Another Question About The Speed Of Light

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Offline JP

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« Reply #50 on: 24/11/2007 17:18:23 »
There's no reason why the Poynting vector should be constant.  In fact, for a plane wave, it has a sinusoidal form: Re{S}~Cos[2(kx-ωt)].  Time-averaging it gets rid of this time dependence, and that's the form that's usually used.

I think the KE/PE in quadrature is going to arise from the KE/PE relations of the propagation medium itself because you're treating the propagation medium as a set of simple harmonic oscillators.  In such propagation, describing the waves in position and in momentum will lead to two expressions out of phase by Pi/2.  You can do an analogous thing with the electromagnetic field, but it's a relation of the field to its derivative, rather than between the E & B fields.

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lyner

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« Reply #51 on: 24/11/2007 19:03:29 »
Does the solution to this conundrum lie in confining the operators to the real parts at each stage in the chain of calculations?
After each 'calculus' operation there are other  products which are, perhaps, 'not really there'. So the E field, generated by the varying B field has to be real - etc.
My maths is so rusty that I couldn't rely on my calculations to prove or disprove anything but perhaps someone could go through the steps, eliminating non-real bits at each stage and see what comes out of it?
I know we blindly accept the complex form of wave representation  in calculations and then say 'just take the real part' at the end. Is it really justifiable? I may be very naive in asking that question.

Else it may be reconcilable by jpetruccelli's ideas (above). It would imply that the higher the refractive index / density of the medium, the more in quadrature the fields would be. Does Maxwell produce that result?
« Last Edit: 24/11/2007 19:05:47 by sophiecentaur »

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Offline syhprum

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« Reply #52 on: 24/11/2007 20:17:14 »
In the real world linear polarized EM waves are a rarity we only meet them when they are generated by lasers or electronics.
When I want to generate circular polarized waves I feed them into a helical antenna with a spacing of 1/4 wavelength between the turns and a circumference of 1 wavelength.
It strikes me that the rate of rotation is very high.
Do these calculations deal with circular polarized waves or only linear.
syhprum

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Offline Soul Surfer

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« Reply #53 on: 24/11/2007 22:50:25 »
Light arrow you surprise me.  You seem to be so familiar with the mathematics yet seem to forget the cyclic properties of the imaginary exponent.  Remember 

exp^iw = cos w +i sin w  if you put your equations in that form it looks perfectly sensible

There is absolutely no problem with the multiplication.
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Offline Soul Surfer

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« Reply #54 on: 24/11/2007 23:02:47 »
Going back to the original equations you will see that  the magnitude of the curl of the magnetic vector  is proportional to the rate of change of the electric vector which for a sinusoidal waveform implies quadrature this is your constant of integration.

similarly  the magnitude if the curl of the electric vector is proportional to the (the negative) of the rate of change of the magnetic vector. 

It is important to understand the physics as well as the mathematics it is the changing electrical fields that create the magnetic fields and the changing magnetic fields that create the electrical ones. 

Alternatively  at a peak point in the electrical or magnetic fields there is a momentary point where the partial derivative with respect to time of the relevant electric or magnetic field is zero. it follows directly from the equations that the value of the corresponding value of the magnetic or electric field must be zero.  This again implies that for a simple propagating plane polarised wave the relative magnitudes of the fields are in quadrature.
« Last Edit: 24/11/2007 23:30:34 by Soul Surfer »
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lyner

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« Reply #55 on: 24/11/2007 23:47:13 »
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n the real world linear polarized EM waves are a rarity we only meet them when they are generated by lasers or electronics.
What about every mf radio and uhf TV broadcast, then?
No only are they common but their phases can be measured a lot easier than light waves!

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lyner

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« Reply #56 on: 24/11/2007 23:54:43 »
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Do these calculations deal with circular polarized waves or only linear.
A circularly / elliptically polarised wave can be regarded as a pair of plane polarised  waves with E vectors at right angles and with their phases in quadrature then exactly the same calculations can be done on each component, independently. Yes, the rate of rotation is high - it's  2pi radians of rotation per cycle.
You can generate good circular polarised waves (on one axis) using crossed dipoles with a 1/4 wavelength delay in the feed to one of them. The off axis polarisation of a helical antenna also goes elliptical.
« Last Edit: 24/11/2007 23:58:09 by sophiecentaur »

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Offline Mr Andrew

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« Reply #57 on: 25/11/2007 03:48:51 »
Soul Surfer, that's exactly what I said!

lightarrow, I don't understand how i represents a phase difference of 90 degrees.
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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Offline syhprum

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« Reply #58 on: 25/11/2007 07:00:27 »
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n the real world linear polarized EM waves are a rarity we only meet them when they are generated by lasers or electronics.
What about every mf radio and uhf TV broadcast, then?
No only are they common but their phases can be measured a lot easier than light waves!
These are of course the EM waves I had in mind as being generated by electronic devices (Vacuum tubes, Klystrons, Transistors etc) but in the wider universe most of the radiation is generated by thermal sources with ill defined frequency and polarisation
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lyner

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« Reply #59 on: 25/11/2007 14:04:03 »
I realised that, afterwards, Syphrum but they are so very measurable that they are well worth discussing and measuring for verification. I did all my electromag learning with radio waves in mind and see it as the most relevant - just a personal view.
Incidentally - a ferrite rod would not be a good measuring tool because of its enormous inductance. To measure E and B fields one would need a very short dipole. feeding a high impedance  and a very small search coil feeding a low impedance. These would not alter the measured phase of any signal, appreciably.

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Offline lightarrow

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« Reply #60 on: 25/11/2007 14:29:23 »
There's no reason why the Poynting vector should be constant. 
Sure, I've never wrote it, infact! I mentioned it just to show how books usually derive the energy of an electromagnetic wave; I was answering to sophiecentaur: he wrote about a way of computing the EM energy which I had never heard before.

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In fact, for a plane wave, it has a sinusoidal form: Re{S}~Cos[2(kx-ωt)].
That's for sure:

Re{EXH}= (1/μ0)Re{(0; 0; Ey*Bz}) = (1/μ0)(0; 0; Re{E0ei(kx - ωt)*(1/c)E0ei(kx - ωt)}) =
= (1/cμ0)E02(0; 0; cos[2(kx - ωt)]).
« Last Edit: 25/11/2007 14:45:14 by lightarrow »

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Offline lightarrow

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« Reply #61 on: 25/11/2007 14:43:27 »
Does the solution to this conundrum lie in confining the operators to the real parts at each stage in the chain of calculations?
Absolutely not, even because otherwise complex calculus would be meaningless; you should think to it as to something that contains all informations that you would obtain with real numbers only, plus other informations.
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After each 'calculus' operation there are other  products which are, perhaps, 'not really there'. So the E field, generated by the varying B field has to be real - etc.
My maths is so rusty that I couldn't rely on my calculations to prove or disprove anything but perhaps someone could go through the steps, eliminating non-real bits at each stage and see what comes out of it?
I know we blindly accept the complex form of wave representation  in calculations and then say 'just take the real part' at the end. Is it really justifiable? I may be very naive in asking that question.
You can make all the computations I made, writing Ey = E0cos(kx - ωt) and using, in case, simple trygonometry relations, as: cos2x = (1/2)(1 + cos(2x)) ecc. It's not difficult, try it.

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Offline lightarrow

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« Reply #62 on: 25/11/2007 15:08:29 »
Light arrow you surprise me.  You seem to be so familiar with the mathematics yet seem to forget the cyclic properties of the imaginary exponent.  Remember 

exp^iw = cos w +i sin w  if you put your equations in that form it looks perfectly sensible

There is absolutely no problem with the multiplication.
If you make the computation with complex vectors, you have to do as I did and no additive constant comes out; if you make the computation with real numbers, the phase difference should come in this way:

Ey = E0cos(kx - ωt)

Bz = (1/c)E0cos(kx - ωt + Φ)

BUT you can't take Φ out of cosine, so no additive constant.

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Offline lightarrow

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« Reply #63 on: 25/11/2007 15:24:22 »
Going back to the original equations you will see that  the magnitude of the curl of the magnetic vector  is proportional to the rate of change of the electric vector which for a sinusoidal waveform implies quadrature this is your constant of integration.
? Make the computation and you'll see this is wrong.

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similarly  the magnitude if the curl of the electric vector is proportional to the (the negative) of the rate of change of the magnetic vector. 

It is important to understand the physics as well as the mathematics it is the changing electrical fields that create the magnetic fields and the changing magnetic fields that create the electrical ones.
In this case we must be more precise:  "it is the time changing of electric field that create the curl of the magnetic field". The curl IS a changing: is a spatial changing; remember how I computed it:
rotE = (=curlE) = (0; 0; ∂Ey/∂x) (all the other terms are zero). If you derive cos(kx - ωt) with respect to time, you'll have:
-ωsin(kx - ωt);
if you derive it with respect to space x, you'll have:
-ksin(kx - ωt)
Where is the phase difference?
« Last Edit: 25/11/2007 15:26:16 by lightarrow »

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Offline lightarrow

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« Reply #64 on: 25/11/2007 15:32:34 »
In the real world linear polarized EM waves are a rarity we only meet them when they are generated by lasers or electronics.
When I want to generate circular polarized waves I feed them into a helical antenna with a spacing of 1/4 wavelength between the turns and a circumference of 1 wavelength.
It strikes me that the rate of rotation is very high.
Do these calculations deal with circular polarized waves or only linear.
As I have already wrote, my calculations deal with linearly polarized waves only. Sincerely I don't know how to make the calc. in the case of circular polarization and if it could come out a phase difference or not in that case.

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lyner

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« Reply #65 on: 25/11/2007 16:17:08 »
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I don't understand how i represents a phase difference of 90 degrees.
from MrAndrew
This is not trivial, particularly in the context of waves. It involves knowing about
1. 'Simple' Trig functions and the rules to differentiate and integrate them.
2. How complex numbers work.
3. How you can represent trig functions in their exponential form.

But a simple argument goes like this.
Multiplying a quantity by -1 reverses its direction - (e.g. -3 times -1 gives you +3 ); on a graph / number line, you have rotated it by 1800
i is the square root of -1, so, using a similar idea to the above, you rotate it half as much - i.e. 900.
Do it twice and you get 1800
i is called an imaginary number, because you can't have 'i grams of  salt' but the answer to many algebraic equations comes out as a complex number containing a 'real part and an 'imaginary' part e.g.  a+ib and it has to be plotted in two dimensions- not just on a number line.

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Offline Mr Andrew

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« Reply #66 on: 25/11/2007 19:58:25 »
Ah, that explanation makes much more sense.  I understand how i = eπi/2...I just didn't see the geometric implications.  Thank you.

lightarrow, here is some math that I hope shows, both with respect to space and time, that E and B are in quadrature:

Take Faraday's Law: ∫E•dl = -∂ΦB/∂t

Differentiating with respect to dl gives: E = -∂2ΦB/∂l•∂t

Now, E is at a maximum or minimum when its derivative (either ∂E/∂t or ∂E/∂l) is zero.

0 = ∂E/∂t [prop] ∫B•dl with no current flowing.  For the integral to be 0, B must be zero.  B is zero when E is at a peak or trough (they are in quadrature since they are waves).

0 = ∂E/∂l [prop] B with no current once again.  Here, B is once again zero when E is at a maximum/minimum (they are in quadrature).

*The assumption that there is no current is justified because light, in this case, is in a vacuum where no current can flow.
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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Offline lightarrow

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« Reply #67 on: 25/11/2007 20:26:44 »
Ah, that explanation makes much more sense.  I understand how i = eπi/2...I just didn't see the geometric implications.  Thank you.

lightarrow, here is some math that I hope shows, both with respect to space and time, that E and B are in quadrature:

Take Faraday's Law: ∫E•dl = -∂ΦB/∂t

Differentiating with respect to dl gives: E = -∂2ΦB/∂l•∂t

Now, E is at a maximum or minimum when its derivative (either ∂E/∂t or ∂E/∂l) is zero.

0 = E/∂t [prop] ∫B•dl
In that formula you have to substitute E with ΦE (flux of the field E through the surface on which border you compute the line integral).
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with no current flowing.  For the integral to be 0, B must be zero.
No. Try to compute that integral in the simple case of a uniform B ≠ 0 and a squared loop line of side L alined with B:
B•dl = B*L + 0 -B*L + 0 = 0.
If the square is on a plane perpendicular to B it's even simpler:
 ∫B•dl = 0 + 0 + 0 + 0 = 0.
Quote
B is zero when E is at a peak or trough (they are in quadrature since they are waves).

0 = E/∂l [prop] B
Where did you find this one?
« Last Edit: 25/11/2007 20:29:42 by lightarrow »

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Offline lightarrow

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« Reply #68 on: 25/11/2007 20:56:02 »
lightarrow, I don't understand how i represents a phase difference of 90 degrees.
Sorry not to have answered you before to this question.
I see sophiecentaur has already answered you, I add something else:

you know that you can represent any complex number z as a point on the cartesian plane, in two ways:

1. z = x + iy;  x is the real part and y the imaginary part and they corresponds to the cartesian coordinates of z in the plane.

2. z = ρe; ρ is the "modulus" (= |z|) and corresponds to the lenght of the arrow that goes from the origin to the point in the plane; θ is the "argument" and corresponds to the angle between the x axis and the arrow.

The second, called "exponential representation" is often more useful, for example when you have to compute the product or the division of two complex numbers z1 and z2:

z1*z2 = ρ1e12e2 = ρ1ρ2ei(θ1+ θ2)

In our discussion the angles θ1 and θ2 can be called the phases of the two complex n.
So, if you multiply a complex number which phase is θ1 by another complex number which phase is θ2, you obtain a third number which phase is the sum of the other 2 and which modulus is the product of the other two.

So, if you have a complex number z which phase is θ and you want to transform it into another complex number with the same modulus but which phase is θ + φ, you just have to multiply z by e:

z*e = ρe*e = ρei(θ+φ)

So, if you want to give it a π/2 phase difference: z*eiπ/2 ecc.

i = eiπ/2 since ρ = 1 and θ = π/2 in that case.

« Last Edit: 25/11/2007 21:02:20 by lightarrow »

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lyner

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« Reply #69 on: 25/11/2007 22:01:38 »
Quote from Batroost
Quote
(1) the absorbtion/re-emission explanation doesn't look to likely in a low-scattering media i.e. why wouldn't a light beam lose direction and/or coherence?
When I was first introduced to em waves we were told of Huygen's construction, involving 'secondary wavelets'; a wavefront of any shape can be regarded as an infinite set of wavelets, starting along the wavefront. (This is just like the idea of diffraction, it seems to me.) The shape of the wavefront, immediately afterwards is given by the sum of all these secondary wavelets. For a plane wave, the secondary wavelets add up only in the forward direction and cancel in all others
The reason for the wave remaining  well behaved as it goes through a medium would be that all the absorbed and re emitted wavelets add up in phase only in the direction given by 'geometrical optics'.  Your concern about the wavefront being destroyed is, actually, groundless.
As with most of Physics, there are usually several valid ways of looking at things.

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Offline Mr Andrew

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« Reply #70 on: 25/11/2007 22:30:50 »
Sorry, there were some errors in my math.

The equation ∂E/∂l = B should be ∂2ΦE/∂l∂t = B and all of the E's in Ampere's Law should be ΦE.  I mixed the vector and integral representations of the law (where curlB [prop] ∂E/∂t when no current flows).

Sorry, I forgot for a second that Faraday's and Ampere's laws had line integrals, not just regular integrals (I am a senior in high school who has not had E&M or Calc II formally-I am teaching myself for now).  Ok, so if a line integral of x is 0, x doesn't necessarily have to be zero.  I'll have to work on this.  If anybody sees where I am going and thinks they can get there without all of the mathematical hiccups and such, go right ahead.  This is really bothering me because, as far as I can tell, lightarrow has provided a valid model for light, and so has Soul Surfer: 
Quote
Alternatively  at a peak point in the electrical or magnetic fields there is a momentary point where the partial derivative with respect to time of the relevant electric or magnetic field is zero. it follows directly from the equations that the value of the corresponding value of the magnetic or electric field must be zero.  This again implies that for a simple propagating plane polarised wave the relative magnitudes of the fields are in quadrature.

Thank you lightarrow for your explaination of the phase changes...as I said, I am not that familiar with the mathematics associated with waves.
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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lyner

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« Reply #71 on: 26/11/2007 12:08:59 »
Is the quadrature or in phase question really relevant?
I am just beginning to wonder if, I have been making the same sort of mistake that A level students make when understanding the momentum of a photon;
"It must be mc, because that's what momentum is".
It requires a serious change of thought for them to get round that one and we may need a similar change of approach for this problem, too..
It may be that, when a wave is traveling at c, you also have to look at it differently and the 'in phase' idea doesn't really contravene anything.  As far as the photon in space is concerned, there is no passage of time so it's energy is not 'pulsing at all and there is no interaction with matter. Putting yourself in a frame where you are traveling alongside the wave / photon stream at c and discussing what you would see  it is probably as dodgy a concept as talking in terms of photons having mass. It certainly must be approached with care.

What I would expect, however, would be for this in-phase situation to change as the wave goes through a medium.  A transmission line could be regarded as a medium because the guided wave is interacting with conduction electrons and we do, actually, find quadrature E and B fields there.
Perhaps there is no conflict at all.

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Offline lightarrow

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« Reply #72 on: 26/11/2007 15:16:06 »
Is the quadrature or in phase question really relevant?
I am just beginning to wonder if, I have been making the same sort of mistake that A level students make when understanding the momentum of a photon;
"It must be mc, because that's what momentum is".
It requires a serious change of thought for them to get round that one and we may need a similar change of approach for this problem, too..
It may be that, when a wave is traveling at c, you also have to look at it differently and the 'in phase' idea doesn't really contravene anything.
I don't have any knowledge at the moment to confirm or confute this idea; my intuition would say there would really be a difference, in the way matter interact with the wave.
Quote
As far as the photon in space is concerned, there is no passage of time so it's energy is not 'pulsing at all and there is no interaction with matter. Putting yourself in a frame where you are traveling alongside the wave / photon stream at c and discussing what you would see  it is probably as dodgy a concept as talking in terms of photons having mass. It certainly must be approached with care.
But you don't need it: fixing a point of space, you will measure different values of the fields if E and B are not in phase; the same if you analyze a standing wave of light between two mirrors.
Quote
What I would expect, however, would be for this in-phase situation to change as the wave goes through a medium.  A transmission line could be regarded as a medium because the guided wave is interacting with conduction electrons and we do, actually, find quadrature E and B fields there.
Perhaps there is no conflict at all.
For the little I've read, I'm pretty sure the relative phase between E and B can change in those situations, as you say, but I don't know exactly which ones; however, in a solid medium, *if it is linear, homogeneous and isotropic* the situation is the same as in the void: you simply have to replace ε0 and μ0 with different values ε and μ, function of the frequency only, so the equations would be exactly the same and so their solutions too.
« Last Edit: 26/11/2007 15:18:52 by lightarrow »

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lyner

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« Reply #73 on: 26/11/2007 15:25:46 »
Quote
But you don't need it: fixing a point of space, you will measure different values of the fields if E and B are not in phase; the same if you analyze a standing wave of light between two mirrors.
Yes, I know. I didn't put what I wanted to say very well. What I meant is that the concept of the energy carried by 'something' going at speed c is probably just as divorced from energy carried by matter as is the concept of its momentum. I didn't mean it doesn't happen - I just meant that perhaps the apparent paradox doesn't matter.

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lyner

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« Reply #74 on: 26/11/2007 15:30:43 »
There certainly is a difference between the em waves interact with matter and the way they interact with other em waves i.e. they don't. We need some different (well informed) input on this, I think.
It is a good subject, though, don't you think?
I was riding home on the bus, earlier this afternoon, watching a squirrel on the pavement and tried to relate what I was seeing  to what we have been discussing.
We have such  compartmentalised lives.

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Offline lightarrow

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« Reply #75 on: 26/11/2007 17:28:45 »
There certainly is a difference between the em waves interact with matter and the way they interact with other em waves i.e. they don't. We need some different (well informed) input on this, I think.
It is a good subject, though, don't you think?
Ah, yes! It was also good for refreshing our studies on electrodynamics!
Quote
I was riding home on the bus, earlier this afternoon, watching a squirrel on the pavement and tried to relate what I was seeing  to what we have been discussing.
We have such  compartmentalised lives.
It's true; it's very difficult nowadays to have a global viewpoint of things.
(A squirrel on the pavement? On the Bus?  [???])

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lyner

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« Reply #76 on: 26/11/2007 18:38:52 »
pavement ≡ sidewalk
me in the bus
squirrel outside
QED

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Offline neilep

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« Reply #77 on: 26/11/2007 20:14:53 »
At the beginning of this thread I hinted at my impending headache!!

I just want to thank you all for it !!.....but also mainly THANK YOU for all your incredible contributions here too !!
Men are the same as women, just inside out !

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Offline Alandriel

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« Reply #78 on: 26/11/2007 20:52:05 »
***** sigh *** I wish I understood the half of it



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Offline lightarrow

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« Reply #79 on: 26/11/2007 21:10:50 »
pavement ≡ sidewalk
me in the bus
squirrel outside
QED
Ah! Ok! [:)]

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Offline syhprum

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« Reply #80 on: 26/11/2007 23:25:24 »
Has a concensus been agreed ? are B & H in phase or quadature ?.
syhprum

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« Reply #81 on: 26/11/2007 23:47:54 »
Don't get us started again.
It probably is, but then again, it may not be.
We're waiting f or someone to explain it better for us, I think.

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Offline Mr Andrew

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« Reply #82 on: 27/11/2007 01:24:26 »
I looked in my physics text (of all places) and I saw both models!?!  The book was University Physics, Ed. 11, by Young and Freedman.  The light waves were in phase when there was a light wave travelling in space but a linearly polarized wave was in quadrature, as well as a standing wave resulting from a reflection from a perfect conductor.  The linearly porlized one I understand but not the standing wave one.  Especially if the light comes in with B and E in-phase.  Anybody else have a copy of this book to look this up?
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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Offline JP

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« Reply #83 on: 27/11/2007 04:36:17 »
Here's another solution that I don't think got mentioned here (though it might have gotten buried):

If you work in the frequency domain in free space, the wave equation becomes the Helmholtz equation:

(Grad2+k2)U(r)exp(-iωt)=0.

Separation of variables (Cartesian) leads to plane wave solutions:

U(r)=exp[i (kr+φ)],

where the phase φ is arbitrary.

Assuming the medium is linear, this is the form of the solution for both electric and magnetic fields, except for a prefactor.  Both fields could have arbitrary phases:

E(r)=E0exp[i (kre)],
B(r)=B0exp[i {krb)].

Now use Faraday's law:

Curl(E)=-∂B/∂t

kxE0exp[i (kre-ωt)]=ωB0exp[i {krb-ωt)]

Not only does this specify that E and B are mutually perpendicular to the plane wave direction k, but this relation is an equality and is true for all r and t.  If we set r=0 and t=0, we get

kxE0exp[i φe]=B0exp[i φb],

where k, was a real vector.  In order for this equation to be satisfied,

φeb.

Therefore the fields are in phase for a linear dielectric. 
« Last Edit: 27/11/2007 04:49:28 by jpetruccelli »

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Offline JP

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« Reply #84 on: 27/11/2007 04:59:26 »
If you're dealing with a medium with absorption or gain, things are different.  Assume E0=E0x.

In an absorption or gain medium, k can pick up a complex part that points in a different direction than its real part.  Let's assume, however, that this particular plane wave has both components pointing in the z direction so that

k=(1+i)z.

Then,
kxE0=y(1+i)E0B0

or

B0=yE0Sqrt(2)/ω exp(i π/2).

In other words, if k has a complex part, which generally arises from an absorption or gain medium, the electric and magnetic fields can be out of phase.

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lyner

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« Reply #85 on: 27/11/2007 18:48:40 »
I spent an interesting hour in the Sussex University library, looking at some 'old friends' - Electromag Theory books. Reading around gave me some insight into the problem. It was the stuff on transmission lines that clinched it.
Without using Maxwell - at least, not explicitly. First, a simple model:
Consider a transmission line with power flowing along it, in the form of an AC signal. Assume it's very long, compared with the wavelength  of the sinusoid  (not necessary but it's nearer the situation of a plane wave through a void).
If the line is terminated (matched) by a resistor  of R ohms where R is the characteristic impedance (root L/C).  All the power flowing on the line will be dissipated; none will be reflected.
The power, dissipated will be V. I .Cos(phase angle). If all the power is dissipated then the phase angle must be zero.  An observer on the line would not know the difference between the line being infinite and the line being terminated properly. If the current and voltage were in phase quadrature, a terminating resistor would not dissipate any  power.

Same argument but, this time imagine a plane em wave hitting a surface of perfect resistance equal to 377Ω (which is the characteristic impedance of free space). All the power will be dissipated - none will be reflected. The current / B field and voltage / E field need to be in phase for all the power to pass across the boundary. If they are not, some will be reflected.

Once you introduce losses into your transmission line - or a complex refractive index (involving losses) into your wave medium - this phase quadrature is lost.

My ideas involving PE and KE were, clearly at odds with reality where EM is concerned.

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Offline Soul Surfer

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« Reply #86 on: 27/11/2007 23:54:54 »
I must admit that looks pretty convincing and I must revise my mental model in future.  I went back to my more recent text books to check it out and try to understand where I went wrong in the first place and see that it probably came from the near field excitation of a half wave dipole radiator which the text book states is very different from the far field where the magnetic and electrical fields are in phase.
Learn, create, test and tell
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« Reply #87 on: 28/11/2007 12:31:02 »
Yes - the half wave dipole is, essentially, a resonant structure and the energy sloshes up and down in the form of time quadrature fields. These dominate until you get away from locality of the radiator.
I think it's a fantastic idea that, when you 'match' an antenna to its feeder, the transmitter just 'sees' a resistance - the radiation resistance- which represents the power being dissipated into space.
This has been an interesting thread - I was loth to disagree with the Maxwell's equation solution (who do I think I am, for God's sake!?) but, until there's a reasonable explanation in words, I find it difficult to get the feeling of understanding.  I feel better about it. now.

Thanks for everyone's contributions.
« Last Edit: 28/11/2007 12:32:41 by sophiecentaur »

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Offline syhprum

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« Reply #88 on: 28/11/2007 13:52:45 »
Yes this is what I read in Sterlings texbook 60 years ago.
syhprum

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lyner

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« Reply #89 on: 28/11/2007 15:57:44 »
I think it was in a latin textbook, too!
Quod erat demonstrandum?
I'm afraid I can only go back to the '60s for my electromag education.
(My spell checker didn't like the second line)
« Last Edit: 28/11/2007 16:07:32 by sophiecentaur »

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Offline Mr Andrew

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« Reply #90 on: 28/11/2007 22:54:44 »
Alas, it seems that the E and B fields of light are in phase in empty space.  But, why does a standing EM wave (created when a wave reflects of of a perfect conductor) have E and B fields that are out of phase, even in quadrature?
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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lyner

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« Reply #91 on: 29/11/2007 23:27:51 »
It's because there is little or no power dissipated on reflection. For that to happen the cos(phase) factor must be near zero i.e quadrature. This quadrature condition is due to the 'boundary conditions' imposed at the reflecting (conducting)  surface. The E field is zero at the surface (short circuit) and the B field (caused by the current induced in the surface) is a maximum - that's quadrature. 
Standing waves in a resonant metal cavity are only possible where the distances between reflecting walls are whole numbers of half wavelengths. The small difference between the actual phase and 90 degrees will be due to finite resistance in the walls and will result in dissipation / decay of the energy in the standing wave.
« Last Edit: 29/11/2007 23:29:43 by sophiecentaur »

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« Reply #92 on: 30/11/2007 08:06:57 »
Alas, it seems that the E and B fields of light are in phase in empty space.  But, why does a standing EM wave (created when a wave reflects of of a perfect conductor) have E and B fields that are out of phase, even in quadrature?
The answer of sophiecentaur is correct; however remember that in a standing wave you don't have one single wave propagating, but two of them, propagating in opposite directions and interfering each other; it's this interfering resultant non-propagating wave which has fields in quadrature.

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lyner

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« Reply #93 on: 01/12/2007 00:08:30 »
The quadrature condition must apply on any reflecting surface for any wavelength, because there can never be any E field at the conducting surface - it implies that there must be current on the surface which modifies the B field near the surface.
You could look upon your 'two waves' as an infinite set of waves going backwards and forwards between the end boundaries.
A standing wave will only exist where the separation of the boundaries corresponding to an integral number of half waves. That is only satisfied for a 'comb' of frequencies for which all these waves interfere constructively. At other frequencies, the quadrature parts add up to zero.
Any loss mechanism in the system will introduce a small trace of in-phase E and B wave which allows dissipation of energy to occur and the standing wave to decay. To maintain the standing wave, power must be supplied (with E and B in phase)  to make up for the loss.

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Offline lightarrow

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« Reply #94 on: 01/12/2007 12:50:47 »
The quadrature condition must apply on any reflecting surface for any wavelength, because there can never be any E field at the conducting surface - it implies that there must be current on the surface which modifies the B field near the surface.
You could look upon your 'two waves' as an infinite set of waves going backwards and forwards between the end boundaries.
A standing wave will only exist where the separation of the boundaries corresponding to an integral number of half waves. That is only satisfied for a 'comb' of frequencies for which all these waves interfere constructively. At other frequencies, the quadrature parts add up to zero.
Any loss mechanism in the system will introduce a small trace of in-phase E and B wave which allows dissipation of energy to occur and the standing wave to decay. To maintain the standing wave, power must be supplied (with E and B in phase)  to make up for the loss.
Sincerely I haven't understood completely what you have written (and with this I'm not asserting that you can be wrong); anyway, if you send a linear polarized monochromatic EM wave to a perfectly reflecting surface, both incident and reflected wave have E and B in phase.

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lyner

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« Reply #95 on: 03/12/2007 23:27:59 »
This is difficult and my earlier post was garbled - sorry.
What I am saying is, basically, that the standing wave consists of many waves flowing up and down the cavity (or at least earlier and earlier portions of the same wave that entered in the first place). It's only when the cavity size is right that you actually see the standing wave. If there were no losses, energy would go up and down the cavity between for ever. It couldn't die down but there would be no nodes or antinodes because no peaks or troughs in the wave would coincide (except for resonance).
Take a realistic, lossy, example in which there is a permanent low power source to keep the energy topped up. Follow the injected signal down the cavity. At the end there will be some power dissipated - corresponding to some in-phase E and B and  some power reflected - corresponding to some in phase E and B. In addition, because of the currents induced in the end, there will be some local quadrature E and B fields - the induced current in the ends will be proportional to dE/dt and this will cause a quadrature B (proportional to the Current) field near the end. The in phase fields propagate down the cavity and back, carrying power and induce more currents in the end walls and more induced quadrature B field. In the steady state, when the injected signal is at a resonant frequency, the quadrature fields will add constructively and be much greater than the in phase fields (related by the Q factor of the cavity) which are only due to the lost power.
I can't be sure (no textbook here) but, for a large spacing between the ends, I would expect the phase relationship in the middle to be nearer zero(?) because the , essentially 'end effects' would be less.
« Last Edit: 04/12/2007 11:52:29 by sophiecentaur »

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Offline lightarrow

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« Reply #96 on: 04/12/2007 13:53:05 »
This is difficult and my earlier post was garbled - sorry.
What I am saying is, basically, that the standing wave consists of many waves flowing up and down the cavity (or at least earlier and earlier portions of the same wave that entered in the first place). It's only when the cavity size is right that you actually see the standing wave. If there were no losses, energy would go up and down the cavity between for ever. It couldn't die down but there would be no nodes or antinodes because no peaks or troughs in the wave would coincide (except for resonance).
Take a realistic, lossy, example in which there is a permanent low power source to keep the energy topped up. Follow the injected signal down the cavity. At the end there will be some power dissipated - corresponding to some in-phase E and B and  some power reflected - corresponding to some in phase E and B. In addition, because of the currents induced in the end, there will be some local quadrature E and B fields - the induced current in the ends will be proportional to dE/dt and this will cause a quadrature B (proportional to the Current) field near the end. The in phase fields propagate down the cavity and back, carrying power and induce more currents in the end walls and more induced quadrature B field. In the steady state, when the injected signal is at a resonant frequency, the quadrature fields will add constructively and be much greater than the in phase fields (related by the Q factor of the cavity) which are only due to the lost power.
I can't be sure (no textbook here) but, for a large spacing between the ends, I would expect the phase relationship in the middle to be nearer zero(?) because the , essentially 'end effects' would be less.
I think you are are right. This show us how fields can be different from that situation to that of an EM wave propagating in the void. It's quite amazing.

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Offline McQueen

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« Reply #97 on: 09/12/2007 08:16:39 »
There are some things about light that baffled physicists even before QM came along and made it  more complicated.  Suppose you are stationary and a light bulb is switched on in front of you, if the speed of light is measured it will be found to be 300,000 Km/sec. Next if you move towards the light source at a speed of  100,000 Km/sec what will the speed of light be ? According to Galilean transformations it should be 400,000 Km/sec. (i.e the speed of light (300,000 Km/sec) plus our own speed (100,000 Km/sec). But it isn’t, the speed of light is still 300,000 Km/sec.

Commonsense tells us that this can’t be true,  that if the photons in the light beam are rushing towards us at 300,000 Km/sec and we are rushing towards the light at 100,000 Km/sec they should appear to be traveling at the speed at which they were emitted plus our own speed of 300,000 Km/sec plus 100,000 Km/sec = 400,000 Km/sec. This is what should happen, but it doesn’t.

Next consider the opposite situation. Suppose that the light bulb is standing still, and this time we are moving away from it at 100,000 Km/sec What will the velocity of the photons measure now ? It should be 200,000 Km/sec. But it isn’t it is still 300,000 Km/sec.

All this was proved by Michelson and Morley in 1887. Imagine the confusion.  So what happened in the end. How was everything solved? Well it wasn’t, instead Einstein made the  speed of light a postulate.  The constancy of the speed of light instead of being a puzzle is now taken for granted as just being constant, it is a postulate.

The only possible solution to the constant speed of light may lie in the existence of an ether and the manner in which light propagates. see this thread . The description of how light propagates as given in this thread may hold the key to its constant speed.
« Last Edit: 09/12/2007 08:19:27 by McQueen »
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lyner

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« Reply #98 on: 09/12/2007 12:46:38 »
I think it is totally delusional to expect to 'understand exactly'  anything.
Our whole life is based on some sort of picture we 'see' in our brains and, between what happens and our awareness of it there are layers of 'metaphor'.
Our minds / brains are, basically, pragmatic. We build our image of the world using 'postulates',  internally generated or learned.
If you hope to improve on the idea of the constant speed of light then you have to supply, not just a "wouldn't it be nice to think of it as" description. Yo have to build a whole edifice which is at least as large and as self-consistent as the existing ideas.
If you want an aether then build a proper body of Science around the idea and show that your model predicts what we actually observe - IN DETAIL and with proper evidence.
Half-arsed theories about specific areas of Science are two a penny.
You can have any private ideas you like but, for people to take you seriously, you need to publish a lot more supporting evidence. Without that, you are not going to convince me or many others.

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Offline lightarrow

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« Reply #99 on: 09/12/2007 14:01:11 »
The only possible solution to the constant speed of light may lie in the existence of an ether and the manner in which light propagates.
This ether would be stationary with respect to what?