How does the Lorentz boost change if we introduce transformation to the minkowski metric

Question

Let's say we have the Lorentz boost given by the $ \Lambda^\mu_\nu$ in the Minkowski metric $diag\{1,-1,-1,-1\}$. Now if I do a transformation on the Minkowski metric such that the new metric is $diag\{1,-1,-1,-1/\tau^2\}$. For this transformation I know the Jacobian $\frac{\partial \widetilde{x}^\alpha}{\partial x^\mu}$.

So will the Lorentz boost in the new coordinate system be given by -->

$\widetilde{\Lambda}^\alpha_\beta = \frac{\partial \widetilde{x}^\alpha}{\partial x^\mu} \frac{\partial x^\nu}{\partial \widetilde{x}^\beta}\Lambda^\mu_\nu $

Because when I do the above transformation on the Minkowski boost and then compare it with the boost by definition in the transformed coordinates, they don't match?

Please see these links for more information about this transformation .. I uploaded the image instead of writing it all here.. About the metric and About the transformation

Answer 1

Although the Lorentz boosts have "tensor indices", they are not defined as tensors on the space, but as elements of the isometry group.

Let $M,M'$ be two (pseudo-)Riemannian manifolds with metric tensors $g,g'$.

Given a diffeomorphism $\phi : M\to M'$, you could let the isometry $\Lambda : M\to M$ (which, on $\mathbb{R}^{1,3}$ would be given by $\Lambda(x)^\mu = \Lambda^\mu_\nu x^\nu + a^\nu$ for some translation vector $a$ - recall that the isometry group of Minkowski space is the Poincare group) transform into $\Lambda' := \phi\circ\Lambda\circ\phi^{-1} : M'\to M'$, whose map on the tangent spaces would indeed be given by your formula.

If $\phi$ is an isometry in the sense that $g(v,w) = g'(\mathrm{d}\phi(v),\mathrm{d}\phi(w))$, then $\Lambda'$ is an isometry of $M'$, so this provides also an isomorphism of the isometry groups of $M$ and $M'$ by $\Lambda \mapsto \Lambda'$.