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P: 665
I just realized that my early post had some error in it, so I fix it here:

 Quote by Altabeh I think you mean that if there is a direct coordinates transformation to make a Minkowski spacetime out of a given metric $$g_{\mu\nu}(x^{\alpha})$$ with a non-vanishing second derivative, let n be 4 and $$\alpha=0,..,3$$, for the sake of convenience, then that spacetime is not curved, this is true in the sense that $$g_{\mu\nu}(x^{\alpha})\rightarrow \eta_{\mu\nu}$$ can be obtained through an explicit coordinates transformation $$x^{\alpha}\rightarrow \bar{x}^{\alpha}$$. For example, $$d\bar{s}^2 = d\bar{t}^2 +2\bar{x}^2d\bar{t}d\bar{x}-(1-\bar{x}^4)d\bar{x}^2,$$ has a non-vanishing second derivatives wrt $$\bar{x}$$, and it can be made Minkowski through $$\bar{x}=x$$ and $$\bar{t}=t-x^3/3$$ at every point. This last bold-face is so important in our observation of flat spacetimes and distinguishing them with the curved ones.
Now according to this, I think you are able to answer the questions huddling in your mind:

 Quote by avirab da2 + db2 + dc2 is flat. f(b)da2 + db2 + dc2 is not flat if f(b) has a non-zero second derivative. That's it. This metric is interesting for several reasons
Take the metric

$$ds^2=(1+ax)^2dt^2-dx^2-dy^2-dz^2,$$

where $$a$$ is some constant. $$(1+ax)^2$$ has a non-zero second derivative wrt x, but the metric is flat!!

 If there are two metric functions, eg f(b)da2 + h(a)db2 + dc2, then in theory there can be coordinate transformations which make these 'cancel' each other, so that they are both 1.
This is not right unless one hits

 It of course depends on what functions f and h are
.

 But if h = 1, and f has a non-zero 2nd derivative, there's no way to make the metric flat.
I made it flat! Look at the example above!

 I am looking at some intuitive 'reason' that one can give for this (that there's no way that this one metric function f(b) can be made to be 1 via a coordinate transformation), it can only be spread around to functions in the other parts of the metric.
You just need to have the Riemann tensor vanished to get a flat spacetime; this can't always be understood from the metric directly. For instance, the above metric has non-vanishing Christoffel symbols but is not curved since the Riemann tensor is non-zero. These exceptional cases do not let us distinguish flat spacetimes from curved ones. In GR, one uses the geodesic coordinates system to make sure that the metric gets flat locally, because once we understand that Christoffel symbols vanish, then the Riemann tensor is necessarily zero but the inverse is not true always, as you can see!

 And similarly for the connection, if there is only one non zero or only one 'much larger' than all the others, which are negligibly small, as is the case for the classical limit of Schwarzschild.
Only locally ture! Why don't you draw a tiny attention to my notes? I said that Schwarzschild connections (Christoffel symbols) do not vanish so the spacetime is curved everywhere and tends to be Minkowski (flat) at large distances (r-->oo) from the source; leading connections to vanish locally because they are of the dimension 1/r outside the source of gravitational field!

AB