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I'm looking for a rather intuitive explanation (or some references) of the difference between the metric of a curved space-time and the metric of non-inertial frames.

Consider an inertial reference frame (RF) with coordinates $\bar x^\mu$, in flat spacetime $\eta_{\mu \nu}$ (Minkowski metric).

  1. If I have well understood, on one hand, I can go to an accelerated RF by change of coordinates $x^\mu(\bar x)$. The metric is given by: $$\tag{1}g_{\mu \nu}(x) = \frac{\partial \bar x^{\alpha}}{\partial x^{\mu}} \frac{\partial \bar x^{\beta}}{\partial x^{\nu}} \eta_{\alpha \beta}$$

  2. On the other hand, I know that a curved space-time with metric $q_{\mu \nu}$ cannot be transformed to Minkowski $\eta_{\mu \nu}$ by coordinate transformation. In other words there does NOT exist any coordinate $x^\mu(\bar x)$ such that (in the whole coordinate patch): $$\tag{2}q_{\mu \nu}(x) = \frac{\partial \bar x^{\alpha}}{\partial x^{\mu}} \frac{\partial \bar x^{\beta}}{\partial x^{\nu}} \eta_{\alpha \beta}\qquad \leftarrow \text{(does not exists in curved space)}$$

So far, everything is more or less ok... But my question is:

  1. What is the difference between $q_{\mu \nu}$ and $g_{\mu \nu}$? I mean, in both cases a particle would "feel" some fictitious forces (in which I include the weight force due to the equivalence principle).

  2. What physical situation can $q_{\mu \nu}$ describe and $g_{\mu \nu}$ cannot?

I additionally know that by change of coordinates $q_{\mu \nu}$ is locally Minkowski. But still, I can't see clearly the difference.

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  • $\begingroup$ Although the question is of a different nature, the answer in here (physics.stackexchange.com/q/11806) might be of help. Briefly, general relativity has equations that strongly depend on the curvature tensor. Since this is a tensor, in a curved space-time you cannot change coordinates to make it to "disappear" globally. $\endgroup$ – Phoenix87 Jan 5 '15 at 15:39
  • $\begingroup$ Thank you, I'll have a look, but it doesn't look like a very intuitive explanation from a physical point of view since the question starts as "I'm a mathematician with hardly any knowledge of physics". $\endgroup$ – Worldsheep Jan 5 '15 at 15:44
  • $\begingroup$ Could one perform an infinite number of frame transformations to accomplish this goal? I know it's not practical, but it's one way to (foolishly?) handle acceleration with special relativity. $\endgroup$ – honeste_vivere Jan 5 '15 at 18:31
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Gravity is a gauge theory. Gauge transformations are diffeomorphisms (coordinate changes) described by your equations. Therefore, the space of all possible metrics (the moduli space) is the quotient of the space of all $g_{\mu \nu}$ over these coordinate changes.

So your $g_{\mu \nu}$ can be set to $\eta_{\mu \nu}$ by some coordinate transformation. It means that they belong to the same equivalence class.

On the other hand, $q_{\mu \nu}$ belongs to another equivalence class. It can be seen by computing the Riemann curvature tensor. For any $g_{\mu \nu}$ it should be zero, but not for $q_{\mu \nu}$.

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  • $\begingroup$ Thank you for your answer, but physically, what is the difference between q and g? $\endgroup$ – Worldsheep Jan 5 '15 at 18:39
  • $\begingroup$ They describe different geometries. $g$ describes the usual flat space-time; $q$ is curved and therefore possesses some interesting features (for example, the interior angles of a triangle do not necessarily add to 180 degrees). What else do you want to hear, isn't it enough to consider them different? $\endgroup$ – Prof. Legolasov Jan 6 '15 at 2:39
  • $\begingroup$ Yeah... I think everything is starting to be a bit clearer now! $\endgroup$ – Worldsheep Jan 6 '15 at 13:52
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OP's question (v2) seems partially caused by imprecise use of the word local:

  1. If OP's eq. (1) holds locally in a neighborhood $U\subseteq M$, then there exists coordinates in $U$ such that the metric $g_{\mu\nu}$ becomes on Minkowski-form in $U$, and then the (Levi-Civita) Riemann curvature tensor $R^{\sigma}{}_{\mu\nu\lambda}$ vanishes in $U$, or eqivalently, the manifold $M$ is by definition flat in $U$. The implications also hold in the opposite direction, after possibly going to a smaller neighborhood $V\subseteq U$.

  2. For an arbitrary point $p\in M$ on a Lorentzian manifold $(M,g)$, there exist Riemann normal coordinates in a sufficiently small coordinate neighborhood $U\subseteq M$ of the point $p$ such that the metric $g_{\mu\nu}$ becomes on Minkowski-form with vanishing (Levi-Civita) Christoffel symbols $\Gamma^{\lambda}_{\mu\nu}$ locally in the point $p$ (but not necessarily in the punctured neighborhood $U\backslash\{p\}$ and the manifold $M$ is not necessarily flat in $U$). In particular, the (Levi-Civita) Riemann curvature tensor $R^{\sigma}{}_{\mu\nu\lambda}$ does not necessarily vanish at $p$.

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