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  How to prove explicitly that by including Dirac fermions into the Einstein-Hilbert action we make torsion to be non-zero?

+ 5 like - 0 dislike

Recently I've heard the statement that by including Dirac fermions into the Einstein-Hilbert action we make torsion be non-zero, so that is one of problem of quantum gravity. How to prove that explicitly? Intuitively it's somehow connected with the form of Dirac action in curved spacetime (which includes vierbein), but I don't know how to demonstrate it directly.

Maybe it can be done by assuming Christoffel symbols and metric as independent quantities and then by variation of action by Christoffel symbol? In result I'll get some equation for Christoffel symbol, then I'll add to one equation the another one with indices permutation for getting equation on torsion tensor. If the tensor-free part of equation will Benin-zero, then torsion isn't zero. Is this thinking right?

This post imported from StackExchange Physics at 2015-01-06 21:35 (UTC), posted by SE-user PhysiXxx

asked Jan 4, 2015 in Theoretical Physics by PhysiXxx (45 points) [ revision history ]
edited Jan 6, 2015 by Dilaton
More on torsion: physics.stackexchange.com/q/137738/2451

This post imported from StackExchange Physics at 2015-01-06 21:35 (UTC), posted by SE-user Qmechanic

1 Answer

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First of all this is not a quantum gravity problem and to be fair not a problem. We take a torsion free theory because torsion is not observed as far as I know. Furthermore you do not need Dirac fermions in order that torsion is not vanishing! Also you can have fermions and arrange them such that torsion vanishes! Simply by introducing the cotorsion in your vierbein connection and write your new connection as 

$\tilde{\omega}_{\mu}^{ab} = \omega_{\mu}^{ab} - K_{\,\,\, \mu}^{ab}$


$$K_{\,\, \mu \nu}^{\rho} = (\mathcal{T}_{\,\, \mu \nu}^{\rho} + \mathcal{T}_{\mu \nu}^{\,\,\,\, \rho} + \mathcal{T}_{\nu \mu}^{\,\,\,\, \rho})/2$$

Of course $C$ is the torsion tensor defined as $\mathcal{T}_{\, \mu \nu}^{\rho} = 2\Gamma_{[\mu \nu]}^{\rho}$. Then $\tilde{\omega}_{\mu}^{ab} $ only depends on the vierbein $e_a^{\, \mu}$ as

$$ \tilde{\omega}_{\mu}^{ab} = e^{a\rho}\partial_{[e}e_{\rho]}^{b} + \frac{1}{2}e^{a\rho}e^{b\sigma}e_{c\mu}\partial_{[\sigma}e_{\rho]}^{c} - (a \leftrightarrow b) $$

The cotorsion does not contribute to the Lorentz transformations at all ( of $\Omega_{\mu}$ actually, too much background I need to write to make it precise). It turns out that you can choose $\omega_{\mu}^{ab}$ such that the torsion vanishes in a gravitational theory that contains spinorial fields.


But to the point, on why fermions induce torsion in GR. First of all let us remember that GR is invariant under the big group $GL(4,\mathbb{R})$ that people sometimes call it "symmetry of the world" (yes I have heard it indeed). The not necessarily symmetric Christoffel connection $\Gamma$ then belongs to its associated algebra which is $\mathfrak{gl}(2,\mathbb{R})$. Good. One important thing we require that any realistic theory of gravity has is the metric compatibility condition, i.e. $\nabla_{\lambda}g_{\mu \nu} = 0$, something you will most certainly agree upon. Remember that this covariant derivative involves a lot of Christoffel symbols and it acts on a vector field $V^{\mu}$ as $\nabla_{\nu}V^{\mu} = \partial_{\nu}V^{\mu} \Gamma_{\lambda \nu}^{\mu}V^{\lambda}$. Another thing we have to remember is what is the infinitesimal length in GR. This is given by the usual $ds^2 = g_{\mu \nu} dx^{\mu} dx^{\nu}$ but we can also use the famous "vierbeins" $e_{a}^{\mu}$ which form an orthonormal frame that transforms under both  $GL(4,\mathbb{R})$  and the connected component of the Lorentz group $SO(3,1)$. Then the metric can be written in therms of the vierbeins as $g_{\mu \nu}=\eta_{\mu \nu}e_{a}^{\mu}e_{b}^{\nu}$. Now, this metric is transforming as a usual two-ranrk tensor under $GL(4,\mathbb{R})$ and as a scalar under $SO(3,1)$. In correspondence with the connection of the metric we also need a connection for this new orthtonormal frame. This is given by the connection $\omega_{\mu}^{ab}$. It turns out that its possible to define the covariant derivative of any field representation of $SO(3,1)$ using  $\omega_{\mu}^{ab}$, for example for a collection of scalar fields we get that the covariant derivative of acts as $$D_{\mu}\phi^I = \partial_{\mu}\phi^I + (\Omega_{\mu})_{J}^{I}\phi^J $$. Here, we have introduced a new tensor $\Omega_{\mu} \equiv -\frac{i}{2}\omega_{\mu}^{ab}S_{ab}$. Of course $S_{ab}$ are the usual spinor matrices which are given by $S_{ab}=\Sigma_{ab}/2 = \frac{i}{4}[\gamma_{a},\gamma_{b}] $ and $\gamma_{a}$ are the usual Dirac $\gamma$-matrices that form a Clifford algebra. In any case we will be intrested in the spinor representation of the Lorentz group and how the covariant derivative acts on it: $$D_{\mu}\psi = \partial_{\mu}\psi - \frac{i}{4}\omega_{\mu}^{ab}\Sigma_{ab}\psi.$$ The curious thing is on the way the $\Omega_{\mu}$ transforms. It turns out that in order to transform properly the new connections $\omega{\mu}^{ab}$ must obey

$$\omega_{\mu}^{ab} \to {\omega'}_{\mu}^{ab} = \omega_{\mu}^{ab} + \lambda_{\,\,c}^{a}\omega_{\mu}^{cb} + \lambda_{\,\,c}^{b}\omega_{\mu}^{ca} - \partial_{\mu}\lambda^{ab}$$

where $\lambda_{ab}=-\lambda_{ba}$ are the usual transformation parameters (the ones we find in the exponential). Very good, now that we have established the basics lets go to the main point. The covariant derivative $\nabla_{\mu}$ needs to be generalized in order to incorporate the new connections. The new covariant derivative can be written as $\mathcal{D}_{\nu}$ and acts on the vierbeins as

$$\mathcal{D}_{\nu}e^{a\mu} = \partial_{\nu}e^{a\mu} + \Gamma_{\rho \nu}^{\mu}e^{a\rho} + \omega_{\nu}^{ab}e_b^{\mu} =0$$

and this condition determines the form of the spin connection $\omega_{\mu}^{ab}= g^{\rho \nu}e_{\rho}^{a}\nabla_{\mu}e_{\nu}^b$ with thisoption not being unique though. How does torsion come into the game? Well, let us see what happens with the Riemann curvature tensor first. The curvature tensor should transform as a $GL(4,\mathbb{R})$ tensor, as usual, and as a second rank Lorentz tensor now. This can be written in terms of the  new connection as

$$R_{\,\,\, \mu \nu}^{ab} = \partial_{\mu}\omega_{\nu}^{ab} +\omega_{\nu}^{ac}\omega_{\nu c}^{\,\,\,\, b}  - \partial_{\nu}\omega_{\mu}^{ab} - \omega_{\nu}^{ac} \omega_{\mu c}^{\,\,\,\,b} \tag{Curv}$$

simply by projecting out 2 components of a 4-rank tensor as following $R_{\, \sigma \mu \nu}^{\rho} = e_{a}^{\, \rho}e_{b \sigma}R_{\,\, \mu \nu}^{ab}$. If you replace this to  into (Curv) you will get back the usual Riemann tensor. Now note that we have everything we need in order to formulate a gravitational theory that includes Lorentz spinors (which transform as world scalars). The catch is, as we said in the beginning, that the Christoffel symbols might not be symmetric. The antisymmetric part of the Christoffel symbol transforms as a $GL(4,\mathbb{R})$ tensor and is written as

$$\mathcal{T}_{\, \mu \nu}^{\sigma} = 2\Gamma_{[\mu \nu]}^{\sigma} \tag{Tor} $$

Then, if the right hand side is non-vanishing the Einstein tensor also becomes assymetric and as a result the stress-energy-momentum tensor. 

answered Jan 7, 2015 by conformal_gk (3,625 points) [ revision history ]
edited Jan 9, 2015 by conformal_gk

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