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Alistair Hosie
11-05-2010, 04:08 PM
In the world of darkness it is possible to project light.Is it possible to project in the world of light darkness? I say no actually this is one of the first posts I put here many years ago when I was just a boy.C'mon Evan put me out of my misery this is a serious question and I have a feeling someone here maybe you will be able to explain why or why not.I say personally that darkness is the absence of light .and cannot be created or rather projected like a torch beam, where there is light. Alistair
__________________

TGTool
11-05-2010, 04:49 PM
You're obviously excluding politicians when you assert that darkness can't be created or projected. :D

MichaelP
11-05-2010, 04:53 PM
Darkness projected in the world of light is called "shadow". :)

beanbag
11-05-2010, 05:15 PM
Light comes from particles call photons. You only "see" light because those photons bounce off something and then go into your eye. So in a dark room (absence of photons) you can project light, e.g. a flashlight shoots out photons. If you have a lighted room with photons already bouncing around, you can't remove them because there is nothing that can "suck" them out.

However, just because there is light, does not mean you can see it. For example, in a dark field, you can aim the flashlight away from you and into the sky. It still shoots out a beam of light, but because that beam of light does not reflect back into your eye, you can't see it. But if there is smoke around due to a welding incident, the beam of light will reflect off the little particles and into your eye. Then you can see the beam of light.

MuellerNick
11-05-2010, 05:27 PM
If you have a lighted room with photons already bouncing around, you can't remove them because there is nothing that can "suck" them out.

Wait a moment! I'll just switch on my black hole and will be back in a moment %&$)(&carrier lost

beanbag
11-05-2010, 05:30 PM
Wait a moment! I'll just switch on my black hole and will be back in a moment %&$)(&carrier lost

Looks like the LHC worked as some expected.

Evan
11-05-2010, 05:52 PM
It is possible to project a beam that will cause light to vanish. There are many catches to this statement but it is based on the phenomenon of intereference. When a photon illuminates a surface at a particular point if a photon of the same energy but shifted in time by one half of a wavelength reaches that same point at the same time the result is darkness. This happens even though the surface would have been illuminated if either photon alone reaches the same point but when both do then instead of twice the illumination we have none.

Light behaves according to the laws of probability that govern quantum mechanics. This leads to some interesting unexpected effects. There is one in particular that is very surprising.

A polarizing filter as found in polarized sunglasses allows only light that is vertically polarized to pass. When light of random polarization is reflected from a horizontal surface the horizontally polarized component is reflected with much higher efficiency that the vertically polarized portion. The probability of reflection changes with the change in angle between vertical and horizontal.

If you take a filter such as a lense from a pair of polarized sunglasses it passes the most light in the vertical orientation and the least in the horizontal. That is how they reduce glare.

The light that makes it through the lens is predominantly vertically polarized. If you then place another filter following the first but rotate it 90 degrees then nearly all the light is blocked by the second lens since it has a very low probability of transmitting vertically polarized light.

This is as we would expect and makes perfect sense. Keep in mind that the polarized lenses do not change the polarity of the electromagnetic waves that make up a photon. They are only selective for a particular orientation.

Now the mystery occurs. If you place a third lens between the pair that together block all the light and you orient the third lens at 45 degrees then suddenly some light will pass through all three lenses. Remove the middle lens and it does not.

Why?

Vertically polarized light has a zero chance of making it through a horizontal polarizer. But, it has a 50/50 chance of making it through a 45 degree polarizer. The middle polarizer therefor passes half the light that made it through the first lens. The light that makes it through the first and the middle lens is predominantly polarized at 45 degrees. It has a 50/50 chance of making it through the last lens. So, with just two lenses at 90 degrees there is a zero probability that light may pass through both. However, placing a third lens between those two alters the probability to allow 1/2 x 1/2 x 1/2 which equals 1/8 of the light will make it through the three filters but none passes just two.

Alistair Hosie
11-05-2010, 06:36 PM
Well done and I'll ponder on that tomorrow just off to bed I knew Evan would know.Actually an eye surgeon friend in germany and I taklked about this at length many years ago off to sleep hope I didn't wake you Evan LOL Alistair

MuellerNick
11-05-2010, 06:54 PM
But, it has a 50/50 chance of making it through a 45 degree polarizer.

The chance is cos(45) = 0.707
etc ...


Nick

beanbag
11-05-2010, 07:09 PM
Keep in mind that the polarized lenses do not change the polarity of the electromagnetic waves that make up a photon. They are only selective for a particular orientation.


I would disagree with this statement. The output of a polarizer is polarized light in the new orientation, with the amplitude/intensity reduced according to the angle. (The amplitude is the projection along the new direction)



Vertically polarized light has a zero chance of making it through a horizontal polarizer. But, it has a 50/50 chance of making it through a 45 degree polarizer.

The way you wrote the above two statements makes it sound like the output is still vertically polarized light with half the intensity.
Rather, the output of the 45 degree polarizer is 45 degree polarized light with half the intensity. (sqrt 2 decrease in amplitude for the 45 degrees, and then square again to get intensity out of amplitude ---> 1/2 intensity)



The light that makes it through the first and the middle lens is predominantly polarized at 45 degrees.
That's right, it changed the vertically polarized light into 45 degree.

I think you understand polarization fine- I just think you didn't explain certain parts well.

mmm, now that I think about it some more, the way you wrote it has some validity in the single photon / particle representation of light.

beanbag
11-05-2010, 07:11 PM
The chance is cos(45) = 0.707
etc ...


Nick

Evan is right, kind of. You square the amplitude to get intensity, i.e. "chance" or probability.

beanbag
11-05-2010, 07:25 PM
Well done and I'll ponder on that tomorrow just off to bed I knew Evan would know.Actually an eye surgeon friend in germany and I taklked about this at length many years ago off to sleep hope I didn't wake you Evan LOL Alistair

hey, I was the one who explained it, give credit where credit is due :D

Your Old Dog
11-05-2010, 07:38 PM
I haven't read the entire thread:

Alistair, I'll buy your "I say personally that darkness is the absence of light." If you can tell me who or what is casting the truly huge shadow in the universe? LOL

Optics Curmudgeon
11-05-2010, 07:42 PM
hey, I was the one who explained it, give credit where credit is due :D
And incorrectly, at that. Evan is right, polarizers do not change the polarization. You need a waveplate to do that.

Joe

Optics Curmudgeon
11-05-2010, 07:44 PM
Or a Faraday rotator, or Pockels cell, or Kerr cell....

Evan
11-05-2010, 07:48 PM
I would disagree with this statement. The output of a polarizer is polarized light in the new orientation, with the amplitude/intensity reduced according to the angle. (The amplitude is the projection along the new direction)



Heh. There is no difference between that interpretation and mine. Either way the same amount of light makes it through and the probabilities work out the same. For more on this the authoritative work still remains "QED, Quantum Electro Dynamics, By Richard Feynman. "

beanbag
11-05-2010, 08:01 PM
And incorrectly, at that. Evan is right, polarizers do not change the polarization. You need a waveplate to do that.

Joe

What's the output from the 45 degree polarizer, then?

Evan
11-05-2010, 08:11 PM
Remember that the input is random polarized light. The photons such as sunlight are evenly distributed at all angles. The ability of a photon to pass through a polarizer is governed by probability so one half of photons will pass through. They retain the same polarization that they had to begin with. That also applies to all angles of polarization, not just 45 degrees. The output of the 45 degree polarizer consists of photons of all polarizations including 45 degrees. At no angle is the probability zero in respect of the first polarizer. At no angle is the probability of that output zero in respect of the third polarizer.

PeteF
11-05-2010, 08:19 PM
In the world of darkness it is possible to project light.Is it possible to project in the world of light darkness? I say no actually this is one of the first posts I put here many years ago when I was just a boy.C'mon Evan put me out of my misery this is a serious question and I have a feeling someone here maybe you will be able to explain why or why not.I say personally that darkness is the absence of light .and cannot be created or rather projected like a torch beam, where there is light. Alistair
__________________

It depends on what you mean by "project", however something you may be familiar with where areas of "darkness" are created are optical flats. If you're not familiar the usual searches should explain enough.

Pete

beanbag
11-05-2010, 08:23 PM
The output of the 45 degree polarizer consists of photons of all polarizations including 45 degrees.

By that logic, if you put another 45 degree polarizer behind the first 45 degree polarizer, such that they are aligned, it would cut the intensity another 50%.

Evan
11-05-2010, 08:32 PM
Sorry, I just logged in to correct myself.

"The output of the 45 degree polarizer consists of photons of all polarizations including 45 degrees from the light that made it through the first polarizer"

beanbag
11-05-2010, 09:05 PM
Sorry, I just logged in to correct myself.

"The output of the 45 degree polarizer consists of photons of all polarizations including 45 degrees from the light that made it through the first polarizer"

My question still stands, though...

What is the difference between:

natural light --> 0 deg polarizer --> 45 deg -->output

vs

natural light --> 0 deg polarizer --> 45 deg --> 45 deg (aligned) -->output

G1K
11-05-2010, 10:08 PM
And where does Lucas, the prince of Darkness, fit into all of this ?



R

Yes, I went there...

philbur
11-05-2010, 11:06 PM
Actually he gave his usual long-winded dialogue, but also as usual it mostly had nothing to do with the question.

In any case for complete destructive interference you have to make the energy disappear, then you have to answer - where did it disappear to?

Phil:)


I knew Evan would know.

Evan
11-05-2010, 11:13 PM
What is the difference between:

natural light --> 0 deg polarizer --> 45 deg -->output

vs

natural light --> 0 deg polarizer --> 45 deg --> 45 deg (aligned) -->output


In theory (perfect polarizers) the probability function for that orientation has remained open so as long as you make no attempt to quantify the polarization of any of the photons then they will have a 100 percent chance of going through the second identically aligned polarizer. If you make any attempt by any means to actually measure the polarization of any of the photons then the probability function collapses and the result will be the same as if the first 45 degree polarizer did not exist.

In fact, if you even set up an experiment that somehow measures the polarization of photons that have already passed through all of the filters then the result will change as the function will collapse backward in time.

Stranger yet if you set up the experiment so that you could find out the polarization of individual photons but never actually turn on that part of the experiment the probability function collapses anyway. Some people will recognize that this is simply a variation of the two slit interference experiment. The "rules" are the same.

In the quantum world you cannot observe or measure anything individual without affecting the outcome. As long as you deal only in probabilities then the probabilistic behaviour is exhibited.

Evan
11-05-2010, 11:16 PM
then you have to answer - where did it disappear to?


I said there were catches. It appears nearby as an area of double brightness. Regardless, the area of darkness DOES exist.

PeteF
11-05-2010, 11:19 PM
Actually he gave his usual long-winded dialogue, but also as usual it mostly had nothing to do with the question.

Ha ha, I must admit I did have a giggle at that comment. How do you spell obfuscation? ;)

Evan
11-05-2010, 11:24 PM
How do you spell obfuscation

Q U A N T U M- M E C H A N I C S

Don't feel bad if it makes no sense to you. Even Feynman admitted that he didn't really understand how it works.

PeteF
11-05-2010, 11:32 PM
No Evan, your spiel on polarising had zero/zilch/nada/big fat NOTHING to do with the question. This was pointed out by somebody else, not me old mate! I was just having a good old laugh.

I managed to answer the question in one line AND give an example a machinist may be familiar with. Don't feel bad if you don't understand the question Evan ;)

Umm .. Ohh - Bee - few ... no no wait!

philbur
11-05-2010, 11:46 PM
Before it is possible to speculate on the answer you need to define darkness. Are we talking the absence of electromagnetic radiation detectable by the human eye, or something else. Also you have to define what magnitude of the absence of light qualified it as darkness.

Phil:)


In the world of darkness it is possible to project light.Is it possible to project in the world of light darkness? __________________

philbur
11-05-2010, 11:59 PM
Only if you obsruct all other sources, but then if you can do that you didn't need the obstructive interference in the first place.

Phil:)


Regardless, the area of darkness DOES exist.

philbur
11-06-2010, 12:01 AM
But of course you do.;)

Phil:)


Even Feynman admitted that he didn't really understand how it works.

RB211
11-06-2010, 01:07 AM
If you can do it to sound waves, why not light waves?

Evan
11-06-2010, 01:13 AM
No Evan, your spiel on polarising had zero/zilch/nada/big fat NOTHING to do with the question.

Perhaps not to you. How much do you know about quantum mechanics? The concepts involved are required to gain some sort of understanding of interference phenomena and how probability describes the results. The polarization example can be performed using ordinary items that many people have laying about. The two slit experiment which illustrates the same principles of the probability of interference is not something that is readily performed.

Evan
11-06-2010, 01:16 AM
Only if you obsruct all other sources, but then if you can do that you didn't need the obstructive interference in the first place.


Nonsense. Interference phenomena can be demonstrated using optical flats and randomly polarized white light.

darryl
11-06-2010, 02:18 AM
If you define light as the electromagnetic radiation that the eye is capable of responding to, and if you accept that said radiation is made up of photons, and further if you accept that photons must enter the eye before there is any perception of light, then when no photons enter the eye, there is darkness.

If no photons enter the eye from any direction, there is the perception of darkness. If photons enter the eye from every direction within its field of view except for one spot, then you have what could be called a ray of darkness. If you had a photon detection device and placed it anywhere within that ray, it would not detect any photons with a direction that is towards your eye, even though there may be photons travelling in other directions through that area. If you were to be able to stand off to the side and attempt to look through that ray of darkness sideways, you would see the light that wouldn't have entered your eye from your original position. In essence you would be proving that there is still light in that ray of darkness, otherwise as you look at it sideways, you would see what appears to be a completely black zone. It would look like a jet black rod, or a cone. You would be able to pass your hand through it, but that portion of your hand within that zone would be jet black.

You can look towards a hole in the wall at night and see darkness. You can walk about the room and from anywhere you can see that hole, you will see darkness. It could be said that a ray of darkness is entering your eye, but if that's true, the room must be full of rays of darkness emanating from that hole in the wall. The room is full of darkness, and also full of light. We know it's full of light because in the direction of that hole we perceive darkness- the absence of light from that direction. Because we can change our point of view and still perceive darkness from the direction of that hole, we have to assume that there is no ray of darkness, only a boundary area where light doesn't come through (the hole).

Then of course, we must question whether this does really represent darkness. It might be night time, but there will be stray photons making their way through that hole anyway. Starlight, moonlight, some guy waving a flashlight around as he's casing your place- maybe the amount of photons from any source is too low to be above your eyes threshold for seeing- so this raises another question, that of how you define what is light. If it's of such a low level that your eye doesn't register it, is there any light? And is that darkness?

Put your head into a light-proof container and seal all around with light proof material. Essentially you now have 360 degrees of darkness, though you might 'see' stick figures and other anomalies. Do these things kill the perception of darkness?

To me, a ray of darkness would be something you could look at from any angle and it would look like a black object. You couldn't see through it, lengthwise or sideways. If you passed anything through it, you would lose sight of that part that was within it. It might look like a wire, a cylinder, or a cone, and it would have a straight axis. When you view a ray of light, that's how you see it, although in that case, whenever you pass anything through it, you see that thing enhanced by the light. The ray of darkness would hide anything within it. I have never known of any such a thing.

beanbag
11-06-2010, 03:39 AM
In theory (perfect polarizers) the probability function for that orientation has remained open so as long as you make no attempt to quantify the polarization of any of the photons then they will have a 100 percent chance of going through the second identically aligned polarizer. If you make any attempt by any means to actually measure the polarization of any of the photons then the probability function collapses and the result will be the same as if the first 45 degree polarizer did not exist.



I think what you said in a roundabout way is that the second polarizer makes no difference between the two outputs. Hence why I said that after the first 45 deg polarizer, the light is now polarized at 45 deg, as it can pass thru the second polarizer no problem. Your setup AFAIK can be described completely classically, so there is no need to invoke QM, or god forbid QED.

I don't agree with Mr. Curmudgeon's assertion that "polarizers do not change the polarization.", but I will wait to see why he thinks I am wrong. I think that polarizers DO change the polarization, but they reduce some of the amplitude in the process. IOW, the 45 deg polarizer takes vertically polarized light at the input and converts it into dimmer 45 degree light.

Evan
11-06-2010, 04:33 AM
so there is no need to invoke QM, or god forbid QED.


It isn't a matter of need. I originally suggested interference phenomena as an example of Alistair's "dark ray". That phenomenon is correctly described by QED. QED isn't difficult to understand. Feynman is (was) such a skilled teacher that he presents it so well that it almost makes sense. What he didn't know he predicted and later discoveries are consistent with his explanations. I am not sure what you consider to be "classical" explanations of the various phenomena that accompany the interactions of light. Some are useful approximations but break down utterly in the face of various limits. QED deals with the limit conditions correctly and is both internally and externally consistent. QED and Quantum mechanics are simply faces of the same coin. Since photons in the range of visible light exhibit both wave and particle behaviour one cannot simply ignore one side of the duality to make explanations appear simpler.

Ian B
11-06-2010, 05:24 AM
Light is the result of particles called photons. These bounce around a room when a light switch is switched on. It stands to reason, therefore, that when the light is switched off, the light bulb then emits photoffs, the 'anti' light particles that cancel out any errant photons still remaining.

Proof of this is that the room then remains in a state of darkness until the photons return.

I believe that photoffs are more commonly referred to as 'dark matter'...

Ian

beanbag
11-06-2010, 05:41 AM
I originally suggested interference phenomena as an example of Alistair's "dark ray". That phenomenon is correctly described by QED.

It is also described just fine using wave mechanics of classical E&M, with no need to mention "probabilities" nor individual particles.



QED isn't difficult to understand. Feynman is (was) such a skilled teacher that he presents it so well that it almost makes sense.


Maybe what you mean to say is that some concepts in QED are not difficult to understand. To actually work thru the math and equations to make quantitative predictions is rather difficult.



I am not sure what you consider to be "classical" explanations of the various phenomena that accompany the interactions of light.
Some are useful approximations but break down utterly in the face of various limits.


That's totally true, but your experiment isn't one of them. However, it is one of the "classic" cases where you can very slowly and carefully use QM principles (with a minimum of math) to get the same answer as somebody else could have gotten much quicker. I found this page:
http://www.informationphilosopher.com/solutions/experiments/dirac_3-polarizers/

where Dirac very slowly and carefully steps thru your setup using QM principles. The point is not really to get into a link battle with you posting reference back and forth. You can just note via the length that a QM treatment involves more than just saying "photon" and "probability" here and there.



Since photons in the range of visible light exhibit both wave and particle behaviour one cannot simply ignore one side of the duality to make explanations appear simpler.

yes they can, if it gives the right answer ;) We like shortcuts, right?

MuellerNick
11-06-2010, 06:04 AM
Evan is right, kind of. You square the amplitude to get intensity, i.e. "chance" or probability.


Intensity, not amplitude. Right.

Nick

Evan
11-06-2010, 07:23 AM
That's totally true, but your experiment isn't one of them.

My understanding of light is based on QED, not "shortcuts".



To actually work thru the math and equations to make quantitative predictions is rather difficult.

The book QED, The Strange Theory of Light and Matter does not demand the ability to work all of the math in order to understand the theory and even use it. That is part of the beauty of the theory and it has never been found to be wrong in any aspect it describes. It was developed because the classical theory fell apart in certain areas and produced meaningless results. One such area is the polarization of light at grazing incidence. That is a direct result of the failure of the classical theory to deal with any time the probabilities of anything approach zero or infinity.

I am not going to post links on this as I have no need. Feel free to look up what you must. I recommend the freely available series of lectures given by Feynman on this topic. Much of his work has been placed in the public domain which I am certain would have pleased him.

philbur
11-06-2010, 08:09 AM
Yes but if the room lights are on then it isn't dark. Don't over-think it.
PHil:)


Nonsense. Interference phenomena can be demonstrated using optical flats and randomly polarized white light.

Evan
11-06-2010, 10:03 AM
That doesn't make sense. It is the light of the room itself that is being used to create dark bands. Don't underthink it.

philbur
11-06-2010, 11:19 AM
So how does this magical device influence light approaching the point of projection directly from walls, ceiling and floor. Did you observe that the bands are only darker than the average, not absolute black. Did you wonder why.

Silly billy.

Phil:)


That doesn't make sense. It is the light of the room itself that is being used to create dark bands. Don't underthink it.

Tony Ennis
11-06-2010, 11:23 AM
QED is a good book. Should be required reading.

philbur
11-06-2010, 11:33 AM
Back to the original point.

So what. It produces an area that is less bright than the surrounding area. If I have other sources of very bright light shining on the interference pattern then you would not see the pattern. Hence my point about the need to block other sources and the recommendation not to over-think it.

Phil:)


Nonsense. Interference phenomena can be demonstrated using optical flats and randomly polarized white light.

Evan
11-06-2010, 02:12 PM
Silly billy.

Ad hominem starts with you.


So how does this magical device influence light approaching the point of projection directly from walls, ceiling and floor. Did you observe that the bands are only darker than the average, not absolute black. Did you wonder why.


It is the difference between a perfectly constructed physical object and one made by human hands. It isn't a deficit in the principle. It has never been a source of wonder to me that humans are imperfect. Your arguments are specious and not to the point.


If I have other sources of very bright light shining on the interference pattern then you would not see the pattern.

?????? The brighter the light the clearer the pattern. All incident light will contribute to the pattern.

I suggest you do some research on white light interferometry. Start by looking up "airy disc" and "telescope"

Optics Curmudgeon
11-06-2010, 02:24 PM
I think what you said in a roundabout way is that the second polarizer makes no difference between the two outputs. Hence why I said that after the first 45 deg polarizer, the light is now polarized at 45 deg, as it can pass thru the second polarizer no problem. Your setup AFAIK can be described completely classically, so there is no need to invoke QM, or god forbid QED.

I don't agree with Mr. Curmudgeon's assertion that "polarizers do not change the polarization.", but I will wait to see why he thinks I am wrong. I think that polarizers DO change the polarization, but they reduce some of the amplitude in the process. IOW, the 45 deg polarizer takes vertically polarized light at the input and converts it into dimmer 45 degree light.

If you operate on the assumption that a "polarizer" is a device that takes randomly polarized light and shifts the E field vectors of all of the photons (or waves, if you wish) into one orientation you might believe there is a rotation happening. Unfortunately, the actual function of a "polarizer" is not visually apparent. What these actually are is an exclusion device, not a conversion device. When randomly polarized light goes through the device (there are many types, Polaroid is only one, but the most commonly seen by the average person) what comes out is all of the polarizations that went in, with the polarizations that are different from the device's E field axis removed to a degree proportional to the cosine of the angle between the photon's (or wave's) field vector and that of the polarizer. Thus, a polarizer oriented to produce "vertically" polarized light actually produces light that is essentially "un-horizontal", with the horizontal component removed. If you produce a polar plot of the probability of a horizontally polarized light quantum passing through a vertically oriented polarizer it will look like a two leafed clover, with the zero amplitude points horizontal. This is why you get most of your input light through a polarizer. There are polarizers that produce highly linear polarizations, but do so by reflection. The off-polarization light that is absorbed by Polaroid filters is reflected by these, and the effect is the inverse. These devices are more difficult to use, and generally wavelength sensitive, so aren't commonly seen. So, polarizers remove light with unwanted polarization, by absorbing or otherwise rejecting it. The vectors of the "survivors" are unchanged. Polarization rotators are completely different, and rely on either natural birefringence or external electric or magnetic fields inducing birefringence in materials. The three polarizer experiment is a classic "puzzlement", used by teachers to illustrate the statistical nature of light by "loading the dice" with the third polarizer. It has nothing to do with the light's polarization being changed by a passive device.

Joe

philbur
11-06-2010, 03:11 PM
I suggest you need to read what has been said, not what you hope has been said. Read it all again.

Phil:)





It is the difference between a perfectly constructed physical object and one made by human hands. It isn't a deficit in the principle. It has never been a source of wonder to me that humans are imperfect. Your arguments are specious and not to the point.



?????? The brighter the light the clearer the pattern. All incident light will contribute to the pattern.

I suggest you do some research on white light interferometry. Start by looking up "airy disc" and "telescope"

John Stevenson
11-06-2010, 03:20 PM
Here's the first page of Stevie Wonders new book:-

The wave patterns of light.













http://homepage.ntlworld.com/stevenson.engineers/lsteve/files/swonder.jpg

.

jugs
11-06-2010, 03:45 PM
Here's the first page of Stevie Wonders new book:-

The wave patterns of light.













http://homepage.ntlworld.com/stevenson.engineers/lsteve/files/swonder.jpg

.

Thanks john, I can see what the girls are arguing about now (there will be tears before bedtime). :D

john
:)

PeteF
11-06-2010, 04:45 PM
Perhaps not to you. How much do you know about quantum mechanics? The concepts involved are required to gain some sort of understanding of interference phenomena and how probability describes the results. The polarization example can be performed using ordinary items that many people have laying about. The two slit experiment which illustrates the same principles of the probability of interference is not something that is readily performed.

BS Evan! Your spiel is akin to a learner driver asking how to stop the car and the "instructor" going into a 30 minute monologue on hydraulic circuits, frictional surfaces and conservation of energy. Never having learned to drive themselves.

Now, one would think a photocopier repairman like yourself would be quite well versed in optics and could provide a simple and concise answer to the question without having to spear off into completely unrelated material and plagiarise an example I gave. Apparently not :rolleyes:

Alistair Hosie
11-06-2010, 05:00 PM
Wow repair man or not Evan is pretty well versed in this subject just as I thought he would be you too pete.I always thought there were so many clever people here and more I am fascinated by this as a child Thanks even though I do understand the theory of polarization or thought I did.Now I am greatly put in my place and feel helpless in terms of my small scientific brain even though I have a Bsc with honours albeit in psychology many,many, years ago:D I am amazed and insecure in my understanding or perhaps my misunderstanding of this.Thanks a million so far Guys and pete sorry I see you are every bit the man when it comes to explaining this I didn't mean to shut you out I am just and was then a bit sleepy Morphine you know :D love you guys big Al.Alistair

darryl
11-06-2010, 06:01 PM
Thanks, John, you made my day!

beanbag
11-06-2010, 06:30 PM
OC,

At least in terms of output intensity vs angle, the way you describe polarizers is exactly the way I expect them to work. I never said that polarizers "rotate" polarization because that would mean it would change the direction of the E field without decreasing the amplitude. I consider the polarization of light to be the direction the E field points. So when the output of a polarizer is a vector with a different amplitude and direction, I call it "changed".

I see from your explanation that you are really trying to avoid treating the light and polarizer classically by just saying that it takes one E-field vector and extinguishes a certain component, resulting in a new vector. Instead, you invoke quantum this-and-that's by saying:

"The vectors of the "survivors" are unchanged. ... The three polarizer experiment is a classic "puzzlement", used by teachers to illustrate the statistical nature of light by "loading the dice" with the third polarizer."

That's fine, I guess. If you want to whip out the QM, I would say that the polarizer is an operator, and it "changes" the wave function of the photons that go thru it. It so happens that if you "look for" photons that have the original polarization you can occasionally find one, due to probabilities and wave function collapse, etc.

I see that QM /statistics can be used to solve this particular problem, but I don't see why classical E&M cannot, and why you and Evan choose to use the QM language which can be confusing unless stated with a lot of care and precision.

PeteF
11-06-2010, 06:39 PM
Al, you have absolutely no right to feel insecure about your knowledge, sadly some here use obtuse "explanations", apparently in an attempt to make themselves appear more knowledgeable and intelligent than reality. The truth of the matter is somebody who REALLY knows their subject will be able to relate a factual explanation in language and examples that a lay-person will understand. One of my favourite maxims is: a professional will make a difficult job look easy, an amateur will make an easy job look difficult.

With regards your question, you may be familiar with active noise cancelling headphones, Bose is a popular manufacturer of these. They are designed to use an additional audio signal to cancel an existing sound and give the illusion of reduced sound at the eardrums of the listener. Sound is a waveform just as is light, and similar principles can be applied to both, or indeed any waveform.

Pete

Alistair Hosie
11-06-2010, 07:04 PM
Pete I love you Ausies My poor uncle Jimmie was torpedoed or his ship was :D by the japanese They were towed into Australia for repairs and then as the repairs were started the second world war ended.He remained there got married to an Ausie shiela and had afew kids till he eventually died in his early sixties .Hhe wanted me to emigrate and go there ,and I had filled in all the forms ( Brits with relatives already there and established could go on a sponsored scheme) my uncle sponsored me and I was all set to go past all the medicals and paperwork when suddenly I was told that after six months I could be called up for Vietnam this was in mid sixties when the war was well established I declined their kind offer and stayed People forget the poor old Beutiful Aussies died for Vietnam man, I stayed home and married my lovely bronwen whom I met a couple of years later.True story two of my sons have been staying there and I hope to see it before I die.Alistair

John Stevenson
11-06-2010, 07:10 PM
True story two of my sons have been staying there and I hope to see it before I die.Alistair

Go tomorrow then you old fart.

The Artful Bodger
11-06-2010, 07:42 PM
He wanted me to emigrate and go there ,and I had filled in all the forms ( Brits with relatives already there and established could go on a sponsored scheme) my uncle sponsored me and I was all set to go past all the medicals and paperwork when suddenly I was told that after six months I could be called up for Vietnam this was in mid sixties when the war was well established I declined their kind offer and stayed People forget the poor old Beutiful Aussies died for Vietnam man, I stayed home and married my lovely bronwen whom I met a couple of years later.True story two of my sons have been staying there and I hope to see it before I die.Alistair

According to a recent survey Australia is the second best country in the world to live in, NZ is third so I can only imagine the Skippies cheated again.;) (Best country according to the survey is Norway).

The Artful Bodger
11-06-2010, 07:51 PM
Go tomorrow then you old fart.

Excellent advice to Johno, as they would call him down there, here is an example for you:-

http://farm2.static.flickr.com/1150/5152774270_0e9fca1578.jpg

Australia as seen from your tourist bus!!:D

Ed.
11-06-2010, 08:19 PM
Al,
With regards your question, you may be familiar with active noise canceling headphones, Bose is a popular manufacturer of these. They are designed to use an additional audio signal to cancel an existing sound and give the illusion of reduced sound at the eardrums of the listener. Sound is a waveform just as is light, and similar principles can be applied to both, or indeed any waveform.

Pete
Hi Pete, I don't agree with the last sentence in your statement. As I understand it, sound is the transmission of energy from one particle to the next and light is actually a photon particle, both have waveforms. So in the case of sound, if there are no further particles to transmit that energy then the sound stops ie: vacuum, where as light needs no additional particles to hit to continue its travel, therefore they use different principles, so if you put a light bulb and a sound speaker inside the center of a small sealed metal box, light would be stopped from escaping whereas the sound would still be heard, muffled but still heard as the sound energy would pass though the walls and air but the light would be stopped by the boxes steel walls. Do the same again but use a transparent glass box and form a vacuum inside and the light will pass through to the outside but the sound would not. So they may have some similar principles but not all, so you have to apply different principles to different waveforms. Just my 2 cents worth

Ed.
11-06-2010, 08:26 PM
Excellent advice to Johno, as they would call him down there, here is an example for you:-

http://farm2.static.flickr.com/1150/5152774270_0e9fca1578.jpg

Australia as seen from your tourist bus!!:D

:D :D Love that image, so true in so many places on Oz, I am wondering how many members looking at that image are wondering what the heck does that mean!

Ed.
11-06-2010, 09:00 PM
Here are two links of actual photo's for those who still don't get it :D

http://travel.webshots.com/photo/1487314027079303117MZEivo

http://images.travelpod.com/users/rbisset/rtw05-06.1144141980.straight.jpg

PeteF
11-06-2010, 09:01 PM
Ed, I'm glad you agree they are similar principles. You are absolutely correct they are different types of waves, however I sought a simple explanation and analogy that could be (hopefully) be understood by somebody previously completely unfamiliar with the concept. The principles are similar. At the end of the day this is a machinist BB, to immediately spear off into relatively complex discussion on physics is of little or no value to the OP. When my wife asks me "Is it going to rain today?" She wants to know whether or not to hang the washing out, the most valuable answer in that scenario is "Yes", "No", or "Maybe", not a 30 minute session of dew-point, types of clouds, chaos theory, or pressure gradients. A bunch of meteorologists on the other hand may have other ideas ;)

Pete

Ed.
11-06-2010, 09:18 PM
Ed, I'm glad you agree they are similar principles. You are absolutely correct they are different types of waves, however I sought a simple explanation and analogy that could be (hopefully) be understood by somebody previously completely unfamiliar with the concept. At the end of the day this is a machinist BB, to immediately spear off into relatively complex discussion on physics is of little or no value to the OP. When my wife asks me "Is it going to rain today?" She wants to know whether or not to hang the washing out, the most valuable answer in that scenario is "Yes", "No", or "Maybe", not a 30 minute session of dew-point, types of clouds, chaos theory, or pressure gradients. A bunch of meteorologists on the other hand may have other ideas ;)

Pete
No Prob Pete, I do know what you mean, but this section is good like that, it seems to transform seemingly simple questions into the ridiculous blown out of proportion, dissected to death, twisted and manipulated, out of context responses. ( I think I covered them all :D ) so unless you double check your wording and every possible meaning it will get out of hand. But it does result in a good laugh most of the time.

I just wanted to clarify that point about the "similar principles" without getting into other esoteric things like gravitational and magnetic waves, neutrinos etc. :D

PeteF
11-06-2010, 09:28 PM
No, quite right Ed. Besides when I first saw the initial question I had a bet with another member about how many pages this would go to. I've got 10 bucks riding on it going to at least 10 pages, and I was afraid it wouldn't make it! :D I'm now quietly confident my 10er is safe

beanbag
11-06-2010, 09:30 PM
The vectors of the "survivors" are unchanged.

Ya know, what I really dislike about this interpretation is that it paints a picture where the photons that exit a polarizer are "unchanged but lucky". Next I would ask, "ok, what happens if you put another aligned polarizer right behind that?". Now you will have to explain to poor Alistair how this entire batch of photons doesn't have to try its luck again, and goes thru the second polarizer completely unscathed.

BTW, I know (I think...) the classical and QM solution to this situation, but I would like to hear your explanation based on "unchanged" polarizations.

beanbag
11-06-2010, 09:38 PM
The book QED, The Strange Theory of Light and Matter does not demand the ability to work all of the math in order to understand the theory and even use it.

Ohhh, THAT QED.

I thought you meant THIS one:

http://books.google.com/books?id=xt-Vvhloo8YC&printsec=frontcover&dq=quantum+electrodynamics+feynman&hl=en&src=bmrr&ei=6AHWTIToLoq0sAPg3Y2OCw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CC8Q6AEwAA#v=onepage&q&f=false

Ed.
11-06-2010, 09:54 PM
No, quite right Ed. Besides when I first saw the initial question I had a bet with another member about how many pages this would go to. I've got 10 bucks riding on it going to at least 10 pages, and I was afraid it wouldn't make it! :D I'm now quietly confident my 10er is safe

Hey Pete, Don't sweat on it, (I am still chuckling away here), I think you are going to win that $10. If by some strange twist of fate it doesn't get close, you could always rev it up a bit.... neutrinos anyone? :D :D

Evan
11-07-2010, 02:38 AM
Now, one would think a photocopier repairman like yourself would be quite well versed in optics and could provide a simple and concise answer to the question without having to spear off into completely unrelated material and plagiarise an example I gave. Apparently not

Now you are the second one in with the ad hominems.

See ya.

Further, I haven't clue what "example" you refer to. If you mean the 3 polarizer example then I suggest you search the distant past archives for this same example by me years ago.

I saved you the trouble.

http://bbs.homeshopmachinist.net/archive/index.php/t-25721.html

John Stevenson
11-07-2010, 05:40 AM
Hey Pete, Don't sweat on it, (I am still chuckling away here), I think you are going to win that $10. If by some strange twist of fate it doesn't get close, you could always rev it up a bit.... neutrinos anyone? :D :D

No I don't like that artificial sweetener, I have cut down on sugar but still need a one spoon hit per cup.

recoilless
11-07-2010, 08:09 AM
Now you are the second one in with the ad hominems.

See ya.

Further, I haven't clue what "example" you refer to. If you mean the 3 polarizer example then I suggest you search the distant past archives for this same example by me years ago.

I saved you the trouble.

http://bbs.homeshopmachinist.net/archive/index.php/t-25721.html

Just a simple question for you, and others who possibly do it.
When you quote a previous post, you don't leave the writer's name in the quote, why? Only because I like to see who the post is attributed to am I asking. I suppose if it's the guy directly above and its a serve and volley exchange it wouldn't be needed. But, at times, it's a quote from a previous page.

Just wanting to know, and nothing personal.

Thanks.

Evan
11-07-2010, 10:42 AM
I never use the quote button so it isn't automatic. If I think there may be some doubt as to who wrote something I try to remember to attribute it and per your question I shall do so more often. I figure that the person that I am quoting can recognize what they wrote. I tend also to forget that most people are using the default setting of 20 posts per page so it isn't as easy to just scroll back to read the entire post I quoted from.

Your concern is noted and I will attribute more often.

MuellerNick
11-07-2010, 12:20 PM
I'm now quietly confident my 10er is safe

OK, two more pages to go. If you need supprt, PN me because I stop reading after 4 pages, I like that rule. :D

There is a very simple explanation to that -completely unrelated to the initial question- describe phenomenon.

Any yes, you can transmit a beam of darkness. Very simple! Most of the US-households do have that item. Yes, they even store darkness.

Nick

lazlo
11-07-2010, 12:23 PM
Any yes, you can transmit a beam of darkness. Very simple! Most of the US-households do have that item.

You mean a Wife? I have one of those -- she transmits a beam of darkness when I come home from the pub at 2:00 AM :)

John Stevenson
11-07-2010, 12:42 PM
You mean a Wife? I have one of those -- she transmits a beam of darkness when I come home from the pub at 2:00 AM :)

Your wife's name is Lucas ?

MuellerNick
11-07-2010, 12:57 PM
she transmits a beam of darkness when I come home from the pub at 2:00 AM

Ah! A beam of darkness and a beam of silence. After too many Jim Beam. :D


Nick

dockrat
11-07-2010, 02:10 PM
Your wife's name is Lucas ?

Well THAT broke me up!!!!! lol

Pete F
11-07-2010, 02:26 PM
I want to help out the other Pete F (the one without the space in the middle of his name) from the other side of the world win his bet, so I'm posting...

I think I might have to check out Feynman's book. My Quantum Mechanics prof sucked, and did a really poor job of presenting the topic conceptually. He did look and act like Doc Brown from Back to the Future though, which at least made class somewhat entertaining :D.

-Pete

PeteF
11-07-2010, 03:00 PM
I want to help out the other Pete F (the one without the space in the middle of his name) from the other side of the world win his bet, so I'm posting...

I think I might have to check out Feynman's book. My Quantum Mechanics prof sucked, and did a really poor job of presenting the topic conceptually. He did look and act like Doc Brown from Back to the Future though, which at least made class somewhat entertaining .

-Pete

OMG, I feel like Mini-me! :eek:

MCS
11-07-2010, 03:59 PM
I never bet, but some cases deserve support.

beanbag
11-10-2010, 08:06 PM
Sorry Pete, 3 days without a reply, so this thread is officially dead. There wasn't enough animosity to keep it going.

However, I will add that I thought about the notion that "a polarizer does not change the photon polarization", and I am now more sure that it is wrong.

From a QM perspective there are only two basis states for a photon, namely |v> and |h> which mean vertical and horizontal respectively. They are sort of like a coordinate system if u like, written in QM notation. A photon wavefunction (the thing that completely describes a photon) is a combination of these two coordinates, such as a photon polarized at 45 degrees= .71|h> + .71|v> (can also be written as a column vector (.71,.71).

The polarizer in QM is an operator that takes the photon wavefunction at the input and changes it at the output. It is a 2x2 matrix, sort of like the one you would use to do a coordinate rotation. To construct the matrix for the vertical polarizer, I only need two pieces of info:

1. The polarizer lets vertically polarized photon thru
2. It blocks horizontal ones.

Skipping a little math, the operator is:
0 0
0 1

So for incoming photons of ANY polarization, cos(T)|h> + sin(T)|v>, the result if you (left) multiply the operator on the wavefunction is always sin(T)|v> only, which means that all outgoing photons have vertical polarization ONLY. The sin(T) part is only an amplitude, and if it is less than one, means that sometimes you will see the photon get thru, and sometimes it will get absorbed by the polarizer.

In the case of the 45 degree polarizer, the operator is
.5 .5
.5 .5

and ALL outgoing photons have the form constant*(|h> + |v>), i.e. they are ALL polarized at 45 degrees. Note that this is NOT the same as saying that half the photons are vertically polarized, and half are horizontally polarized. IF that were true, then I could put another polarizer at 135 degrees after that, then half again of the photons would go thru. In fact, none will go thru. You can't just take the photon wavefunction and break it up, saying that this part belongs to these photons, and that part belongs to some others.

Statements about probabilities, effects of measurements, and wave function collapse are not really relevant here until I bust out the <?| notation part of the maths. IOW, the state of a photon is well defined even if you don't make any measurements.

This experiment isn't really all that amazing if you just think about it like deflecting the path of a stream of water using two gentle bends instead of one big bend that causes all kinds of turbulence and splashing.

Anyway, I assume the sound of silence is everybody nodding in consent. ;)

Evan
11-10-2010, 08:20 PM
Statements about probabilities, effects of measurements, and wave function collapse are not really relevant here until I bust out the <?| notation part of the maths. IOW, the state of a photon is well defined even if you don't make any measurements.


There are many experiments that actually change the macroscopic result depending on whether the quantum state of the photons are measured or not. It goes much further than that too. The result may change if you are able to measure the state but don't. In other words, just having the capability to determine state makes a difference even if the state isn't observed. Also, it has been demonstrated that introducing a method that gives the capability to measure state after the fact will change the result in the past. There is no propagation delay associated with this and there is no time limit either.

PeteF
11-10-2010, 08:23 PM
[I]
Sorry Pete, 3 days without a reply, so this thread is officially dead. There wasn't enough animosity to keep it going.

I mentioned Evan's previous occupation, presuming it would involve optics that he could explain without involving chaos theory or the precise definition of the Big Bang. Instead apparently it was an insult, so with him off sulking in the corner I shot myself well and truly in the foot, as I missed an easy 5 pages there! :D

Still, we didn't specify a time period to achieve the somewhat lofty goal, so without further ado I'll see your claim of nonsense and raise it one factor of BS.

You think I'm bluffing don't you :cool:

edit: Ah, in the time it takes me to post I've been pipped at the post ... again. I shall therefore clear the floor.

Evan
11-10-2010, 08:41 PM
Instead apparently it was an insult, so with him off sulking in the corner ....

Funny, I assumed that you were ashamed of your statement that I plagerised your example of something.

I still have no idea of what you mean but it obviously wasn't the polarizer experiment since I posted that 2 years before you were a member. Would you care to explain what you did mean?

PeteF
11-10-2010, 08:49 PM
Ah no Evan, it was my one line explanation in plain English, including the example of an optical flat, the latter you later mentioned without credit as if it was in fact your flash of brilliance.

Evan
11-10-2010, 08:54 PM
I don't recall you mentioning an optical flat. But then, it wasn't a flash of brilliance either. It is simply a very well known phenomenon.

I don't even recall you posting before I gave that explanation. I frequently miss posts because they do not show up in the time order that they were made. Sometimes I go back to look for something and discover that one or two posts are now visible that weren't there the last time I replied to something. This isn't a rare occurrence. It happens frequently so I assume it has something to do with caching systems used by my satellite service provider.

PeteF
11-10-2010, 09:04 PM
I don't recall you mentioning an optical flat. But then, it wasn't a flash of brilliance either. It is simply a very well known phenomenon.

I don't even recall you posting before I gave that explanation. I frequently miss posts because they do not show up in the time order that they were made. Sometimes I go back to look for something and discover that one or two posts are now visible that weren't there the last time I replied to something. This isn't a rare occurrence. It happens frequently so I assume it has something to do with caching systems used by my satellite service provider.

Aha :rolleyes: I posted it on page 2 Evan. That's either one hellishly slow feed you've got going there or comprehension isn't something you're claiming to be an expert in. As difficult as the latter is to imagine.

I do agree however it is an excellent example to use on a machinist's BB ... even if I do say so myself! :p

Evan
11-10-2010, 09:08 PM
Since this thread is only 3 pages so far your bet is still in danger. Consider also what may happen if I leave my reply window open for 20 minutes because I was sidetracked by some other event and then post the reply.

My connection at this time of day (evening) is slower than dialup. The browser frequently times out.

PeteF
11-10-2010, 09:16 PM
Since this thread is only 3 pages so far your bet is still in danger. Consider also what may happen if I leave my reply window open for 20 minutes because I was sidetracked by some other event and then post the reply.

Not on the default setting in my browser it isn't. At this post in my browser it's 9 pages and counting so not much more to endure.

Yes I shall give your little problem the appropriate amount of consideration, both now and in future. Since you're clearly not aware, perhaps this is a good opportunity to mention browsers have a refresh feature Evan. Something I personally take full advantage of when I return to my computer, but hey, that's just me.

PeteF
11-10-2010, 09:21 PM
YES!!! 10 pages, like taking candy from a baby really.

Of course it's apparently dependent of browser/settings, but as the sole arbiter of this wager I have the final say and no further correspondence will be entered into. :p

Optics Curmudgeon
11-10-2010, 10:14 PM
Well, I thought it was dead, so was happy to walk away, but...

IOW, the state of a photon is well defined even if you don't make any measurements.
The tremor felt in the Munich area earlier today wasn't an earthquake, but Heisenberg rolling in his grave. I don't know the original attribution, but years ago I saw on the door of an astronomy professor as small note that read: "complex problems always have simple, easy to understand, wrong answers". So I ask: if polarizers rotate polarization why is there such an assortment of optical devices whose sole purpose is to rotate or otherwise modify polarization. Is it an industry conspiracy, a big ripoff? No need to spend big bucks on a Faraday rotator, just use a polarizer!

Joe

beanbag
11-10-2010, 10:24 PM
So I ask: if polarizers rotate polarization why is there such an assortment of optical devices whose sole purpose is to rotate or otherwise modify polarization. Is it an industry conspiracy, a big ripoff? No need to spend big bucks on a Faraday rotator, just use a polarizer!

Joe

Tell me where the QM I posted is wrong.
The devices you mentioned rotate polarization without losing any amplitude.
A polarizer changes polarization but loses amplitude in the process.

Evan
11-10-2010, 10:53 PM
perhaps this is a good opportunity to mention browsers have a refresh feature Evan. Something I personally take full advantage of when I return to my computer, but hey, that's just me.


As you are clearly not aware that doesn't do a bit of good if your service provider is using a transparent proxy cache. For instance, if I post a new thread and there are no replies the view count does not change from 1 until either a reply is posted or until the clock passes the next hour. Apparently they have the cache set to update hourly. There nothing I can do from this end to change that. As soon as they refresh THEIR cache then the view count updates accordingly. Then, if no further posts are made it doesn't update again for another hour. This will also be the same for all other threads but I don't notice it as easily. That will probably also apply to other changes.

The cache is most likely in the satellite itself as it cuts down on the latentcy.

MichaelP
11-10-2010, 11:39 PM
Beanbag, I'm not an expert in physics, of course, but as far as I remember from my middle school years, polarizers are filters: they allow through only correspondingly polarized portion of light and attenuate the rest (bell curve). They don't change polarization. Then, since only a portion of light gets through, its intensity drops after the polarizer.

PeteF
11-10-2010, 11:40 PM
Well that's certainly extraordinarily ironic Evan, because that very "ISP cache problem" that apparently prohibited you from seeing my post some 5 hours before your revelation, doesn't seem to stop you from engaging in tit-for-tat exchanges for which you are infamously known for ... including this very exchange between the two of us! Indeed many a time I have wondered if you ever sleep at all, as no sooner has somebody's post gone up, even within reasonable hours of my own time zone, than I see there's a reply from you!

beanbag
11-11-2010, 12:05 AM
Beanbag, I'm not an expert in physics, of course, but as far as I remember from my middle school years, polarizers are filters: they allow through only correspondingly polarized portion of light and attenuate the rest (bell curve). They don't change polarization. Then, since only a portion of light gets through, its intensity drops after the polarizer.

If that were true, then Evan's 3 polarizer example wouldn't work.

Also, look at post 22.

Evan
11-11-2010, 12:27 AM
Well that's certainly extraordinarily ironic Evan, because that very "ISP cache problem" that apparently prohibited you from seeing my post some 5 hours before your revelation, doesn't seem to stop you from engaging in tit-for-tat exchanges for which you are infamously known for ... including this very exchange between the two of us! Indeed many a time I have wondered if you ever sleep at all, as no sooner has somebody's post gone up, even within reasonable hours of my own time zone, than I see there's a reply from you!


Look up "traffic shaping", especially as to how it applies to satellite internet service. The quality of service and the methods used vary from minute to minute. I have no control over it and if something that I should be seeing doesn't show up then there is no way for me to know that. Traffic shaping is very detailed right down to the level of service provided to each customer depending on their usage patterns both long term and over the last few minutes. It even varies connection speed according to what type of connection is made and the type of content being provided. For instance, peer to peer connections are given lowest priority and are throttled way down most of the time. Windows update service runs at the highest speed. Anything else will vary according to rules that the provider does not disclose.

The longer you continue this conversation the more apparent it becomes that you do not know what you are attempting to discuss.

As for tit for tat exchanges, I think you have assumed that crown. Just look up your recent postings and see how many are contentious posts NOT directed at me.

PeteF
11-11-2010, 12:40 AM
Oh FFS :rolleyes:


Look up "traffic shaping", especially as to how it applies to satellite internet service. The quality of service and the methods used vary from minute to minute. I have no control over it and if something that I should be seeing doesn't show up then there is no way for me to know that.

Righto Evan :rolleyes:

MichaelP
11-11-2010, 01:13 AM
Also, look at post 22.
In the second scenario you'll get less total intensity at the output just because the polarizers are neither ideal nor identical.

However, the second polarizer will drop incoming light intensity much less than the first one. This is just because there will be less for the second polarizer to reject (attenuate).

beanbag
11-11-2010, 01:20 AM
In the second scenario you'll get less total intensity at the output just because the polarizers are neither ideal nor identical.

So what if they are ideal polarizers?

MichaelP
11-11-2010, 01:29 AM
If they're ideal and identical, the second polarizer will cause no change in either light composition or intensity.

beanbag
11-11-2010, 01:42 AM
If they're ideal and identical, the second polarizer will cause no change in either light composition or intensity.

Right. So how does the light go unscathed thru the 3rd polarizer if the 2nd didn't "rotate" it into the correct configuration? (albeit with some intensity loss)
What is the polarization of the light after the first zero degree polarizer?
What is the polarization after the second?

MichaelP
11-11-2010, 02:18 AM
Sorry, I don't follow. Could you describe the setup?

In general, if you use multiple ALIGNED polarizers and they're ideal and identical, only the first polarizer will change light composition and intensity. The rest will play no role at all.

In real life, of course, it will be different.

beanbag
11-11-2010, 02:36 AM
Sorry, I don't follow. Could you describe the setup?

In general, if you use multiple ALIGNED polarizers and they're ideal and identical, only the first polarizer will change light composition and intensity. The rest will play no role at all.

In real life, of course, it will be different.

ok, I'm going to change the setup a little bit :
natural light -> zero degree polarizer -> zero degree polarizer -> 45 degree polarizer -> 45 degree polarizer.

What is the polarization (and intensity) of light after each polarizer?

Would you agree that if the polarization of light matches that of the polarizer, it will go thru unscathed?

Evan
11-11-2010, 04:59 AM
In general, if you use multiple ALIGNED polarizers and they're ideal and identical, only the first polarizer will change light composition and intensity. The rest will play no role at all.


The difference between an ideal polarizer and one that isn't is only in the amount of stray randomly polarized light that is permitted through. The polarizing effect is still exactly the same for the photons that don't bypass the filter effect altogether. Even the chance of being filtered or not is described accurately by the probability function at each polarizer according to it's real world effectiveness as a polarizing filter.

A polarizer is a FILTER. The meaning of the word is very specific in this respect. A filter only allows or doesn't allow the passage of light based on the characteristics of the individual photons. It doesn't change those characteristics, only the overall numbers of photons are changed according to the properties of the filter. In this experiment the filter may be evaluated by considering the probabilities of transmitting a single photon at a time.

This where the independence of the quantum effects from time comes into play. Interference effects actually do not depend on the time of arrival of individual photons. Interference effects and polarization are one and the same since they depend on the orientation of the electromagnetic field of a photon. An interference pattern in the classic two slit experiment will be formed even if the photons arrive on the target one at a time and even if the time difference is measured in years. (note: this experiment has been performed)

At the quantum level probability operates even on individual photons. It is not restricted to the overall behaviour of masses of photons.

beanbag
11-11-2010, 05:45 AM
A polarizer is a FILTER. The meaning of the word is very specific in this respect. A filter only allows or doesn't allow the passage of light based on the characteristics of the individual photons. It doesn't change those characteristics, only the overall numbers of photons are changed according to the properties of the filter.

The only part of the photon that the polarizing "filter" leaves alone is the wavelength.

Evan
11-11-2010, 06:09 AM
Sorry, that isn't the interpretation consistent with the properties either of the filter or the photon. If the filter were a rotator then the probabilities wouldn't be additive and that is NOT consistent with either experiment or theory. If the filter were to change the polarization then the probabilities would be effectively "reset" after passage and that IS consistent with what happens if the filter is replaced with a rotator. You can test this using lenses from RealD glasses. They are 1/4 wave rotator plates, not linear polarizers. I grabbed half a dozen to play with when we went to see Avatar and the difference is both obvious and predictable according to probability.

jugs
11-11-2010, 06:52 AM
Post 83 - Sorry Pete, 3 days without a reply, so this thread is officially dead. There wasn't enough animosity to keep it going.



:eek: Oh yes there is :D

john
:)

beanbag
11-11-2010, 07:08 AM
Sorry, that isn't the interpretation consistent with the properties either of the filter or the photon. If the filter were a rotator then the probabilities wouldn't be additive and that is NOT consistent with either experiment or theory. If the filter were to change the polarization then the probabilities would be effectively "reset" after passage and that IS consistent with what happens if the filter is replaced with a rotator. You can test this using lenses from RealD glasses. They are 1/4 wave rotator plates, not linear polarizers.

I appreciate your efforts to explain, but your language is confusing to me. When you mention "probabilities" and "reset", it sounds like you are taking some measurement of the photon, which will obviously screw up the rest of the experiment. Mathematically, you are inserting yet another operator between the polarizers, and the measurement "forces" the photon into one of the eigenstates of the measurement operator, e.g. number, so the photon is forced to choose whether it is "there or not". I mentioned earlier that one is perfectly able to specify the attributes of the photon using only the |?> notation (including mixed states and less than unity amplitude) and that there is no need to whip out any intermediate measurements.

Also, I don't know why you and OC keep thinking that I am confusing a polarizer with a "rotator" when I have already pointed out their differences.

Finally, I have already posted on the math, or QM, aspects of polarized light and polarizers. It turns out to pretty much exactly match the "Jones calculus (http://en.wikipedia.org/wiki/Jones_calculus)" matrix representation of dealing with polarizers and rotators. Take the Jones vector for vertically polarized light, left multiply by the Jones matrix for 45 degree polarizer, result is Jones vector for 45 degree light times .71 amplitude. I hope you can focus on the math part of it in future responses so there is less chance for confusion or ambiguity.

Ed.
11-11-2010, 07:59 AM
YES!!! 10 pages, like taking candy from a baby really.
:p
Hey PeteF, I hope you enjoy the $10, and you had doubts that it wouldn't make the 10 pages??

MichaelP
11-11-2010, 10:12 AM
ok, I'm going to change the setup a little bit :
natural light -> zero degree polarizer -> zero degree polarizer -> 45 degree polarizer -> 45 degree polarizer.

What is the polarization (and intensity) of light after each polarizer?

Would you agree that if the polarization of light matches that of the polarizer, it will go thru unscathed?

If they're ideal and identical, the chain effect will be the same as if we'd choose the following shorter chain:

Nat. light->zero degree polarizer->45 degree polarizer

The intensity drops after each different degree polarizer. Bell curve of polarization centered around zero degree after the first polarizer(s). Small fraction of it (only part of one peripheral of the bell= the areas where their bell curves intersect)- after the second one(s).

Note: "ideal polarizer" in my interpretation is not the one with a line instead of bell curve, but rather one that doesn't change intensity of the light more than its bell curve dictates. A fully transparent one or true to its bell curve, so to speak.

I'll, certainly, agree with your statement that "if the polarization of light matches that of the polarizer, it will go thru unscathed".

P.S. The only reason I participated in the convertion was to point that polarizers do not CHANGE polarization per se. They just select a portion of light that fits into their bell curve and allow it through. Frequency analogs of such filters would be color (light) filters or radio frequency filters. Red color filter doesn't change the color of the incoming light. It jst allows certain spectrum of it to pass through. Same as with you tuner: it doesn't change whole radio band, it just selects a part of it to allow you to listen to your favorite station. The rest is severely attenuated.

That's really all I wanted to say. :)

I'm not capable of discussing quantum physics or relativism here, so let me skip this portion. I fully accept a probability that we get into the realms of the world where "things exists, but not really". I, certainly, don't want to display the kind of attitude described on the door of the astronomy professor. :) Little knowledge is a danderous thing, so let me get back to my soapbox.

Willy
11-11-2010, 10:53 AM
YES!!! 10 pages, like taking candy from a baby really.

Of course it's apparently dependent of browser/settings, but as the sole arbiter of this wager I have the final say and no further correspondence will be entered into. :p

Sorry Pete, but since it was your post that put this thread "over the top" so to speak, the results are invalid.

After much deliberation and heated discussions with the judges, they have determined that they cannot award you the $10.

Their reasoning is that it was your post, two actually, that put this thread well into the "ten page" category.
They said, and I quote, "bringing the finally tally into and beyond the ten page threshold will be considered as a self fulfilled prophecy therefore the results of this wager are null and void".

It was also suggested by the committee that as an act of salvation that it would only be "the right thing to do" if you were to send me the $10 as an act of self redemption.:D

Alistair Hosie
11-11-2010, 11:50 AM
Wow the first time I posted this many years ago it was popular then, and it appears to be drawing lots of intereresting replies even now.I am watching and trying to learn something .:DAlistair

beanbag
11-11-2010, 04:32 PM
If they're ideal and identical, the chain effect will be the same as if we'd choose the following shorter chain:

Nat. light->zero degree polarizer->45 degree polarizer

I'll, certainly, agree with your statement that "if the polarization of light matches that of the polarizer, it will go thru unscathed".

P.S. The only reason I participated in the convertion was to point that polarizers do not CHANGE polarization per se.

Don't worry, there is no need to discuss quantum, as I said a long time ago in this thread, but someone managed to bring it up.

Since you admitted that "if the polarization of light matches that of the polarizer, it will go thru unscathed", then you will have to admit that the light that is between and after the two 0 degree polarizers is 0 degrees, and the light between and after the two 45 degree polarizers is 45 degrees. So what did the first 45 degree polarizer do to turn the 0 degree light into 45 degree light? :D

MichaelP
11-11-2010, 06:22 PM
It DID NOT TURN anything. It simply filtered out everything beyond its bell curve and allowed the rest to get through.

If I pick some cleaver people (IQ from 120 up) out of a group of 300, I'll get, let's, say 23. If you further need to select the elite with IQ above 140, you'll get 4 out of those 23.

Does it mean we converted dumb into smart? No, we just rejected those who didn't fit our requirements.

beanbag
11-11-2010, 06:31 PM
It DID NOT TURN anything. It simply filtered out everything beyond its bell curve and allowed the rest to get through.

If I pick some cleaver people (IQ from 120 up) out of a group of 300, I'll get, let's, say 23. If you further need to select the elite with IQ above 140, you'll get 4 out of those 23.

Does it mean we converted dumb into smart? No, we just rejected those who didn't fit our requirements.

OK, go back to Evan's original example, then. How is it that adding MORE filters now lets MORE light get thru the setup?

MichaelP
11-11-2010, 06:39 PM
Do you mean the three polarizers example? I really need to think about it further. It's somewhat similar to the famous "if you walk half the distance every day, you'll never reach the destination".

PeteF
11-11-2010, 06:53 PM
Ed and Willy, sorry but I'm laying on my banana lounge in Bermuda, getting fat on the spoils of victory. No no no, my posts weren't blatant bribes to influence the outcome, I've listed them as "political donations", and indeed claimed them as a tax deduction in the best traditions of western politics. Prost! ... burp!

Let's face it though, it's all too easy when one of the lead characters in the charade continues to poke a stick at a long-forgotten point and is willing to go on record, and take numerous posts to invent the scenario that, no it's not that he missed reading a person's post (easily done, so why not just admit that), it was his internet provider's fault. Meanwhile, while the National Standards Lab is developing a new standard, reported to be called the Evanametre, as the defacto measurement of internet latency, and is the time taken (in milliseconds) for a post to appear on this BB before a contrary, self-fulfilling, argumentative, or (the preferred) particularly condescending reply is bestowed upon us mere mortals by residents of Williams Lake. While internet latency typically hovers around 5 Evanometres, in a quite remarkable coincidence, in this particular thread it blew out to 5 hours, so of course it was an original example, how dare I suggest otherwise! Unfortunately that still leaves the minor inconvenience of having to explain how, amongst all this remarkable and highly convenient "internet caching", there was a quote inserted by the author of said fiction from an intervening post. I'm almost willing to go double or nothing, that the great oracle Himself will dismiss that as another "well known" phenomena, despite it having about the same probability as sending a manned mission to the centre of the sun and there finding a giant Snow White and her seven dwarves camped around singing Kumbaya while toasting marshmallows!!

beanbag
11-11-2010, 06:55 PM
Sorry, I didn't follow Evan's examples. What was it in plain language?

Look at post #7

natural light -> 0 deg polarizer -> 90 deg polarizer

vs.

natural light -> 0 deg polarizer -> 45 deg polarizer -> 90 deg polarizer

Evan: "So, with just two lenses at 90 degrees there is a zero probability that light may pass through both. However, placing a third lens between those two alters the probability to allow 1/2 x 1/2 x 1/2 which equals 1/8 of the light will make it through the three filters but none passes just two."

Beanbag: My explanation would be that the middle 45 deg polarizer "turned" the polarization of light

John Stevenson
11-11-2010, 06:58 PM
Surely marshmellows will melt at the centre of the sun before they toast ?

Other than that i follow you 100% :rolleyes:

MichaelP
11-12-2010, 03:32 PM
Look at post #7

natural light -> 0 deg polarizer -> 90 deg polarizer

vs.

natural light -> 0 deg polarizer -> 45 deg polarizer -> 90 deg polarizer

Evan: "So, with just two lenses at 90 degrees there is a zero probability that light may pass through both. However, placing a third lens between those two alters the probability to allow 1/2 x 1/2 x 1/2 which equals 1/8 of the light will make it through the three filters but none passes just two."

Beanbag: My explanation would be that the middle 45 deg polarizer "turned" the polarization of light
Well. Evan is correct on what will happen (but not the math since the first filter will, with all probability, cut much more than 1/2 of the ambient light ).

The light reappearans will be brought by the intermediate 45-degree polarizer due to a different vectors decomposition. (Sorry for my language. I'm not in a math or physics field).

Something similar to the vectors of forces that affect a body resting on a tilted plane.