Necessity of equivalence principle

Is the equivalence principle necessary to formulate general relativity or is it possible to formulate general relativity without it?

• I have no idea what you are asking. Commented May 24 at 5:27
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Commented May 24 at 7:25
• There is no question. Commented May 24 at 8:51
• Sorry I edit It Commented May 25 at 4:26

There are several equivalence principles, the weak, Einstein and strong EPs, and general relativity is based on the Einstein equivalence principle. This states:

• the weak EP is valid.

• The outcome of any local non-gravitational experiment is independent of the velocity of the freely-falling reference frame in which it is performed (local Lorentz invariance).

• The outcome of any local non-gravitational experiment is independent of where and when in the universe it is performed (position invariance).

GR is also a metric theory:

• Spacetime is endowed with a symmetric metric.

• The trajectories of freely falling test bodies are geodesics of that metric.

• In local freely falling reference frames, the non-gravitational laws of physics are those written in the language of special relativity.

So strictly speaking the Einstein equivalence principle and the geodesic equation are separate assumptions. However it is widely accepted, though I don't think it has been formally proven, that any theory satisfying the Einstein EP must be a metric theory. So the answer to your question is that the geodesic equation is a necessary part of GR and you cannot formulate GR without it.

If you are interested in learning more about this I can strongly recommend the Living Reviews in Relativity article The Confrontation between General Relativity and Experiment.

• Your treatment here is definitely much more in-depth than something I would be able to provide, but I'd like to point out that we would not have to particularly assume the geodesic equation if we are fine with doing semiclassical approximation on curved backgrounds. This is because geodesics satisfy Euler-Lagrange equations, and so the semiclassical approximation to path integral formulations on curved backgrounds will automatically select for geodesics as the classical trajectories. Commented May 24 at 6:31
• +1, But what about the tetrad Palatini formulation of General Relativity, in order to be compatible with Dirac fermions? Do geodesics differ whether we are treating scalar, spinor, or vector particles? Commented May 24 at 8:32
• @JeanbaptisteRoux Particles with spin in general do not follow geodesics. The spin will interact with the curvature to produce the MPD force (en.wikipedia.org/wiki/Mathisson–Papapetrou–Dixon_equations). Commented May 24 at 8:46
• As a clarification, freely falling test bodies following geodesics is not an input assumption for GR. It follows the Einstein equation. Commented May 24 at 8:52

It's possible to take different assumptions to start formulating general relativity, but any assumptions which reproduce the standard theory will imply the equivalence principle, because it's just true in GR. For example, you could formulate GR by just saying:

Spacetime is a Pseudo-Riemannian manifold whose metric extremizes the action $$S = \int d^n x \sqrt{-g} \left(\frac{1}{16 \pi G_N}R + \mathcal{L}_{matter}\right)$$

The equivalence principle is baked into the assumption that spacetime is a Pseudo-Riemannian manifold, because any such manifold looks like Minkowski space if you zoom in far enough (which is the content of "a freely falling frame feels no gravity locally"). The assumption I made above is pretty technical though, and people usually prefer to start with the more qualitative equivalence principle (as well as the other usual axioms) and derive the above conclusions. But they are equivalent.