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

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?

• 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? – ACuriousMind Mar 17 '15 at 22:34
• 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. – pyramids Mar 17 '15 at 22:36
• @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 :) – physicist Mar 17 '15 at 23:07
• 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. – G. Bergeron Mar 19 '15 at 18:55
• @G.Bergeron I thought boost transformation should transform just like any two indices tensor. I was wrong I think, as ACuriousMind explained. – physicist Mar 19 '15 at 21:37

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 $\Lambda \mapsto \Lambda'$.