[Physics] Lorentz Transformations Vs Coordinate Transformations

coordinate systemsdifferential-geometrygeneral-relativitylorentz-symmetryspecial-relativity

I'm really confused about Lorentz transformations at the moment. In most books on QFT, Special Relativity or Electrodynamics, people talk about Lorentz transformations as some kind of special coordinate transformation that leaves the metric invariant and then they define what they call the Lorentz scalars. But from my point of view (which is somehow grounded in a background from differential geometry), scalars and the metric, which is a tensor, are invariant under any "good" coordinate transformation and that's a lot more than just Lorentz transformations, so I don't see why there's a special role for the Lorentz transformations in special relativity. Saying that the metric is invariant under Lorentz transformations is non-sense to me, because indeed it should be under any type of coordinate transformation if it's a well defined metric on a Minkowski manifold.

It seems to me that Lorentz transformations should be relating observers (frames) and not coordinate systems – that would make more sense to me, but usually people mix both concepts as if they were exactly the same. I'd like to understand what it means when one says that some scalar is Lorentz invariant. If someone could clarify me this conceptual confusion, I would be really grateful.

A general diffeomorhpism is not an isometry. Or rather, it can be made into an isometry. Consider smooth manifolds $M$ and $N$, with metrics $g$ and $h$. Let $\phi:M\rightarrow N$ be a diffeo. We say that $\phi$ is an isometry if $g=\phi^*h$.

But now, let's forget about $h$. We define it instead as $$h=(\phi^{-1})^*g.$$

Then $(M,g)$ and $(N,h)$ are automatically isometric as (semi-)Riemannian spaces.

With this said, consider $(M,g)$ to be Minkowski spacetime. Let $\phi:M\rightarrow M$ be a diffeo. Let $X,Y$ be vector fields. Obviously, it is true, that $$(\phi^{-1})^*g(\phi_*X,\phi_*Y)=g(X,Y),$$ so applying a diffeo to every object on the manifold will preserve relations. But is it true that $$g(\phi_*X,\phi_*Y)=g(X,Y)?$$ Or alternatively, $$(\phi^{-1})^*g(X,Y)=g(X,Y)?$$

No. In general it is not true. Those transformations for which $\phi^*g=g$ in Minkowski spacetime are PoincarĂ© transformations. The (homogenous) linear ones are Lorentz-transformations. This concludes my answer, but here is a (hopefully) illuminating aside.

Although this is in a slightly different context, here is an example, where the difference between isometries or general invertible & structure-preserving transformations make a difference:

Consider local Lorentzian geometry using local (possibly anholonomic) frames. What is the minimum information needed to give the local geometry exactly?

For coordinate frames: The metric components $g_{\mu\nu}$.

For completely general frames: The metric components $g_{ab}$, and the relationship between any one coordinate frame and the general frame, which is $e^\mu_a$ or $\theta^a_\mu$ ($\theta^a=\theta^a_\mu dx^\mu$, $e_a=e^\mu_a\partial_\mu$).

For orthonormal frames: The relationship between any one coordinate frame and the orthonormal frame. Why? Because if $\theta^a_\mu$ is given, then $g_{\mu\nu}=\eta_{ab}\theta^a_\mu\theta^b_\nu$.

So you can see, that despite the fact that all frames are just tools, and they don't have physical/geometric relevance, and thus all frames are equally good, specifying a frame and demanding it to be orthonormal actually gives a metric! There is valuable information content in the fact that a frame is orthonormal, and this info is lost if we use a general frame.

We can of course cast this notion in the language of transformation by noting that given an initial frame, a system of orthonormal frames can be constructed by demaning that two valid frames differ by a Lorentz transformation: $e_{a'}=\Lambda^a_{\ a'}e_a.$ So, despite the fact that any $GL(n,\mathbb{R})$-valued transformation is a good frame transformation, the Lorentz-valued transformations are special. A system of frames for which Lorentz-transforms allowed only specifies a metric uniquely. The associated statement in modern, invariant differential geometry would be that any reduction of the frame bundle $F(M)$'s $GL(n,\mathbb{R})$ into a generalized orthogonal group uniquely yields a semi-Riemannian metric.