Showing that Lorentz transformations are the only ones possible

Blackforest

Ok, well-done and -explained (thanks). But all this concerns only special relativity. Where do you see that the question asked by the OP (and recalled by friend) is imposing linearity? For me it only imposes the Christoffel's work; see the other discussion "O-S model of star collapse" post 109, Foundations of the GTR by A. Einstein and translated by Bose, [793], (25). My impression (perhaps false) is that SR is based on a coherent but circular way of thinking including "linearity" for easy understandable historical reasons. The preservation of a length element (which is the initial question here) does not impose a flat geometry. Don't you think so?

TrickyDicky

I find this confusing, if you start by assuming a spacetime structure that admits bijections from ℝ4 into ℝ4 (that is E^4 or M^4) as the underlying structure because it seems natural to you, you are already imposing linearity for the transformations that respect the relativity principle. This leaves only the two posible transformations you comment below. The second postulate of SR is what allows us to pick which of the two is the right transformation.

But if you follow this path it is completely superfluous to prove anything about mapping straight lines to straight lines to get the most general transformation that does that and once you have it restrict it to the linear ones with a plausible physical assumption, since you are already starting with linear transformations.
Just a minor correction the Lorentz transformations are locally isomorphic to the restricted group.

Fredrik

How am I "already imposing linearity"? I'm starting with "takes straight lines to straight lines", because that is the obvious property of inertial coordinate transformations, and then I'm using the theorem to prove that (when spacetime is ℝ⁴) an inertial coordinate transformation is the composition of a linear map and a translation. I don't think linearity is obvious. It's just an algebraic condition with no obvious connection to the concept of inertial coordinate transformations.

Right, if we add that to our assumptions, we can eliminate the Galilean group as a possibility. But I would prefer to just say this: These are the two theories that are consistent with a) the idea that ℝ⁴ is the underlying set of "spacetime", and b) our interpretation of the principle of relativity as a set of mathematically precise statements about transformations between global inertial coordinate systems. Now that we have two theories, we can use experiments to determine which one of them makes the better predictions.

How is that a correction? It seems like an unrelated statement.

Blackforest

The experiment for the actual discussion here is the Morley and Michelson experiment.

Intuitively (I am not a specialist) this means that that isomorphism holds true only locally (on short distances around the observer). There is not really a global inertial coordinate system (except on the paper, in theory). And (as far I understand the generalized version of the theory) this is a crucial point. Among others things, this was forcing us (Weyl's work) to introduce the concept of parallel transport and of connection.

TrickyDicky

The assumption of a spacetime that is globally R^4(not just locally wich is the weaker asumption) means your underlying geometry is flat(Minkowskian, Euclidean), do you agree?
Given that space, the transformations that leave inertial coordinates invariant in the sense of SR first postulate must automatically be linear transformations, do you agree? Maybe this is not as obvious to see as I think, but I I think it is correct.

Well, It just seemed important to make more precise that the isomorphism you were talking about is local.

Fredrik

And pre-relativistic classical mechanics. It concerns all theories with ℝ⁴ as spacetime. I think it's pretty cool that there are only two such theories that are consistent with a straightforward interpretation of the principle of relativity.

Someone who tries to argue that a transformation that satisfies the OP's condition must be a Lorentz transformation has probably already assumed that spacetime is ℝ⁴, and that the theory will involve global (i.e. defined on all of spacetime) inertial coordinate systems. That a transformation between two global inertial coordinate systems is a bijection and takes straight lines to straight lines is just a consequence of the definition of "global inertial coordinate system". The 4-dimensional version of the theorem I stated and proved in #98 shows that a bijection that takes straight lines to straight lines is affine (i.e. a composition of a linear map and a translation). So when we begin to consider the OP's condition, it's already a matter of determining which affine maps satisfy it. And the condition implies that 0 is taken to 0, so there's no translation involved, i.e. the transformation is linear.

I don't think there's anything circular about it. It's perhaps naive to think that we should be able to use ℝ⁴ as our spacetime, and talk about global inertial coordinate systems. But it makes sense to first find all such theories, and then ask what other theories are worth considering. I might take a look at that problem when I have worked out all the details of the ℝ⁴ case.

Fredrik

I don't agree. We don't have a geometry at that stage, because until we have chosen an inner product (or something similar), ℝ⁴ is just a set. (And in the case of Galilean transformations, we will never define anything like an inner product on ℝ⁴). The lines that we call "straight" are straight in the Euclidean sense, but we're not considering them because they're straight in the Euclidean sense, but because they describe motion with a constant velocity. We don't need an inner product to see that they do.

They must automatically be affine maps, but it takes a non-trivial theorem* to see that, and you specifically said that there's no need to prove that theorem.

*) This theorem is essentially "the fundamental theorem of affine geometry", stated in terms of vector spaces instead of affine spaces.

But it's not. This is the 1+1-dimensional version of what I said, with all the details made explicit: For each K>0, the group ##G_K=\{\Lambda(v)|v\in (-c,c)\}##, where ##c=1/\sqrt{K}## and
$$\Lambda(v)=\frac{1}{\sqrt{1-Kv^2}}\begin{pmatrix}1 & -Kv\\ -v & 1\end{pmatrix}$$ is isomorphic to the restricted Lorentz group.

There's nothing local about this. In fact, when K=1, this group is the restricted Lorentz group, and the isomorphism is the identity map.

TrickyDicky

Sorry, aren't we asuming inner product spaces? how can we even talk about transformation matrices otherwise?



Well, it's not with your assumption of flat inner product space, but if you consider general manifolds the restricted Lorentz group is locally isomorphic to the Lorentz group.

Fredrik

I'm not even mentioning matrices until later in the argument, after I've determined that we're dealing with linear operators. To associate a matrix with a linear operator, we only need a basis.

TrickyDicky

You wanted to prove that linear transformations are the only ones possible if one wants use rigorously the first postulate of SR, you bring a R^4 vector space because you consider natural the assumption that the space must be globally R^4, not just locally like in general manifolds, and in this space you need to perform matrix multiplications like:##T(x)=\Lambda x## that looks like a matrix product to me so we are starting with an R^4 vector space with an inner product structure, no? That is called a Euclidean structure IMO.

Samshorn

Here's a web page that talks about how Einstein and others justified the linearity of the transformations, and the extra assumptions necessary to exclude linear fractional transformations: http://www.mathpages.com/home/kmath659/kmath659.htm

Fredrik

I'm not using the principle of relativity to prove that they're linear. The notation ##T(x)=\Lambda x+a## doesn't mean that ##\Lambda## is a matrix at this point. It only means that I'm using the standard convention to not write out parentheses when the map is known to be linear. We don't need an inner product to associate matrices with linear operators. We only need a basis for that. If U and V are vector spaces with bases ##A=\{u_i\}## and ##B=\{v_i\}## respectively, then the ij component of ##T:U\to V## with respect to the pair of bases (A,B) is defined as ##(Tu_j)_i##. The matrix associated with T (and the pair (A,B)) has ##(Tu_j)_i## (=the ith component of ##Tu_j##) on row i, column j.

* Spacetime is a structure with underlying set M.
* We intend to use curves in M to represent motion.
* There's a special set of curves in M that we can use to represent the motion of non-accelerating objects.
* M can be bijectively mapped onto ℝ⁴.
* A coordinate system on a subset ##U\subset M## is an injective map from U into ℝ⁴.
* A global coordinate system on M is a coordinate system with domain M.
* A global inertial coordinate system is a global coordinate system that takes the curves that represent non-accelerating motion to straight lines.
* If x and y are global coordinate systems, then ##x\circ y^{-1}## represents a change of coordinates. I call these functions coordinate transformations. When both x and y are global inertial coordinate systems, I call ##x\circ y^{-1}## an inertial coordinate transformation. (I'm getting tired of saying "global" all the time).
* These definitions imply that an inertial coordinate transformation is a bijection that takes straight lines to straight lines.
* The fundamental theorem of affine geometry tells us that this implies that inertial coordinate transformations are affine maps.
* This implies that an inertial coordinate transformation that takes 0 to 0 is linear.
* The principle of relativity tells us among other things that the set of inertial coordinate transformations is a group.
* This group has a subgroup G that consists of the proper and orthochronous inertial coordinate transformations that take 0 to 0.
* We interpret the principle of relativity as imposing a number of other conditions on G.
* Since the members of G are linear (we know this because they are affine and take 0 to 0), we can write an arbitrary member of G as a matrix. (This requires only a basis, not an inner product, and all vector spaces have a basis).
* The conditions inspired by the principle of relativity determine a bunch of relationships between the components of that matrix.
* Those relationships tell us that the group is either the restricted Galilean group without translations, or isomorphic to the restricted Lorentz group. (Restricted = proper and orthochronous).
* This implies that the group of all inertial coordinate transformations is either the Galilean group or the Poincaré group.
* We therefore define spacetime as a structure that has ℝ⁴ as the underlying set, and somehow singles out exactly one of these two groups as "special".
* A nice way to define a structure that singles out the Poincaré group is to define spacetime as the pair (ℝ⁴,g), where g is a Lorentzian metric whose isometry group is the Poincaré group.
* There's no equally nice way to handle the Galilean case. I think we either have to define spacetime as (ℝ⁴,G,g), where G is the Galilean group and G the metric on "space", or define it as a fiber bundle. (An ℝ³ bundle over ℝ, where each copy of ℝ³ is equipped with the Euclidean inner product). The former option is ugly. The latter is difficult to understand, unless you already understand fiber bundles of course.

friend

I'm given to understand that

dτ²=dt²-dx² = dt'²-dx'²

when (t',x') are the Lorentz transformation of (t,x).

Perhaps it's instructive to consider in what circumstances dτ should want to be considered invariant wrt to coordinate changes. Maybe those requirements are the driving force behind the necessity of the Lorentz transformations.

For example, the most obvious use of dτ is in the calculation of the line integral,

[tex]\int_{{\tau _0}}^\tau {d\tau '} = \tau - {\tau _0}[/tex]
which is the length of a line measured in terms of segments marked off along the length of the line. Then, of course, we can always place this line in an arbitrarily oriented coordinate system and express τ in term of those coordinates.

So the question is, when do we want to use the coordinates (t,x), and when would we want τ-τ₀ to be invariant wrt to those coordinates?

Usually, we specify a curve in space by parameterizing the space coordinates with an arbitrary variable, call it "t". But since the x and t coordinates are arbitrarily assigned, the length of the curve can depend on the (t,x) coordinates. But if you specify that the length of the curve is invariant, then this requires the Lorentz transformations between coordinate systems.

But what requires the length of the curve to be invariant? Perhaps if we have a more fundamental requirement like

[tex]\int_{{\tau _0}}^\tau {f(\tau - {\tau _0})d\tau } = a[/tex]
this will require the length of τ-τ₀ to be invariant wrt to coordinate changes in (t,x). For example, maybe [itex]{f(\tau - {\tau _0})}[/itex] might be a probability distribution along a path so that its integral along the path must be 1 in any coordinate system.

Did I get this all right? I would appreciate comments. Thank you.

micromass

Why do you think we need inner products to define matrix products??

TrickyDicky

No, it's not needed, I thought Fredrik was assuming Euclidean geometry but he wasn't.

Fredrik

I have some concerns about this part. Maybe there is some circularity in the argument after all. It doesn't seem obvious* that the "special" curves in spacetime that represent non-accelerated motion should include curves that correspond to infinite speed in some inertial coordinate system. If we leave them out, then what I call an inertial coordinate transformation will be a map that takes finite-speed straight lines to finite-speed straight lines. Of course, inertial coordinate transformations in SR (i.e. Poincaré transformations) can take infinite-speed lines to finite-speed lines and vice versa. If inertial coordinate transformations can't do this, there's no relativity of simultaneity. So if we leave out the infinite-speed lines from the start, we will come to the conclusion that there's only one possibility: The group is the Galilean group. (Hm, maybe there will actually be infinitely many possibilities, distinguished by what exactly they're doing to infinite-speed lines).

Do we have a reason to include infinite-speed lines other than that we know what we want the final answer to be?

*) Recall that the main reason why we need spacetime to include that special set of curves is that they (or at least some of them) are to represent the motions of "observers" that are minimally disturbed by what's being done to them. (An "observer" here is not necessarily conscious. It could be a measuring device).

strangerep

This sort of thing is one reason why I prefer to start from inertial observers defined as those that feel no acceleration. If one finds the maximal dynamical group applicable to the zero-acceleration equations of motion, the problematic case you mentioned can be handled by taking a limit afterwards.