# Physical interpretation of Parity?

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.
I'll try to write something now, although I can not spend much time into details yet.

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.

In case somebody has Weinberg's volume 1 QFT they may glance at eq (2.6.16) which is exactly what I am saying : the parity of a one-particle state remains the same after a combination of boosts and rotations.

Hans de Vries
Gold Member
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.
Humanino, The confusion is caused by me because I used the wrong wording.
So, a parity reversal of a reference frame will move events from the future to
the past and visa versa in another reference frame but it will never reverse the
arrow of time. Similar, a time reversal in one reference frame will change spatial
coordinates in another reference frame but it will not change the parity. This is
what I meant to acknowledge in my previous post.

Regards, Hans

...
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).

arivero
Gold Member
I do not think I understood that. The mixed product is an example of 3-form. Can you elaborate ?
Yes it is a 3-form and it changes sign, it is odd under parity.

I mean, it seems that 0 forms and 2 forms are even under parity, while 1 forms and 3 forms are odd under parity. If it is so, the we should not expect a generic volume form (a n-dimensional pseudscalar) to be invariant under parity.

Hmm, about the mistake "helicity vs parity"... the thing that we built by multiplying all the gamma matrices in a given dimension, is it helicity, chirality or parity? I believe to remember that this operator is trivial (or doesn't exist really) in odd euclidean dimension.

Hans de Vries
Gold Member
Thanks Humanino, It's good to hear that people are actually reading my book Regards, Hans

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.
I'll try to write something now, although I can not spend much time into details yet.

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.
But that seems like answering: "relativity is only concerned with continuous symmetries because Lie groups only handle continuous symmetries"

While Lie groups are powerful tools, I don't feel it necessary to demand relativity is defined with these tools. Nor do I consider the fact that Lie groups cannot be used with Parity making the description any more physical or physically intuitive.

Maybe I'm just misunderstanding your answer, or I didn't make my question clear.
Does anyone else care to provide some insight?

Relativity is explained to school kids as:
1) physics looks the same in all inertial frames
2) there is a finite velocity that is the same in all inertial frames

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. 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?

arivero
Gold Member
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:

...

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.

...

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?

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).

Then, looking again to the equations with some rigour, the proofs of invariance were very dependent not only of relativity, but also of the *continuous* properties of the wavefunction, of analicity and holomorphy. So guess that they are not going to work with discrete symmetries. Still, CPT was proved to be a symmetry.

It could be interesting to note the origin of the symmetries. T appears because of relativity: the euclidean version has not a specific T coordinate separated from other rotations and reflections. And P appears in the d-1 dimensional part, after "fixing" or "forgetting" about the T coordinate. And, as only odd-dimensional spaces have a discrete Parity reflection (I guess?) then also P in 4 D is specific of the minkowskian metric, and it disappears in euclidean 4D space.

EDIT: the last parragraph insinuates a way to a different introduction: to use Euclidean space plus "complex time". Then both P and T should appear from the algebra of complex time.

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.
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.

The simplest answer to why the C is broken, well in the massless-neutrino standard model, there is no right handed neutrino to couple to. Of course it explains "how" and not "why".

turin
Homework Helper
I just wanted to comment on the Relativity issue. Relativity does not require everything to be invariant under Lorentz transformations. For example, a (three) vector generally does change under a Lorentz transformation. It is OK for some things to change under the Lorentz transformation (even if physically measureable). It is the physical laws that Relativity requires to be invariant, not arbitrary physical properties such as position, direction, ... and parity.

If you define the Lorentz transformations to form a Lie group, then they preserve parity (something about continuous exponentiation from the Lie algebra that cannot achieve a det=-1). However, if you define the Lorentz transformations as the transformations that leave the metric invariant, then some of these transformations violate parity (because the metric is quadratic, and so parity does not change the metric).

There seems to be some confusion in this thread about what exactly each of us means when we say "parity". By parity, I mean either the transformation of spatial reflection itself, or the eigenvalue of such a transformation (either +/-1). Note, this is distinct from the related concepts of helicity and chirality, which are also distinct from each other. I will explain:

I will refer to parity as "even" or "odd", chirality as "right" or "left", and helicity as "up" or "down". Note that the two labels in each pair are mutually exclusive, but otherwise arbitrarily assigned. For example, there is no fundamental meaning of a right-handed electron vs. a left-handed electron, and the labeling is ultimately inherited from the original idea of instrinsic spin and the right-hand-rule. With that said, I may have my convention backward compared to the standard convention. Anyway, here is the basic idea of how these concepts are related:

even = right + left
odd = right - left
right = even + odd
left = even - odd
up = right with momentum + left against momentum
down = right against momentum + left with momentum

Even states do not change under parity, whereas odd states change sign under parity. Right and left states are swapped under parity. (You can check the above relationships to verify that this is consistent.) Thus parity cannot be allowed in the group of transformations of Relativity, because it changes a physical law: for example, the chirality of the beta particle produced in a beta decay.

arivero
Gold Member
I noticed that most of my contributions in this thread are wrong in diverse details, misreadings, misconceptions, etc... I would prefer to delete all of them, it is not possible, so let me alert to the reader to not to believe any of my comments here. I will try to straight some comment on volume forms, parity, chirality, minkowski and euclidean, but it will be elsewhere and "elsewhen".