[lightarrow]

You could see it quite easily on a cheapo oscilloscope, I should think. Don't know why I didn't do it. myself, when I had the chance - I used to have access to a Range Rover, kitted out as a mobile lab and did frequent field strength and other measurements  at all frequencies - including 198kHz.
I guess it was all so 'obvious' at the time that I didn't need to prove it for myself.

Lightarrow - I will have to look at your stuff in detail. Did you look at the link on my previous post?

Yes; I think it's wrong what he says, but I have to study it in more detail.

My first reaction is that, when you integrate your E, do you not get an 'i' multiplier for your B?

You are perfectly right! But I forgot the same 'i' when I computed ∂E_y/∂x, so they cancel each other. Now I will correct my previous post. Thank you for your correction.

[lightarrow]

www.play-hookey.com/optics/transverse_electromagnetic_wave.html
I've just spent ages trying to find a reference. Here it is.
It says it better than I have.

http://www.shef.ac.uk/physics/teaching/phy205/lecture_18.htm
9^th row from below:
<<This equation can only be satisfied if a=0 (i.e. E and B are in phase)>>

[syhprum]

I am now convinced that remote from the antenna B & H are in phase but is there not a difference close to the antenna before the electro magnetic wave becomes established.
I believe the text book I referred to was by 'Sterling' but I can find no reference to it but I recall drawings of a vertical antenna with the current running up and down and a horizontal circular magnetic field spreading out from it

[lightarrow]

I am now convinced that remote from the antenna B & H are in phase but is there not a difference close to the antenna before the electro magnetic wave becomes established.
I believe the text book I referred to was by 'Sterling' but I can find no reference to it but I recall drawings of a vertical antenna with the current running up and down and a horizontal circular magnetic field spreading out from it

The computation I made, indeed, is valid in the void, in regions of space which don't contain sources (that is, charges or currents); near an antenna, which is a source, E and B can be not-in phase.

[lyner]

I think it was Sterling's book, amongst others, that I saw the  time quadrature thing.
However, the above maths and the Sheffield,  'Maxwell'  lecture look ok - I can't really argue with it.
Perhaps someone could help me with how em waves appear to differ from  other waves, in which the periodicity of the PE is in quadrature with the KE - giving a constant flow of energy. It seems such a fundamental idea that I can't just let it go without a good reason.
I shall have to go to a hotel in Droitwich** with a borrowed oscilloscope and see for myself if I can't get satisfaction! Perhaps we could all have a party there!!

**Home of the 198kHz main UK transmitter

[syhprum]

I live about 200km from Droitwich, have a decent car, a 150 MHz two channel oscilloscope and plenty of spare time between now and 17/12/07.
Make the necessary arrangement's and the ferrite rod antenna.
PS

I think Rugby on 60 KHz would have been better but I believe it no longer operates, there is a 77.5KHz DCF77 German station near Frankfurt where the beer would be better

[sohail]

I thought that when for instance light travels through a perspex prism it changes velocity and that's why the process of refraction takes place. So the speed of light depends on the media it's travelling through and therefore is not always constant.

Here's a quote from wiki which kinda backs me up and also elaborates further:

'Light traveling through a medium other than a vacuum travels below c as a result of the time lag between the polarization response of the medium and the incident light. However, certain materials have an exceptionally high group index and a correspondingly low group velocity. In 1999, a team of scientists led by Lene Hau were able to slow the speed of a light pulse to about 17 metres per second;[8] in 2001, they were able to momentarily stop a beam.[9]

In 2003, Mikhail Lukin, with scientists at Harvard University and the Lebedev Institute in Moscow, succeeded in completely halting light by directing it into a Bose–Einstein condensate of the element rubidium, the atoms of which, in Lukin's words, behaved "like tiny mirrors" due to an interference pattern in two "control" beams.'

[syhprum]

No the light always travels at 'c' but it hangs around being adsorbed by atoms and then remitted

[Batroost]

And another quote from Wiki:

It is sometimes claimed that light is slowed on its passage through a block of media by being absorbed and re-emitted by the atoms, only traveling at full speed through the vacuum between atoms. This explanation is incorrect and runs into problems if you try to use it to explain the details of refraction beyond the simple slowing of the signal.

The alternative explanation offered involves a 'mixed' wave of electromagnetism and mechanical oscillation within the material. This sems more credible to me... as:

(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?

(2) it ignores the wide range of wavelengths over which media (such as glass) are continuously transmissive whereas absorbtion tends to occur at relatively well defined frequencies.

[lightarrow]

I think here is explained very well (Physics Forums FAQ):
http://www.physicsforums.com/showthread.php?t=104715

quote:
<< Do Photons Move Slower in a Solid Medium?

Contributed by ZapperZ. Edited and corrected by Gokul43201 and inha

This question appears often because it has been shown that in a normal, dispersive solid such as glass, the speed of light is slower than it is in vacuum. This FAQ will strictly deal with that scenario only and will not address light transport in anomolous medium, atomic vapor, metals, etc., and will only consider light within the visible range.

The process of describing light transport via the quantum mechanical description isn't trivial. The use of photons to explain such process involves the understanding of not just the properties of photons, but also the quantum mechanical properties of the material itself (something one learns in Solid State Physics). So this explanation will attempt to only provide a very general and rough idea of the process.

A common explanation that has been provided is that a photon moving through the material still moves at the speed of c, but when it encounters the atom of the material, it is absorbed by the atom via an atomic transition. After a very slight delay, a photon is then re-emitted. This explanation is incorrect and inconsistent with empirical observations. If this is what actually occurs, then the absorption spectrum will be discrete because atoms have only discrete energy states. Yet, in glass for example, we see almost the whole visible spectrum being transmitted with no discrete disruption in the measured speed. In fact, the index of refraction (which reflects the speed of light through that medium) varies continuously, rather than abruptly, with the frequency of light.

Secondly, if that assertion is true, then the index of refraction would ONLY depend on the type of atom in the material, and nothing else, since the atom is responsible for the absorption of the photon. Again, if this is true, then we see a problem when we apply this to carbon, let's say. The index of refraction of graphite and diamond are different from each other. Yet, both are made up of carbon atoms. In fact, if we look at graphite alone, the index of refraction is different along different crystal directions. Obviously, materials with identical atoms can have different index of refraction. So it points to the evidence that it may have nothing to do with an "atomic transition".

When atoms and molecules form a solid, they start to lose most of their individual identity and form a "collective behavior" with other atoms. It is as the result of this collective behavior that one obtains a metal, insulator, semiconductor, etc. Almost all of the properties of solids that we are familiar with are the results of the collective properties of the solid as a whole, not the properties of the individual atoms. The same applies to how a photon moves through a solid.

A solid has a network of ions and electrons fixed in a "lattice". Think of this as a network of balls connected to each other by springs. Because of this, they have what is known as "collective vibrational modes", often called phonons. These are quanta of lattice vibrations, similar to photons being the quanta of EM radiation. It is these vibrational modes that can absorb a photon. So when a photon encounters a solid, and it can interact with an available phonon mode (i.e. something similar to a resonance condition), this photon can be absorbed by the solid and then converted to heat (it is the energy of these vibrations or phonons that we commonly refer to as heat). The solid is then opaque to this particular photon (i.e. at that frequency). Now, unlike the atomic orbitals, the phonon spectrum can be broad and continuous over a large frequency range. That is why all materials have a "bandwidth" of transmission or absorption. The width here depends on how wide the phonon spectrum is.

On the other hand, if a photon has an energy beyond the phonon spectrum, then while it can still cause a disturbance of the lattice ions, the solid cannot sustain this vibration, because the phonon mode isn't available. This is similar to trying to oscillate something at a different frequency than the resonance frequency. So the lattice does not absorb this photon and it is re-emitted but with a very slight delay. This, naively, is the origin of the apparent slowdown of the light speed in the material. The emitted photon may encounter other lattice ions as it makes its way through the material and this accumulate the delay.

Moral of the story: the properties of a solid that we are familiar with have more to do with the "collective" behavior of a large number of atoms interacting with each other. In most cases, these do not reflect the properties of the individual, isolated atoms.>>

[Batroost]

Nice. []

[syhprum]

I sometimes feel like one of the panelist on the Stephen Fry quiz program "Q".
I offer the commonly held answer to a question only to have it demolished by greater experts.

[Batroost]

I know the feeling.

But sometimes Mr Fry is just plain wrong; his 'electricity' themed programme made a few errors anyway.

[lyner]

whereas absorbtion tends to occur at relatively well defined frequencies.

Why should the frequencies be 'well defined'?  As I keep saying - solids are not Hydrogen atoms. If there are enough possible changes of energy state, you can have as many frequencies as you like - in fact a continuum.  In a solid, this is the case. If the medium is glass, for instance, then there must be  just the right conditions for exitation of electrons and re-radiation without significant loss - i.e transparency with a delay mechanism. It would be necessary for the effective conductivity to be very low or you would get reflection at the surface and no transmission.

Lightarrow:

http://www.shef.ac.uk/physics/teaching/phy205/lecture_18.htm
9th row from below:
<<This equation can only be satisfied if a=0 (i.e. E and B are in phase)>>

I hate to disagree (he lied) with established thinking but the expressions used in all the texts contains the complex form of the wave. Can we be sure that the relevant part has been chosen in producing the final answer? My maths is not good enough to be certain but there seems to be a loophole here. Not all solutions have reality in maths.
But, more to the point, can you answer my objection on the grounds that free em  waves  appear to be fundamentally different from all other waves in how they transport the energy?
Even an electric wave travels along an LC delay line with a phase difference between volts and current - or E and B fields on a transmission line. What happens when it gets to the end of a transmission line and encounters a dipole radiator? Can there be a sudden hiccup in the phase of one of the fields?
I am confused.

[syhprum]

It is an interesting case a delay line or coaxial cable terminated by an antenna, the first thing that occurs to me is that the cable would have a propagation velocity of less than "c" in which case B and H would have the 90° phase difference but what happens in a vacuum insulated cable where PV would equal c would this still apply or would Maxwell rule ?

[Soul Surfer]

That could be the case of a cable but it is not true of a waveguide within which the propagation group velocity is lower than the velocity of light.  In this context waveguides are probably a better model than coaxial cables.

[lyner]

The coax cable situation just means that there is no lower cutoff frequency.  You could substitute your coax for a balanced pair. You still have a TEM wave and the phases are in quadrature. Then, when you get to the antenna, everything is supposed to change. How, why?
Also, I don't see why the velocity should be a problem, in any case. Maxwell takes into account complex refractive indices.

[lightarrow]

I sometimes feel like one of the panelist on the Stephen Fry quiz program "Q".
I offer the commonly held answer to a question only to have it demolished by greater experts.

It's the best way to learn something... []
Remember that only a few months ago I would have given the same your answer to the behaviour of light in solid matter.  []
Your answers are not trivial...

[lightarrow]

Lightarrow:

http://www.shef.ac.uk/physics/teaching/phy205/lecture_18.htm
9th row from below:
<<This equation can only be satisfied if a=0 (i.e. E and B are in phase)>>

I hate to disagree (he lied) with established thinking but the expressions used in all the texts contains the complex form of the wave. Can we be sure that the relevant part has been chosen in producing the final answer? My maths is not good enough to be certain but there seems to be a loophole here. Not all solutions have reality in maths.

Yes, physical solutions must be real, not complex; infact we have to take the real part of the complex number, at the end of the computation:

E_y(physical) = Re[E₀e^{i(kx - ωt)}] = E₀cos(kx - ωt)

B_z(physical) = Re[(E₀/c)e^{i(kx - ωt)}] = (E₀/c)cos(kx - ωt).

Of course this doesn't change the relative phase between E and B.

But, more to the point, can you answer my objection on the grounds that free em  waves  appear to be fundamentally different from all other waves in how they transport the energy?
Even an electric wave travels along an LC delay line with a phase difference between volts and current - or E and B fields on a transmission line. What happens when it gets to the end of a transmission line and encounters a dipole radiator? Can there be a sudden hiccup in the phase of one of the fields?
I am confused.

In the void there isn't anything that can retard B with respect to E or vice-versa. Inside matter, let's say that you generate an E first; this moves charges (let's say electrons) and this movement generate B; you see that E acts immediately, but since charges have to accelerate, that is they have an inertia (mechanical and electromagnetical), the field B cannot arrive immediately. That is a simplicistic explanation that probably won't work in all cases of phase-difference, but can however give you an idea of what can happen.

[Mr Andrew]

I think I can prove that E and B are in quadrature using Maxwell's Theory of Electromagnetism:  A changing electric field generates a magnetic field of a strength that is proportional to the change in the E field.(Ampere's Law).  A changing magnetic field generates an electric field of a strength that is proportional to the change in the B field. (Faraday's Law).  Therefore, the greatest E field strength (the peak/trough of the E field wave) occurs when the B field is changing most rapidly (when the field strength is 0) and visa versa for a changing E field.  In other words, the E and B fields in light must oscillate in quadrature.  Was that clear?  I'm not quite sure how to reconcile this with lightarrow's calculations but they both seem valid.