EM radiation strength wrt time

[artis]

So I see EM radiation , visible spectrum including being portrayed as two sinewave vectors each perpendicular to other one being that of the E field the other being the B field, the field is carried by photons, let's take the visible spectrum as an example.
So having a specific frequency the radiation has a specific wavelength , another property from frequency is the energy that each individual photon carries at that frequency.

Here is my question, how can each photon carry the same energy at any given frequency if a sinewave has peaks and then closer to the so called "zero crossing" it has very little energy if any?

Or am I confusing this with how electrons move in a wire in the case of AC when at peaks they have their maximum potential while near midpoint they have almost no energy? Does this only work for currents and charged particles with mass and isn't applicable to massless photons ?

 

[BvU]

Hello again,

Familiar with the Poynting vector ? Stands for the directional energy flux.

And yes, a sine wave has zeroes, but just like with noise the amplitude of the EM wave is the determining factor.

Photons are complicated beasts. We claim we can actually see them, but I wonder...

 

[Dale]

how can each photon carry the same energy at any given frequency if a sinewave has peaks and then closer to the so called "zero crossing" it has very little energy if any?

A photon is not a point on a classical EM wave. You cannot think of photons as definite pieces of a classical wave at all, they use a substantially different formalism. They don’t have definite positions or even paths.

 

[artis]

I do know from reading that photons cannot be treated as points and even particles with mass like electrons are hardly ever bullet like.

I have heard about Poynting vectors and I can visualize the fields that they represent and how they are transverse to one another which I assume is the basis on how polarizer's work by canceling one of the waves while letting it's 90 degrees shifted "friend" through although the more complex maths behind them are beyond me.My idea here was this, in a wire that passes AC 50Hz current the electron motion aka current follows the field and when the field is zero the current is zero.
Now let's go to photons, they too have frequency (can have any within the EM frequency spectrum) so even though a photon doesn't have a precise structure or position it still has a quanta of energy that it carries and the energy of an individual photon is proportional to it's frequency just like the energy of an electron at any given instant in an AC cycle is proportional to the voltage at that moment in the cycle.

Here is my misunderstanding, if low frequency AC electrons could come out of the wire and fly through air then at the point where sine wave has its zero crossing those electrons would have no potential or kinetic energy (apart from average thermal kinetic energy)
I see light and other EM radiation also shown as a sine wave and photons are the particles that carry this radiation so do photons also experience "zero energy" at regular intervals along their path or does this doesn't apply to them and no matter at which point in space a photon is observed (hit a metal plate for example, photoelectric effect) it will always have the same energy given by its frequency?

 

[Dale]

if low frequency AC electrons could come out of the wire and fly through air then at the point where sine wave has its zero crossing those electrons would have no potential or kinetic energy (apart from average thermal kinetic energy)

I am not sure if this is correct. It would depend largely on the details of the movement of the electrons in the air. I am sure that you could contrive a situation where it is true, but also situations where it is false. You might be better off sticking to more clearly defined situations like electrons in normal wires.

so do photons also experience "zero energy" at regular intervals along their path

Again, a photon does not have a definite path along which to experience zero energy. Also, photons also don’t have a definite energy except when they interact with matter. And even then depending on the interaction itself the energy may still not be definite.

However, even though energy is not always defined, it is conserved.

 

[artis]

Ok let's keep the electrons in the wire for the example so there is a moment periodically at any frequency when there is zero current so at that moment the electrons have only as much energy as the thermal energy of the wire.

But I feel for photons there is no moment when they have zero energy even though they have frequency and any sinewave goes through a "zero crossing" inevitable and does so periodically. Although I feel that talking about EM radiation from a sinewave viewpoint would mean to talk about wave properties of "light" instead of individual photons would that be more correct? Because the wave experiences these peaks and midpoints but what then the individual photons experience which make up that wave?PS. So a photon has no mass, experiences no time, has no physical structure and now you say has no definite energy, almost feels like it might not exist and only appears out of "nowhere" once it strikes something along it's path?
If photon has no definite energy even coming from a monochromatic source then why do we measure it having a specific energy corresponding to its wavelength?
Wasn't this the whole photoelectric effect and beginning of quantum mechanics?

 

[Dale]

Ok let's keep the electrons in the wire for the example so there is a moment periodically at any frequency when there is zero current so at that moment the electrons have only as much energy as the thermal energy of the wire.

Yes, for a simple resistive alternating current every time the voltage goes to zero the current also goes to zero and therefore so does the power.

So a photon has no mass, experiences no time, has no physical structure and now you say has no definite energy, almost feels like it might not exist and only appears out of "nowhere" once it strikes something along it's path?

Now you are starting to get it. Except again it doesn’t have a definite path (it goes along all possible paths).

If photon has no definite energy even coming from a monochromatic source then why do we measure it having a specific energy corresponding to its wavelength?

The energy and the wavelength and the frequency all do have a definite relationship to each other. None of them are definite until an interaction, but if you measure one you know all the others too.

The energy is not definite because the energy has an uncertainty which is related to the lifetime of the transition.

 

[artis]

So unlike other elementary particles that we can detect through various detectors and deduce their physical properties , the photon was simply a term coined by physicists to explain why EM field interacts with physical matter (like a metal plate) and does so in a specific manner which resembles distinct packets of energy as the corresponding effects on physical particles like emitted electrons with specific KE were used as the proof of that?So we know how the EM field works and how it can be emitted and received etc but truly deep down we have no deeper understanding of what makes it work apart from the properties that the field have that can be described based on the fields interaction with physical matter (as any kind of detector and probe is a physical element) ?

If this is the case then why physicists draw a distinction between the so called "virtual photons" and "real" ones.? I have read that virtual photons are the ones used in case the EM field is static like from a point charge while real photons have frequency and are used when the EM field is changing in time like AC of any frequency.
So in both cases they are photons aka a figure of speech just that the properties of the field change from one case to the other so we assign them a bit different name?

 

[tech99]

how can each photon carry the same energy at any given frequency if a sinewave has peaks and then closer to the so called "zero crossing" it has very little energy if any?

I suppose that if you could ride the beam of light you would find the peaks travel along with you.
In a conductor, the electrons have energy due to their position and motion (analogous to KE and PE); not like photons which are a packet of energy.

 

[Dale]

So unlike other elementary particles that we can detect through various detectors and deduce their physical properties , the photon was simply a term coined by physicists to explain why EM field interacts with physical matter (like a metal plate) and does so in a specific manner which resembles distinct packets of energy as the corresponding effects on physical particles like emitted electrons with specific KE were used as the proof of that?

This is heading in the right direction, but of course there are some subtleties. First, all quantum particles share some of the weirdness of photons, like not having a definite path. But photons have some additional weirdness. One is that photons are bosons whereas ordinary particles are fermions, this makes a huge difference in their behavior. Another is that certain operators like position that are defined for massive particles (even if their value is uncertain) are not defined at all for massless particles.

but truly deep down we have no deeper understanding of what makes it work

I 100% disagree with this. We have a complete understanding of EM. That understanding is called QED (quantum electrodynamics). With it we are able to accurately predict every known EM behavior and interaction.

The weirdness is not due to a lack of knowledge, but simply to the fact that photons themselves are weird. We understand them perfectly and they are weird, but their weirdness is quantifiable and systematic and logical. They can be accurately and completely described, but the language for that description is QED and not ordinary English.

 

[sophiecentaur]

I suppose that if you could ride the beam of light

Only Bosons can do that.

 

[artis]

Dale I wasn't thinking that we do not understand the EM field or can't predict what it does sure we can , I was referring to the fact that we can only "interact" with that field through other means like emitted particles from the photoelectric effect or effects the field has on matter and our knowledge of the field comes from those interactions, would you agree to this?But I'm still wondering why there is this difference because both light and low frequency AC in my wall socket are EM fields , they just have different frequencies/wavelengths and one travels along a metal wire as its medium while the other travels through air or vacuum. So both are sinewaves, for low frequency AC the force it can exert on a charged particle goes to zero as the voltage goes to zero but that also means that at this moment in time the field itself is zero.

Now if we take the same photoelectric effect then applying an EM field to the metal plate (above threshold frequency) will result in electron emission but would I be able to draw a sinewave from the emitted electron energies as in at peaks their energies are higher and closer to midpoint their energies are lower ? From what i understand it doesn't work that way as for EM radiation of a certain frequency all the emitted electrons have the same KE?
If this is so then I wonder why photons only have the energy that corresponds to the peak value of a given frequency sinewave ? because normally for low frequency AC the energy associated with the wave constantly changes in time but it seems to me this is not the case for photons.

 

[davenn]

But I'm still wondering why there is this difference because both light and low frequency AC in my wall socket are EM fields , they just have different frequencies/wavelengths and one travels along a metal wire as its medium while the other travels through air or vacuum.

No you seem to not know/realise that the EM field associated with a wire is not in the wire, the wire is not it's medium.. The EM field is on the outside of the wire, the wire is only a waveguide to take the energy in the EM field from one place to another.

There is also EM radiation out from the wire into the air/space/etc.
If there wasn't, then radio antennas would not workDave

 

[Dale]

If this is so then I wonder why photons only have the energy that corresponds to the peak value of a given frequency sinewave ?

This isn’t actually correct. Write it down. The energy of a photon has no correspondence with the peak value of a classical wave amplitude.

The quantum state that most closely resembles a classical wave is called a coherent state. You may want to read about a coherent state a bit to get a better understanding:

https://en.m.wikipedia.org/wiki/Coherent_states

would I be able to draw a sinewave from the emitted electron energies as in at peaks their energies are higher and closer to midpoint their energies are lower ?

Look at figure 1 in the Wikipedia article. Can you see the answer to your question there? Figure 1 represents three coherent states corresponding to different classical amplitudes. What do you notice? Read the whole article and think a bit.

Notice that the coherent state is not an eigenatate of the photon number nor is it an eigenstate of the energy, so neither the energy nor the photon number are definite. The photon number, in particular, has an uncertainty relationship with the phase. How can that uncertainty relationship address your question?

 

[artis]

davenn I misspoke , yes sure a low frequency field can exist outside a wire (military long range submarine radio signals for example) it;s just that a wire as a medium greatly reduces the loss the field otherwise would have traveling from source to receiver is what I meant to say.as to what Dale said, ok I'll be honest I looked up coherent states etc also looked at the wiki article and because I suck at maths I had a hard time understanding things and the heavy quantum mechanics jargon didn't help either.
Let me try to put this into my own words and explanation and you tell me how close I am, is that ok?

1) As for the photoelectric effect I think I know why for any given frequency only electrons with specific KE will be emitted because only the photons at or near the sine peaks are powerful enough to give enough energy to emit an electron? The ones closer to the midpoint are not powerful enough (or speaking in quantum terms don't have definite energy?) and so just by looking at the emitted electrons I will not be able to fully reconstruct the incoming photon EM sine but just determine the maximum energy of the photons on that sine wave by which I will be able to calculate the frequency/wavelength and then based on frequency make a mathematic reconstruction of the sinewave?

2) So because in quantum mechanics all particles have certain probabilities about their momentum and position I take that for photons in an EM field the probability of their energy changes with respect to where we look at the sinewave, so if we "probe" closer to the peaks we have a high probability of encountering photons with specific energy which corresponds to the frequency of the field but as we go lower from the peaks closer to the midpoint the photons have less and less definite energy?

Clearly the energy of the field is highest at the peaks and lowers as one goes closer to midpoint as can be seen in a microwave oven where if one takes out the rotating "dish" so called "hot spots" form and next to them places of no heating which translate to the peaks and "zero crossings" of the standing EM field.

 

[Dale]

only the photons at or near the sine peaks are powerful enough to give enough energy to emit an electron?

No! There is no correspondence whatsoever between the energy of a photon and the amplitude of the classical wave. None, nada, zero, zilch, irrelevant, unrelated, immaterial, unconnected, separate. I don’t know how more clear I can be on this.

The amplitude of the classical wave is wholly and completely unrelated to the energy of the photons. It is related to the number of photons, not their energy.

Within a given coherent state the number of photons is uncertain as is the phase. As you get a lower and lower energy state (smaller classical amplitude) the uncertainty in your number of photons decreases as the uncertainty in your phase increases, making determining the “peak” and the “zero” more uncertain.

I take that for photons in an EM field the probability of their energy changes with respect to where we look at the sinewave,

No, see above.

Clearly the energy of the field is highest at the peaks and lowers as one goes closer to midpoint as can be seen in a microwave oven where if one takes out the rotating "dish" so called "hot spots" form and next to them places of no heating which translate to the peaks and "zero crossings" of the standing EM field

This is all classically correct. The problem is when you try to map your correct classical understanding to quantum mechanics. You frankly don’t have the background yet. I would recommend forgetting everything you think you know about photons and waiting until you have the background before tackling them again.

At this point I fear that the more we discuss in the absence of that background the more you will need to unlearn. In any case, based on our discussion above I don’t feel that I personally am helping you gain an understanding and I am probably making things worse, so I will back off and leave future responses to others who may be better able to communicate the issues.

 

[artis]

If quantum mechanics were a person and I were a judge I would make it serve life in prison without parole on the grounds of "first degree confusion" , but as a judge I couldn't pass the sentence because of the "uncertainty" of evidence supporting it. :D

No Dale no need to back out, so now you said it " the photon energy is unrelated to the sine amplitude whether it be peaks or zero crossings" ,
So the only thing I think I know for sure is that the classical wave frequency/wavelength relates directly to the energy of the photons, increase the frequency means decreasing the wavelength results in higher energy photons.So going slowly here tell me when you said this

The amplitude of the classical wave is wholly and completely unrelated to the energy of the photons. It is related to the number of photons, not their energy.

how is the number of photons related to the amplitude of the classical sinewave?
I would like to say more photons at peaks less at midpoints but then I would probably be wrong.PS. Now after you explained I think I kind of got the picture where you said I need to look at the wiki diagram 1. So for low amplitude wave it is impossible to determine the photon energies because they are spread out in a "cloud of probabilities" from midpoint to peak which made the sinewave look "out of focus" and spread out in the wiki picture.
But then let me ask does this apply to any frequency classical wave both lower and higher and just relates to the amplitude of the wave? Because we can have very high frequency wave with low amplitude.

 

[davenn]

it;s just that a wire as a medium greatly reduces the loss the field otherwise would have traveling from source to receiver is what I meant to say.

That's still wrong
please reread my previous post ... The wire is not a medium for the EM energy, it is a form of waveguide only

 

[artis]

Ok I understand , in a sense the EM field can travel through everything from vacuum to concrete just that in each medium it will have different "resistance" to its travel (for lack of a better word) so we use wires as waveguides to minimize the loss over distance traveled.

 

[davenn]

hi artis

Ok I understand ,

I'm not sure that you do

in a sense the EM field can travel through everything from vacuum to concrete just that in each medium it will have different "resistance" to its travel (for lack of a better word)

that is a completely different and unrelated subject ... "How different materials impede the propagation of an EM wave/signal". It comes into play when discussing cable insulation (dielectric) materials ... see below.

so we use wires as waveguides to minimize the loss over distance traveled.

no, if that were the case, there would be no need for radio communications.

Just to clarify again ... the EM does not travel in the wire, it travels along the outside of the wire in the air, insulation etc that surrounds the wire

A wire without any insulation has the least loss. This is shown by the speed of the EM wave Approx 95% compared to it's speed in free space.
Surround the wire with an insulation layer ( a dielectric) and its speed decreases, how much depends on the type of insulation (dielectric).
A solid plastic and the EM propagation speed can drop to as low as 55 - 60% that of free space.
As the dielectric gets "closer to air" foamed ( lots of air bubbles) etc the losses decrease

here's a foamed dielectric between inner and outer conductors

in this next pic there are variations all the way from foamed to just spiral/spacers
In that case, the dielectric is mostly air with just a little bit of Teflon as a spacer. Something has to be there to keep
the inner and outer conductors apart

The EM is traveling in between the inner and outer conductors

Dave

 

[davenn]

You may also be thinking about the reason for a hollow centre conductor ?

3 main reasons
1) if it is a tube rather than solid, then less copper is needed ... cheaper to produce
2) added to that, due to "skin effect" ( google it) current only flows on the outside of the conductor so there is no need for a solid conductor
3) copper is reasonably heavy, so less copper = less weight = easier to handleDave

 

[artis]

So could we say that the wires serve as sort of "support bases" for the field in a sense that the field traveling along the wire excites electric current in the wire and that current in turn excites the field further , so for a given length if the field has to excite electrons in a material with higher resistance then the field uses more of its energy per that length and so we say the conductor has larger losses while another conductor with lower resistance will be able to let more EM energy per the same length.

yup thank God I already have went through this so I know why they make hollow tubes at RF.
For lower frequencies simply twisting many small wires together is sufficient as when I made and smps used litz wire for the traffo.Somehow I still have to get back to my original still not understood question but I am fine with these points of advice in between.

 

[artis]

Even though I feel some people are reluctant to further aid me with my quest I will ask for one confirmation while I'm reading the more complicated parts so that at least we get the "basics" covered.

So for electron current in a wire where one can determine the energy for individual electrons at any given instant of the sine wave and those energies change with respect to the sinewave.
Instead for photons the only thing one can deduce from a sinewave (the EM classical wave) is their energy which is directly related to the frequency/wavelength of the sine, and that's it. ?PS. I'm still reading about the coherent states and how that relates to what I'm trying to understand here which got Dale frustrated because of me guessing wrongly.

 

[Dale]

So for electron current in a wire where one can determine the energy for individual electrons at any given instant of the sine wave

Well, technically electrons don’t even exist in circuit theory. Circuit theory uses the continuum approximation and only uses voltage and current (charge can then be indirectly defined as the time integral of current). But other than that your statement is right: the electrical energy of an infinitesimal bit of charge is classically well defined.