EM Waves: Understanding w/o Quantum Mechanics?

[barnaby]

Another thing that I've never really understood...

As fas as I'm concerned, waves need a medium to travel through, but electromagnetic waves seem not to need one at all. I just can't visualise them as oscillations in anything.

It sort of helps if I visualise EM waves as photons, because then they're particles in their own right... but that gives rise to a whole lot of physics I simply am not equipped mathematically or conceptually to tackle yet (I started AS Physics this year).

Is it necessary to go into quantum mechanics to understand electromagnetic waves?

EDIT: I'm reading this: http://hyperphysics.phy-astr.gsu.edu/hbase/forces/funfor.html#c3 at the moment, but a lot of it is slightly over my head.

Another question that I forgot to ask when I initially posted this thread was basically the same, but for actual electrical/magnetic attraction between magnets and coils of wire and whatnot...

 

[Dale]

No, you really don't need QM to understand a very wide range of EM phenomena, just Maxwell.

You are right, EM waves do not need a medium to travel through. In a simplified sense the oscillating E field sustains the oscillating B field which sustains the oscillating E field ...

 

[barnaby]

No, you really don't need QM to understand a very wide range of EM phenomena, just Maxwell.

You are right, EM waves do not need a medium to travel through. In a simplified sense the oscillating E field sustains the oscillating B field which sustains the oscillating E field ...

Come again?

 

[cepheid]

Come again?

He means that if you look at two of Maxwell's Equations:

A time varying magnetic field induces an electric field:

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$

A time varying electric field induces a magnetic field (assume zero current density)

$$\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$

So the oscillating fields E and B continue to generate/sustain each other, explaining why an EM wave can propagate through free space and exist far from the source that generated it.

The fact that either can exist as a wave can be shown more explicitly by starting with these two first order PDE's and decoupling them to produce a single second order PDE in terms of either ONLY B or ONLY E (your choice). This second order equation turns out to be a vector wave eqn, i.e. both E and B are governed by a wave eqn in free space.

 

[Kurdt]

If you've just started AS physics you might find the Maxwell equations hard to follow since you won't have the mathematical tools necessary to understand the equations. cepheid's post basically explains how the mathematics was done to predict that light was a wave that could propagate through free space without a medium.

 

[barnaby]

You're right. I have no idea where to start with partial differential equations. As far as I know, ordinary differential equations don't crop up until FP1 of Further Mathematics, which I won't do till next year.

$$\epsilon_0$$ - that's the 'permittivity of free space', right?

Ah well, curiosity killed the cat I s'pose, I'll have to wait till undergraduacy.

Thanks a lot, though.

 

[Danger]

Ah well, curiosity killed the cat I s'pose,

No, curiosity killed one cat; the other learned from the experience and continued on in an alternate time-line.

 

[Kurdt]

$$\epsilon_0$$ - that's the 'permittivity of free space', right?

That is correct.

If you're really interested there is no harm in reading ahead as long as it doesn't interfere with other studies.

 

[barnaby]

Can you suggest any books (other than the Feynman Lectures) which might help me do that?

 

[Kurdt]

If you want to read ahead on the mathematics then a good undergraduate first year maths book is 'Guide 2 Mathematical Methods' by John Gilbert and Camilla Jordan. This covers most of the basic mathematics you will need during a university physics course.

AS far as pop sci EM books go I'm not sure what's out there. Perhaps someone else will have a better idea.

 

[Dale]

Come again?

I'm sorry about that. Let me try again. When you look at a wave on the water you notice that it changes in both space and time. If you take a picture you see that there are variations in space at that instant in time when you took the photo. If you look at a single piece of something floating you see that it bobs up and down, the variations in time at a single point in space. So fundamentally a wave is a variation in space and time.

Now, to make things simple let's consider for a bit a 1-D wave like a vibrating string. In a string there is basically two places to put energy, one is the kinetic energy or how fast a given piece of string is moving, the other is the elastic potential energy or how much the string is deformed from rest. The kinetic energy is related to the changes in time and the elastic energy is related to the changes in space and so a wave propagates along the string.

Let's turn to the equations provided by cephid

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$
Don't worry about the funny math symbols at this point. 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.

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

So, just like the string and the waves on the water we have changes in space and time. And just like the string we have two places for energy to go (electric field or magnetic field). 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.

I hope that helps.

 

[robphy]

Try my visualization
http://physics.syr.edu/~salgado/software/vpython/EMWave-Maxwell.py
from
physics.syr.edu/~salgado/software/vpython[/URL]

You'll need to install VPython.
The above visualization is the basis of a poster presentation that I'm giving at the upcoming AAPT meeting: http://www.aapt.org/scheduler/wm2008/IndexResult.cfm?Code=PST2-31

 

[cepheid]

Sorry for neglecting to take into account the background of the OP (essentially high school physics). I wasn't familiar with the UK system.

 

[kcodon]

Back to the original question from Barnaby:

As fas as I'm concerned, waves need a medium to travel through, but electromagnetic waves seem not to need one at all. I just can't visualise them as oscillations in anything.

This is also what physicists back in the day used to think...they called it the aether, and it was the medium that EM traveled 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 traveling 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. For example if you look it up, some put forward "aether entrainment" where the aether is kind of caught up by mass, so in the case of the MM experiment, the aether around the Earth would be still relative to the earth, so there would obviously be the null result by MM. However aether entrainment was disproved by other experiments by using lead blocks etc to entrain the aether etc, and also the simple fact that if the aether was entrained, then the aether around the Earth would be still relative to use, but would be moving very quickly relative to the aether in other parts of the solar system and galaxy...and this would lead to some seriously whacked views of the stars that we don't observe.

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.

Now you move onto wave particle duality, for in your case it would be lovely if light was photons, but you'll find out that light sometimes exhibits properties only explainable by wave nature. Yup I've been through your thoughts before...I too will take AS and A2 next year.

Kcodon

 

[barnaby]

Yep, that's exactly my trouble - I understand that Michelson-Morley disproved the 'aether', but I just don't really understand the physics behind what the truth is. It's conceptually difficult for me to visualise a magnetic 'field' - with a wave of, say, water, you can see what's carrying it - the 'field of water'. Similarly with a wave in a string.

But, as far as I can see, there's nothing to a magnetic/electric field - there's no tangible 'electromagnetism' through which the waves can travel.

What I seem to be learning from this thread (and I really can't thank you all enough for taking the time to explain this, despite the fact I'm clearly not getting any further) is that *somehow* an electric/magnetic field is generated, a fluctuation in which generates a magnetic/electric field, the generation of which causes a fluctuation.. which causes a generation...

Or am I still completely off?

 

[kcodon]

Yup I've had the same problems too. At the moment I'm trying to figure out how waves and photons are related with respect to the "sphere" of emission, with radius=ct. I don't know if you may find it helpful, but look at

https://www.physicsforums.com/showthread.php?t=203224

it may help - some of it is over my head but some of it is useful. I even asked the same question you are looking at...I didn't know whether EM waves were "ripples" in an EM field, or if they were self propagating. I believe the answer was self propagating. Interestingly though, the magnetic field is actually the equivalent to the electric field taking relativity into account, so I don't believe you can say one generates the other, which generates the other. In one of the posts cbacba says that due to traveling at the speed of light, time must be stopped so the wave can not actual oscillate itself...it merely appears to, or something along those lines. Maybe I will ask that...

Kcodon

 

[Claude Bile]

Remind yourself what is meant by an E-field. It is the force a +1 C charge will feel if placed at some point in space (call it "r"). A sinusoidally varying E field simply means a +1 C charge, when placed in the path of that field, will feel a sinusoidally varying force.

There is no "substance" to E (or B fields for that matter) if you can accept that force can be transmitted through a vacuum - then it shouldn't be a big leap to accept that EM fields, which are really just "force-maps", can also be transmitted through a vacuum.

Claude.

 

[kcodon]

Claude I have a question if you don't mind, related to what you just posted.

With the sinusoidally varying amplitude of electric field, does it vary strength, but always remain in one direction so to speak; or does the field vary from positive to negative - i.e. because the sinusoidal graph goes from a positive maxima to a negative one. I do not see how an electric field can be caused to change polarity...

Kcodon

 

[ZapperZ]

Now you move onto wave particle duality, for in your case it would be lovely if light was photons, but you'll find out that light sometimes exhibits properties only explainable by wave nature. Yup I've been through your thoughts before...I too will take AS and A2 next year.

Kcodon

This actually isn't true.

In QM, you do not need separate or different formulation to describe all of the behavior of light. There's only one consistent description of it. So both the apparent wave-like and particle-like behavior can be describe by just ONE consistent formulation.

Now, the reason why we continue to go back to using the wave equation, for example, to describe phenomena like diffraction, double-slit, etc. is because it is easier to handle when you are dealing with ordinary light (i.e. large number of photons). If we force you students to deal with all this using purely QM, you'd be barely alive by the time we are done with you. It isn't as trivial, as can be seen from the Marcela's paper in EJP. Thus, we continue to use a simple wave-picture to deal with what we historically designate as wave-like behavior. But don't confuse this with there being no single formulation (thus, no "duality") for all the behavior of light.

P.S. this has been covered in our FAQ as well.

Zz.

 

[Claude Bile]

Claude I have a question if you don't mind, related to what you just posted.

With the sinusoidally varying amplitude of electric field, does it vary strength, but always remain in one direction so to speak; or does the field vary from positive to negative - i.e. because the sinusoidal graph goes from a positive maxima to a negative one. I do not see how an electric field can be caused to change polarity...

Kcodon

A linearly polarised E field in an EM wave will cycle from a "positive" maximum to a "negative" minimum. The terms positive and negative though are arbitrary, whether something is positive or negative depends only on your choice of coordinates - the important thing is that there is a change in direction.

Claude.

 

[kcodon]

My apologies then Barnaby for telling you mistruths.

However I myself aren't comfortable with them, and ZapperZ I have read the FAQ and it tells me nicely that the nature of light can be described as one "thing":

This formulation (be it via the ordinary Schrödinger equation, or the more complex Quantum Electrodynamics or QED), describes ALL characteristics of light – both the wave-like behavior and the particle-like behavior.

However what is the one "thing", if not a wave and not a particle? I have a feeling its just a mathematical representation and can have little physical property attributed to it...? I've read Feynman QED and light is treated as just a particle isn't it; quoting Feynman: "Quantum Electrodynamics resolves this wave-particle duality by saying that light is made of particles (as Newton originally thought), but the price of this great advancement of science is a retreat by physics to the position of being able to calculate only the probability that a photon will hit a detector, without offering a good model of how it actually happens" This is still a mathematical representation isn't it (sum of histories etc). If there is a physical interpretation I would love to hear it, otherwise could you maybe explain briefly in laymans terms what Schrodingers equation actually represents? Thanks.

A linearly polarised E field in an EM wave will cycle from a "positive" maximum to a "negative" minimum. The terms positive and negative though are arbitrary, whether something is positive or negative depends only on your choice of coordinates - the important thing is that there is a change in direction.

Claude.

Ok that's what I thought but how does one get a changing direction for electric field from a charge that remains the same? I.e. if we generate EM radiation by an alternating current, the electric field of the electrons in the wire will always be attractive to a positive charge outside it, whether they are alternating sideways or not...??

Kcodon

 

[ZapperZ]

However I myself aren't comfortable with them, and ZapperZ I have read the FAQ and it tells me nicely that the nature of light can be described as one "thing": However what is the one "thing", if not a wave and not a particle? I have a feeling its just a mathematical representation and can have little physical property attributed to it...? I've read Feynman QED and light is treated as just a particle isn't it; quoting Feynman: "Quantum Electrodynamics resolves this wave-particle duality by saying that light is made of particles (as Newton originally thought), but the price of this great advancement of science is a retreat by physics to the position of being able to calculate only the probability that a photon will hit a detector, without offering a good model of how it actually happens" This is still a mathematical representation isn't it (sum of histories etc). If there is a physical interpretation I would love to hear it, otherwise could you maybe explain briefly in laymans terms what Schrodingers equation actually represents? Thanks.

I hate to tell you this, but your "wave" and "particle" description are also nothing more than "mathematical representation" as well!

Regardless on whether this is true, the fact that it can all be described by only one description means that I don't have to switch gears to describe one aspect of it versus another. This clearly isn't a behavior of something that has a "dual" personality.

Zz.

 

[kcodon]

I hate to tell you this, but your "wave" and "particle" description are also nothing more than "mathematical representation" as well!

Ok that is an excellent point and I take my hat off to you for that one!

Regardless on whether this is true, the fact that it can all be described by only one description means that I don't have to switch gears to describe one aspect of it versus another. This clearly isn't a behavior of something that has a "dual" personality.

Ok then ZapperZ I ask you how do you envisage a photon, apart from its mathematical properties? I mean in the way one can envisage a particle and a wave (well I just realized this is hard to do after reading another thread). Or do you see it as one of those questions where the answer is irrelevant so to speak?

Could you maybe shed some light on the question I asked Claude in my last post?

Kcodon

 

[Claude Bile]

Ok that's what I thought but how does one get a changing direction for electric field from a charge that remains the same? I.e. if we generate EM radiation by an alternating current, the electric field of the electrons in the wire will always be attractive to a positive charge outside it, whether they are alternating sideways or not...??

Kcodon

Ah, but it is the transverse attraction/repulsion that is significant here, not the longitudinal component. This is because the vast majority of wave-emitters are dipoles (or higher order "poles"), which look electrically neutral from afar.

Claude.

 

[kcodon]

Ah, but it is the transverse attraction/repulsion that is significant here, not the longitudinal component. This is because the vast majority of wave-emitters are dipoles (or higher order "poles"), which look electrically neutral from afar.

Ok silly me, the wire is neutral isn't it

Ok so if we had an electric dipole that was rotating, would this generate EM radiation? I.e. if we placed the positive test charge near the rotating dipole, in its plane, it would experience a repulsive and attractive force depending on where the dipole was in its oscillation. Then we would have the oscillating electric field, and thus magnetic, and hopefully EM radiation? Does some form of dipole such as this set up in the wire with alternating current, such that the positive test charge experiences attraction and repulsion?

If answer is yes, then must a dipole always be present...you referred to "vast majority"??

Thanks,

Kcodon

 

[Claude Bile]

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.

 

[kcodon]

Cool banana's that's 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 traveling 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...

Thanks anyways,

Kcodon

PS I have no idea about polarization etc so the idea of the torque is way over my head lol.

 

[kcodon]

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

 

[rbj]

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$
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.

$$\nabla \times \mathbf{B} = \mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}$$
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.") don't know 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 traveled 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 traveling 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 $v$ from left to right and you measure the speed of some sound coming from your left, you will measure it to be $2v$ 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 $1/ \sqrt{ \epsilon_0 \mu_0 }$?

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 $c/2$ 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 $1/ \sqrt{ \epsilon_0 \mu_0 }$? how is it different for you or for the observer that is traveling along with the flashlight at $c/2$? 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 $1/ \sqrt{ \epsilon_0 \mu_0 }$.


that's my spin on it.

 

[Claude Bile]

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 traveling 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...

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.