The text I'm reading claims that in an infinitesimally small region of the manifold, we can always find observers that locally see the metric as either Minkowski, $\eta_{\mu\nu}$, or Euclidean, $\delta_{\mu\nu}$. The observer seeing a Minkowski metric makes sense to me, if we look at a small enough region the curvature is small enough that the metric reduces locally to that of flat space, at least that's what I think. But finding the metric to be Euclidean implies $diag(1,1,1,1)$, unless I am wrong about that. My question is in what physical situation could an observer see a Euclidean metric?
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$\begingroup$ Which text? The first "or" should be an "xor": The manifold is Euclidean xor Lorentzian. Not both. Possible duplicate: physics.stackexchange.com/q/129187/2451 $\endgroup$– Qmechanic ♦Commented Mar 31, 2020 at 15:15
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$\begingroup$ Hmmm. What's the text? I guess they may mean that if all velocities are relatively small, then we get Galilean relativity. $\endgroup$– PM 2RingCommented Mar 31, 2020 at 15:16
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$\begingroup$ The notes for my GR course, perhaps text was a poor choice of wording, the exact quote is "For this, let us assume that we start with an infinitesimally small region of our manifold. In this region one can find a set of ‘privileged’ observers that locally see the metric as trivial (the LIF for GR). This means that in this coordinate system the metric is diagonal, with +1 or −1 components (two examples are δµν or ηµν) and has vanishing first derivatives" $\endgroup$– CharlieCommented Mar 31, 2020 at 15:17
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$\begingroup$ I think that the text could be referring to that the spacetime metric is locally Minkowskian $ diag(1,-1,-1,-1) $ and then the projected 3D space-type for an instant $ t=t_0 $ is euclidean $ (1,1,1) $ $\endgroup$– Raúl Aparicio BustilloCommented Mar 31, 2020 at 15:22
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$\begingroup$ Maybe, the notes have consistently used the (-,+,+,+) signature though. Also the indices on the Kronecker delta metric are spacetime indices which unless the notation has been confused suggests the entire metric is the identity. $\endgroup$– CharlieCommented Mar 31, 2020 at 15:26
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He can't see a Euclidean metric, except by restricting to space.
I think the writer did not mean what he appears to say, as quoted in the comment. I think he only intended to give two examples of diagonal metrics, not to give two examples of metrics which an observer can actually see.
answered Mar 31, 2020 at 15:58
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$\begingroup$ That makes more sense, so there is definitely no local frame in which the metric is the 4d identity? $\endgroup$– CharlieCommented Mar 31, 2020 at 16:06
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1$\begingroup$ Absolutely no possible example. Remember, it is a fundamental principle that the laws of physics are the same for all observers. It follows that any inertial observer can see a locally Minkowski metric. The metric signature refers to the sign of the eigenvalues; we must always have one for time and three for space, and opposite signs for time and space. $\endgroup$ Commented Mar 31, 2020 at 16:13
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$\begingroup$ Great, question answered, thank you :) $\endgroup$– CharlieCommented Mar 31, 2020 at 16:14