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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.

Do these calculations deal with circular polarized waves or only linear.

Quoten 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!

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)].

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.

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 sensibleThere is absolutely no problem with the multiplication.

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.

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.

I don't understand how i represents a phase difference of 90 degrees.

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}/∂tDifferentiating with respect to dl gives: E = -∂^{2}Φ_{B}/∂l•∂tNow, 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

lightarrow, I don't understand how i represents a phase difference of 90 degrees.

(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?

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.

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.

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.

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.

pavement ≡ sidewalkme in the bussquirrel outsideQED

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 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.

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.

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.