# Speed and light

• blacksheepdork

#### blacksheepdork

My physics teacher says that when light enters a more dense medium it slows down, this is fine but then he states that when light leaves the same medium it again speeds up without an outside force acting on it. why is this? why does it seem to prove Newton wrong?

My understanding is that the tiny time delay is due to absorption and re-emission of photons in the medium and not an actual slowing down of the photons themselves.

so your saying that the "apparent" increase in velocity is actually due to the refraction of the light beam?
is there any way you could give me some example?

It's because the photon in a medium is doing a gigantic game of pinball, bouncing off of hundreds of thousands of atoms on its way forward. It isn't going as fast (to us) because it isn't traveling in a straight line.

Welcome to the forums, BTW

i realize why it slows down when it goes through the medium. where i get lost is when my teacher says that once it leaves, it again speeds up... there's no outside force acting on it and yet it speeds up. why?

Because it isn't speeding up. It's just going in a straight line once it leaves.

EDIT: If you need anything more precise than that to quell your mind, I'll have to leave it to the physicists here for a better explanation.

[/engineer]

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Originally posted by blacksheepdork
i realize why it slows down when it goes through the medium. where i get lost is when my teacher says that once it leaves, it again speeds up... there's no outside force acting on it and yet it speeds up. why?

Try Thinking of a photon as a car speeding down the road at 60mph. On the open road(vacuum) it travels at at a constant 60mph.

Now it reachs a section where there are a number of stoplights (a dense medium). It still travels at 60 between each stop light, but has to pause at each light. The pauses causes its average speed through this section to drop below 60.

once it is clear of the stoplights, its average speed returns to 60 mph.

With light traveling through a medium it is a little different in that the light isn't stopped, but absorbed by the Atom (At this time, the photon ceases to exist) then after a short delay, another photon is emitted by the atom. which travels at c until it encounters an atom itself. Once the light has left the medium, it is free to travel without these little interuptions, and does so at its natural speed of c.

So, wouldn't the photons be refracted into many different directions? If we shone a laser, or other concentrated light beam that could reach far into space, hypothetically, would we not be able to see it in space for the photons would be deflected into random directions and not be powerful enough for us to see it?

Is that why a light beam will fade away after some distance, even if it isn't shown as a cone? Interesting, I completely forgot about all of that.

Greetings !

I'm not exactly certain of what your question
is in your last message, Funkee. If you're
talking about a beam of light then it is
comprised of many photons traveling at roughly
the same direction. The more focused the beam
is the longer will the photons travel relativly
close to each other (in vacuum), but eventually
the difference in their direction, no matter
how small, will cause the beam to spread over
a very large area - so that it would no longer
be easily visible if you happen to stand in its
path.

The reason for the difference is the fact that
each photon is generated by a different atom
or separate accelerated electron and the direction
of the emmited electromagnetic fluctuation (photon)
is never exactly the same (not theoreticly for an
infitesimal distance, at least).

Live long and prosper.

Yes. Essentially the question was answered in the previous posts, it was one of those crazy posts after a week of no sleep. It just makes sense, and I had never really thought about it. I just wanted to make sure I understood it.

Originally posted by enigma
It's because the photon in a medium is doing a gigantic game of pinball, bouncing off of hundreds of thousands of atoms on its way forward. It isn't going as fast (to us) because it isn't traveling in a straight line.

In a medium, right? In a vacuum, it would be a straight line right? Why would the photon's activity in relation to surrounding particles be an effect that is relative to a certain observer? isn't the effect universal?

Theoretical explanation

Assume the photon is a bipartite system, a positron and an electron orbiting each other around a central point moving in a straight line, held together by mutual attraction, kept apart by orbital energy which would be a function of the original emission velocity.

Entering a medium of greater density, meaning more foreign particles per unit volume, should introduce more load on the system energy, reducing the orbital separation and deflecting the straight line path. If the planes of entry and exit are parallel, reentering the medium of lesser density should reverse the process, approximating the original path, offset and with a greater loss of energy than would have been experienced if the same distance had been traversed in the original medium.

A wave explanation would probably be similar.

It's an interesting notion. The orbital separation distance of the system should alter in the new medium, changing the color perceived by a viewer within the medium. The energy variation and path offset should be a function of the distance between the entry and exit surfaces. Measurements at varying distances should provide useful data. They probably already have.

If the path of the light is visible, it is due to random reemission on contact with foreign particles, as is true of anything we see. With large numbers of individual photon systems, reemission scatters the light in all directions. The individual photon systems that remain relatively coherent comprise the offset beam that exits the medium. They are influenced by the increased force fields in the denser medium but have no direct contact with foreign particles.

In a medium, right? In a vacuum, it would be a straight line right? Why would the photon's activity in relation to surrounding particles be an effect that is relative to a certain observer? isn't the effect universal?

"To us" in his post meant that we're computing net velocity, not instantaneous velocity.

Hurkyl

Originally posted by Funkee
So, wouldn't the photons be refracted into many different directions? If we shone a laser, or other concentrated light beam that could reach far into space, hypothetically, would we not be able to see it in space for the photons would be deflected into random directions and not be powerful enough for us to see it?

Is that why a light beam will fade away after some distance, even if it isn't shown as a cone? Interesting, I completely forgot about all of that.
You are exactly correct. A laser in space is invisible unless you shine it into your eye.

It doesn't really violate Newton's laws because it's not really following them in the first place, lol. If you are talking about inertia here, that is a concept you need to throw out for light. Intertia, actually, is a measurement of mass. Light has no mass to speak of. For this reason, light always travels the same speed. The reason it "slows down" in a medium is because its progress is interfered by giant pylons (atoms) that keep pushing it off course. It still travels at light-speed, but it isn't taking the shortest route through the material, and therefore the net velocity is lower than "light-speed."

What I never quite understood was why light slows down as it passes through glass, yet somehow goes through it in what seems like a straight line. Hope somebody here can clear that up. (Obviously, the zig-zag ping-pong game of photons we're talking about doesn't apply to the entire beam of light. I would just expect the light to dissipate, though, not continue in an overall straight line.)

Originally posted by CJames
It doesn't really violate Newton's laws because it's not really following them in the first place, lol.

True, true, really true! (Hint: interferences)

Well, what happens is that the energy of the photons gets absorbed by electrons... the easiest direction for electrons to rerelease that energy is (via conservation of momentum) to send one or more new photons traveling the same direction as the old one. If individual atoms have very little freedom of motion, conservation of momentum virtually forces the atoms to transmit the light in the same direction it was originally going. However, in materials with a lot of freedom of motion (say... gases), the atoms have the freedom to be deflected and can send photons off in new and interesting directions.

Hurkyl

So if photons have no mass, how does a solar sail work?

Let me guess, transfer of momentum via particle emission from the sun? Not photons at all?

Thank you Hurkyl, that was very helpful. By the way, I was always curious how the momentum of light is measured, since it obviously has nothing to do with mass. Plus SR changes up the meaning of velocity too. Therefore, mv is completely out of the question.

No Alias, it actually really does have something to do with the momentum carried by light. I have no idea how to measure the momentum of light though...as I just said lol.

The straightforward derivation is as follows:

Starting from the relativistic equation for classical velocity and momentum:

p = &gamma m v
where &gamma = 1 / (1 - (v/c)2)1/2

we see that if we let m go to zero that p goes to zero, but if we let v go to c, then p goes to infinity. If we let m go to 0 and v go to c simultaneously, then we get the indeterminate form

p = 0 * &infin

which suggests that some other trick might yield meaningful results.

The total energy of an ordinary particle is:

E = &gamma m c2

Rearranging the equation a bit yields:

E2 = (pc)2 + (mc2)2

We've eliminated &gamma from the equation, and in doing so removed all singularities from the formula. We've discovered the trick! Now, when we let m = 0 and v = c simultaneously, the above formula yields:

E = pc

So if we know the energy of a photon, we simply divide by c to get its momentum.

To be entirely rigorous, I'd have to point out that the above is a derivation for zero mass light speed particles and not necessarily photons (which are electromagnetic wave packets), and assumes continuity of the total energy equation at v = c. To be entirely rigorous, the proof would have to come from the field equations for electromagnetism (say... Maxwell's equations for relativistic EM).

Hurkyl

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Originally posted by Hurkyl
Well, what happens is that the energy of the photons gets absorbed by electrons... the easiest direction for electrons to rerelease that energy is (via conservation of momentum) to send one or more new photons traveling the same direction as the old one. If individual atoms have very little freedom of motion, conservation of momentum virtually forces the atoms to transmit the light in the same direction it was originally going. However, in materials with a lot of freedom of motion (say... gases), the atoms have the freedom to be deflected and can send photons off in new and interesting directions.

Hurkyl
In the same reason, a metal must be transparent for light as a glass .
Is it so?

Metal is a bad example, because of its unusual electron arrangement (in comparison to other materials)...

But for opaque objects in general, what I've read is that due to phase lag you get 100% destructive interference inside the material, rather than the case for transparent materials where you get nearly 100% destructive interference, but the energy piles up and makes its way through eventually.

Hurkyl

Originally posted by Hurkyl
Metal is a bad example, because of its unusual electron arrangement (in comparison to other materials)...

But for opaque objects in general, what I've read is that due to phase lag you get 100% destructive interference inside the material, rather than the case for transparent materials where you get nearly 100% destructive interference, but the energy piles up and makes its way through eventually.

Hurkyl
Ok. The big amount of free electrons which has chaotic moving in metal must unpredictably change the paths of all photons. This means that a surface of the metal can diffuses the light only. But the best mirrors are got exactly from the metal . They reflects nearly 100% photon and, herewith, angle of the fall strictly corresponds to the angle of the reflection. So?

I don't know the properties of electron gases, so I can't really say much about them, except that I doubt the existence of an accurate description of their properties without having to resort to QM (specifically the QM consequences of indistinguishability of particles).

Anyways, the way to look at reflection is that if the easiest way to release the energy (retransmit a photon in original direction) is not allowed, then the next easiest way will be predominant! The second easiest way to release the energy is, for surface atoms, perfect reflection. A perfect reflection means that the surface atom only picked up a deflection directly into the material, which minimizes the disturbance of neighboring surface atoms.

Hurkyl

Thanks Hurkyl, I've always been curious about some of this stuff. I enjoyed that derivation.

Another thing I'm curious about is pigmentation. I know the whole textbook response, certain colors of light are absorbed, and others reflected. I also know this has to do with the electron configuration of the atom (hence spectroscopy.) What I don't know, is where the absorbed energy ends up going. The electron doesn't just hold it forever. Also, why are the reflected frequencies not reflected "the next easiest way" with most opaque materials?

I think that they are with most opaque materials; just most opaque materials don't have a (nearly) perfectly flat surface that allows a sharp reflected image.

Anyways, the energy that is kept eventually gets rereleased as heat (blackbody radiation).

Hurkyl

I think that they are with most opaque materials; just most opaque materials don't have a (nearly) perfectly flat surface that allows a sharp reflected image.
That would make sense. So, essentially, anything that can be smoothed down to a flat surface will eventually have mirror-like properties, correct?

Well, I imagine not everything can be smoothed down so nicely.

Hurkyl

Unfortuntatly much of the above thread is on shakey ground. The direction of emitted photons is TOTALLY RANDOM, at the level of the photon there is NO angle of reflection is equal to angle of incidence. That is due to entirely different phenomena.

I would strongly recommend that if you are interested in getting a good explaniation of this that you read Fynmans QED. It is a very small book written for the non scientist.

If light travels at light speed and can go any faster,what if there was a force repelling photons from going any faster.the more energy you apply to a photon it still doesn't go faster,what if this energy is transferred into sideways motion instead of forward motion,making photons travel as particles in curved paths.the repelling force would oppose forward motion,so like putting two like sides of two magnets together you feel the forces and as you try to make them touch they slide back and forth in opposition.so the photons do the same thing.vibrate up and down in opposition to forward motion.light would then take energy that should make it go faster and change its frequency to a higher one,thus photons all at different frequencies when next to each other,scatter each other because there out of phase to one another,and there own up and down sideways forward motion accounts for frequencies and wave properties in one,all because of forward repulsion to motion.

Hey Blacksheepdork

look what you started! the world is full of questions and answers! Boy does the physics teacher have a thing or two to learn also.

Actually, that's an interesting question, Integral.

I had been implicitly presuming the same thing; I was talking about how "easy" it is to reemit the energy in a particular direction, but kept tacit the law of large numbers which (kinda sorta) states that the net result is predominately the easiest one.

But... some of the interesting QM experiments seem to say otherwise. For instance, in the famous experiments involving half-silvered mirrors, photons are said to either be transmitted through the mirror, or to be perfectly reflected via incidence = reflection...

I guess (I think) it falls down to the fact that the photon is really the center of probability for a wave packet, so by definition it follows perfectly the expected behavior.

Hurkyl

Well, in a vacuum a photon isn't following that random a path, right? Obviously, it's not following a simple angle of incidence, since it's actually a complex wavefunction, but each imaginary particle in the wavefunction I would expect reflects the way you expect it. That's how it seems to me anyway. Not that I wouldn't be interested in reading that book, but it doesn't feel to me like this entire thread is on shaky ground...'course that's me

Take care.