Dirac equation and general covariance

1. Apr 26, 2008

jdstokes

According to the principle of general covariance, the form of equations should be independent of the coordinates chosen. In general relativity, this is implemented by expressing laws of physics as tensor equations.

In physics equations are often expressed in index notation, which allows Lorentz covariance to be checked by implementing the Lorentz transformation on the components. An example is the Dirac equation $(\mathrm{i}\gamma^\mu \partial_\mu - \bar{m})\psi(x)$.

I prefer to think of tensors as multilinear functions from the tangent space and its dual to the real numbers for two reasons. Firstly it avoids the need to explicitly check for Lorentz covariance and secondly it generalizes to be generally covariant when the spacetime is curved.

How would one express the Dirac equation in coordinate independent fashion so as to make it generally covariant (or even just Lorentz covariant?). The first thing one would have to deal with is the abuse of notation $\gamma^\mu \partial_\mu$. Secondly one would have to think about where the $\psi(x)$ actually lives on the manifold (it's clearly not a smooth section of the tangent bundle like $\partial_\mu$).

Would anyone be able to shed some light on this?

2. Apr 27, 2008

lbrits

In order to deal with spinors in GR, you first realize that the symmetry group $$GL(N,R)$$ has no spinor representations, whereas $$SO(N,R)$$ does. Using the principle of equivalence, we can speak of spinors in local inertial frames, or, equivalently, introduce an orthonormal frame (veilbein) at each point on the tangent space and talk about spinors in that basis. The veilbein $$e_\mu^a$$ has a tangent index $$\mu$$ and a Lorentz index $$a$$ and connects the two spaces. Things with Lorentz indices act like scalars from the tangent point of view, whereas things with tangent indices act like scalars from the Lorentz point of view.

Spinors have Lorentz indices, and so they act like scalars under diffeomorhisms. The relevant derivative is $$\partial_\mu$$. But since Lorentz transformations at different points on the manifold need to be correlated, you need to introduce a connection known as the "spin connection" that is the analog of the Christoffel symbols, but for Lorentz indices.

I've sketched the basic idea here, but I've written an extensive overview of the subject which I invite you to take a look at ( http://www.physics.thetangentbundle.net/wiki/Quantum_field_theory/fermions_in_curved_space [Broken] http://www.mathematics.thetangentbundle.net/wiki/Differential_geometry/spin_connection [Broken]).

Last edited by a moderator: May 3, 2017
3. Apr 27, 2008

jdstokes

Let's forget curved spacetime for a moment and concentrate on Minkowski space. In this context, Dirac spinors are defined as quadruples of complex-valued functions which transform according to the spinorial representation of SO(3,1).

Is it possible to give a coordinate-free definition of Dirac spinors, in much that same way that a vector would be defined as a linear function from the cotangent space to $\mathbb{R}$, or better as a smooth section of the tangent bundle?

You could simplify this question even more by asking how to do the above for ordinary 2-dimensional spinors?

Last edited: Apr 27, 2008
4. Apr 27, 2008

lbrits

The best place to look would be under spin bundles. Firstly, $$\gamma^a\partial_a$$ is coordinate invariant once you start thinking in terms of abstract index notation. Secondly, you could probably concoct some sort of mapping between spinors and derivatives, noting that a vector is the tensor product of two spinors, i.e.,

$$\partial_\mu \rightarrow \frac{\partial}{\partial x^{\alpha \dot\beta} }$$. (I don't know if you're familiar with dotted/undotted notation). The two indices transform under $$SU(2)$$ and the complex conjugate of that representation.

I'm also sure this has been done before. If you are really interested, feel free to hound me some.

5. Apr 27, 2008

lark

There's a handy Wikipedia page http://en.wikipedia.org/wiki/Dirac_equation" [Broken] which makes these things clear.
Laura

Last edited by a moderator: May 3, 2017
6. Apr 28, 2008

schieghoven

I found it easiest to construct the theory in terms of (Weyl) 2-spinors - these have 2 complex components and transform according to SL(2). SL(2) is the set of 2-by-2 complex matrices with unit determinant. There are two distinct ways to do this: if $$R \in SL(2)$$, then a weyl spinor \psi could transform as
$$\psi \rightarrow R \psi$$​
or as
$$\psi \rightarrow (R^\star)^{-1} \psi$$​
.
These are called Left-handed or Right-handed Weyl spinors, respectively. (This is what is meant by the Lorentz group having 2 distinct representations in SL(2)). A Dirac spinor is a pair of one LH Weyl spinor and one RH Weyl spinor.

SL(2) arises because its Lie algebra is isomorphic to that of the Lorentz group SO(1,3) - they both have 6 generators corresponding to a closed subalgebra of 3 infinitesimal rotations, and 3 infinitesimal boosts. This gives a geometric interpretation to spinors: infinitesimal lorentz transformations correspond one-to-one with infinitesimal SL(2) transformations of spinors.

This is possible on curved spacetime as well. In this case we assume a general metric g_ij with Minkowskian signature, and define the symmetry group in terms of this metric. There are 6 generators, which we denote J_ab with antisymmetry in the two indices. The Lie-algebra structure is characterised by the bracket
$$[J_{ab},{J_{ij}] = g_{ai} J_{bj} + g_{bj} J_{ai} - g_{aj} J_{bi} - g_{bi} J_{aj}$$​
and it is in fact possible to choose 2-by-2 matrices from sl(2) that satisfy this bracket... if you like I can post some further details about how to do this. Anyway the short answer is that the symmetry group corresponding to the general metric g_ij is also SL(2), and has LH and RH representations in much the same way as in flat space. In this way it's possible to define spinors at each point of a curved manifold.

So, I think that spinors require more structure on the manifold than vectors... vectors can be defined independently of the metric, but spinors require a metric.

I think this is a fairly complicated topic, so I hope this condensed summary of results doesn't just cloud the water further.... :-)

Dave

7. Apr 28, 2008

jdstokes

Interesting.

What I'm taking in from all of this is that the spinor fields should be regarded as making up a new fiber bundle where the fiber is a 4-dimensional complex vector space instead of the ordinary real vector space used for vectors.

The spinors are postulated to transform under Lorentz transformations according to the SL(2) representation of the Lorentz group, but remain invariant under more general coordinate transformations.

This is all well and good, but in the tangent bundle for ordinary vectors, the transformation rule under Lorentz/general transformations is not postulated but follows as a corollary of the fact that the vectors are smooth derivations of the base space.

It seems to me that there should be a corresponding functional interpretation of spinors which leads directly to their (postulated) transformation law. Is such an interpretation known or has been investigated?

I'm sick and tired of hearing that spinors are defined as sets of numbers transforming according to some transformation law. I want a coordinate-independent, set-theoretic definition.

Last edited: Apr 28, 2008
8. Apr 28, 2008

jdstokes

How should one interpret conjuctions such as $\gamma^\mu \partial_\mu$ and $\partial_\mu \psi$ of objects belonging to different fiber bundles?

9. Apr 28, 2008

schieghoven

I agree, this is the most powerful and lucid definition of tangent vectors on a C^inf manifold. I've also thought about the question of whether there is a comparably elegant definition for spinors, but unfortunately no, I couldn't come up with anything nor locate anything in the literature. Tell me if you find something.

However, this idea of transformation laws is a central part of gauge theory; it's not peculiar to the question of fermions on a curved spacetime. For instance, the Higgs doublet is a field with 2 complex components; it's postulated to transform as scalar w.r.t lorentz or coordinate transformations, and additionally according to a separate U(1)xSU(2) gauge symmetry. As far as I can see, there really isn't any underlying mechanism or property of the manifold that would produce this structure, nor can you characterise the Higgs field in any other way except by how it transforms under a gauge transformation.

Dave

10. Apr 29, 2008

jdstokes

I have not been able to find a definition which satisfies me yet. I think something close might be buried in `An introduction to spinors and geometry with applications in physics' by Benn and Tucker.

They first define a Clifford algebra C(V,g) for an inner product space (V,g) by quotienting the tensor product space T(V) in such a way that the resulting space is finite dimensional and the Clifford product satisfies $\{v,u\} = g(v,u)\; \forall u,v \in V$. This is analagous to the construction of the space of exterior forms, by taking T(V) and modding out elements of the form $a\otimes x\otimes x \otimes b$ where x is a vector in V and a,b in T(V) and then endowing with the exterior product.

The regular representation consisting of endomorphisms of C(V,g) induces an irreducible representation of the Clifford algebra on its minimal left ideals (whatever that means). These are called the spinor representations. The minimal ideal is called the space of spinors.

Given a manifold with metric, one has the exterior bundle. With a Clifford algebra associated to each fiber, this forms a Clifford bundle.

One can develop a Clifford calculus just like exterior calculus. Except this time for some reason we have the possibility that inhomogeneous differential forms (sums of differential forms with different degrees) will arise when we Clifford multiply differential forms together. This is obviously going to be useful when we come to express certain laws of physics such as the Dirac equation.

A spinor bundle is just a smooth assignment of minimal left ideals to the fibers of the Clifford bundle.

Spinor fields are sections of spinor bundle. At each point in the manifold, the Fiber of the clifford bundle at p is isomorphic to an algebra of matrices, where matrices are determined by the choice of cotangent basis vectors as follows

$\hat{e}^a \leadsto \{ \gamma^a,\gamma^b\} = 2g^{ab}$.

I don't have the time to fully decode all of this due to impending coursework so I'll have to leave this for future study.

In any case, what all this shows is that the choice of gamma matrices is dictated by a choice of basis. Thus it makes sense for gamma matrices to transform under a change of basis (ie Lorentz boost). In curved space, there is no finite-dimensional representation of the general group of transformations. So make a choice of basis for each point on the manifold, then introduce a set of vierbiens indexed by each basis vector so that the contraction of the vierbien with the gamma matrix transforms smoothly between points.

Last edited: Apr 29, 2008
11. Apr 29, 2008

George Jones

Staff Emeritus
Benn and Tucker is an interesting book.

A somewhat more mainstream presentation is given in Chapter 19 (Dirac Equation) in The Geometry of Physics: An Introduction by Theodore Frankel.

12. Apr 30, 2008

schieghoven

I don't see what the advantage of Clifford algebras are. It's still just representation theory... It seems much more straightforward to construct the irreducible representations of the Lorentz group, and thereby go directly to spinors. In addition, I can't think of any application where inhomogeneous sums of differential forms would be useful, yet these form a central part of the theory of Clifford algebras.

Dave

13. May 1, 2008

jdstokes

I believe the Dirac equation is an example of an equation in physics which is essentially an inhomogeneous sum of differential forms: $\gamma^\mu \partial_\mu \psi$ has different degree to $\bar{m}\psi$, yet they appear as a sum.

14. May 1, 2008

schieghoven

No, neither of those terms can be represented by a differential form or any sort of sum of differential forms. This is evident from the fact that a differential form of any order is invariant under a rotation about any axis by 2\pi. Both $\gamma^\mu \partial_\mu \psi$ and $\bar{m}\psi$ change sign under such a rotation. The two quantities transform the same way under infinitesimal lorentz transformation, and that is the key consistency test for adding them together.

Spinors take some getting used to, particularly after the dogmatically geometric construction of general relativity. I've gradually come to think of spinors as a generalisation of the complex scalar field. A complex scalar field (such as a Klein-Gordon field) $\phi = |\phi| e^{i\alpha}$ also has quite some ambiguity in its definition: you can redefine it by a complex phase which varies arbitrarily in spacetime. (...equivalent to introducing an electromagnetic potential. This is the electromagnetic "U(1) gauge covariance"). So it's really impossible to give physical meaning to the phase of a complex field, since it can be freely redefined without changing the physical consequences of the theory. In the same way, the complex numbers which comprise a spinor don't necessarily have a direct geometrical meaning, since they can also be redefined locally, according to their symmetry group, SL(2). (...equivalent to introducing a 'spin connection').

I'd be happy to hear if anyone has an opinion on this interpretation...... :-)

Last edited: May 1, 2008
15. May 1, 2008

George Jones

Staff Emeritus
The exterior algebra for an n-dimensional vector V has dimension 2^n, as does the Clifford algebra for V if there is a bilinear form g for V. So, the exterior algebra and Clifford algebra are isomorphic as vector spaces, but not as associative algebras. A particular vector space isomorphism is chosen to identify elements of the two algebras, and is used to define a Clifford product on the exterior algebra. The vector space of the exterior algebra together with its new Clifford product is isomorphic as an associative algebra to the Clifford algebra. Since spin transformations act on the Clifford algebra, they also act on the exterior algebra via the chosen isomorphism.

16. May 1, 2008

schieghoven

Hmmm, interesting. But the spinors $\psi$ themselves are not elements of the Clifford algebra. The elements of the Clifford algebra transform as $A \rightarrow SAS^{-1}$ for spin transformations S. It is possible to define spinors by requiring that they transform as $\psi \rightarrow S\psi$, but this distinguishes them from the geometrical quantities that can be constructed within the framework of either the exterior algebra or the Clifford algebra. It does not provide any further geometrical interpretation of spinors, except that that it defines them by requiring that they transform according to the fundamental representation of the Lorentz group.... and this, of course, is simply the standard definition anyway.

Thanks for the comment, I had forgotten some of the details of Clifford algebras.

Dave

17. May 2, 2008

jdstokes

As George has already pointed out, the Clifford algebra can be regarded as the exterior algebra with a different product. In this way the Dirac equation can be regarded as an inhomogeneous sum of differential forms.

My understanding of spinors is that they are merely irreducible representations of the Clifford algebra. This is consistent with the definition of space of spinors as a minimal left ideal of the Clifford algebra which transforms irreducibly under the regular representation of the Clifford algebra.

For more on why spinors are minimal left ideals of the Clifford algebra see

http://www.valdostamuseum.org/hamsmith/clfpq.html [Broken]

Indeed, some amount of identification via isomorphismss is necessary for all this stuff.

Last edited by a moderator: May 3, 2017
18. May 2, 2008

schieghoven

Well, given that 'minimal left ideal' is simply a fancy name for describing the set of matrices which look like
$\psi = \begin{pmatrix} \psi_1 &0&0&0\\ \psi_2 &0&0&0\\ \psi_3 &0&0&0\\ \psi_4 &0&0&0 \end{pmatrix}$​
I don't see any huge improvement over defining spinors as column vectors, namely
$\psi = \begin{pmatrix} \psi_1 \\ \psi_2 \\ \psi_3 \\ \psi_4 \end{pmatrix}$​
The fact that the first can be decomposed into a sum of matrices which represent differential forms is not particularly useful. It can't even be done in a Lorentz-covariant fashion, because the regular representation of the Clifford algebra $\psi \rightarrow S\psi$ does not represent a Lorentz transformation of the differential forms. Differential forms all transform like $A \rightarrow SAS^{-1}$.

Dave