Is this definition of Lorentz transformations correct?

Consider 3+1 dimensional space-time manifold $M$. Let $v,u$ are two vectors of the vector bundle and $g$ be a metric on $M$.

Now we can define the inner product in this vector bundle $(u,v)=u^{\mu}v_{\mu}=g_{\mu\nu}u^{\mu}v^{\nu}$. Under any coordinate transformation, this inner product is invariant(because (u,v) is a scalar and also because the transformation of $g,u,v$ cancels properly).

Let's say now, we do a coordinate transform from $x$ to $x^{'}$ (this is just selecting another chart according to my understanding). Then $(u,v)=g_{\mu\nu}u^{\mu}v^{\nu}=g_{\mu'\nu'}u^{\mu'}v^{\nu'}=g_{\mu'\nu'}u^{\mu}v^{\nu}\Lambda^{\mu'}_{\mu}\Lambda^{\nu'}_{\nu}$

This is true for any $u,v$ so, $g_{\mu\nu}=g_{\mu'\nu'}\Lambda^{\mu'}_{\mu}\Lambda^{\nu'}_{\nu}$. Here $\Lambda^{\nu'}_{\nu}=\frac{\partial x^{\mu'}}{\partial x^{\mu}}$.

Now if we need to find $\Lambda$ that does not change components of g as a set, which means ${g_{00}, g_{01}, ..., g_{33}}={g_{0'0'}, g_{0'1'}, ..., g_{3'3'}}$, we get the Lorentz transformations. Which is group $O(1,3)$ in this case.

In brief, Lorentz transformations are coordinate transformations that do not change the components of the metric (or we can say line element $ds^2$ looks the same).

Could someone comment on the accuracy of the above description?


1 Answer 1


This is not the case. There are no such transformations which in general leave the components of an arbitrary metric unchanged (except the trivial one such that $\Lambda^{\mu}{}_{\nu} = \delta^\mu_{\nu}$). Lorenz transformation do not satisfy the restrictions you impose. As a simple example, take the FLRW metric and apply a Lorentz transformation. You will easily see the components are not the same.

Perhaps you mean to be talking about Special Relativity, where the metric is fixed to be the Minkowski metric and the spacetime in flat?

  • $\begingroup$ You can assume a flat metric without loss of generality(because we are considering a point in space-time). Since g is symmetric, it is diagonalizable with a suitable basis. Under such a basis, the matric becomes locally flat. $\endgroup$
    – htr
    Mar 15, 2023 at 17:57
  • $\begingroup$ No you cannot, unless you instead mean to talk about local Lorentz transformations acting in the tangent space. If so, you should probably edit the question. $\endgroup$
    – Eletie
    Mar 15, 2023 at 17:58
  • $\begingroup$ This long thread contains a few nice answers that demonstrate that in flat spacetime the Lorentz transformations are the only ones that preserve the metric $ds^2\,.$ I agree with Eletie that when matter is involved this is not the case. An interesting example of a metric in flat spacetime that looks a bit different than the one we know is the Rindler metric. It could be a nice exercise to deduce the Lorentz transformations from the invariance of that. $\endgroup$
    – Kurt G.
    Mar 15, 2023 at 18:03
  • $\begingroup$ @Eletie, the Chart transition maps that I mentioned are local right? $\endgroup$
    – htr
    Mar 15, 2023 at 18:04
  • $\begingroup$ The chart transition maps are not local in the sense of a singular point, they transform patches on manifold, while flatness only holds (generally) at a singular point after correct choice of coordinates $\endgroup$ Mar 15, 2023 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.