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In Linearized gravity one can perform coordinate transformations

$$x^\mu \rightarrow x'^\mu=x^\mu+\xi^\mu(x)~~~~~~\text{with the condition } \Biggl|\frac{\partial \xi^\mu(x)}{\partial x^\nu}\Biggr|\ll1$$

It seems to me that the condition prevents me from rotating my coordinate system. If I wanted to rotate around the $x^3$-axis the transformation for $x^1$ would be

$$x'^1=x^1\cos\alpha-x^2\cos\alpha=x^1+(x^1\cos\alpha-x^2\cos\alpha-x^1)=:x^1+\xi^1(x)$$

So

$$\frac{\partial\xi^1}{\partial x^1}=\cos\alpha-1$$

which is not necessarily small. This seems very weird to me, because I feel like it should not make a difference how I rotate my coordinate system without severe consequences for my theory...

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  • $\begingroup$ Well, your condition is satisfied for sufficiently small angles. I'm not especially knowledgeable about linearized gravity, but perhaps it is rotational invariant only to first order. This seems intuitively plausible. $\endgroup$
    – Geoffrey
    Sep 8, 2020 at 17:12

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Yes, one can rotate coordinate system in linearized gravity (and perform Lorentz boosts and finite translations). Just these transformations would not be infinitesimal diffeomorphisms.

Remember, that linearized gravity is Poincare invariant field theory in addition to the gauge symmetries of infinitesimal diffeomorphisms.

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  • $\begingroup$ Ah yes, I understand. I was under the wrong impression that the transformations I wrote down, were the only ones that are possible. Thanks! :) $\endgroup$ Sep 9, 2020 at 6:54

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