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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 !!!




 

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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« 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.
 

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...
 

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


 

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
 

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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?
 

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.
 

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.
 

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 »
 

Offline lightarrow

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Another Question About The Speed Of Light
« 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.
 

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.
 

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.
 

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.
 

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.
 

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!
 

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.
 

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 »
 

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 
 

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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 »
 

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 »
 

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)>>
 

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
 

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.
 

lyner

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Another Question About The Speed Of Light
« 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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