Direction of Radiation Emitted from Atoms

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Offline Mr Andrew

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« on: 10/12/2007 16:38:32 »
Light is a directional wave...it is a linear wave which propagates in a specific direction.  What in the atom determines which direction it emits radiation in?  Is it dipole moment or something to do with the nucleus?  I can't seem to find anything on the internet.  Anyone have an idea?
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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Offline Soul Surfer

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« Reply #1 on: 11/12/2007 09:05:48 »
In general for atoms in a gas there is no particular direction in which light is emitted but for example in a gas laser when there is already radiation of the same wavelength going in a particular direction atoms that are on the verge of emitting radiation will do it in the same (or possibly the opposite) direction ie along a single line.  That is how lasers work.  In solids the situation may be more complex and molecular structure can affect the propagation of light but this is usually seen in absorption because solids don't emit light unless there is an unusually high level of excitation as in solid state lasers  which are very structured and have mirrored reflectors like gas lasers
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lyner

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« Reply #2 on: 11/12/2007 17:51:11 »
Quote
Light is a directional wave...it is a linear wave which propagates in a specific direction.
Not strictly true. It is, essentially, a spherical wave. It travels in the same speed in all directions. All 'rays' cover a range of angles, however small. The wavefront is spherical because each part of it has traveled the same distance after a given time.
As Soul  Surfer says, atoms don't have a directional light pattern. You can use many sources, radiating in phase, as do the atoms in a laser, or a number of radiating dipoles with the same radio frequency signal to produce a directional beam. Alternatively you can use a pinhole, lens or tube to select or direct the energy to go in a particular  direction. The wavefront is still spherical but its energy is restricted to a narrow range of directions.

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Offline JP

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« Reply #3 on: 11/12/2007 20:42:34 »
Quote
Light is a directional wave...it is a linear wave which propagates in a specific direction.
Not strictly true. It is, essentially, a spherical wave. It travels in the same speed in all directions. All 'rays' cover a range of angles, however small. The wavefront is spherical because each part of it has traveled the same distance after a given time.

Both are true.  A light field can be written in terms of summing a bunch of plane waves (directional waves) or it can be written as the sum of a bunch of spherical waves emanating from each source point.

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Offline Mr Andrew

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« Reply #4 on: 11/12/2007 22:09:19 »
Here's my problem with the spherical wave model:  when a wave (which has a photon of energy) is detected it loses energy.  Since it can only have integer multiples of hf in energy, it loses all of its energy.  Therefore, if I put a detector one meter from a light source, none of that light would ever reach me if I stand on the other side of the light source but 1.1 m away.  However, two people can see the same light bulb from different distances...how does that work?

-I know you can model light like a spherical waves but are they really like that or are they directional and it is only when there are many of them that you can treat light like a spherical wave?
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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Offline JP

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« Reply #5 on: 11/12/2007 23:03:34 »
Classically light is a wave.  If you put a detector in the way, it will catch all the energy from the wave that would pass through the space it occupies.  Two people can see the light bulb from different distances because they are both absorbing different parts of the wave.  If I stood right behind someone else, they would absorb all the light coming to me, and I wouldn't see the light bulb.

If you go to quantum mechanics, the light comes in packets which move about according to probability waves.  for each photon that gets emitted, each detector has a probability of seeing it that's proportional to the amount of probability wave it intercepts.  If we both look at source that emits a single photon, only one of us will see it.  However, as lots of photons are emitted, we'll each see some in proportion to the probability we expect to see. 

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lyner

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« Reply #6 on: 11/12/2007 23:36:49 »
Quote
Light is a directional wave...it is a linear wave which propagates in a specific direction.
Not strictly true. It is, essentially, a spherical wave. It travels in the same speed in all directions. All 'rays' cover a range of angles, however small. The wavefront is spherical because each part of it has traveled the same distance after a given time.

Both are true.  A light field can be written in terms of summing a bunch of plane waves (directional waves) or it can be written as the sum of a bunch of spherical waves emanating from each source point.
I don't know where the plane wave analysis is used - I only know of Huygens (secondary wavelet) construction, which is, effectively, standard diffraction treatment. In standard diffraction methods, the intergrals used assume that a wavefront consists of a line of sources and these are summed at a subsequent point or surface. Where does the plane wave analysis take its origin for the plane waves? It would have to be from infinitely wide sources and it would be difficult to describe what happens at an aperture / object. It would be interesting to see how it's done.
When you are far enough away to treat the source  as a point then parts of the wave which are in  phase (i.e. wavefronts) must be equal distances from the source - that makes the wavefront spherical. However you choose to analyse it (Huygens or other), the spherical nature of the resulting wavefront must exist unless you say the speed is different in different directions.
'Rays' as such, don't  / can't really exist - they are just constructional tools - like magnetic lines of force.

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Offline Soul Surfer

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« Reply #7 on: 11/12/2007 23:52:12 »
All electromagnetic radiation behaves in the same way.  if you want to see the statistical and classical wave behaviour it id best to look at radio and reasonable intensities of light where the many photons emitted randomly in all directions from the source produce spherical wavefronts with limited coherency.  to appreciate what is really going on in the quantum world look at much higher energy gamma ray emission from an excited nucleus.  Here to the radiation can be emitted in any direction but in doing so the nucleus also reacts because of the momentum of the emitted energy.
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Offline JP

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« Reply #8 on: 12/12/2007 01:38:56 »
@ sophiecentaur:  The thesis work I'm doing right now uses a lot of plane wave analysis of fields.  The way it works is that you look at any field satisfying the Helmholtz equation.  If you solve the equation in Cartesian coordinates, you get plane-waves as a basic solution.  If you don't care about evanescent components of the field (i.e. the exponentially decaying components that decay by a few wavelengths from the source), you can write any field as a sum of weighted plane waves traveling in all different directions.  If you're mathematically inclined, the plane waves form a "complete basis" for describing fields.  The mathematics is also analogous to working in "spatial frequency" (i.e. the Fourier spatial transform of the field).

You're right that an aperture/object gets somewhat difficult.  You end up having to take a Fourier transform of the object in order to see how it influences the plane waves.  However, for propagation in free space, it's often quite a bit easier to deal with plane waves instead of diffraction integrals.  And it provides a nice framework for defining "rays." 

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Offline lightarrow

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« Reply #9 on: 12/12/2007 13:15:54 »
Quote
Light is a directional wave...it is a linear wave which propagates in a specific direction.
Not strictly true. It is, essentially, a spherical wave. It travels in the same speed in all directions. All 'rays' cover a range of angles, however small. The wavefront is spherical because each part of it has traveled the same distance after a given time.

Both are true.  A light field can be written in terms of summing a bunch of plane waves (directional waves) or it can be written as the sum of a bunch of spherical waves emanating from each source point.
Ok, however, quantum mechanically, light is made of photons and if there are only a few photons (per unit time ecc.) in the region of space considered, you can't use that description anylonger and the directionality of light is lost (in the absence of specific apparatus/equipments).

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lyner

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« Reply #10 on: 12/12/2007 20:14:37 »
1. Lightarrow: don't you just need to replace 'intensity' with 'probability' when you are dealing with small numbers? The same pattern will apply at a distance.

2. jpetruccelli: what is the application for your work? It sounds interesting.

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Offline JP

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« Reply #11 on: 12/12/2007 21:04:59 »
2. jpetruccelli: what is the application for your work? It sounds interesting.

My particular area of research involves a creature called a "Wigner function:" http://en.wikipedia.org/wiki/Wigner_function.  It basically allows you to take any optical field that satisfies the Helmholtz equation, and map it onto a set of rays by specifying the weight assigned to each ray.  It has some funny properties (such as some rays having negative weight), but these properties are required in order to describe interference and diffraction. 

The purpose of this is to take an optical field, do one (relatively simple) computation to assign weight to a bunch of rays, and then use ray-tracing (which is much more efficient than diffraction integrals) to propagate the field through whatever optical system we want.  It's also a nice way of deriving an exact ray-optical picture of field propagation by starting from Maxwell's equations.

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Offline Mr Andrew

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« Reply #12 on: 13/12/2007 02:13:49 »
Ok, so light is a spherical probability wave (it has no direction but outwards)...it is spread out over a spherical surface and any detector on that surface has a probability of detecting that photon inversely proportional to the size of the surface.  That is all well and good and easy to picture but what is it about light that makes it behave that way?  What mechanism decides where the light should be absorbed?  Probabilities only come into play when we don't know enough about the initial conditions to predict what outcome will prevail.  If I have two cups on a table and I put a coin under one while you are not looking, there is a 50% chance that the coin will be under the cup you choose.  However, if you peak over your shoulder you know with absolute certainty where the coin is even though there are still two cups and only one coin.  Therefore, can we really rule out light having a directionality?  The only systems we ever deal with are systems with a lot of photons emitted in all directions.  Dividing our chances of absorbing x amount of photons at position P by the number of photons there are gives a probability of absorbing an individual photon at that position...but isn't that just like the coin and the cup?-I don't know in what direction (under what cup) the photon (coin) is but I know the probability that I am going to absorb it.
--Life is the greatest experiment that any person will ever conduct.  It should be treated with the same scientific method as any other experiment.

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Offline JP

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« Reply #13 on: 13/12/2007 03:46:13 »
Probabilities only come into play when we don't know enough about the initial conditions to predict what outcome will prevail.  If I have two cups on a table and I put a coin under one while you are not looking, there is a 50% chance that the coin will be under the cup you choose.  However, if you peak over your shoulder you know with absolute certainty where the coin is even though there are still two cups and only one coin.

In quantum mechanics, even if you know as much as possible about the initial conditions, you can't predict the outcome exactly.  In other words, a quantum mechanical coin is under both cups at the same time, with probability of 50% until you peek.  No matter how early you try to peek into the system, it's got a 50% chance of showing up in either cup.

If there is a directionality, it shows up in the probability being higher that photons will be detected in a certain direction.  This would be like if your coin was loaded to land head up more often.  You couldn't tell this from one flip alone, but if you take a bunch of flips, you'd notice the pattern.  Similarly, you need to measure a bunch of photons in order to determine if your atom is emitting a field in a given direction.

But all this is confusing the question.  You want to look at the classical field direction emitted by the atom, which is an averaging over a lot of photons.  In that case, if you assume the atom is a small dipole, you probably could direct the light by somehow forcing the atom's dipole moment to align in a particular direction.  You would almost certainly get a pretty wide beam of light, since there's quantum fluctuations if you're dealing with a single atom, but you could probably end up relating the direction of the light to the polarization of the atom.

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Offline lightarrow

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« Reply #14 on: 13/12/2007 08:42:58 »
1. Lightarrow: don't you just need to replace 'intensity' with 'probability' when you are dealing with small numbers? The same pattern will apply at a distance.
I think it's the same. Intensity is the power of electromagnetic radiation going through a unit area; you can measure this power counting the number of photons hitting that area in the unit time.

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lyner

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« Reply #15 on: 13/12/2007 10:22:30 »
Whichever way you look at this, there are some difficult concepts. You can't get away with a simple, overall classical approach (of course) so you need to introduce, either some 'super-directivity' of the atom as an emitter of waves or the concept of quantum entanglement, so that only one of the possible target systems gets the photon energy.
My background of classical em treatment of antennae makes me very conscious of needing to treat the atom as a well behaved antenna, where possible. This implies a very broad beam with, perhaps, a few nulls in its pattern. I am then, as a consequence,  stuck with the other problem about what happens at the reveiving end of the event and there you have to bring in some quantum oddities. But  entanglement is already a factor which is acknowledged to apply in other quantum circumstances, so  I am not really rocking the boat by introducing it here.
If people insist on photons as little bullets then they have to suggest why a radiator would not be subjected to classical wave considerations which apply to a 'small' (or even 'not incredibly huge') system.

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lyner

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« Reply #16 on: 13/12/2007 10:33:30 »

But all this is confusing the question.  You want to look at the classical field direction emitted by the atom, which is an averaging over a lot of photons.  In that case, if you assume the atom is a small dipole, you probably could direct the light by somehow forcing the atom's dipole moment to align in a particular direction.  You would almost certainly get a pretty wide beam of light, since there's quantum fluctuations if you're dealing with a single atom, but you could probably end up relating the direction of the light to the polarization of the atom.
Yes, a very broad directivity and the direction of the (dipole axis)  nulls would be the only control that you would have. No narrowly directed beams available here.

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lyner

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« Reply #17 on: 13/12/2007 10:41:29 »
Ok, so light is a spherical probability wave (it has no direction but outwards)...it is spread out over a spherical surface and any detector on that surface has a probability of detecting that photon inversely proportional to the size of the surface.  That is all well and good and easy to picture but what is it about light that makes it behave that way?  What mechanism decides where the light should be absorbed? 
My argument is that, until  the uncertainty of which target system gets the photon energy is resolved, all you can discuss is the probablity (as in all quantum problems). Once there is an outcome (i.e. one system gets the energy) , that is communicated to all of space and no other outcome is possible (i.e. no one else gets the energy). This idea is ok, I think, because it applies in many  situations.

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« Reply #18 on: 13/12/2007 10:46:34 »
Not strictly true. It is, essentially, a spherical wave. It travels in the same speed in all directions. All 'rays' cover a range of angles, however small. The wavefront is spherical because each part of it has traveled the same distance after a given time.
As Soul  Surfer says, atoms don't have a directional light pattern. You can use many sources, radiating in phase, as do the atoms in a laser, or a number of radiating dipoles with the same radio frequency signal to produce a directional beam. Alternatively you can use a pinhole, lens or tube to select or direct the energy to go in a particular  direction. The wavefront is still spherical but its energy is restricted to a narrow range of directions.

OK, so this works from a quantum perspective.

Forgetting for the moment about the particulate perspective, how does the work from the relativistic perspective?  According to relativity, an EM wave is nothing but a manifestation of a contraction of space happening perpendicular to the direction of motion of a charge - so in which direction is the charge moving?  To have a spherical wavefront, would not the charge must somehow be moving in 3 dimensions at once?  Is this feasible?

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Offline lightarrow

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« Reply #19 on: 13/12/2007 12:33:23 »
Forgetting for the moment about the particulate perspective, how does the work from the relativistic perspective?  According to relativity, an EM wave is nothing but a manifestation of a contraction of space happening perpendicular to the direction of motion of a charge - so in which direction is the charge moving?  To have a spherical wavefront, would not the charge must somehow be moving in 3 dimensions at once?  Is this feasible?
Can you explain this? (It's new to me!)

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Offline lightarrow

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« Reply #20 on: 13/12/2007 12:39:20 »
Ok, so light is a spherical probability wave (it has no direction but outwards)...it is spread out over a spherical surface and any detector on that surface has a probability of detecting that photon inversely proportional to the size of the surface.  That is all well and good and easy to picture but what is it about light that makes it behave that way?  What mechanism decides where the light should be absorbed?
Good question. They are debating about it from ~ 80 years ago, but the answer is still "you can't know". As sophiecentaur wrote, you can only know the probability of detecting, not where the photon will be detected. If the wave optics computation tells you that in every point of a 1 m2 screen there is (e.g.) the same value of light intensity and you send one photon at a time, after many detections you will see a uniform pattern of light, but it's completely impossible (according to the present QM description) to predict which detector will "click" first, which second ecc.
« Last Edit: 13/12/2007 12:51:11 by lightarrow »

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« Reply #21 on: 13/12/2007 12:53:06 »
My argument is that, until  the uncertainty of which target system gets the photon energy is resolved, all you can discuss is the probablity (as in all quantum problems). Once there is an outcome (i.e. one system gets the energy) , that is communicated to all of space and no other outcome is possible (i.e. no one else gets the energy). This idea is ok, I think, because it applies in many  situations.
Faster than light's speed? Doesn't it seem "quite strange"? [:)]
« Last Edit: 13/12/2007 12:54:37 by lightarrow »

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Offline lightarrow

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« Reply #22 on: 13/12/2007 12:59:14 »
Whichever way you look at this, there are some difficult concepts. You can't get away with a simple, overall classical approach (of course) so you need to introduce, either some 'super-directivity' of the atom as an emitter of waves or the concept of quantum entanglement, so that only one of the possible target systems gets the photon energy.
I haven't understood this concept. Can you explain better?

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« Reply #23 on: 14/12/2007 15:30:58 »
What I mean by a 'super directive' radiator is ok, I assume, so you have a problem with the other idea(?)
Quantum entanglement is an idea which deals with two (or more) quantum systems, which are in unknown states but where the states are mutually connected - for instance, the spin of two electrons with, otherwise, identical quantum numbers. You don't know the spin of either electron - so it could be either value. If you measure / force the spin of one electron, you have immediately determined its state and that instantly determines the state of the other one. You can take two such systems miles apart and, once you have revealed/ resolved/measured the state of one, the other's condition is, instantly, determined and it will behave according to its newly determined state. The information about this is, magically, you might say, communicated through space. It's like a pair of Schroedinger cats in two boxes; when one is found alive, the other is, suddently dead and vice versa.
The principle is used in quantum computers - and they actually work - so I think we need to accept it as a valid idea.
http://en.wikipedia.org/wiki/Quantum_entanglement
Once you accept the principle, it allows you to accept my choice of ways of looking at the photon thing. If you can't go along with that, the bullet theory is preferable - despite the anti - classical behaviour it implies.

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« Reply #24 on: 14/12/2007 15:43:23 »
  To have a spherical wavefront, would not the charge must somehow be moving in 3 dimensions at once?  Is this feasible?
the radius of this sphere is equal to the distance from the source; the further you get from the source, the more nearly equal are the distances from each part of the source (it can be regarded as  a point source, eventually) and, as the wave spreads out in all directions at c, the wavefront will be a sphere - same radius in all directions. For a large radius, of course, this is more or less a plane in the region of the detector.
The above only refers to the phase - it does not imply omnidirectionality of amplitude. In fact, your  (valid, at first sight) objection is dealt with by allowing / insisting that there must be nulls in some directions in the radiation pattern. There is a throrem, somewhere, which deals with just that point - can't remember where I read it, tho.


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Offline lightarrow

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« Reply #25 on: 14/12/2007 20:10:25 »
What I mean by a 'super directive' radiator is ok, I assume, so you have a problem with the other idea(?)
No, I haven't understood why you say that to explain which detector will reveal the photon, you have to introduce high directivity or entanglement. I know what is entanglement, but it's not clear to me how this concept would explain where the photon will be reveald on the screen. I would have called "hidden variables"; did you mean something like that?

Personally, I like Rovelli's (and others) "Relational" interpretation:
http://arxiv.org/abs/quant-ph/0604064
Essentially, it says that you can't separate a quantum system's properties from the quantum system which measures them.
« Last Edit: 14/12/2007 20:25:09 by lightarrow »

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lyner

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« Reply #26 on: 15/12/2007 00:06:20 »
I suppose what I am saying is that either there is a decision made at the source about the direction it sends the photon or in the choice between detectors as to which one detects it.
An analogy would be a person having two (acoustic) speaking tubes, going to two destinations. He could choose, randomly, to speak down either tube( first option) OR, there could be an electronic circuit connecting (almost instantaneously) the two remote earpieces and controlling which of them was open at any one time (second option).  The result would be, in both cases, that one, but not both of the listeners could hear the message - on a random basis. You would have no way of knowing which system was at work if you couldn't get inside the system and examine it more closely.
I favour the second option because I am not happy with an option which requires extremely non-classical behaviour of a wave. The second has its problems, too, of course but it has precedents. Is that clearer - or do you just not agree with my distinction between the two?

Quote
can't separate a quantum system's properties from the quantum system which measures them.
I guess that could be it, then. The whole system could be, at its simplest, one emitter and two possible detectors. Once one detector has detected the source, it has resolved the uncertainty so the other receiver cannot detect it. In a real situation, this could imply a 'total' web of entanglement for all possible detectors - or everything. . . . but why not?
btw, it seems that your link is restricted to subscribers.

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« Reply #27 on: 15/12/2007 02:25:43 »
Forgetting for the moment about the particulate perspective, how does the work from the relativistic perspective?  According to relativity, an EM wave is nothing but a manifestation of a contraction of space happening perpendicular to the direction of motion of a charge - so in which direction is the charge moving?  To have a spherical wavefront, would not the charge must somehow be moving in 3 dimensions at once?  Is this feasible?
Can you explain this? (It's new to me!)

Sorry, I probably did not explain myself very well.

Magnetism is a manifestation of special relativity acting on a moving electric field.  Thus electromagnetism is also a manifestation of relativity acting upon an oscillating electric field.

The point is that for special relativity to be applied, you must have something that has a velocity, which means it has a vector (one dimension, not three).  A spherical wave front has no vector.

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Offline lightarrow

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« Reply #28 on: 15/12/2007 08:17:03 »
Forgetting for the moment about the particulate perspective, how does the work from the relativistic perspective?  According to relativity, an EM wave is nothing but a manifestation of a contraction of space happening perpendicular to the direction of motion of a charge - so in which direction is the charge moving?  To have a spherical wavefront, would not the charge must somehow be moving in 3 dimensions at once?  Is this feasible?
Can you explain this? (It's new to me!)

Sorry, I probably did not explain myself very well.

Magnetism is a manifestation of special relativity acting on a moving electric field.  Thus electromagnetism is also a manifestation of relativity acting upon an oscillating electric field.

The point is that for special relativity to be applied, you must have something that has a velocity, which means it has a vector (one dimension, not three).  A spherical wave front has no vector.

Yes, it has.
Spherical wave:
E = (E0/r)ei(kr - ωt)

k = wave vector; it has the direction of the propagating wave in every point of space; its modulus is: k = 2π/λ; λ = wavelenght.

ω = 2πf; f = frequency; in the void: ω = ck

r = position vector; from the source of the wave (source point) to the position where you want to compute the fields (field point).

E0 is a vector with constant modulus and perpendicular to k.
« Last Edit: 15/12/2007 18:32:15 by lightarrow »

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Offline lightarrow

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« Reply #29 on: 15/12/2007 08:21:58 »
[...]
btw, it seems that your link is restricted to subscribers.
Doesn't it work if you click on "pdf", for example?
Try if you can see the document in .pdf directly from this link:
http://arxiv.org/PS_cache/quant-ph/pdf/0604/0604064v3.pdf

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Offline lightarrow

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« Reply #30 on: 15/12/2007 10:20:56 »
I suppose what I am saying is that either there is a decision made at the source about the direction it sends the photon or in the choice between detectors as to which one detects it.
An analogy would be a person having two (acoustic) speaking tubes, going to two destinations. He could choose, randomly, to speak down either tube( first option) OR, there could be an electronic circuit connecting (almost instantaneously) the two remote earpieces and controlling which of them was open at any one time (second option).  The result would be, in both cases, that one, but not both of the listeners could hear the message - on a random basis. You would have no way of knowing which system was at work if you couldn't get inside the system and examine it more closely.
I favour the second option because I am not happy with an option which requires extremely non-classical behaviour of a wave. The second has its problems, too, of course but it has precedents. Is that clearer - or do you just not agree with my distinction between the two?
Ok, now it's clear what you intended. However your second explanation seems problematic to me: you can decide to construct the detecting apparatus much later than the signal has been sent (e.g. we constructed light detectors billions of years after than light has been sent from very distant stars). How can you imagine a previous "connection" between source and detector?

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« Reply #31 on: 15/12/2007 11:09:58 »
That's why I made my subsequent remark that everything must be connected to everything else by this quantum-entangling net.
Constructing the detectors would involve connecting to the net at the time of construction. Bringing a detector into existence would involve slotting it into this net; in fact 'existence' as such, would mean connection into the net.
The decision/probability wave would, effectively, be spreading out from the source and the region where the photon would be most likely to be detected would be spreading outwards at what we call the speed of light. The entanglement net would be over / near the surface of this (distorted, because of interaction with matter) sphere.

The more I think about this, the more I see it necessary to have something like this mechanism or else you have to decide that the actual size of a photon is much less than the system it interacts with. If it has larger dimensions, then you still have to explain what happens when it makes a choice between two adjacent atoms. When it has chosen one, what does it do to tell the others that it has chosen? Does it 'swerve' at the last minute towards the one it has chosen?

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« Reply #32 on: 15/12/2007 18:12:34 »
That's why I made my subsequent remark that everything must be connected to everything else by this quantum-entangling net.
Constructing the detectors would involve connecting to the net at the time of construction. Bringing a detector into existence would involve slotting it into this net; in fact 'existence' as such, would mean connection into the net.
The decision/probability wave would, effectively, be spreading out from the source and the region where the photon would be most likely to be detected would be spreading outwards at what we call the speed of light. The entanglement net would be over / near the surface of this (distorted, because of interaction with matter) sphere.

The more I think about this, the more I see it necessary to have something like this mechanism or else you have to decide that the actual size of a photon is much less than the system it interacts with. If it has larger dimensions, then you still have to explain what happens when it makes a choice between two adjacent atoms. When it has chosen one, what does it do to tell the others that it has chosen? Does it 'swerve' at the last minute towards the one it has chosen?
Let's see if I have understood: you say that a preexisting net connects the location of the source S to a specific location in space D, so that, if a detector is subsequently put in that location D, that will be the detector which will detect the photon and no others. Did I understand your idea?

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« Reply #33 on: 15/12/2007 18:26:26 »
Spherical wave:
E = E0(r/r2)ei(kr - ωt).
Sorry, I should have written:
E = (E0/r)ei(kr - ωt)
where E0 is a vector perpendicular to k.
I have corrected my previous post.
« Last Edit: 15/12/2007 18:30:18 by lightarrow »

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lyner

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« Reply #34 on: 15/12/2007 21:47:51 »
Quote
if a detector is subsequently put in that location D, that will be the detector which will detect the photon and no others.
That's about what I meant; once a photon has been detected then nothing else can detect it. The uncertainty is only resolved once the photon has actually been detected so the information about its detection must emanate at that time, to everything else, instantly ( or very much quicker than c).

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Offline JP

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« Reply #35 on: 16/12/2007 05:30:18 »
Spherical wave:
E = E0(r/r2)ei(kr - ωt).
Sorry, I should have written:
E = (E0/r)ei(kr - ωt)
where E0 is a vector perpendicular to k.
I have corrected my previous post.

I think there's an error in that equation.  Spherical waves should be spherically symmetric, and can't depend on a vector r, but only on the radial position, r=|r|.  What's usually called a spherical wave should be written as

E=E0/r  exp[i(kr-ωt)],
(where I think E0 is perpendicular to the outward radial direction at every point, but I'd have to double check).

What you've written is a plane wave traveling in direction k and with an amplitude decaying as 1/r. 

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« Reply #36 on: 16/12/2007 18:09:04 »
Spherical wave:
E = E0(r/r2)ei(kr - ωt).
Sorry, I should have written:
E = (E0/r)ei(kr - ωt)
where E0 is a vector perpendicular to k.
I have corrected my previous post.

I think there's an error in that equation.  Spherical waves should be spherically symmetric, and can't depend on a vector r, but only on the radial position, r=|r|.  What's usually called a spherical wave should be written as

E=E0/r  exp[i(kr-ωt)],
(where I think E0 is perpendicular to the outward radial direction at every point, but I'd have to double check).

What you've written is a plane wave traveling in direction k and with an amplitude decaying as 1/r. 

You are right.
I intended to write the equation of a spherical wave wich source is not in the origin of coordinates, but I made a mistake.

If the source of the spherical waves is in r0 and r is the position vector, then the equation should be (I hope it's correct now!):

E = E0/|r - r0| ei(k|r - r0| - ωt)

where |a| indicates the modulus of a vector a.

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« Reply #37 on: 16/12/2007 20:03:25 »
Now you've sorted your vectors out, what do think of the entanglement net thing?
btw, what's wrong with having a vector r?
It's pointing in the right direction and it's the right length,  - what more do you want?
The E vector would be normal to it. Or is my maths too sloppy?

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« Reply #38 on: 17/12/2007 04:11:50 »
Now you've sorted your vectors out, what do think of the entanglement net thing?
I just want to make sure I follow you correctly: you're saying that you place two detectors around an atom, and release a photon.  You then entangle the photon and the detectors such that you can get either a photon/detector "hit" on detector 1, or you can get one on detector 2, but not both?

Quote
btw, what's wrong with having a vector r?
It's pointing in the right direction and it's the right length,  - what more do you want?
The E vector would be normal to it. Or is my maths too sloppy?

If you're talking about the spherical/plane wave discussion above, the two cases arrive from different assumptions.  Both solutions start from the Helmholtz equation (which is the wave equation for light of a single frequency).  If you assume nothing in particular about the system and solve it in Cartesian coordinates, the basic components of your field will be plane waves, each of which have a given direction vector k and their polarization is perpendicular to that vector.  You can make fields by adding a bunch of these together.

If you assume your system is spherically symmetric, two of your three dimensions drop out (you can rotate up/down and left/right as much as you want, and nothing will change).  Because of this symmetry, the fields being radiated out from the source are going to look identical in all directions, so they can only depend on the distance, r, you are from the source (a scalar quantity).  The basic solution under this symmetry is the spherical wave that lightarrow explained above.

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« Reply #39 on: 17/12/2007 12:33:43 »
Now you've sorted your vectors out, what do think of the entanglement net thing?
You also wrote that detection information will travel faster than light to all the other detectors. How would this happen?

Anyway, I have a different interpretation. Maybe the EM wave is really present in every location of space and so simultaneously on every detector, but only one of them will "click" because in that instant, statistically, it has the (slightly) greatest sensitivity of all of them. Increasing the intensity of light will increase the probability that a detector will "click" and so you will have a greater frequency of detection on the screen = greater number of photons arrived in the unit time.
« Last Edit: 17/12/2007 12:48:07 by lightarrow »

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lyner

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« Reply #40 on: 18/12/2007 18:59:04 »
But how does one detector know that the photon is not available because another one has just taken all its energy? Th two detectors could be in different galaxies, in different directions but the same distance from the  source. They must be, in some way, aware of each other,  quasi-instantaneously.
I don't have a problem with this idea of entanglement - it's no worse than some of the other ideas in modern Physics. I am warming to it. . . .
« Last Edit: 18/12/2007 19:01:17 by sophiecentaur »

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lyner

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« Reply #41 on: 18/12/2007 19:30:18 »

If you're talking about the spherical/plane wave discussion above, the two cases arrive from different assumptions.  Both solutions start from the Helmholtz equation (which is the wave equation for light of a single frequency).  If you assume nothing in particular about the system and solve it in Cartesian coordinates, the basic components of your field will be plane waves, each of which have a given direction vector k and their polarization is perpendicular to that vector.  You can make fields by adding a bunch of these together.

If you assume your system is spherically symmetric, two of your three dimensions drop out (you can rotate up/down and left/right as much as you want, and nothing will change).  Because of this symmetry, the fields being radiated out from the source are going to look identical in all directions, so they can only depend on the distance, r, you are from the source (a scalar quantity).  The basic solution under this symmetry is the spherical wave that lightarrow explained above.
Yes - it makes more sense to talk of a spherical wave.

The entanglement idea seems, to me, to be more and more essential for any of this to work at all - if there is to be a transition from quantum - small scale and discrete energy - and classical - large scale and a continuum of  energy.

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« Reply #42 on: 18/12/2007 22:23:20 »
But how does one detector know that the photon is not available because another one has just taken all its energy? Th two detectors could be in different galaxies, in different directions but the same distance from the  source. They must be, in some way, aware of each other,  quasi-instantaneously.

I send, from here on Earth, a single photon away in the space. Somwhere, at a great distance, just one of the bilions of billions of potential detectors will receive it. Very strange, it's true. But now I ask you this question: how do you know that you have sent exactly one photon, before detecting it?
 
This is how I see it (and I'm aware I can have an high probability to be wrong):
The photon exists in a detector only. You switch on a light source here and then you decrease its intensity so that just one detector at a time will "click". That's the ONLY thing we know for sure. Where are the photons?

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« Reply #43 on: 18/12/2007 22:36:32 »
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The photon exists in a detector only.
Yes; I see your point.
You can only be sure it WAS there when you saw the detector click.
However, you can very much detect individual gamma ray photons.
You could know that a gamma photon had been emitted by looking for an appropriate  nuclear decay and detecting the beta or alpha emissions, without actually detecting the gamma photon. So I am not sure that you are correct for all cases.

As for your 'single photon' scenario. Why would you need to know it was just one photon? You could be sure there was at least one photon.
In any case, we mustn't fall into the trap of asking "what exactly is happening?" We know there is not a complete answer to this , nor will there ever be.

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« Reply #44 on: 18/12/2007 23:40:10 »
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The photon exists in a detector only.
Yes; I see your point.
You can only be sure it WAS there when you saw the detector click.
However, you can very much detect individual gamma ray photons.
You could know that a gamma photon had been emitted by looking for an appropriate  nuclear decay and detecting the beta or alpha emissions, without actually detecting the gamma photon. So I am not sure that you are correct for all cases.

As for your 'single photon' scenario. Why would you need to know it was just one photon? You could be sure there was at least one photon.
In any case, we mustn't fall into the trap of asking "what exactly is happening?" We know there is not a complete answer to this , nor will there ever be.


Your is an interesting critic.
I'm now wondering what can we say about how a detector works in detail. I know that you have to tune its amplification so that it's not too low or too high; in the last case the detector will "click" even with the source switched off!

Someone says this is due to quantum fluctuations of the EM field in the void (don't know if it's the correct explanation, however). In this case maybe the detectors must be tuned to register the same average energy which is sent from the source? So the fact only one detector will click would be a mere consequence of that tuning? I'm in a deep speculation here.
« Last Edit: 18/12/2007 23:42:30 by lightarrow »

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« Reply #45 on: 19/12/2007 02:39:08 »
I've thought about this for a while, and I think it's going to be far easier to explain this as collapse of the wave function rather than entanglement.  The problem might lie in trying to talk about photons, which are so strange.  If you're doing the famous 2-pinhole experiment with electrons and you don't put a detector at the pinholes, the electrons "pass through" both pinholes at once (they're in a superposition of a state passing through pinhole 1 and a state passing through pinhole 2).  The mathematical notation for these two states being added is:

a|1>+b|2>

  Now let's say you turn on detectors at the 2 pinholes.  If you detect the electron at a given pinhole, it immediately collapses to a state of only passing through that pinhole.  If you measure it at pinhole 1 (and there's a probability of |a|2=a*a of doing so), it collapses as:

a|1>+b|2> → |1>

The photon's a bit funkier than an electron because you don't start with a particle and quantize it.  You start with a field and define the photon as it's "minimal excitation," but the same formalism should hold.  In other words, you should be able to write the photon as the sum of states hitting all detectors plus some extra states that will not interact with either detector:

a|1>+b|2>+∑|states missing the detectors>

If either detector gets a "hit," then it collapses to the state striking that detector.  In other words, if it hits detector 1, this state collapses as:

a|1>+b|2>+∑|states missing the detectors> → |1>

You can probably manage a description of things using entanglement, but I believe it would be extremely cumbersome to deal with mathematically.  You'd end up having to entangle every possible state of the photon with every possible state of the detector, which is going to be a lot of states.

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« Reply #46 on: 19/12/2007 10:27:41 »
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a|1>+b|2>+∑|states missing the detectors> → |1>
That notation expresses my idea well. The only thing is that it doesn't explain what is happening when it is 'resolved' by one detector..
The fact that  it collapses into   a|1 → |1 needs, somehow, to be communicated to all the detectors which were included in the first expression, instantly (?). The collapse must be instantaneous. I am happy with that if you are.

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lyner

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« Reply #47 on: 19/12/2007 10:46:59 »
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I'm now wondering what can we say about how a detector works in detail. I know that you have to tune its amplification so that it's not too low or too high; in the last case the detector will "click" even with the source switched off!
You are implying a model which sounds a bit too much like an 'amplifier' for me - I look upon it as a more passive device with, as you say 'tuning' involved.
The idea of resonance was given to me way back in Uni and works for me. I well remember a mechanical experiment with coupled resonators exchanging energy back and forth via their loose coupling; the beat between their natural frequencies corresponding the rate of  the energy passing back and forth.
When I think of an atom emitting a photon, I picture a similar process, with an oscillation ( in the form of an electron as a standing wave) coupling to an em wave (the photon). I picture the reverse process for absorption. Because of the quantum nature of the components of the model, it only happens once, unlike the coupled pendulums. The 'length' of the photon would relate to the time for the energy transfer from atom to em wave, which would, classically, be influenced by the Q factor  (impedance of  reactive components inside the atomic oscillator/ resistance corresponding to the coupling factor).  The Q factor varies for different atoms, I believe; the widths of different spectral lines would show this.

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« Reply #48 on: 19/12/2007 13:15:14 »
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I'm now wondering what can we say about how a detector works in detail. I know that you have to tune its amplification so that it's not too low or too high; in the last case the detector will "click" even with the source switched off!
You are implying a model which sounds a bit too much like an 'amplifier' for me - I look upon it as a more passive device with, as you say 'tuning' involved.

My question is: In the case of one "click" at a time of the detector, how do you know that you have detected exactly the energy that has been sent from the source and not a quantum fluctuation of the void?

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lyner

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« Reply #49 on: 19/12/2007 15:40:22 »
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quantum fluctuation of the void
You have introduced another idea here.
If you are suggesting that an emitted photon's energy is, somehow, absorbed into a general 'bank' of energy, which can turn up somewhere else in a position and time which are related by c, then that's fair enough. It's quite possible that you couldn't tell the difference. This depends on how complete the new idea / model is and how near it fits measurements to-date.

I think you are bringing in an irrelevant argument about effectively  'knowing which photon you detected'. QM doesn't let us assert anything like that but there are plenty of experiments which could link a significant number of received photons with a particular source; take a lamp and a light cell down a coal mine and you can be pretty well sure where the photons all came from! By that I mean that a photon model  which we have been discussing would reasonably indicate that the energy from the source was the energy detected. That's as much as you could say, though. Your alternative model could possibly give an explanation.
What I am aiming for is a model which is as close as possible to a classical one. This  corresponds to a reductionist attitude which characterises most of our Science. On a largely qualitative basis, and without violating much of the quantitative stuff I am familiar with, the model in my mind does this - allowing wave theory, Newton, Fourier and others to apply where possible and for the QM ideas to tend to the classical, in the limit. The only problem is the entanglement idea (just one small step for the brain??!!)
As soon as you enter into the Zero Point Energy world and quantum fluctuations, it gets harder and harder to keep one foot on firm ground (the Science of things we can measure).  Connecting the two worlds is beyond me at the moment but I keep reading.
I may argue very hard at times but I am always open to well founded suggestions.

The main thing that worries me about quantum fluctuations is the open ended possibility of even more loony discussions with fewer and fewer contributors on whom we can rely.