Dirac Equation Covariance

thehangedman

Is the Dirac Equation generally covariant and if not, what is the accepted version that is?

For general coordinate changes beyond just Lorentz, how do spinous transform?

andrien

Covariance of dirac eqn can be proved,see here from 14 to 22
http://www.mathpages.com/home/kmath654/kmath654.htm
Infinitesimal lorentz transformation as it acts on Dirac-spinors is
S=1-(i/4)Δω^μvσ_μv,General case is much complicated.

Bill_K

Lorentz covariance of the Dirac Equation is easy, but for general covariance you do have to modify the equation slightly, and in particular you need to use a set of basis vectors (a tetrad or vierbein) at each point. See Wikipedia's page on "The Dirac Equation", and in particular the section on that page, "Dirac Equation in Curved Spacetime". (Similar remarks apply whether the spacetime is curved or just the coordinate system.)

dextercioby

Chapter 13 of Wald's book maybe ?

andrien

General theory of spinors is treated in book'the theory of spinors' by cartan.

Demystifier

Under general coordinate transformations, spinors transform as scalars (i.e., its components do not transform at all). The thing which transforms non-trivially (as a vector) is the gamma matrix.

The confusing thing is that Lorentz transformation is a special case of a general coordinate transformation, but we know that spinors do transform under Lorentz transformations (as spinors, i.e., "square roots" of vectors). Isnt't it in contradiction with the claim that spinors transform as scalars under GENERAL coordinate tranformations, which includes Lorentz transformations?

The answer is that the transformation of spinors is not a fact of nature, but is a matter of a mathematical convenience. (After all, spinor fields cannot be observed.) In other words it is a matter of DEFINITION, and there are two different definitions of spinor transformation. In one definition they transform as spinors under Lorentz transformations, while in ANOTHER definition they transform as scalars under general coordinate transformations.

The important fact is that the transformation of PHYSICAL quantities does not depend on the definition. For example, the Dirac current transforms as a vector in BOTH definitions. In one definition it is due to the transformation of spinors, while in another definition it is due to the transformation of the gamma matrix.

For more details see also Sec. 8.3.3.3 of
http://arxiv.org/abs/1205.1992

Bill_K

As pointed out in the Wikipedia article and elsewhere, the description of spinors requires the specification of a local basis (tetrad) at each point in spacetime. They aren't affected by coordinate transformations, but they are affected by the Lorentz transformations of the local basis.

Demystifier

Yes. But in flat spacetime one can choose the local tetrad at each point such that it does NOT depend on the point. For this (and only for this) special choice, the Lorentz transformation of the local basis can equivalently be interpreted as the Lorentz transformation of spacetime coordinates. Such an interpretation corresponds to the usual (e.g., Bjorken Drell) view of spinors in particle physics.

Thus, to understand spinors in curved spacetime, to a certain extent one must first UNLEARN what one learned about spinors in flat spacetime. Otherwise, one may remain confused.

samalkhaiat

Let [itex]G[/itex] be a group of coordinate transformation. Assume that for all [itex]g \in G[/itex], we have the following action on the spin group [itex]SL(2,\mathbb{C})[/itex]:
[tex]
G( \psi_{\alpha}(x)) \equiv \bar{\psi}_{\alpha}(\bar{x}) = S_{\alpha}{}^{\beta}(g) \psi_{\beta}(x)
[/tex]
If we take [itex]g = \Lambda \in SO(1,3)[/itex], then the relation [itex]SO(1,3) \simeq SL(2, \mathbb{C})[/itex] means that [itex]S(\Lambda)[/itex] forms an irreducible non-unitary matrix representation of [itex]SL(2, \mathbb{C})[/itex].

Now if we take [itex]g = \frac{\partial \bar{x}}{\partial x} \in GL(4, \mathbb{R})[/itex], then a theorem by Cartan with very long proof shows that [itex]S^{\beta}{}_{\alpha}(g) = \delta^{\beta}_{\alpha}[/itex]. This means that the group of general coordinate transformations (the manifold mapping group), which is bigger that [itex]GL(4, \mathbb{R})[/itex], acts trivially on spinors.

Sam

haushofer

As I see it, you have curved indices and flat indices. The flat indices indicate objects in the tangent space. The curved indices are the indices where the gct's act on. They can be converted into each other by using the vielbein. In that sense there is no confusion; a spinor doesn't have flat or curved indices, and transforms under LT's in the tangent space per definition in the spinoral representation. Of course, LT's are a subgroup of gct's, but you usually act with the LT's in the tangent space.

So in that sense I don't really see the confusion; it's a matter of distinguishing between flat and curved indices.

haushofer

Because you can choose the vielbein as

[tex]
e_{\mu}{}^A = \delta_{\mu}{}^A
[/tex]

and thus don't need to distinguish between flat and curved indices anymore. This amounts to saying that the tangents space at a point in Minkowski space can be chosen as Minkowski space itself.

haushofer

A modern and nice treatment of this is given in Nakahara's book on geometry and topology in physics. The quick way to say it is that first, gamma matrices [itex]\gamma^A [/itex] have flat indices, and you need to convert them with the vielbein into curved indices


[tex]
\gamma^{\mu} \equiv e^{\mu}{}_A \gamma^A
[/tex]

in order to make the Dirac equation gct-invariant. Second, you need to define covariant derivatives on spinors. As spinors only exist in the tangent space, you need to define a covariant derivative D there, which is accomplished with the spin connection. So you should also replace the partial derivative on psi with the covariant one.

Because spinors don't have curved indices, they transform as scalars under gct's.


This shows that in dealing with spinors in curved spacetime one really needs the vielbein instead of the metric.

Demystifier

I agree. Within the theory of spinors in curved spacetime alone, there is no confusion. But my point is that the confusion may arise if one first learns spinors in flat spacetime (e.g., as in Bjorken Drell or any standard particle-physics QFT textbook), and THEN starts to learn spinors in curved spacetime. For that reason, I would propose to change how spinors are teached in flat spacetime, by introducing tetrads from the very start. Or even better, I would propose to change how special relativity is teached, by insisting from the very start that Lorentz transformation is a special case of general coordinate transformation.

dextercioby

I've never seen a treatment of spinors in flat spacetime (especially for the purpose of QFT) using the methods of modern differential geometry (the theory of fiber bundles). Even Wald brings in the 'heavy artillery' only for curved spacetime, but at the price of keeping spinors as classical objects, which is, of course, nonsensical.

Demystifier

Dextercioby, does it mean that you would agree with my proposal in #13?

dextercioby

Yes, the formalism of tetrads is useful for general spinor theory, but unfortunately unnecessary for the purpose of QFT which uses only the algebraic + functional analytic properties of spinor fields (operator-valued distributions) and spinor fields are derived from the representation theory of the restricted Poincaré group.

Anyways, there are no spinor(ial field)s in a classical theory, either SR or GR, so the geometric theory of spinors is necessary in SUGRA or other theories of quantum gravity, for example.

Demystifier

Such a view of spinors may also be used to demystify the "fact" that spinors change sign when rotated by 2 pi. With such an approach spinors are spacetime scalars, so they actually don't change at all when rotated by any angle. They have non-trivial transformation properties (including the change of sign for a rotation by 2 pi) only with respect to the ABSTRACT INTERNAL "space" carrying a complex 2-dimensional representation of the group SO(1,3).