[math-fun] amazing computer-created black hole videos
https://www.youtube.com/watch?v=Qg6PwRI2uS8 https://www.youtube.com/watch?v=ENd8Sz0AFOk searching youtube for "black hole SXS" finds more. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincaré sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
I don't know enough about optics to know whether this is correct or not, but I can believe that a negative solution might be of the form `polarising filters can only apply orientation-preserving transformations to the Poincaré sphere, whereas the antipodal map is orientation-reversing'. Sincerely, Adam P. Goucher
Sent: Monday, June 01, 2015 at 9:10 PM From: "Eugene Salamin via math-fun" <math-fun@mailman.xmission.com> To: math-fun <math-fun@mailman.xmission.com> Subject: [math-fun] A polarization puzzle
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincaré sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
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The time has come to reveal the answer to this little puzzle concerning the optics of polarized light. If nobody objects, I'll post the solution tomorrow. -- Gene
A monochromatic beam of light may be resolved as the sum of two linearly polarised beams in perpendicular planes, their phases displaced by some constant angle p . The device must reverse this displacement, delaying (say) one linear component by angle 2 p . Now consider an individual photon in such an input beam. The corresponding output phase must somehow be smeared between values at separation 2 p , in general impossibly. So sticking my neck out, I conclude that *** Salamin's demon (or genie?) is nonexistent. *** Caveat: my regrettable ignorance concerning elementary physics in general and optics in particular has been only barely perceptibly ameliorated by wrestling with this engaging problem. But even if I have (once again) blown it above, at least I now understand circular polarisation: an ultimately simple business which previously appeared ineffably mysterious. Fred Lunnon On 6/1/15, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincaré sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
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This whole discussion has exposed an area of deep ignorance in my physical knowledge. I find myself now not sure whether polarization is an ensemble property of a whole cohort of photons, or whether it is something one can attribute to a single photon, and how much "state" there is at each level. After about ten minutes I declared myself utterly incapable of thinking about the problem, and if someone has an elementary reference to point me at, that would be greatly appreciated. (Or, if somebody could hazard a math-fun-level explanation of the basic concepts, that would be even better!) On Thu, Jun 4, 2015 at 12:00 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
A monochromatic beam of light may be resolved as the sum of two linearly polarised beams in perpendicular planes, their phases displaced by some constant angle p . The device must reverse this displacement, delaying (say) one linear component by angle 2 p .
Now consider an individual photon in such an input beam. The corresponding output phase must somehow be smeared between values at separation 2 p , in general impossibly. So sticking my neck out, I conclude that
*** Salamin's demon (or genie?) is nonexistent. ***
Caveat: my regrettable ignorance concerning elementary physics in general and optics in particular has been only barely perceptibly ameliorated by wrestling with this engaging problem. But even if I have (once again) blown it above, at least I now understand circular polarisation: an ultimately simple business which previously appeared ineffably mysterious.
Fred Lunnon
On 6/1/15, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincaré sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
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Join the club! I found this link enlightening, but it's surely only a start --- particularly regarding quantum-theoretical aspects (which I haven't even attempted to address yet). https://en.wikipedia.org/wiki/Polarization_%28waves%29 WFL On 6/4/15, Allan Wechsler <acwacw@gmail.com> wrote:
This whole discussion has exposed an area of deep ignorance in my physical knowledge. I find myself now not sure whether polarization is an ensemble property of a whole cohort of photons, or whether it is something one can attribute to a single photon, and how much "state" there is at each level. After about ten minutes I declared myself utterly incapable of thinking about the problem, and if someone has an elementary reference to point me at, that would be greatly appreciated. (Or, if somebody could hazard a math-fun-level explanation of the basic concepts, that would be even better!)
On Thu, Jun 4, 2015 at 12:00 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
A monochromatic beam of light may be resolved as the sum of two linearly polarised beams in perpendicular planes, their phases displaced by some constant angle p . The device must reverse this displacement, delaying (say) one linear component by angle 2 p .
Now consider an individual photon in such an input beam. The corresponding output phase must somehow be smeared between values at separation 2 p , in general impossibly. So sticking my neck out, I conclude that
*** Salamin's demon (or genie?) is nonexistent. ***
Caveat: my regrettable ignorance concerning elementary physics in general and optics in particular has been only barely perceptibly ameliorated by wrestling with this engaging problem. But even if I have (once again) blown it above, at least I now understand circular polarisation: an ultimately simple business which previously appeared ineffably mysterious.
Fred Lunnon
On 6/1/15, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincaré sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
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While we're at it, Feynman figured out something about polarization of light and complex numbers that he seemed to view as the best illustration of why complex numbers are "real". An attempt at a lay explanation appears in Jim Ottaviani and Leland Myrick's 2011 graphic biography "Feynman". I couldn't make sense of it. Jim Propp On Thu, Jun 4, 2015 at 10:48 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Join the club! I found this link enlightening, but it's surely only a start --- particularly regarding quantum-theoretical aspects (which I haven't even attempted to address yet).
https://en.wikipedia.org/wiki/Polarization_%28waves%29
WFL
On 6/4/15, Allan Wechsler <acwacw@gmail.com> wrote:
This whole discussion has exposed an area of deep ignorance in my physical knowledge. I find myself now not sure whether polarization is an ensemble property of a whole cohort of photons, or whether it is something one can attribute to a single photon, and how much "state" there is at each level. After about ten minutes I declared myself utterly incapable of thinking about the problem, and if someone has an elementary reference to point me at, that would be greatly appreciated. (Or, if somebody could hazard a math-fun-level explanation of the basic concepts, that would be even better!)
On Thu, Jun 4, 2015 at 12:00 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
A monochromatic beam of light may be resolved as the sum of two linearly polarised beams in perpendicular planes, their phases displaced by some constant angle p . The device must reverse this displacement, delaying (say) one linear component by angle 2 p .
Now consider an individual photon in such an input beam. The corresponding output phase must somehow be smeared between values at separation 2 p , in general impossibly. So sticking my neck out, I conclude that
*** Salamin's demon (or genie?) is nonexistent. ***
Caveat: my regrettable ignorance concerning elementary physics in general and optics in particular has been only barely perceptibly ameliorated by wrestling with this engaging problem. But even if I have (once again) blown it above, at least I now understand circular polarisation: an ultimately simple business which previously appeared ineffably mysterious.
Fred Lunnon
On 6/1/15, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincaré sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
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Spin is a property of individual quantum particles. It arises from considering angular momentum in a quantized setting. There are spin-0 particles like pions, spin-1/2 particles like electrons, protons, and neutrons, spin-1 particles like photons, spin-3/2 particles like Delta baryons, and spin-2 particles like gravitons. We can also treat collections of particles like nuclei as having spin. A spin-0 particle has only one polarization. A spin-1/2 particle has two polarizations, +1/2 = "up" and -1/2 = "down". The Stern-Gerlach experiment shoots a beam of electrons through a magnetic field at a detector; you get two bright dots. A spin-1 particle has three polarizations, -1, 0, 1. In general, a spin s particle has 2s+1 polarizations: -s, -s+1, ..., s-1, s. Pauli predicted the neutrino in part because a neutron can decay via the beta process to a proton and an electron---but that meant the total spin for the neutron should be 1, not 1/2 as had already been observed. Pauli added the neutrino so that the spin of a neutron plus a neutrino would equal the spin of a proton plus an electron. In quantum mechanics, we have a Hilbert space with one dimension for each possible outcome. A photon is a spin-1 particle, but outside of a waveguide, we never see the 0 polarization, so the state of an individual photon lives in a 2-dimensional Hilbert space. We can write the polarization state of a free photon as a complex-weighted sum of the left and right circular polarizations: a|L> + b|R>. The condition |a|^2 + |b|^2 = 1 reduces the four degrees of freedom (two per complex coefficient) to three; this is what's known as the Riemann sphere. The linear polarizations are |H> = (|L> + i|R>)/sqrt(2) and |V> = (|L> - i|R>)/sqrt(2). The inner product of the two is denoted <H|V>, where <H| = <L| - i*<R| denotes the conjugate transpose of |H>. The result is (1*1 + (-i)*(-i))/2 = 0, so they're orthogonal, whereas <H|H> = <V|V> = (1*1 + (-i)*i)/2 = 1. On Thu, Jun 4, 2015 at 6:38 AM, Allan Wechsler <acwacw@gmail.com> wrote:
This whole discussion has exposed an area of deep ignorance in my physical knowledge. I find myself now not sure whether polarization is an ensemble property of a whole cohort of photons, or whether it is something one can attribute to a single photon, and how much "state" there is at each level. After about ten minutes I declared myself utterly incapable of thinking about the problem, and if someone has an elementary reference to point me at, that would be greatly appreciated. (Or, if somebody could hazard a math-fun-level explanation of the basic concepts, that would be even better!)
On Thu, Jun 4, 2015 at 12:00 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
A monochromatic beam of light may be resolved as the sum of two linearly polarised beams in perpendicular planes, their phases displaced by some constant angle p . The device must reverse this displacement, delaying (say) one linear component by angle 2 p .
Now consider an individual photon in such an input beam. The corresponding output phase must somehow be smeared between values at separation 2 p , in general impossibly. So sticking my neck out, I conclude that
*** Salamin's demon (or genie?) is nonexistent. ***
Caveat: my regrettable ignorance concerning elementary physics in general and optics in particular has been only barely perceptibly ameliorated by wrestling with this engaging problem. But even if I have (once again) blown it above, at least I now understand circular polarisation: an ultimately simple business which previously appeared ineffably mysterious.
Fred Lunnon
On 6/1/15, Eugene Salamin via math-fun <math-fun@mailman.xmission.com> wrote:
Here is a puzzle concerning the optics of polarized light. Every state of polarization has its opposite. For linear polarization, it's linear but rotated 90 degrees. For circular polarization, it's circular with opposite helicity. For general elliptic polarization, it's elliptic with the ellipse rotated 90 degrees, and the helicity reversed. On the Poincaré sphere, opposite states of polarization are represented by diametrically opposite points. The puzzle is to construct an optical device that reverses the polarization state. For any input, the output is the opposite polarization. Or, prove that it can't be done. -- Gene
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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
participants (7)
-
Adam P. Goucher -
Allan Wechsler -
Eugene Salamin -
Fred Lunnon -
James Propp -
Mike Stay -
Warren D Smith