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Author Topic: How can photons be localised if the universe is expanding?  (Read 47247 times)

Offline JP

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How can photons be localised if the universe is expanding?
« Reply #75 on: 12/02/2009 21:34:11 »
Are you saying that a holographic image is a 'solvable system' then?
I would have thought that it's not?
In a way it seems similar to the idea of 'synergy', that a 'whole' becomes more than its parts.
At least those that we are able to describe.

You have to be very careful in what you mean by "solvable."  An optical transmission hologram is "solvable" in the sense that all it takes to view it is to illuminate it with the appropriate reference beam. 
 

Offline yor_on

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How can photons be localised if the universe is expanding?
« Reply #76 on: 12/02/2009 21:42:39 »
'phased out' Terminated gradually
(Can't help that, blame them English, and my slightly twisted sense of humour:)

Vph = c/n
phase velocity = lights speed in a vacuum, divided by the index of refraction.
And the index of refraction would then be the change of speed and wavelength.

And wavelength is the distance following the propagation between two points in the same phase in consecutive cycles of a wave
Like if you see a sea wave and look from crest to crest. That will then be the wavelength.

As compared to frequency which decides the number of occurrences within a given time period (seconds).
All of the sea waves passings measured, using time, from an arbitrary point chosen (a buoy f.ex).

And would that then be equal to 'photons per time period' interacting with a material?
As a lower wavelength should be equal to 'fewer photons/per time period' if seen as particles?
Or:)

Thanks Lightarrow:)


 

Offline yor_on

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How can photons be localised if the universe is expanding?
« Reply #77 on: 12/02/2009 21:44:57 »
Now why are you doing this to me JP.
I was innocent before, now I'm stepping into paths where there well might be thygers.
Will I fall of the edge?

------

What fascinates me (amongst other things:) is how far we have traveled using linear ('solvable') mathematics.
So the the reference beam can be seen as a linear solution in itself?

Or is it the whole 'system' (reference beam + hologram) that becomes solvable by it?
I'm not entirely sure how you mean? If so it reminds me a little of how fermions can be 'translated' into bosons by magnetic confinements, that then change their 'properties/spin' and allow them to be introduced into a BEC.

Or do you mean that the reference beam can be seen as a linear system in itself?
No offense meant JP.
« Last Edit: 12/02/2009 22:20:14 by yor_on »
 

Offline JP

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How can photons be localised if the universe is expanding?
« Reply #78 on: 12/02/2009 23:23:16 »
Linear is a very precise mathematical term.  You really need to be asking is this system linear with respect to this type of input.  The question, "Is a hologram linear?" doesn't really mean much unless you provide more details. 

Another way to use the term linear, that you might be going for, is to ask if the underlying physics is modeled by linear differential equations.  In the case of optical holograms, the answer is yes, since Maxwell's equations are linear differential equations.

However, there are plenty of cases where the underlying physics is based on linear differential equations, but the problem at hand can't be easily solved.  Quantum mechanics is based on linear differential equations, for example.
« Last Edit: 12/02/2009 23:25:06 by jpetruccelli »
 

Offline yor_on

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How can photons be localised if the universe is expanding?
« Reply #79 on: 13/02/2009 12:27:57 »
JP I can see how one can have two non linear systems and using them together get a linear solution to a problem.
So I'm not that sure where linear ends (sort of, if it ever does:) and non linear takes over.
So yes, it depend on your choice of input.

But my impression is that all linear systems is just a 'sub species' of non linear mathematics.
If you look at the universe it is a non linear system, as far as I get it. Our Earth is a non linear system too.

But as in all mathematics every nook of it, when looking closer, seems to 'grow':)
And that I think is true for linear mathematics too.
And it is a fascinating thought that we can use linear math to describe and solve problems involving non linear systems.
 
 

Offline lightarrow

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How can photons be localised if the universe is expanding?
« Reply #80 on: 13/02/2009 12:42:01 »
'phased out' Terminated gradually
(Can't help that, blame them English, and my slightly twisted sense of humour:)

Vph = c/n
phase velocity = lights speed in a vacuum, divided by the index of refraction.
And the index of refraction would then be the change of speed and wavelength.

And wavelength is the distance following the propagation between two points in the same phase in consecutive cycles of a wave
Like if you see a sea wave and look from crest to crest. That will then be the wavelength.

As compared to frequency which decides the number of occurrences within a given time period (seconds).
All of the sea waves passings measured, using time, from an arbitrary point chosen (a buoy f.ex).
Ok.

Quote
And would that then be equal to 'photons per time period' interacting with a material?
As a lower wavelength should be equal to 'fewer photons/per time period' if seen as particles?
The number of photons of a light's beam passing through a specific surface has nothing to do with their frequency or their wavelenght, unless you fix the beam's power; if you fix it, then, lower wavelenght (= greater frequency = greater sinle photon's energy) correspond, as you say, to lower number of photons passing through the surface per unit time. The computation is very simple: beam's power going through the surface = energy per unit time passing there = (number of photons per unit time passing there)*(single photon's energy). So, if that power is 3mW = 3*10-3W and you have red photons with λ = 663nm = 6.63*10-7m --> photon's energy = hν = hc/λ = 6.63*10-34*3*108/6.63*10-7 = 3*10-19J So the number N of photons passing per unit time is: 3*10-3W/3*10-19J = 1016 photons per second. With blue photons with λ = 464nm you have instead (464/663)*1016 = 0.7*1016 photons per second.

But if you don't fix the beam's power, the number of photons cannot be determined changing the wavelenght.
« Last Edit: 13/02/2009 17:10:36 by lightarrow »
 

Offline yor_on

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How can photons be localised if the universe is expanding?
« Reply #81 on: 13/02/2009 15:43:21 »
This is very interesting, and you know what you are doing to me now?
Giving me an even bigger headache:::)))

This is the way I innocently looked at it before before meeting you 'wolfs' of the deep mathematics:)
And that surely describes you Lightarrow and that other dangerous person jpetruccelli.
Are you going to tell me that the Earth isn't flat too, huh.
No way, the edge is near.

You say that "lower wavelenght (= greater frequency = greater single photon's energy)"
But isn't a lower wavelength a 'red shifted' one?
That is, as a 'wave-system' having their 'crests' placed further away from each other in time?
And a higher 'blue shifted' wave is one where the 'crests' sits very near each other?

So I am using the wrong descriptions here?

By fixing the beams power I presume you mean its 'energy content'.
And that will then not be related to the wavelength? But then it can't have to do with the frequency either, can it?

If one say that the frequency is -a- wavelength repeatedly observed over time when passing a point?
What is the difference then? You can pick the wavelength out of any frequency it seems to me.
Can you give me a example of one single wavelength observed 'on its own' in spacetime?

When we measure the wavelength, don't we just 'lift it out' as a part of a longer 'frequency'?
And how do one then define that 'energy content', if neither waves singly or 'frequency's' defines it?

----

I could look it up, and I probably should. But it's so much more fun treating it this way.
If I get what you say right? If you have two waves, containing the same wavelength and frequency, they still can have different energy contents?

« Last Edit: 13/02/2009 16:00:52 by yor_on »
 

Offline lightarrow

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How can photons be localised if the universe is expanding?
« Reply #82 on: 13/02/2009 17:43:51 »
This is very interesting, and you know what you are doing to me now?
Giving me an even bigger headache:::)))

This is the way I innocently looked at it before before meeting you 'wolfs' of the deep mathematics:)
And that surely describes you Lightarrow and that other dangerous person jpetruccelli.
Are you going to tell me that the Earth isn't flat too, huh.
No way, the edge is near.

You say that "lower wavelenght (= greater frequency = greater single photon's energy)"
But isn't a lower wavelength a 'red shifted' one?
No, red shifted means greater wavelenght.

Quote
That is, as a 'wave-system' having their 'crests' placed further away from each other in time?
And a higher 'blue shifted' wave is one where the 'crests' sits very near each other?
Yes. Where is the problem? Lower wavelenght means less distance so the crests are near each other! So it's blue shifted and so higher frequency.

Quote
So I am using the wrong descriptions here?

By fixing the beams power I presume you mean its 'energy content'.
And that will then not be related to the wavelength? But then it can't have to do with the frequency either, can it?
This is an "emblematic" example of the importance of precise terms in physics; if you say 'energy content' you could interpret this phrase as "energy of a single photon", in which case you would be right; but you're not! The reason is that in this case you have to interpret it as "intensity" of the beam. Both "beam intensity" and "power going through a unit surface" are *precisely defined* physical concepts; with these kinds of concepts only, you can say if you are saying correct things or not. It's more boring, maybe, but it's the only way to discuss about physics (it's not a critic, just an advice  ;)).

Quote
If one say that the frequency is -a- wavelength repeatedly observed over time when passing a point?
What is the difference then? You can pick the wavelength out of any frequency it seems to me.
We are assuming to be in the void, or the things get more complicated, ok? Then, if we are in the void, light's speed = signal velocity = phase velocity = c = constant and invariant; then frequency is the number of crests passing through a specific fixed point of space and this number is proportional to the wave's speed and inversely proportional to wavelenght: the greater the wavelenght, that is the distance between two crests, the more time you have to wait to detect the second crest (because the wave's speed doesn't depend on wavelenght, in this case) and so you detect less crests per unit time, so you have a lower frequency.

Quote
Can you give me an example of one single wavelength observed 'on its own' in spacetime?
What do you mean with 'on its own'?

Quote
When we measure the wavelength, don't we just 'lift it out' as a part of a longer 'frequency'?
First, a wave can be longer or shorter, because it's a distance, but a frequency cannot be "longer"; in case it's "higher" or "greater"; second, what does 'lift it out' mean?

Quote
I could look it up, and I probably should. But it's so much more fun treating it this way.
If I get what you say right? If you have two waves, containing the same wavelength and frequency, they still can have different energy contents?
Yes, in the sense of different beam's powers. If you think about it a little bit, it's quite obvious: you can have a blue-light beam which doesn't even make you close your eyes if looked directly, and an infrared laser beam who makes a hole through 1cm of steel in a few seconds...
Which has more "energy content"?  ;)
In the first case you have very energetic *single* photons, but the number of photons emitted per unit time (= passing through a surface per unit time) is very low; in the second case it's the opposite: low energy single photons but a much higher number of them per unit time. As I said, the beam's power is the product of the two things: (energy of a single photon)*(number of photons per unit time) and this product is much greater in the second case.
You can treat it classically, without any need of photons: in the first case you have an high frequency wave with low amplitude (the height of the crests), in the second, low frequency and high amplitude; in this case however frequency doesn't matter, you don't need to know it; you compute the beam's power from amplitude only.

So, *fixing the wavelenght*, amplitude (classical analysis) is proportional to the number of photons per unit time (quantistic analysis).
« Last Edit: 13/02/2009 18:06:50 by lightarrow »
 

Offline yor_on

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How can photons be localised if the universe is expanding?
« Reply #83 on: 13/02/2009 19:45:25 »
Ok, I think we have a language collision here.
'Lower' as I saw it meant lower waves, and so 'stretched out' in time/space.

So I got confused:)
Still, we are talking about the same thing as far as I can see.

----

I wrote "Can you give me an example of one single wavelength observed 'on its own' in spacetime?"
I meant that I can't visualize a wave as something consisting of only 'crest to crest'.

As I see it, a wave is always something more than just that 'crest to crest', and only when looked at as particles can you describe it as on 'its own'. I might be wrong though, but then I really would like you to point me to a experiment showing and explaining it.


And yes, there is some 'definition problems' here:)

A frequency is what you get out of your flashlight right.
And it can be of longer or shorter duration, don't you agree? Depending on how long you need that light. Also it goes back to my question if you can define 'crest to crest' as something 'existing' at all?

That's what I meant by "When we measure the wavelength, don't we just 'lift it out' as a part of a longer 'frequency'?"

Am I that unclear when I write about it Lightarrow?
So does that mean that we have photons of different inherent 'energy level' then?
There is no singly defined 'light quanta' if I get your last explanation correct?
As 'photons' is the smallest constituent I know of.

----------
You write "you can have a blue-light beam which doesn't even make you close your eyes".
And what we see as the blue light here is its frequency, right? Would it have a very low amplitude if so?
Could that amplitude then be seen as something corresponding to the amount of photons (particle wise?)

---
With the blue-light beam you speak of an high frequency wave with low amplitude.
You seem to say that the frequency is what is corresponding to 'particle/photon density' if I get it right.
Then it's not the amplitude like I thought?

"In the second case it's the opposite: low energy single photons but a much higher number of them per unit time."
And that would be "an infrared laser beam who makes a hole through 1cm of steel in a few seconds..."
And here it will be of ", low frequency and high amplitude;"

But then you say that "in this case however frequency doesn't matter, you don't need to know it; you compute the beam's power from amplitude only." Which seems to support my first assumption of it being the amplitude after all.
Waves is a confusing concept for me:)

-----------
But looking at the wiki (photons), it states that "The Maxwell wave theory, however, does not account for all properties of light. The Maxwell theory predicts that the energy of a light wave depends only on its intensity, not on its frequency; nevertheless, several independent types of experiments show that the energy imparted by light to atoms depends only on the light's frequency, not on its intensity" http://en.wikipedia.org/wiki/Light_quantum#Historical_development which seems to support my former assumption of frequency being the thing that relates to energy, and so make that assumption I made of a BEC and redshift to be connected more or less correct?

And that concept seems to give support to the idea of 'waves' not moving at all taken from 'mathpages'?
http://www.mathpages.com/HOME/kmath210/kmath210.htm
« Last Edit: 13/02/2009 21:56:46 by yor_on »
 

Offline lightarrow

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How can photons be localised if the universe is expanding?
« Reply #84 on: 14/02/2009 16:24:52 »
Ok, I think we have a language collision here.
'Lower' as I saw it meant lower waves, and so 'stretched out' in time/space.
Try to guess why physics and mathematics uses a precisely defined language  ;)

Quote
So I got confused:)
Still, we are talking about the same thing as far as I can see.

----

I wrote "Can you give me an example of one single wavelength observed 'on its own' in spacetime?"
I meant that I can't visualize a wave as something consisting of only 'crest to crest'.

As I see it, a wave is always something more than just that 'crest to crest', and only when looked at as particles can you describe it as on 'its own'. I might be wrong though, but then I really would like you to point me to a experiment showing and explaining it.
I still can't understand what you mean; remember however that you cannot observe a photon from its frame of reference, because it hasn't (if this is what you intended with on 'its own').
An electromagnetic wave is nothing else than a field of vectors E and B in space which varies from point to point, fixing the time, and in the time, fixing the point; "field of vectors E and B in space" means just that you can detect a specific value of the electric or the magnetic force on still charges (in the case of electric force) or moving charges (magnetic force) put in that specific point.
The EM wave is not "something material with the shape of a wave" travelling through space, as in the case, e.g., of sea waves.

Quote
And yes, there is some 'definition problems' here:)

A frequency is what you get out of your flashlight right.
And it can be of longer or shorter duration, don't you agree? Depending on how long you need that light.
If you have a stoboscopic lamp, for example, the frequency of the flashes is the number of flashes per unit time (that is, per second); if this number is greater, the period, that is the interval of time between a flash and the next one, is lower: T =1/ν  where T = period and ν = frequency.

Quote
Also it goes back to my question if you can define 'crest to crest' as something 'existing' at all?
Can't understand what you mean. If you have a sinusoidal EM wave, in the point (0,0,0) (computed in metres) you have an electric field directed along the y axis and which value is 7.1 Newton/Coulomb and in the point (0,0,5) you have the electric field with the same value of 7.1 Newton/Coulomb and same direction, then the "crest to crest" distance = wavelenght is 5 metres. Where is the problem?

Quote
That's what I meant by "When we measure the wavelength, don't we just 'lift it out' as a part of a longer 'frequency'?"

Am I that unclear when I write about it Lightarrow?
So does that mean that we have photons of different inherent 'energy level' then?
But you already know that photons can have different energy because of the different frequency of the wave.

Quote
There is no singly defined 'light quanta' if I get your last explanation correct?
As 'photons' is the smallest constituent I know of.
Light quanta in the sense of "an absolute minimum quantity of energy", there is not, infact. Light "quanta" means that *given the frequency ν of the EM wave*, there is a minimum value of the energy that you can detect from the wave: E = h*ν. But if you don't fix the frequency, then this minimum doesn't exist, in the sense that it's not a non-zero number but tends to zero (when the frequency tends to zero).

Quote
You write "you can have a blue-light beam which doesn't even make you close your eyes".
And what we see as the blue light here is its frequency, right? Would it have a very low amplitude if so?
Exactly.

Quote
Could that amplitude then be seen as something corresponding to the amount of photons (particle wise?)
Yes, in the sense that: lower amplitude = less photons. Remember that this equation is valid only if you fix the frequency.

Quote
With the blue-light beam you speak of an high frequency wave with low amplitude.
You seem to say that the frequency is what is corresponding to 'particle/photon density' if I get it right.
Then it's not the amplitude like I thought?
No, the frequency in this case has *nothing to do* with the number of photons; the frequency here is an attribute *of the wave*. If you want to talk of "the number of photons passing through an area per unit time" then you should avoid the term "frequency" in this context, or you will get a mess.

Quote
"In the second case it's the opposite: low energy single photons but a much higher number of them per unit time."
And that would be "an infrared laser beam who makes a hole through 1cm of steel in a few seconds..."
And here it will be of ", low frequency and high amplitude;"

But then you say that "in this case however frequency doesn't matter, you don't need to know it; you compute the beam's power from amplitude only." Which seems to support my first assumption of it being the amplitude after all.
I've already written this, however I'll repeat it: you can compute the beam's power in two different ways, either one or the other; the first is classically, just with the wave's amplitude; the second is quantistic and in this case you need to know 2 things, not one: the wave's frequency AND the number of photons passing through an area per unit time.

Quote
But looking at the wiki (photons), it states that "The Maxwell wave theory, however, does not account for all properties of light. The Maxwell theory predicts that the energy of a light wave depends only on its intensity, not on its frequency; nevertheless, several independent types of experiments show that the energy imparted by light to atoms depends only on the light's frequency, not on its intensity" http://en.wikipedia.org/wiki/Light_quantum#Historical_development which seems to support my former assumption of frequency being the thing that relates to energy, and so make that assumption I made of a BEC and redshift to be connected more or less correct?

And that concept seems to give support to the idea of 'waves' not moving at all taken from 'mathpages'?
http://www.mathpages.com/HOME/kmath210/kmath210.htm
Yor_on, if you just want to compute the beam's power, you don't need to know the frequency, you don't even need to know the names "quanta" or "Planck constant" and so on, you only need classical electromagnetic theory (Maxwell's equations) that is, the wave amplitude.

The quantistic properties of light comes out when you want to describe *others* situations:

1. Black body radiation
2. Photoelectric effect
3. Compton effect
4. Entropy of the EM radiation
5. Anti-bunching effect
...
« Last Edit: 14/02/2009 16:36:50 by lightarrow »
 

Offline yor_on

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How can photons be localised if the universe is expanding?
« Reply #85 on: 14/02/2009 16:54:18 »
I will need to read you again Lightarrow:)
But that's no news, you are quite good at putting my ideas on their head(s).

Just as a 'by side' while reading up on your thoughts here:)
http://www.sciencenews.org/view/generic/id/36794/title/Photons_caught_in_the_act
Sounds interesting, I thought?

----------

Lightarrow, I'm of two minds when it comes to energy of a photon (or more, 'minds' that means:).
To me it would make sense if all photons was of the same quantified energy level internally.

And the difference seen would be an aspect of 'time' and 'frames of reference)
That's also one of the reasons why I try to understand the concept of frames and where the 'delimits' of that concept may be. Also I like to relate everything to the 'spacetime' I personally observe, and so I might at times seem rather, ah, ever seen a tortoise promenading? He's real fast :) compared to some on this site:) *whistles as he innocently takes stock of himself, finding his 'pace of mind' quite appropriate, eh, to himself that is:)*

You see, as this guy(gal?) at mathpages stated you might be able to see photons/waves as not moving at all. And if you do so you will have to ask yourself how that goes together with the concept of photons/waves being of different 'energy' internally, as seen in their interactions with you observing. I have no answer for that. But it seems as a easier concept if they was seen as being of the same 'light quanta' ?

But yes, I like to think that I, at times :), can see the more 'main stream' definition too. On the other hand, what defines a boson is also that strange possibility of it 'superimposing' itself on other bosons, right? So I'm definitively of more than one mind here.

But I still haven't answered your questions:)
« Last Edit: 14/02/2009 17:37:02 by yor_on »
 

Offline lightarrow

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How can photons be localised if the universe is expanding?
« Reply #86 on: 14/02/2009 19:52:51 »
I will need to read you again Lightarrow:)
But that's no news, you are quite good at putting my ideas on their head(s).

Just as a 'by side' while reading up on your thoughts here:)
http://www.sciencenews.org/view/generic/id/36794/title/Photons_caught_in_the_act
Sounds interesting, I thought?

----------

Lightarrow, I'm of two minds when it comes to energy of a photon (or more, 'minds' that means:).
To me it would make sense if all photons was of the same quantified energy level internally.
"Internally" it has no meaning, if, with "internally" you mean "in a frame of reference where the photon is at rest". If instead you mean "its minimum energy" then it's zero, because a particle's minimum energy is mc2 where m is the mass, and a photon's mass is zero.

Take any particle, with mass or without mass; if it's total energy is E, then its associated frequency is E/h. This quantity depends on the frame of reference, so it will have a minimum value. The energy is minimum when you don't have kinetic energy but only rest energy (mc2). So its minimum frequency = its invariant frequency = mc2/h.

Quote
And the difference seen would be an aspect of 'time' and 'frames of reference)
That's also one of the reasons why I try to understand the concept of frames and where the 'delimits' of that concept may be. Also I like to relate everything to the 'spacetime' I personally observe, and so I might at times seem rather, ah, ever seen a tortoise promenading? He's real fast :) compared to some on this site:) *whistles as he innocently takes stock of himself, finding his 'pace of mind' quite appropriate, eh, to himself that is:)*
The problem is no much of "speed" but of "taking the right path" of reasonings.

Quote
You see, as this guy(gal?) at mathpages stated you might be able to see photons/waves as not moving at all.
Don't let fascinating words confuse you and study the principles, it's better!  ;)
 

Offline yor_on

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How can photons be localised if the universe is expanding?
« Reply #87 on: 14/02/2009 20:22:07 »
Where would that frame be Lightarrow?
"in a frame of reference where the photon is at rest"
That statement should be understood as 'observed from a frame moving at the same velocity relative the photon', shouldn't it?

A photon don't have any mass defined to it, but it do have a momentum, correct?
And it do have some sort of energy.

Kill me, I wrote charge here:)

-------

Do you agree with this reasoning?
Its an answer to what happens when you 'shoot a ray of light through the prism'.

---Quote--

If photons were massive, we would indeed have a big problem here!

However, they are massless and the energy of a single photon is related
only to its frequency.  The frequency is unchanged during all the
transitions.

Energy = (Plancks constant) * (frequency)

It is also possible to re-write the energy (or frequency) in terms of
the velocity and wavelength of the light. The velocity changes during
the transitions, going from fast to slow and back to fast as the photons
leave the substance.  However, the wavelength of the light also changes
by the same factor during the transitions going from long to short and
back to long.

E = (Planck's constant) * (Velocity) / (wavelength)

   = (Planck's Constant) * (Velocity/Index of Refraction) /
(Wavelength/Index of Refraction)

The Index of Refraction cancels itself out and you are left with the same
before and after any transition.  All in all, the energy remains constant.


---------------------
Michael S. Pierce
Materials Science Division
Argonne National Laboratory
=========================================================End quote===========

It seems reasonable to me?

Although I have a little trouble with 'the energy of a single photon is related only to its frequency'.
Does photons seen as one particle have a 'frequency'?

Shouldn't it just be 'the energy of photons is related only to their frequency, if described as a wave' if so?
As I still don't get how you define/describe the equivalence of one single photon to a wave?

---

A massless 'particle' like the photon is defined (amongst other properties:) by its momentum, which seems to act similar to mass in that it can 'push' on 'matter'.

Matter is a unique state to me, clearly defined by the Pauli exclusion principle who states 'that any two fermions can't share/occupy that same quantum state simultaneously'. So they have a very strict 'structure' keeping their 'shape' in space and time, with the uncertainty principle (HUP) defining the limitations of observing that 'structure'.

--

By changing fermions (particles) as Helium four atoms half-integer spins into integer spins you can get the same spin as a Boson, who always seems to be of 'whole' spins. If you create a strong magnetic field and then line up those atoms together those atoms will 'pair up' to each other thereby creating 'whole' spins becoming like photons inside a BEC, right.

So by expending a little energy you can get some types of particles to act as bosons.
How much energy do you need to create stable particles?
http://www.physics.ucdavis.edu/~chertok/CMS/Interviews/sacbee2-cmsstory.pdf


« Last Edit: 14/02/2009 21:50:11 by yor_on »
 

Offline lightarrow

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How can photons be localised if the universe is expanding?
« Reply #88 on: 15/02/2009 10:35:24 »
Where would that frame be Lightarrow?
"in a frame of reference where the photon is at rest"
That statement should be understood as 'observed from a frame moving at the same velocity relative the photon', shouldn't it?
Say things well: not 'observed from a frame moving at the same velocity relative the photon', but 'observed from a frame moving at the same velocity of the photon' and so at a relative velocity of zero respect to the photon. But such a frame doesn't exist; whatever your speed, you will always see the photon as travelling at c.

Quote
A photon don't have any mass defined to it, but it do have a momentum, correct?
And it do have some sort of energy.
Yes, and so?

Quote
Do you agree with this reasoning?
Its an answer to what happens when you 'shoot a ray of light through the prism'.

---Quote--

If photons were massive, we would indeed have a big problem here!

However, they are massless and the energy of a single photon is related
only to its frequency.  The frequency is unchanged during all the
transitions.

Energy = (Plancks constant) * (frequency)

It is also possible to re-write the energy (or frequency) in terms of
the velocity and wavelength of the light. The velocity changes during
the transitions, going from fast to slow and back to fast as the photons
leave the substance.  However, the wavelength of the light also changes
by the same factor during the transitions going from long to short and
back to long.

E = (Planck's constant) * (Velocity) / (wavelength)

   = (Planck's Constant) * (Velocity/Index of Refraction) /
(Wavelength/Index of Refraction)

The Index of Refraction cancels itself out and you are left with the same
before and after any transition.  All in all, the energy remains constant.
Exactly. This shows you that the wavelenght decreases when the wave enters the glass, but frequency remains the same (as I wrote you in another post, but you almost didn't believe it  :)).

Quote
It seems reasonable to me?

Although I have a little trouble with 'the energy of a single photon is related only to its frequency'.
Does photons seen as one particle have a 'frequency'?

Shouldn't it just be 'the energy of photons is related only to their frequency, if described as a wave' if so?
As I still don't get how you define/describe the equivalence of one single photon to a wave?
You are beginning to understand why it's not possible to simply say "light is an EM wave" or "light is a collection of moving particles"...
Yor_on, if things here were simple, they even hadn't the need to create a new theory, that is Quantum Mechanics and the need to derange all the previously known physics...
*No one* knows how a photon is made or what is made of; we only know that it has an energy proportional to the frequency of the EM wave and just a few of other things. When we talk of the "photon's frequency" we are actually talking of the "EM wave frequency".

Quote
A massless 'particle' like the photon is defined (amongst other properties:) by its momentum, which seems to act similar to mass in that it can 'push' on 'matter'.
No, it's not similar to mass, this is a property of energy, not of mass; you have an exchange of momentum every time energy (in whatever form) is exchanged between two systems. The fact that at school they told you momentum is m*v is because they couldn't write you the correct (but more complicated) formula: p = Sqrt[(E/c)2 - (mc)2]. This formula is valid even if m = 0: in this last case you get p = E/c.

Quote
Matter is a unique state to me, clearly defined by the Pauli exclusion principle who states 'that any two fermions can't share/occupy that same quantum state simultaneously'. So they have a very strict 'structure' keeping their 'shape' in space and time, with the uncertainty principle (HUP) defining the limitations of observing that 'structure'.

By changing fermions (particles) as Helium four atoms half-integer spins into integer spins you can get the same spin as a Boson, who always seems to be of 'whole' spins. If you create a strong magnetic field and then line up those atoms together those atoms will 'pair up' to each other thereby creating 'whole' spins becoming like photons inside a BEC, right.

So by expending a little energy you can get some types of particles to act as bosons.
How much energy do you need to create stable particles?
http://www.physics.ucdavis.edu/~chertok/CMS/Interviews/sacbee2-cmsstory.pdf
Sorry, I don't know the answer to this question.
 

Offline yor_on

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How can photons be localised if the universe is expanding?
« Reply #89 on: 15/02/2009 13:30:48 »
How do you talk about 'EM wave frequency 'without there being any 'charge' to the photon?
As for the duality, I knew about it Lightarrow. It's just that I like to look at it in a 'particle way' too:)

I wrote "A massless 'particle' like the photon is defined (amongst other properties:) by its momentum, which seems to act similar to mass in that it can 'push' on 'matter'." upon which you answered "No, it's not similar to mass, this is a property of energy, not of mass;"

Strange, What are you then seeing as transferring into 'mass' inside that 'perfectly reflecting box'?
Wherein we have that photon bouncing. It can't be the energy, can it. If it was, then the photon should interact with 'matter' and so be 'exchanged' into another photon of 'lower energy level' ad infinitum until it disappears (mainstream wise). If one look at it as particles then they should be absorbed by this material (glass f ex.) and then constantly released as a new photon, then, in the very end, the photon should die.

Or can one see photons as 'particles' bouncing off perfectly reflecting surface of 'matter'?
Without ever interacting and so loosing its energy?
 
When looked on as a wave inside a perfectly reflecting sphere (better than a cube, don't you agree?) then the wave just should continue to bounce though, it seems to me. And the only thing acting on this sphere from the wave(photon) should be its 'momentum'. That is if we assume the wave being perfectly reflected without any interference etc.

But then there is the problem of this 'momentum' losing something at every bump? Light will always travel at 'c' in a vacuum so let's assume that there is a perfect vacuum inside that sphere. So can this original 'momentum' get lost? If we see waves as being of different frequencies and being able to translate into those other frequency's then the answer seems to be, a yes, right?
As 'momentum' is directly connected to energy.

But it still seems to me that 'momentum' is what gives the 'mass'?

As for when the photon 'is at rest' we seem to agree then:)
Namely, never.

--

Me writing "As I still don't get how you define/describe the equivalence of one single photon to a wave?" then that is a question I've asked others before, in other forums, in other ways, without getting a satisfactory answer.
So looking back at it, I do think I understood it before this Lightarrow.
But thanks all the same.
« Last Edit: 15/02/2009 14:55:18 by yor_on »
 

Offline yor_on

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How can photons be localised if the universe is expanding?
« Reply #90 on: 15/02/2009 16:23:56 »
So how is it possible for one wavelength to have different frequency's

Wavelength = what you see looking on a sea wave from crest to crest in time.
Frequency = same wavelength as above, repeated definitely/indefinitely in time.

Looking at it like this it seems very strange right:)
That a wavelength isn't the same as the frequency.
But it gets its explanation when we look at what I didn't write about here.
Namely time and different density.

When that light passes through that prism it encounters a new sort of 'density'.
As glass is transparent we know that most of the light gets through, but we also knows that it gets 'slowed down'.
'Slowed down' here means that the same frequency (waves), if we placed our 'reference point/buoy' inside that prism letting those waves interact with the prisms density/atoms, would need a longer time passing past our 'reference point/buoy'.

And when waves on sea take a longer time past that buoy, what do we call that? A calmer sea with a slower frequency, right.
And as time hasn't slowed down the only thing left is the frequency of those waves, although their wavelength still is the same.

Or is it?
Won't the wavelength also appear longer in time, just as the frequency?
So the proportionality between a given frequency and its 'individual' wavelength(s) is still the same, don't you agree:)
The only thing changed here is the medium it travels through, and the time we observe it taking.

So can the same wavelength have different frequencies then?
Yes, and No :)

To me it seems that it should be seen as a 'compression' or 'decompressive/expandable' effect not affecting the inherent proportionality.
A little like an accordion when played :).

The formula used describing this is: Speed of medium c = wavelength lambda times frequency f.
Where lambda is just another expression for wavelength. And f is frequency.
So lambda is λ
So 'Speed of light inside a medium' is written... c = λ * f ...

And wavelength then is the distance between repeating units of a propagating wave of a given frequency.

« Last Edit: 15/02/2009 16:45:30 by yor_on »
 

Offline lightarrow

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How can photons be localised if the universe is expanding?
« Reply #91 on: 15/02/2009 17:47:16 »
How do you talk about 'EM wave frequency 'without there being any 'charge' to the photon?
What is a "charge" to the photon?

Quote
As for the duality, I knew about it Lightarrow. It's just that I like to look at it in a 'particle way' too:)

I wrote "A massless 'particle' like the photon is defined (amongst other properties:) by its momentum, which seems to act similar to mass in that it can 'push' on 'matter'." upon which you answered "No, it's not similar to mass, this is a property of energy, not of mass;"

Strange, What are you then seeing as transferring into 'mass' inside that 'perfectly reflecting box'?
Wherein we have that photon bouncing. It can't be the energy, can it. If it was, then the photon should interact with 'matter' and so be 'exchanged' into another photon of 'lower energy level' ad infinitum until it disappears (mainstream wise). If one look at it as particles then they should be absorbed by this material (glass f ex.) and then constantly released as a new photon, then, in the very end, the photon should die.

Or can one see photons as 'particles' bouncing off perfectly reflecting surface of 'matter'?
Without ever interacting and so loosing its energy?
 
When looked on as a wave inside a perfectly reflecting sphere (better than a cube, don't you agree?) then the wave just should continue to bounce though, it seems to me. And the only thing acting on this sphere from the wave(photon) should be its 'momentum'. That is if we assume the wave being perfectly reflected without any interference etc.

But then there is the problem of this 'momentum' losing something at every bump? Light will always travel at 'c' in a vacuum so let's assume that there is a perfect vacuum inside that sphere. So can this original 'momentum' get lost? If we see waves as being of different frequencies and being able to translate into those other frequency's then the answer seems to be, a yes, right?
As 'momentum' is directly connected to energy.

But it still seems to me that 'momentum' is what gives the 'mass'?
I haven't said that "momentum is energy", I said that there is a transfer of momentum when there is a transfer of energy; it's not the same thing. When energy moves from one system to another, there is momentum, that's all. No need of mass. Concerning light, you can see it classically (electromagnetic waves have momentum) or quantistically (photons have momentum but no mass); the result is the same: momentum but no mass.
 

Offline lightarrow

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« Reply #92 on: 15/02/2009 17:56:27 »
So how is it possible for one wavelength to have different frequency's

Wavelength = what you see looking on a sea wave from crest to crest in time.
Frequency = same wavelength as above, repeated definitely/indefinitely in time.

Looking at it like this it seems very strange right:)
That a wavelength isn't the same as the frequency.
But it gets its explanation when we look at what I didn't write about here.
Namely time and different density.

When that light passes through that prism it encounters a new sort of 'density'.
As glass is transparent we know that most of the light gets through, but we also knows that it gets 'slowed down'.
'Slowed down' here means that the same frequency (waves), if we placed our 'reference point/buoy' inside that prism letting those waves interact with the prisms density/atoms, would need a longer time passing past our 'reference point/buoy'.

And when waves on sea take a longer time past that buoy, what do we call that? A calmer sea with a slower frequency, right.
And as time hasn't slowed down the only thing left is the frequency of those waves, although their wavelength still is the same.

Or is it?
Won't the wavelength also appear longer in time, just as the frequency?
So the proportionality between a given frequency and its 'individual' wavelength(s) is still the same, don't you agree:)
The only thing changed here is the medium it travels through, and the time we observe it taking.
If you want an analogy with matter waves, you have to take two different media, water and something else, ground for example; when the wave goes from water to ground, the frequency is the same but the wavelenght varies according to the different phase velocity in the new medium: if the velocity increases, the wavelenght increases. Remember that frequency and wavelenght are inversely proportional only if they refer to the same medium, not to different media (as I had already written).

Quote
So can the same wavelength have different frequencies then?
Yes, and No :)

To me it seems that it should be seen as a 'compression' or 'decompressive/expandable' effect not affecting the inherent proportionality.
A little like an accordion when played :).

The formula used describing this is: Speed of medium c = wavelength lambda times frequency f.
Where lambda is just another expression for wavelength. And f is frequency.
So lambda is λ
So 'Speed of light inside a medium' is written... c = λ * f ...

And wavelength then is the distance between repeating units of a propagating wave of a given frequency.
Speed of light inside a medium is not c but another value c/n (but you already know it...  :))
 

Offline yor_on

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How can photons be localised if the universe is expanding?
« Reply #93 on: 15/02/2009 18:07:28 »
You are right in that 'c' is defined as the speed of light in a vacuum.
Here I would say that it is used to define 'light' in general, not defining its medium.
http://wiki.answers.com/Q/How_can_the_same_wavelength_have_different_frequencies.

So Lightarrow?
Are you saying that my explanation is wrong?
It's not perfect, but it comes as near as I need to be, to see the idea.
What exactly do you see as wrong in it?
 

Offline lightarrow

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« Reply #94 on: 15/02/2009 18:17:10 »
You are right in that 'c' is defined as the speed of light in a vacuum.
Here I would say that it is used to define 'light' in general, not defining its medium.
http://wiki.answers.com/Q/How_can_the_same_wavelength_have_different_frequencies.

So Lightarrow?
Are you saying that my explanation is wrong?
It's not perfect, but it comes as near as I need to be, to see the idea.
What exactly do you see as wrong in it?
The relation Vph = λ*f must be interpreted *in a specific medium*: λ in that medium and f *in the same medium* NOT λ in a medium and f in another medium! If you change the medium, you have to change the phase velocity Vph which is written in that relation.

Example1: medium = glass, n=1.5. Vph = c/1.5 = 200,000 km/s. If the wave has a frequency of 5*1014Hz, than the wavelenght is *fixed* and you can't vary it, λglass = (2*108m/s)/5*1014Hz = 4*10-7 = 400nm

Example2: the first medium is void and the second is glass, n=1.5. The wave has still a frequency of 5*1014Hz, so the wavelenght in the void is: λvoid = (3*108m/s)/5*1014Hz = 600nm.

The frequency is the same in the two cases, but λ is different. If λ were the same, then the frequency would have been different.
« Last Edit: 15/02/2009 18:31:34 by lightarrow »
 

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How can photons be localised if the universe is expanding?
« Reply #95 on: 15/02/2009 18:44:49 »
I asked "How do you talk about 'EM wave frequency 'without there being any 'charge' to the photon?"
Then you ask "What is a "charge" to the photon?"

As far as I've understood there are no experimental proof for photons having a charge?
And that's why I wondered about from where the photons 'EM wave frequency' came.
As an EM wave is supposed to have a charge, as I saw it?

And the second paragraph of yours, wherein you state that you never have said that "momentum is energy" I totally agree.
It's me, questioning the idea of energy being converted to mass inside that perfectly reflecting sphere.

It was as I wrote 'A massless 'particle' like the photon is defined (amongst other properties:) by its momentum, which seems to act similar to mass in that it can 'push' on 'matter'.' And then reading your answer
" No, it's not similar to mass, this is a property of energy, not of mass; "

That started me wondering anew about how momentum/energy was transferred by  photons inside that 'box/sphere'.



 

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How can photons be localised if the universe is expanding?
« Reply #96 on: 15/02/2009 18:53:44 »
Reading you about my explanation i see that you feel that I've mixed different mediums?

I've used a wave inside a prism, passing a thought up 'buoy', and then I draw a parallel to what one can see at a normal sea?
So is my explanation wrong then?

You are free to correct it, using words, explaining the concept behind this mathematics then:)

------

Reading us again Lightarrow I get a distinct feeling that if we both had your mathematic knowledge it would be somewhat easier to agree, or agree to disagree too perhaps::))
But as it is, you have it and I don't, which, when using words, seems to lead us to misinterpretations at times?

Then again, one should be able to set words to ones concepts and ideas, don't you agree?
And so, I try:)




« Last Edit: 15/02/2009 21:32:44 by yor_on »
 

Offline lightarrow

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« Reply #97 on: 16/02/2009 15:19:11 »
I asked "How do you talk about 'EM wave frequency 'without there being any 'charge' to the photon?"
Then you ask "What is a "charge" to the photon?"

As far as I've understood there are no experimental proof for photons having a charge?
And that's why I wondered about from where the photons 'EM wave frequency' came.
As an EM wave is supposed to have a charge, as I saw it?
No. Who gave you the permission to think this?  :)   Light, photons, have nothing to do with electric charges. Light is the mediator of the force between two or more charges, which are the sources of the force (and of the field).

Quote
And the second paragraph of yours, wherein you state that you never have said that "momentum is energy" I totally agree.
It's me, questioning the idea of energy being converted to mass inside that perfectly reflecting sphere.

It was as I wrote 'A massless 'particle' like the photon is defined (amongst other properties:) by its momentum, which seems to act similar to mass in that it can 'push' on 'matter'.' And then reading your answer
" No, it's not similar to mass, this is a property of energy, not of mass; "

That started me wondering anew about how momentum/energy was transferred by  photons inside that 'box/sphere'.
In the same way as light transfers momentum when it reflects off a mirror: in a (coventional) mirror light interacts with the free electrons of the metal which covers the glass; the wave's electric field interacts making these electrons oscillate along the field, up and down (I assume the wave propagates horizontally); then the magnetic field acts on the moving electrons through the Lorentz force, pushing them ahead, along the wave's direction. Note that I've only made a classical description here, not used QM at all.
 

Offline lightarrow

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« Reply #98 on: 16/02/2009 15:41:09 »
Reading you about my explanation i see that you feel that I've mixed different mediums?

I've used a wave inside a prism, passing a thought up 'buoy', and then I draw a parallel to what one can see at a normal sea?
So is my explanation wrong then?

You are free to correct it, using words, explaining the concept behind this mathematics then:)
Yor_on, I have two difficulties here: the first is that I'm not extremely good in english language, and the second that you express what you want to say with intuitive, non-rigorous concepts; the result is that I barely understand what you mean every time, and sometimes I really don't know what to answer you for this reason.  :)
This said, I try to anser to what it could be your problem, but I'm not sure at all I will answer what you really wanted to know...
To understand what is the frequency of a wave, you don't have to move inside the medium (material or non-material) and you don't have to measure any speed of the wave: you only have to stay in a specific point inside the medium, or put a detector there, and measure how many times you detect an effect in 1 second. Example with sea waves: you are on the sea, in a fixed point, inside a boat; it's dark and you can't see anything, you can only hear the sound of the water colliding with the boat; you notice that this sound repets  periodically in the time; you take a chronometer and you count 1 of these sounds every 2 seconds. Then the frequency is 1/2 = 0.5 Hz. It would be exactly the same with an EM wave's frequency, you only would use a different detector.

Quote
Reading us again Lightarrow I get a distinct feeling that if we both had your mathematic knowledge it would be somewhat easier to agree, or agree to disagree too perhaps::))
But as it is, you have it and I don't, which, when using words, seems to lead us to misinterpretations at times?
Then again, one should be able to set words to ones concepts and ideas, don't you agree?
And so, I try:)
The best we can do is to use the same language; since we are talking about physics, the language is that of the physics. Unfortunately it doesn't exist a dictionary English/Physics - Physics/English and to understand the language we have first to understand the meanings, that is the concepts (this is also true for many words of a different language, for example from italian to english ecc., but in most of the cases a dictionary Italian/English - English Italian is enough; with physics it's not so).
« Last Edit: 16/02/2009 15:45:50 by lightarrow »
 

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How can photons be localised if the universe is expanding?
« Reply #99 on: 16/02/2009 17:46:21 »
Yes Lightarrow, I agree to what you write. 'Language' can be hell to interpret at times.
When I was 'discussing' wavelength contra frequency, I meant that my conclusion was, that even though you can say that one wavelength have different frequencys, the 'proportionality' between the wavelength and the frequency remains the same.

Is that wrong?

You just 'stretched' that original frequency by putting it through the prisms 'density' and so making the wave crests become wider apart in that wave, if i understood it right.

And in a way that's unintuitive, as one might expect the opposite:)

For example, if I exchange that 'wave' and the prism, and instead try to describe it by a spring and some thick mud. Now I'll try to press that spring through that mud.
What would happen is that the spring would 'compact' by being pressed through that mud, not 'expand', right:)

So trying to see light as some sort of 'spring' meeting a higher density (mud) doesn't work very well.
But seeing it in terms of density and time does, to me that is:)

If one accept that light will be slowed down by a higher density then it will need more time traversing that density. And those waves will then, in that medium, be observed as changing to a 'greater' frequency as it moves through.

And with 'greater' I mean that the crests and troughs between the waves will have a greater distance between them (red shift) when seen passing a static point of observation inside that medium.

Am I seeing this correct Lightarrow?


----------

Ahh, but here we have Swedish/English/Italian and then Italian/English/Swedish.
Can you see the possibilities inherent here:)

It's a bleeding miracle that we agree on anything:::)))
I love it :)

Especially when you share your mathematics, then the mix gets explosive...
(Yes, I'm joking Lightarrow:). You are one of the few taking the time to explain your mathematical concepts)
« Last Edit: 16/02/2009 17:57:07 by yor_on »
 

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