# Why the metric tensor?

Inspired by my previous question Variation of gamma and Christoffel (mostly from AccidentalFourierTransform's comment), I was wondering why the metric tensor is used in physics to define a spacetime as opposed to any other tensor. It makes sense that the metric defines a notion of distance and corresponds to the generalization of the newtonian potential so in that sense it is fundamental.

Along those lines however, it seems more fundamental/easier to construct physics based on the tetrads $e^{a}_\mu$ instead. According to this paper variations with respect to the tetrad generalizes the Einstein equations (eq 122) and also has the stress-energy tensor proportional to the tetrad (eq 127). What makes the metric particularly "privileged" as to be the basis of GR (at least in elementary presentations)?

• There ought to be a joke about an Imperial tensor somewhere here... – Kyle Kanos Jun 20 '18 at 20:46
• While you can always use the tetrad field locally one thing to remember is that not all spacetimes admit a global section of it. – Slereah Jun 22 '18 at 6:27

You have correctly identified that the metric $g_{\mu\nu}$ is not the only exclusive field to characterise gravitation. In particular, as you mentioned, the vielbein $e^a_\mu$ can be used. It is particularly useful to work with the vielbein when computing curvatures by hand, using Cartan's equation.

This is because we can usually read off connections from $\mathrm de + \omega \wedge e = 0^\dagger$ and compute the curvature tensor from the standard relation,

$$R = \mathrm d \omega + \omega \wedge \omega.$$

When it comes to supergravity, the vielbein is crucial. The Einstein supergravity multiplet consists of the components, $e^a_\mu$, $\Psi_{\mu\alpha}, \bar\Psi^{\dot\alpha}_\mu, A_\mu,B,\bar B$.

Furthermore, for a general supermanifold, it is quite common to work with the supervielbein, which consists of,

$$E_A = (E_\mu,E_\alpha,\bar E^{\dot\alpha})$$

such that for a point $p\in\mathbb R^{p|q}$, $\{E_A |_p\}$ is a standard basis for $T_p(\mathbb R^{p|q})$.

$\dagger$ In index notation, $\mathrm de^a + \omega^a_b \wedge e^b = 0$ is a linear system of equations to solve for $\omega^a_b$, where we have abbreviated $e^a = e^a_\mu \mathrm dx^\mu$ and $\omega^a_b = \omega^a_{b\mu}\mathrm dx^\mu$.

• Could you expound on your reference to Cartan's equation and the differential form notation? I am new to both and don't know what you are trying to explain in that part. Otherwise this answer is clear. – Quantumness Jun 20 '18 at 20:47
• @Quantumness See edit. – JamalS Jun 20 '18 at 20:57