Lorentz Transformations Vs Coordinate Transformations

Question

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.

Answer 1

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.

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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 Poincaré transformations. The (homogenous) linear ones are Lorentz-transformations. This concludes my answer, but here is a (hopefully) illuminating aside.

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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.

Answer 2

You are, I guess, discussing special relativity. In that case, the most natural geometrisation is to postulate that spacetime $\mathcal{S}$ is an real affine space of dimension 4 with a quadratic form of signature $(+,-,-,-)$. Everything follows, among which the two fundamental aspects to study:

• the group $\mathcal{P}$ of affine transforms leaving the quadratic form invariant (called the Poincaré group by physicists);
• there is an infinite family of frames where the quadratic form has a matrix $\mathrm{diag}(+1,-1,-1,-1)$. Any two such frames are mapped onto each others by an element of $\mathcal{P}$.

The first point is about what physicists would call an active transform whereas the second one would be called a passive transform.

You asked about Lorentz transforms: as always for a group of affine transforms, $\mathcal{P}$ is the semi-direct product of the subgroup of translations and of a group $\mathcal{L}$ of linear transforms on the vector space $S$ associated with $\mathcal{S}$. Then $\mathcal{L}$ is called the Lorentz group.

Note, to finish, if it was not totally obvious, that this is totally similar to affine Euclidean geometry and isometries: the only difference is the signature of the quadratic form, which is positive definite, and of course that the dimension is 3 and not 4.

Answer 3

First off: not all coordinate transformations preserve the metric. As a simple example, consider $\mathbb R^2$ transformed under \begin{align} x'&=x+y\\ y'&=y, \tag 1 \end{align} which does not preserve the usual norm's diagonal structure. Whenever we're interested in some geometric structure, typically there will be some transformations that respect it and some transformations that don't, and we limit our interest to the former. For euclidean space and linear transformations, that's what orthogonal transformations do.

Lorentz transformations are the direct analogue for the Minkowski metric: there's plenty of transformations that don't respect it (like the Galilean frame transformations, that look exactly like $(1)$ above) and a restricted set of "good" transformations that do respect the metric. The set of the latter is, by definition, the set of Lorentz transformations, and it's as crucial a tool as orthogonal transformations are for the study of euclidean space.

Answer 4

Looking at your question i guess i have a simple answer.if any two observers are in a specific type of coordinate frame (cartesian polar.....e.t.c)and they want to know the energy momentum position and velocity then they will use lorentz transform to find out each others position velocity energy momentum.but if one observer is in cartesian frame and the other is in polar then they must take also the coordinate transform too from polar to cartesian or vice versa.we often confuse between coordinate frame and reference frame.there is a subtle difference.coordinate frame such as cartesian polar cylindical system.but reference frame is observers frame.we can quantify reference frame by using any type of coordinate frame

Answer 5

What you are forgetting is that in Minkowski space time, there is a kind of special coordinates ,the so called observer coordinates or normal coordinates in differential geometry and we can show, that, there is a one to one correspondence between this coordinates and an inertial observer. The point here is to have conscience that components of tensors in this special coordinates, are equal to what the observer associated with this coordinates, measures. So changing this special coordinates is the same as changing observers.

As example suppose $F=F_{\mu \nu}dx^{\mu}dx^{\nu}$ is the electromagnetic tensor in one of this special coordinates associated with some observer $O$. Then we know that for example $E_1=F_{11}$ is the first component of the electric field. If we change coordinates we have $F=F_{\mu \nu}dx^{\mu}dx^{\nu}=F'_{\mu \nu}dx'^{\mu}dx'^{\nu}$. Now if $F'_{\mu \nu}$ obtained by Lorentz coordinates transformation we know that another observer $O'$ associated with this coordinates, will measure the first component of the electric field as $E'=F'_{1 1}$.

So instead of using all the machinery of differential geometry, physicists uses this coordinates approach which is more easy.