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nelufar
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I would like to know how can parity be physically interpreted ?
nelufar said:I would like to know how can parity be physically interpreted ?
I was recently trying to develop a chrial projection operator for photons, but I could not do it for the reason to which you allude. I was basically trying to use something like [itex]F\pm{}i\epsilon{}F[/itex] and extract from it a projector on [itex]A[/itex]. (Never mind the details.) I'm not very strong in group (representation) theory, so I wasn't sure how to interpret the results of my calculations, but apparently photons necessarily carry both chiralities in equal proportion, and there's no way to avoid this (and this in turn is related to the fact that the chirality violating part is a total divergence). I would very much like to hear anything that you can tell me about how a (say SU(2)) vector boson can violate parity. I think that it is plausible what you say, that the nonabelian nature of the group can allow for the violation of chirality (because the divergence issue relates to the gauge invariance).fermi said:To violate parity, one either needs fermions, or one needs to couple the electromagnetic tensor to its dual. The latter is a pure divergence we usually throw away, but in some cases it can have physical effects, especially for non-abelian theories.
Not produced in Co-60 decays, as far as I'm aware.clem said:Turin, what is weird?. Aren't there many more right handed than left handed people?
humanino said:How can it be interpreted ? Very well thank you
Take the image in a mirror, or reflection through a point in 3D : they differ only by a reflection through a straight line, which is also a rotation by pi around the line, but space is taken isotropic.
A scalar is just a scalar, so nothing happens to it upon reflection. But I think that a pseudoscalar is really a triple-cross-product, so what happens to the vectors affects what happens to the pseudoscalar.arivero said:How does the mirror work with scalar versus pseudoscalar?
Hey, arivero I know you know mirror symmetry. How is the algebraic concept of parity better at providing a geometrical picture for a pseudoscalar ?arivero said:How does the mirror work with scalar versus pseudoscalar?
Almost. I was actually thinking something like AxBxC for a pseudoscalar. I think that your version is correct.humanino said:... an example of pseudoscalar is a mixed product, which is the scalar product of a vector with the vector product of two vectors, and is also what turin wanted to refer to I believe.
arivero said:Ah, I think I see the problem: In a odd number of dimensions, parity transformation of a vector amounts to a change of sign of the vector, but this is not true in an even number of dimensions. Consider for instance the 2 dimensional vector (1,7). The vector (-1,-7) is a rotation, not a parity transformation. The vector (1,-7) is a parity transformation.
So while for odd dimensions your argument proves that the pseudoscalar changes sign, it is not straightforward to build the same argument for even dimensions.
Actually, are we sure that for even dimension the pseudoscalar changes sign?
Maybe I should have insisted more on the fact that the definition of parity is taking the image in a mirror, or just forget all together the mention of point inversion. Taking the image in a mirror means reflection through an hyperplane, which is indeed inverting only one coordinate, no matter the number of dimensions. It inverts the coordinate perpendicular to the mirror and leaves the coordinate along the mirror unchanged.humanino said:Take the image in a mirror, or reflection through a point in 3D
Yep, but there was more people in the threadhumanino said:I was specific when I defined parity .
humanino said:I'm pretty sure a pseudoscalar does change sign whichever dimension. As long as one can define an orientation for base vectors, reversing one vector will switch the sign of the elementary volume
I do not think I understood that. The mixed product is an example of 3-form. Can you elaborate ?arivero said:Can we look at parity as a grading in p-forms? I mean, in 3D, 0-forms are scalars, 1-forms are vector, 2-form are pseudovector, 3-forms are pseudoscalar, and only the 1- and 3- objects happen to change sign.
If it is so, then parity should no change the sign of the elementary volume in an even dimension. So I was in doubt.
humanino said:I have the vague feeling that there is a confusion between helicity and parity here. If parity were not Lorentz invariant, we would not label particle states with parity. Usually when we talk about Lorentz transformations, we include only those Lorentz transformations which are connected with the identity, which form the restricted Lorentz transformations, orthochronous and proper. We separate the four discrete connected components of the full Lorentz group by introducing additional parity and time inversions. Parity leading to a different, unconnected component, of the full Lorentz group, is a topological quantum number. It is quite possible that I do not understand what Hans is saying, but the fact that helicity and parity coincide only for massless particles hints to me that he is referring to helicity, and I agree that helicity is not Lorentz invariant for massive particles.
If I restrict my Lorentz transformations to spatial rotations and boosts, then all transformations have determinant one and the same signature. They cannot change parity, by construction.
I would like to understand this, and I would appreciate if Hans can elaborate. I am only saying on the top of my head without any explicit representation, so I may be confused.
That is the part that confuses me. While I accept that we only keep the continuous transformations as part of relativity, I never got a good physical feeling for why. So I just have to accept it by rote which is unfortunate.humanino said:... which form the restricted Lorentz transformations, orthochronous and proper. We separate the four discrete connected components of the full Lorentz group by introducing additional parity and time inversions.
It is not very satisfying that to the question "how to change the sign of the determinant by a Lorentz transformation continuously deformable to the identity", I get the answerHans de Vries said:Parity inversion, like time reversal, is not a Lorentz invariant operation since parity inversion in one reference frame involves a partial time-reversial in any other reference frame (with the exception of light like objects such as the chiral components)
This is not a fair answer. I am still not convinced that a confusion between helicity and parity is not the reason I do not understand. If I am wrong, I would appreciate if you could show me more details.Hans de Vries said:it's just to obvious to even mention for many readers here.
When you talk about spatial and time coordinates, that will work for momentum, but for instance that will not work for the components of spinor. We can talk about scalars, vectors, tensors and their pseudo-friends entirely in terms of spinors, since we can construct from spinors quantities that transform as scalars, vectors, tensors or their pseudofriends. For a spinor parity will be defined using only spinor components. The reason helicity is not conserved in a Lorentz transformation is because momentum is involved in the definition of helicity. Instead of convincing me that parity is not conserved by Lorentz transformation, your answer only confirms to me that you are confusing helicity and parity.I'm just referring to that you can not manipulate the spatial coordinates
or the time coordinate alone in isolation with also manipulating the other
I'll try to write something now, although I can not spend much time into details yet.JustinLevy said:That is the part that confuses me. While I accept that we only keep the continuous transformations as part of relativity, I never got a good physical feeling for why.
humanino said:Hans, I am trying to work my head around equations because I still can not understand thisIt is not very satisfying that to the question "how to change the sign of the determinant by a Lorentz transformation continuously deformable to the identity", I get the answer
This is not a fair answer. I am still not convinced that a confusion between helicity and parity is not the reason I do not understand. If I am wrong, I would appreciate if you could show me more details.
When you talk about spatial and time coordinates, that will work for momentum, but for instance that will not work for the components of spinor. We can talk about scalars, vectors, tensors and their pseudo-friends entirely in terms of spinors, since we can construct from spinors quantities that transform as scalars, vectors, tensors or their pseudofriends. For a spinor parity will be defined using only spinor components. The reason helicity is not conserved in a Lorentz transformation is because momentum is involved in the definition of helicity. Instead of convincing me that parity is not conserved by Lorentz transformation, your answer only confirms to me that you are confusing helicity and parity.
Thank you, I appreciate the clarification. I think you write things rigorously, as in your book, so I usually read your posts with care and benefit from them. As a result, I thought I misunderstood badly parity and I spent quite some time today going back to the basics (which is never a bad thing).Hans de Vries said:...
Yes it is a 3-form and it changes sign, it is odd under parity.humanino said:I do not think I understood that. The mixed product is an example of 3-form. Can you elaborate ?
But that seems like answering: "relativity is only concerned with continuous symmetries because Lie groups only handle continuous symmetries"humanino said:I'll try to write something now, although I can not spend much time into details yet.That is the part that confuses me. While I accept that we only keep the continuous transformations as part of relativity, I never got a good physical feeling for why.
The tools we use to study Lie algebra representations do not know about the "large" topological structures of the corresponding Lie group. In Lie algebra theory, to obtain a finite transformation one exponentiate the generators times the parameters of the transformation. Another way to say, when the (parameters of the) transformation (are) is infinitesimal, the generators appear just as linear terms exp[i x G ] = 1 + i x G + ... where x is the parameter (angle) and G the generator. We study representations of Lie groups using the Lie algebra, the commutation relations between the generators [Gi,Gj]. For a complicated non-trivial topology, one will obtain representations which can be decomposed into simple representations.
The short answer is essentially : the tools from Lie algebra give us only the connected part to the identity (because det [ exp^{ i * x * G } ] =1) so when we have parts which are not continuously connected with identity (such as when we have time reversal or parity), we obtain additional discrete topological quantum numbers which are dealt with individually.
At the school kid level, only to appeal to experiment. This is, we truly believed, for some years, that the equations were invariant under parity. And then we found P violation. And we believed that they were still invariant under T and CPT, and then we found CP violation (so if CPT is preserved, T is not).JustinLevy said:But that seems like answering: "relativity is only concerned with continuous symmetries because Lie groups only handle continuous symmetries"
...
Relativity is explained to school kids as:
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But that implies parity should be conserved according to relativity, since if I apply a parity transformation to an inertial coordinate system, you get another inertial coordinate system.
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So that is not correct.
My question is essentially,
Is there a better way to present relativity to school kids that is technically correct, yet still gives an easy physical understanding / physical intuition?
I probably have misunderstood the question then. My answer was to "why we treat the continuous and discrete parts separately". It was purely technical and quite independent of physics, and it was not about the physics of a breaking in Nature of the discrete part.JustinLevy said:But that implies parity should be conserved according to relativity, since if I apply a parity transformation to an inertial coordinate system, you get another inertial coordinate system.