# Why do we study teleparallel gravity if it is equivalent to general relativity?

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 a metric, teleparallel gravity uses tetrad fields defined in the 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.

• for one, the separation of gravity from inertia is nice, which allows the definition of an energy-momentum tensor density for the gravitational field Commented Jun 1, 2018 at 16:30
• The real question is: Why has Teleparallel Gravity not replaced standard GR yet? It is much clearer wrt. separating inertial and gravitational effects, energy-momentum of the field, and being a gauge theory of translations.
– user257090
Commented Jun 6, 2020 at 15:03

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.