Is this section of the wikipedia page for gamma matrices wrong?

[atyy]

@stevendaryl: Have you noticed that Sam has corrected the typo in his post #42? He writes that choosing a frame DOES NOT change the geometrical nature of an object. Originally it read that choosing a frame changes the geometrical nature of an object, which was not what he intended.

 

[stevendaryl]

@stevendaryl: Have you noticed that Sam has corrected the typo in his post #42? He writes that choosing a frame DOES NOT change the geometrical nature of an object. Originally it read that choosing a frame changes the geometrical nature of an object, which was not what he intended.

I'm not sure how I read it, originally, but I understand and agree with the corrected claim. My point is that if \psi is a spinor--a geometric object--and \Psi is its representation in some particular frame, F, then \Psi is NOT a geometric object, it's a matrix of numbers (or actually, it's a matrix-valued function of spacetime).

It's analogous to the distinction between energy and rest-mass. Energy is a component of a 4-vector. Rest mass is the value of that component in a particular frame (the frame in which the particle is at rest). Energy changes under a Lorentz-tranformation. Rest mass does not.

 

[PeterDonis]

Rest mass is the value of that component in a particular frame (the frame in which the particle is at rest). Energy changes under a Lorentz-tranformation. Rest mass does not.

Some people would object to this way of phrasing it, because the word "component" tends to imply "frame-dependent", so it seems like you're saying that something frame-dependent is not frame-dependent.

A less contentious way of phrasing it would be: energy is the 0-component of a 4-vector, the 4-momentum of the particle; rest mass is the invariant length of that 4-vector. The invariant length is obviously a frame-independent geometric object. In the frame in which the particle is at rest, it can be shown that the 0 component of the 4-momentum vector is equal to its invariant length.

 

[stevendaryl]

Some people would object to this way of phrasing it, because the word "component" tends to imply "frame-dependent", so it seems like you're saying that something frame-dependent is not frame-dependent.

A less contentious way of phrasing it would be: energy is the 0-component of a 4-vector, the 4-momentum of the particle; rest mass is the invariant length of that 4-vector. The invariant length is obviously a frame-independent geometric object. In the frame in which the particle is at rest, it can be shown that the 0 component of the 4-momentum vector is equal to its invariant length.

Well, yes, that's a way of looking at it, but it doesn't generalize in the way that I want it to. Here's another way of looking at:

Let U be the 4-velocity of a massive particle. Then if P is the momentum vector for the particle, then m = U_\mu P^\mu is a scalar, and it happens to equal P^0 in a frame in which the particle is at rest.

But more generally,

1. Let e_0, e_1, e_2, e_3 be any collection of 4 linearly independent vectors such that the first is timelike and the rest are spacelike.
2. Let V be any 4-vector.
3. Let e^j be a set of corresponding co-vectors satisfying e^j_\mu e_k^\mu = \delta^j_k (Note: j and k refer to which vector, while \mu refers to which component).
4. Define V(j) = e^j_\mu V^\mu

These are 4 scalars, not components of a 4-vector. However, if e_j are chosen to be basis vectors for some coordinate system, then V(j) = V^j holds in that coordinate system.

 

[samalkhaiat]

I'm not an expert, but I think I know enough to understand an argument based on it. Just write down what you think to be a good explanation, and if necesary, I can ask you some additional questions.

I think, one you should raise the following question first. That is, what is it that make Lorentz group admits spinor representation? Spinors are the defining rep. of the simply connected group SL (2 , \mathbb{C} ) not the non-simply connected Lorentz group SO(1,3). However, a fact and an accident come to the rescue. The fact: Lorentz group is connected, so one can relate it to a unique (up to isomorphism) covering group, denoted by Spin (1,3), which is i) simply connected and ii) the homomorphism \rho : Spin(1,3) \rightarrow SO(1,3) is analytic and locally 1-to-1 which means that Lorntz group and its covering group have isomorphic Lie algebras. The rest of the story is of course known to every body (I hope). That is, i) how to show that SL(2, \mathbb{C} ) is a double-covering group of Spin(1,3) (the accident), ii) determine all IRR’s of its algebra, and iii) identified as IRR’s of the Lie algebra of SO(1,3). The job is done and we can do QFT.
This programme can not be implemented in curved spacetime because the relevant group is GL(n , \mathbb{R} ). One might ask why? After all, like Lorentz group, GL(n) is non-compact and non-simply connected, so why we cannot do what we did in the Lorentz case? Cartan proved that GL(n) does not admit spinor representation. Since the proof is long and hard(ish) I refer you to Cartan book so that I save myself form embarrassment. In simple descriptive way, I can say: On top of its natural representation (i.e., scalar, vector, and tensor), Lorentz group (due to the accidental isomorphism Spin ( 1 , 3 ) \sim SL( 2, \mathbb{C} ) ) manages to borrow the spinor representation from its covering group. GL(4) is not as lucky because the representations its universal covering group consist of scalar, vector and tensors only. Therefore, GL(4) does not gain any new representations from its covering group. For this reason, we say that GL(4) does not act in the index-space of spinors.

I think, I answered (using spin complex and bundles) similar question some time ago in here. I will try to locate that thread and post a link to it.

Sam

 

[PeterDonis]

Let U be the 4-velocity of a massive particle. Then if P is the momentum vector for the particle, then m = U_\mu P^\mu is a scalar, and it happens to equal P^0 in a frame in which the particle is at rest.

Yes, but that's because $P = m U$, or perhaps a better way of saying it would be $U = P / |P|$, i.e., $U$ is just $P$ normalized to be a unit vector. And the rest frame is just the frame in which $U$ is the timelike basis vector. In other words, all you're saying is that, for any timelike vector, I can find a coordinate chart in which the 0 component of the vector is its invariant length, and all other components are zero.

But more generally, ...

Yes, this is all just a way of saying that the components of a vector in a particular coordinate chart are just contractions of that vector with the set of basis vectors (or covectors) of the chart. But, again, if the key point is what's invariant and what isn't (i.e., what's a frame-independent geometric object and what isn't), then the term to focus on, IMO, is "contraction", not "component".

 

[stevendaryl]

Yes, this is all just a way of saying that the components of a vector in a particular coordinate chart are just contractions of that vector with the set of basis vectors (or covectors) of the chart. But, again, if the key point is what's invariant and what isn't (i.e., what's a frame-independent geometric object and what isn't), then the term to focus on, IMO, is "contraction", not "component".

Well, sort-of, but it's clear that given any coordinate system and any vector V there exists 4 contractions giving the 4 components of that vector in that coordinate system.

Similarly (and this is how the discussion got started), if \psi is a Dirac spinor, and F is a frame, then we can certainly come up with 4 additional spinors \psi_1, \psi_2, \psi_3, \psi_4 such that the 4 complex numbers \Psi_j defined by \Psi_j = \bar{\psi}_j \psi returns the components of \psi in frame F.

The four numbers \Psi_j are Lorentz scalars, even though they happen (not coincidentally) to be equal to components of a Dirac spinor in one particular coordinate system.

 

[atyy]

This programme can not be implemented in curved spacetime because the relevant group is GL(n , \mathbb{R} ). One might ask why? After all, like Lorentz group, GL(n) is non-compact and non-simply connected, so why we cannot do what we did in the Lorentz case? Cartan proved that GL(n) does not admit spinor representation. Since the proof is long and hard(ish) I refer you to Cartan book so that I save myself form embarrassment. In simple descriptive way, I can say: On top of its natural representation (i.e., scalar, vector, and tensor), Lorentz group (due to the accidental isomorphism Spin ( 1 , 3 ) \sim SL( 2, \mathbb{C} ) ) manages to borrow the spinor representation from its covering group. GL(4) is not as lucky because the representations its universal covering group consist of scalar, vector and tensors only. Therefore, GL(4) does not gain any new representations from its covering group. For this reason, we say that GL(4) does not act in the index-space of spinors.

Would it be correct to say that this is the reason we must use tetrads to incorporate fermions in curved spacetime?

 

[dextercioby]

Hi atyy, yes and usage of the weak form of the equivalence principle. Actually, the discussion is more complex and the geometric view cannot be ignored and it's very interesting to see how geometry mixes with standard harmonic analysis. The full description of spinor(s) (fields) is neatly done in terms of the so-called spinor bundles and frame bundles in the famous chapter 13 of Wald's book and to more extent in other books (Bleecker, Lawson).

 

[samalkhaiat]

I think, I answered (using spin complex and bundles) similar question some time ago in here. I will try to locate that thread and post a link to it.

Post #7 and 2
www.physicsforums.com/showthread.php?t=240240

Very remarkable and surprising, I looked through my posts on PF and couldn't locate that thread. But when I tried Google, I found it

 

[samalkhaiat]

Would it be correct to say that this is the reason we must use tetrads to incorporate fermions in curved spacetime?

Correct. However, the tetrads (veirbein) are necessary objects for any differentiable manifold. They follow from the fact that the spacetime manifold, considered as a topological space, is locally (homeomorphic to \mathbb{R}^{(1, n - 1)}) flat Minkowskian.

Sam

 

[Demystifier]

Samalkhait, thank you for your answer which was quite clear to me. I have a few additional questions.

For this reason, we say that GL(4) does not act in the index-space of spinors.

So does it mean that, under general coordinate transformations, spinors transform as scalars?

And if your answer is yes (which I think would be the correct answer), can we think of Lorentz transformations as nothing but a special case of general coordinate transformation?

And when we interpret Lorentz transformations in that way, then can we say that, in this interpretation at least, spinors transform as scalars under Lorentz transformations?

 

[samalkhaiat]

So does it mean that, under general coordinate transformations, spinors transform as scalars?

And if your answer is yes (which I think would be the correct answer),

In order to stress the fact that spinors do not belong to GL(n), it is better to say that Lorentz spinor is treated as “scalar” in curved space.

can we think of Lorentz transformations as nothing but a special case of general coordinate transformation?

All real matrix groups are subgroups of GL(n, \mathbb{R}) including \mbox{SL} ( n , \mathbb{R}) and \mbox{Spin} (n).

And when we interpret Lorentz transformations in that way, then can we say that, in this interpretation at least, spinors transform as scalars under Lorentz transformations?

You take spinor from flat space and treat it as “scalar” in curved space and then you conclude that spinor is Lorentz scalar?
This is false argument unless you tell people how you define “your” Lorentz scalar and spinor.

The usual definition of Lorentz scalar and Lorentz spionr are the following:

Lorentz scalar is an object with vanishing spin matrix, i.e. under Lorentz transformation, a scalar field transforms by the identity matrix:
\phi_{ i } ( x ) \leftarrow \bar{ \phi }_{ i } ( \bar{ x } ) = \delta^{ j }_{ i } \phi_{ j } ( x ) .
In the representation theory, we say that Lorentz scalar belongs to the representation space V^{ ( 0 , 0 ) }.

Lorentz (bi)spinor is an object with non-vanishing spin matrix \Sigma^{ \mu \nu }, i.e. under Lorentz transformation, Dirac spinor field transforms as
\psi ( x ) \rightarrow \psi^{ ' } ( \bar{ x } ) = \exp ( - \frac{ i }{ 2 } \omega_{ \mu \nu } \Sigma^{ \mu \nu } ) \psi ( x ) .
In the representation theory, we say that (bi)spinor belongs to the representation space V^{ ( 0 , 1/2 ) } \oplus V^{ ( 1/2 , 0 ) } .

Suppose (for the sake of argument) that \psi is Lorentz scalar, then \partial_{ a } \psi is Lorentz vector.

Now, Einstein’s EP tells you that the derivative of scalar is covariant vector in curved space. So, \partial_{ a } \psi \rightarrow \partial_{ \mu } \psi, i.e., there is no need for connection!

Suppose (as you say) that \gamma^{ a } is Lorentz vector. Thus, in curved space, we will have the contra-variant vector \gamma^{ \mu }.

Thus, you would conclude (in 3 seconds) that Dirac equation i \gamma^{ \mu } \partial_{ \mu } \psi = 0, is generally covariant. Of course, we know this is not true.

We also know, It took 30 years to figure out the correct form of Dirac equation in GR.