# Lorentz Transformations Vs Coordinate Transformations

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.

• "Lorentz transformations should be relating observers (frames) and not coordinate systems" is the argument made by Léon Brillouin in his 1970 book "Relativity Reexamined". Brillouin was strongly influenced by P.W Bridgman in this critique (e.g. (1962) "A sophisticates primer of Relativity".) Such arguments may lead one to reject space-time and to wonder if the "intrinsic" speed of light at an arbitrary point in space, is actually different somehow to the "measured" speed of light, made by moving a ref frame to the same arbitrary point in space. Discussion of this may be off topic here. – James Arathoon Sep 8 '17 at 1:38

I think the basic reason not to identify Lorentz transformations with general coordinate transformations(which is the Diff(M) symmetry in GR in the absence of a background spacetime symmetry and therefore it is considered a gauge symmetry)is the presence of a constant curvature background of which the Poincare group is the isometry group and the Lorentz transformations the subgroup of it that fixes the origin. I'm not sure what you mean by Special relativity not relating frames, that's actually what boosts do.

A lorentz scalar is just an invariant quantity constructed from other Lorentz invariant objects like 4-momentum or magnitudes of 4-vectors.

An observer is a wordline $$\gamma(\lambda)$$ with a choice of basis $$e_0(\lambda)$$,$$e_1(\lambda)$$,$$e_2(\lambda)$$ $$e_3(\lambda)$$ where these basis $$e_{\mu}(\lambda)$$ are basis to the tangent space ($$T_{\gamma(\lambda)}M$$) where the observer passes, such that: $$e_0(\lambda)=v_{\gamma}$$

where $$v_{\gamma}$$ are the tangent vector to the curve $$\gamma$$ and

$$g(e_{\mu},e_{\nu})=\eta_{\mu \nu}=\begin{bmatrix} 1 \\ & -1 & & & \\ & & -1 & & \\ & & & & -1 \\ \end{bmatrix}$$
Suppose we have to observes $$(\gamma,e_\mu)$$ and $$(\gamma',e'_\nu)$$ with $$\gamma(0)=\gamma'(0)$$ since both $$e_\mu$$ and $$e'_\nu$$ are basis of $$T_{\gamma(0)}M$$ we should have $$e_\mu=\Lambda^{\nu}_{\mu}e'_\nu$$ Now since $$g(e_\mu,e_\nu)=g(\Lambda^{a}_{\mu}e'_a,\Lambda^{b}_{\mu}e'_b)=\eta_{\mu \nu}$$ so $$\Lambda^{a}_{\mu}\Lambda^{b}_{\mu}\eta_{ab}=\eta_{\mu \nu}$$ That is $$\Lambda \in O(1,3)$$, in other words Lorentz transformations relates observers frames at the same point.

The confusion arises is most book is because of this.
If $$v \in T_{p}M$$ there is a unique geodesic $$\gamma$$ such that $$\gamma'(0)=v$$ and $$\gamma(0)=p$$

Let $$\gamma$$ be this curve, the exponential map is defined by
$$\exp_p :T_{p}M \to M$$ such that $$\exp_p(v)=\gamma(1)$$

So we associated each vector $$v \in T_{p}M$$ to a point $$q=\gamma(1) \in M$$. If $$M$$ is a Minkowski sapce , this map is one to one.

Now let $$e_{\mu}$$ be a orthonormal basis in $$T_{p}M$$ than $$v=x^{\mu}e_{\mu}$$ , so we can associated each point $$q \in M$$ the components of the vector $$v$$ in the basis $$e_\mu$$. This coordinates are called normal coordinates.

Since we have $$e_\mu=\Lambda^{\nu}_{\mu}e'_\nu$$ than $$v=x^{\mu}e_{\mu}=x^{\mu}\Lambda^{\nu}_{\mu}e'_\nu=x'^{\nu}e'_\nu$$.

It is not said in physics books ,but it is supposed that the coordinates associated with every observable are normal coordinates so if their coordinates transforms like this $$x'^{\nu}=x^{\mu}\Lambda^{\nu}_{\mu}$$ these observers are related by a Lorentz transformation

Because invariance is meant strictly, think isometry, rather than transforming by pullback, which physicists sometimes call covariance. In other words, invariance of the metric means that in an appropriately chosen basis both the metric components with respect to this basis and the basis elements themselves are invariant.

For an example consider the metric on $$R^2$$ with respect to global coordinates $$x,y$$, $$g=dx^2+dy^2$$. The components with respect to $$\{dx,dy\}$$ are $$g_{\mu\nu}=\delta_{\mu\nu}$$, $$\mu,\nu=1,2$$. Under the isometry $$x \to x+a$$, $$y\to y+b$$ both $$\{dx^\mu\}$$ and $$g_{\mu\nu}$$ are invariant.

Instead under the diffeomorphism $$x \to x+y$$, $$y \to x-y$$ we have $$dx \to dx^\prime=dx + dy$$, $$dy \to dy^\prime=dx - dy$$, $$g_{\mu\nu}\to g^\prime_{\mu\nu}=(1/2)\delta_{\mu\nu}$$ while of course $$g_{\mu\nu} dx^\mu dx^\nu=g^\prime_{\mu\nu} dx^{\prime \mu}dy^{\prime \mu}$$ because, as you say, the metric is a tensor, hence its value does not depend on the chosen coordinate system.