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

[Demystifier]

OK, let us start with eq(53). If $\Psi$ is a SCALAR (as you defined it to be), then it should transforms trivially (by the identity matrix) under the Lorentz group, i.e., Lorentz transformations should not mix the components of $\Psi$:
$$\Psi^{ ' } = ( S^{ - 1 } \psi )^{ ' } = I_{ 4 \times 4 } ( S^{ -1 } \psi ) = S^{ - 1 } \psi . \ \ \ (1)$$
This is clearly inconsistent with (52) unless $S = S^{ -1 } = I_{ 4 \times 4 }$ which we know it is not. Therefore, either eq(52) is wrong (which is not wrong) or eq(53) does not define a scalar.

You say that (52) is not wrong, but actually the catch is that there IS a trouble with (52). Namely, the expression $(S^{ -1 }\psi)'$ suggests that you need to transform not only $\psi$, but also $S^{ -1 }$. The confusing thing here is that $S^{ -1 }$ is itself an (inverse) transformation, so what does it mean to transform the transformation? For that reason, even though (52) seems right, formally it does not have a proper mathematical form.

What does it mean? It means that (52) is valid only in ONE Lorentz frame (the one in which $\Psi=\psi$), which is why your result above is inconsistent with (52). Nevertheless, all other equations in this Appendix are valid in all Lorentz frames (and have the proper mathematical form), which makes the whole theory consistent because after (53) there is no further reference to Eq. (52).

Eq. (52) is just a more elegant (but perhaps also more confusing) way to write the second equation in (28). The only purpose of the confusing equation (52) is to justify Eq. (53). But unlike (52), equation (53) is valid in all frames so should not be confusing.

 

[stevendaryl]

What occurs to me about this business of spinors being scalars is that it's similar to frame fields in GR. A frame field is a set of 4 vector fields, written $\vec{e}_0, \vec{e}_1, \vec{e}_2, \vec{e}_3$ with $\vec{e}_0$ timelike and the others spacelike. For any vector $\vec{A}$, and for any frame field, we can compute a set of 4 numbers:
$A_i = \vec{e}_i \cdot \vec{A}$. These 4 numbers are scalars, not components of a 4-vector, since they don't transform under a coordinate transformation.

The thing that's confusing is that you can certainly collect the 4 numbers into a vector-like object $\vec{\tilde{A}} = (A_0, A_1, A_2, A_3)$. But it's not a vector, since its components are scalars. So what is it?

 

[Demystifier]

What occurs to me about this business of spinors being scalars is that it's similar to frame fields in GR.

Indeed, in GR it is impossible to define a spinor transformation under general coordinate transformations. For that reason, in GR the transformation rules are redefined, such that "spinors" transform as scalars and gamma matrices as components of a vector under general coordinate transformations.

If samalkhaiat was right (which fortunately he is not), then it would be impossible to treat spinors in GR.

A frame field is a set of 4 vector fields, written $\vec{e}_0, \vec{e}_1, \vec{e}_2, \vec{e}_3$ with $\vec{e}_0$ timelike and the others spacelike. For any vector $\vec{A}$, and for any frame field, we can compute a set of 4 numbers:
$A_i = \vec{e}_i \cdot \vec{A}$. These 4 numbers are scalars, not components of a 4-vector, since they don't transform under a coordinate transformation.

The thing that's confusing is that you can certainly collect the 4 numbers into a vector-like object $\vec{\tilde{A}} = (A_0, A_1, A_2, A_3)$. But it's not a vector, since its components are scalars. So what is it?

These are a collection of 4 scalars. Unfortunately, most GR textbooks do not explain these subtleties very well. A good exception is the Wald's General Relativity textbook, where he introduces two kinds of indices thus avoiding the confusion.

 

[stevendaryl]

These are a collection of 4 scalars. Unfortunately, most GR textbooks do not explain these subtleties very well. A good exception is the Wald's General Relativity textbook, where he introduces two kinds of indices thus avoiding the confusion.

Actually, it occurs to me that it's kind of an "internal symmetry" like isospin. The 4-tuple of scalar values can certainly be treated as a vector, but it's a vector that lives in a different space than the tangent vectors of spacetime.

 

[Demystifier]

Actually, it occurs to me that it's kind of an "internal symmetry" like isospin. The 4-tuple of scalar values can certainly be treated as a vector, but it's a vector that lives in a different space than the tangent vectors of spacetime.

Yes, that's also a correct way to think of it. But I have to slightly correct your terminology. The tangent space (at a point on a curved manifold) is a FLAT space. So the internal space is the tangent space, which should be distinguished from the "physical" curved space.

When the physical space is flat (which is the case in special relativity), the things are more confusing because both physical and tangent space are flat, so it is more difficult to distinguish them. Yet, there is a difference because on the physical space you are allowed to use curved (e.g. polar) coordinates, while on the internal tangent space you are only allowed to use flat coordinates because otherwise you cannnot define spinor transformations in the internal space.

 

[atyy]

What occurs to me about this business of spinors being scalars is that it's similar to frame fields in GR. A frame field is a set of 4 vector fields, written $\vec{e}_0, \vec{e}_1, \vec{e}_2, \vec{e}_3$ with $\vec{e}_0$ timelike and the others spacelike. For any vector $\vec{A}$, and for any frame field, we can compute a set of 4 numbers:
$A_i = \vec{e}_i \cdot \vec{A}$. These 4 numbers are scalars, not components of a 4-vector, since they don't transform under a coordinate transformation.

The thing that's confusing is that you can certainly collect the 4 numbers into a vector-like object $\vec{\tilde{A}} = (A_0, A_1, A_2, A_3)$. But it's not a vector, since its components are scalars. So what is it?

In line with Demystifier's comment about the tangent space, the Dirac equation in curved spacetime is written with orthonormal tetrads, eg. Eq 5.1 and 5.2 of http://arxiv.org/abs/1108.3896.

At the risk of making things even more complicated, I would like to know what the Dirac equation looks like in generalized coordinates. The defining relation for the gamma matrices is:

$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu}$

where $\eta^{\mu \nu}$ is the Minkowski metric for a cartesian basis. What I'm wondering is whether you can do the Dirac equation in an arbitrary basis by letting the defining relation be:

$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 g^{\mu \nu}$

where $g^{\mu \nu}$ is the metric in this basis. However, the way that people write the Dirac wave function is:

$\left( \begin{array}\\ \Psi_1(x^\mu)\\ \Psi_2(x^\mu)\\ \Psi_3(x^\mu) \\ \Psi_4(x^\mu)\end{array} \right)$

and there seems to be no connection between the basis $x^\mu$ used to describe the components $\Psi_j(x^\mu)$ and the basis used to describe the gamma matrices. So even if spherical coordinates (for example) are used to describe the components, people continue to use a cartesian basis for the gamma matrices.

That seems kind of inconsistent, because a Lorentz transformation is just a special case of a coordinate transformation, and in that particular special case, people definitely make sure to transform the coordinates and the gamma matrices together.

Bill_K comments on this and why tetrads are used to write the Dirac equation in curved spacetine in post #2 of https://www.physicsforums.com/showthread.php?t=668131.

A related thread about the Dirac equation under general coordinate transformations was https://www.physicsforums.com/showthread.php?t=653985, which has several nice comments like haushofer's post #10.

 

[samalkhaiat]

Well, certainly a column vector of 4 complex numbers can be a Lorentz scalar, in the sense that those 4 numbers are not changed under a Lorentz transformation. So if you have a 4-component spinor $\psi$ that transforms in such-and-such a way under Lorentz transformations, and you have a particular frame $F$, then the 4 quantities $\Psi$ which are the components of $\psi$ in frame $F$ are 4 (Lorentz) scalars.

It's the same sort of thing as in the definition of invariant mass. Energy is not a scalar, it is a component of a 4-vector. But if you fix a frame (for example, the rest frame of the particle), then the energy as measured in that frame is a Lorentz scalar.

May be you need to recall the following fact in geometry.

“Geometrical objects of a given space are classified by the representations of the global symmetry group of that space”.

Poincare’ group is the global symmetry group of Minkowski space-time (prove it if you have time!)
Thus, there exists a one-to-one correspondence between objects in Minkowski space and representations of Poincare’ group. The dictionary is given as follow

1) Trivial Rep.:
$$( \mbox{ I } , a ) \leftrightarrow \mbox{ scalar } : \phi ( x )$$
2) Defining Rep.:
$$( \Lambda , a ) \leftrightarrow \mbox{ vector } : V^{ \mu } ( x )$$
3) Bispinor Rep.:
$$( \sqrt{ \Lambda } , a ) \leftrightarrow \mbox{ Dirac field } : \ \psi_{ i } ( x )$$
4) Spinorvector Rep.:
$$( \sqrt{ \Lambda } \otimes \Lambda , a ) \leftrightarrow \mbox{ R-S field } : \ \psi^{ \mu}_{ i } ( x )$$
5) Rank-n Tensor Rep.:
$$( \Lambda \otimes \Lambda \otimes \cdots \otimes \Lambda , a ) \leftrightarrow \mbox{ Rank-n Tensor field } : \ T^{ \mu_{1} \mu_{2} \cdots \mu_{n} } ( x )$$
And the list continues. The point now is this: choosing a frame DOES NOT change the geometrical nature of an object. This is because the representation theory of Poincare’ group does not care less about your frame. For example: evaluating vector, say, in the “lab-frame” or in the “bathroom-frame” does not turn it into non-vector. It is as simple as the saying: there exists no frame where an electron becomes a meson.

Possibly a clearer way to think about it: Let $\psi_A, \psi_B, \psi_C, \psi_D$ be four different spinors such that in the lab frame, their conjugates have the representations:

$\bar{\psi}_A = (1,\ 0,\ 0,\ 0)$
$\bar{\psi}_B = (0,\ 1,\ 0,\ 0)$
$\bar{\psi}_C = (0,\ 0,\ 1,\ 0)$
$\bar{\psi}_D = (0,\ 0,\ 0,\ 1)$

Then for any $\psi$, the following four numbers are Lorentz-scalars:

$\Psi_1 = \bar{\psi}_A \psi$
$\Psi_2 = \bar{\psi}_A \psi$
$\Psi_3 = \bar{\psi}_A \psi$
$\Psi_4 = \bar{\psi}_A \psi$

If $\psi$ satisfies the Dirac equation, i.e. Dirac spinor, then
$$\Psi_{ 1 } = \psi_{ 1 } , \Psi_{ 2 } = \psi_{ 2 } , \cdots \Psi_{ 4 } = \psi_{ 4 } . \ \ (1)$$
These will be used below.

Then you can put the 4 numbers together into a "scalar bi-spinor":

The above mentioned list does not contain this “scalar bi-spinor”. Therefore, it does not exist: No such object exists in Klein’s geometry and I have never heard of it. We have, however, a scalar-(iso)spinor such as the K-meson doublet $( K^{ + } , K^{ 0 } )$ and, we also have a scalar-(iso)vector like the $\pi$ meson triplet $( \pi^{ + } , \pi^{ - } , \pi^{ 0 } )$. Here, the word “scalar” refers to the behaviour under Poincare’ group which does not mix different members of the doublet or the triplet. And, the words isospinor and isovector refer to the behaviour of the doublet and the triplet under the action of the internal iso-spin group $SU(2)$, i.e., not space-time spinor and vector.

$\left( \begin{array}\\ \Psi_1 \\ \Psi_2 \\ \Psi_3 \\ \Psi_4\end{array} \right)$

This column of scalars is constructed to be equal to the representation of $\psi$ in the lab frame.

Substituting (1) in this column, we find
$$\Psi = \psi$$
Thus, if $\psi$ transforms as a bi-spinor, then $\Psi$ transforms as bi-spinor too.
So, you went through all that business of constructing rows and columns and ended up writing
$$\mbox{ DIRAC } = \mbox{ dirac } .$$
I am impressed. May be, you need to go through the following simple exercise:
You are given the following set of four numbers
$$\psi_{ 1 } = 1, \ \psi_{ 2,3,4 } = 0 .$$
Show that
$$\Psi = \left( \begin{array} { c } 1 \\ 0 \\ 0 \\ 0 \\ \end{array} \right)$$
is not Lorentz scalar. Determine its form in an arbitrary Lorentz fram.
Solving this exercise will help you a lot, so do it.

Sam

 

[samalkhaiat]

You say that (52) is not wrong, but actually the catch is that there IS a trouble with (52). Namely, the expression $(S^{ -1 }\psi)'$ suggests that you need to transform not only $\psi$, but also $S^{ -1 }$. The confusing thing here is that $S^{ -1 }$ is itself an (inverse) transformation, so what does it mean to transform the transformation? For that reason, even though (52) seems right, formally it does not have a proper mathematical form.

What does it mean? It means that (52) is valid only in ONE Lorentz frame (the one in which $\Psi=\psi$), which is why your result above is inconsistent with (52). Nevertheless, all other equations in this Appendix are valid in all Lorentz frames (and have the proper mathematical form), which makes the whole theory consistent because after (53) there is no further reference to Eq. (52).

Eq. (52) is just a more elegant (but perhaps also more confusing) way to write the second equation in (28). The only purpose of the confusing equation (52) is to justify Eq. (53). But unlike (52), equation (53) is valid in all frames so should not be confusing.

If $S \in SO(1,3) , \mbox{ and } \ \psi \in \mathbb{ C } ( 2^{ 2 } )$, then $S^{ - 1 } \in SO(1,3) \ \mbox{ and } S^{ - 1 } \psi \in \mathbb{ C } ( 2^{ 2 } )$. In other words, if $\psi$ is a bi-spinor, then $\Psi = S^{ - 1 } \psi$ is also a bi-spinor.

Sam

 

[samalkhaiat]

Indeed, in GR it is impossible to define a spinor transformation under general coordinate transformations. For that reason, in GR the transformation rules are redefined, such that "spinors" transform as scalars and gamma matrices as components of a vector under general coordinate transformations.

If samalkhaiat was right (which fortunately he is not), then it would be impossible to treat spinors in GR.

I don’t think so my friend, you are totally confused about this. The “transformations” of the gammas in curved space come not from the gammas themselves but from the vielbein fields. Seeking compact notations and convenience, some people choose to redefine the gammas by writing
$$\gamma^{ \mu } ( x ) = e^{ \mu }{}_{ a } ( x ) \ \gamma^{ a },$$
where, $\mu = 0 , 1, .. , 3$ is world index and $a = 0 , 1 ,.. , 3$ is local Lorentz index. You could do every thing without even introducing $\gamma^{ \mu } ( x )$.
The generally covariant action is
$$A = \int d^{ 4 } x \ | e | e^{ \mu }{}_{ a } ( x ) \left( i \bar{ \psi } ( x ) \gamma^{ a } \nabla_{ \mu } \psi ( x ) \right) ,$$
where
$$\nabla_{ \mu } \psi = \partial_{ \mu } \psi + \frac{ 1 }{ 8 } \Omega_{ \mu a b } ( x ) [ \gamma^{ a } , \gamma^{ b } ] \psi ,$$
and $\Omega^{ \mu }_{ a b } ( x )$ is the spin-connection.
You want more, I am able to tell you “almost” any thing you need to know about this subject.


Sam

 

[stevendaryl]

The point now is this: choosing a frame does change the geometrical nature of an object.

Okay, I explained already why you were wrong. If $\psi_A, \psi_B, \psi_C, \psi_D$ are 4 different spinors, and $\psi$ is another spinor, then do you agree that the four numbers below are all Lorentz scalars:

$\Psi_1 = \bar{\psi}_A \psi$
$\Psi_2 = \bar{\psi}_B \psi$
$\Psi_3 = \bar{\psi}_C \psi$
$\Psi_4 = \bar{\psi}_D \psi$

Do you agree with that claim? If not, why not?

[edit]

Note: the four scalars above can be put into a column vector $\Psi$, but this $\Psi$ is NOT a spinor in the sense of representations of the Lorentz group. That's because its four numbers are Lorentz SCALARS. $\Psi$ does NOT get transformed by a Lorentz transformation. So your conclusion, that because $\psi = \Psi$ in the lab frame (meaning the components are equal) then $\psi = \Psi$ in every frame, is wrong. If $\Psi$ were a spinor, that would be true. But it's NOT a spinor---it's a collection of 4 Lorentz scalars.

What you're saying is analogous to saying:

In the rest frame of a particle, the 4-momentum $P$ can be represented by the row vector $V = (mc, 0, 0, 0)$. Since $P=V$ in one frame, they must be equal in all frames. Therefore, in every frame, $P = (mc, 0, 0, 0)$.
​

The reasoning is wrong, because $V$ is not a vector in the sense of representations of the Lorentz transforms. $V$ is 4 numbers that happen to be equal to the four numbers in the vector $P$ in the rest frame of the particle.

 

[atyy]

At the risk of making things even more complicated, I would like to know what the Dirac equation looks like in generalized coordinates. The defining relation for the gamma matrices is:

$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu}$

where $\eta^{\mu \nu}$ is the Minkowski metric for a cartesian basis. What I'm wondering is whether you can do the Dirac equation in an arbitrary basis by letting the defining relation be:

$\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 g^{\mu \nu}$

where $g^{\mu \nu}$ is the metric in this basis. However, the way that people write the Dirac wave function is:

$\left( \begin{array}\\ \Psi_1(x^\mu)\\ \Psi_2(x^\mu)\\ \Psi_3(x^\mu) \\ \Psi_4(x^\mu)\end{array} \right)$

and there seems to be no connection between the basis $x^\mu$ used to describe the components $\Psi_j(x^\mu)$ and the basis used to describe the gamma matrices. So even if spherical coordinates (for example) are used to describe the components, people continue to use a cartesian basis for the gamma matrices.

That seems kind of inconsistent, because a Lorentz transformation is just a special case of a coordinate transformation, and in that particular special case, people definitely make sure to transform the coordinates and the gamma matrices together.

This article puts fermions in curved spacetime with tetrads, and uses tetrads and the "constant gamma matrices" to define "spacetime gamma matrices" which do have the form of your guess $\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 g^{\mu \nu}$. http://particlephd.wordpress.com/2009/02/07/the-spin-connection/

 

[Demystifier]

You want more, I am able to tell you “almost” any thing you need to know about this subject.

Just one thing. How does the spinor transform under general coordinate transformations?

 

[samalkhaiat]

Just one thing. How does the spinor transform under general coordinate transformations?

How much do you know about the representation theory of $GL(n)$ and its Lie algebra?

 

[Demystifier]

How much do you know about the representation theory of $GL(n)$ and its Lie algebra?

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.

 

[stevendaryl]

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.

It seems to me that the details of how spinors transform under various coordinate transformations is orthogonal to Sam's unresolved complaints about the alternative way of doing the Dirac equation. It seems to me that the alternative way is clearly equivalent to the usual way, and there is nothing about the details of transformations that could change that. Those details factor through the argument, and don't affect it in any way.

The confusion is from failing to distinguish between a spinor, which is a geometric object, and a column matrix, which is the representation of that spinor in a particular frame.

Pick two frames, $F$ and $F'$. Let $\psi$ be a spinor described by the Dirac equation. Let $\Psi$ be the representation of $\psi$ in frame $F$ and let $\Psi'$ be the representation in $F'$. Let $\partial_\mu$ mean differentiation with respect to $F$ coordinates, and $\partial'_\mu$ mean differentiation with respect to $F'$ coordinates. Then the usual covariance of the Dirac equation tells us the following two equations hold:

1. $(-i \gamma^\mu \partial_\mu + m) \Psi = 0$
2. $(-i \gamma^\mu \partial'_\mu + m) \Psi' = 0$

Now, letting $\Lambda$ be the transformation matrix relating the two coordinate systems, we can rewrite: $\partial_\mu = \Lambda^\nu_\mu \partial'_\nu$. So equation 1 becomes:

$(-i \gamma^\mu \Lambda^\nu_\mu \partial'_\nu + m) \Psi = 0$

If we define $\Gamma^\nu$ to be $\gamma^\mu \Lambda^\nu_\mu$, then we have:

1. $(-i \Gamma^\nu \partial'_\nu + m) \Psi = 0$
2. $(-i \gamma^\mu \partial'_\mu + m) \Psi' = 0$

In equations (1) and (2), $\Psi$ and $\Psi'$ are matrices, not spinors. In frame $F'$, you can use either equation; they are both valid equations, and they are equivalent, in the sense that the following equations hold:

1. $U(\Lambda) \gamma^\mu U^{-1}(\Lambda) = \Lambda^\mu_\nu \gamma^\mu$
2. $\Psi' = U(\Lambda) \Psi$
3. $\Gamma^\nu = \Lambda^\mu_\nu \gamma^\mu$

(for the appropriate matrix $U(\Lambda)$)

In these equations, you can view (1) as the defining equation for $U(\Lambda)$. Equations (2) and (3) are then just the definitions of what we mean by $\Psi'$ and $\Gamma^\mu$

So the only aspect of this alternative way of doing the Diract equation that is not true by definition is equation (1) for the matrix $U(\Lambda)$. The details of $U$ are important for relating the two approaches, but the details are unimportant for seeing that the two approaches are equivalent.

 

[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.