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I know that in general relativity, unlike in electromagnetism, the field tensor must be dependent on metric tensor second derivatives, since I can always find a reference system in which its first derivatives are null.

I don't really get this. I mean, I guess it is due to the equivalence principle that I can find a locally flat reference system so that the metric tensor is constant and its first derivatives are zero. But if they were zero, wouldn't also its second derivatives be zero?

Qmechanic
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PsycoPulcino
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    You can't transform away intrinsic curvature. –  Jun 20 '17 at 13:06
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    This is explained in point 2 of my Phys.SE answer here. – Qmechanic Jun 20 '17 at 13:18
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    The partial first derivatives are zero at a point P – Avantgarde Jun 20 '17 at 13:40
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    It is indeed due to the equivalence principle that you can do this, but the cancellation of the first derivatives of $g$ is at one point only. So it's akin to a local extremum of a function of one variable: like the curve $y=x^2$ at the point $x=0$. Move away from your special point a distance $\epsilon$ and, unless your spacetime is truly flat, the first derivatives are in general nonzero. – Selene Routley Jun 20 '17 at 13:41

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Example, when you throw an object straight upward, when object reach the maximum height, the displacement is constant, the velocity(first derivative) is zero, but the acceleration (second derivative) is not.

Tonduea
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