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### Author Topic: Another Question About The Speed Of Light  (Read 40173 times)

#### JP

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##### Another Question About The Speed Of Light
« 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.

#### lyner

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##### Another Question About The Speed Of Light
« 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 »

#### syhprum

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##### Another Question About The Speed Of Light
« 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.

#### Soul Surfer

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##### Another Question About The Speed Of Light
« 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.

#### Soul Surfer

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##### Another Question About The Speed Of Light
« 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 »

#### lyner

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##### Another Question About The Speed Of Light
« 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.
No only are they common but their phases can be measured a lot easier than light waves!

#### lyner

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##### Another Question About The Speed Of Light
« 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 »

#### Mr Andrew

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##### Another Question About The Speed Of Light
« 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.

#### syhprum

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##### Another Question About The Speed Of Light
« Reply #58 on: 25/11/2007 07:00:27 »
Quote
n the real world linear polarized EM waves are a rarity we only meet them when they are generated by lasers or electronics.
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

#### lyner

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##### Another Question About The Speed Of Light
« 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.

#### lightarrow

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##### Another Question About The Speed Of Light
« 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 »

#### lightarrow

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##### Another Question About The Speed Of Light
« 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.
Quote
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.

#### lightarrow

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##### Another Question About The Speed Of Light
« 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.

#### lightarrow

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##### Another Question About The Speed Of Light
« 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.

Quote
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 »

#### lightarrow

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##### Another Question About The Speed Of Light
« 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.

#### lyner

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##### Another Question About The Speed Of Light
« Reply #65 on: 25/11/2007 16:17:08 »
Quote
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.

#### Mr Andrew

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##### Another Question About The Speed Of Light
« 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 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 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.

#### 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 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).
Quote
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 B
Where did you find this one?
« Last Edit: 25/11/2007 20:29:42 by lightarrow »

#### lightarrow

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##### Another Question About The Speed Of Light
« 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.

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 »

#### lyner

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##### Another Question About The Speed Of Light
« 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.

#### Mr Andrew

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##### Another Question About The Speed Of Light
« 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 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.

#### lyner

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##### Another Question About The Speed Of Light
« 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.

#### lightarrow

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##### Another Question About The Speed Of Light
« 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 »

#### lyner

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##### Another Question About The Speed Of Light
« Reply #73 on: 26/11/2007 15:25:46 »
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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.

#### lyner

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##### Another Question About The Speed Of Light
« 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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##### Another Question About The Speed Of Light
« Reply #74 on: 26/11/2007 15:30:43 »