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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

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    $\begingroup$ Why do you think the Lorentz boost "transforms" at all? It is, intrinsically, an element of a group, not a vector or tensor on the space, so why should it transform in any way? $\endgroup$
    – ACuriousMind
    Mar 17, 2015 at 22:34
  • $\begingroup$ A transformation that changes the metric only in one spatial dimensions seems strange. If you get a length contraction, I (naively) expect there to be some time dillation, too. $\endgroup$
    – user73762
    Mar 17, 2015 at 22:36
  • $\begingroup$ @ACuriousMind that's very interesting comment.I was treating a certain Lorentz boost which I can compute, to be a simple tensor of two indices. And so when I transform the coordinates with a certain Jacobean then then the boost matrix should also transform in a similar manner. is my understanding wrong? By the way this is a research question not a homework exercise :) $\endgroup$
    – physicist
    Mar 17, 2015 at 23:07
  • $\begingroup$ You were right, I talked too fast and supposed wrongly another transformation. Yours does preserve length, but I don't understand why you expect the transformed boost to have the same functionnal form as the boost when expressed relative to the new coordinates: it is not the same transformation. $\endgroup$ Mar 19, 2015 at 18:55
  • $\begingroup$ @G.Bergeron I thought boost transformation should transform just like any two indices tensor. I was wrong I think, as ACuriousMind explained. $\endgroup$
    – physicist
    Mar 19, 2015 at 21:37

1 Answer 1

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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 $\Lambda \mapsto \Lambda'$.

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