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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?  ???)
 

lyner

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

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 !!
 

Offline Alandriel

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


 

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! :)
 

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 ?.
 

lyner

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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.
 

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?
 

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 »
 

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.
 

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.
 

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.
 

lyner

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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 »
 

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.
 

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 »
 

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?
 

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 »
 

Offline lightarrow

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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.
 

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.
 

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.
 

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 »
 

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.
 

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 »
 

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.
 

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?
 

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