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## Main Question or Discussion Point

I would like to know how can parity be physically interpreted ?

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I would like to know how can parity be physically interpreted ?

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arivero

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You mean, geometrically?I would like to know how can parity be physically interpreted ?

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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.

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I do understand what you explained, but when we say that the parity is not conserved in weak interaction, how this can be explained?

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Meir Achuz

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I've always had trouble with this myself.

I think I've traced it back to the sloppy way relativity is explained to school kids:

1) physics looks the same in all inertial frames

2) there is a finite velocity that is the same in all inertial frames

Because of this (well, the first one at least), people often make arguments like "but the universe doesn't care where you place the origin", "the universe doesn't care which direction you define as the z axis" ... and so I thought "the universe doesn't care if you change the sign of your z labels". After all, if I apply a parity transformation to an inertial coordinate system, you get another inertial coordinate system. So from the sloppy relativity above, I just implicitly assumed things like parity symmetry and time reversal symmetry ... and since all physics I learned for quite awhile fit with this, it seemed fine.

I realize now that relativity only includes continuous transformations, but because of that upbringing, I still have trouble settling this with my expectations. How can empty space possibly care if I change the signs of all z labels?

I accept it, but it doesn't sit well with my intuition, so any "intuition correcting" new viewpoints on how to look at all this would be appreciated.

One friend offered me this: Relativity is best thought of as requiring certain local properties of spacetime and theories. And only continuous symmetries make any sense applied "locally". I can kind of see this, but I'm not sure if this is right.

---

P.S. I can't help but think that if things turned out different, and parity and time reversal were exact symmetries, that people would be claiming relativity predicts them.

I think I've traced it back to the sloppy way relativity is explained to school kids:

1) physics looks the same in all inertial frames

2) there is a finite velocity that is the same in all inertial frames

Because of this (well, the first one at least), people often make arguments like "but the universe doesn't care where you place the origin", "the universe doesn't care which direction you define as the z axis" ... and so I thought "the universe doesn't care if you change the sign of your z labels". After all, if I apply a parity transformation to an inertial coordinate system, you get another inertial coordinate system. So from the sloppy relativity above, I just implicitly assumed things like parity symmetry and time reversal symmetry ... and since all physics I learned for quite awhile fit with this, it seemed fine.

I realize now that relativity only includes continuous transformations, but because of that upbringing, I still have trouble settling this with my expectations. How can empty space possibly care if I change the signs of all z labels?

I accept it, but it doesn't sit well with my intuition, so any "intuition correcting" new viewpoints on how to look at all this would be appreciated.

One friend offered me this: Relativity is best thought of as requiring certain local properties of spacetime and theories. And only continuous symmetries make any sense applied "locally". I can kind of see this, but I'm not sure if this is right.

---

P.S. I can't help but think that if things turned out different, and parity and time reversal were exact symmetries, that people would be claiming relativity predicts them.

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turin

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The following thought experiment both explains parity and how it is experimentally violated by the weak part of the electroweak force:

Cosider a current loop in the xy plane. Does this impose a physical distinction between the +z direction and the -z direction? No. Careful: don't confuse your convention for the sign of charge with a physical distinction! For example, you could say that the current is in the clockwise direction because it is composed of electrons travelling in the counterclockwise direction. But you could get the same current with positrons going in the clockwise direction. Then, just change your unphysical sign convention for charge, and you see that the choice of clockwise vs. counterclockwise is physically arbitrary.

Now, place an unstable atom that decays through beta-decay in the middle of the loop of charge. Does this impose a physical distinction between the +z direction and the -z diretcion? YES! Why? Because the weak force doesn't couple to righthanded-handed neutrinos. Therefore, the beta particles will prefer one z direction over the other. The more deeply I think about this, the weirder it seems. It is a very profound statement about the universe!

Cosider a current loop in the xy plane. Does this impose a physical distinction between the +z direction and the -z direction? No. Careful: don't confuse your convention for the sign of charge with a physical distinction! For example, you could say that the current is in the clockwise direction because it is composed of electrons travelling in the counterclockwise direction. But you could get the same current with positrons going in the clockwise direction. Then, just change your unphysical sign convention for charge, and you see that the choice of clockwise vs. counterclockwise is physically arbitrary.

Now, place an unstable atom that decays through beta-decay in the middle of the loop of charge. Does this impose a physical distinction between the +z direction and the -z diretcion? YES! Why? Because the weak force doesn't couple to righthanded-handed neutrinos. Therefore, the beta particles will prefer one z direction over the other. The more deeply I think about this, the weirder it seems. It is a very profound statement about the universe!

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Unlike the fundamental bosons (such as Higgs, photon, or gluons), massive fermions transform under a reducible representation of the Lorentz group L=SO(3,1). This representation is (1/2,0) + (0,1/2). Despite being reducible, this is a faithful representation of L, because L is simple and non-compact. Something even more amazing to me is the fact that the massless fermions need not transform like that: they can transform by only (1/2, 0) or by (0, 1/2) (either of which is no longer a faithful representation of L.) This is the result of decoupling in the Dirac equation. Both of these representations actually refer to the chirality of fermions and not to the parity, but the two are closely related. As was mentioned in one of the answers earlier, if you have fermions of one chirality without the other, you will violate parity automatically. But it is quite possible to violate parity even if all fermions come in both chiralities. This is because the fermion vector current (which couples to vector gauge bosons) can be formed from one chirality alone: [itex]\psi^{\bar}[/itex][itex]__{L}[/itex][itex]\gamma_{\mu}\psi__{L}[/itex]. Here L is for the left chirality, but the right chirality works just as well. This is also true for the axial vector current. [In contrast the scalar, pseudo-scalar, and tensor currents require both chiralities to be present.] Therefore, even if both chiralities exist, some vector bosons may only couple to the left handed fermions, and not couple to their right handed partners. This amazing fact just falls out of the representations of the Lorentz group. Indeed, this is what happens in the Standard Model, or in its extensions.

So the parity violation is made possible by the representation of the Lorentz group the fermions transform under, and by the vector currents you can build out of them. Our daily experience with the space-time does not prepare us to what kind of odd things fermions can do. Parity violation was very counter-intuitive (to me) until I learned about the fermions and their representations under L. Indeed it all comes from the Lorentz group if you have fermions.

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Meir Achuz

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Turin, what is weird?. Aren't there many more right handed than left handed people?

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turin

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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).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.Turin, what is weird?. Aren't there many more right handed than left handed people?

One of my professors once mused aloud to me the physical origin of the handedness of the DNA helix. (We were messing around with chiral crystals.) Apparently, there are various molecules that are vital to life with one chirality, but the identical chemical composition can be poissonous if the chirality is reversed. The biological explanation is straightforward: one chirality is compatible with DNA (a chiral molecule) whereas the other is not. So then, one wonders why DNA should be chiral. Why don't we have a near-equal mixture of both chiralities of DNA in our bodies to maximize the use of both chiralities of environmental molecules? Well, it turns out that amino acids are produced naturally in one particular chirality. To explain that, some hypothesize the existence of chiral radiation in the earlier stages of the universe that preferentially destroyed one chirality of amino acid, suggesting that the radiation itself is chiral. That is what I think is weird; a fundamental violation of chirality programmed into the universe itself. And it still occurs today (in beta decays, among other "fundamental" phenomena).

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arivero

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arivero

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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.

How does the mirror work with scalar versus pseudoscalar?

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turin

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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.How does the mirror work with scalar versus pseudoscalar?

EDIT:

Actually, no, that's not quite right.

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Hey, arivero I know you know mirror symmetry. How is the algebraic concept of parity better at providing a geometrical picture for a pseudoscalar ?How does the mirror work with scalar versus pseudoscalar?

Anyway, 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. An easy geometrical picture of a mixed product is as an oriented volume, for instance the right or left volume of a triad of base vectors, represented as a determinant. The mirror image of my right hand is my left hand. So the mixed product of my thumb, index and middle fingers, which is the oriented volume subtended by those three fingers, changes sign in the mirror.

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turin

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Almost. I was actually thinking something like... 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.

Anyway, I think that we both agree about the basic idea: a pseudoscalar, φ, is actually a trilinear function of three vectors, {

What concerns me (re the edit in my previous post) is that this only works for an odd number of dimensions (even if you replace the cross product with the wedge product), because the pseudoscalar should change sign under parity in any number of dimensions, right? And the dot product of the cross product idea that you mention is further restricted to exactly three dimensions, right? I guess that's OK since I think that the OP is looking for an intuitive geometrical understanding, but I seem to be missing something more fundamental about parity and pseudoscalars. For example, I don't think that this says anything about intrinsic parity. But again, I suppose that it is sufficient for an intuitive geometrical understanding.

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arivero

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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?

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?

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I was specific when I defined parity as a reflection in a mirror, that the inversion through a point works only in 3D (only in odd dimension spaces)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.Take the image in a mirror, or reflection through a point in 3D

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.

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arivero

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Yep, but there was more people in the threadI was specific when I defined parity .

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.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

If it is so, then parity should no change the sign of the elementary volume in an even dimension. So I was in doubt.

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Hans de Vries

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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)

How many textbooks do not naively handle parity inversion and time-

reversal as the extra symmetry operations, after just having handled

the general Lorentz/Poincaré transformation, without mentioning

or even realizing this ???

Regards, Hans

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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.

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I do not think I understood that. The mixed product is an example of 3-form. Can you elaborate ?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.

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Hans de Vries

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I'm just referring to that you can not manipulate the spatial coordinates

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.

or the time coordinate alone in isolation with also manipulating the other

but maybe it's just to obvious to even mention for many readers here.

I did happen to to hit on the work this guy JCYoon again last night who has

this "amazing crusade" of "Lorentz violation of the Standard Model" here:

http://www.jcyoon.com/phpBB/index.php

I couldn't convince him a while a ago that the chiral components as the

two components of the fermion propagator are still light like for a fermion

with mass. Individually they still propagate at c even though the fermion

as a whole doesn't.

I happen to introduce the Dirac equation in my book by showing this for the

1D case in the first few sections:

http://physics-quest.org/Book_Chapter_Dirac.pdf

Regards, Hans

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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.... 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.

That is what I was referring to in https://www.physicsforums.com/showpost.php?p=2599932&postcount=7". Can someone help provide some intuition / a better physical understanding of why we only consider the continuous transformations in relativity?

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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 answerParity 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.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

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