# What is the physical significance of locally Minkowskian but not necessarily geodesic coordinates?

Consider an event $P$ in spacetime and a coordinate system $x_\mu$ such that at $P$:

1. $g_{\mu\nu}=\eta_{\mu\nu}$

2. $g_{\mu\nu,\sigma}=0$, or equivalently the Christoffel symbols vanish at $P$.

$x_\mu$ thus constitutes a locally Minkowskian geodesic coordinate system at $P$ and hence represents a local inertial frame (LIF). We can then write the metric at $P$ as $ds^2=dx_1^2-dx_2^2-dx_3^2-dx_4^2$ and read off immediately the physical signficance of the coordinates: $dx_1$ is the time as measured by this LIF and $dx_2,dx_3$ and $dx_4$ are ruler distances along three orthogonal space directions in this LIF.

My question is: Consider a different coordinate system $x_{\mu'}$ (using the primed index notation) such that condition 1. holds but condition 2. is not known to hold*. We can still write the metric at $P$ in the form $ds^2=dx_{1'}^2-dx_{2'}^2-dx_{3'}^2-dx_{4'}^2$, but what physical signficance, if any, do the $dx_{\mu'}$ have now?

*I believe this is the case whenever one encounters a general diagonal metric and tries to interpret its coordinates by 'locally scaling' the coordinates so that the metric at a particular event is $\eta_{\mu\nu}$.

• I may have misunderstood your question but doesn't condition one immediately imply condition two? – Rumplestillskin Aug 12 '17 at 8:26
• I wasn't aware this is the case, could you prove it? Condition 1 is a statement only about the form of the metric at P, it doesn't say what the metric looks like away from P. – Andrew Aug 12 '17 at 8:47
• Coordinates do not bear any physical significance in general, as they can be transformed at will without changing the physics (the gauge freedom of GR). I wouldn't expect to be able to attach a meaning to general coordinates, which are just a mathematical tool to describe a metric (representatives of equivalence classes). – Phoenix87 Aug 12 '17 at 10:22
• You can do that, and in that case Christoffel symbols might not be necessarily zero, and you'll ultimately have some nonzero $R_{\mu \nu \sigma \rho}$. But note that we have exactly the freedom we need to fix all the components of first derivatives of $g_{\mu \nu}$ at $P$. And so for simplification, we just set all of those to zero. The second derivatives are never all zero, however – Avantgarde Aug 12 '17 at 13:29
• I'm aware the freedom exists to set all the first derivatives of the metric to zero. I suppose what motivated me to this question was being presented with a general diagonal metric, and then having to read off which part of the metric represented LIF time and LIF distance etc. It seems most books usually get the metric into Minkowski form by scaling all the coordinates at P but never bother to check that this transformation makes the Christoffel symbols vanish. – Andrew Aug 12 '17 at 13:41