What mathematical constraints does the equivalence principle impose on GR?

Question

I'm trying to determine exactly how the equivalence principle affects GR mathematically (rather than conceptually). I found this StackExchange post which more or less says that the equivalence principle is what allows you to represent the gravitational field with a metric tensor. However, I'm not really convinced that's the case: in Kaluza-Klein theory you can represent EM (together with gravity) in terms of a metric tensor, and there's no EM equivalence principle. In fact, I'm pretty sure the dilaton scalar field in Kaluza-Klein theory violates the equivalence principle so it cannot be fundamental to the concept of "geodesics in spacetime" (since this is what Kaluza-Klein theory is).

Another answer to that same question touches on this and mentions something about the equivalence principle implying that there's no torsion tensor. I was wondering if someone could expand on this. Does this mean that the fact the metric connection is torsion-free is a consequence of the equivalence principle?

Answer 1

• The dynamics of the scalar and vector fields can be described in a manner consistent with the special theory of relativity. Given the fact that Newtonian gravity is described by a gravitational potential $\phi_N (t, x)$, which satisfies the Poisson equation $\nabla^2 \phi_N = 4\pi G\rho_m$ (where $\rho_m$ is the mass density), it might seem that one could construct a theory for gravity consistent with special relativity by suitably generalizing the Poisson equation for the gravitational potential. It turns out, however, that this is not so straightforward. The natural description of the gravitational field happens to be completely different and is intimately linked with the geometrical properties of the spacetime. The mathematical possibility of such a geometric description is intimately connected to the principle of equivalence.

Principle of equivalence: The trajectories of particles having the same initial conditions in a given gravitational field must be independent of the properties of the particle (such as mass etc).

• This immediately puts a constraint that the action for a particle moving in a gravitational field should have mass of the particle as an overall scaling factor, and only then the equations of motion would be independent of $m$. Indeed, in weak gravity limit we have the action as $$A = -mc^2\int_{\tau_1}^{\tau_2}d\tau \left(1+\frac{\phi}{c^2}\right)$$

• In Newtonian gravity, the kinetic and potential energy terms of the Lagrangian $L = \frac{1}{2}mv^2 − m\phi$ are proportional to the mass of the particle $m$ and hence $m$ has no influence on the equations of motion for a particle. Therefore, the trajectories $q(t)$ of material particles (with the same initial conditions) will be independent of the properties of the particle and will depend only on the gravitational potential $\phi(x_i)$. We expect this feature to hold in any relativistic generalization of the description of gravity.

Principle of equivalence: The mechanical experiments involving gravity cannot distinguish between a small, closed, box located near the surface of Earth (where the acceleration due to gravity is $g$) and another similar box located in interstellar space which is moving with an acceleration $g$.

• Put other way, around any event $P$ we choose a locally inertial frame in which the laws of special relativity are valid. Let us assume that a given metric tensor varies over a region of size $L$ in the sense that, for a typical component of the metric tensor, $\partial{g}/{g} ≈ L^{−1}$. The principle of equivalence assures us that, within such a box, gravitational effects should vanish to $O(L)$ accuracy.

• Basically to treat $g$ is a constant over a small region of length $l$ (suppose the size of the box) is equivalent to asking for an approximately constant Christoffel symbols $\Gamma_{bc}^{a}$ (or affine connections) with their first-order derivatives being zero.

$$\Gamma_{bc}^{a} = 0, \quad \frac{\partial\Gamma_{bc}^{a}}{\partial x^{\mu}} = 0$$