In what way is a reflected photon degraded?

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Offline simplified

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In what way is a reflected photon degraded?
« Reply #50 on: 13/11/2011 15:55:10 »
The photons can lose energy when they bounce off things, but the effect is usually only noticeable when the things are small.
http://en.wikipedia.org/wiki/Compton_scattering
λ'-λ=h(1-cosΘ)/mc
Then exact energy of photon(after collision) can be calculated:
              E'=Emc²/[E(1-cosΘ)+mc²]
m - mass of Mike'S mirror(or sail)

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Offline JP

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In what way is a reflected photon degraded?
« Reply #51 on: 13/11/2011 16:15:11 »
Yor_on, coherent states do talk about both photons as particles and classical waves and provides the bridge between them.  Its not a way of getting around it.  Its a way of linking the two ideas in a rigorous way that makes it clear that both models are accurate and compatible.

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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #52 on: 13/11/2011 16:30:11 »
I don't know what you're arguing now? I think of coherent states as not being 'waves' although they use that analogue. They are a function of indeterminacy to me, is that what you don't like? Maybe it would be better if you defined your position? Then I can see how you mean.
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Offline CZARCAR

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In what way is a reflected photon degraded?
« Reply #53 on: 13/11/2011 17:17:11 »
why not consider light as a screw/tornado action instead of a wave? & how might clockwise & counterclockwise effect?

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Offline simplified

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In what way is a reflected photon degraded?
« Reply #54 on: 13/11/2011 17:55:40 »
why not consider light as a screw/tornado action instead of a wave? & how might clockwise & counterclockwise effect?
Then light cannot be polarized,but it can be.

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Offline JP

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In what way is a reflected photon degraded?
« Reply #55 on: 13/11/2011 18:05:30 »
Yor_on, you keep talking about waves and photons and say that coherent states "preclude" our definition of waves and photons.  You're misunderstanding what coherent states are, since they don't preclude those definitions.

The problem is that when you think of quantum theory of an electron, for example, the electron as a particle and the electron as a wave are two ways of looking at the same thing.

When you think of light, you have light as a photon particle, light as a photon wave or light as a classical wave.  The photon particle and photon wave are two ways of looking at the same thing.  The classical wave is not the same thing as a photon.  The classical wave is one very special case that can be built from photons, but you can build other things from photons that cannot be realized with classical models. 

The coherent state is the way of building a bridge linking photons to classical waves. 

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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #56 on: 13/11/2011 20:44:58 »
Hmm, preclude as in defining a theoretical proposition, joining our definition of photons and waves was what I meant. English is dangerous :)
« Last Edit: 13/11/2011 20:46:44 by yor_on »
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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #57 on: 14/11/2011 02:07:25 »
Can you define a interaction as taking place, but changing nothing for those that interacts? How, doesn't it 'cost' something to interact?

What would differ a interaction in such a definition from a particle never interacting. And looking at it as 'information', what information can be exchanged in such a definition of interactions without 'costs'?
« Last Edit: 14/11/2011 02:09:56 by yor_on »
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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #58 on: 14/11/2011 10:17:55 »
Macroscopically "A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. In reality, any macroscopic collision between objects will convert some kinetic energy to internal energy and other forms of energy, so no large scale impacts are perfectly elastic. However, some problems are sufficiently close to perfectly elastic that they can be approximated as such."

relative One-dimensional relativistic.

"Since the total energy and momentum of the system are conserved and the rest mass of the colliding body do not change, it is shown that the momentum of the colliding body is decided by the rest masses of the colliding bodies, total energy and the total momentum. The magnitude of the momentum of the colliding body does not change after collision but the direction of movement is opposite relative to the center of momentum frame."

So this one defines it as possible for a particle to keep the momentum in a collision? Either that or someone need to correct them :) And now we have some magic in the air, at a macroscopic plane none can expect it, but on a quantum level it's perfectly okay to hit as many objects you like as long as "the total energy and momentum of the system are conserved and the rest mass of the colliding body do not change."

Well, what more can you define to a particle moving relative some other? Momentum and mass (rest mass), and as neither is expected to change we now made it a possibility. Isn't that a circular logic?

Hmm, okay JP corrected me in that a momentum is like a velocity in that it has a magnitude and a direction, so as we get a change of direction the momentum can not be defined as the same. But we still have a unchanged magnitude, don't we?
=

Let us add to it ""Collisions" in which the objects do not touch each other, such as Rutherford scattering or the slingshot orbit of a satellite off a planet, are elastic collisions. In atomic or nuclear scattering, the collisions are typically elastic because the repulsive Coulomb force keeps the particles out of contact with each other."

So a slingshot orbit of a satellite off a planet, is a elastic collision? So can it gain or lose momentum by such a slingshot orbit? If it can, is it then still 'elastic'? I think NASA uses slingshots? I also think it bleeds of momentum  (magnitude of energy or 'kinetic energy') of whatever it slingshot around?
« Last Edit: 14/11/2011 14:34:43 by yor_on »
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Offline JP

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In what way is a reflected photon degraded?
« Reply #59 on: 14/11/2011 14:20:45 »
Momentum has a magnitude and direction.  If the magnitude stays the same, but the direction changes, then momentum has changed.

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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #60 on: 14/11/2011 14:26:45 »
Ah yes, sorry about that, but it's the magnitude I'm discussing here. Or maybe it is the 'kinetic energy'? Tell me which one you think is most correct JP.
=

Maybe I should make that clearer above.
« Last Edit: 14/11/2011 14:29:16 by yor_on »
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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #61 on: 14/11/2011 14:46:38 »
Let us summarise it then. As something have a 'elastic collision' it has the same momentum going 'away' from whatever it collided with, as it had when going towards it, as I get it? Because if you mean that the 'speed' will differ after the collision then that is also part of the momentum, and its kinetic energy, isn't it?

I'm getting confuseder here :)
What the he* is 'unchanged' for that particle?
=

Eh, the same momentum just means that you, ignoring direction, will find it to have a same speed and 'magnitude of energy'. As I get it. The direction shouldn't really matter for this, should it? Nah.
« Last Edit: 14/11/2011 14:50:28 by yor_on »
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Offline JP

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In what way is a reflected photon degraded?
« Reply #62 on: 14/11/2011 16:25:56 »
Let us summarise it then. As something have a 'elastic collision' it has the same momentum going 'away' from whatever it collided with, as it had when going towards it, as I get it?

Nope.  An elastic collision means the total kinetic energy of all colliding objects is the same after the collision as it was before.  The momentum can be split up in different ways.  For an example of this, check out the animated images on the wiki page: http://en.wikipedia.org/wiki/Elastic_collision 

The only time the object will recoil with approximately the same speed and opposite direction as it had originally is when the object its hitting has a much, much larger mass.  As an example, consider that you roll a billiard ball at something else.  If you roll it at a wall, it will bounce off at roughly the same speed and opposite direction as it had originally.  If you roll it at another billiard ball, it won't.
« Last Edit: 14/11/2011 16:29:31 by JP »

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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #63 on: 14/11/2011 21:28:03 »
Well, I'm not arguing the angles here. I'm just wondering what "The magnitude of the momentum of the colliding body does not change after collision................. but the direction of movement is opposite relative to the center of momentum frame." means?

(And the ... is mine, for emphasising my point.)

I don't get that first part at all JP? And 'magnitude' here I presume to be a statement of that 'kinetic energy'.
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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #64 on: 14/11/2011 21:33:47 »
I like your definition better "An elastic collision means the total kinetic energy of all colliding objects" as I presume that to be when considering a whole 'system' the collider and what it collides with?
=

Or did you also mean that the collider will keep its 'kinetic energy' untouched by the collision, and, here angles can be used to argue the absurdity, possibly :) Then again, If it really does, can it then be a 'interaction'? And waves interacts too, don't they? maybe I will see how it is thought to be later, hopefully.
==

How about this. When a wave pass a medium as glass, it 'collides' several times doesn't it?
But there they define the result as a 'higher frequency/energy', not as unchanged? And assume it just pass one atomic layer, will it then be unchanged.

When it comes to a 'photon' passing that medium you have a opposite effect, it losing 'momentum/kinetic energy' as I understands it?

Seems most define elastic collisions to gases and 'scattering', as "interactions of sub-atomic particles which are deflected by the electromagnetic force." And there they keep their 'kinetic energy'. Gamma rays from a nuclear explosion produce high energy electrons through Compton scattering though, which to me clearly implies that they interact.
« Last Edit: 14/11/2011 21:59:59 by yor_on »
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Offline JP

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In what way is a reflected photon degraded?
« Reply #65 on: 14/11/2011 23:22:42 »
I don't get that first part at all JP? And 'magnitude' here I presume to be a statement of that 'kinetic energy'.

Momentum is a vector.  A vector has a length (magnitude) and direction.  Momentum can change if either its magnitude or direction changes.  For momentum to be conserved, when you add up all the vectors of all the interacting particles before and after the collision, the total momentum vector has to have the same magnitude and same direction.  If all you know is its magnitude, you don't have enough information to describe the collision.

Kinetic energy is different than momentum.  Its a scalar quantity, so it doesn't have a direction.

Collision problems in physics usually involve figuring out as much information as you can, and then using conservation of kinetic energy and conservation of momentum to set up 3 or 4 separate equations to solve for the unknown values. 

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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #66 on: 15/11/2011 14:19:43 »
I have no problem with definitions that expect the values for a whole 'system' to stay the same. Without the conservation laws everything would become weird. What I'm not getting are definitions in where you expect the kinetic energy belonging to a colliding particle to be the same before as after a collision? But it's also a matter of how you look at it I presume. If it mean that the difference is so small that no one expect it to make a difference I wonder if they're right. At some scale I think everything is balanced against everything, like a ever spreading sphere of relations, and if it is so then even very small differences will build up, as I guess that is :)
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Offline JP

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In what way is a reflected photon degraded?
« Reply #67 on: 15/11/2011 15:50:00 »
What I'm not getting are definitions in where you expect the kinetic energy belonging to a colliding particle to be the same before as after a collision?

You don't expect this in general.  It's usually an approximation made to get a "good enough" answer. 

What usually happens is that a light particle is hitting a very heavy particle.  The heavy particle barely moves because of its mass, while the light particle bounces back with nearly the same speed it originally had.  Because it barely moves, you can approximately say that it has zero kinetic energy.  If it's an elastic collision, then the kinetic total energy of the particles after the collision has to be the same as the total energy before.  Since the bigger particle gets approximately zero kinetic energy after the collision, the smaller particle keeps the same kinetic energy.

This is obviously just an approximation, though, since if you make that assumption, then momentum isn't conserved.  The heavy particle has zero momentum both before and after, but the light particle's momentum changes direction.  Whether it's "good enough" depends on the calculation you're trying to do.

You can check out a simulation here: https://www.msu.edu/~brechtjo/physics/airTrack/airTrack.html 
If you set the mass of the blue cart to 100, you can see that it barely moves, while the red card bounces back with roughly the same speed it had coming in.

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Offline Geezer

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In what way is a reflected photon degraded?
« Reply #68 on: 15/11/2011 19:10:38 »
At some scale I think everything is balanced against everything, like a ever spreading sphere of relations, and if it is so then even very small differences will build up, as I guess that is :)

It will be. But some small amount of energy will always end up going down the entropy drain (and probably contribute to the ultimate demise of our Universe [:D]).
There ain'ta no sanity clause, and there ain'ta no centrifugal force ĉther.

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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #69 on: 17/11/2011 14:33:02 »
Yeah, that's a other one I'm wondering about. The conversation laws define it as transformations, nothing ever getting 'lost'. Entropy seems to defines a ground state as 'heat' ? If I got it right. But for every transformation, if we assume some logic chain from 'high energies' to 'low energies' there seems to me to be something irrecoverably lost? Would that be radiation? And if so, at what time scale does it disappear? The last one sounds weird, especially as I'm not that happy any more over 'virtual particles' but as 'energy tags down', what is it 'tagging down'?
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Offline JP

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In what way is a reflected photon degraded?
« Reply #70 on: 17/11/2011 17:27:12 »
Energy is always conserved in collisions.  But energy can change from one form to another.  The details of where the energy ends up depends on the situation, but some energy generally ends up as heat in the surroundings, and can't be recovered.

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Offline Geezer

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In what way is a reflected photon degraded?
« Reply #71 on: 17/11/2011 19:32:28 »
Yeah, that's a other one I'm wondering about. The conversation laws define it as transformations, nothing ever getting 'lost'. Entropy seems to defines a ground state as 'heat' ? If I got it right. But for every transformation, if we assume some logic chain from 'high energies' to 'low energies' there seems to me to be something irrecoverably lost? Would that be radiation? And if so, at what time scale does it disappear? The last one sounds weird, especially as I'm not that happy any more over 'virtual particles' but as 'energy tags down', what is it 'tagging down'?

Yes, entropy does seem to result in a sort of ground state. The interesting thing about it is that it seems to be irreversible, but I'll probably derail this topic if I go any further!
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Offline JP

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In what way is a reflected photon degraded?
« Reply #72 on: 17/11/2011 20:08:48 »
Ground state is a bit of a misnomer for it, though, since ground state in physics generally means the lowest energy state available to a given quantum mechanical system.  Entropy does provide an irreversible loss of energy to the environment, though, which can establish a background energy level that can't be harnessed to do work.

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Offline yor_on

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« Reply #73 on: 18/11/2011 12:21:37 »
Quite right Geezer. "The interesting thing about it is that it seems to be irreversible" but heat is usable energy, you can collect it and 'reinforce' it. And assume that you do that until you reach? What? A state where you can't collect more heat? Now, where did that excess of heat we collected and transformed go?
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Offline yor_on

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« Reply #74 on: 18/11/2011 12:41:36 »
The question is, is there a difference between 'usable energy transformations' and unusable? And what exactly is that difference? Heat is not a good enough answer to it, neither is 'energy' in itself, as I see it :)

We talk about symmetries and the 'breaking points' in them, as where something change into something different. That is in physics related to the 'energies described'. So, either you can assume that 'energy' as some weird state by itself never disappear, only transforms. But then you have to explain how we can use it to get 'work done'.

Or you can assume that there must be something more to a transformation than 'symmetry breaks'. I know, we don't define transformations as that really, but if I assume that nothing gets lost, then all transformations seems, to me, to manipulate states of equilibrium.

Even if I assume that everything is those kind of equilibriums and symmetries I still have to see why we can manipulate them, by what?
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Offline JP

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In what way is a reflected photon degraded?
« Reply #75 on: 18/11/2011 14:11:02 »
Symmetry breaking doesn't really have much to do with this, nor do collisions.  It all comes down to understanding thermodynamics, which is beyond this discussion.  Perhaps we can start another thread with a discussion of why sometimes energy can be used to do work and sometimes it can't.

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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #76 on: 18/11/2011 16:11:59 »
Yeah, it's vague. But it's a valid question.

"A reflected photon imparts pressure to the reflecting surface.
This would imply that work has been done.
The photon has not been absorbed but it has been degraded by the loss of energy in some form.
In what way is the photon degraded by the loss of energy?" By Mike.

Can easily be seen to cover that question. And if we think entropy and thermodynamics is a branch of physics then we can use it in a mixed environment too. If that's not possible then it's not a branch of physics at all, which would be weird.
« Last Edit: 18/11/2011 16:37:46 by yor_on »
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Offline JP

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In what way is a reflected photon degraded?
« Reply #77 on: 18/11/2011 16:36:01 »
Yeah, it's vague. But it's a valid question.

It is indeed a valid question, which is why many brilliant physicists spent years working out the mathematics of thermodynamics and statistical mechanics in order to answer it.

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Offline yor_on

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« Reply #78 on: 18/11/2011 16:38:23 »
Sorry JP :) We collided there.
(the postings I mean)
=

Maybe you could say that the 'energy' becomes indeterministic in that it 'tags down'?
Or?
« Last Edit: 18/11/2011 16:46:25 by yor_on »
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Offline JP

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In what way is a reflected photon degraded?
« Reply #79 on: 18/11/2011 19:08:20 »
Yor_on, you're asking for very broad answers to things which don't have them.  Thermodynamics and statistical mechanics talk about very precise cases in which energy or information gets lost in an irreversible way in a system.  Your broad question about things being "degraded" or "tagging down" need to be put in much more precise terms for anyone to answer them scientifically.

However, I can give you a very hand-waving answer despite all that.  Imagine making a movie of a collision process.  If you can play the movie backwards and it still seems like a likely collision to you, then information/energy wasn't irreversibly lost in the collision.    If you can definitely tell which way was forward, then something was lost. 

For example, two billiard balls colliding is pretty reversible.  But the break shot in a game of pool is a pretty irreversible process. 

In the case of a single photon reflecting elastically off a mirror, it is reversible.  It would be just as plausible for a mirror to be moving, a photon to strike it, and the reflected photon carry away the energy of the mirror as it is for a photon to strike a stationary mirror and cause it to start moving.

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Offline yor_on

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« Reply #80 on: 19/11/2011 00:57:39 »
In what way would that matter? Are you thinking that as we can't define the process as going 'one way' (time reversibility) there is no 'tagging down'? Or is there something else you mean? As for why we have to look at it as a 'system' I find it a very nice conceptual tool. The question though, is where the 'energy' is thought to go, as 'heat' or as ?

Then, on the other hand, this is assuming that whether you look at it from one 'direction', or the other, we still are discussing objects colliding, not 'systems'.
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Offline JP

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In what way is a reflected photon degraded?
« Reply #81 on: 19/11/2011 01:15:00 »
Two particles colliding elastically is a system that happens to be reversible, so nothing is lost.  You don't have to lose energy to heat in a collision.

I'm not sure what "tagging down" means.

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« Reply #82 on: 20/11/2011 17:24:08 »
Ah, just meaning a 'interaction' :)

So let's see, we got three ways of a 'interaction' then?

1. Losing 'energy', measured as a 'photon'
2. Gaining 'energy', measured as a 'wave'

(Both of those passing a medium as glass, or water.)

3. Neither, a interaction without anything happening, the 'elastic collision' in where we still see 'something' changing its direction at the 'rebound'. And that measured as 'waves', and 'photons'?

How about a experiment in where you 'split a photon (or wave, pick your choice :). Sending half back, letting half pass through some 'mirror'. Don't we call that a entanglement? Meaning that they are the exact same, except in their spin/polarisations (opposite)?

Can they be the same if I use any of those three definitions above?
3. right?

But the 'photon/wave' passing that mirror then, it should relate to either 1 or 2, shouldn't it?
« Last Edit: 20/11/2011 17:38:27 by yor_on »
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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #83 on: 20/11/2011 17:34:23 »
The question is actually one of me trying to see what a 'wave function' is. If the 'wave function' is a local phenomena, belonging to something 'propagating' then it becomes tricky for me. If a 'wave function' is a expression of the surrounding parameters defining its 'existence' then there's 'nothing there', except the conservation laws.

Maybe :)
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Offline JP

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« Reply #84 on: 21/11/2011 01:12:40 »
Ah, just meaning a 'interaction' :)

So let's see, we got three ways of a 'interaction' then?

No, you have an unlimited number of ways of having an interaction, so long as energy and momentum are conserved for the entire system.

Energy in = energy out
Momentum in  = momentum out

You're free to have any kind of interaction you want, so long as those hold (and potentially other conservation laws such as angular momentum, charge, etc.)

An elastic collision is one special case in which all the energy remains as kinetic energy.

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« Reply #85 on: 21/11/2011 01:47:13 »
That one you should expand on JP.

And you're treating it as a 'system'. It's somewhat of a paradigm shift from treating it as objects 'interacting', isn't it? Not that I see it as wrong, actually I'm thinking along similar lines myself in my totally unscientific way :) discussing all interactions/outcomes as definitions from where and how we measure.

It makes more sense to look at it as 'systems', and becomes easier to define (conservation laws). But it also binds into how to see a entanglement for me. And there my question (as above) still would be how 'identical' those photons would be, in that entanglement where one 'part' passes the mirror and the other gets reflected.
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Offline JP

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In what way is a reflected photon degraded?
« Reply #86 on: 21/11/2011 03:31:07 »
That one you should expand on JP.

And you're treating it as a 'system'. It's somewhat of a paradigm shift from treating it as objects 'interacting', isn't it? Not that I see it as wrong, actually I'm thinking along similar lines myself in my totally unscientific way :) discussing all interactions/outcomes as definitions from where and how we measure.

You're right.  A system is a set of particles that are interacting.

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Offline yor_on

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In what way is a reflected photon degraded?
« Reply #87 on: 21/11/2011 22:41:02 »
Yes, but as with using the conservation laws to describe it you treat the whole 'system' as one entity, and that's my point there.
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In what way is a reflected photon degraded?
« Reply #88 on: 21/11/2011 22:46:15 »
Conservation laws hold whether you know about the whole system or not.  They're just not terribly useful for computing anything if you can't keep track of where all the energy went.

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In what way is a reflected photon degraded?
« Reply #89 on: 22/11/2011 01:42:53 »
I lose you there? Either you are treating each object by itself, and then define a collision and 'energies etc' for each one, relative that collision. Or you define it as a 'system' in where the 'whole system' is unchanged.

If you mean that you do treat it as individual objects meeting, although following conservation laws, each one exchanging 'whatever' with each other, then I think what I wrote before stands about the types of interactions described. And in the example I'm speaking of photons/waves passing a mirror, and getting 'split'. I think it is right? And it doesn't add up for me. Maybe you can point out where I go wrong there?
« Last Edit: 22/11/2011 01:44:54 by yor_on »
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In what way is a reflected photon degraded?
« Reply #90 on: 22/11/2011 02:40:22 »
I think we probably agree and it's just a language issue.  I'm using system to mean a collection of interacting particles.  Each individual collision within that system should obey a set of conservation laws.

I'm not sure I follow your point about a photon and a mirror.  If you assume energy is conserved in the sense that Original Energy of Photon = Final Energy of Photon + Final Kinetic Energy of Mirror, and the same with momenta, then the photon will redshift, losing energy, and the mirror will move after the collision.  This is consistent with the idea that the photon gives up a little energy and momentum to the mirror. 

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Offline yor_on

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« Reply #91 on: 22/11/2011 13:28:54 »
Well, it all goes together for me :) I was looking at my understanding of entanglements JP, and I'm not sure if I understand it at all. The same goes for a 'wave collapse' and how/where we would/could expect it. In reality, whatever I now mean by that, 'photons' and 'waves' should be 'the same as usual' as I see it, whether 'entangled' or not. 

So I used the idea of a 'mirror' of some kind, EM or otherwise to describe a situation in where you 'entangle' radiation by letting half pass through, the other half to 'bounce' back. Then I looked at the descriptions of how that light passing through the mirror would behave in a measurement comparing it to the idea of a 'elastic collision' as one might assume the light 'bouncing'. Finally I asked if they the would be 'identical'?

I mean, the polarisation/spin is definitely 'opposite', I'm sure of that one, so they are entangled in that motto. But if I got it right they should give us a different 'energy/momentum', also depending on how you measure, still not being 'identical' if my assumptions is right.

Those things all go into each other, don't they?
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In what way is a reflected photon degraded?
« Reply #92 on: 22/11/2011 14:35:45 »
I think you're confusing the matter a lot by jumping to advanced quantum mechanics without fully understanding classical collisions, including classical waves.  You really do need to build up from basics before advanced QM. 

Anyway, you can understand the effects you're describing purely classically.  Consider a wall with a bunch of holes in it and balls you're throwing at it.  If the ball hits the wall and bounces off, it will lose some energy and the wall will recoil a bit.  If the ball passes through a hole in the wall, it will continue on with the same energy and momentum as it initially had and the wall won't be effected.

Now, consider the case of a classical light wave hitting a 50% reflective mirror.  50% of the reflects off the mirror, which transfers some energy and momentum to the mirror while the light loses energy and changes direction, while 50% of the light passes through without causing the mirror to gain energy or momentum anything to the mirror.

If those two cases make sense, then the jump to QM is pretty simple.  If you send 1 photon through the mirror, then you're 50% likely to measure a final state that is: photon reflects + mirror recoils and 50% likely to measure the effect photon transmitted + mirror does not recoil.

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In what way is a reflected photon degraded?
« Reply #93 on: 22/11/2011 17:32:19 »
Ah well JP.

I'll let it rest for now.
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