Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: DoctorBeaver on 19/12/2008 09:28:13
-
My little beaver brain is working overtime again. This is what its murky depths have churned out this time.
As has been explained elsewhere, for a photon travelling at c time has no meaning. In fact, time doesn't really exist for it. It therefore follows that a photon could be everywhere in the universe at the same time. But the universe is expanding, so there is now more "everywhere" for it to be in than there was at any time in the past. How can the photon be everywhere at once if everywhere is getting bigger?
[???]
-
Its not just photons beaver quanum mechanics implies that every particle and quantum is everywhere in the universe over all time. its just that the probability of being in most places other than where it is, is rather small!!!
-
Ian - I am aware of that, but I think you have missed my point; to whit, the infinite time dilation aspect of travelling at c.
-
its just that the probability of being in most places other than where it is, is rather small!!!
Could we change that to "its just that the probability of appearing in most places other than where conventional Science would predict, is rather small!!!"
?
The existence of a quantum particle 'everywhere' is fine by me. Anything to avoid trying to specify where it 'is, exactly' rather than where it has an effect. At c, a photon is, effectively anywhere because it can 'choose to' reveal itself, instantly in any one of a number of places, depending on its probability function (the diffraction pattern with which it's associated), because it exists independently of time.
-
To answer your question the universe is not getting bigger from the edges its getting bigger from all over and that is why the cosmic microwave bacground radiation which was like starlight when it was generated is now srtetched right out into the microwave region.
-
Sophie I think you've got it a bit wrong there It depends a bit who's time you are talking about the photon doesn't "know anything about time between it being created in an interaction and it being destroyed in another interection. from our point of view it can have been created as part of the cosmic microwave background flash a few hundred thousand years after the big bang travelling a long way from a specific direction and arriving at our detector and being absorbed at a clearly definable location.
-
I know about the expansion not being from the edges.
Let me try saying this a different way. If nothing stops a photon from travelling then, theoretically, it could eventually actually have been at all points in the universe. Time has no meaning for a photon. Therefore it would have been at all those points simultaneously. But as the universe is expanding (and for this thought experiment it doesn't really matter how that expansion is happening) there are more points being created all the time that the photon could be at.
-
SS
You don't appear to be disaggreeing with what I wrote. The production and destruction of a photon will be observed in 'our' space / time. The conditions which determine when and where are determined by the experimental setup (including, if you like, the BB). Because it is not 'experiencing time' the photon can know instanly (from our viewpoint) where it needs to have started and where it needs to finish its journey between source and detector. CBR would behave just the same as light from a light bulb in this respect.
In fact what I am saying is that many of the difficulties and 'paradoxes' can be resolved by looking at the problem this way and not thinking of a photon as being at a point, ever.
DrB
Your problem with points in between being generated by expansion is dealt with because the only thing that matters is where / when the ends of the experiment are. The photon doesn't have to 'be' anywhere in between.
-
SC - there is no experiment involved in what I'm saying; just a photon - any photon - travelling at c. It doesn't get measured in any way. Put yet another way, how does that photon "view" the expansion of the universe if time doesn't exist as such?
-
If you're observing anything it can be an experiment. You can't have a photon unless it's produced by something and you have no idea where it is or when it is unless it is absorbed by something. Of course it is 'measured' at both ends. The atom it interacts with has measured it, effectively, when it decides if it has the right energy for an interacton and the one which chucked it out used a 'measured' amount of energy to make it with. We are considering a testable hypothesis in this discussion!
How does it 'view' the expansion? That's almost not a valid question because it would have to see a rate of change of displacement which implies dividing a distance by time. Dividing by, effectively, zero will give the impression of an 'infinite' or indeterminatly rapid step change between the emission and absorption events. But, as the photon doesn't really need to exist except at each end of the process, why should it (or we) care?
How's that, Dr B?
-
SC
What you said...
If you're observing anything it can be an experiment.
What I said...
It doesn't get measured in any way.
Right, forget that intelligent observers exist. Early enough in the universe's history there weren't any anyway.
A photon gets created somewhere, somehow - could be the CMBR or something else; it doesn't matter. Off the photon goes on its journey through the universe without interacting with anything else (unlikely, but theoretically possible). If spacetime is closed then there is the possibility that it will travel through every point in space - actually travel through every point. From the perspective of the photon, such a journey would take zero time. But, in fact, by the time the photon has indeed made its journey, the universe has expanded. So, is it true to say that from the photon's point of view, the universe has expanded in zero time?
Or, if you prefer to think of it this way, take a photon emitted from the Sun and heading towards the Earth. It takes roughly 8 minutes in our time, but zero time to the photon due to the infinite time dilation. In the 8 minutes of our time that elapse, the relative positions of the Sun and the Earth has changed. How does that change "look" to the photon in zero time? How can something happen if there is zero time for it to happen in?
-
I can see that you are still insisting that, somehow, the photon (like a bullet) is speeding on its way from a to b. If you look at it that way then I can see there may be a problem. But why does it have to exist between a and b? The only places and times that you can prove it's there are at each end of 'its' journey.
There is a great difference between a photon and all other so-called particles. An electron can make a journey and make its presence felt by many systems on the way - giving a little (in principle, measurable) nudge to every charge it passes. But a photon only interacts at each end of its 'journey'. As far as I can see, then, it is not necessary to consider what it is getting up to on the way - it gets up to nothing, in fact. I also hold that the 'it', to which we refer , is just the energy, which will behave according to the wave with which it's associated. It can't / needn't be considered as a particle except where it is interacting.
You seem to be over concerned about the changing distance between source and detector - why should that concern an object which doesn't really exist on the journey? What about the situation where source and detector happen to be approaching each other at the same speed that the space between them is expanding - wouldn't that situation be just as difficult to explain?
My way of looking at it is not to say "Off the photon goes on its journey through the universe without interacting with anything else" because, for a start, you can only say that it hasn't interacted yet. I say it is potentially everywhere at all times between its creation and its absorption. It's journey is only described once it has actually completed the two interactions.
There is, of course, a very finite time involved in the interactions at each end because they are, essentially, resonances which take time to build up. As far as the photon is concerned, the two ends could be one atom's width or the whole universe apart.
The bullet picture makes it extremely difficult to reconcile the effects of diffraction - requiring a message to travel instantly all around the Universe to tell all other atoms that the photon has arrived 'here' and can't be seen anywhere else. But, even in the simplest, two slits, experiment, something has to tell each photon that it is part of a number of others which must, somehow sort themselves out into an interference pattern - which will always occur when there are 'enough of them'. It's a nightmare this way round; stick to the wave interpretation.
-
For what I'm trying to get at, it doesn't matter whether it's a particle or a wave. So, let's say that "...the wave could propogate through every point in space..." and be throughout the whole of space at the same time. Space is still expanding and there is still zero time for the photon.
-
OK but so what?
Are you after finding out what the photon 'experiences'? What do you mean by 'the photon'?
If ,as I suggest, it doesn't exist except where it interacts then it's not 'like anything' for the photon. There isn't a photon for it to be like for.
It's a bit like sending a TV picture over a link. The picture exists at each end but it doesn't travel across the link in the form of a picture. It would not make sense to talk in terms of the picture during the transit time of the signal - all you could say would be that the link characteristics could affect what the picture turned out like.
Or. I have this idea and you are receiving it (and understanding what I say - or gnashing your teeth about that idiot the other end). The text, the binary signal from the keyboard, the wi fi signal from computer to router, etc etc are not the IDEA, they only carry a representation of the idea.
For me, the photon is just what 'goes on' during the interactions each end. It would be meaningless to discuss its existence except where / when it exists.
-
That sounds as if you're saying that in between the photon being emitted and being detected it doesn't exist anywhere. From everything I've read, that is simply not the case. Also, I've seen it proposed in physics text books that an electron can be everywhere at once.
-
Yes - I'm saying just that. Or, rather, that it has no meaning or reason for considering it as existing. I suppose it's a bit a a Zen thing - like "when a tree falls in the forrest, does it make a sound if there's no one to hear it?". But that is a bit of Zen nonsense because there are plenty of things for a falling tree to interact with and there is no interaction for a photon to have on the way.
Yes- I've read the same thing but that doesn't mean it's the best way to look at things. There isn't a 'right answer' to any of this stuff. All you can do is get a theory together which predicts most events as accurately as possible.
The wave theory gives the right answer about where to expect light to turn up (it gives a probability distribution) and the particle theory gives a good answer about how the quanta of energy in the wave will interact (the quantum of energy and the momentum).
Until I thought out this way of looking at it I had a terrible problem reconciling the two approaches. I have yet to hear a coherent (no pun intended) way of explaining what happens, in terms of photons, during diffraction. And you have to remember that every 'photon' follows the rules of diffraction in every circumstance! Diffraction is always there.
-
So the expansion of the universe has no effect on photons?
-
Yes expansion has an effect.
If you consider the point the photon was emitted and the point it is detected (presuming that the two are separated by large (intergalactic) distances) then the red-shift of the photon is a measure of the expansion between the two points (after we subtract the gravitational red-shift/blue-shift).
-
Yes. That is an energy matter. Another problem for the 'traveling photon' model. The photon would need to have both energies at the same time because it could not be changing 'with time'.
-
The problem with photons (I think) is there is no non-mathematical language which can
be applied. You cannot construct a spacetime diagram for a photon (well, you can, but you just end up with the y-axis, or you have to construct a Minkowski diagram with ct as the y-axis and the photon as a 45 degree plot). Therefore it is meaningless to talk about time from a photon's POV, unless you are doing a translation...
-
As has been explained elsewhere, for a photon travelling at c time has no meaning.
Your basic premise is flawed. Time has no meaning without the observable context of mass. Photons have mass(albeit infinitesimally small mass). This is the reason why black holes can capture light....if photons had no mass they would pass by black holes. Also effects like gravitational lensing would not be observed if photons had no mass.
"It therefore follows that a photon could be everywhere in the universe at the same time."
Time does exist for photons (because they have mass) which is why they cannot accelerate past the speed of light in a vacuum.
"But the universe is expanding, so there is now more "everywhere" for it to be in than there was at any time in the past."
Is it really? All of this conjecture is based on the Big Bang theory, the red shift and the premise that the Doppler effect of sound translates into the Doppler effect of light. In fact it doesn't and cannot be proven to. Why you ask? Because we have yet to discover a way to accelerate light past the speed of light.
When you decelerate sound the frequency becomes lower. When you decelerate light the frequency becomes lower (observable shift toward the red spectrum).
When you accelerate sound waves the frequency becomes higher. When you accelerate the speed of light what happens? The answer is no one know because you cannot accelerate the speed of light.
This explains why we see a red shift in the universe and no corresponding blue shift. And it explains why scientist incorrectly came to the conclusion that all matter originated from one point.
What science hasn't explained is why we have yet to discover the universe's edge. The point at which we can no longer see mass. Everywhere we have looked there has been mass.
I believe that the reason we haven't discovered the edge of the universe is because there is no edge and that the universe is infinite.
Other problems with the big bang theory is that the rate of expansion of the universe according to theory had to be improbably precise in order for the atoms to coalesce into the galaxies we have today. Any infinitesimally small increase would have resulted in a universe of scattered atoms. Any infinitesimally small decrease would have resulted in all the matter coagulating into one massive lump. Many groups point to this improbability as evidence of a Prime mover. I point to it and say that it is evidence of a flawed theory.
-
As has been explained elsewhere, for a photon travelling at c time has no meaning.
Your basic premise is flawed.
It's not my basic premise. Please take up the issue with Richard Feynman, Stephen Hawking & Lisa Randall as it was in books by them that I read it first.
Because we have yet to discover a way to accelerate light past the speed of light.
That statement is a total nonsense. If you accelerate light past the speed of light then the speed of light has increased and you would have to accelerate it again to get past it which will increase it... ad infinitum.
-
I should also like to point out that the title of this thread has been changed and no longer reflects my initial question.
-
Yes expansion has an effect.
If you consider the point the photon was emitted and the point it is detected (presuming that the two are separated by large (intergalactic) distances) then the red-shift of the photon is a measure of the expansion between the two points (after we subtract the gravitational red-shift/blue-shift).
But, surely, that is only from an observer's POV. The photon itself won't "notice" it.
-
If, indeed, it's "there" to notice anything.
-
If, indeed, it's "there" to notice anything.
Well, yes.
-
I can see that you are still insisting that, somehow, the photon (like a bullet) is speeding on its way from a to b. If you look at it that way then I can see there may be a problem. But why does it have to exist between a and b? The only places and times that you can prove it's there are at each end of 'its' journey.
There is a great difference between a photon and all other so-called particles. An electron can make a journey and make its presence felt by many systems on the way - giving a little (in principle, measurable) nudge to every charge it passes. But a photon only interacts at each end of its 'journey'. As far as I can see, then, it is not necessary to consider what it is getting up to on the way - it gets up to nothing, in fact. I also hold that the 'it', to which we refer , is just the energy, which will behave according to the wave with which it's associated. It can't / needn't be considered as a particle except where it is interacting.
You seem to be over concerned about the changing distance between source and detector - why should that concern an object which doesn't really exist on the journey? What about the situation where source and detector happen to be approaching each other at the same speed that the space between them is expanding - wouldn't that situation be just as difficult to explain?
My way of looking at it is not to say "Off the photon goes on its journey through the universe without interacting with anything else" because, for a start, you can only say that it hasn't interacted yet. I say it is potentially everywhere at all times between its creation and its absorption. It's journey is only described once it has actually completed the two interactions.
There is, of course, a very finite time involved in the interactions at each end because they are, essentially, resonances which take time to build up. As far as the photon is concerned, the two ends could be one atom's width or the whole universe apart.
The bullet picture makes it extremely difficult to reconcile the effects of diffraction - requiring a message to travel instantly all around the Universe to tell all other atoms that the photon has arrived 'here' and can't be seen anywhere else. But, even in the simplest, two slits, experiment, something has to tell each photon that it is part of a number of others which must, somehow sort themselves out into an interference pattern - which will always occur when there are 'enough of them'. It's a nightmare this way round; stick to the wave interpretation.
SC I read your thoughts with great interest, there are more than you sharing this view.
Still, what does that make the sun?
Isn't that a source of 'light' aka photons?
If there is no 'distance' involved,why do they need a source relative us measured as at a distance?
I'm sorry if I sound 'meta-physical' here, but if they are this in woven' in spacetime, why do 'they' (mostly) get created from 'sources' we measure as being 'distant' in space?
No disrespect meant SC, you are all interesting, I'm just trying to get a 'grip' of how you and others see it:)
-
As has been explained elsewhere, for a photon travelling at c time has no meaning.
Your basic premise is flawed. Time has no meaning without the observable context of mass. Photons have mass(albeit infinitesimally small mass). This is the reason why black holes can capture light....if photons had no mass they would pass by black holes. Also effects like gravitational lensing would not be observed if photons had no mass.
"It therefore follows that a photon could be everywhere in the universe at the same time."
Time does exist for photons (because they have mass) which is why they cannot accelerate past the speed of light in a vacuum.
Gravitational lensing and the reason photons cant escape blackholes is due to space and the path that photons take through it being bent. Gravity has no direct affect on a photon.
-
My little beaver brain is working overtime again. This is what its murky depths have churned out this time.
As has been explained elsewhere, for a photon travelling at c time has no meaning. In fact, time doesn't really exist for it. It therefore follows that a photon could be everywhere in the universe at the same time. But the universe is expanding, so there is now more "everywhere" for it to be in than there was at any time in the past. How can the photon be everywhere at once if everywhere is getting bigger?
[???]
Localization is at a cost though. The Uncertainty inherent in matter, requires it's path be complimentary. Sure, you can localize a particle, if you are willing to give up your time to observe an infinite amount of paths simultaneously.
It would require impossible odds.
-
This is one of the many things making photons fascinating.
But we have sources for them.
We have interactions with them at different places in spacetime.
Even though they may change 'entity' in those interactions.
We can still construct a 'particle path' from them.
That will follow everything we know about light.
For us to have that 'shimmering weave' :) of 'light' hiding everywhere.
We don't seem to need any 'sources' at all.
And our ideas of how energy transforms seems childish if so.
As this potential is everywhere, not needing any 'source' at all to my eyes.
But it might be right.
A 'field' of light overlaying a 'field' of time overlaying a 'field' of space all seemingly centered around matter.
And no distance?
It is like we really have two different realities coexisting at all time.
In one there is three dimensions and time creating distance motion space light and matter.
All of them entities by themself.
And then we have ?
-
yor_on
You say that all those photons 'come from' the Sun. I would agree that photon interactions take place between atoms in the Sun and in your eyes. Between those places, the light behaves like a wave - it is refracted and diffracted on the way. Why can't that be enough?
Why is the idea of Photons being like little bullets such an attractive one, I wonder?
-
Yes SC, I see some as coming from the Sun.
Most of them we see in fact:)
And I have no problem with accepting them to be waves, if traveling in spacetime.
What I have problem with is the idea that they only will exist as a 'interaction'.
To me that imply, if that idea is correct of course:) that they would exist independent of any source.
Accepting that view, all our ideas of spacetime seems to collapse.
Then what are what we call sources?
We have built spacetime around concepts as distance motion 'c' etc.
All of those seems questionable, to say the least, if we allow for photons to be at 'all paths'.
Our descriptions 'falters' if you see how I think:)
----------
Then again.
We can send coherent laser light in a very straight path, can't we?
And if we measure that specific wave length outside what we might call its 'path' it won't be there, right?
So light have a path, doesn't it?
Radiation by incoherent light should be able to be treated as the sum of all coherent 'wavelengths' there might be.
And if so, sun light also will have a 'path' in spacetime it seems to me.
Am I right?
-
Let us take the idea of superimposing photons upon each other.
Photons are bosons and can be superimposed as much as you like without taking up any space.
Treated as waves there are experiments proving that.
They can either cancel each other out, or build on each other.
This is called 'Constructive or Destructive Interference'.
If we can do it when they are treated as waves what hinders us from doing it when treating them as particles?
Why would it violate HUP (Heisenberg uncertainty principle) when treated as particles but not as waves?
Am I getting this wrong?
If I assumed that it was possible, as the HUP doesn't forbid it, (as I see it:)?
It only states that we as observers won't be able to prove all parameters, if observing.
That's not the same as stating that under no circumstances, observed or not observed, can all parameters for a given 'system/particle/object' be existing in spacetime at the same time, is it?
Is it possible to superimpose 'photons' on each other?
And will they then have more energy as measured from a point of impact?
-
Yes SC, I see some as coming from the Sun.
Most of them we see in fact:)
And I have no problem with accepting them to be waves, if traveling in spacetime.
What I have problem with is the idea that they only will exist as a 'interaction'.
To me that imply, if that idea is correct of course:) that they would exist independent of any source.
Accepting that view, all our ideas of spacetime seems to collapse.
Then what are what we call sources?
Sources of electromagnetic field.
We have built spacetime around concepts as distance motion 'c' etc.
All of those seems questionable, to say the least, if we allow for photons to be at 'all paths'.
Our descriptions 'falters' if you see how I think:)
Actually, we have built spacetime around tangible concepts as objects with mass.
Then again.
We can send coherent laser light in a very straight path, can't we?
And if we measure that specific wave length outside what we might call its 'path' it won't be there, right?
So light have a path, doesn't it?
Make this simple experiment: put a coin perpendicular to the light's beam, the coin's area greater than the beam's area. Will the coin stop photons from reaching a point after it? According to your view the answer should be yes, but actually it's not, because of diffraction.
-
Are you talking about coherent light or incoherent light here Lightarrow?
But my question wasn't about what would happen if you put matter in its path.
It was if one could prove that the lasers light had a certain path in space.
And from there ask if ordinary light also might have a defined path in spacetime.
As for building spacetime, didn't that start a long time ago:)
With Newton and Einstein, Feynman appearing relatively recently?
It's all a question of definitions of course.
I still have trouble seeing photons as something just appearing as 'interactions'.
---------
But yes, I can see what you mean here.
Still, to me that would be the next question:)
As we first have prove that there are no 'paths' as I see it.
So, do lasers have a 'specific' path in spacetime or not.
-
Are you talking about coherent light or incoherent light here Lightarrow?
No difference at all, in principle.
You get diffraction all the time with all em waves. It's just a matter of degree.
I still have trouble seeing photons as something just appearing as 'interactions'.
I think you ('one') must try to think through the true consequences of treating light as photons 'on the way'.
I realise that there is a huge photon culture because of the scale of things for visible light but any treatment must encompass all wavelengths of em energy. I can see why, because working out diffraction integrals (even just approximately) is a lot less friendly than drawing lines from a to b and saying that the energy 'goes that way' except when it occasionally gets diffracted.
Do / can you possibly think that a 1500m radio broadcast consists of a shower of little bullets - how big would they be? Would they be the same 'size' as gamma photons?
-
Good question SC.
I can give you one question with a twist:)
Do you think you and everything you use/meet today is made of light?
Explain that, no theories.
Just those experiments proving we are light.
What do you think:)
-------------
I haven't said that I don't believe in the result of the two slit experiments?
It's just that I do believe in them :)
And that matter and light, at least looking from where we are, is fundamentally differently expressed.
One of the reasons why I'm not happy with 'many paths' is that it seems to make us all 'probabilities' in a statistical universe.
And considering our self structuring capabilities, as well as entropy and the arrow of time we have?.
To me it seems that we are more than 'probabilities'.
-
Are you talking about coherent light or incoherent light here Lightarrow?
Both.
But my question wasn't about what would happen if you put matter in its path.
It was if one could prove that the lasers light had a certain path in space.
And from there ask if ordinary light also might have a defined path in spacetime.
Yes, but only statistically, in the sense that most of the field can be localized in space, in some cases as this one. If it wasn't so, optical geometry wouldn't have existed...The conditions which the fields must satisfy for the optical geometry approximation to be correct are:
1. Wavelenghts much smaller than all the other physical dimensions involved in the system.
2. Distance between stops/slits and detector screen not much greater than the stop's/slits' dimensions (or you will get a diffraction/interference pattern the same).
As for building spacetime, didn't that start a long time ago:)
With Newton and Einstein, Feynman appearing relatively recently?
It's all a question of definitions of course.
Remember that spacetime is not simply a mathematical object, but a physical one; to define it you need reference frames, physical objects, physical clocks.
I still have trouble seeing photons as something just appearing as 'interactions'.
A photon must be a physical object.
1.How would you define the existence of a physical object without ever using the word "measure"?
2.Can you measure a photon between source and detector?
-
Hhmmm :)
How about this then:)
If our Sun got quenched.
If now light do have a path, isn't there something called spontaneous particle creation?
That, it is said, the light enjoys itself 'playing' with, even though it goes back to being a wave very quickly?
If that is true, can't then light be seen as a particle having a geodesic path following spacetime.
And if that is true, as well as space containing 'in itself' a huge amount of energy.
And if Feynman is correct in his interpretation of possibilities innumerous for that 'waves/photons' path.
And if I'm right in wanting to allow that wave the properties of a particle?
Will then all be dark here as that sun goes out?
Why?
In my view it's quite simple.
The Sun went out, what do you expect?
From the other view it seems more of a question of probabilities, not discounting the one, wherein the 'light' still will shine with or without a 'source', even if it would be of extremly low probability.
Or am I thinking wrong here?
You could say that even without Feynman's 'many paths', spacetime still would have that possibility.
But to me that concept makes it so much more probable, as it invites us to a spacetime in where all paths are taken at all times, not caring for the 'geometry' we observe versus matter.
Ah the headache:)
"
A photon must be a physical object.
1.How would you define the existence of a physical object without ever using the word "measure"?
2.Can you measure a photon between source and detector? "
How do you mean by 'physical object'?
Only at its moment of 'interaction' or at all times?
1.Yep :)
Lovely question that one.
As well as difficult to answer.
Everything we do or experience comes from some kind of interaction with our surroundings.
So you are definitively right in that this (observing) is the first and last 'proof' of something existing outside ourselves.
Even though it is more of an archetype than a true 'objective evidence', it is still what we build upon.
But we also expect 'cause and effect' (times arrow) to work even without us observing.
For example you filling a bath tube, leaving it for a while and when coming back finding it overflowing.
That is also an archetype, but only at QM level can it be said to be 'broken'.
(Ah, times arrow I meant, not the bath tube:)
So going from those definitions I believe(?:) that this path reasonably could be seen to exist even without any observable interactions.
2. No.
(But see 1.)
(That is if you by measuring, mean observing it 'passively', as an object of its own (a ball)?)
But if you by 'measure' mean.
If it will interact with any 'obstacle' placed in, what I see as, its 'path'?
Then..
Yes, it will.
-----
Another problem I have is with dimensions:)
Take string theory.
Is it eleven dimensions in M theory?
When we start with one dimension.
Nothing special, just trying to see it here in 'spacetime' (in time).
Will we ever see it?
Nope.
Two dimensions then?
Yep, from some angles it will exist, from others it won't.
Have we ever observed anything like that here?
Nope.
What scientists define as 'two dimensional' today is just an 'approximation' of it.
Like a lattice wherein you hinder atoms from moving any other way than back and forth.
Is that two-dimensional?
Use my 'test' in space time and see for your self.
Will you at any angle observe this lattice as 'disappearing?'
And if your answer is no.
Well, then it can't be two-dimensional (as I define it).
What we have here is three dimensions and time.
In this spacetime we find 'matter' and 'time', 'space' and 'light'.
Most of the other 'things' we have defined, as gravity, motion and distance, seems to me to come from them.
Acceleration being a special cause.
We know it is so.
And that it 'works'.
Otherwise I wouldn't be able to write this.
I'll stop here for now:)
Btw: Anyone feeling that they do understand what a 'dimension' is?
You are very welcome to explain how you see it:)
Although DB might want that explanation in a different thread?
-----
(Maybe 'time' and 'space' also fall under matter and light?:)
Could 'time' exist without matter?
As without matter where would 'space' be?
That would be 'time' on its own, and time is a 'relation' right?
And yes, I'm just wondering, nothing more :)
-----------
-
Hhmmm :)
How about this then:)
If our Sun got quenched.
If now light do have a path, isn't there something called spontaneous particle creation?
That, it is said, the light enjoys itself 'playing' with, even though it goes back to being a wave very quickly?
If that is true, can't then light be seen as a particle having a geodesic path following spacetime.
Sorry, I don't understand what you want to say.
And if that is true, as well as space containing 'in itself' a huge amount of energy.
And if Feynman is correct in his interpretation of possibilities innumerous for that 'waves/photons' path.
And if I'm right in wanting to allow that wave the properties of a particle?
Will then all be dark here as that sun goes out?
See up. Could you explain what you want to say in a different way ?
Why?
In my view it's quite simple.
The Sun went out, what do you expect?
From the other view it seems more of a question of probabilities, not discounting the one, wherein the 'light' still will shine with or without a 'source', even if it would be of extremly low probability.
Or am I thinking wrong here?
You could say that even without Feynman's 'many paths', spacetime still would have that possibility.
But to me that concept makes it so much more probable, as it invites us to a spacetime in where all paths are taken at all times, not caring for the 'geometry' we observe versus matter.
See up.
Ah the headache:)
"
A photon must be a physical object.
1.How would you define the existence of a physical object without ever using the word "measure"?
2.Can you measure a photon between source and detector? "
How do you mean by 'physical object'?
Only at its moment of 'interaction' or at all times?
Let's make an example. I do the following statement: "everything you see is due to the interaction with your eyes or your camera, ecc. of little angels with blue wings and green faces. They are destroyed at the impact with your eyes or your detectors."
Do you think it's a scientific statement? Let's imagine that you say "No". Then I'll argue that my angels explains very well what you see and the properties of light (tell me a light's property and I'll explain you how the "theory of Angels" describes perfectly that property... [;)]).
But we also expect 'cause and effect' (times arrow) to work even without us observing.
For example you filling a bath tube, leaving it for a while and when coming back finding it overflowing.
Since this is a much more difficult question, for the moment I prefer to say that it's a different situation, because:
1. We are talking about macroscopic objects.
2. We are talking about objects with non-zero mass.
Remember that the fact photons don't have a position operator in QM (as, e.g., electrons, have) is essentially due to the fact they are massless. It is this absence of such an operator that makes it impossible to ascribe them a precise position in space when not detected.
About string theory, M theory and the rest, sorry but I don't know much about them...
-
Well, :(At least it seemed logic to me:)
Lightarrow, as far as I understand we start everything by archetypes, or do you see it differently?
One of the first archetypes needed to be defined is if 'anything' exist at all outside ones mind.
If your answer is a yes to that one, the rest of what we call 'objectivity' 'observations' 'measurements' will follow and make sense as 'outside phenomena'.
Not hinging on us being here at all.
The second one I used was the arrow of time, and I called that one a archetype too :)
And that is as we without knowing why it does it, or how it does it, observes it as having an arrow.
And that we expect it to continue to do so, at least macroscopically.
And from those two 'postulates' I find it reasonable to expect photons to be able to have a 'path' in spacetime, even without either you nor me observing.
(And no, I don't see light as little angels, but I like the idea, who knows:)
Reminds me a little of that 'postulate' wherein the question was something like.
'How many angels can dance on a needles 'point'':)
As for the beginning of what I wrote, I agree.
It needs to be much clearer.
--------
Give me some 'time' on that one, please:)
-
Do you think you and everything you use/meet today is made of light?
Explain that, no theories.
Just those experiments proving we are light.
I've been lurking here following this thread with great interest. I've been compiling evidence FOR (http://photontheory.com/TheEvidence.html) and AGAINST (http://photontheory.com/EvidenceAgainst.html) So far the evidence for that far outweighs the evidence against. It seems that since that is such a restrictive case it should be easy to dispose of.
-
Do you think you and everything you use/meet today is made of light?
Explain that, no theories.
Just those experiments proving we are light.
By 'light' I take it you mean electromagnetic waves(?).
The answer is No, in any case. The waves are only there due to interactions between charge systems, involving mass (thus excluding 'light' which is massless).
How can one explain without theory? "I feel that because it makes sense to me" is no use as an explanation.
-
Well, :(At least it seemed logic to me:)
Lightarrow, as far as I understand we start everything by archetypes, or do you see it differently?
Archetypes? Do you mean "basic concepts"?One of the first archetypes needed to be defined is if 'anything' exist at all outside ones mind.
This is an interesting questions. Harry Palmer (creator of the Avatar course) says: "Let's start to take things away of the universe. Planets, then stars, then all the matter, dust and clouds and atoms or particles. Then even time and space,ecc. What remains? Our awarness. If we took away even our awarness, who could prove there is still something in the universe?"
Anyway this is more phylosophy than physics, at least now (maybe one day we will know something more about awarness).
-
As I understands it an 'archetype' is something so 'obvious' that we more or less take it as granted.
Like one and one makes two.
I read somewhere that most animals know the difference between one and two and 'many'.
Sometimes they can be wrong though.
The 'archetypes' I mean, well, the animals too:)
I mean?
Is one and one still two??
It can, as I see it, easily become three:)
-
Is one and one still two??
I think you will like this (hope to have said it well in english);
<<People are of 10 categories: those who know binary numbers and those who don't>> [:)]
-
Yep:)
-
Apropos nothing at all:)
Considering my inclination for logic..
"If Logic is a systematic method for getting the wrong conclusion... with confidence.
Then surely statistics is a systematic method for getting the wrong conclusion... with 95% confidence."
Or something to that order?
-
My little beaver brain is working overtime again. This is what its murky depths have churned out this time.
As has been explained elsewhere, for a photon travelling at c time has no meaning. In fact, time doesn't really exist for it. It therefore follows that a photon could be everywhere in the universe at the same time. But the universe is expanding, so there is now more "everywhere" for it to be in than there was at any time in the past. How can the photon be everywhere at once if everywhere is getting bigger?
[???]
I've been thinking about this and I think that something I added as a late post in another topic may have a bearing here too, because the premise that time has no meaning for a photon may not be correct.
From that earlier thread:
Although we can move through three spatial dimensions, at any point in time the movement vectors for all three spatial directions can be summed to a single vector. Thus movement is essentially in a single direction and can be expressed by a single value, just as when we drive heading North-West we don't say we are driving West at x mph and North at y mph; we just use the summed vector.
With movement induced time dilation, the same thing is happening, except this time the two vectors being summed are the summed spatial movement vector and the temporal movement vector. The reason we get time dilation is because it is the sum of these two vectors, spatial and temporal, which cannot exceed 'c', so as the spatial movement vector increases, the temporal vector must decrease. With zero spatial movement then, we move temporally at 'c', which in turn implies we have zero length in that direction
Now, if we are moving at 'c' in the temporal dimension, have no length in that direction, and we are aware of time passing, then it would seem plausible that when we see a photon moving at 'c' though our spatial dimensions we are seeing a two-dimensional object moving along it's temporal dimension; this being our third spatial dimension, in which it has no length.
-
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??
And if so then 'expansion' is no obstacle as that 'wave' then will cover 'everything' we can observe.
But if it are able to do this, our ideas of distance seems wrong to me.
They will only work at a macroscopic plane.
-
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??
It's the opposite [:)]
λ = Vph/ν
λ = wavelenght
Vph = phase velocity
ν = frequency
Since frequency doesn't vary for the wave entering the material, and phase velocity decreases, wavelenght decreases.
-
If you look at the light as a wave, what is it that 'breaks it' in a BEC?
But if it is a wave it will have a 'frequency' until it 'disappear' Lightarrow.
There is no possibility of treating it as not having that to my eyes, as long as we have time.
A straight line is no wave.
There will be a wave stretching out into?
But it can't be seen as being 'everywhere'.
That seems a to far reaching conclusion to me.
The probability function here just state that 'it's gone out to lunch' and is not present.
But it will still be defined to the same location in spacetime.
--
(Thought of it as you describing 'waves height' here:)
((Yep, and a bad cold it is too::))
_______
Ok reading you once more, you're saying the opposite right.
That as the wave is slowed down its wavelength decreases?
--
But how is that possible?
When we talk about redshift we define it as being of lower 'energy' having a longer wavelength.
And this is what we do in a BEC? We 'steal' the waves energy, redshifting it until it 'disappears'.
So how can the wavelength decrease?
I'm reading you wrong somehow.
(got a cold, can I blame that:)
-
Actually this gets me back to the question if photons inherently can be of different light quanta (energy).
If they as I believe only are of 'one' 'energy size' then what happens here?
When we 'steal' their energy?
We can't steal it 'gradually', not if it is a decided 'energy size'.
(It should be easy to test what is right here.
We just need to 'freeze' one photon:)
Ah, that was somewhat of a joke..
You see, we can say that we are 'slowing' photons down by taking away some of them (particle wise).
But that should then mean that all photons will be annihilated when we say they are 'standing still'.
doesn't make sense, does it:)
Worth thinking about.
-
Is there, in fact, a red shift? There doesn't appear to me to be any reason for a change in frequency under these circumstances. When going through any other transparent medium the frequency (colour) stays the same. In extreme cases, you wouldn't see any light if that happened.
Is that where your confusion arises?
-
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'?
Even if I looked on it in terms of 'reducing probabilities'?
And said that as the probability becomes 'zero' the waves can't be present any more.
But then we won't need suns either (sources), if this is correct.
Well, they will still be the greatest probability, but it wrecks havoc with a lot of processes we take for granted.
--------
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'?
But why should I need to do so?
Would that mean that only 'certain' manipulations are allowed in a BEC?
Those specifically 'wavelike' experiments might not work then?
And entanglement where you mix waves through a beamsplitter then?
Calling those entangled might work when dealing with waves.
But if seen as 'particles' instead, why do they get 'entangled' at all.
It is the spin we are discussing here.
-
But doesn't it have to be treated as a red shift?
Why?
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'?
Photon's energy = hν and the frequency doesn't change...
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'?
There is no need. You have an electromagnetic field which varies in the time in every point of the BEC, but that doesn't propagate in space, as a stationary wave does. Where is the problem?
-
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.
-
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?
Phase velocity (as I had written).
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 what does "red shifted" means for you? If the medium is the same, then phase velocity is the same, then frequency is inversely proportional to wavelenght; but if you consider wavelenght and frequency in a medium, and then wavelenght and frequency in *another* medium, you can't say that the first wavelenght must be inversely proportional to the second frequency! It's meaningless. Equations says everything. λ = Vph/ν. What's the problem with it?
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?
Why should they lose their energy? Does a light beam from the Sun lose its energy when it enters through your window (assuming no absorption from the glass)?
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.
Light in a material medium behaves in a completely different way than light in the void, not only because the speed is different. Infact, saying "light moves in a material medium" is actually misleading. If physicists in the books, instead of talking of "light propagating inside the material", talked of "an electromagnetic field propagating inside the material", it would be less fascinating but probably less misleading. Inside a material, you just have many atoms, each of which oscillate electrically. That's all. You don't need anything else. Think about the OLA that you see in soccer stadiums made by people; there isn't conceptually any difference between the OLA and an EM wave travelling inside a material medium. What you see "propagating" in an OLA is nothing else than the phase of the oscillation; if people where not syncronized in that way, but in another, you wouldn't see anything propagating.
-
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.
Then those makes sense of the universe to me:)
Even though 'frames' is something I'm not sure of the definition of yet.
Also I've seen temperature as something having a direct relation to 'energy'.
And temperature being a description of 'movements' inside a molecule or a wave, and therefore of particles (photons) too.
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?
-
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.htm
Is it this phenomena you are referring to Lightarrow.
--------
So the 'wave' becomes an infinitesimal line of 'dots', each 'dot' representing a 'crest' of that wave, and the group velocity disappears?
I think i can see your point, I will need to think about this, ah, later:)
-
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.htm
Is it this phenomena you are referring to Lightarrow.
Exactly. Phase velocity could be just "an artefact". Another example: do you know those yellow lamps which sometimes they have put on the border of a dangerous bent in a road? They are syncronized to switch on in a sequence: first the lamp n.1, then the n.2, immediately close to the first, and so on; in this way you have the sensation that a light is moving along the border of the bend; of course, nothing is *physically* moving, do you agree with it? What problem in imagine to syncronize those lamps so that at t=0 the first lamp switches on and then at t=1 second, the last lamp, 1 million of km away switches on? No problem at all. The flash of light has travelled for 1 million km in 1 second. Do something physical propagated at that speed? No. That's the same with phase velocity.
For a long time physicists believed that group velocity, that is the velocity of a group of waves taken as a single entity, would represent the "real" physical parameter. Then it was discovered that even group velocity can be > c!
What instead really counts is what is called "signal velocity" which however could be very difficult to compute or even to identify. Anyway, signal velocity is always < c.
Don't believe that the subject of waves is an easy one. (Maybe this is the main reason some people prefer to think of light only in terms of particles [:)]).
-
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.
Ok but you don't have to change the medium: if you change frame of reference you have to stay in the void (or in another medium, but fixed); then you can relate wavelenght with light's frequency, with photon's energy and with the relative speed of the two frames.
So when photons is placed in a BEC their 'temperature' as well their internal 'movements' disappear I thought?
It's impossible to localize a photon, in the void or (as far as I know) in a BEC. About the photon temperature it's a meaningles concept in physics.
" 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?
Imagine that at t=0 the electric field of the wave is maximum for atoms n. 1, 101, 201, 301, ecc.
Then you can identify the crests of the wave at t=0. Snap another picture at t=1. The atoms at which the electric field is maximum are the same as before or not? If not, the crests has moved, so the phase has moved, so phase velocity is ≠ 0; if they are, phase velocity = 0.
Don't know where is the photon, or if it still exist or not.
-
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?
-----
Does this make better sense?
I've changed my questions to stop 'chopping it up'?
-
I can take a shot at this since I've done a bit of reading up on these different velocities.
The signals you're worrying about here are what are called "analytic." This means that all the information about the entire signal is contained in every tiny bit of the signal. If you had perfect instruments, you could measure an infinitely small chunk of the signal and immediately know the entire analytic signal. What this means is that (theoretically) you have all the information in the signal as soon as its tip reaches you. The signal velocity is therefore the time it takes this tip of the signal to reach you. This can never exceed the speed of light or information could move faster than light.
Group velocity is essentially the speed at which the "shape" of the pulse travels. If you choose your pulse's shape right and pick the proper medium to send it through, you can make the peak of the pulse travel ahead faster than c, but you can never make the very tip of the pulse move ahead faster than c. It's fairly tough to do this, which is why people thought for a long time that the group velocity was the signal velocity.
The poorly-drawn diagram should illustrate this. The tip of the pulse is the red bit at the start. You can change the shape of the pulse behind so that the peak comes sooner or later by adjusting the group velocity.
[diagram=402_0]
Finally, there's good evidence to suggest that for (classical) light signals, the signal velocity is always c. However, the tip of the signal is usually so small that you can't actually measure it with any real-world detectors, so the practical signal velocity ≤ c.
-
I've seen studies where pulses of light tunneling through a barrier seemed to move faster than light. But what was actually happening was that the light pulse position was measured from the centre of the pulse. The input pulse was wider than the output pulse. The centre of the shorter pulse exited the barrier before the centre of the input pulse arrived at the barrier.
So to me it seemed no mystery at all.
-
jpetruccelli, thanks for your explanation.
Although I need more (always:)
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?
Am I correct here?
-
Reading that alternative point of view from 'mathpages' I find that easier to encompass.
If one allow a wave to be seen as series of changes in different substances/materials as we are used to measure them, then their 'absence' in 'space' (vacuum/absence of matter) just becomes another part of its geometry, just like its 'inherent energy', 'virtual particles' 'photons' and 'spontaneous particle creation'.
Then the question seems to be not what we measure, but more what we are 'missing' if you see my drift.
Aether presumed some sort of 'substance' with some sort of 'density/resistance' if I understood it right? But this phenomena we are discussing here seems more to have with 'limitations of observing' combined with 'times arrow'?
Although, if this was correct it should imply either that there is an 'arrow' even for subatomic processes or that spacetime itself force some 'causality flow' to a larger system (spacetime) that we won't notice when studying 'delimited' versions of subatomic systems?
---
Or that there is a 'arrow' to something that then will be one 'step' further down.
Passing both particles and photons, a coherence to 'spacetime' itself, not even noticing our differentiation.
Yep, a damned mystic, that's what I am:)
-
The idea that every 'bit' of that wave would contain the information about the whole wave sounds very holographic to me?
Do you mean holomorphic? If so, they're quite similar. Holomorphic generally means that the function is defined over a region in the complex plane, and that knowing the complex derivative at a point is enough to reconstruct the function anywhere in that region. More often a signal is defined over some real variables, but you can relate it by extending those variables to the complex plane.
I presume we are talking about a analog waveform here, so 'bit' could as easily be exchanged for 'slice', right?
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.
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?
Phase velocity is the speed of a single frequency of the signal. For light, this is a plane-wave component of the field, which exists over all space at all time. So these plane waves themselves can't send information--they already exist everywhere. An arbitrary pulse can be written in terms of these plane waves by using Fourier transforms. If you put a non-analytic point in your pulse to send information with it, you end up needing a whole bunch of plane waves to describe it (I'm pretty sure you need infinitely many of them, but I'd have to double check the math). In that case, the non-analytic point, which is composed of a bunch of single-frequency plane waves, travels at the signal velocity. The phase velocity isn't terribly meaningful in this case, since you don't really end up caring about the single-frequency plane waves.
So to summarize, there are 3 velocities you asked about:
Signal velocity: the speed the signal transmits information. For light, this is generally the speed at which "kinks" in the signal propagate.
Group velocity: the speed at which the bulk of the "pulse" appears to propagate. This can be greater than c, but the "kink" at the start of the pulse is still limited by c, and doesn't have to move with the group velocity.
Phase velocity: the speed at which a single frequency component of a signal propagates. This is useful when decomposing signals into plane waves, but I don't think it has much application in analyzing the propagation of information via a pulse.
-
I presume we are talking about a analog waveform here, so 'bit' could as easily be exchanged for 'slice', right?
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.
Now you're giving me problems Mr jpetruccelli:)
But you have written a very clear description all the same.
And that takes some work to do, so thanks for it.
Now, on to my questions.
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?
I know that there are ideas of our universe where one treat it as a 'bit' or as some say 'pixel' universe representing 1:s or 0:s from the 'bottom up" and sees it all as 'information'.
Is this what this idea represents?
Holomorphic?
Nice word:)
To me, if you say that you by analyzing one part of a system can get information about the whole system, then that seems a holographic principle.
I will have learn what differs them, could you describe it so that I could get a inkling to the difference between those descriptions?
That as holography is easily proved by verifiable experiments, even though every 'slice' will have lost some information it still will give us a visual representation of a 'whole' system/image.
On the other hand, 'slices' in holographic 'representations' will be non linear descriptions, and therefore not 'limited' into bits if I'm correct.
I like the way you explain phase velocity, and I'm sure it makes a lot of sense mathematically.
To summarize I get an impression that you are telling me that to get a 'linear solution' (1+1=2) of waves, we will need to break them down into bits. And as I think you say, the sampling frequency is what will define the precision then.
But how one expect a 'bit' to be a holographic representation of a whole system is harder for me to understand?
A bit is one isolated state of a binary system and as you say it represents either 'on' or 'off'?
I'm learning.
Slowly:)
-
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?
I think this is quickly going far from the original question, so if anyone wants to split this thread, it's probably a good idea.
I'm not a signals expert, but the point is that both the digital and analog signals I'm talking about are waves. A simple representation of a digital signal is just an wave that can only be amplitude 0 or amplitude 1. An analog signal has a continuous range of amplitudes it can take on. But if you send either signal as a wave waves the basic ideas of waves should apply.
As to holographic. I believe I misinterpreted what you were asking about before. Holographic has to do with encoding information from somewhere such that every point of your recording receives information from every point in your original image.
The analytic/holomorphic concept I'm talking about is a particularly nice behavior of "smooth" functions such that you need only see a small piece of a function to know its broader behavior. If you have an analytic function, then you need only see a piece of it to get the entire thing, which means if someone is sending an analytic signal at you, you (theoretically) get all your information along with the tip of it. I'm not concerned with how much information each function carries, just that all of that information resides at the tip.
Assuming you're dealing with waves that can be pieced together from analytic functions (and this is the case for classical electromagnetic fields, but I assume it holds for waves in general), each section carries its information in its tip (and at every other point in the section). That's why it makes sense to define the signal speed to be the speed at which the tip of a signal travels.
I've completely swept under the rug all considerations about information content as well as practical considerations of how to physically send/detect signals. In practice, your information content is going to be limited by how you encode and send your waves. Plus, measuring the very tip of a wave with infinite accuracy is impossible, so you'll be limited in how much information you can get from it.
-
Nice explanation again friend.
As you said we seem to move a little apart from the original question though.
We'll let it rest for a while then?
I'm sure it will come up again though:)
---------
Just one more Q.
I see holographic information as a non linear description/system.
Even though they make it by rotational linear polarization the result, as it is caught as 3D-image, will be non linear.
Would you say that 'Holomorphic functions ' describes a linearly or non linearly system?
But suddenly I'm not so sure about holographic images?
It's defined by the way waves reflects and diffracts, so can one say that such a image is 'indefinable' or should one say that the information density contained is 'finite' and therefore a solvable problem, that is linear?
But that is wrongly thought of me, right:)
Non linearity isn't about the amount of information in that 3D-image.
It's about if the solution of it mathematically is possible as a finite description?
finite here is should be a single clear answer, not indefineable.
I think? :)
---
You're right.
This thread will lead into the Jungle:)
-
This might be another question of the kind of language we're using to describe things. When I think of linear problems, it has to do with how the mathematics treats your input variables. If you have a linear operation, L, acting on some system, S1 and some system S2, then L(S1+S2)=L(S1)+L(S2). In other words, L acting on the sum of the systems equals the sum of L acting on each system independently. Additionally, if you multiply a system by a certain number, c, then L(cS)=cL(s).
In this language, you need to specify the process and the system you're concerned. The process of recording a hologram is generally nonlinear with respect to the input field, since its usually the square of the fields that gets recorded. Playing back a hologram is usually linear with respect to the input field since you just multiply the hologram by the reference field.
Reconstructing function at any point from a given patch isn't linear in the function's variables.
Linear systems are usually desirable because solving them is usually much simpler, and asking a computer to crunch the numbers is often much faster.
-
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?
A detailed explanation would require complicated drawings. Group velocity, in already existing EM waves, is an "artefact" just as phase velocity is. It is not only if you consider a so short pulse of light that you can be sure all of the EM perturbation can be identified with the very pulse; in that case group velocity is also signal velocity. Group velocity is an artefact because it represents the speed at which the "shape" of the wave packet travels; this shape moves just in virtue of phase shifting among the various pure components of the packet, so it's nothing physical who moves.
I still enjoy the way you methodically explain your point(s), but, does this mean that phase velocity soon will be 'gone' from physics?
What do you mean? Phase velocity is ω/k and both are important parameters and also phase speed is important because is directly related to index of refraction.
-
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."?
-----
Yes, JP (jpetruccelli) I knew that a linear equation is a solvable one :)
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.
--------
But I liked that other description wherein a wave is just a representation of the 'time quality' of causality.
Then it is like those 'traces' you see inside that 'fog chamber' where particles gets observed.
image for image put together to make a description in time.
It's both easier to accept and much more complex as it is about all movements, not only waves.
But now people will tell me that time is 'events' again:)
Nope, it's not.
This phenomena, if seen like I do (for the moment:), is more about the qualities of space / vacuum.
Also it's about what we define as movement generally.
We do it using the arrow of time that we observe macroscopically.
That's what define us all here.
So you won't be able to define any movements without using time as a component.
I don't really care if you try to 'obscure' time by changing the name to 'events'.
If there is a 'time like' component then its name doesn't matter.
In fact the idea of time being 'events' seems to create more problems than solutions.
Like needing a 'time aether' binding those 'events' etc.
Then what we see as a wave demands a 'causality' at least macroscopically.
You could see it as a 'sea of happenings' in that vacuum wherein 'spacetime' just 'put forward' those happenings that fit our macroscopic reality / observations (arrow of time).
Or like there is a causality arrow hidden there inside 'spacetime' as a whole.
If I by 'spacetime' also define and take into consideration those processes that we don't really can quantify.
Like virtual particles operating outside HUP.
-
I got the impression before that you meant that phase velocity was phased out
What does it mean?
:) in exchange for signal velocity Lightarrow?
That's why I asked. I haven't used that description 'signal velocity' before.
So you can say you have learnt something new [:)]
Btw: What do you mean by Phase speed being "directly related to the index of refraction."?
Vph = c/n where n = 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.
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.
-
'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:)
-
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.
-
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.
-
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.
-
'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.
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.
-
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?
-
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.
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.
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 [;)]).
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.
Can you give me an example of one single wavelength observed 'on its own' in spacetime?
What do you mean with 'on its own'?
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?
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).
-
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
-
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 [;)]
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.
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.
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?
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.
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).
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.
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.
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.
"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.
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
...
-
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:)
-
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.
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.
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! [;)]
-
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
-
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.
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?
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 [:)]).
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".
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.
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.
-
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.
-
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.
-
How do you talk about 'EM wave frequency 'without there being any 'charge' to the photon?
What is a "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'?
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.
-
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).
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... [:)])
-
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?
-
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.
-
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:)
-
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).
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.
-
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.
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).
-
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)
-
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?
Yes and no. If you want to consider the case of a single wavelenght having different frequencies, you are necessarily talking about different media in which the wave travels; then the proportionality is valid only in every medium but not both at the same time; nevertheless how can wavelenght be (inversely) proportional to frequency if one wavelenght correspond to 2 different frequencies? Inversal proportionality between frequency f and wavelenght λ means: f = k/λ, where k is a constant, so, if f1 = k/λ1 then you must have f2 = k/λ2 so if f1 ≠ f2 then λ1 must necessarily be ≠ λ2.
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.
Wait a moment! Don't know which is the example I made that you are referring to, but when a light wave goes from the void to glass, then the waves are not stretched, they are compressed, infact the wavelenght decreases.
The frequency instead stays the same.
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:)
Yes, it's so.
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.
No, the frequency stays the same, but wavelenght decreases, as I said.
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?
No, when light enters glass, the distance between the crests decreases (the "spring" is more compacted). Ok, let's see if I can make a more simple example. Think about soldiers who run, one after the other, at a precise time delay: let's say that when a soldier passes by a point in space, the next one will always pass after exactly 1 second, wherever they are; now you observe them running on a road; you measure the distance between two soldiers and you find exactly 8 metres; now you are able to compute the speed of this soldier's "wave": you know that the next soldier will arrive after 1 second, so he has covered 8 metres in 1 second --> speed = 8m/s.
Now you observe them running in the sand; theyr distance is 5 metres --> speed (= phase speed, as you now have understood) = 5m/s
The frequency of the wave didn't change because there is still 1 second delay between a soldier and the next one, but the wavelenght decreased, they have been "compacted". Incidentally you can observe the same in a F1 race: if the time delay between two cars is the same, they are "compacted" where they go slower (in the bends, for example) and "stretched" where they go faster. It could be strange, but with light it's exactly the same (think: not even need of relativity [:)])
An interesting thing: Let's have a road and, parallel and next to it, a strip of sand (like a road immediately next to a beach); if you want that your soldiers will go, in the least time possible, from a point A in the middle of the road to another point B, in the middle of the beach and not under (or up) the point A, the the soldiers have to make a path which is exactly the one you can compute with optical geometry and Snell's law of refraction!
http://www.iop.org/activity/education/Projects/Teaching%20Advanced%20Physics/Vibrations%20and%20Waves/Images%20300/img_tb_4456.gif
http://en.wikipedia.org/wiki/File:Snells_law_wavefronts.gif
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)
Ah, yes, I totally agree here [:)].
-
Awh, there goes my whole plan down the drain, so you say that the light when coming into the glass actually will get a more compressed frequency.
So when we have two frames. Frame_A and frame_B moving apart from each other.
And we on frame_A is observing a light beam coming at us from frame_B.
The we will notice a red shift, right.
And that red shift, if visualized, will then as observed have a greater :) distance between its crests and troughs?
But that shouldn't then be seen as a 'weaker energy' per time unit when the light finally is 'hitting' our frame?
If I look at it as I'm used to (particles) i think of them as getting 'spread out' in time as I'm moving from them and therefore 'weaker' per time unit.
And if we were moving those frames towards each other, and observe the light as being blue shifted.
Then that doesn't mean that our frame will receive a larger amount of 'energy' per time unit?
As that then, to me, would mean that those photons (particles) will get compressed as observed per time unit.
As that is why I want those phreaking waves to become of greater (is it 'magnitude' I should use here?) distance between their crests and troughs:) when moving through that prism?
When I think of them (waves) meeting any density (the prism) I see them as becoming 'depleted' of energy in their interactions with the electron clouds and therefore, as I thought, also 'red shifted' (stretched out in time so to speak:).
To me it seems strange to say that they become compressed as I associate that with blue shift.
Do you see how I think here?
Awhhh...
I think I will call on Manuel now:)
----
If one look at waves as getting compressed in that prism, how do one explain the 'photon wikis' statement that the 'energy' of a wave is directly correlated to its frequency?
"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.
For example, some chemical reactions are provoked only by light of frequency higher than a certain threshold; light of frequency lower than the threshold, no matter how intense, does not initiate the reaction. Similarly, electrons can be ejected from a metal plate by shining light of sufficiently high frequency on it (the photoelectric effect);
the energy of the ejected electron is related only to the light's frequency, not to its intensity.[26]"
http://en.wikipedia.org/wiki/Photon#Historical_development
Waves drive me wild...
Anybody wanna sponsor a ticket to Hawaii?
I heard they have great waves there:)
-----
If I think of it as photons hitting electron clouds inside that prism I can see how they will interact and energize those clouds, some of the energy of those photons will 'dissapear' as heat etc, but I'm not sure how they will get compressed in time. It seems like if we have a specific amount of 'balls' rolling through :) they all will be 'braked' the same amount inside that prism and became fewer or/and of lesser energy when coming out on the other side.
So yes, you are right, but it still wrecks havoc with my former ideas of their 'expression' as waves.
Am I wrong in how I see them as red and blue shifted relative those two frames then, when they meet my frame_A?
As blueshift to me is a higher frequency and more energy per time unit, and redshift the opposite?
And of course :) that wiki?
-------
Ah, here he are:)
- I know nothing.
- - Noothing I say::))
(So now we are two... Manuel, and me:)
-
Awh, there goes my whole plan down the drain, so you say that the light when coming into the glass actually will get a more compressed frequency.
NOT FREQUENCY! IT'S WAVELENGHT WHICH IS REDUCED! THE FREQUENCY STAYS THE SAME! How many times have I to write it? [xx(]
So when we have two frames. Frame_A and frame_B moving apart from each other.
And we on frame_A is observing a light beam coming at us from frame_B.
The we will notice a red shift, right.
Ok.
And that red shift, if visualized, will then as observed have a greater :) distance between its crests and troughs?
Yes.
But that shouldn't then be seen as a 'weaker energy' per time unit when the light finally is 'hitting' our frame?
Yes, but not only because of this! The amplitude is lower, too. Quantistically (without considering the amplitude) you also have *less photons* passing per unit time, in addition to have a reduced energy of every single photon.
If I look at it as I'm used to (particles) i think of them as getting 'spread out' in time as I'm moving from them and therefore 'weaker' per time unit.
Yes, see up.
And if we were moving those frames towards each other, and observe the light as being blue shifted.
Then that doesn't mean that our frame will receive a larger amount of 'energy' per time unit?
Yes.
As that then, to me, would mean that those photons (particles) will get compressed as observed per time unit.
Yes, in the sense I wrote up.
As that is why I want those phreaking waves to become of greater (is it 'magnitude' I should use here?) distance between their crests and troughs:) when moving through that prism?
1. What happens to waves when they enter through a prism has *nothing to do* with changing frame of reference! The energy of the wave *doesn't vary at all* inside the prism. The fact it has a smaller wavelenght has nothing to do with the energy, but only with the fact that phase velocity is smaller. Have I clarified it now? (Please, tell me yes [:)])
2. You MUST use the proper terms: the distance between two next crests is *wavelenght*.
When I think of them (waves) meeting any density (the prism) I see them as becoming 'depleted' of energy in their interactions with the electron clouds and therefore, as I thought, also 'red shifted' (stretched out in time so to speak:).
To me it seems strange to say that they become compressed as I associate that with blue shift.
Do you see how I think here?
Awhhh...
I think I will call on Manuel now:)
----
If one look at waves as getting compressed in that prism, how do one explain the 'photon wikis' statement that the 'energy' of a wave is directly correlated to its frequency?
THE FREQUENCY STAYS THE SAME! IT'S WAVELENGHT WHICH IS REDUCED!
"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.
This phrase is wrong, if you don't interpret it in the correct way; you have to specify what does it mean and in which conditions. It refers to photoelectric effect, not to an EM wave in general. It also explain it, it says: "the energy imparted by light to atoms", it doesn't say: "the energy of the EM wave".
-
Awh:)
Don't kill me now::))
I told you I was sloow.
---------
The wavelength becomes smaller between crests, ok.
My thick head sees those 'wavelengths' as the 'parts' making up a 'frequency' when put together Sir.
*Hiding under a table*
"(Please, tell me yes [:)])"
Ah, when you put it like that... [B)] [B)]
(yep, browbeaten to a pulp, that's me:)
Who can refuse such an earnest request.
So... Maybe:)
-----
I need to make a picture of it in my head sort of Lightarrow.
Don't give up on me now, we are closing in on the culprit I think (I think?)
----------
Those dastardly waves.
""They seek him here, they seek him there, they seek that, ah, wave everywhere."
As a obnoxious friend of mine once wrote:)
--------
I wrote this about redshift.
'But that shouldn't then be seen as a 'weaker energy' per time unit when the light finally is 'hitting' our frame?'
And your answer was.
" Yes, but not only because of this! The amplitude is lower, too. Quantistically (without considering the amplitude) you also have *less photons* passing per unit time, in addition to have a reduced energy of every single photon."
So the frames (_A and _B) moving away from each other in this scenario also, somehow, reduces the 'energy' inherent (if I may use that expression) in those 'single' photons. That you will need to explain to me.
-
The wavelength becomes smaller between crests ok.
My thick head sees those 'wavelengths' as the 'parts' making up a 'frequency' when put together Sir.
*Hiding under a table*
The wavelength can have closer crests and the time spent traversing the media can be greater. It requires wavelength plus time to produce frequency. [:)]
-
So is this a description without a time variable?
It's me not getting the concept here, I know [:I]
-
--
It's like there is two simultaneous descriptions.
One treating the idea of a wavelength, not considering it/them 'building up' to a frequency.
Another treating just a frequency.
What exactly is seen as a energy in a compressed wave.
I think I might get it?
What we are discussing here is not so much the waves 'compression' as the relation between a wave and the medium through which it traverse, am I right?
-
The wavelength becomes smaller between crests, ok.
My thick head sees those 'wavelengths' as the 'parts' making up a 'frequency' when put together Sir.
As Vern have already said (don't want to say that I already said this too, because it would be the second or third time I say that "I've already said it" [:)]) wavelength is not enough to "make up" frequency, you *also* need the speed of propagation of waves, that is phase speed. I also made the example of the soldiers, which are evenly delayed in time and that this time delay *doesn't vary* and so *the frequency doesn't vary* while their spacing, and so the wavelenght *does* vary, when they move from a medium to another and so when their (phase) velocity varies. What's wrong with that example?
"(Please, tell me yes [:)])"
Ah, when you put it like that... [B)] [B)]
(yep, browbeaten to a pulp, that's me:)
Who can refuse such an earnest request.
So... Maybe:)
-----
I need to make a picture of it in my head sort of Lightarrow.
Don't give up on me now, we are closing in on the culprit I think (I think?)
----------
Those dastardly waves.
""They seek him here, they seek him there, they seek that, ah, wave everywhere."
As a obnoxious friend of mine once wrote:)
--------
I wrote this about redshift.
'But that shouldn't then be seen as a 'weaker energy' per time unit when the light finally is 'hitting' our frame?'
Here, again, you are talking of different frames of reference, not about light going through a glass, are you aware of this?
And your answer was.
" Yes, but not only because of this! The amplitude is lower, too. Quantistically (without considering the amplitude) you also have *less photons* passing per unit time, in addition to have a reduced energy of every single photon."
So the frames (_A and _B) moving away from each other in this scenario also, somehow, reduces the 'energy' inherent (if I may use that expression) in those 'single' photons. That you will need to explain to me.
Ok. The fact that a single's photon's energy (what you say 'energy' inherent, which is not correct, because it reminds "intrinsic" which is wrong, because a photon has zero intrinsic energy because intrinsic means without considering kinetic energy) varies from a ref. frame to another is because the wavelenght, and so the frequency of the EM wave varies (be careful here! Now, I *can* say: "the wavelenght, and so the frequency", because we are considering *one only* medium, so phase velocity is constant, so frequency here *is* inversely proportional to wavelenght and so, one frequency ↔ one wavelenght); furthermore (but you can't understand this without knowing Maxwell's equations AND relativity) the electric and the magnetic field amplitudes (= "strenght" of the fields, to say it in a more simpistic way) varies: for a light source which is approaching, the fields increases, if it recedes, the field decreases.
Quantistically: we have said that a single photon's energy is reduced (a source which is receding); why also, the number of photons passing (or emitted, or absorbed, or detected) in the unit time is reduced too? That's very simple: exactly in the same way as if you shooted away balls (or anything else) at a constant rate (in your frame of ref.) and with constant speed to a friend still on Earth while receding from him: he will receive less balls per unit time, with respect to the case in which you don't move at all.
-
You say "Here, again, you are talking of different frames of reference, not about light going through a glass, are you aware of this?"
Yes, but I see our observation of that light inside that prism as two frames too, one is what happens inside that prism, the other frame is us outside, observing it.
As I said before, I'm not really sure on where the the definition of frames are valid in physics:)
But ok, we have the velocity, which is a measure of distance in time units, right?
And if we are talking about a massless photon that should be momentum?
Then we have the wavelength, which is a measure of distance between crests in time.
And frequency which is a measure of oscillating ( = vibrating ) 'periodic' waves which seems to mean waves that follow one another in regular succession.
Those periodic waves can be split into transverse and longitudinal waves, and here we come in on Verns favorite electromagnetic description of photons :) i think? And also on Lightarrows, as seen of lately, desperate tries to pommel some sense into my poor head relating to waves universe and all.
'Normal' electromagnetic waves can be seen as propagating (moving:) in a transverse manner.
a regular up-and-down pattern (sinusoidal) in which the 'vibration' or motion is perpendicular to the direction the wave is moving.
'Perpendicular' here means, seen as a straight line at right angles to another line , and the idea is like what you will see when you use a whip or if you look on how a snake travels on the ground.
And a longitudinal wave is a wave just like a sound wave, and that kind of waves spreads out in circles from a center point, like a siren f ex. and the 'vibration' of that motion moves in the same direction as the 'circle' of sound itself (360 degrees).
---
Btw: Siren should be understood as a warning signal that is a loud wailing sound.
And not as a woman who is considered to be dangerously seductive...
(not that I would mind though:)
------
But now the real headache starts:)
For we have been speaking of frequency as having a oscillating (vibrating) property, right?
But there are some property's that aren't related to those oscillating (whips & snakes, if transverse) properties at all, according to Lightarrow (and other Wizards and occult materials I've now consulted in my hunt for the holy grail, yes, the hermetic arts is near:)
So open your eyes now and hearken Romans Swedes and assorted other nationalities...
One of them 'non oscillating' is wavelength, (distance between crests), and yet another is called amplitude.
(height of wave, as seen from a thought 'middle line' splitting horizontally trough from crest).
Although wavelength isn't related to frequency mathematically, it still plays an important role together with amplitude when listening to sound.
And it is here I kind of lose it:) To me wavelengths seems a very simple concept where lots of the same sort of them, following each other in time, definitely will give a frequency.
But according to this they don't have any oscillating (vibrating) property's, and to give a description of them we need a new arcane concept. Here it comes, period...
Yep, that's it.
'Period' in a transverse wave (think snake:) is the time it will take for one 'cycle' from trough to crest and back to trough.
So with any 'oscillator' frequency will be the amount of time (seconds usually) it took to finish a cycle.
And in a longitude wave it will be the distance (interval) in time between those concentric circles of sound, or that stone you threw into the pond and the distance between those circles that you see expanding in the water.
But here we are discussing transverse waves, and in a 'oscillator' the frequency will be the number of cycles per second, and in wave motion (snake:) the number of waves that pass a given point per second, expressed in Hertz (Hz) after Heinrich Rudolf Hertz.
And so far so good, but I still haven't found out how wavelength suddenly is exchanged for period?
As wavelength is a description of crest to crest one might say that it is a metaphysical 'band aid' for explaining how waves compresses while periods is more like a 'three dimensional' description of something 'real' just as 'frequency' could be seen as, well, ah, as they have a direct relation.
And now I will hide behind the couch again :)
-
You say "Here, again, you are talking of different frames of reference, not about light going through a glass, are you aware of this?"
Yes, but I see our observation of that light inside that prism as two frames too, one is what happens inside that prism, the other frame is us outside, observing it.
As I said before, I'm not really sure on where the the definition of frames are valid in physics:)
No, we talk of different frames of reference, in relativity, when we are still or we are moving with respect to the physical system under analysis. When light enter a prism, we are still fixed in our chair in the laboratory, we don't suddenly move ourselves! You would measure a difference in the wave's wavelenght propagating in only one medium, only if you would change your speed with respect to the wave source (it means that you or the wave source (the lamp, the laser, ecc.) at a certain moment start moving). The fact light wave changes wavelenght when entering a prism is *all another physical principle* and for this reason the energy doesn't change, the frequency doesn't change (as, instead, in the previous case of two different frame of reference moving with respect to each other). Can I say that you have "integrated" this fact?
If you still have doubts on this, I go towards suicide! [:-'(]
But ok, we have the velocity, which is a measure of distance in time units, right?
And if we are talking about a massless photon that should be momentum?
No, momentum is all another thing. If you want to compute a photon's speed making the limit of the speed of a massive particle, when mass m goes to zero, you can do it! :
a massive particle's momentum is
p = m*v/Sqrt[1-(v/c)2] that is: p2 = m2v2/[1-(v/c)2]
on the other hand, from the special relativity relation between momentum and energy we have:
p2 = E2/c2 - m2c2
equating the two:
m2v2/[1-(v/c)2] = E2/c2 - m2c2
after various passages, we can extract v2:
v2 = (E2 - m2c4)/[m2c2 + (E2 - m2c4)/c2]
now we make the limit for m who goes to zero:
lim v2 =
m→0
= lim (E2 - m2c4)/[m2c2 + (E2 - m2c4)/c2] =
m→0
= E2/(E2/c2) = c2
So we have proved that any massless particle, with non-zero energy, must move at c.
Then we have the wavelength, which is a measure of distance between crests in time.
And frequency which is a measure of oscillating ( = vibrating ) 'periodic' waves which seems to mean waves that follow one another in regular succession.
Ok, however remember that, to talk of frequency, these waves *must* move, or you won't measure a periodic effect *in a fixed point of space*; to talk of wavelenght, you don't need to know if they are moving or not, because you don't measure an effect on time, but on space, fixed the time; have you "taken" this? [:)]
Those periodic waves can be split into transverse and longitudinal waves, and here we come in on Verns favorite electromagnetic description of photons :) i think? And also on Lightarrows, as seen of lately, desperate tries to pommel some sense into my poor head relating to waves universe and all.
'Normal' electromagnetic waves can be seen as propagating (moving:) in a transverse manner.
Even "no-normal" ones [:)]. Light waves cannot have longitudinal components, or they would have a mass (and they don't have...)
a regular up-and-down pattern (sinusoidal) in which the 'vibration' or motion is perpendicular to the direction the wave is moving.
'Perpendicular' here means, seen as a straight line at right angles to another line , and the idea is like what you will see when you use a whip or if you look on how a snake travels on the ground.
And a longitudinal wave is a wave just like a sound wave
...in air; (in a solid it can have transverse components).
and that kind of waves spreads out in circles from a center point
why, transverse waves can't spread out in circle? They can too.
, like a siren f ex. and the 'vibration' of that motion moves in the same direction as the 'circle' of sound itself (360 degrees).
---
Btw: Siren should be understood as a warning signal that is a loud wailing sound.
And not as a woman who is considered to be dangerously seductive...
Damn! For a moment I had hoped about her! [:)]
(not that I would mind though:)
------
But now the real headache starts:)
For we have been speaking of frequency as having a oscillating (vibrating) property, right?
Again, remember that this oscillation happens *in the time* fixed the space, otherwise you can't talk of frequency.
But there are some property's that aren't related to those oscillating (whips & snakes, if transverse) properties at all, according to Lightarrow (and other Wizards and occult materials I've now consulted in my hunt for the holy grail, yes, the hermetic arts is near:)
So open your eyes now and hearken Romans Swedes and assorted other nationalities...
One of them 'non oscillating' is wavelength, (distance between crests), and yet another is called amplitude.
(height of wave, as seen from a thought 'middle line' splitting horizontally trough from crest).
Although wavelength isn't related to frequency mathematically,
maybe you intended "physically".
it still plays an important role together with amplitude when listening to sound.
And it is here I kind of lose it:) To me wavelengths seems a very simple concept where lots of the same sort of them, following each other in time, definitely will give a frequency.
No; if "the snake" doesn't move, you still have wavelenght (distance between its crests or its "curls"), but you don't have any frequency at all.
-
Ah, we agree on the Sirens :)
I will reread you, and hopefully come up with something (driving you to new heights of despair:) soon Lightarrow.