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Author Topic: a better understanding of photons. . . .  (Read 3077 times)

lyner

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a better understanding of photons. . . .
« on: 28/11/2007 14:32:41 »
physicsforums.com/archive/index.php/t-32102.html
If you read this, or at least some of it, you may be enlightened or you may be confused; particularly if you have always regarded photons as little bullets.
It certainly comes to no real conclusions - but can you?

There are a lot of mis conceptions about the little devils (or are they little?).
It seems clear that the only 'packet-like' thing about them is the amount of energy each one is associated with.
The old 'corpuscular' theory of light is responsible for a lot of the problems we seem to have in visualising what is going on. Heisenberg spoils most of the pictures of them which we carry in our minds.
How big are they?
How long do they last?
How is the energy carried?
Comments please, you budding Physicists.


 

Offline Mr Andrew

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a better understanding of photons. . . .
« Reply #1 on: 28/11/2007 23:04:35 »
I have said before, in the thread on gravitons and elsewhere, that photons are simply units of energy equal to hν joules where h is Planck's constant and ν is the frequency of the light.  As for whether they are waves or particles or both, I think they are neither.  I believe that light is a wave with a definite amount of energy (one photon) and when we detect the wave the photon of energy is transferred to the detector in some way (like excitation of electrons in atoms).  How big are they?  They have no spatial dimensions, just like joules do not.  They are units of energy.  How long do they last?  They do not have duration but we can say that light waves have a photon of energy from the instant they are created to the instant that we detect them.  How is this photon of energy carried?  In the oscillations of the EM fields in free space or another material.

Anyone else have a different answer?
 

lyner

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a better understanding of photons. . . .
« Reply #2 on: 29/11/2007 23:43:23 »
If an atom in a gas, decays and emits a photon and there are two other atoms, on either side of the emitting atom, each of the neighbours is likely to absorb the photon. So, the photon, before it committed itself to being absorbed by one particular atom, was spreading out in all directions. It would have had a 'radiation pattern' equivalent to a short dipole - ring doughnut shaped; circular around an equatorial plane and 1-sin(angle) pattern in the North South direction.
Not like a bullet at all, until the process has actually happened. The probability pattern of a particular atom in a particular direction being the one to absorb the wave is given by the classical wave pattern of the radiating atom.
Trying to explain phenomena using photons rather than waves can, at times, be extremely lumpy!
I can never understand why people are so in love with photons and want to use them to explain everything , at all costs. Anyone would think that em waves were REALLY photons.
They certainly have their place - but only when it's appropriate.
I don't know of many lens designers who use particles to explain and design their optical systems.
 

Offline JP

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a better understanding of photons. . . .
« Reply #3 on: 30/11/2007 19:04:20 »
I think a lot of the confusion comes from drawing an analogy between the Schrodinger equation and the Helmholtz equation in optics.  The Schrodingr equation arises when you try to calculate the quantum behavior of particles by treating them in a wavelike manner.  Since all light is waves, the Helmholtz equation begins by treating light as a wave.  It so happens that these two equations have the same form.  Since they're mathematically equivalent, you can get the same "classical" bullet-like trajectories out of them.  For the Schrodinger equation, these represent the paths of the particles in the classical limit.  For the Helmholtz equation, these represent light "rays," which are artificial constructs used to model light fields in certain limits. 

Photons show up when you actually quantize the electromagnetic field, which requires more rigorous mathematics than the Schrodinger (or Helmholtz) equation.
 

Offline lightarrow

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a better understanding of photons. . . .
« Reply #4 on: 30/11/2007 22:08:04 »
Trying to explain phenomena using photons rather than waves can, at times, be extremely lumpy!
I can never understand why people are so in love with photons and want to use them to explain everything , at all costs.
(Maybe Freud would have something to say about it  ;))

Quote
Anyone would think that em waves were REALLY photons.
They certainly have their place - but only when it's appropriate.
I don't know of many lens designers who use particles to explain and design their optical systems.
 

Offline lightarrow

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a better understanding of photons. . . .
« Reply #5 on: 30/11/2007 22:32:48 »
physicsforums.com/archive/index.php/t-32102.html
If you read this, or at least some of it, you may be enlightened or you may be confused; particularly if you have always regarded photons as little bullets.
It certainly comes to no real conclusions - but can you?

There are a lot of mis conceptions about the little devils (or are they little?).
It seems clear that the only 'packet-like' thing about them is the amount of energy each one is associated with.
The old 'corpuscular' theory of light is responsible for a lot of the problems we seem to have in visualising what is going on. Heisenberg spoils most of the pictures of them which we carry in our minds.
How big are they?
According to the present knowledge, the photon dimension doesn't exist, which doesn't mean that it's zero, but that the concept itself doesn't exist! Someone says a photon's dimension can really have a physical meaning: inside a box with reflecting walls, for example, a photon's dimension is that box's dimension; however, to introduce such concepts the theory itself should be modified (don't know how, hope he will find it).

Quote
How long do they last?
For us, between emission and absorption; for them...zero seconds, since (cdt)2 = dx2 (proper time = dτ = (1/c)sqrt[(cdt)2 - dx2]).

Quote
How is the energy carried?
It depends on the ref. frame
 

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a better understanding of photons. . . .
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