[lightarrow]

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.

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; E_y*B_z}) = (1/μ₀)(0; 0; Re{E₀e^{i(kx - ωt)}*(1/c)E₀e^{i(kx - ωt)}}) =
= (1/cμ₀)E₀²(0; 0; cos[2(kx - ωt)]).

[lightarrow]

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.

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 E_y = 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]

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:

E_y = E₀cos(kx - ωt)

B_z = (1/c)E₀cos(kx - ωt + Φ)

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

[lightarrow]

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.

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; ∂E_y/∂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?

[lightarrow]

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]

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]

Ah, that explanation makes much more sense.  I understand how i = e^πi/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 = -∂Φ_B/∂t

Differentiating with respect to dl gives: E = -∂²Φ_B/∂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.

[lightarrow]

Ah, that explanation makes much more sense.  I understand how i = e^πi/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 = -∂Φ_B/∂t

Differentiating with respect to dl gives: E = -∂²Φ_B/∂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 Φ_E (flux of the field E through the surface on which border you compute the line integral).

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.

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?

[lightarrow]

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^iθ; 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 = ρ₁e^iθ₁*ρ₂e^iθ₂ = ρ₁ρ₂e^{i(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^iφ:

z*e^iφ = ρe^iθ*e^iφ = ρe^i(θ+φ)

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

i = e^iπ/2 since ρ = 1 and θ = π/2 in that case.

[lyner]

Quote from Batroost

(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]

Sorry, there were some errors in my math.

The equation ∂E/∂l = B should be ∂²Φ_E/∂l∂t = B and all of the E's in Ampere's Law should be Φ_E.  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: 

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]

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]

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.

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.

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.

[lyner]

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]

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.

[lightarrow]

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!

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?  [])

[lyner]

pavement ≡ sidewalk
me in the bus
squirrel outside
QED

[neilep (OP)]

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 !!

[Alandriel]

***** sigh *** I wish I understood the half of it

[lightarrow]

pavement ≡ sidewalk
me in the bus
squirrel outside
QED

Ah! Ok! []