# The role of Lorentz tranformations

My questions concern the role of Lorentz transformations in Special Relativity and General Relativity, as described in the following fragment of the series of GR lectures: https://www.youtube.com/watch?v=iFAxSEoj6Go&list=PLFeEvEPtX_0S6vxxiiNPrJbLu9aK1UVC_&index=14&t=0s#t=102m50s  (lecture 13 of International Winter School on Gravity and Light 2015 by Frederic Schuller, can be found also here: https://gravity-and-light.herokuapp.com/lectures)

In short, it says that:

1. The role of Lorentz transformations is exactly the same in SR and GR.
2. Namely, Lorentz transformations relate the frames of any two observers at the same point $$p \in M$$ and as such are the change of the basis of the tangent space at $$p$$, $$T_p M$$.

3. Therefore, it is conceptually wrong to think of them as acting on the points of the spacetime manifold $$M$$ as transforming $$x^\mu \to x'^\mu = \Lambda^\mu_\nu x^\nu$$.

Here are my questions:

1. Is there any physics textbook that follows consistently this way of thinking? People usually use $$x^\mu \to x'^\mu$$ as the formula for the Lorentz transformations without mentioning that this is in any way improper.

2. How to think about the Lorentz invariance of laws, e.g. of the Maxwell equations? It was historically an important observation that the Maxwell equations are not Galilei invariant but Lorentz invariant, which led to the construction of SR. But checking the invariance of the equations amounts to checking how the equations behave when we change $$x^\mu \to x'^\mu$$ $$-$$ at least this was always presented to me in this way.

3. The transformation $$x^\mu \to x'^\mu$$ also seems to be used in the derivation of Noether's theorems.

4. If Lorentz transformations take place in the tangent space and translations take place in the spacetime, then how does it make sense to talk about the Poincaré group that encompasses them all?

## 3 Answers

The local Lorentz transformation are acting on the tangent space $$T_p(M)$$ to the curved GR manifold at each point $$p$$. The idea of a "tangent space" is a formal way of ascribing a flat space to the neighbourhood of $$p$$ in which we are so close to $$p$$ that we don't notice the curvature. This is in the same way that when we draw of map of a town, we do not notice that the surface of the Earth is a sphere and not an infinite plane. The $$x^\mu$$ are coordinates in this neighbourhood. Each point $$p$$ has it's own neighbourhood with their origin at $$p$$. Although the neighbourhoods of nearby $$p$$'s will overlap, when we get sufficiently far away we can no longer maintain the convenient fiction that we are in a flat space.

The equivalence principle says that each point $$p\in M$$ has a sufficiently small neighbourhood in which we don't notice the curvature and so, for example, the flat space Maxwell equations can be used. These flat-space equations are Lorentz invariant, so each point has its own attached group of Lorentz tranformations that act on the local coordinates $$x^\mu$$ just as they do in SR.

• But tangent space is associated to one point, so how can I translate any manipulations on it to other points on spacetime? By assuming that they are associated with the same tangent space? Also, the lecturer stressed that tangent space is a vector space and spacetime is a manifold without the vector space structure - does it make any problem for such a translation? – wiktoria Sep 15 '19 at 19:09
• And what you said seems to be against the first claim of the lecturer that I listed, namely that the meaning of Lorentz transformations is the same in SR and GR. – wiktoria Sep 15 '19 at 19:15
• The flat tangent space is "soldered" to the curved manifold by the "solder form". There is a wikipedia page with that title about how it works. The only difference between SR and GR is that in the former you can take the neighbourhood of Minkowski space that is as being treated as a vector space as large as you like. In GR you are restricted to mapping only a small region of the curved manifold 1-1 onto the vector space. – mike stone Sep 15 '19 at 19:34

Therefore, it is conceptually wrong to think of [Lorentz transformations] as acting on the points of the spacetime manifold M

It's definitely wrong to apply the Lorentz transformation to coordinates on some arbitrary pseudo-Riemannian manifold, as the output will be meaningless. If the manifold is flat in some region, and your coordinates on that region are Minkowskian, then it's not wrong and it's sometimes useful.

If "conceptually wrong" means "pedagogically ill-advised" then I think it's conceptually wrong to base your understanding of special or general relativity on Lorentz transformations at all. We don't understand Euclidean geometry in terms of Cartesian coordinate transformations because we have an evolved intuitive sense of how it works that doesn't require coordinates. It's better to try to adapt that intuition to spacetime. As a consequence of its intrinsic structure, certain mappings of points to points in the Euclidean plane (resp. spacetime) take valid compass-and-straightedge constructions (resp. systems evolving in a way that's permitted by the laws of physics) to other valid constructions (resp. other valid histories). As a special case of that, if you define a certain type of coordinate system, and write one of your mappings in terms of those coordinates, it may have the form of a Cartesian/Lorentz transformation. But the universe doesn't care about coordinates or Lorentz transformations as such, just about the intrinsic structure of the thing that you're trying to mathematically describe.

Notably, the laws of physics appear to be local, and don't treat even a proton, much less a spaceship, as a single conceptual unit, so any Lorentz transformation that acts outside of a differential neighborhood is in some sense going beyond the scope of the laws of physics. Nonlocal Lorentz transformations work (when they do) only "by accident." But they're still useful, and you should still use them; you just shouldn't assume that the universe uses them.

People usually use $$x^\mu \to x'^\mu$$ as the formula for the Lorentz transformations without mentioning that this is in any way improper.

They're either talking about Minkowski coordinates on a flat region of spacetime or about a vector in a tangent space. They could plausibly use $$x$$ for either one.

How to think about the Lorentz invariance of laws, e.g. of the Maxwell equations?

It's really just rotational invariance. There are probably a lot of different ways that you could understand or formalize the symmetries of an arbitrary Riemannian manifold with an ordinary positive-definite metric, and all of those carry over to pseudo-Riemannian manifolds. The latter only seem more complicated because we don't have an evolved intuition for mixed signatures like we do for the +++ signature.

(Actually mixed signatures are theoretically more complicated in some ways – e.g. the group of rotations isn't compact – but I think that for the purposes of your question it doesn't matter.)

The transformation $$x^\mu \to x'^\mu$$ also seems to be used in the derivation of Noether's theorems.

I don't know anything about most of Noether's theorems, but the famous one called "Noether's theorem" doesn't depend on Lorentz symmetry; it works also in Newtonian mechanics for example.

If Lorentz transformations take place in the tangent space and translations take place in the spacetime, then how does it make sense to talk about the Poincaré group that encompasses them all?

In general it doesn't. The Poincaré group is the isometry group of Minkowski spacetime. A de Sitter or AdS or FLRW spacetime has a different isometry group. A realistic spacetime like large-scale FLRW with a bunch of randomly placed stars has no nontrivial isometries at all. Since the laws of physics are local, there is no physically meaningful difference between the highly symmetric spacetimes and the nonsymmetric ones.

• "Conceptually wrong" in the mathematical sense: according to the lecturer, Lorentz transformations act on the tangent space, not on the spacetime manifold, so applying them to the spacetime manifold would be mathematically meaningless. On the spacetime manifold act (all) diffeomorphisms and, the lecturer claims, there is no point in restricting them to the class of linear transformations, because always all diffeomorphisms are allowed. – wiktoria Sep 17 '19 at 16:59

General relativity has two symmetries, namely:

• Local Lorentz symmetry, which is more or less similar to Yang-Mills gauge symmetry. It's about transformation at the same space-time point and in general relativity the "gauge symmetry" is the local Lorentz symmetry. For example, the components of a spinor would change under local Lorentz transformation, even through the underlying coordinate stays put. Note that local Lorentz transformation does not change the metric $$g_{\mu\nu}$$.

• Diffeomorphism symmetry, which does involve space-time coordinate transformation $$x^\mu \to x'^\mu$$. For example, the components of a spinor would not change under diffeomorphism transformation, since a spinor is a zero-form and diffeomorphism transformation only changes the components of non-zero-forms. Note that diffeomorphism transformation does change the metric $$g_{\mu\nu}$$, since the metric tensor can be loosely regarded as the "multiplication" of two tetrad one-forms (see here).

In special relativity, the metric is fixed to Minkowskian metric, which effectively breaks both the Local Lorentz symmetry and the diffeomorphism symmetry. However, there is a residual symmetry, i.e. the global Lorentz symmetry, which combines the partial local Lorentz symmetry and the partial diffeomorphism symmetry. This combination explains why the Lorentz symmetry in special relativity involves both rotation in spinor space (leftover from local Lorentz symmetry) and rotation in coordinate $$x^\mu \to x'^\mu = \Lambda^\mu_\nu x^\nu$$ (leftover from diffeomorphism symmetry). For example, both the components of a spinor AND the underlying coordinate would change under the global Lorentz transformation.

The "combination" mentioned above is facilitated by choosing a uniform frame field (vierbein or tetrad), which effectively pegs the "Roman" index (pertaining to the Gamma matrices and local Lorentz symmetry) to the "Greek" index (pertaining to the p-forms and diffeomorphism symmetry).

• Why are you saying that in Special Relativity local Lorentz symmetry is only partial? Which part of it is valid and which is not? – wiktoria Sep 16 '19 at 17:19
• @wiktoria, in SR the Lorentz rotation has to be global, i.e. the rotation angle has to be the same at each spacetime point. The coordinate dependent rotation would necessitate a Lorentz/spin connection (which is absent in SR) for the proper definition of a covariant derivative. – MadMax Sep 16 '19 at 18:30