# Can space-time be defined by the requirement that the physical laws are simple?

When I was student I was told that time is defined by the requirement that the physical laws are simple. For example, in classical mechanics time can be defined by the requirment that the velocity of an isolated body is constant. One could generalize this approach by assuming that space-time is defined by the requirement that the physical laws are simple. This can be easily expressed in mathematical language as follows.

In general relativity the universe is represented by the triple $(M, g, T)$ where $M$ is a four-dimensional differentiable manifold, and $g$ and $T$ are a Lorentzian metric and a tensor field on $M$ satisfying Einstein's equation.

According to the above approach, we could say that the "true" physical reality of the universe is represented by the pair $(M, T)$, while the metric $g$ emerges as an appeareance from the requirement that evolution appears simple, i.e., that $g$ and $T$ satisfy Einstein's equation.

From the mathematical point of view, this approach poses for example the following problems: does any tensor field $T$ admit a metric $g$ such that Einstein's equation is satisfied? To what extent a tensor field $T$ univocally determines the metric $g$?

Does this approach make sense, at least from the mathematical point of view?

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Already the first paragraph is completely bogus, so before I continue reading, let me point this out and prompt you to make some sense of it. You can't define time by the requirement that velocity is constant. Without time, there is no concept of velocity at all. Time is the most fundamental notion in all of physics and is never defined a posteriori. In particular, in GR, we require that the manifold be Lorentzian, so again there is an implicit Minkowsian notion of space-time. There is no way to get around this. Also, there is no need to get around this (besides cheap philosophy...). –  Marek Jun 16 '11 at 15:20
Let X be the configuration space of a classical N-particle system. Assume that a curve on X, i.e., the image of a map from R to X, is given. Any map with that image corresponds to a different definition of time. The correct definition of time corresponds to the maps which satisfy Newton's law, if any. –  bgalvan Jun 16 '11 at 15:57
For what it's worth, the first paragraph makes sense to me. –  Ted Bunn Jun 16 '11 at 18:12
@bgalvan: correct definition? It is easy to measure time in units which have arbitrary functional dependence on your Newtonian time. This notion of time is equally good and yet it would not satisfy Newton's first law, etc. –  Marek Jun 16 '11 at 19:07
@Marek: correct definition?? If the measurement of time which arbitrary functional dependence is equally good as any other, why all people measure time such in a way that Netown's law is satisfied? –  bgalvan Jun 16 '11 at 20:32

If $T$ is "non-degenerate" (at every point as a linear transformation from $T_pM \to T_p^*M$), then modulo some issues with the constraint equations $g$ can be solved from $T$, at least locally in time, after prescribing it in a compatible way on a space-like hypersurface. See this article by De Turck.
On the other hand, the prescription of initial/boundary data may be crucial. Roughly speaking the prescribing Ricci curvature equation is analogous to the inhomogeneous wave equation with a source term. In which case the hyperbolic nature of the equations shows that for any initial configuration there exists a different evolution. So it may be that this will lead to multiple different metrics compatible with your $T$ (indeed consider the case where you restrict your manifold to be $M = \mathbb{R}^4$; the global nonlinear stability of Minkowski space, combined with classical results on classical results on the existence of solutions to the vacuum constraint equations shows that on $M = \mathbb{R}^4$ there are many incongruent (since for these other solutions the Weyl tensor is non-vanishing) solutions to Einstein's equation when $T = 0$; I would expect something similar for the case of prescribed $T$).