Medium of EM Waves?

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  • #26
Claude Bile
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Indeed such a dipole will radiate. A test charge would experience a force from a wave emitted by such a dipole (more than that, it would also experience torque due to the fact that the polarisation of the output wave is rotating).

A said "vast-majority" as a disclaimer. Some radiating mechanisms (nuclear radiation is an obvious one) do not require an oscillating dipole.

Claude.
 
  • #27
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Cool banana's thats neat!

However when you say it will cause a test charge to feel a force, I take it this is not due to the actual alternating electric field around the dipole is it? I mean at some distance close to the rotating dipole, the test charge will be in its electric field, and will thus feel force not due to the EM. Does this mean that when the test charge is far away, outside of the influence of the electric field from the rotating dipole, that it will feel a force from the EM radiation? And that it will feel the same force from the EM radiation no matter how far it is from the source, as long as it absorbs the same number of photons? (which I assume it doesn't...however in the case of one photon?) NB I take it the test charge must absorb the EM waves to experience their force?

Just checking too...we are talking of the same plane of rotation for the dipole. I mean like a helicopter blades and the tip of one blade is positive, the other negative.

And finally I seem to recall that accelerating charges emit radiation...I see how this works in the case of a rotating dipole (centripetal acceleration), however what about a linearly accelerating dipole, or even a linearly accelerating charge? I do not see how an oscillating electric field is caused from this??? Hopefully this is not relativity, though I fear it may be, as I read somewhere that while a charge may appear to emit radiation as it is accelerating relative to one observer, if you are travelling at a uniform velocity relative to the charge, you see no radiation. Still where oscillating electric fields come from I do not know. This causes confusing problems...energy conservation and stuff... :confused:

Thanks anyways,

Kcodon

PS I have no idea about polarization etc so the idea of the torque is way over my head lol.
 
  • #28
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Oh and I just thought too...if the rotating dipole radiates (very slowly?), then I assume it loses energy and thus slows down. However does this not violate conservation of angular momentum? Or do photons have angular momentum...is this their spin?

Kcodon
 
  • #29
rbj
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[tex] \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} [/tex]
The left-hand side describes spatial changes in the electric field and says that they are equal to how the magnetic field changes in time.

[tex] \nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t} [/tex]
Similarly with this one. Now, spatial changes in the magnetic field are equal to how the electric field changes in time.

The end result is that the "medium" for an electric wave is a magnetic wave and viceversa, or an electromagnetic wave propagates without a medium. It is self-contained and self-sustaining.
hmmm. i hadn't heard it put that way. ("that the "medium" for an electric wave is a magnetic wave and viceversa.") dunno yet what i think about it.

Back to the original question from Barnaby: This is also what physicists back in the day used to think...they called it the aether, and it was the medium that EM travelled through. However the Michelson Morley experiment disproved the aether. The experiment went along the lines that if there was an all pervasive medium like the aether, then our earth motion in the solar system would mean that we would be travelling at different relative speeds to the aether, so then we should detect different speeds for the EM to propagate (as the EM was limited to a finite speed in the aether). MM detected no change in the speed of light, and thus concluded that there was no aether.

I myself questioned this, cause I also didn't like the idea of self-propagating waves, but there seems to be no way around there being no aether.
...
The most sensible, killer blow for aether is that if EM travels through the aether as one would expect from modern physics, it would have to be seriously hard, light and dense, which it obviously isn't cause otherwise we'd kinda notice it. Unless of course you believe that EM doesn't travel through the aether as we would expect, in which case you will be in a whole mess trying to sort yourself out.
i think that Einstein likely knew of the MM experiment and results, but for Einstein there was little choice that Nature had in the matter.

there is a good and reasonable theoretical reason that we would expect the aether to not exist (Alfred was likely the first to have the insight): The primary physics is that there is no reason to prefer one inertial observer ("preferring" such an observer might mean to declare such an inertial observer to be stationary from some absolute frame of reference) over another inertial observer but at a different constant velocity than the first.

consider the propagation of sound, for instance. if the wind is steady and blowing across your face at some velocity [itex]v[/itex] from left to right and you measure the speed of some sound coming from your left, you will measure it to be [itex]2v[/itex] faster than if it came from your right. that is because you are moving relative to the "aether" (air) medium that carries the sound wave. but there is no such medium for light or any other E&M wave. it's a a changing E field that is causing a changing B field which is causing a changing E field which is causing a changing B field which is causing a changing E field, etc. that propagation of an E field and B field disturbance, which, when you solve Maxwell's Equations, has a velocity [itex] 1/ \sqrt{ \epsilon_0 \mu_0 } [/itex]?

so then, how do we tell the difference between a moving vacuum and a stationary vacuum? what would it mean for a vacuum to be whizzing past your face at 0.99c? is there any meaningful reason why we would notice? if we can't tell, if there really is no difference between a moving vacuum and a stationary vacuum, that such a concept is really meaningless, then whether the light that you are measuring originated from a flashlight mounted on a rocket moving past you at [itex] c/2 [/itex] or from a stationary (relative to you) flashlight, how does that change the fact that a changing E field is causing a changing B field which is causing a changing E field which is causing a changing B field which is causing a changing E field, etc.? that propagation of an E field and B field disturbance, which has velocity [itex] 1/ \sqrt{ \epsilon_0 \mu_0 } [/itex]? how is it different for you or for the observer that is traveling along with the flashlight at [itex] c/2 [/itex]? whether you are holding the flashlight or moving past it at high velocity, Maxwell's Eqs. say the same thing regarding the nature of E&M in the vacuum and you will both measure the speed of that propagation to be [itex] 1/ \sqrt{ \epsilon_0 \mu_0 } [/itex].


that's my spin on it.
 
  • #30
Claude Bile
Science Advisor
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However when you say it will cause a test charge to feel a force, I take it this is not due to the actual alternating electric field around the dipole is it? I mean at some distance close to the rotating dipole, the test charge will be in its electric field, and will thus feel force not due to the EM. Does this mean that when the test charge is far away, outside of the influence of the electric field from the rotating dipole, that it will feel a force from the EM radiation?
A test charge some distance away, won't feel the effect of a given oscillation instantly - this is a reflection of the fact that EM waves have a finite speed.
And that it will feel the same force from the EM radiation no matter how far it is from the source, as long as it absorbs the same number of photons? (which I assume it doesn't...however in the case of one photon?) NB I take it the test charge must absorb the EM waves to experience their force?
Yes, photon flux is proportional to irradiance - so two charges sitting in points of space with equal irradiance will feel the same magnitude of force. Photons do not have to be absorbed to exert a force, for example a wave transmitted through a transparent medium, will cause that medium to polarise momentarily, thus the medium experiences a force, even though the photon continues on its merry way.
Just checking too...we are talking of the same plane of rotation for the dipole. I mean like a helicopter blades and the tip of one blade is positive, the other negative.
Yes.
And finally I seem to recall that accelerating charges emit radiation...I see how this works in the case of a rotating dipole (centripetal acceleration), however what about a linearly accelerating dipole, or even a linearly accelerating charge? I do not see how an oscillating electric field is caused from this???
An oscillating field would not be caused by something like this, the field would not be sinusoidal, it would be some other shape (Gaussian maybe). Using Fourier analysis, we can decompose any waveform into its spectrum, so while the total field may not strictly be oscillatory, we can describe any field as a sum of oscillatory components.
Hopefully this is not relativity, though I fear it may be, as I read somewhere that while a charge may appear to emit radiation as it is accelerating relative to one observer, if you are travelling at a uniform velocity relative to the charge, you see no radiation. Still where oscillating electric fields come from I do not know. This causes confusing problems...energy conservation and stuff... :confused:
No! Any inertial frame of reference will "see" the charge being accelerated.
PS I have no idea about polarization etc so the idea of the torque is way over my head lol.
Polarisation is not complicated, it is just the direction the electric field pushes the hypothetical test charge. A rotating polarisation will cause the test charge to move in a circular (or corkscrew) motion.
Thanks anyways,

Kcodon
No problem!

Claude.
 
  • #31
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A test charge some distance away, won't feel the effect of a given oscillation instantly - this is a reflection of the fact that EM waves have a finite speed.

Yes, photon flux is proportional to irradiance - so two charges sitting in points of space with equal irradiance will feel the same magnitude of force.
Ok let me reword my question. I am trying to determine whether the EM wave itself can be thought of as simply a ripple in the EM field of the initial source. This would explain why a lesser force is felt at a distance (EM field is weaker i.e. 1/d^2 and EM radiation is less probable/weaker by 1/r^2...they are equivalent), and seems a much simpler view of things. Or is it that the EM radiation is seperate from the field itself. This is why I asked the following:
However when you say it will cause a test charge to feel a force, I take it this is not due to the actual alternating electric field around the dipole is it? I mean at some distance close to the rotating dipole, the test charge will be in its electric field, and will thus feel force not due to the EM. Does this mean that when the test charge is far away, outside of the influence of the electric field from the rotating dipole, that it will feel a force from the EM radiation?
In other words, is the EM radiation a seperate field, or the same field as that of the source (changing as the sources does). If so what evidence for this is there? Do you understand my question? Sorry I'm just struggling to put it into words, even though in my head the concept is so simple :rolleyes:

Yes, photon flux is proportional to irradiance - so two charges sitting in points of space with equal irradiance will feel the same magnitude of force.
So this implies that the force felt by one photon (i.e. same irradiance) will get weaker as radius gets bigger? Kind of goes with what I was saying above.

Photons do not have to be absorbed to exert a force, for example a wave transmitted through a transparent medium, will cause that medium to polarise momentarily, thus the medium experiences a force, even though the photon continues on its merry way.
Ok this is interesting. Somehow the photon gives up energy to cause the force, yet travels away with the same energy? Or in the case of the transparent medium, is this to do with the resonance in the phonons etc, so energy is still conserved?

An oscillating field would not be caused by something like this, the field would not be sinusoidal, it would be some other shape (Gaussian maybe). Using Fourier analysis, we can decompose any waveform into its spectrum, so while the total field may not strictly be oscillatory, we can describe any field as a sum of oscillatory components.
Ok I looked up Fourier analysis and it looks a bit over my head...but are you saying that the shape of the field (whatever it may be) of a linearly accelerating partice, can be represented as the sum of other fields that area sinusoidal? That seems like cheating almost, if you know what I mean. Mind you how else you manage to physically get a sinusoidally oscillating EM field from a linearly accelerating charge I have no idea...if anyone else does it would be much appreciated if you shared it.

Hmmm I have just read some stuff from other posts, and it looks pretty confusing as to why accelerating charges emit radiation:

https://www.physicsforums.com/showthread.php?t=183074"

(I like the idea stated here by Andrew Mason that charges can only be accelerated by an EM force etc)

And then also it gets confusing with relativity and gravity etc:

https://www.physicsforums.com/showthread.php?t=160533"

No! Any inertial frame of reference will "see" the charge being accelerated.
I said uniform velocity relative to the (accelerating) charge...so you are accelerating also...not an inertial frame.

Polarisation is not complicated, it is just the direction the electric field pushes the hypothetical test charge. A rotating polarisation will cause the test charge to move in a circular (or corkscrew) motion.
Thanks for the heads up...I looked into polarisation and you're right it wasn't too challenging :smile:

I was considering starting a new thread about accelerating charges and radiation, however it doesn't actually look like anyone knows, so I think I'll leave it. Thinking about it you can kind of intuitively see how the strength of the electric field changes...it "builds up" in the direction of acceleration (I'm just guessing here), but not how a sinusoid is generated.

Thanks again for your help Claude,

Kcodon
 
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  • #32
Claude Bile
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An EM wave is part of the EM field, yes.

Regarding the force applied by a photon - The quantity "Photon flux" is in units of photons per second per unit area. If two charges experience the same Photon flux, they feel the same amount of photons on average over time. Obviously there will be an element of Quantum randomness here - note that the relationship between Photon flux and Irradiance is a classical relation.

If we want to look at the effect of single photons, then we need to move into the QM regime. In this case, photons have an energy that is dependent on frequency, and a momentum that is dependent on wavelength.

With regard to photons in a medium - When the photon "uses" energy to polarise a medium, where does the energy go? Back into the EM field! (The energy you use to pull apart two opposite charges goes into the EM field that is generated as a result) While we can (and often do) decompose the total field external field and the internal response field, the distinction is irrelevant from the photons point of view. What matters is the presence of an EM field oscillating at some frequency, and that is preserved in this case.

Onto EM radiation caused by accelerating charges - I agree that it is highly confusing! It's one of those topics I try to veer away from, as my own experience in sorely lacking - the only case I am reasonably confident in discussing is the case of an oscillating dipole, and only then from a phenomenological perspective.
Kcodon said:
I said uniform velocity relative to the (accelerating) charge...so you are accelerating also...not an inertial frame.
My apologies, I must have misread something - Once your reference frame is non-inertial, then you need to invoke GR (and all bets are off as far as I'm concerned, GR is another one of those topics I know little about).

Claude.
 
  • #33
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Can you suggest any books (other than the Feynman Lectures) which might help me do that?
Read Maxwell's papers.

Or, better yet, read some Faraday, and his wonderful lines of force :D



"An EM wave is part of the EM field, yes."
by Claude

To make things nice and simple for this guy, try to perceive the EM field radiating from the source outward in all directions in a 2D plane.
Of course this field extends forever and it gets weaker at the inverse square of the distance. However, the strength of the field (the density of the photons if you will) can change depending on the location of the source.

A simple experiment you can do to try this is get a magnet, and some iron fillings. At a certain distance there is a point where the magnetic attraction of the magnet is strong enough to attract the iron to one of it's poles. Once you find the distance needed to start attracting them you can move the magnet there quickly. If you notice it takes time for the filings to sense that the magnet is there. That is because the strength of the field is changing due to the position of the source, there are more little info photons moving to the iron to say, Hey, There is a magnet there, and it is exerting a force on you. Thus, the iron responds and attracts itself to the magnet.
 
  • #34
81
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An EM wave is part of the EM field, yes.
Ok I feel so amazingly silly at the moment...yet so enlightened!! These few words have basically revolutionized my view of EM radiation and it seems so much simpler now - thank God! Thanks so much Claude for persisting!

With regard to photons in a medium - When the photon "uses" energy to polarise a medium, where does the energy go? Back into the EM field!
Touche! Oh this is a good day!

Haha yup I think I will steer clear of linearly accelerating charges, for the meantime anyways. For now I suppose I move on to the next question...back to the topic of the thread to some degree...what is an electric "field". I can imagine there are mathematical descriptions, but I don't think they stem from a physical description...more so they create a variety of interpretations. I think this then leads to the "what is charge" question, and into a whole lot of metaphysics that will likely get messy...especially for a laymen like me, without the math skill to understand the mathematical descriptions.

To Riogho
If you notice it takes time for the filings to sense that the magnet is there.
Somehow I get the feeling that the speed of light would make this a somewhat miniscule time :smile: However I appreciate the description of the EM field.

Thanks again to all,

Kcodon
 

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