

They can be measured by high school students.
This is essentially the problem with "Dark knight".
He hasn't a clue.

[hamdani yusuf]

The oscillation in the E field IS (underline IS ) an observable,  the wave function is not.    This is the sense in which I would think that the oscillations in the E field are an observable.

How would you detect Electric field directly?
In case of radio frequency, we can observe its effects on electronic components at the receiver. But it's not a direct observation either. We rely on the assumptions of how those components work.
At lower/subsonic frequency, we can observe its effect on an electrically charged metal ball, like in the video I posted previously.
At infrared frequency, it can induce vibration on some atoms tied by chemical bonds.
At visible and ultraviolet frequency, it can induce vibration on electrons tied to atomic nuclei.

[Eternal Student (OP)]

Hi.

How would you detect Electric field directly?

   That's kind of the main point of starting the thread.  I don't know of a good way to do this for light, although some reasonable ideas have been presented and I have a few more ideas now.   Thank you very much to everyone who has written something.   
   So one reasonable option is something like that suggested in post #103,  get in a space rocket and reduce the frequency down to something you can measure directly.  Another reasonable suggestion is an experiment like the one you ( @hamdani yusuf ) suggested involving IR absorption in CO₂ molecules.

    In principle, an Electric field should be that which creates a force on a charge, it's that simple - go straight from the definition of what an E field should be.   So, to be very specific it should be measured as a force acting on an object that is just due to charge.  You could replace the test object with some other material (make it more massive, less permeable to magnetism, a different colour, not having any Hydrogen atoms or whatever...)  and the force would be the same provided the charge was the same.   To put a numerical quantity on it, the field strength is the force (in Newtons) per unit charge (in Coulombs) and it has a vector property - the direction of the force on a positive charge.
   So, working from that definition, the most simple experiment that directly shows an oscillating E field is present would be one where you can measure an oscillating force on a charge and clearly demonstrate it was only due to the charge. 
    Moving it up one small level of assumption then, an experiment where you could see a small charged blob being moved about and oscillating as if it is being driven by an oscillating E field - that would be a reasonable and fairly direct measurement of an oscillating E field  (for most people).
   As you state,  we tend to increase the number of assumptions when the E field gets small and especially when the frequency of oscillation increases.  It's not guaranteed that the effect on an electrical component is really directly measuring or responding to an E field,  however there's not that many assumptions being made.   E.g. sticking a pole up in the air while radio waves pass across it and measuring an oscillating current being produced is reasonably direct measurement of an oscillating E field being there - let's say it's only another level or two up on the ladder of assumptions you might be asked to make.
   By the time you're up to visible light frequencies there's a stack of assumptions that are usually made.

Best Wishes.

[hamdani yusuf]

Basically, you can measure EM wave in any frequency, as long as you have quick enough rectifier.

[paul cotter]

Hi Eternal Student, in your post #102 you alluded to having missed some posts- probably just as well. In the early days of investigations on the photoelectric effect I am sure a strictly monochromatic light source was not being used. If frequency mixing was occurring then any source with two or more components would have delivered a+b to trigger electron emission even when the highest frequency( incident ) had energy below that of the metal's work function. Analysis of the phenomenon would have been intractable.

[alancalverd]

How would you detect Electric field directly?
In case of radio frequency, we can observe its effects on electronic components at the receiver.

Actually not. We detect the current induced in the receiving aerial by the magnetic component.

[Petrochemicals]

Hi.

I don't know whether you have the capabilities, or more to the point whether it would work but you could try slowing light down.

   Yes we could,  we could put the light in a dense medium.   However, this affects the wavelength and speed only.  The frequency of oscillations would be unchanged.
   It's the frequency, or "the speed of oscillation" if you prefer, that makes it so hard to directly measure the oscillation in the E field.   We don't seem to have a lot of equipment that can respond to such rapid fluctuations in the E field.

    However we could try something similar.  In fact, you've made a first class suggestion when I think about.   We don't need to change the speed, we can just change the frequency.
   Emit some visible light,  then get in space rocket and move fast.    The light should have a relativistic Doppler shift and then all we have to observe is the oscillation in the E field of some radio waves (which we can more or less do to everyones satisfaction, just stick a pole in the air to act as an radio antenna and directly measure an oscillating current flow).

Best Wishes.

You could make your own doppler machine, various moving parts to stretch the frequency. But as you state this would be measuring radio waves rather than light directly.

[Bored chemist]

If you have a really bright light source, like an exploding universe, and you wait until the expansion has stretched out the EM radiation to the microwave region of the spectrum, then you can measure that frequency directly using a big enough antenna, an amplifier and a frequency meter.
Is that cheating?

[paul cotter]

Alancalverd,i disagree. Reciprocity in transmitting/receiving tells us that while a tx aerial generates both H&E fields so will an rx aerial respond to both.

[alancalverd]

So a charge moving towards the receiver will induce a current in the antenna? And we don't need to rotate our TV antenna to match the polarisation of the transmitter? Hmm. Need to think about that one.

[Bored chemist]

So a charge moving towards the receiver will induce a current in the antenna? And we don't need to rotate our TV antenna to match the polarisation of the transmitter? Hmm. Need to think about that one.

That's not what he said, was it?

How would you know if you were lining up the magnetic field or the electric one?

[Eternal Student (OP)]

Hi.

Basically, you can measure EM wave in any frequency, as long as you have quick enough rectifier.

    Surely not.   The wires which made the dipole aerials are shiny reflective things.   If you just shine visible light on an aerial then it just reflects off it,  it doesn't interact with the aerial to generate a current in it.   Meanwhile, most X-rays or gamma rays can just go straight through without any significant interaction along the way.
   The limitation isn't just the diodes or rectifier, it's the interaction with the aerial.

Actually not. We detect the current induced in the receiving aerial by the magnetic component.

   I'm in general agreement with @paul cotter ,  it's both Electric and magnetic field. 

Is that cheating?

   No more so than getting in a space rocket and using a Doppler shift to bring the frequency down.  So yes, it's cheating, you aren't really measuring an oscillation in e-m radiation of visible light frequency but it is suggesting that the oscillations would have been there in the visible light even if the equipment you have wasn't fast enough to find it.  So it's OK for the bigger picture that an oscillation is there and is theoretically observable.

In the early days of investigations on the photoelectric effect I am sure a strictly monochromatic light source was not being used....

   Probably true.   However, the multiple sources and frequencies of light probably weren't positioned in the right places, they wouldn't have been kept in phase anything like as well as some lasers,  they wouldn't have been polarised in the right way  etc.   By random chance a few rays of light might combine so as to make a photon of the right frequency appear at an atom of the metal and there could have been a few electrons released.  The equipment may not have been good enough to identify that from general noise.  You would not have been able to eliminate the possibility that a cosmic ray from space, or a high energy photon from a decaying piece of granite inside the building, had just come in and hit the metal.   That sort of thing is going to happen quite often and is going to be much of the "noise" that would have present in the experiment.  I mean, on a hot day when a thunderstorm is due the air itself will be quite ionised anyway even before you start liberating electrons from the surface of the metal.
    As I implied in an earlier post,  I'm not entirely sure what will happen with the arrangement of lasers around an atom.  However,  @Colin2B was fairly sure and seemed to be suggesting that experiments have already been done like this.  In my limited experience on this forum, Colin2B is rarely so wrong as to imagine experiments that were never actually done.
    Let's think about this in a different way:  A positron and an electron can annihilate to produce a pair of photons, that's everyday physics now with the PET scanners they use in hospital.  We think that during the early moments of the universe the opposite happened, pairs of photons combined to form some matter, this was the earliest synthesis of matter.  Now, if photons can combine to make particles of ordinary matter, then it's really not asking a lot for photons to combine and just make a new photon with a higher amount of energy.  I'm just an ordinary person and I like a nice simple model with photons as particles just as much as the next guy.  However, according to an entirely quantum mechanical model for light like QED or even QFT,  there are no particles, it's all just waves in an underlying field.  It shouldn't be too surprising that those waves can combine under certain circumstances.  We just choose to interpret that as something that was identifiable as individual photons emitted by the laser combining to form a new higher energy photon at the atom.
   As I mentioned before, the wave function describing the oscillations in an E and B field for light is NOT the quantum mechanical wave function for light.  However, I find it quite interesting that it is a step closer to the QM wave function.  For the atom surrounded by lasers, if we can arrange it so that the E and B fields oscillate correctly at the atom, then we seem to have also arranged the QM wave oscillations so that a superposition will happen and we can get something that looks like a photon of a higher energy.   That, I think, is what @Colin2B  was alluding to in his comments about the photon being detected at the atom.

Best Wishes.

Postscript:  New posts have come in.   

So a charge moving towards the receiver will induce a current in the antenna?

  Yes it will.  However, it would be a one-off or DC current you could observe.  The free charges in the aerial will move so as to oppose the E field that is created by the approaching charge and there will have to be a significant difference in the number density of free charges on the conductor (the aerial) at different places.  It's especially evident when comparing the surface closest to the moving charge with the surface furthest away from the moving charge.  You would get an AC current provided you move the charge to and fro.  You know this @alancalverd, you've answered questions about conductors in an E field elsewhere and and at other times on this forum.

[paul cotter]

Completely off topic; I just want to briefly say what a pleasure all of you have provided an old fogey like myself in such stimulating ( for me, certainly ) debate.

[Colin2B]

If you have a really bright light source, like an exploding universe, and you wait until the expansion has stretched out the EM radiation to the microwave region of the spectrum, then you can measure that frequency directly using a big enough antenna, an amplifier and a frequency meter.
Is that cheating?

I was going to suggest the same, but can’t find an example of optical to microwave redshift. Haven’t had a lot of time, will give so e thought next week.

@Eternal Student
I’ve been away with limited wifi.  Because threads like this shift focus frequently assumptions are made and the direction changes which means original brief comments can be misunderstood, but will expand after tomorrow.
Interesting topic which has resulted in a weaving of different threads, some of which I’ll remove when I have time ie DarkKnight

[hamdani yusuf]

Actually not. We detect the current induced in the receiving aerial by the magnetic component.

The components used in this video aren't magnetic.

Basically, you can measure EM wave in any frequency, as long as you have quick enough rectifier.