Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: neilep on 19/11/2007 15:27:17
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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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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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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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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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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...
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 (http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html)
I guess it's all relative [;D]
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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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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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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.
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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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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.
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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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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.
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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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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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www.play-hookey.com/optics/transverse_electromagnetic_wave.html (http://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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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!
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'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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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)).
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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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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?
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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 ∂Ey/∂x, so they cancel each other. Now I will correct my previous post. Thank you for your correction.
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www.play-hookey.com/optics/transverse_electromagnetic_wave.html (http://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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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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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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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
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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
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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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No the light always travels at 'c' but it hangs around being adsorbed by atoms and then remitted
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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.
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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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Nice. [:)]
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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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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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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.
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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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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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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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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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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:
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.
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.
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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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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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I think we must hold a vote on this matter I have been converted I side with lightarrow
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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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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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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 S•N; • = scalar product, is the power per unit area of the EM wave flowing through that surface.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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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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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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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)eiΦE0ei(kx - ωt)
so the phase factor would become a multiplicative constant, not an additive one.
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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.
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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!
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There's no reason why the Poynting vector should be constant. In fact, for a plane wave, it has a sinusoidal form: Re{S}~Cos[2(kx-ωt)]. Time-averaging it gets rid of this time dependence, and that's the form that's usually used.
I think the KE/PE in quadrature is going to arise from the KE/PE relations of the propagation medium itself because you're treating the propagation medium as a set of simple harmonic oscillators. In such propagation, describing the waves in position and in momentum will lead to two expressions out of phase by Pi/2. You can do an analogous thing with the electromagnetic field, but it's a relation of the field to its derivative, rather than between the E & B fields.
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Does the solution to this conundrum lie in confining the operators to the real parts at each stage in the chain of calculations?
After each 'calculus' operation there are other products which are, perhaps, 'not really there'. So the E field, generated by the varying B field has to be real - etc.
My maths is so rusty that I couldn't rely on my calculations to prove or disprove anything but perhaps someone could go through the steps, eliminating non-real bits at each stage and see what comes out of it?
I know we blindly accept the complex form of wave representation in calculations and then say 'just take the real part' at the end. Is it really justifiable? I may be very naive in asking that question.
Else it may be reconcilable by jpetruccelli's ideas (above). It would imply that the higher the refractive index / density of the medium, the more in quadrature the fields would be. Does Maxwell produce that result?
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In the real world linear polarized EM waves are a rarity we only meet them when they are generated by lasers or electronics.
When I want to generate circular polarized waves I feed them into a helical antenna with a spacing of 1/4 wavelength between the turns and a circumference of 1 wavelength.
It strikes me that the rate of rotation is very high.
Do these calculations deal with circular polarized waves or only linear.
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Light arrow you surprise me. You seem to be so familiar with the mathematics yet seem to forget the cyclic properties of the imaginary exponent. Remember
exp^iw = cos w +i sin w if you put your equations in that form it looks perfectly sensible
There is absolutely no problem with the multiplication.
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Going back to the original equations you will see that the magnitude of the curl of the magnetic vector is proportional to the rate of change of the electric vector which for a sinusoidal waveform implies quadrature this is your constant of integration.
similarly the magnitude if the curl of the electric vector is proportional to the (the negative) of the rate of change of the magnetic vector.
It is important to understand the physics as well as the mathematics it is the changing electrical fields that create the magnetic fields and the changing magnetic fields that create the electrical ones.
Alternatively at a peak point in the electrical or magnetic fields there is a momentary point where the partial derivative with respect to time of the relevant electric or magnetic field is zero. it follows directly from the equations that the value of the corresponding value of the magnetic or electric field must be zero. This again implies that for a simple propagating plane polarised wave the relative magnitudes of the fields are in quadrature.
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n the real world linear polarized EM waves are a rarity we only meet them when they are generated by lasers or electronics.
What about every mf radio and uhf TV broadcast, then?
No only are they common but their phases can be measured a lot easier than light waves!
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Do these calculations deal with circular polarized waves or only linear.
A circularly / elliptically polarised wave can be regarded as a pair of plane polarised waves with E vectors at right angles and with their phases in quadrature then exactly the same calculations can be done on each component, independently. Yes, the rate of rotation is high - it's 2pi radians of rotation per cycle.
You can generate good circular polarised waves (on one axis) using crossed dipoles with a 1/4 wavelength delay in the feed to one of them. The off axis polarisation of a helical antenna also goes elliptical.
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Soul Surfer, that's exactly what I said!
lightarrow, I don't understand how i represents a phase difference of 90 degrees.
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n the real world linear polarized EM waves are a rarity we only meet them when they are generated by lasers or electronics.
What about every mf radio and uhf TV broadcast, then?
No only are they common but their phases can be measured a lot easier than light waves!
These are of course the EM waves I had in mind as being generated by electronic devices (Vacuum tubes, Klystrons, Transistors etc) but in the wider universe most of the radiation is generated by thermal sources with ill defined frequency and polarisation
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I realised that, afterwards, Syphrum but they are so very measurable that they are well worth discussing and measuring for verification. I did all my electromag learning with radio waves in mind and see it as the most relevant - just a personal view.
Incidentally - a ferrite rod would not be a good measuring tool because of its enormous inductance. To measure E and B fields one would need a very short dipole. feeding a high impedance and a very small search coil feeding a low impedance. These would not alter the measured phase of any signal, appreciably.
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There's no reason why the Poynting vector should be constant.
Sure, I've never wrote it, infact! I mentioned it just to show how books usually derive the energy of an electromagnetic wave; I was answering to sophiecentaur: he wrote about a way of computing the EM energy which I had never heard before.
In fact, for a plane wave, it has a sinusoidal form: Re{S}~Cos[2(kx-ωt)].
That's for sure:
Re{EXH}= (1/μ0)Re{(0; 0; Ey*Bz}) = (1/μ0)(0; 0; Re{E0ei(kx - ωt)*(1/c)E0ei(kx - ωt)}) =
= (1/cμ0)E02(0; 0; cos[2(kx - ωt)]).
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Does the solution to this conundrum lie in confining the operators to the real parts at each stage in the chain of calculations?
Absolutely not, even because otherwise complex calculus would be meaningless; you should think to it as to something that contains all informations that you would obtain with real numbers only, plus other informations.
After each 'calculus' operation there are other products which are, perhaps, 'not really there'. So the E field, generated by the varying B field has to be real - etc.
My maths is so rusty that I couldn't rely on my calculations to prove or disprove anything but perhaps someone could go through the steps, eliminating non-real bits at each stage and see what comes out of it?
I know we blindly accept the complex form of wave representation in calculations and then say 'just take the real part' at the end. Is it really justifiable? I may be very naive in asking that question.
You can make all the computations I made, writing Ey = E0cos(kx - ωt) and using, in case, simple trygonometry relations, as: cos2x = (1/2)(1 + cos(2x)) ecc. It's not difficult, try it.
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Light arrow you surprise me. You seem to be so familiar with the mathematics yet seem to forget the cyclic properties of the imaginary exponent. Remember
exp^iw = cos w +i sin w if you put your equations in that form it looks perfectly sensible
There is absolutely no problem with the multiplication.
If you make the computation with complex vectors, you have to do as I did and no additive constant comes out; if you make the computation with real numbers, the phase difference should come in this way:
Ey = E0cos(kx - ωt)
Bz = (1/c)E0cos(kx - ωt + Φ)
BUT you can't take Φ out of cosine, so no additive constant.
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Going back to the original equations you will see that the magnitude of the curl of the magnetic vector is proportional to the rate of change of the electric vector which for a sinusoidal waveform implies quadrature this is your constant of integration.
? Make the computation and you'll see this is wrong.
similarly the magnitude if the curl of the electric vector is proportional to the (the negative) of the rate of change of the magnetic vector.
It is important to understand the physics as well as the mathematics it is the changing electrical fields that create the magnetic fields and the changing magnetic fields that create the electrical ones.
In this case we must be more precise: "it is the time changing of electric field that create the curl of the magnetic field". The curl IS a changing: is a spatial changing; remember how I computed it:
rotE = (=curlE) = (0; 0; ∂Ey/∂x) (all the other terms are zero). If you derive cos(kx - ωt) with respect to time, you'll have:
-ωsin(kx - ωt);
if you derive it with respect to space x, you'll have:
-ksin(kx - ωt)
Where is the phase difference?
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In the real world linear polarized EM waves are a rarity we only meet them when they are generated by lasers or electronics.
When I want to generate circular polarized waves I feed them into a helical antenna with a spacing of 1/4 wavelength between the turns and a circumference of 1 wavelength.
It strikes me that the rate of rotation is very high.
Do these calculations deal with circular polarized waves or only linear.
As I have already wrote, my calculations deal with linearly polarized waves only. Sincerely I don't know how to make the calc. in the case of circular polarization and if it could come out a phase difference or not in that case.
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I don't understand how i represents a phase difference of 90 degrees.
from MrAndrew
This is not trivial, particularly in the context of waves. It involves knowing about
1. 'Simple' Trig functions and the rules to differentiate and integrate them.
2. How complex numbers work.
3. How you can represent trig functions in their exponential form.
But a simple argument goes like this.
Multiplying a quantity by -1 reverses its direction - (e.g. -3 times -1 gives you +3 ); on a graph / number line, you have rotated it by 1800
i is the square root of -1, so, using a similar idea to the above, you rotate it half as much - i.e. 900.
Do it twice and you get 1800
i is called an imaginary number, because you can't have 'i grams of salt' but the answer to many algebraic equations comes out as a complex number containing a 'real part and an 'imaginary' part e.g. a+ib and it has to be plotted in two dimensions- not just on a number line.
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Ah, that explanation makes much more sense. I understand how i = eπi/2...I just didn't see the geometric implications. Thank you.
lightarrow, here is some math that I hope shows, both with respect to space and time, that E and B are in quadrature:
Take Faraday's Law: ∫E•dl = -∂ΦB/∂t
Differentiating with respect to dl gives: E = -∂2ΦB/∂l•∂t
Now, E is at a maximum or minimum when its derivative (either ∂E/∂t or ∂E/∂l) is zero.
0 = ∂E/∂t [prop] ∫B•dl with no current flowing. For the integral to be 0, B must be zero. B is zero when E is at a peak or trough (they are in quadrature since they are waves).
0 = ∂E/∂l [prop] B with no current once again. Here, B is once again zero when E is at a maximum/minimum (they are in quadrature).
*The assumption that there is no current is justified because light, in this case, is in a vacuum where no current can flow.
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Ah, that explanation makes much more sense. I understand how i = eπi/2...I just didn't see the geometric implications. Thank you.
lightarrow, here is some math that I hope shows, both with respect to space and time, that E and B are in quadrature:
Take Faraday's Law: ∫E•dl = -∂ΦB/∂t
Differentiating with respect to dl gives: E = -∂2ΦB/∂l•∂t
Now, E is at a maximum or minimum when its derivative (either ∂E/∂t or ∂E/∂l) is zero.
0 = ∂E/∂t [prop] ∫B•dl
In that formula you have to substitute E with ΦE (flux of the field E through the surface on which border you compute the line integral).
with no current flowing. For the integral to be 0, B must be zero.
No. Try to compute that integral in the simple case of a uniform B ≠ 0 and a squared loop line of side L alined with B:
∫B•dl = B*L + 0 -B*L + 0 = 0.
If the square is on a plane perpendicular to B it's even simpler:
∫B•dl = 0 + 0 + 0 + 0 = 0. B is zero when E is at a peak or trough (they are in quadrature since they are waves).
0 = ∂E/∂l [prop] B
Where did you find this one?
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lightarrow, I don't understand how i represents a phase difference of 90 degrees.
Sorry not to have answered you before to this question.
I see sophiecentaur has already answered you, I add something else:
you know that you can represent any complex number z as a point on the cartesian plane, in two ways:
1. z = x + iy; x is the real part and y the imaginary part and they corresponds to the cartesian coordinates of z in the plane.
2. z = ρeiθ; ρ is the "modulus" (= |z|) and corresponds to the lenght of the arrow that goes from the origin to the point in the plane; θ is the "argument" and corresponds to the angle between the x axis and the arrow.
The second, called "exponential representation" is often more useful, for example when you have to compute the product or the division of two complex numbers z1 and z2:
z1*z2 = ρ1eiθ1*ρ2eiθ2 = ρ1ρ2ei(θ1+ θ2)
In our discussion the angles θ1 and θ2 can be called the phases of the two complex n.
So, if you multiply a complex number which phase is θ1 by another complex number which phase is θ2, you obtain a third number which phase is the sum of the other 2 and which modulus is the product of the other two.
So, if you have a complex number z which phase is θ and you want to transform it into another complex number with the same modulus but which phase is θ + φ, you just have to multiply z by eiφ:
z*eiφ = ρeiθ*eiφ = ρei(θ+φ)
So, if you want to give it a π/2 phase difference: z*eiπ/2 ecc.
i = eiπ/2 since ρ = 1 and θ = π/2 in that case.
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Quote from Batroost
(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?
When I was first introduced to em waves we were told of Huygen's construction, involving 'secondary wavelets'; a wavefront of any shape can be regarded as an infinite set of wavelets, starting along the wavefront. (This is just like the idea of diffraction, it seems to me.) The shape of the wavefront, immediately afterwards is given by the sum of all these secondary wavelets. For a plane wave, the secondary wavelets add up only in the forward direction and cancel in all others
The reason for the wave remaining well behaved as it goes through a medium would be that all the absorbed and re emitted wavelets add up in phase only in the direction given by 'geometrical optics'. Your concern about the wavefront being destroyed is, actually, groundless.
As with most of Physics, there are usually several valid ways of looking at things.
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Sorry, there were some errors in my math.
The equation ∂E/∂l = B should be ∂2ΦE/∂l∂t = B and all of the E's in Ampere's Law should be ΦE. I mixed the vector and integral representations of the law (where curlB [prop] ∂E/∂t when no current flows).
Sorry, I forgot for a second that Faraday's and Ampere's laws had line integrals, not just regular integrals (I am a senior in high school who has not had E&M or Calc II formally-I am teaching myself for now). Ok, so if a line integral of x is 0, x doesn't necessarily have to be zero. I'll have to work on this. If anybody sees where I am going and thinks they can get there without all of the mathematical hiccups and such, go right ahead. This is really bothering me because, as far as I can tell, lightarrow has provided a valid model for light, and so has Soul Surfer: Alternatively at a peak point in the electrical or magnetic fields there is a momentary point where the partial derivative with respect to time of the relevant electric or magnetic field is zero. it follows directly from the equations that the value of the corresponding value of the magnetic or electric field must be zero. This again implies that for a simple propagating plane polarised wave the relative magnitudes of the fields are in quadrature.
Thank you lightarrow for your explaination of the phase changes...as I said, I am not that familiar with the mathematics associated with waves.
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Is the quadrature or in phase question really relevant?
I am just beginning to wonder if, I have been making the same sort of mistake that A level students make when understanding the momentum of a photon;
"It must be mc, because that's what momentum is".
It requires a serious change of thought for them to get round that one and we may need a similar change of approach for this problem, too..
It may be that, when a wave is traveling at c, you also have to look at it differently and the 'in phase' idea doesn't really contravene anything. As far as the photon in space is concerned, there is no passage of time so it's energy is not 'pulsing at all and there is no interaction with matter. Putting yourself in a frame where you are traveling alongside the wave / photon stream at c and discussing what you would see it is probably as dodgy a concept as talking in terms of photons having mass. It certainly must be approached with care.
What I would expect, however, would be for this in-phase situation to change as the wave goes through a medium. A transmission line could be regarded as a medium because the guided wave is interacting with conduction electrons and we do, actually, find quadrature E and B fields there.
Perhaps there is no conflict at all.
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Is the quadrature or in phase question really relevant?
I am just beginning to wonder if, I have been making the same sort of mistake that A level students make when understanding the momentum of a photon;
"It must be mc, because that's what momentum is".
It requires a serious change of thought for them to get round that one and we may need a similar change of approach for this problem, too..
It may be that, when a wave is traveling at c, you also have to look at it differently and the 'in phase' idea doesn't really contravene anything.
I don't have any knowledge at the moment to confirm or confute this idea; my intuition would say there would really be a difference, in the way matter interact with the wave.
As far as the photon in space is concerned, there is no passage of time so it's energy is not 'pulsing at all and there is no interaction with matter. Putting yourself in a frame where you are traveling alongside the wave / photon stream at c and discussing what you would see it is probably as dodgy a concept as talking in terms of photons having mass. It certainly must be approached with care.
But you don't need it: fixing a point of space, you will measure different values of the fields if E and B are not in phase; the same if you analyze a standing wave of light between two mirrors.
What I would expect, however, would be for this in-phase situation to change as the wave goes through a medium. A transmission line could be regarded as a medium because the guided wave is interacting with conduction electrons and we do, actually, find quadrature E and B fields there.
Perhaps there is no conflict at all.
For the little I've read, I'm pretty sure the relative phase between E and B can change in those situations, as you say, but I don't know exactly which ones; however, in a solid medium, *if it is linear, homogeneous and isotropic* the situation is the same as in the void: you simply have to replace ε0 and μ0 with different values ε and μ, function of the frequency only, so the equations would be exactly the same and so their solutions too.
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But you don't need it: fixing a point of space, you will measure different values of the fields if E and B are not in phase; the same if you analyze a standing wave of light between two mirrors.
Yes, I know. I didn't put what I wanted to say very well. What I meant is that the concept of the energy carried by 'something' going at speed c is probably just as divorced from energy carried by matter as is the concept of its momentum. I didn't mean it doesn't happen - I just meant that perhaps the apparent paradox doesn't matter.
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There certainly is a difference between the em waves interact with matter and the way they interact with other em waves i.e. they don't. We need some different (well informed) input on this, I think.
It is a good subject, though, don't you think?
I was riding home on the bus, earlier this afternoon, watching a squirrel on the pavement and tried to relate what I was seeing to what we have been discussing.
We have such compartmentalised lives.
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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!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? [???])
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pavement ≡ sidewalk
me in the bus
squirrel outside
QED
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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 !!
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***** sigh *** I wish I understood the half of it
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fwww.duiops.net%2Fseresvivos%2Fgaleria%2Fperros%2FBulldog%2520With%2520Headache.jpg&hash=fa775a523c063b77bbc6bf2b8211fc59)
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pavement ≡ sidewalk
me in the bus
squirrel outside
QED
Ah! Ok! [:)]
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Has a concensus been agreed ? are B & H in phase or quadature ?.
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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.
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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?
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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 (k•r+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 (k•r+φe)],
B(r)=B0exp[i {k•r+φb)].
Now use Faraday's law:
Curl(E)=-∂B/∂t
kxE0exp[i (k•r+φe-ωt)]=ωB0exp[i {k•r+φb-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,
φe=φb.
Therefore the fields are in phase for a linear dielectric.
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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)E0=ωB0
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.
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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.
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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.
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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.
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Yes this is what I read in Sterlings texbook 60 years ago.
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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)
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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?
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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.
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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.
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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.
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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.
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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.
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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.
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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 (http://www.thenakedscientists.com/forum/index.php?topic=11766.25). The description of how light propagates as given in this thread may hold the key to its constant speed.
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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.
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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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This ether would be stationary with respect to what?
Since the ehther would occupy the whole of the Universe why would it have to be stationary to anything? If you have been to a firework display then surely you would have seen that the light extends to a far greater distance than the actual display. This would be a similar situation to having a Virtual photon ether, how could you have a Big Bang without it being accompanied by light, which because of its nature would permeate every corner of the expanding Universe..
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This ether would be stationary with respect to what?
Since the ehther would occupy the whole of the Universe why would it have to be stationary to anything? If you have been to a firework display then surely you would have seen that the light extends to a far greater distance than the actual display. This would be a similar situation to having a Virtual photon ether, how could you have a Big Bang without it being accompanied by light, which because of its nature would permeate every corner of the expanding Universe..
Ok, since you don't understand even basic physics, I ask you in another way: Does Earth move with respect to ether? At which speed and in which direction?
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Ok, since you don't understand even basic physics, I ask you in another way: Does Earth move with respect to ether? At which speed and in which direction?
It depends on what kind of ether you are talking about. A virtual photon ether would be completely permeable to matter, it would possess latent electromagnetic properties which are activated when it comes into contact with real photons. Therefore this kind of an ether, remember it permeates the whole of space, would therefore be stationary with respect to the earth.
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Ok, since you don't understand even basic physics, I ask you in another way: Does Earth move with respect to ether? At which speed and in which direction?
It depends on what kind of ether you are talking about. A virtual photon ether would be completely permeable to matter, it would possess latent electromagnetic properties which are activated when it comes into contact with real photons. Therefore this kind of an ether, remember it permeates the whole of space, would therefore be stationary with respect to the earth.
Ok, so Earth is a preferential reference frame in the universe. Was your previous name "Ptolemy", by chance? [:)]
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Certainly, the Greeks didn't take a lot of notice of experimentation.
I think you are 'banging your head against a brick wall', lightarrow.
Is there an Italian equivalent of that expression?
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Or "flogging a dead horse" perhaps.
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If there was a ether then should we not expect it to behave in a similar manner to the Cosmic Background Radiation.
IE, Infinite, coming from every direction at once and moving.
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Certainly, the Greeks didn't take a lot of notice of experimentation.
I think you are 'banging your head against a brick wall', lightarrow.
Is there an Italian equivalent of that expression?
The same, but without specifing "brick" [:)]
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Or "flogging a dead horse" perhaps.
Maybe there is a corresponding in italian, but I've never heard.