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Details of Lightwaves
19 years 7 months ago #13272
by kcody
Replied by kcody on topic Reply from Kevin Cody
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />I gave up on this a while back because the two slit experiment makes the details of the photon unclear. The matter may never be known and that is because the photon is too small.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
IMO the two slit experiment is proof that light is a wave. Streams of particles would travel in a straight line, and couldn't interfere like that. Waves of water passing under under a multi-support bridge do exactly that, most noticeably if the waters are still.
I've heard it said elsewhere, I think in TVF's book, that the photon is emitted locally when the wave encounters a boundary in its medium, ie an atom. That makes sense to me. Put a piece of plywood on the side of that bridge opposite where the wave originates, and watch particles of water go flying from the points of constructive interference.
For now, I'm more interested in the still-propagating wave.
- Kevin
<br />I gave up on this a while back because the two slit experiment makes the details of the photon unclear. The matter may never be known and that is because the photon is too small.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
IMO the two slit experiment is proof that light is a wave. Streams of particles would travel in a straight line, and couldn't interfere like that. Waves of water passing under under a multi-support bridge do exactly that, most noticeably if the waters are still.
I've heard it said elsewhere, I think in TVF's book, that the photon is emitted locally when the wave encounters a boundary in its medium, ie an atom. That makes sense to me. Put a piece of plywood on the side of that bridge opposite where the wave originates, and watch particles of water go flying from the points of constructive interference.
For now, I'm more interested in the still-propagating wave.
- Kevin
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19 years 3 months ago #13454
by proton
Replied by proton on topic Reply from Wayne Harrison
More of a general question really,but if a short segment of an oscillating field rotates tranversely at right angles to the direction of propogation at speeds approaching c(is this possible?) and that "photon" travels at c then out of phase should not make any difference since there would be no phase difference in any wave.
proton
proton
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19 years 3 months ago #13499
by proton
Replied by proton on topic Reply from Wayne Harrison
Another question,Sapphire light clocks using microwave reflections as their timing mechanisms,fine for improving net speed but any ideas on application to universal constants and their "changing" values.
proton
proton
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19 years 3 months ago #13500
by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
In a typical oscillator circuit, the current is 90deg out of phase with the applied voltage. Since the magnetic field is proportional to current, it too would be 90deg out of phase with the applied voltage. Reducing the oscillator to molecular size, you would have electrons going in and out of a tiny antenna. The potential of the antenna would be max or min when the current is zero. So my intuition tells me the textbook is correct.
I think it was Plank who said, in so many words, that a photon is either a wave or a particle, depending on which you are looking for. Look for a wave and you'll find a wave; look for a particle and you'll find a particle. The two-slit experiment works for waves or particles. For particles, it is a probability distribution; for waves it is an interference pattern.
The laser clock is ideal for understanding how time is relative to velocity. The light reflects back and forth between the ends of the laser, emitting one pulse of light each time it arrives at a particular end (not at both ends). To a first observer, moving with the clock, it takes the same amount of time for light to go from end to end in either direction. To a second observer, watching the clock pass at high speed, the light takes longer to go from back to front than from front to back. The total distance traveled by the light during one clock pulse is twice the length of the laser plus the distance it travelled during one pulse. Therefore, the clock runs slower in the second observer's reference frame.
Of course, actually seeing the clock pulses is another matter; you have to add the time it takes for light to travel from the clock to the observer, which brings in the doppler shift. So the second observer will see the clock running faster as it approaches and slower as it goes away; but correcting for the doppler shift will reveal that the clock is running at the same slow rate the whole time.
I think it was Plank who said, in so many words, that a photon is either a wave or a particle, depending on which you are looking for. Look for a wave and you'll find a wave; look for a particle and you'll find a particle. The two-slit experiment works for waves or particles. For particles, it is a probability distribution; for waves it is an interference pattern.
The laser clock is ideal for understanding how time is relative to velocity. The light reflects back and forth between the ends of the laser, emitting one pulse of light each time it arrives at a particular end (not at both ends). To a first observer, moving with the clock, it takes the same amount of time for light to go from end to end in either direction. To a second observer, watching the clock pass at high speed, the light takes longer to go from back to front than from front to back. The total distance traveled by the light during one clock pulse is twice the length of the laser plus the distance it travelled during one pulse. Therefore, the clock runs slower in the second observer's reference frame.
Of course, actually seeing the clock pulses is another matter; you have to add the time it takes for light to travel from the clock to the observer, which brings in the doppler shift. So the second observer will see the clock running faster as it approaches and slower as it goes away; but correcting for the doppler shift will reveal that the clock is running at the same slow rate the whole time.
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- cosmicsurfer
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19 years 3 months ago #14187
by cosmicsurfer
Replied by cosmicsurfer on topic Reply from John Rickey
For the doppler shift corrections to work the relativity in a given space must be continuous and the same, otherwise for instance if a force field or electromagnetic packet such as found in FTL tunneling were to break the symmetry I think that the doppler shift would no longer be valid. Then the speed of light medium would be operating with in the boundaries of a warped space and may not be synchronous with the surrounding relativity any longer. Hence the clock might be operating then under new rules. However, from our frame of reference the Universe appears isotropic and the same everywhere as one relativity, but other forces that are unseen and hyperdimensional could defy our known rules of relativity.
John
John
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19 years 3 months ago #14116
by Thomas
Replied by Thomas on topic Reply from Thomas Smid
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by kcody</i>
<br />
An unreleased high-school-level physics text I'm reading makes the claim that not only are EM waves transverse, but that the electric and magnetic components will always be in planes 90 degrees apart; and will always be 90 degrees out of phase.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
For travelling EM waves, the electric field E and magnetic field B are always *in* phase. This is because the direction of propagation is proportional to the cross product ExB. If E and B would be 90 degress out of phase, the direction of propagation would not be constant but changing sign all the time.
E and B are phase shifted by 90 deg however for *standing* EM waves. This is because the average of ExB has to be zero here.
(see scienceworld.wolfram.com/physics/ElectromagneticWave.html for more on this).
www.physicsmyths.org.uk
www.plasmaphysics.org.uk
<br />
An unreleased high-school-level physics text I'm reading makes the claim that not only are EM waves transverse, but that the electric and magnetic components will always be in planes 90 degrees apart; and will always be 90 degrees out of phase.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
For travelling EM waves, the electric field E and magnetic field B are always *in* phase. This is because the direction of propagation is proportional to the cross product ExB. If E and B would be 90 degress out of phase, the direction of propagation would not be constant but changing sign all the time.
E and B are phase shifted by 90 deg however for *standing* EM waves. This is because the average of ExB has to be zero here.
(see scienceworld.wolfram.com/physics/ElectromagneticWave.html for more on this).
www.physicsmyths.org.uk
www.plasmaphysics.org.uk
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