
051012, 02:45 AM #1
If spacetime is modeled as the torsor of translations, rotations and changes of the
Thinking back to this old theme of mine: http://www.sciforums.com/showthread....56#post2039656
I was trying to get back into the fundamental mathematical framework of relativity (Galilean and Special).

If spacetime is modeled as the torsor of translations, rotations and changes of the standard of rest that preserve the law of inertia, what corresponds to the Lorentz transform?
The question is a little illformed. Questions/comments welcome.
If translations is the Lie group then spacetime is 4dimensional and every pair of events has an associated element that translates between them.
The law of inertia requires existence of inertial paths, "straight" lines through spacetime where is constant in . So for any unit time, , we have the event separation that is naturally associated with an element of the translations.
For any two velocities, there should be an element of the rotations (say SO(3)) that makes them parallel. What is needed to complete the picture is a way to "translate" between velocities that are parallel. Some choices are SO(1,1), and Galilean boosts.
Questions. Is SO(3,1) a product of SO(3) and SO(1,1) ?
Is this effort doomed due to misunderstanding?

051012, 04:33 PM #2
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 2,395
IMHO you need to go more fundamental than that, rpenner. The mathematical framework is modelling reality, but in itself it isn't fundamental.
I'll assume we're talking about a simple boost in the x direction for now: it's a trigonometric tilt in a mathematical space. You can call it a rotation, but that doesn't get to the bottom of it. A Lorentz transform occurs when you change your state of motion, which alters the results of your use of motion to measure space and time. To appreciate this, set the mathematics aside for a moment, and get down to the fundamentals of light moving through space. Remember that clocks "clock up" local motion and show you some cumulative display you call the time. Now take a look at the wellworn expression for a spacetime interval in flat Minkowski spacetime:
It's related to Pythagoras' theorem, used in the Simple inference of time dilation due to relative velocity. We've got two parallelmirror light clocks, one in front of me, the other which we've sent with you on an outandback trip. I see the light moving like this ǁ in my local clock and like this /\ in the moving clock. Treat one side of the angled path as a rightangled triangle, and the hypotenuse is the lightpath where c=1 in natural units, the base is the speed v as a fraction of c, and the height gives the Lorentz factor γ = 1/√(1v²/c²). Notice there's no literal time flowing in these clocks, merely light moving at a uniform rate through the space of our SR universe. From which we plot straight worldlines through the abstract mathematical space we call Minkowski spacetime. Also note that the underlying reality behind the invariant spacetime interval between the start and end events is that the two lightpath lengths are the same. Macroscopic motion comes at the cost of a reduced local rate of motion, because the total rate of motion is c. Hence the minus in front of the t. It really is that simple. The Lorentz transform occurs when you change x from zero to some nonzero positive value. Your worldline is initially vertical, it ends up tilted. But it is restricted, it can only tilt so far. That's when the hypotenuse flatlines and your worldline is at 45 degrees.
It's a lightpath length. The light can take many paths, but all the path lengths are the same. It doesn't matter how you move through space, our departure is event 1 and our subsequent meeting is event 2, and the total light path length between your parallel mirrors is the same as between mine. Think of events as collisions and you can't go far wrong.
Forget about the law of inertia. I'll tell you about inertia another day. There's a symmetry to it that for some weird reason just isn't in the textbooks, as if E=mc² never happened. Bizarre.
No, you just have to start from the bottom, and make sure you understand the xboost situation and tilting your worldline before you start swivelling it too. Understanding the mathematics isn't enough. You have to understand the why of it.

051012, 08:35 PM #3
Clearly you haven't understood the intent or the language. When I say rotations, I am talking about rotations, as in Euclidean rotations in 3 spatial dimensions.
The law of inertia becomes the law of straight lines in the model. This part, I think I understand is that for a given line, an element of the Lie group of translations transforms the line into itself.

051112, 06:56 AM #4
Watch out, Farsight is going to dazzle us with his grasp of the aforementioned mathematical framework!
Still have all the answers but can't reply to any questions, aye Farsight?
Because you're so familiar with all those textbooks, right? Come on Farsight, do you really think anyone is going to buy the "I'm well read in the physics literature!" front you're trying to put up?
But you don't understand the maths and your 'whys' are just baseless assertions based on personal preference and, quite frankly, ignorance.

051112, 03:50 PM #5
If is the model of spacetime, and is the set of all straight lines in it, and we accept as axiomatic that all straight lines in admit at least two distinct points as members, that any two points uniquely identify an element of the translation group, T, and that all points along a straight line may be obtained as images of a point by a suitable translation in the same direction, then it follows that any line is associated with a subgroup of T for which the line is invariant.
This is a generalization of "straight" for T could be SO(3), could be an ordinary 2sphere, and so L would be the set of great circles. Almost certainly we want T to be a real Lie group.
But right now, I seem to be in a definition muddle. If is the model of spacetime, does it follow that any follow that any point is the image of a unique translation of another point? There are assumptions that a line can be extended indefinitely in both directions, and that a straight line in the model corresponds to uniform motion in spacetime. Should I nail T down to be a connected real Lie group, or will it be sufficient for my purposes to suppose merely that it is a group. I think I need to sketch out a map of where I am going with this and in my next post attempt to start from scratch again.

051312, 01:36 AM #6
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 2,395
rpenner: make sure you read Light is heavy by van der Mark and 't Hooft. (That's not the 't Hooft by the way). The paper will help you to understand inertia. Then read The Other Meaning of Special Relativity by Robert Close. This will help you to understand the underlying physics. It's rather trivial actually, schoolboy stuff. Once you have this understanding, you won't be "lost in maths" any more.

051312, 02:14 AM #7
Because it's helped you so much.
Except that it isn't just getting 'lost in maths', the sorts of mathematical things Rpenner is mentioning are central constructs within much of mathematical physics and provide a precise and concrete formal description of structures seen in nature.
For example, Lie groups underline symmetries and thus allow us to compute conserved quantities. The same groups describe how particles should interact with one another and themselves and not just in a "Gluons interact with gluons but photons don't interact with photons". Instead you can compute actual behaviours of particles to a very accurate level which can then be tested.
And making precise predictions is an important part of physics, as you keep whining about when it comes to string theory. The irony of you saying such things, along with throwing out comments like "I'll explain inertia to you one day", while simultaneously being unable to provide a single quantitative prediction is not lost on other people.
But feel free to show I'm mistaken in my assessment of your claims. If you can 'explain inertia' and what not perhaps you could provide everyone with a working model of some phenomena pertaining to inertia? Perhaps you'd like to derive conservation of linear momentum from your 'explanation'? After all, the use of Poincare invariance in the mathematical formalism Rpenner is talking about, which you consider to be 'getting lost in maths' provides such a prediction from first principles. So before you start giving advice to others perhaps you could show you're at least on their level.

051512, 01:09 PM #8

051512, 01:30 PM #9
Actually, I was hoping to be able the answer the question myself, but I'm scraping the limits of my understanding and will probably have to grow in knowledge to complete it. My algebra and physics books are weak on Lie algebras and Lie groups.
Right now, I'm favoring:
"If spacetime is modeled as the torsor of the group generated by the Lie algebra of Euclidean translations, Euclidean rotations and changes of the standard of rest that preserve the law of inertia and generally preserve the straightness of lines, what corresponds to the Lorentz transform?"
What I think you wind up with is a test theory of a local Galilean or Lorentzian manifold and then you bracket it with experimental results to show that a local Lorentzian manifold is the only physical result. However,
 If I can prove the straightness of lines follows from the law of inertia, I can drop that assumption.
 I need to get a better handle on Lie algebra notation, so that I can talk about the products of the Lie algebras corrsponding to R, R^3, SO(3) and the "change of the standard of rest"
 I think if I drop Euclidean from translations and rotations, the preservation of straight lines (which is defined in terms of translations) might not be strong enough. But there's a chance that dropping Euclidean will be interesting on its own

051512, 03:02 PM #10
If is a model of spacetime, and is the set of all straight lines in that model, such that if is an element of an indexed set of generators for the algebra of translations, that there exists some nonzero linear combination of generators associated with the line such that and .
Last edited by rpenner; 051512 at 04:08 PM. Reason: mathbb > mathcal

051512, 04:19 PM #11
For all points in a line, like x in L, the nonzero translation algebra member t associated with the line transforms x into a point in the line, but any of the generators which is linearly independent of this t associated with the lines transforms x into a point off the line.
Sorry for the notation, but I am still working with elements of a basis and don't have a way to express some of my thoughts without subscripts at this time.

051612, 01:50 AM #12
.... OK, you know you are doing it wrong when you prove isn't a Lie algebra. ... grumble.

051712, 07:16 AM #13
Feel free to start another thread Farsight. I'm looking forward to learning a bit about curvature now.
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