Another Question About The Speed Of Light

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Offline neilep

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Another Question About The Speed Of Light
« on: 19/11/2007 15:27:17 »
Ok.....Now I need some 'pigeon english ' explanations here...

1: How come the speed of light is constant ?...is there no fluctuation at all ?

2:
...regarding the above question abut the constant speed of light(and this is where I probably will get a headache).....but...surely light has to accelerate up to C yes ?.....no ?.....How can something travel at a speed without having originally been traveling below that speed ?

THANK YOU so much for my impending headaches !!!


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Offline lightarrow

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« Reply #1 on: 19/11/2007 16:22:06 »
Ok.....Now I need some 'pigeon english ' explanations here...

1: How come the speed of light is constant ?...is there no fluctuation at all ?

2:
...regarding the above question abut the constant speed of light(and this is where I probably will get a headache).....but...surely light has to accelerate up to C yes ?.....no ?.....How can something travel at a speed without having originally been traveling below that speed ?

THANK YOU so much for my impending headaches !!!
1. No fluctuations, according to the present measurement technology; some theories should predict a dependence of light's speed from frequency, in case photons are not completely massless. Did you mean this or the fact light's speed is independent on the frame of reference?
2. Is light, or photons, a body which starts at v = 0 and reaches v = c? If this is true, then the acceleration would last for such a little time that we are still not able to measure it. No one is able to establish that light behaves in that way or not, yet; it  could be something that doesn't accelerate at all, but which exists only at that speed.

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lyner

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« Reply #2 on: 19/11/2007 18:09:47 »
Waves (all sorts)  don't build up speed but they do build up amplitude as they start. If they changed speed, their wavelength would have to change and I don't think that has ever been observed (except, of course, when moving from one medium to  another). I think that would violate boundary conditions.

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Offline lightarrow

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« Reply #3 on: 19/11/2007 18:57:28 »
Waves (all sorts)  don't build up speed but they do build up amplitude as they start. If they changed speed, their wavelength would have to change and I don't think that has ever been observed (except, of course, when moving from one medium to  another). I think that would violate boundary conditions.
Ok, but electrons (e.g.) are waves too...

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Offline Alandriel

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Another Question About The Speed Of Light
« Reply #4 on: 19/11/2007 19:44:41 »
erm...... *puts up hand shyly*

in a vacuum yes, it is constant.... but.....

I read in one of Hawkins' books that when the speed of light was calculated that 'they' (who I don't know) took an average (how - no idea either) and that there were seasonal differences... ay ay ay


Measurements are a tricky business anyways...

Quote
This defines the speed of light in vacuum to be exactly 299,792,458 m/s.  This provides a very short answer to the question "Is c constant": Yes, c is constant by definition!

However, this is not the end of the matter.  The SI is based on very practical considerations.  Definitions are adopted according to the most accurately known measurement techniques of the day, and are constantly revised.  At the moment you can measure macroscopic distances most accurately by sending out laser light pulses and timing how long they take to travel using a very accurate atomic clock.  (The best atomic clocks are accurate to about one part in 1013.)  It therefore makes sense to define the metre unit in such a way as to minimise errors in such a measurement.

The SI definition makes certain assumptions about the laws of physics.  For example, they assume that the particle of light, the photon, is massless.  If the photon had a small rest mass, the SI definition of the metre would become meaningless because the speed of light would change as a function of its wavelength.  They could not just define it to be constant.  They would have to fix the definition of the metre by stating which colour of light was being used.  Experiments have shown that the mass of the photon must be very small if it is not zero (see the FAQ: What is the mass of the photon?).  Any such possible photon rest mass is certainly too small to have any practical significance for the definition of the metre in the foreseeable future, but it cannot be shown to be exactly zero--even though currently accepted theories indicate that it is.  If it wasn't zero, the speed of light would not be constant; but from a theoretical point of view we would then take c to be the upper limit of the speed of light in vacuum so that we can continue to ask whether c is constant.

Previously the metre and second have been defined in various different ways according to the measurement techniques of the time.  They could change again in the future.  If we look back to 1939, the second was defined as 1/84,600 of a mean solar day, and the metre as the distance between two scratches on a bar of platinum-iridium alloy held in France.  We now know that there are variations in the length of a mean solar day as measured by atomic clocks.  Standard time is adjusted by adding or subtracting a leap second from time to time.  There is also an overall slowing down of the Earth's rotation by about 1/100,000 of a second per year due to tidal forces between the Earth, Sun and Moon.  There may have been even larger variations in the length or the metre standard caused by metal shrinkage.  The net result is that the value of the speed of light as measured in m/s was slowly changing at that time.  Obviously it would be more natural to attribute those changes to variations in the units of measurement than to changes in the speed of light itself, but by the same token it is nonsense to say that the speed of light is now constant just because the SI definitions of units define its numerical value to be constant.

http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html


I guess it's all relative  [;D]



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Offline syhprum

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« Reply #5 on: 19/11/2007 20:02:36 »
An old text book I had described the radiation from an antenna as starting of with the magnetic and electrostatic fields in phase and needing a quarter wavelength to settle down into proper electromagnetic radiation
syhprum

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lyner

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« Reply #6 on: 19/11/2007 22:13:43 »
Quote
Ok, but electrons (e.g.) are waves too...
OK you've thrown down the gauntlet. I had to think for a millisecond about this one.Here goes.

The energy equation E = hf explains this one away.
As you give the electron more KE, its frequency increases. The frequency of a photon, however, stays the same so its speed would, reasonably, stay the same. The frequency of the electron wave would stay constant if its speed were constant.
How's that?

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Offline Mr Andrew

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« Reply #7 on: 20/11/2007 03:56:11 »
Keep in mind that electrons are not travelling waves, but standing waves around the nucleus of an atom.  They have no speed relative to the nucleus.
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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lyner

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« Reply #8 on: 20/11/2007 13:03:53 »
But the De Broglie wavelength idea applies to moving objects as well. Bear that in mind when you walk through a door and find you have been diffracted - it wasn't just the 11 pints of heavy you drank.

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Offline lightarrow

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« Reply #9 on: 20/11/2007 15:43:50 »
Quote
Ok, but electrons (e.g.) are waves too...
OK you've thrown down the gauntlet. I had to think for a millisecond about this one.Here goes.

The energy equation E = hf explains this one away.
As you give the electron more KE, its frequency increases. The frequency of a photon, however, stays the same so its speed would, reasonably, stay the same. The frequency of the electron wave would stay constant if its speed were constant.
How's that?
How do you know that "The frequency of a photon, however, stays the same" and doesn't vary in an extremely tiny fraction of second? The only difference between electron and photon, apart from the charge, could be simply that the photon is much lighter; this is just a speculation of course.

What I mean is that if you put an electron in a 100 KV electric field, it accelerates up to its final speed in such a small time that it would be problematic to notice; if all electrons were generated in that way, you would say they don't accelerate at all, but they born at that final speed (and frequency).

So, I don't see that the answer to the initial problem stays in the difference between waves and particles; I could be wrong, anyway.
« Last Edit: 20/11/2007 15:48:08 by lightarrow »

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Offline lightarrow

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« Reply #10 on: 20/11/2007 16:10:54 »
An old text book I had described the radiation from an antenna as starting of with the magnetic and electrostatic fields in phase and needing a quarter wavelength to settle down into proper electromagnetic radiation
"In phase" with respect to what? E and M fields are always in phase with respect to each other in an EM wave.

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Offline syhprum

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« Reply #11 on: 20/11/2007 16:39:57 »
I thought that there was a 90° phase difference between the peak intensity of the magnetic field and that of the electrical field in an electromagnetic wave.
syhprum

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Offline lightarrow

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« Reply #12 on: 20/11/2007 18:23:23 »
I thought that there was a 90° phase difference between the peak intensity of the magnetic field and that of the electrical field in an electromagnetic wave.
Actually it's a frequent mistake, even among student of physics. The 90° angle is between the E and B vectors: if an EM wave goes away from you and E is vertical up, then B is horizontal right.

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lyner

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Another Question About The Speed Of Light
« Reply #13 on: 20/11/2007 22:12:53 »
Ah no!
The E and B fields are in quadrature phase AND direction.
As with all waves, the energy flow is shared between a potential energy (E ) and kinetic energy (B). The two energies add up to a constant. (sin squared plus cos squared). If it were not this way, the energy would arrive in 'dollops' and not smoothly.
Pressure and velocity in a sound wave are obviously in quadrature - peak pressure when gas is stationary. It just has to be the same idea with em waves.
If you look in every A level textbook (and even some degree texts), the diagram is wrong. In the more advanced books they get it right.
There is no surprise there because the proper diagram is very hard to draw, compared with the wrong one.

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lyner

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« Reply #14 on: 20/11/2007 23:11:22 »
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.

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Offline Mr Andrew

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« Reply #15 on: 20/11/2007 23:45:58 »
I'm from the U.S. and I was just wondering how the school system works in the UK.  A level, degree?  Degree is pretty obvious but what about the letters?  How far do they go--E, F, Z?  Just curious.

Wow!, I have never even heard of light being like that but it makes so much sense!  The energy is constant (hν), just like it is constant in a harmonic oscillator (kA2/2).  That too makes much sense!  Oh, here come ideas for the graviton thread....  You'll see them there soon!
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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lyner

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« Reply #16 on: 21/11/2007 10:04:36 »
'In the old days', we had O - ordinary level exams at 16yrs ( after first, second ----- fifth form) to prepare / select pupils for sixth form, then A -advanced level after two years in the sixth form.
O level was taken by fewer than half of pupils and, others took a CSE (certificate of secondary education) In 1985, the GCSE (general certificate of secondary education) replaced O level and CSE and this is the standard School exam.
A level remained, but has recently been re-structured in two halves; AS (Advanced Subsidiary) is taken first, after one year in 6th form, and is treated as a qualification on its own (giving 'half points' for university entrance). Students usually take 4 AS exams and often drop one subject and do 3  'A2' subjects to bring them to full A level.
A level is the basic University entrance qualification. There are, now, many more University places and the standard of A level, in many subjects is different. Physics AS is laughable, in many ways, compared with first year A level, even 8 years ago.

HOWEVER, in parallel with this system there is another 'vocational' path with a number of qualifications like NVQ ( National Vocational Qualification). They may count as equivalents in some cases.

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Offline lightarrow

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Another Question About The Speed Of Light
« Reply #17 on: 21/11/2007 11:48:09 »
Ah no!
The E and B fields are in quadrature phase AND direction.
As with all waves, the energy flow is shared between a potential energy (E ) and kinetic energy (B). The two energies add up to a constant. (sin squared plus cos squared). If it were not this way, the energy would arrive in 'dollops' and not smoothly.
Pressure and velocity in a sound wave are obviously in quadrature - peak pressure when gas is stationary. It just has to be the same idea with em waves.
If you look in every A level textbook (and even some degree texts), the diagram is wrong. In the more advanced books they get it right.
There is no surprise there because the proper diagram is very hard to draw, compared with the wrong one.
If you want I can prove they are in phase, but it's a lot of computations (probably there are simpler ways but I don't know them at the moment).

Maxwell's equations in the void:

divE = 0
rotE = -∂B/∂t
divB = 0
rotB = (1/c2)∂E/∂t

using the last:

∂(rotB)/∂t = (1/c2)∂2E/∂t2

using the second and the last:

rot(rotE) = rot(-∂B/∂t) = -∂(rotB)/∂t = -(1/c2)∂2E/∂t2
--> grad(divE) - nabla2E = -(1/c2)∂2E/∂t2

But divE = 0 (first equation) so:

nabla2E - (1/c2)∂2E/∂t2 = 0

This is the famous equation of waves.
In the particular case of plane, monocromatic waves travelling along the x direction, the solution is:

Ex = Ez = 0
Ey = E0ei(kx - ωt)

where ω = 2πf; f = frequency
k = ω/c

So:

rotE = (∂Ez/∂y - ∂Ey/∂z; ∂Ex/∂z - ∂Ez/∂x; ∂Ey/∂x - ∂Ex/∂y) =

= (0; 0; ikEy)

But rotE = -∂B/∂t

so, integrating in dt e considering that there are no static fields:

B = (0; 0; -ik∫Eydt) = (0; 0; -ik(-1/iω)Ey) = (0; 0; (k/ω)Ey) =  (0; 0; (1/c)Ey)

So: Bx = By = 0

Bz = (1/c)E0ei(kx - ωt)

So:

1. B is at 90° with respect to E

2. the amplitude of B is (1/c)*amplitude of E

3. B and E have the same phase (that is (kx - ωt)).
« Last Edit: 21/11/2007 18:58:40 by lightarrow »

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Offline syhprum

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« Reply #18 on: 21/11/2007 17:25:03 »
I am still puzzled as to wether the magnetic field and the electrical field are in phase or not !.
It seems logical to me as the induced voltage in a circuit is proportional to the rate of change of the magnetic field there should be a 90° difference but as we are dealing with a wave propagating at 'c' no doubt they get pulled into line.
If I lived close to a high power LF transmitter I would be tempted to try to make some experimental verification using a ferrite antenna and an open wire antenna 
syhprum

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lyner

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« Reply #19 on: 21/11/2007 18:08:16 »
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?
My first reaction is that, when you integrate your E, do you not get an 'i' multiplier for your B?
What is your reaction to the 'conservation of energy flow' idea? To me, that sounds like a clincher. It applies to all other sorts of wave, so why not em waves?
« Last Edit: 21/11/2007 18:12:42 by sophiecentaur »

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Offline lightarrow

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« Reply #20 on: 21/11/2007 18:56:08 »
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.
Quote
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 ∂Ey/∂x, so they cancel each other. Now I will correct my previous post. Thank you for your correction.
« Last Edit: 21/11/2007 18:59:27 by lightarrow »

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Offline lightarrow

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« Reply #21 on: 21/11/2007 20:37:04 »
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
9th row from below:
<<This equation can only be satisfied if a=0 (i.e. E and B are in phase)>>

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Offline syhprum

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« Reply #22 on: 21/11/2007 20:56:37 »
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
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Offline lightarrow

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« Reply #23 on: 21/11/2007 22:02:51 »
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.

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lyner

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« Reply #24 on: 21/11/2007 22:28:45 »
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
« Last Edit: 22/11/2007 17:24:30 by sophiecentaur »

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Offline syhprum

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« Reply #25 on: 21/11/2007 22:52:18 »
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
« Last Edit: 21/11/2007 23:03:21 by syhprum »
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Offline sohail

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« Reply #26 on: 21/11/2007 23:01:12 »
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.'




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Offline syhprum

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« Reply #27 on: 21/11/2007 23:06:24 »
No the light always travels at 'c' but it hangs around being adsorbed by atoms and then remitted
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Offline Batroost

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« Reply #28 on: 22/11/2007 19:53:49 »
And another quote from Wiki:

Quote
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.

Never express yourself more clearly than you are able to think.

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Offline lightarrow

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« Reply #29 on: 22/11/2007 20:04:09 »
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.>>

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Offline Batroost

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« Reply #30 on: 22/11/2007 20:32:22 »
Nice. [:)]
Never express yourself more clearly than you are able to think.

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Offline syhprum

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« Reply #31 on: 22/11/2007 20:34:38 »
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.
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Offline Batroost

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« Reply #32 on: 22/11/2007 20:39:36 »
I know the feeling.

But sometimes Mr Fry is just plain wrong; his 'electricity' themed programme made a few errors anyway.
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« Reply #33 on: 22/11/2007 23:21:56 »
Quote
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:
Quote
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.

« Last Edit: 22/11/2007 23:24:19 by sophiecentaur »

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Offline syhprum

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« Reply #34 on: 23/11/2007 11:26:46 »
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 ?
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« Reply #35 on: 23/11/2007 11:56:36 »
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.
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« Reply #36 on: 23/11/2007 13:12:35 »
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.

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Offline lightarrow

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« Reply #37 on: 23/11/2007 13:48:39 »
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...

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Offline lightarrow

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« Reply #38 on: 23/11/2007 14:14:25 »
Lightarrow:
Quote
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:

Ey(physical) = Re[E0ei(kx - ωt)] = E0cos(kx - ωt)

Bz(physical) = Re[(E0/c)ei(kx - ωt)] = (E0/c)cos(kx - ωt).

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

Quote
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.
« Last Edit: 23/11/2007 14:17:50 by lightarrow »

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Offline Mr Andrew

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« Reply #39 on: 23/11/2007 19:31:53 »
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.
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Offline lightarrow

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« Reply #40 on: 23/11/2007 20:05:52 »
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)..

Not exactly.

Faraday's Law:

rotE = -∂B/∂t

--> the line integral of E along a closed loop is equal to minus the time variation of the magnetic flux through that loop.

Ampere's Law in the void:

rotB = (1/c2)∂E/∂t

--> the line integral of B along a closed loop is equal to (1/c2) the time variation of the electric flux through that loop.


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Offline syhprum

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« Reply #41 on: 23/11/2007 20:10:39 »
I think we must hold a vote on this matter I have been converted I side with lightarrow
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« Reply #42 on: 23/11/2007 22:18:59 »
Lightarrow's sums must be right. (He IS pretty reliable in this direction)
SO there is a problem with my interpretation of the energy flow in a wave.
No one has argued with my idea that all other waves consist of PE and KE variations which are in phase quadrature. So put me straight: is not the E field a potential form of energy and is not the B field a dynamic / kinetic form of energy? The fact that they are transverse and spacially in quadrature makes the em different from other waves, I admit but em waves follow the general principle of phase quadrature when guided by a wire / wires / waveguide so what is the difference, once they get launched into space (or a medium; it all can't change just because you have the occasional molecule of air in the way)?
Help me with the physical interpretation of this - there must be one which can reconcile my misgivings - unless the in-phase idea is wrong (god I would so like it to be wrong! Just think of all the  experts being gutted! Even Lightarrow!!!)

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Offline syhprum

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« Reply #43 on: 23/11/2007 22:30:21 »
I think the difference lies in currents in conductors dragging electrons around and currents in the void being free of that in-pediment.
In my original question I spoke of the B & H phase difference in the currents in the antenna and them falling into phase when they got a quarter wavelength into the void.
I seem to recall that is what Sterling was telling us in his 1930's textbook
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« Reply #44 on: 24/11/2007 01:29:32 »
Lightarrow's sums must be right. (He IS pretty reliable in this direction)
SO there is a problem with my interpretation of the energy flow in a wave.
No one has argued with my idea that all other waves consist of PE and KE variations which are in phase quadrature. So put me straight: is not the E field a potential form of energy and is not the B field a dynamic / kinetic form of energy?
Sincerely I have never heard anything like that, in an EM radiation, but, who knows; what I know is that energy in an EM wave can be computed from the Poynting vector S = EXH where X means vectorial product: if N is the versor normal to a certain surface then SN; • = scalar product, is the power per unit area of the EM wave flowing through that surface.
Quote
The fact that they are transverse and spacially in quadrature makes the em different from other waves, I admit but em waves follow the general principle of phase quadrature when guided by a wire / wires / waveguide so what is the difference, once they get launched into space (or a medium; it all can't change just because you have the occasional molecule of air in the way)?
Help me with the physical interpretation of this - there must be one which can reconcile my misgivings - unless the in-phase idea is wrong (god I would so like it to be wrong! Just think of all the  experts being gutted! Even Lightarrow!!!)
However let's remember that near a cicuit E and B can be independent one of the other: E comes from charges and B comes from currents and you can make a circuit that will generate the E that you want and the B that you want; you can't do it in an EM wave: E and B are one *dependent* on the other, infact one is generated from the other and not from specific charges/currents configurations.
Anyway, let someone more expert come here and tell us the truth...

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Offline Mr Andrew

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« Reply #45 on: 24/11/2007 03:51:07 »
lightarrow, I am aware of the mathematical formulation of the two laws that I referred to.  They mean exactly what I said, except that I didn't mention that the fields that these changes in E or B create are in a direction perpindicular to the direction of the wave's propagation (if the left was nabla•E or nabla•B then the fields generated by the changing fluxes would be in the same direction as c).  I took it as a given that the fields were perpindicular to the direction of propagation as light is a transverse wave.  I know that your calculations are correct (they appear many different places and I have taken the time to check the math, assuming that rot(rotV))=grad(divV)-nabla2V is true, they work out), my problem is that this solution seems to be just as infallible as yours, yet only one can be right.

Does anybody see what's wrong here?
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« Reply #46 on: 24/11/2007 08:52:11 »
I think that I have the solution to the problem of phase. Light arrow has not quite completed the analysis.  He has shown that the electric fields and magnetic fields stay in the same phase with respect ot each other but he has not proved exactly what that phase is  (There is always a constant in the integration ) to do that you have to go back to the original equations to determine the constant.  If you do that you will find that is where the phase difference lies.
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« Reply #47 on: 24/11/2007 13:26:23 »
I think that I have the solution to the problem of phase. Light arrow has not quite completed the analysis.  He has shown that the electric fields and magnetic fields stay in the same phase with respect ot each other but he has not proved exactly what that phase is  (There is always a constant in the integration ) to do that you have to go back to the original equations to determine the constant.  If you do that you will find that is where the phase difference lies.

But what you say it's not possible: to have a phase difference Φ between E and B, you should have:

Ey = E0ei(kx - ωt)

Bz = (1/c)E0ei(kx - ωt + Φ) = (1/c)eE0ei(kx - ωt)

so the phase factor would become a multiplicative constant, not an additive one.

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« Reply #48 on: 24/11/2007 14:26:04 »
lightarrow, I am aware of the mathematical formulation of the two laws that I referred to.  They mean exactly what I said, except that I didn't mention that the fields that these changes in E or B create are in a direction perpindicular to the direction of the wave's propagation (if the left was nabla•E or nabla•B then the fields generated by the changing fluxes would be in the same direction as c).  I took it as a given that the fields were perpindicular to the direction of propagation as light is a transverse wave. I know that your calculations are correct (they appear many different places and I have taken the time to check the math, assuming that rot(rotV))=grad(divV)-nabla2V is true, they work out), my problem is that this solution seems to be just as infallible as yours, yet only one can be right.

Does anybody see what's wrong here?

Let's take Faraday's Law:

rotE = -∂B/∂t

integrating on a surface σ:

∫rotE•dσ = -(∂/∂t)∫B•dσ

using Stokes theorem:

E•dl = -(∂/∂t)ΦB

where: the first integral is computed on the edge of the surface σ, dσ = Ndσ where N is the versor of the element of area dσ and dl is the line element of that edge; ΦB is the total flux of B through σ.

Now let's take as σ a disk with radius r, so the edge is the circle of radius r; if B is uniform through the disk and perpendicular to it, for cylindrical symmetry the field E in the circle is tangent to it and with uniform intensity E and we can write:

2πrE = -πr2∂B/∂t  -->

--> E = -(r/2)∂B/∂t

For a sinusoidal time-varying B, that is, B = B0sin(ωt) we have:

E = -(r/2)ωB0cos(ωt)

and so E is 90° out of phase in time, with respect to B, as you say.

Note however the key hypotesis: if B is uniform through the disk; in our case instead, B is not uniform (it varies sinusoidally also with space and not only with time), so we have to use Faraday's Law in the differential form:

rotE = - ∂B/∂t

and so we have to compute also the spatial derivatives of E; for this reason the "i" factor comes in both members of that equation and so the phase is the same (I remind you that a factor "i" equals a 90° phase difference: i = eiπ/2).

However, in the post where I computed E and B for an EM wave, I should have said that I took a plane polarized wave (because that's more simple); I'm not sure of what could come out if the wave wouldn't be that way, actually.


About:

rot(rotV))=grad(divV)-nabla2V

it's not very simple to prove it, but it's a well known mathematical rule; you can find it in (good) electrodynamics books.
« Last Edit: 24/11/2007 14:44:32 by lightarrow »

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« Reply #49 on: 24/11/2007 15:08:26 »
Lightarrow
...
What is your reaction to the 'conservation of energy flow' idea? To me, that sounds like a clincher. It applies to all other sorts of wave, so why not em waves?
I have thought about it but I don't know how to solve this; I don't know if this flow of energy does really have to be constant or not and what it could mean. Interesting question however!