Light is ...

siarad · Feb 13, 2005

On reflection I think I was a bit tsunami-ised. Light is a TEM wave, & me a designer! so what I said was based upon an error. For light to behave as you say it must travel in rays which could be either wave or photon I guess.

xmo1 · Feb 28, 2005

It would be good to have some quality URL's to augment this discussion. This topic may be worth a trip to the library for me, so I could see some of the detection equipment, and get a better feel for the mechanics of light. How much of physics and astronomy depend on the understanding of light! I read about Tesla in my childhood, and the reading left me with a never ending feeling that we (as science) missed something important (an observation, a force or element). So I'm always looking for it, if only by deductive reasoning. I think that was what Tesla was doing, and I think he was on the verge of finding it. <div class="Discussion_UserSignature"> DenniSys.com </div>

xmo1 · Mar 1, 2005

Thank you. I am doing it. <div class="Discussion_UserSignature"> DenniSys.com </div>

aetherius · May 11, 2005

I am curious about the shape and movement of the photon and wave and found this thread. Let me see if I can explain my questions with a hypothetical. Start with a photon emitter in the exact center of a sphere. Next, assume that I could cover the entire sphere with photon detectors. Then, I emit a single photon from the emitter. Question 1: Would the photon be detected at only one emitter? (Assuming all are equidistant from the emitter.) Question 2: If the answer to Q1 is yes--If I replaced the detectors with people that have the ability to detect the light of a single photon, how many would see the "flash" of light associated with the single photon emission? In other words, will the people (or detectors) be able to see light from the single photon because they all detect the wave? Or, does the detection of the wave by a single person/detector cause the wave to collapse so that no others can detect it?

aetherius · May 11, 2005

Thanks ranur. It was in reading your posts in the thread on matter that raised my curiosity. When the photon is emitted at the center of the sphere does its associated wave propogate uniformly in all directions? If yes then does that imply that the volume of my hypothetical sphere and the volume of the photon wave are approximately the same at the instant the photon is detected?

aetherius · May 11, 2005

Thank you, your explanation was very clear but I'm having trouble comprehending, "The answer to this is..No. It exists in a sphere at a known distance from the emitter." Let me describe my mental image and then maybe you can see where I'm having trouble. When the photon is emitted I picture a tiny bubble like sphere that originates at the point of emission. I don't picture a photon traveling in any specific direction. Rather, I see the tiny photon sphere expanding uniformly in all directions at C. At the instant a point on the expanding sphere contacts a detector, the detector registers a detection. If that does not make sense then the rest of this is perhaps meaningless, but... I see the surface of the expanding photon sphere as rippled instead of smooth. The peaks and troughs of the ripples define the wavelength of the photon.

i_think · May 12, 2005

"I see the tiny photon sphere expanding uniformly in all directions at C. At the instant a point on the expanding sphere contacts a detector, the detector registers a detection." Does the expanding photon sphere or “shell” represent all possible positions of the photon at x units of time after emission, and contact with a detector the collapse of the probability wave? I guess that would explain the non-local behaviour of a photon, but if I picture a steady stream of photon shells expanding from the Sun, they have no way of avoiding an encounter with Mercury and collapsing before reaching any other planets further out. Only photons shells reflected off Mercury and then Venus would reach the Earth, and we would not even see the Sun directly. Sorry, but perhaps I don’t understand the idea correctly.

aetherius · May 12, 2005

Does the expanding photon sphere or “shell” represent all possible positions of the photon at x units of time after emission, and contact with a detector the collapse of the probability wave? Yes, that's what I was thinking. ...they have no way of avoiding an encounter with Mercury and collapsing before reaching any other planets further out. Exactly, that is the problem that I think wave-particle duality deals with. The wave only collapses upon detection. A simple encounter with Mercury would not count as "detection". The portion of the expanding sphere that's blocked by Mercury would simply fill back in as the sphere expands beyond Mercury. I think this view is consistent with Saiph's explanation earlier in this thread.-- When an atom's electrons de-excite, they emit electromagnetic radiation, in all directions, in the form of a spherical wave. However when one "detects" wavefronts of quantum phenomena, you detect it at only a single point. This is the source of why single photons interfere with themselves. When they are traveling unobserved, they do so as a wave, and interact with their surroundings as waves. However, when the wavefront impinges upon a detector, you detect only a single particle (the photon) in a single spot. The rest of the wave is not there. You've collapsed the wavefront as it's usually explained. However, I don't have complete grasp of the concept so I may be missing something.

siarad · May 13, 2005

<blockquote>In reply to:<hr />When the photon is emitted at the center of the sphere does its associated wave propagate uniformly in all directions?” The answer to this must be.. Yes.<hr /></blockquote> That seems to answer why a photon passes through two slits. My problem now is can the detection occur at any point of the sphere & if so does the sphere then collapse. There seems to be no way to measure this speed of collapse but if it was large enough maybe it's instantaneous therefore grossly exceeding C but as it no longer contains information would be acceptable.

siarad · May 13, 2005

That sounds logical as moving at C there is no time or space, I think, which can only exist during time. I'm glad my hair is thinning as thinking of this is making my head hot <img src="/images/icons/laugh.gif" /> <img src="/images/icons/wink.gif" />

Saiph · May 17, 2005

I think: Here's where it can get complicated, you're dealing with two different pictures. If you wish to view the model entirely as a wave, there is no issue, as the spreads out, the portion that hits mercury reflects (and is dampened, as it isn't a perfect reflector) and any light from that portion we observe appears to be emitted by mercury. The rest, that did not impede upon mercury, appears to come from the sun. This can be visualized by having a calm basin of water, with a single obstacle in it (besides the edges). Make a wave (drop something in) and you'll see it radiate, then bounce off the obstacle. Those reflected waves appear to center on the obstacle. Now, if you wish to view the model entirely as a particle, you have a slightly different picture: The sun emits a mass of particle called photons. These spread out. Those that hit mercury reflect (or are absorbed), and appear to come from it. Those that don't, appear to come from the sun. Now, the problem is: Which is it? The answer: Both. Experiments will show photons (and only photons) emitted from the sun. And they will show waves, and only waves emitted from the sun (to date no experiment has ever been created that can predict both the wave nature and particle nature of light at the same time, and directly). So we have...both. And here's where your question comes in. If one views the wave as a "probability" wave, not as a real thing, then one is created for each "photon" emitted by the sun. This probability wave does not collapse every time it is interacted with, as that is only a probability. When in reaches out to mercury, there is a probability it will be detected. If it is not detected there, the probability wave continues outwards. Probability waves can reflect without collapsing as well, and this alters the overall shape of the wave. Since the probability of the wave is tied into its shape (where th <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

i_think · May 17, 2005

Wow, thanks Saiph! I never expected such a complete answer that I could actually understand.

Saiph · May 17, 2005

No problem. That comes from having to beat my head on the subject for quite some time myself. I also love answering those types of questions, because it forces me to think it through and find a way to explain it in a coherent fashion. I.e. I learn it best by teaching (which is true for almost everyone). <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

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