# The Naked Scientists Forum

### Author Topic: In what way is a reflected photon degraded?  (Read 22561 times)

#### 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)

#### 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.

#### 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.

#### 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?

#### 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.

#### 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.

#### 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 »

#### 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 »

#### 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 »

#### 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.

#### 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 »

#### 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 »

#### 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 »

#### 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'.

#### 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 »

#### 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.

#### 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 :)

#### 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.

#### 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 ).

#### 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'?

#### 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.

#### 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!

#### 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.

#### yor_on

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##### In what way is a reflected photon degraded?
« 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?

#### yor_on

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##### In what way is a reflected photon degraded?
« 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?

#### The Naked Scientists Forum

##### In what way is a reflected photon degraded?
« Reply #74 on: 18/11/2011 12:41:36 »