I have at hand the book by Clifford Will, "Theory and Experiments in Gravitational Physics", and the following Living Reviews in Relativity article. He quotes the Einstein Equivalence Principle (EEP) as follows [Will2006, 3.2]:

  1. "mass" is proportional to "weight", hence trajectories of freely falling "test" bodies are independent of their masses, structures and compositions; (Newton EP, or Week EP)
  2. local non-gravitational test experiments are independent of the velocity of the freely falling reference frame in which they're performed. That is, Special Relativity (as well as QED and so on) is locally valid; (Local Lorentz Invariance, LLI)
  3. these experiments are also independent of where and when in the Universe they're performed. (Local Position Invariance, LPI)

Then Will claims that

it's possible to argue convincingly that if EEP is valid, than gravity must be a "curved spacetime" phenomenon. [...] The only theories that can fully embody EEP are those that satisfy the postulates of "metric theories of gravity", which are

  1. The spacetime manifold has a symmetric metric $g_{\mu\nu}$;
  2. trajectories of freely falling test bodies are geodesics of $g_{\mu\nu}$;
  3. locally, the non-gravitational laws of physics are those of Special Relativity, QED and so on...

Ok, this claim is more than reasonable and experimental tests of EEP made with metric theories have yield optimal results and confirmations. But, is it rigorous or exhaustive? Is there the possibility of a technical fault with testing EEP with metric theories? I mean, can I test this claim at the same time as EEP, and vice versa? Is teleparallelism an alternative to the introduction of a metric?

Though manifolds are a very general concept, and hence saying that the spacetime can be described by a manifold is too general to be wrong, embedding a metric is natural but more subtle and not so obvious to me asserting that gravity can be interpreted with its curvature, as a result of EEP. Can someone convince me that it's the only reasonable thing to do or, on the contrary, show me some alternatives?


2 Answers 2


Is teleparallelism an alternative to the introduction of a metric?

Teleparallel gravity still comes with a metric - just take the tetrad field as orthonormal basis and there it is. The main difference between GR and teleparallelism is that the former uses curvature, the latter torsion to model gravity. According to Kleinert, there's actually a type of gauge symmetry between these theories and there exists an infinite number of intermediate descriptions in terms of both curvature and torsion.

That is, Will's claim that EEP necessarily implies a curved spacetime is not true with teleparallel gravity.

As far as I can tell, your interpretation is correct. Note that some of the features that are due to the geometric structure of GR need to be imposed dynamically in the teleparallel formulation:

The equations of motion of test particles in case of GR are given by geodesic equations, whereas teleparallelism uses force equations. This means in contrast to GR, teleparallelism allows extensions where the weak equivalence principle does not hold and inertial and gravitational mass differ.

Similarly, according to this paper I just googled, local Lorentz invariance is not an intrinsic feature of teleparallel theories, but a consequence of the choice of action that differs from the Einstein-Hilbert action by a boundary term only.

  • $\begingroup$ Thank you for your contribution, Christoph! What I meant to ask was if teleparallelism does include EEP without requiring a metric whose geodesics describe gravity. That is, Will's claim that EEP necessarily implies a curved spacetime is not true with teleparallel gravity. Is my interpretation right? $\endgroup$ Nov 29, 2013 at 17:04
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    $\begingroup$ @AstoundingJB: see edit $\endgroup$
    – Christoph
    Nov 29, 2013 at 17:49

There is a sense in which metric theories of spacetime are "general".

I simplify to four dimensions, but the argument generalizes to higher dimensions.

Consider a particle whose path is parameterized by four coordinates $x^{a} = (t(s),x(s),y(s),z(s))$. We wish to describe the motion of the particle, given that at s=0, each of these functions has a known value and their first derivatives have a known value.

Then, we are stuck with an equation of the form:

$${\ddot x}^{a} = f({\rm x's})^{a}$$

From here, you're stuck adding actual structure to the problem. I'm going to think on this a bit more, but leave this here, and see if comments help build this up.

EDIT: after comments:

Well, once you have a metric, then you can parametrize the arcs by arc length subtended, and the first derivative is trivial. Then, if you have a uniquely determined path, every point in your phase space has a predictable trajectory. Then, while your IVBP does not necessarily need to be phrased in terms of metric and connection, there is always a metric and connection that is equivalent to your IVBP -- you can always just restate your $f^{a}$ as some connection, and solve for a "metric". This will break down if the geometric principle fails -- if differently massed particles with no back-reaction follow different curves.

  • $\begingroup$ Hi Jerry, thank you for the answer! Well, geometrically a metric is needed to introduce the notion of distance (at least locally). Then, in fact the notion of trajectory needs a metric and your argument is surely right. But this can't actually help me to understand Will's claim that EEP implies a metric for the spacetime whose curvature is the gravity... $\endgroup$ Nov 29, 2013 at 16:48
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    $\begingroup$ @AstoundingJB: provided an edit. The basic idea is that any formulation you come up with will be equivalent to a metric theory, even if the language you use (such as, for instance, generic Newtonian mechanics) isn't necessarily that of a metric theory. $\endgroup$ Nov 29, 2013 at 18:38

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