0 Members and 2 Guests are viewing this topic.
Ok another thought.If we say that a photon if red shifted, that is, as seen from our frame of observation being of a lesser 'energy' content.We also say that its wave is longer, right.So in a BEC where we see photons as frozen/still that wave should be?Infinite??
But doesn't it have to be treated as a red shift?
If we are discussing it from a wave perspective?If I don't see it as a redshift of those waves, then how should I see it SC?In what other way can one describe a wave losing its 'energy'?
There is an option where I don't look at a BEC as consisting of 'stopped waves' at all.But just as consisting of stopped 'particles'?
Lightarrow If you look at light as waves and leave the duality aside for a moment.What would you say is the properties of light getting slowed in a BEC?
The way I understand it, the only way we differ between the 'energy content' of waves. Is that we say that the more 'red shifted' a wave is, the less 'energy' that wave will have relative us.
And 'energy' is a definition of the probable movement/fluctuations inside a observed object as well as its spatial acceleration or relative motion as observed from another frame of reference.
And if that is wrong I don't know how to see waves:)And really would like to see how you see it.Do you see another way of defining how waves lose their 'energy' in a BEC?
Or lose 'energy' generally as observed from our 'frame of reference'?So what happens with that 'wave'?That's where I get 'stuck'? (for the moment:)----------Are you seeing it as a wave(s) inside a BEC contains its 'energy content' in some way? Even though all 'motion' as seen from our frame of reference stops?Won't I then need to question red shift as a definition of relative 'energy contents' if so?Then red shift is an spatial effect, as when observed from 'another' frame, special for that part of 'spacetime' that we can observe.And having nothing to do with its 'energy content'.And the idea of 'frames of reference' then needs a clearer definition yet, as you can 'release' redshifted waves of light (flashlight) in your accelerating rocket as well as in any other 'frame' defined macroscopically seen?Am I seeing waves wrong?And red shift too perhaps?And 'frames of reference?No big surprise:)----If i extrapolate on that idea it seems to say that no matter how we define a wave, relative us observing, they have a 'gold standard' and are all 'the same', containing the same 'energy'.And then frames becomes a very difficult concept to me.Not that it wasn't before too.And us saying that light should loose 'energy' by being 'stopped' is wrong.Which should mean that what we call 'temperature' and 'energy' is two different concepts.
Looking at the definitions of phase velocity I also found this."If we imagine the wave profile as a solid rigid entity sliding to the right, then obviously the phase velocity is the ordinary speed with which the actual physical parts are moving. However, we could also imagine the quantity "A" as the position along a transverse space axis, and a sequence of tiny massive particles along the x axis, each oscillating vertically in accord with A0 cos(kx - wt). In this case the wave pattern propagates to the right with phase velocity vp, just as before, and yet no material particle has any lateral motion at all. This illustrates that the phase of a traveling wave form may or may not correspond to a particular physical entity. It's entirely possible for a wave to "precess" through a sequence of material entities, none of which is moving in the direction of the wave. In a sense this is similar to the phenomenon of aliasing in signal processing. What we perceive as a coherent wave may in fact be simply a sequence of causally disjoint processes (like the individual spring-mass systems) that happen to be aligned spatially and temporally, either by chance or design, so that their combined behavior exhibits a wavelike pattern, even though there is no actual propagation of energy or information along the sequence."http://www.mathpages.com/HOME/kmath210/kmath210.htmIs it this phenomena you are referring to Lightarrow.
Lightarrow, thanks for answering, I'm just trying to see what you see.I've always thought of red shift as a relation between two frames of reference.Also that there exist a light quanta of a exact value.
So when photons is placed in a BEC their 'temperature' as well their internal 'movements' disappear I thought?
" phase velocity (physics) The velocity of a point that moves with a wave at constant phase. Also known as celerity; phase speed; wave celerity; wave speed; wave velocity. "Where is the velocity of that waves 'point' in a BEC?
The idea that every 'bit' of that wave would contain the information about the whole wave sounds very holographic to me?
I presume we are talking about a analog waveform here, so 'bit' could as easily be exchanged for 'slice', right?
But what about the phase velocity then? It seems to be seen as a specific 'wave crest', undulating up and down, of a larger sets of connective wave, as we follow it in time.That then describes that one 'wave crest' inside this larger system of waves. And the group velocity then will be an 'average' description of the waves propagation.And as that also could break 'c' in 'materials' we have defined a new definition called 'signal velocity' to keep the theory intact?
I don't think so. A bit is usually represented by a wave being in one of two states, being "on" and "off" respectively. Think for a moment of going from "off" to "on", i.e. sending a 01 signal. Your wave goes from 0 amplitude to 1 amplitude, which is a discontinuous jump--or a non-analytic point. This jump contains the information about your signal and it's the thing that propagates at the signal velocity. In reality things can get more complicated if you want to determine on/off by sampling your signal at even time intervals. Then you could send "0000..." by just an unchanging wave of amplitude 0. Still, the signal velocity would be limited by c.
If you treat a analogue signal bit wise you will take away 'information' from it, won't you?So by doing so you transform it into a 'simpler' solution?
Lightarrow, I will need to give this a lot more thought before I try to have a own 'view' on it, as this need a proper understanding, and, as I'm sloow:)But in the meantime, I would appreciate a explanation of what you see as the difference between group and signal velocity.To me it seems like a change of names? As I understand them both to describe the process by which no 'information' can propagate faster than 'c' in space?
I still enjoy the way you methodically explain your point(s), but, does this mean that phase velocity soon will be 'gone' from physics?
I got the impression before that you meant that phase velocity was phased out
in exchange for signal velocity Lightarrow?That's why I asked. I haven't used that description 'signal velocity' before.
Btw: What do you mean by Phase speed being "directly related to the index of refraction."?
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.
'phased out' Terminated gradually(Can't help that, blame them English, and my slightly twisted sense of humour:)Vph = c/nphase 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?
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 an 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'?
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?
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.
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
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_actSounds 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.
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.
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 relatedonly to its frequency. The frequency is unchanged during all thetransitions.Energy = (Plancks constant) * (frequency)It is also possible to re-write the energy (or frequency) in terms ofthe velocity and wavelength of the light. The velocity changes duringthe transitions, going from fast to slow and back to fast as the photonsleave the substance. However, the wavelength of the light also changesby the same factor during the transitions going from long to short andback 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 samebefore and after any transition. All in all, the energy remains constant.
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
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'?
So how is it possible for one wavelength to have different frequency'sWavelength = 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 λSo '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.
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?
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'.
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:)