3
$\begingroup$

We know there are many modified theories of gravity, like $f(R)$ gravity, $f(T)$ gravity, and alternative theory of gravity, that is teleparallel gravity. Currently I am studying this theory. Teleparallel gravity is equivalent to general relativity but with different geometry, instead of metric, teleparallel is using tetrad field defined in tangent space.

Here, I just don't understand why we study teleparallel gravity if that theory is equivalent to general relativity? is there any case or physics phenomenon where teleparallel gravity can explain better than general relativity? for example, $f(T)$ gravity is proposed to explain about dark matter and dark energy.

$\endgroup$
  • $\begingroup$ for one, the separation of gravity from inertia is nice, which allows the definition of an energy-momentum tensor density for the gravitational field $\endgroup$ – Christoph Jun 1 '18 at 16:30
  • 2
    $\begingroup$ This is a strange question. You study teleparallel gravity, but you ask everyone else why. $\endgroup$ – MBN Jun 4 '18 at 14:07
5
$\begingroup$

There are different teleparallel gravities, if you noticed in the literature. The one which is equivalent is called Teleparallel Equivalent of General Relativity (TEGR) and it is a particular action choice that makes it equivalent.

If you decompose the variables, the metric $g_{\mu\nu}$ and the affine connection $\Gamma_{\mu\nu}^\alpha$ on the manifold into tetrad, $e_\mu^a$ which is the potential for translation symmetry, and spin connection, $\omega_{\mu} {}^a {}_b$ which is the potential for linear transformations, then you obtain the following equivalence between the Ricci scalar with respect to the Levi-Civita connection and Torsion tensor: $$ \det(e) \hat{R} = \det (e) \left( \frac14 T^{abc} T_{abc} + \frac12 T^{abc} T_{bac} - T^a T_a \right) + 2 \partial_\mu \left[ \det(e) T^\mu \right] $$ where $\det(e)$ is the determinant of the tetrad, $T^{abc}$ is the Torsion tensor, $T_a$ is the trace of the Torsion tensor, and $\hat{R}$ is the Ricci scalar with respect to the Levi-Civita connection, $\hat\omega_{\mu}^{ab}$, not the affine one which is zero for teleparallel gravity (cf. Weitzenböck connection).

Therefore, if your action is as follows: $$ \mathcal{S}_{TEGR} = \int d^4 x \det (e) \left( \frac14 T^{abc} T_{abc} + \frac12 T^{abc} T_{bac} - T^a T_a \right) + \mathcal{S}_{matter} $$ it will be exactly equivalent to General Relativity up to a total derivative.

Instead, if you even choose different coefficient for the torsion-squared terms, it will be both phenomenologically and dynamically different.

The advantage of teleparallel gravity is that you can build a guage theory for gravity in curvature-flat spacetime since the spin connection vanishes identically. Nevertheless, the geometry does not have trivial geodesics, instead the curved world lines would still exist because of twirly features of the geometry.

$\endgroup$
1
$\begingroup$

I think TEGR is more beautiful than GR when you couple gravity to spinor, leading to zero spin connection. So it is not equivalent to GR when spinors are involved at the quantum level.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.