Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: MikeS on 31/10/2011 08:16:19
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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?
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Work is done only if the surface actually moves. In most cases of most surfaces the collisions are perfectly elastic the surface does not move as a whole and so no energy is lost. The same would be true for grains of dust etc. Even when photons of light are reflected or scattered by individual atoms the atoms are much heavier than the momentum of the photons and scattering is usually elastic see Rayleigh Scattering however this may not be the case particularly for higher energy photons and energy will be lost see Compton scattering.
An interesting aside. In general for higher energy photons like gamma rays the reaction of the emitting nucleus causes a spreading of the wavelength however there are some cases where the nucleus is effectively locked into the structure by quantum effects and the energy of the photon is incredibly precise and this can be used for very accurate measurements foe example the gravitational red shift caused by the earth's gravity. this is called The Mossbauer effect.
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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?
In your case:
E -lost energy
m' - zero mass of photon = hν/c²
m - mass of surface
c - speed of light
E=2(m')²c²/m
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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
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Thanks guys but I don't think your replies really answers the question "In what way is the photon degraded by the loss of energy?"
A photon can not loose energy by slowing down and as I understand it, it is not red-shifted. So what is happening? Could it be that some photons are being reflected with no energy transfer and some are being absorbed?
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if a photon loses energy it is red shifted or more accurately absorbed and remitted with a lower energy with the result that the particle it hit moves off with the energy deficit in a direction associated with the absorbed and emitted photons
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I'm not sure I get this one? Are you suggesting that photons don't get absorbed by a mirror but instead are 'reflected'? There are two ways to describe it, and using a wave picture is the oldest one and also the one we easiest recognize. The other is a particle picture in where the photons get absorbed by the mirrors atoms, most releasing new photons, that over a broad distribution presents us with the angles of reflection we associate with a wave. Although some of the explanations to how a 'photon' is expected to do so seem to be connected to the the way a wave interfere, quenching and reinforcing, via 'frequencies' which, as far as I know, no 'photon' is expected to have in a particle picture, frequency that is. Another way to define their angles 'reflecting' might be the momentum they have and how that gets 'mirrored' by the atom creating new photons.
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I was referring to MikeS's statement a photon is not red shifted when it is involved in nonelastic scattering. You seem to have a hangup about the wave/particle duality it is a true duality photons always have the full properties of both. Can you not think of a wave packet? true as the photon energy rises it is easier to think of them as particles. Processes inside the limits of quantum uncertainty cannot be observed it is only the results that matter
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If it's a perfectly reflecting surface, presumably there would be no loss. The photon being absorbed and re-emitted at the same frequency and the surface recoiling in the process. This would imply a gain of energy, so it can't happen. But what is happening? Is the photon red-shifted by being reflected by a perfectly reflecting surface? Can a photon be absorbed and re-emitted by a perfectly reflecting surface? If the surface is perfectly reflecting how can it absorb the photon? If it's a perfectly reflecting surface does the photon impart any energy to that surface?
Thanks
Soul Surfer
Having read your previous replies I can't see how Rayleigh Scattering could be the answer as it applies to transparent objects.
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Consider a largely but not perfectly reflecting heavy surface. If a photon strikes the surface it can be either reflected or absorbed by the surface. A photon has both energy and momentum if it is absorbed the energy in the photon warms the surface which could cause another photon of lower energy to be emitted. If it is reflected the photon is unchanged except that its direction of travel has changed and therefore its momentum has changed so the law of conservation of momentum applies and some momentum has been given to the surface. Note the momentum of a photon is very much less than its energy (a factor of the speed of light). An illustration of this process is given in the behaviour of the Crookes radiometer. as the pressure in the tube is reduced from lowish to a hard vacuum.
Why do you think that Rayleigh scattering only applies to "transparent " objects?
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Consider a largely but not perfectly reflecting heavy surface. If a photon strikes the surface it can be either reflected or absorbed by the surface. A photon has both energy and momentum if it is absorbed the energy in the photon warms the surface which could cause another photon of lower energy to be emitted. If it is reflected the photon is unchanged except that its direction of travel has changed and therefore its momentum has changed so the law of conservation of momentum applies and some momentum has been given to the surface. Note the momentum of a photon is very much less than its energy (a factor of the speed of light). An illustration of this process is given in the behaviour of the Crookes radiometer. as the pressure in the tube is reduced from lowish to a hard vacuum.
Why do you think that Rayleigh scattering only applies to "transparent " objects?
Soul Sufer
Thanks for reply. I had been trying to find information on the Crooks radiometer but couldn't as I didn't know what it was called. Interesting, I always thought it operated directly by radiation pressure but apparently its a heat engine.
I didn't know anything about Rayleigh scattering so did a quick internet search and it seemed to only be happening in transparent mediums like gasses.
"If it is reflected the photon is unchanged except that its direction of travel has changed and therefore its momentum has changed so the law of conservation of momentum applies and some momentum has been given to the surface."
But "if some momentum has been given to the surface" then surely, the photon can not be "unchanged". It would have to be re-emitted at a lower frequency in order to conserve energy. Or, if it were reflected unchanged, then it can't have transferred any energy to the reflecting surface. ??????
We know that the reflected photon does not loose energy (red-shift). So does it impart momentum to the reflecting surface? Or is it only absorbed photons that impart momentum?
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Even individual atoms are too heavy to allow a normal light photon to loose a significant amount of its total energy and momentum
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Even individual atoms are too heavy to allow a normal light photon to loose a significant amount of its total energy and momentum
But my point is any transference of energy and momentum, no matter how small must show as a change in the photons energy for the law of conservation of energy to hold.
This is a quote about solar sails.
"How does light push a solar sail?
Photons, which are "particles" of light, bounce off the reflective material of the sail. (Newton's Third Law of Motion states that for every action there is an equal and opposite reaction.) The reaction here causes a change in momentum, pushing the sail and accelerating the spacecraft. A photon reflecting off the mirror-like surface of a solar sail gives the sail a double kick -- a push equal to twice the photon's momentum (one push from the sail stopping the photon and one from it reflecting the photon and accelerating it away)."
http://www.discoversolarenergy.com/solar/sails.htm
The photon is reflected essentially unchanged, it still has the same energy. So where has the energy imparted to the sail come from?
If we bounce light backwards and forwards between two parallel mirrors it does not loose energy by becoming red-shifted and yet the mirrors get a slight push on every reflection.
Here is another article on solar sails.
"Sailing Motion
Sailing motion is determined by theta, the angle between the solar radial and the ship's total force vector. A positive theta generally adds energy and causes outward motion. A negative theta generally reduces energy and causes inward motion. The projection of the acceleration vector onto the velocity vector determines the actual change in energy in the local gravitational field. A sail does not extract energy from or put energy into the reflected light to accomplish its sailing.
If the reflection does not extract energy from or inject energy into the photons, how does the sailing ship gain or lose energy? The reflected photons have the same energy flux they had prior to the interaction, but a different momentum vector. It is this altered momentum vector that gives the ship an accelerating force that allows it to work against the gravitational field to gain or lose energy within the field. The absorbed photons are the energy lost from the impinging flux. The absorbed energy is re-radiated from the sail, with some helping and some hindering the ship’s motion."
http://sail.quarkweb.com/light.htm
I believe this is saying that the altered momentum vector of the photons is where the force that causes the acceleration comes from. In other words the force comes from the change in direction (through rebound) of the photons (Newtons third law...) That still doesn't explain why the sail can gain energy if the reflected photons don't loose any.
Does anyone know of any experimental evidence confirming that reflected photons do exert a 'push'?
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p=mc
2p=2mc
(2mc/mc²)*100%≈0% of energy [:P]
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Mike, if the photon gives up some of it's energy, it will rebound with a smaller frequency. Generally they give so little energy to atoms or molecules that it's negligible. With a solar sail, radiation pressure is also tiny, but there is a huge number of photons bouncing off it for a long period of time and no friction to slow it down, so these tiny pushes keep adding up.
The push of reflected photons can be measured, and I know of a bunch of experiments that measure it in one way or another, usually in a lab with laser light. There is already one functioning solar sail, as well: http://en.wikipedia.org/wiki/IKAROS
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Mike, if the photon gives up some of it's energy, it will rebound with a smaller frequency. Generally they give so little energy to atoms or molecules that it's negligible. With a solar sail, radiation pressure is also tiny, but there is a huge number of photons bouncing off it for a long period of time and no friction to slow it down, so these tiny pushes keep adding up.
JP
I understand all of the above but I don't understand how a reflected photon can give up none of its energy (see my last post above) but still give a small push to the reflecting surface. This would seem to imply an increase of energy which can't be the case as it would break the law on conservation of energy.
This makes me question whether a reflected photon can impart any momentum to the reflecting surface, or is it only absorbed photons that can do this?
Any experiment to measure the push of a reflected photon must involve a significant number of photons and can we be certain that all the photons have been reflected not absorbed? As no mirror is perfect we can't be certain. If an experiment were able to confirm that the push was twice as much for a reflected photon as an absorbed photon then it would be evidence confirming the hypothesis. I have not been able to find any evidence of such an experiment. If you know of any links I would appreciate you posting them.
It's been proven that solar sail technology is possible and we are told it is due to the push given by reflected photons bouncing off the mirror and the mirror recoiling but how can we be certain that it is not due to absorption?
(Aside. If it is from absorption then there could also be a small contribution from the forces that power a Crookes radiometer. Presumably (?) if a solar sail works by reflection then it is not a heat engine but if it works by absorption then it is.)
Thanks for your time.
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Well, yeah :)
I'm afraid I still think of it as a 'duality', although I can see why people jump from a photon to 'frequency', it's not the same to me. It's a weird subject though.
Both answers correct in century-old optics dilemma. (http://physicsworld.com/cws/article/news/41873)
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Mike, if the photon gives up some of it's energy, it will rebound with a smaller frequency. Generally they give so little energy to atoms or molecules that it's negligible. With a solar sail, radiation pressure is also tiny, but there is a huge number of photons bouncing off it for a long period of time and no friction to slow it down, so these tiny pushes keep adding up.
The push of reflected photons can be measured, and I know of a bunch of experiments that measure it in one way or another, usually in a lab with laser light. There is already one functioning solar sail, as well: http://en.wikipedia.org/wiki/IKAROS
I know when the solar sail can be very effective.We should send a rocket to a comet.Then the rocket should travel on the comet till the maximal speed of the comet. Then the rocket should fly up from the comet and release a solar sail. A high-speed object receives more energy of photons, therefore such rocket can travel to the nearest stars. [:)]
Maybe the rocket can recieve the same speed without comet. [:-\]
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Mike, if the photon gives up some of it's energy, it will rebound with a smaller frequency. Generally they give so little energy to atoms or molecules that it's negligible. With a solar sail, radiation pressure is also tiny, but there is a huge number of photons bouncing off it for a long period of time and no friction to slow it down, so these tiny pushes keep adding up.
JP
I understand all of the above but I don't understand how a reflected photon can give up none of its energy (see my last post above) but still give a small push to the reflecting surface. This would seem to imply an increase of energy which can't be the case as it would break the law on conservation of energy.
It should give up some of its energy, and its frequency does change.
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If light bouncing back and forth between two mirrors looses its energy (as generally believed) by becoming red-shifted then this should be visible and measurable. But in multiple reflections there does not appear to be any red-shift, instead the light gradually becomes dimmer. This would seem to indicate that that the light is not loosing energy by being reflected but by absorption as the mirrors are not perfectly reflecting.
If anyone knows of any links to experimental evidence, one way or another, please post them. Thanks
My own feelings are that reflected light does not impart momentum to the mirror and so is not red-shifted.
Light reflected by two perfectly reflecting mirrors would essentially last unchanged indefinitely, were it not for photons escaping through quantum tunnelling. I appreciate this view may be wrong but would like to see the evidence.
More confusion here
http://www.physicsforums.com/archive/index.php/t-3647.html
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I don't think there ever will be perfectly reflecting mirrors Mike. But assume there was, then they would have to 'reflect' perfectly too. Maybe that's the reason there can't be :)
Anything interacting with something else must impart a momentum, and lose some of its 'energy' in form of? Heat? and maybe 'something else' too? I don't know there. I'm still too hung up on the idea of usable 'energy' versus 'used unusable energy' to really make up my mind on that one. But if it 'interacts' it will lose 'energy'.
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If photon gives more momentum to a mirror,then this photon gives less energy to the mirror. [;)]
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Mike, I can't tell you about any experiments but I can tell you that energy and momentum are conserved. That's a fundamental law, even in quantum mechanics.
Light does impart momentum to the mirror by conservation of momentum. If the mirror moves as a result of that momentum, then the photon has to redshift. I suspect this redshift can be thought of as the doppler shift, since the reflected photon is coming off a moving mirror. Obviously the amount the mirror moves from one photon is infinitesimal, so the redshift is as well.
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Mike, I can't tell you about any experiments but I can tell you that energy and momentum are conserved. That's a fundamental law, even in quantum mechanics.
Light does impart momentum to the mirror by conservation of momentum. If the mirror moves as a result of that momentum, then the photon has to redshift. I suspect this redshift can be thought of as the doppler shift, since the reflected photon is coming off a moving mirror. Obviously the amount the mirror moves from one photon is infinitesimal, so the redshift is as well.
Yes,then imparted momentum < 2p
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That's a tricky one. In my book it loses 'energy' by losing momentum. To assume otherwise is to define momentum as something else than 'energy'. But momentum will transfer a 'energy' if we define it as the ability to do work, and transform. So, I don't think you can do that, differ them.
"In empty space, the photon moves at c (the speed of light) and its energy and momentum are related by E = pc, where p is the magnitude of the momentum vector p. This derives from the following relativistic relation, with m = 0
E2 = p2c2 + m2c4."
Also take a look at How Does the Total Energy of a Particle Depend on Momentum. (http://galileo.phys.virginia.edu/classes/252/energy_p_reln.html)
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Which then, if we apply it on the concept of a time less, mass less, photon implies that photons then can 'change energy' intrinsically :) (when 'reflected' from a mirror) I don't think so.
When we instead look at it as a wave?
"Minkowski's formulation, on the other hand, seems more natural from the point of view of quantum mechanics. As light slows down inside a medium its wavelength also decreases, but quantum mechanics tell us that shorter wavelengths are associated with higher energies, and therefore higher momenta. In fact, Minkowski's approach suggests that the momentum of a single photon of light increases by a factor n as it passes through a medium. This result can also be supported by strong theoretical arguments, among them one that considers what happens when an atom moving at some speed through a medium absorbs a photon and experiences an electronic transition."
And as a wave it should also interact, as it will pass glass in and back out, and so get a 'higher momentum' :) Eh, in a ordinary mirror that is.
But assume just a (perfectly) reflecting surface, no glass involved. Then, if it is true that wave can't lose 'energy' due to reflection we can let it reflect between two mirrors into infinity, so, will it lose momentum then? If we now assume that momentum is different from the concept of 'energy'?
Einstein used E=mc2, there mass is assumed to be 'E = energy'. Which in physics is the same 'm' you multiply the velocity by, to find a momentum (p). Light use "m" because of its energy, not (invariant) rest mass.
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That's a tricky one. In my book it loses 'energy' by losing momentum. To assume otherwise is to define momentum as something else than 'energy'. But momentum will transfer a 'energy' if we define it as the ability to do work, and transform. So, I don't think you can do that, differ them.
I can't, but ability to impart momentum is not ability to impart energy. [:P]
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Energy and momentum are not the same thing by any definition. Momentum is given by three numbers and energy by one number for starters.
Now, under some conditions, you can use conservation of energy/momentum/mass to figure out one from the other, but that doesn't mean they're the same thing.
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Well, I'm saying that momentum becomes 'energy' in its transformation/interaction. Am I wrong?
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Then again, thinking of a 'light wave' and trying to see its momentum becomes really tricky as one prefer to define a momentum to something specific, like a 'photon/particle'. So how do quantum mechanics find the momentum for a standing wave? Its 'energy' alone. And how about a wave existing 'everywhere'? Where is the momentum?
It's a weird subject :)
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What do you mean by its transformation? Isn't that assuming it's becoming something that isn't momentum? It should always stay momentum, and momentum in = momentum out.
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Well, if you want to define it as something not being 'energy', then it has to transform into 'energy' when interacting. As long as I'm thinking right here that is. But I agree, that wasn't the best formulation, interaction is clearer.
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What I mean is that 'momentum' is an expression of energy, and I find it very hard to see it any other way. But then we have interactions, in where the energy expressed will belong to to both particles interacting. When light blue shift/red shift it is a consequence of interactions, that is two 'frames of reference' 'communicating', as in that light being annihilated on your retina, not an expression of a 'photon' changing energy intrinsically.
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Would you expect the momentum to stay the same (for the wave)? Isn't the conservation law implied for the whole 'system', not the reflected wave. That is, the momentum, counting in both mirror and wave, is unchanged?
Or is it something else you mean?
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Would you expect the momentum to stay the same (for the wave)? Isn't the conservation law implied for the whole 'system', not the reflected wave. That is, the momentum, counting in both mirror and wave, is unchanged?
Or is it something else you mean?
was p + 0
then -p + 2p
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Momentum of the whole system stays the same in the interaction.
Energy of the whole system stays the same in the interaction.
That's why you can't transform one to the other. If momentum somehow transformed into energy, you'd violate conservation of both energy (you'd gain some) and momentum (you'd lose some). What can happen is that different parts of the system gain or lose energy/momentum, but the total in the system has to stay constant for both energy and momentum, independently.
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Yes, that's how I understands it. Transform here is my way of saying what you define as, gain or lose 'energy'. But it's the interaction creating it, not some magical, each one by themselves, 'happening' :)
Didn't know you couldn't use the word transform for this btw. Would it be wrong to say that radiation transforms into energy too?
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Perhaps convert would be more acceptable :)
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Well, like everything else in science, precision is king. If you're precise about what you mean by transform, it's fine to use it in a scientific context.
But yes, in interactions particles can transfer energy or momentum. It's possible for this energy and momentum to create new particles as well.
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Also, it's worth checking out these three pages on wikipedia. They go into details about how elastic scattering of light from an electron (i.e. it doesn't lose energy) is just a limit of the more general scattering where the energy loss is so small that it can be neglected:
http://en.wikipedia.org/wiki/Thomson_scattering
http://en.wikipedia.org/wiki/Compton_scattering
http://en.wikipedia.org/wiki/Klein-Nishina_formula
The last link actually provides the quantum model that extends from the elastic limit to the inelastic limit.
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Thinking about it, presumably a photon can not be reflected. To be reflected there would have to be a point at which it was stopped. If not reflected then it must be absorbed and another photon re-emitted.
There seems to be two trains of thought on this.
One, the photon is re-emitted at the same frequency, in which case it cannot have imparted any momentum to the mirror.
Two, the photon imparts momentum to the mirror and another photons is re-emitted at a lower frequency.
Either way the 'reflected' photon does not impart any momentum to the mirror as there is no reflection. The photons impart momentum on absorption and again on re-emission. The more reflective the surface, the higher the frequency of the re-emitted photon.
If the above is correct then solar sails do not work by reflecting light but by absorbing and re-radiating it.
What's wrong with this argument?
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What hurts me head here is the assumption that a interaction can leave what interacts untouched. In this case 'photons' bouncing of a mirror. So maybe you can differ a photons momentum from 'energy', although I don't see how. No, on second thought, I don't think you can. To use the idea of them 'bouncing' and keeping their energy, and then say it is proved by the way photons don't redshift ignore the fact that they are constantly sent out from a source, get annihilated at a sink, and re-emitted depending on their 'energy' relative what they annihilate against. What is even more irritating is 'proving' it using wave functions. It's rather naive to define a duality from one point of view solely, as long as there is no experimental proof of one side of it being secondary and dependent on the first. I won't do that.
I still think of it as a duality, not a wave packet.
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And a photon doesn't have a frequency, only a momentum and a energy. It is waves that have a frequency, and a wavelength. Ah, that also mean that a individual 'photon' can't red shift, only 'waves' do that. And even if considering them over a extended period of time passing some 'point', the photons passing when measured have not red shifted. There must be some underlying definition for how they express the duality, in different situations, but I don't think considering them as 'renormalized wavepackets' with arbitrarily made 'cutoffs' answer that one. There is something else governing the way they express themselves. There is something inherently wrong with any approach that need to lift out the 'wrong' numbers, to get to the right ones, we see experimentally.
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Yor_on, photons do have a frequency. They don't oscillate like classical waves, but they have a property called wavelength. This is necessary, because when you add them up in the right way, this property is what creates a classical wave with a given wavelength. That frequency, f, is also what gives them energy of hf.
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Mike, your reasoning is correct. Reflection, absorption/re-emission and scattering from a mirror are all different words for the same phenomenon. You can't reflect without giving up energy, but the usual assumption is that the energy loss is so small that it doesn't matter.
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I don't know if I agree, I think of it as a duality? Maybe you are thinking of trapping photons in microwave cavities? How about Compton scattering, describing the opposite?
" The Compton effect is introduced in Gasiorowicz (in section 1.3, pages 7-9 in the 3rd edition, or pages 11-13 in the 2nd edition). The text describes the experimental discovery of the effect discovered by Arthur H. Compton - radiation of a given wavelength (in X-rays) sent through a foil was scattered in a manner inconsistent with classical radiation theory.
If one is dealing with elastic scattering,the system can be understood quantitatively as Thomson scattering. However, the Compton effect can be understood as photons scattering inelastically off individual electrons."
Here you start with x-rays, and end with 'photons'. Photoelectrons, Compton and Inverse Compton Scattering (http://venables.asu.edu/quant/proj/compton.html)
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You don't agree with what?
Photons have a frequency, which is true: E=hf after all. If you want to know how this relates to the frequency of a classical wave, check out coherent states. It takes a lot of math to prove it, but you can superimpose photons in such a way to make a classical wave of the same frequency as the photons: http://en.wikipedia.org/wiki/Coherent_states
Also, check my links above if you don't believe me that Compton and Thompson scattering are related. The general quantum scattering problem is quite complex, but you can simplify it by assuming certain things, which is what is done in these two cases. Thompspon scattering assumes that the photon loses so little energy that you can treat it as losing none and still get accurate results.
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I don't agree to it being one or the other, to me radiation seems to define itself through the relations/interactions with what surrounds it in the measurement.
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And no JP, they are related, the link I gave says it too.
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So you're agreeing with me?
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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?
From what I have noticed throughout life is that whenever a force is exerted and deflected it does lose energy in some way. Just as when light is deflected when bent through space as it is gravitationally bent as well considering how far stars are away from us. A case in point would be the stars that we see at night and 'appear' to be in a fixed position in the sky. Considering how far they are away, they may be a bit off from where they appear. The closer the star of course the less it may be off, such as Alpha centauri the closest one to us if I recollect right. The image is accurate of course, it just may not be exactly where we see it in space. As far as my point goes, concerning loss of energy, have you ever known anything in life that that has a 'theoretical' constant velocity that didn't lose some type of momentum or energy when acted upon by some external force that is fixed? One has to apply this formula when taking this aspect into consideration and make the assumption that this is equated in some way.
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I don't know how you see it? Most people today define it from waves, with photons becoming some sort of 'focus' for that. I don't think either one is perfect, on the other hand I can't say how it should work. My own, most unscientific opinion, is that both waves and 'photons' are a description of what relations there are defining them.
That doesn't necessarily demand a underlying 'reality' aka, some 'hidden' background, defining what they really, really, are. But I think it demand another way of looking at them, that make their duality a result of the demands, different in different experiments. So I'm happy with thinking of both as 'real', well, as real as we can define it. Which then means that I'm not convinced of anything :)
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Ah, I think I understand you now. Check out that link I posted above (http://en.wikipedia.org/wiki/Coherent_states). It explains the precise mathematical connection between photons and waves.
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Yep JP. it's better to discuss it from coherent states, that precludes any of our ideas on 'waves and 'photons'' But then I meet another problem. I do like indeterminacy (HUP), but I'm not sure I find 'virtual particles' as convincing. To me they are starting to become a figment of our deep conviction of thingies 'moving and propagating'. And I'm not sure what that is any more. Physics are truly confusing, reminding me of Pandoras box. You may open it, but on your own peril. There has to be a way of describing it not involving 'virtuality' as in something 'moving'.
And to me it must have to do with 'time'.
I prefer indeterminacy to 'virtual particles' those days
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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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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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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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why not consider light as a screw/tornado action instead of a wave? & how might clockwise & counterclockwise effect?
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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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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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Hmm, preclude as in defining a theoretical proposition, joining our definition of photons and waves was what I meant. English is dangerous :)
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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'?
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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?
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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?
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Momentum has a magnitude and direction. If the magnitude stays the same, but the direction changes, then momentum has changed.
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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.
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Maybe I should make that clearer above.
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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?
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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.
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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.
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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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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?
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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.
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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.
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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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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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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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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]).
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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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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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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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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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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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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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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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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.
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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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Sorry JP :) We collided there.
(the postings I mean)
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Maybe you could say that the 'energy' becomes indeterministic in that it 'tags down'?
Or?
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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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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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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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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?
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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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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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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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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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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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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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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?
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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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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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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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Ah well JP.
I'll let it rest for now.
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JP
2 things.
!. After some slow thinking, I get what you mean. It's a paradigm change to me. But I get it. And those years later I think I see what you mean.
Can't promise anything though. I will argue :)
2. I miss you man
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To my eyes you're one of the clearest, thoughtful, and careful thinkers I know of.
Wish I had has you as a teacher, maybe I would have stayed in school.