Torsion tensor in Relativity

Question

While reading Sean Carroll's book on general relativity, I came across something called as a 'Torsion Tensor' which is defined as: $$\Gamma{^\lambda}{_{\mu\nu}} - \Gamma{^\lambda}{_{\nu\mu}} = T{^\lambda}{_{\mu\nu}}$$ When the Christoffel symbol is symmetric in its lower indices, it's known to be torsion-free as it is specified in the book. There's nothing much given about the tensor beyond this other than it's taken as a given in Riemannian geometry. Is there anyway to imagine this torsion or understand it intuitively? Like what do you mean by the term torsion in space time? Squishing of space-time?

Answer 1

Imagining torsion geometrically is not as easy as imagining curvature.

More or less, torsion measures how a curve in the tangent space of a point $x\in M$, obtained by parallel transporting each tangent vector of a close curve in $M$ back to the point $x$, is far from being close as well.

Torsion really plays a role when treating spinors. In fact, it couples with spinors and there you can understand better its physical meaning. For instance you can take a first generalization of GR, called Einstein-Cartan-Sciama-Kibble theory, where we let the Ricci curvature "contain" torsion and we can see that a new field equation arises:

$$Q{^\mu}_{\nu\sigma}=-16\pi\Sigma{^\mu}_{\nu\sigma},$$

where $Q$ is the torsion tensor and $\Sigma$ the spin tensor.

Answer 2

The torsion tensor $T^\lambda_{\mu\nu}$ is defined as the antisymmetric part of the affine connection coefficients $\Gamma^\lambda_{\mu\nu}$ $$T^\lambda_{\mu\nu}\equiv\Gamma^\lambda_{\mu\nu}-\Gamma^\lambda_{\nu\mu}$$ In General Relativity, it is postulated that $T^\lambda_{\mu\nu}=0$. The presence of torsion in affine connection simply implies that the covariant derivative of a scalar field $\phi$ doesn't commute, that is $$\nabla_{[\mu}\nabla_{\nu]}\phi=-T^\lambda_{\mu\nu}\nabla_\lambda\phi$$ For a vector $v^a$ and a covector $w_a$, the following relations are valid: $$\nabla_{[\mu}\nabla_{\nu]}v^\sigma=R_{\mu\nu\lambda}^\sigma v^\lambda-2T^\lambda_{\mu\nu}\nabla_\lambda v^\sigma$$ and $$\nabla_{[\mu}\nabla_{\nu]}w_\lambda=R_{\mu\nu\lambda}^\sigma w_\sigma-2T^\sigma_{\mu\nu}\nabla_\sigma w_\lambda$$ where $R^\sigma_{\mu\nu\lambda}$ is the Riemann tensor.

From these definitions, it follows that torsion measures the amount by which the boundary of a loop fails to close after being parallel transported. Thus, non-zero torsion signifies that a loop made of parallel transported vectors is not closed, i.e., geodesics as extremal lines don't coincide with autoparallels.

Answer 3

The symmetry of the connection is a central assumption of Einstein's theory of gravitation. Without it, the connection could not be uniquely determined from the metric. However I am not aware of a text that explains the experimental basis for the no-torsion assumption. Misner, Thorne, and Wheeler (box 10.2) give a clue saying "torsion does not have any place in a gravity theory based on the equivalence principle." I think they are referring to the Einstein Equivalence Principle (EEP) which holds that the outcomes of experiments in freely falling laboratories must not depend on the velocity of the laboratory.

Non-zero torsion causes tetrads carried in different directions past a given space-time point to rotate relative to one another. To illustrate what that would mean, suppose you carry a non-rotating bicycle wheel, which you know is not rotating because the tension in the spokes is minimized. Then suppose your friend flies past you, carrying a bicycle wheel which they also determined was non-rotating. In a universe with non-vanishing torsion, the wheels could be rotating relative to one another.

To see why, consider a coordinate system in which none of the legs of a tetrad at rest show any apparent rate of change (a tetrad means three orthogonal gyroscopes together with the tangent vector to their world line). Since your coordinate system is chosen so that none of the tetrad legs are changing in time, the connection coefficients with first subscript $t$ vanish, $\Gamma^\alpha_{t\nu} = 0$ for all $\alpha$ and $\nu$. Since the basis vectors are unchanged under time translation, symmetry implies $\Gamma^\alpha_{\nu t} = 0$, in particular that the time component of any vector is unchanged by translation along the space-like directions $\nu=1,2,3$. Due to metric compatibility, vectors maintain their lengths, and since the time components of all vectors are unchanged, it means the space components can only be rotated, for example $\Gamma^z_{xy} = 1$ and $\Gamma^y_{xz} = -1$ (corresponding to a rotation of the $y-z$ plane induced by translation in $x$). Invoking symmetry again, $\Gamma^z_{yx} = 1$ and $\Gamma^y_{zx} = -1$. For translation along $y$ and $z$ to also be rotations, one requires $\Gamma^x_{yz} = -1$ and $\Gamma^x_{zy} = 1$. But that contradicts the symmetry assumption. The conclusion is that symmetry of the connection ensures that all the coefficients vanish if they are zero for translation in any time-like direction.

For this reason, non-zero torsion makes it impossible to construct inertial coordinates in the usual sense around a given point. Worse, if the metric no longer uniquely specifies the connection, one needs to reconsider its significance. If the metric is retained, then one also needs to decide whether test particles follow the geodesics (distance-minimizing paths) or the auto-parallels (parallel transport of the tangent vector) as these are distinct conditions in the presence of torsion. One then also needs to question the roles of gyroscopes in thought experiments, as the symmetry of the connection is what ensures that the apparent torque due to gravito-magnetic (Coriolis-like) forces on the flywheel recovers the parallel transport of a space-like vector. Also the field equations need to be modified because both of the Bianchi identities fail in the presence of torsion.