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The How does the Euclidean metric is the symmetry group of Euclidean space. It includes rotations and translations.

Say I consider an Euclidean space and a time parameter. How does the Euclidean metric (which in canoncial coordinates is $\delta_{ij}$, with $i,j\in \{1,2,3\}$), transform under a Galilean transformation, more specifically a velocity boost $t\mapsto t,x\mapsto x-vt$? In a way, from the three dimensional pov, which doesn't see $t$, this is just a translation.

So here is what I'd like a derivation for:

$$(t\mapsto t'=t,\ x\mapsto x'=x-vt)\ \Longrightarrow\ h_{ij}\equiv\delta_{ij}\mapsto\ ?$$

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1 Answer 1

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The metric doesn't transform at all. The metric stays fixed even in an accelerating coordinate system, with any arbitrary acceleration. This is what it means to say that time is separate from space in Newtonian mechanics.

The structure of Newtonian mechanics, however, includes more than the metric, it includes the acceleration law, that the forces are physical and determined by the configuration of positions and velocities. From this, you see that since only the constant velocity transformation preserves the acceleration, only the Galilean transformations are true symmetries.

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