What would slowed light look like if you overtook it?

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Offline DoctorBeaver

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In some mediums light travels slower than c. In diamond, for instance, its speed is less than 0.5c.

Imagine, then, that an artificial diamond millions of miles long was made and a beam of light fired along its length. If you were outside of the diamond travelling along its length at 0.75c, and the beam was fired before you reached its point of origin, how would the leading edge of the beam appear to you as you caught up with it and overtook it? Would it simply be a case of frequency-shifting until you passed it? I assume it would become invisible once you had gone past it.
« Last Edit: 01/01/2009 22:10:34 by DoctorBeaver »
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Offline LeeE

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What would slowed light look like if you overtook it?
« Reply #1 on: 02/01/2009 00:44:18 »
You wouldn't see it at all.  As pointed out in another thread, you can't detect light until it interacts with something and gives up at least some of it's energy to the detection process.

It seems to me though, that if light is emitted in a vacuum and then enters a transparent medium where it is slowed, the energy must be maintained in some way, for when the light subsequently leaves the medium and resumes it's original speed it is not given energy by the medium to do so.
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What would slowed light look like if you overtook it?
« Reply #2 on: 02/01/2009 07:40:01 »
Hard to say. I'd have to think about this one.
http://www.youtube.com/watch?v=SZGcNx8nV8U

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Offline Soul Surfer

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What would slowed light look like if you overtook it?
« Reply #3 on: 02/01/2009 10:41:34 »
Look at my reply on the slowed light in Bose Einstein condensates for further information.  In effect the photons are continually being absorbed and re-emmitted inside the time limits allowed bt the Heisenberg uncertainty principle effectively delaying their progress.
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Offline DoctorBeaver

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What would slowed light look like if you overtook it?
« Reply #4 on: 02/01/2009 11:22:43 »
You wouldn't see it at all.  As pointed out in another thread, you can't detect light until it interacts with something and gives up at least some of it's energy to the detection process.

Surely, it would interact with particles in the diamond?
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Offline LeeE

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What would slowed light look like if you overtook it?
« Reply #5 on: 02/01/2009 14:47:41 »
You wouldn't see it at all.  As pointed out in another thread, you can't detect light until it interacts with something and gives up at least some of it's energy to the detection process.

Surely, it would interact with particles in the diamond?

Some of the light will be absorbed by the medium, so that proportion of it will be interacting but you'll not be able to see that interaction unless there's some energy left over from the interaction for you to detect (I suppose you could use an external source of energy to monitor the medium for changes but then you'd be detecting the energy from the external source, not the original light beam).  Although I believe there are some materials that can change the wavelength of EMR passing through it, I don't think that diamond is one of them, so if the light leaving the diamond is the same as the light entering it, minus whatever is absorbed by the diamond, then any interactions must be perfectly symmetrical, leaving both the medium and the light unchanged after the interaction.  I think??
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Offline DoctorBeaver

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What would slowed light look like if you overtook it?
« Reply #6 on: 02/01/2009 14:54:46 »
I'm not sure I follow that. If you shine a laser into am ordinary diamond, are you saying no light will emerge?
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Offline lightarrow

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What would slowed light look like if you overtook it?
« Reply #7 on: 02/01/2009 16:24:16 »
In some mediums light travels slower than c. In diamond, for instance, its speed is less than 0.5c.

Imagine, then, that an artificial diamond millions of miles long was made and a beam of light fired along its length. If you were outside of the diamond travelling along its length at 0.75c, and the beam was fired before you reached its point of origin, how would the leading edge of the beam appear to you as you caught up with it and overtook it? Would it simply be a case of frequency-shifting until you passed it? I assume it would become invisible once you had gone past it.
Now you begin to ask too difficult questions!  [:)]

If you are interested in what you see coming from the side of the diamond's bar, assuming you are moving towards the right, then I would say that you shoud probably see a lower frequency (red-shifted) and dimmed light coming out from the left portion of the bar and an higher frequency (blue-shifted) and more intense light coming out from the right one; this because of the fact that every point in the side of the bar can be considered as a source of the same light which enters the bar, because of the little but non-zero scattering of light within the material; when you have overtook the light beam's speed, instead, you shouldn't see light coming from the right portion anylonger, but only from the left.

Note that this would probably be very different from being inside the medium (let's say water, instead of diamond, it's easier...) moving along the bar in the same direction of the beam and looking at the beam directly coming towards us from back or from ahead, because in this case you should see nothing coming from back and coming from ahead, when you reach light's speed inside the medium; when instead you overtake it, you should (maybe) start to see again some light (strongly red-shifted and dimmed) coming from ahead (while still see nothing coming from back) because you are now going "through" the electromagnetic waves present in front of you.
All I've written is higly speculative however because I actually don't know.  [:)]
Good question.  [:)]
« Last Edit: 02/01/2009 16:34:45 by lightarrow »

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Offline DoctorBeaver

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What would slowed light look like if you overtook it?
« Reply #8 on: 02/01/2009 17:22:24 »

If you are interested in what you see coming from the side of the diamond's bar, assuming you are moving towards the right, then I would say that you shoud probably see a lower frequency (red-shifted) and dimmed light coming out from the left portion of the bar and an higher frequency (blue-shifted) and more intense light coming out from the right one; this because of the fact that every point in the side of the bar can be considered as a source of the same light which enters the bar, because of the little but non-zero scattering of light within the material; when you have overtook the light beam's speed, instead, you shouldn't see light coming from the right portion anylonger, but only from the left.


That's more-or-less what I thought. However, would the light to the right be so blue-shifted that it would be invisible? Say, shifted into the gamma ray part of the spectrum? And how could you see any light to the left? You would be travelling faster than it was.
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Offline lightarrow

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What would slowed light look like if you overtook it?
« Reply #9 on: 02/01/2009 19:55:54 »

If you are interested in what you see coming from the side of the diamond's bar, assuming you are moving towards the right, then I would say that you shoud probably see a lower frequency (red-shifted) and dimmed light coming out from the left portion of the bar and an higher frequency (blue-shifted) and more intense light coming out from the right one; this because of the fact that every point in the side of the bar can be considered as a source of the same light which enters the bar, because of the little but non-zero scattering of light within the material; when you have overtook the light beam's speed, instead, you shouldn't see light coming from the right portion anylonger, but only from the left.


That's more-or-less what I thought. However, would the light to the right be so blue-shifted that it would be invisible? Say, shifted into the gamma ray part of the spectrum?
Well, there is no need to go to gamma rays: out of the visible range, in the higher frequencies, you first find ultraviolet, then X-rays, then, at very high energies, gamma rays. Your answer is correct; however, even if light receptors in the retina don't directly perceive those frequencies, actually, especially increasing the frequency, you would start to perceive something in an indirect way: photoluminescence of chemicals inside the eye, then (increasing the frequency) free electrons produced by Compton scattering which recombine with ions or other molecular species, Cerenkov radiation from electrons and then from ions and atoms...a lot of physical phenomena which can generate visible light inside the eye or that can stimulate the discharge of an electrical stimulus from the nerves connected to the eye. It would probably be a sort of "fireworks".

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And how could you see any light to the left? You would be travelling faster than it was.
Focalize your attention on a particular point on the external side of the bar: that point doesn't move but it is a source of a spherical wave of light, expanding in all directions and travelling in the void, so its speed is c.

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Offline DoctorBeaver

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What would slowed light look like if you overtook it?
« Reply #10 on: 02/01/2009 20:18:34 »
OK, I understand now. Thank you, Alberto  [:)]
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Offline yor_on

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What would slowed light look like if you overtook it?
« Reply #11 on: 03/01/2009 09:30:02 »
It is a very interesting question.
It goes into how photons interact with matter (atoms), right.
As I understand it the photons going in will be 'replaced' an enormous amount of times before coming out on the other side.
One proof on that is that light 'bends' (change angle) inside transparent materials, am I right in that?
Because if there was no 'interaction' how could one explain that difference between incoming and outgoing.
Yep, that's me, outgoing at a rakish angle:)

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Also, Isn't it so that what we call the 'slowing' of light, is a result of its interactions with matter?
So in reality light does not slow down, it's only its interactions that take some (space)time?
« Last Edit: 03/01/2009 09:33:26 by yor_on »
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Offline DoctorBeaver

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What would slowed light look like if you overtook it?
« Reply #12 on: 03/01/2009 09:41:38 »
I think you're right on all counts.
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Offline lightarrow

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What would slowed light look like if you overtook it?
« Reply #13 on: 03/01/2009 09:41:43 »
It is a very interesting question.
It goes into how photons interact with matter (atoms), right.
As I understand it the photons going in will be 'replaced' an enormous amount of times before coming out on the other side.
One proof on that is that light 'bends' (change angle) inside transparent materials, am I right in that?
Because if there was no 'interaction' how could one explain that difference between incoming and outgoing.
Yep, that's me, outgoing at a rakish angle:)

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Also, Isn't it so that what we call the 'slowing' of light, is a result of its interactions with matter?
So in reality light does not slow down, it's only its interactions that take some (space)time?
What you say is essentially correct. The slowing down and the bent is a result of light interaction with matter.

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Offline yor_on

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What would slowed light look like if you overtook it?
« Reply #14 on: 03/01/2009 10:13:30 »
In fact, thinking of it the same must be true for a Bose Einstein Condensate.
what they are slowing down by 'bouncing' must create new photons.
And that 'interference' they use at the end, how would that be explained if looking at photons as particles.

What I'm really curious on is if one can 'brake' a original photon, or if it is an infinite amount of interactions creating 'new' photons inside that BEC.
Also how those new photons can become slowed down.
I know that they relative their surrounding will be seen as losing energy each time.

Can a 'original' photon lose its energy?
And still be a photon?

In a BEC they do seem to be able too.

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That is if you look at them 'disappearing' and then coming back again.
Up to that point one could expect it to be about interchanging photons.
But there it seems like it should be the 'original' ones?

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Or is it 'spacetime ' containing all that energy and a 'memory' of what just happened, and then it would be 'new' photons appearing (supposedly) after that 'interference' is withdrawn.

-------

But most probably it is easily explained, even though I don't know how?
« Last Edit: 03/01/2009 10:46:06 by yor_on »
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Offline DoctorBeaver

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What would slowed light look like if you overtook it?
« Reply #15 on: 03/01/2009 10:25:33 »
BEC confuses the hell out of me  [:-\]
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Offline yor_on

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What would slowed light look like if you overtook it?
« Reply #16 on: 03/01/2009 10:57:36 »
You know, even if looked at as waves it still phreaks me out.
First of all, how do one 'brake' a wave?
You can 'quench' it, but 'slow' it down?

----
Let me remind you that all analogues that come to my mind, explaining how waves can be slowed is referring to matter (density), but a light-wave have no density at all, and if seen as only waves I have great problems seeing a momentum too.
----

And secondly, if you quench it?
If one looked at that last instant when all waves 'disappear' inside that BEC.
Would then that be seen as 'quenching'?

But then it seems that nature and us aren't on the same wavelength here:)
Either we have 'some residue' left even when 'quenching' out light.
Or it is truly gone.

A BEC says that it's not gone.
If i get it right, that is:)
« Last Edit: 03/01/2009 11:08:28 by yor_on »
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Offline Bikerman

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What would slowed light look like if you overtook it?
« Reply #17 on: 03/01/2009 11:20:53 »
My (suspect) understanding is that the group velocity is slowed (rather than the phase velocity as in refraction).
Don't know if this helps or not:
http://en.wikipedia.org/wiki/Slow_light

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Offline lightarrow

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What would slowed light look like if you overtook it?
« Reply #18 on: 03/01/2009 15:07:38 »
In fact, thinking of it the same must be true for a Bose Einstein Condensate.
what they are slowing down by 'bouncing' must create new photons.
If you are thinking to free photons which, sometimes, collide with atoms or electrons it's not correct: actually photons are excitations of the electromagnetic field present inside the material, you cannot separate them from the material itself. The photon's speed so are determined by the optical properties of the material, namely the signal velocity of light (at a given frequency). Anyway the details are obscure to me; a complete description requires quantum electrodynamics in condensated states, I imagine.

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And that 'interference' they use at the end, how would that be explained if looking at photons as particles.
See up.

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What I'm really curious on is if one can 'brake' a original photon
No, that's not possible.

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, or if it is an infinite amount of interactions creating 'new' photons inside that BEC.
Why infinite? However you should see it not as an interaction between photons and the material, but as an interaction of the quantized electromagnetic field and the material. So, you can't avoid considering the wave-like properties of the field, in addition to the particle-like ones.

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I know that they relative their surrounding will be seen as losing energy each time.
What you mean?

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Can a 'original' photon lose its energy?
And still be a photon?
For what I know, no, it's not possible, the original photons are probably destroyed and new photons recreated; however, even this statement is probably false in the sense that probably it's not even possible to talk about photons between the initial source and the final detector...

Quote
In a BEC they do seem to be able too.
----------
That is if you look at them 'disappearing' and then coming back again.
Up to that point one could expect it to be about interchanging photons.
But there it seems like it should be the 'original' ones?
Better the first one  [:)]

Quote
Or is it 'spacetime ' containing all that energy and a 'memory' of what just happened, and then it would be 'new' photons appearing (supposedly) after that 'interference' is withdrawn.
-------
But most probably it is easily explained, even though I don't know how?
No one knows the answer, yet...Photons are extremely complicated objects (I don't even know if it's correct to call them "objects"... [:)]).
« Last Edit: 03/01/2009 15:11:50 by lightarrow »

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Offline LeeE

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What would slowed light look like if you overtook it?
« Reply #19 on: 03/01/2009 17:44:30 »
I'm not sure I follow that. If you shine a laser into am ordinary diamond, are you saying no light will emerge?

Nope - exactly the opposite.  Most of the light will re-emerge from the diamond and the light that does emerge will have the same characteristics as the light that entered i.e. frequency.  So apart from the light that's absorbed, and assuming that we're talking about a perfect diamond where there will be very little scattering, hardly any energy will be lost in the transit.  For you to be able to detect the light some the energy from the light must be intercepted by your detectors but if the amount of light that goes in is equal to the amount that comes out, minus the amount that's absorbed, then there's none left over for you to detect.  Sure, some light will be scattered, and you could detect this light as it would not re-emerge with the main beam, but as scattering means that it's direction has changed it will no longer moving, relative to you, as you originally specified, as clearly, it is now moving towards you, or your detector, and not on it's original course.
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Offline yor_on

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What would slowed light look like if you overtook it?
« Reply #20 on: 03/01/2009 17:56:23 »
Lightarrow, I think I stand by my questions as they are :)

Give me a clearer description if you want me to see how you think here.
(Although I actually understand what you say about 'excitations':)

But if I choose to see photons 'particle-wise' and then question the way they will act if so?
Then that also works as far as I know.
It is not forbidden, is it?

Remember that box you described.
What do that concept make them:)

As for 'infinite', of course not, just an expression used as I questioned it.
What I meant there was, more than I ever would like to enumerate, more than the visible stars etc etc :)
But you're right, I should be more exact.

But you see it foremost as 'wave patterned' if I get you right.
I still look at it both ways, with an emphasis on 'particles'.
Where you accept all in form of waves and their interactions, i choose to differ, that as I don't see us as 'light' but as foremost 'matter':)

That doesn't mean that I don't know, or isn't interested in, that approach too.
But firsthand I'm just trying to find a resolution to my own satisfaction, and so I will walk in the wilderness somewhat more:::)))

-------

And yes, I think I might know how you might see photons:)
But I'm stubborn son of a ***
Let us take moon light:)

Coming mainly from the sun according to me, reflected by the moon down to me dreamingly thinking about my next BMW (electrical of course:). Traveling at 'c' mostly, (ah, that would be the Suns light traveling, not the BMW:) quite a distance, caressing my loved ones chrome in about ?nine minutes? from where it left the 'surface' of our Sun.

Would you accept my description as being the 'one and only', or might it just be that there could be another description as valid?
« Last Edit: 03/01/2009 18:13:11 by yor_on »
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Offline DoctorBeaver

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What would slowed light look like if you overtook it?
« Reply #21 on: 03/01/2009 19:59:48 »
LeeE - thank you for clarifying that.
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Offline lightarrow

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What would slowed light look like if you overtook it?
« Reply #22 on: 03/01/2009 23:14:45 »
Lightarrow, I think I stand by my questions as they are :)

Give me a clearer description if you want me to see how you think here.
(Although I actually understand what you say about 'excitations':)

But if I choose to see photons 'particle-wise' and then question the way they will act if so?
Then that also works as far as I know.
It is not forbidden, is it?

Remember that box you described.
What do that concept make them:)
As for 'infinite', of course not, just an expression used as I questioned it.
What I meant there was, more than I ever would like to enumerate, more than the visible stars etc etc :)
But you're right, I should be more exact.

But you see it foremost as 'wave patterned' if I get you right.
I still look at it both ways, with an emphasis on 'particles'.
Where you accept all in form of waves and their interactions, i choose to differ, that as I don't see us as 'light' but as foremost 'matter':)

That doesn't mean that I don't know, or isn't interested in, that approach too.
But firsthand I'm just trying to find a resolution to my own satisfaction, and so I will walk in the wilderness somewhat more:::)))

-------

And yes, I think I might know how you might see photons:)
But I'm stubborn son of a ***
Let us take moon light:)

Coming mainly from the sun according to me, reflected by the moon down to me dreamingly thinking about my next BMW (electrical of course:). Traveling at 'c' mostly, (ah, that would be the Suns light traveling, not the BMW:) quite a distance, caressing my loved ones chrome in about ?nine minutes? from where it left the 'surface' of our Sun.

Would you accept my description as being the 'one and only', or might it just be that there could be another description as valid?

Probably I haven't expressed myself very clearly. First of all, I would like to remark the fact that I actually don't know how it really works...I have never seriously studied QED (Quantum Electro Dynamics); it's not an easy subject, for me.
Second, about photon's behaviour, which is very complicated, believe me, many people who think to have understood it, don't, actually.

I also have to precise again that photons cannot be seen uniquely as waves or as particles and so sometimes, as in this case, a complete description requires both aspects, that is, QED. Of course you can consider just the particle-like behaviour, if you prefer (and, of course, it's not forbidden ), but if you are able to understand all of the photon's behaviour with that only, then make me know, because I would really be interested.

"Remember that box you described. What do that concept make them:)"

I'm not a boy anylonger, I'm becoming slow  [:)] What are you referring at?

About your description of light coming from the Sun and reflecting off the Moon, I don't have anything to say, unless you meant that the same photons which have travelled from the Sun to the Moon are the same who then will travel from the Moon to you (because it's not).

About photons, I'm trying to read this interesting document:
http://stochastix.files.wordpress.com/2008/05/what-is-a-photon.pdf

In one of the articles inside that document, Arthur Zajonc, talking about a book written from Davies, says:
<<Davies uses these and other
problems to argue for a vigorous Copenhagen interpretation
of quantum mechanics that abandons the idea of a “particle as
a really existing thing skipping between measuring devices.”
To my mind, Einstein was right to caution us concerning
light. Our understanding of it has increased enormously in
the 100 years since Planck, but I suspect light will continue to
confound us, while simultaneously luring us to inquire ceaselessly
into its nature.>>
« Last Edit: 03/01/2009 23:49:01 by lightarrow »

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Offline yor_on

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What would slowed light look like if you overtook it?
« Reply #23 on: 04/01/2009 00:22:39 »
I agree, they're not the same photons.

Sorry, I was thinking about the way a photon/wave packet 'travel'.
There are some rather advanced ideas about how they might 'move' or rather 'not move', leaning on Feynman, Fourier transformations and 'De Broglie Matter Waves' amongst others.

I wondered if you too was speculating in those 'orbits', so to speak.
I have a good friend and, if I may say so, real gentleman GoodElf whom is a ardent advocate for this view.
Goggle on him if you're interested.
He is very knowledgeable and as I see it a 'foregoer' in his ideas richness.
As for his math I'm sure you're better equipped to understand it than me.
And that goes for some more people here::))

I got this feeling that you all here have ideas of your own.
But you are cool about it, and others, which I find to be a matter of taste.
And good taste, to my eyes:)

And thanks for the link..
I live for them:)
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What would slowed light look like if you overtook it?
« Reply #24 on: 04/01/2009 11:59:20 »
I'm lost  [???]
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Offline yor_on

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« Reply #25 on: 04/01/2009 12:35:28 »
Bikerman Isn't it group velocity that enables a phase velocity?
So when 'quenching' group velocity you will 'quench' any information remaining?
Can there exist only a phase velocity??

Or am I misunderstanding you?
Or it??

http://en.wikipedia.org/wiki/Group_velocity
« Last Edit: 04/01/2009 12:43:27 by yor_on »
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Offline lightarrow

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« Reply #26 on: 04/01/2009 16:00:33 »
I'm lost  [???]
Where? Ah, yes, if you knew it, you wouldn't be lost... [:)]

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Offline DoctorBeaver

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« Reply #27 on: 04/01/2009 16:07:18 »
I'm lost  [???]
Where? Ah, yes, if you knew it, you wouldn't be lost... [:)]

I got lost when BEC was mentioned
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Offline lightarrow

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« Reply #28 on: 04/01/2009 16:19:28 »
Bikerman Isn't it group velocity that enables a phase velocity?
So when 'quenching' group velocity you will 'quench' any information remaining?
Can there exist only a phase velocity??

Or am I misunderstanding you?
Or it??

http://en.wikipedia.org/wiki/Group_velocity
"Isn't it group velocity that enables a phase velocity?"
This phrase is obscure to me. Do you mean that a single sinusoidal wave cannot exist?
Phase velocity is the speed of a single sinusoidal wave. If there are more than one single wave, then the two or more waves interfere (overlap) producing another shape. This shape travels at a generally different speed (group velocity).

If:

Vph = phase velocity

Vg = group velocity

ν = frequency of a single wave

ω = 2πν = pulsation but is usually called frequency too

λ = wavelenght of a single wave

k = wavenumber = 2π/λ

then you have:

Vph = λν = ω/k  -->  ω = kVph

Vg = dω(k)/dk

The last formula means: if many frequencies propagates in a medium which, in general, can be "dispersive" = the various frequencies has different phase velocities in it = ω depends on k in a non-linear way (see blu formula), then the derivative dω(k)/dk gives the speed at which the shape of all the packet of waves travels.

Edit. The formula Vg = dω(k)/dk is valid of course even when the medium is not dispersive. In this case Vph is a constant so Vg = dω(k)/dk = d(kVph)/dk = Vph.
Example: for light in the void, Vg = Vph = c.
« Last Edit: 05/01/2009 14:59:14 by lightarrow »

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Offline lightarrow

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« Reply #29 on: 04/01/2009 16:22:00 »
I'm lost  [???]
Where? Ah, yes, if you knew it, you wouldn't be lost... [:)]

I got lost when BEC was mentioned
Bose Einstein Condensate. Is a cloud of extremely cold atoms, prepared and confined specifically, through which light propagates at a cyclist's speed. It's quite similar to the fact light travels at lower than c in glass or water.

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Offline Mr. Scientist

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« Reply #30 on: 04/01/2009 16:27:48 »
My answer now, is that you would see it.

I needed some time to think about it. Hard question.
http://www.youtube.com/watch?v=SZGcNx8nV8U

''God could not have had much time on His hands when he formed the Planck Lengths.''

 ̿ ̿ ̿ ̿̿'\̵͇̿̿\=(●̪•)=/̵͇̿̿/'̿'̿̿̿ ̿ ̿̿ ̿ ̿

٩๏̯͡๏۶

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Offline Bikerman

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« Reply #31 on: 04/01/2009 17:07:59 »
Bikerman Isn't it group velocity that enables a phase velocity?
So when 'quenching' group velocity you will 'quench' any information remaining?
Can there exist only a phase velocity??

Or am I misunderstanding you?
Or it??
I was about to compose a length reply but I see that it has already been done above by lightarrow....basically, what he/she said  [:)]

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Offline DoctorBeaver

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« Reply #32 on: 04/01/2009 17:15:56 »
Bikerman Isn't it group velocity that enables a phase velocity?
So when 'quenching' group velocity you will 'quench' any information remaining?
Can there exist only a phase velocity??

Or am I misunderstanding you?
Or it??
I was about to compose a length reply but I see that it has already been done above by lightarrow....basically, what he/she said  [:)]

He. Alberto.
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Offline Bikerman

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« Reply #33 on: 04/01/2009 17:28:00 »
Ahh...OK. I normally make a point of not assuming gender on the internet (more out of courtesy than anything else)  [:)]

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Offline DoctorBeaver

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« Reply #34 on: 04/01/2009 17:58:42 »
Ahh...OK. I normally make a point of not assuming gender on the internet (more out of courtesy than anything else)  [:)]

Same here. Unless their pic has a beard like SpeakerToAnimals at ScienceFile   [:D]
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Offline Bikerman

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« Reply #35 on: 04/01/2009 18:02:04 »
Ahh...OK. I normally make a point of not assuming gender on the internet (more out of courtesy than anything else)  [:)]

Same here. Unless their pic has a beard like SpeakerToAnimals at ScienceFile   [:D]
LOL...yes, a cunning deception she uses :-)

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Offline DoctorBeaver

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« Reply #36 on: 04/01/2009 18:02:59 »
Ahh...OK. I normally make a point of not assuming gender on the internet (more out of courtesy than anything else)  [:)]

Same here. Unless their pic has a beard like SpeakerToAnimals at ScienceFile   [:D]
LOL...yes, a cunning deception she uses :-)

 [:0]

I saw once that you referred to STA as "she" but I assumed it was a slip of the fingers on your keyboard.
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Offline Bikerman

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« Reply #37 on: 04/01/2009 18:05:48 »
Speaker is in fact a she. She describes herself as a 'fat dyke'. She is also, in my opinion, a fabulous teacher and a very clever lady (PhD in particle physics). I have learned a great deal from Speaker and she is one of my favourite internet correspondents.
« Last Edit: 04/01/2009 18:12:48 by Bikerman »

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Offline DoctorBeaver

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« Reply #38 on: 04/01/2009 18:12:05 »
Gawd! I owe her an apology.

I remember your saying that she had a PhD in particle physics.
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Offline Bikerman

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« Reply #39 on: 04/01/2009 18:14:56 »
Gawd! I owe her an apology.

I remember your saying that she had a PhD in particle physics.
It is OK - she often gets mistaken for a 'bloke' and doesn't generally hold a grudge - in fact I think she quite enjoys revealing that she is in fact a large lesbian lady (doesn't half knock the religious bunnies off their stride in debates)

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« Reply #40 on: 04/01/2009 18:16:12 »
Gawd! I owe her an apology.

I remember your saying that she had a PhD in particle physics.
It is OK - she often gets mistaken for a 'bloke' and doesn't generally hold a grudge - in fact I think she quite enjoys revealing that she is in fact a large lesbian lady (doesn't half knock the religious bunnies off their stride in debates)

I can imagine!  [;D]
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Offline yor_on

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« Reply #41 on: 05/01/2009 09:18:23 »
Lightarrow:)
thanks.

I checked it out after you wrote and you are definitely correct.
I just looked at it out of the aspect of FTL.
Thats the problem with not having a good grounding in 'basic physics' I presume.
no matter how 'hungry' one are:)


Especially the way you set up and explained your equation/formula.
It's very refreshing to see it like that.
It makes it that much easier to put into context.

---
Lost DB? :)
To me you seem to 'find yourself' rather quickly::))
« Last Edit: 05/01/2009 09:25:08 by yor_on »
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Offline lightarrow

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« Reply #42 on: 05/01/2009 14:51:29 »
Lightarrow:)
thanks.
You're welcome! For completeness, I have added a little consideration to that post.

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« Reply #43 on: 05/01/2009 15:05:50 »
yor_on
Quote
But if I choose to see photons 'particle-wise' and then question the way they will act if so?
Then that also works as far as I know.
It is not forbidden, is it?
Not 'forbidden' but I should say that you can't demand an answer in those terms. I don't know of a way to explain diffraction (photons, electrons or anything) if you don't take the wave approach. Likewise, I don't see an explanation can be made without thinking about waves here, either.
Why do you want to 'insist'? Standing up in a hammock is not the best way to make love, either.

I could get on my hobby horse here and say "how can you describe something like a photon unless you specify its characteristics in more depth than just calling it a particle?" If you can do that, then you may come up with the answer yourself.
« Last Edit: 05/01/2009 15:08:15 by sophiecentaur »

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« Reply #44 on: 05/01/2009 23:59:32 »
SC :)

I would say that it goes back to what we call matter and light?
And my own desire to understand it, same as yours I guess:)
You say that you are happy with a wave approach, and I agree that it makes a lot of phenomena more easily understood.
Like red shift when climbing up a gravity well. or tunneling.

But I am matter, and so are you, and I see waves every day, but that's not me.
To me there is a difference, so I want to see how far I can get it together:)

It's my mind game, it keeps me from getting bored, and let me discuss with interesting people what I find the most fascinating subject ever, spacetime.

Be warned (s)he who starts to think will easily find it becoming a habit.
Ah, sort of?

And its better than discussing your new car, don't you agree:)



« Last Edit: 06/01/2009 00:01:41 by yor_on »
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« Reply #45 on: 06/01/2009 09:59:26 »
I'm lost  [???]
Where? Ah, yes, if you knew it, you wouldn't be lost... [:)]

I got lost when BEC was mentioned
Bose Einstein Condensate. Is a cloud of extremely cold atoms, prepared and confined specifically, through which light propagates at a cyclist's speed. It's quite similar to the fact light travels at lower than c in glass or water.

I think there's more to it than that. What about the Pauli Exclusion Principle? Don't the fermions in a BEC behave like bosons? That's a point - does the particles' spin change in a BEC?

I'm at a sort-of halfway house with BEC (as with a lot of other things). I've read and understand the low level stuff but I can't understand the mathematical treatments that explain it in depth. I wish there was a site that explained things in more depth but in idiot language (if that's possible).
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« Reply #46 on: 06/01/2009 10:11:39 »
Quote
But I am matter, and so are you, and I see waves every day, but that's not me.

There you go again!
If the electrons inside you were not waves, as well, then why don't they fall into the nuclei of their atoms?

You have to open your mind, `Grasshopper' or you will never come to terms with this stuff. :)

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« Reply #47 on: 06/01/2009 10:18:17 »
Yes Sempai :)

That is in fact a very good question.
If someone would like to do a 'take on it' from a 'particle' perspective I'm all ears.
(quite handy when it comes to flying)

Otherwise I will have to think...
A lot....

--------

But I never said that waves shouldn't be counted in, have I?
If I left that impression, I didn't mean it.

What I am unhappy with is explaining matter as waves and then leave it there.
It reminds me of alchemy in where you define numerous imaginative properties to objects and then from some lofty perspective say 'as above so under'.
It may describe relations better but it does not take a grip on what is the difference between matter and light.
I can't walk through a window pane.

(well I can, but not without breaking it:)
But I am trying to see what unify light with matter.
Like momentum
And spacetime geometry/geodesics.
« Last Edit: 06/01/2009 13:16:03 by yor_on »
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« Reply #48 on: 06/01/2009 14:56:50 »
Bose Einstein Condensate. Is a cloud of extremely cold atoms, prepared and confined specifically, through which light propagates at a cyclist's speed. It's quite similar to the fact light travels at lower than c in glass or water.

I think there's more to it than that. What about the Pauli Exclusion Principle? Don't the fermions in a BEC behave like bosons? That's a point - does the particles' spin change in a BEC?

I'm at a sort-of halfway house with BEC (as with a lot of other things). I've read and understand the low level stuff but I can't understand the mathematical treatments that explain it in depth. I wish there was a site that explained things in more depth but in idiot language (if that's possible).

I tell you as I've (perhaps) understood it: at "normal" densities and temperatures the wavefunctions describing the atoms don't "overlap" in a significant way and so different atoms can be considered as "independent" on each other; every atom is a single entity. At very high densities and very low temperatures (see also Heisenberg principle and tunnel effect) those wavefunction can overlap so much that single atoms become indistinguishable; at that moment a couple, or even a large number of atoms becomes an only quantum system, which total spin is the sum of its constituents spins, so an even number of atoms will have an integer spin and so will become a boson. Now many of these bosons can condense in a single quantum state, that is, a single macroscopic particle. Now you have exquisitely quantum properties in a macroscopic object, not in a single very little particle.
« Last Edit: 06/01/2009 15:00:47 by lightarrow »

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« Reply #49 on: 06/01/2009 15:16:26 »
Grazie, Alberto. I actually managed to understand that!
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