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blacksheepdork

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- Thread starter blacksheepdork
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blacksheepdork

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- #3

blacksheepdork

is there any way you could give me some example?

- #4

enigma

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Welcome to the forums, BTW

- #5

blacksheepdork

- #6

enigma

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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]

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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- #7

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Originally posted by blacksheepdork

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.

- #8

Funkee

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.

- #9

drag

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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 travelling 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).

Does that answer it ?

Live long and prosper.

- #10

Funkee

- #11

RuroumiKenshin

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 travelling 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?

- #12

darl

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.

- #13

Hurkyl

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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

- #14

russ_watters

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You are exactly correct. A laser in space is invisible unless you shine it into your eye.Originally posted by Funkee

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.

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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 disipate, though, not continue in an overall straight line.)

- #16

arivero

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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)

- #17

Hurkyl

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Hurkyl

- #18

Alias

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

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- #21

Hurkyl

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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 c^{2}

Rearranging the equation a bit yields:

E^{2} = (pc)^{2} + (mc^{2})^{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

Starting from the relativistic equation for classical velocity and momentum:

p = &gamma m v

where &gamma = 1 / (1 - (v/c)

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 c

Rearranging the equation a bit yields:

E

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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- #22

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In the same reason, a metal must be transparent for light as a glass .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 travelling 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

Is it so?

- #23

Hurkyl

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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

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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?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

- #25

Hurkyl

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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

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