[DoctorBeaver (OP)]

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.

[LeeE]

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.

[Mr. Scientist]

Hard to say. I'd have to think about this one.

[Soul Surfer]

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.

[DoctorBeaver (OP)]

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?

[LeeE]

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??

[DoctorBeaver (OP)]

I'm not sure I follow that. If you shine a laser into am ordinary diamond, are you saying no light will emerge?

[lightarrow]

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.  []

[DoctorBeaver (OP)]

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.

[lightarrow]

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".

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.

[DoctorBeaver (OP)]

OK, I understand now. Thank you, Alberto  []

[yor_on]

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:)

--------
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?

[DoctorBeaver (OP)]

I think you're right on all counts.

[lightarrow]

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:)

--------
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.

[yor_on]

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.

----------

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?

-------
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?

[DoctorBeaver (OP)]

BEC confuses the hell out of me  []

[yor_on]

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:)

[Bikerman]

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

[lightarrow]

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.

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

See up.

What I'm really curious on is if one can 'brake' a original photon

No, that's not possible.

, 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.

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

What you mean?

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...

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  []

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"... []).

[LeeE]

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.