I Precausality and continuity in 1-postulate derivations of SR

[Enrico]

When constructing mathematical models in theoretical physics, one chooses the math framework(s) that seem best suited to the task. If, one day, some experimental fault or inconsistency emerges than shows Lie groups to be inadequate for current purposes, no doubt someone somewhere will think up something else.

Of course. My aim is just to try to strip down the initial hypotheses as much as possible. That makes the theory more solid, right?

Refusing to knuckle down and do this is one of the most reliable marks of a crackpot.

I'm not refusing to do it, group theory is one of the topics I'd like to study systematically. I had started a few years ago, but life is made by many things, eventually I have devoted my free time to other hobbies. It's a matter of equilibrium, after the whole working day in an office it's not that easy to sit down and study maths. Anyway, the conversation here is a pleasant occasion to share my thoughts with others (rather than being isolated on my own) and get feedback, and I'm trying to get a feeling about topics commonly needed in mainstream theory and their relevance, before - hopefully - facing them systematically.

 

[jbergman]

As I mentioned in my first post, $v$ is the relative velocity between the two frames of reference that we are transforming, so yes, it's a slope.

I think (from some computations) that the continuity hypothesis may be dropped if we assume from the outset that the group be commutative. Does anybody have a justification for commutativity based on the Relativity Principle?

On the other hand, I'd also like to understand whether commutativity is really needed. I tried quite hard to do without it, but such an approach seems unlikely to work. At this point, would be nice to have a counterexample of a non-commutative group obeying the other assumptions of the 1-postulate derivations (and not satisfying continuity in $v$, necessarily).

I'm not sure what you mean by commutative in this context.

The restricted Lorentz group $SO^{+}(1,3)$ is a non-commutative group.

 

[jbergman]

And what about the choice of $v$ as a parameter (in the case with one single spatial coordinate)? I mean, the assumption that, once $v$ be given, the transformation must be uniquely determined. May it be that this assumption automatically implies continuity and precausality, i.e. restricts the group to $SO^\uparrow\left(1,3\right)$?

The single dimension case is much simpler than the 3 dimensional case. For instance, rotations in 1 and 2 dimensions, (SO(1), and SO(2)) form commutative groups but not in SO(3) for 3 dimensions does not.

 

[Enrico]

The single dimension case is much simpler than the 3 dimensional case. For instance, rotations in 1 and 2 dimensions, (SO(1), and SO(2)) form commutative groups but not in SO(3) for 3 dimensions does not.

Yes I was only considering the 1-dimensional case. I have mixed up symbols quite a bit, sorry. This is where my present lack of knowledge in group theory shows up clearly: I'm not sure what the symbol corresponding to $\mathrm{SO}^\uparrow(1,3)$ in one dimension would be.

 

[jbergman]

Yes I was only considering the 1-dimensional case. I have mixed up symbols quite a bit, sorry. This is where my present lack of knowledge in group theory shows up clearly: I'm not sure what the symbol corresponding to $\mathrm{SO}^\uparrow(1,3)$ in one dimension would be.

With one spatial dimension and one time dimension the that would be $\mathrm{SO}^\uparrow(1,1)$.

But, you can't draw conclusions about what happens in four dimensional space time from 2-dimensional space time because as I said commutativity in one case doesn't imply commutativity in the full space so I'm not sure how it helps your argument.

 

[strangerep]

The mention of "commutativity" in the context of this thread arose from the fact that a 1-parameter Lie group is necessarily commutative. Here, it's the group of velocity boosts along an arbitrary, but fixed, direction.

In group-theoretic (1-postulate) derivations of SR, one can simplify some of the computations by using $B(v_1) B(v_2) = B(v_2) B(v_1)\,$, where $B(v)$ denotes a boost transformation along the chosen fixed direction with parameter $v$.

 

[jbergman]

The mention of "commutativity" in the context of this thread arose from the fact that a 1-parameter Lie group is necessarily commutative. Here, it's the group of velocity boosts along an arbitrary, but fixed, direction.

In group-theoretic (1-postulate) derivations of SR, one can simplify some of the computations by using $B(v_1) B(v_2) = B(v_2) B(v_1)\,$, where $B(v)$ denotes a boost transformation along the chosen fixed direction with parameter $v$.

Right. But I thought Enrico was implying that the transformation group for space time is $SO^{+}(1,3)$ because of this commutativity of the 1-parameter subgroups. I don't quite see the connection.

 

[strangerep]

Right. But I thought Enrico was implying that the transformation group for space time is $SO^{+}(1,3)$ because of this commutativity of the 1-parameter subgroups.

I don't think that's what Enrico was implying. (For a bit more context see posts #5, #6 in this thread.) But I should probably let Enrico speak for himself.

 

[Enrico]

I don't think that's what Enrico was implying. (For a bit more context see posts #5, #6 in this thread.) But I should probably let Enrico speak for himself.

As I said, I'm only considering 1 spatial dimension. I did not mean to extend the argument to 3D.

 

[jbergman]

As I said, I'm only considering 1 spatial dimension. I did not mean to extend the argument to 3D.

I finally read the Berzi and Gorini paper so have better context. In the last section of the paper they rule out the rotation group because of precausality.

The question of whether or not commutativity implies continuity seems a bit circular. Because we inferred commutativity from the fact that the image of a one-parameter group is commutative, which is a continuous maps on to a group of transformations.

If you throw out the fact that $f(v)$ is continuous then you have to prove commutativity directly from the principle of relativity, i.e., that $f(v_1)f(v_2)=f(v_2)f(v_1)$.

Continuity seems like a reasonable assumption to me, just as reasonable as homogeneity and isotropy of space. It just says that $f(v_1)$ is near $f(v_2)$ if $v_1$ is near $v_2$.

 

[Enrico]

The question of whether or not commutativity implies continuity seems a bit circular. Because we inferred commutativity from the fact that the image of a one-parameter group is commutative, which is a continuous maps on to a group of transformations.

If you throw out the fact that f(v) is continuous then you have to prove commutativity directly from the principle of relativity, i.e., that f(v1)f(v2)=f(v2)f(v1).

This is my understanding as well.

Continuity seems like a reasonable assumption to me, just as reasonable as homogeneity and isotropy of space. It just says that f(v1) is near f(v2) if v1 is near v2.

Of course it is reasonable. I'm just wondering whether it may be stripped down further. At the moment my opinion is that some continuity assumption is needed. First, in order to prove linearity, continuity at the origin is needed on the transformation formulae, at any fixed $v$ (please see the first part of the Berzi & Gorini paper). Second, it seems to me that continuity at the origin is necessary for the transformation coefficients, as functions of $v$. Maybe it is sufficient to impose continuity (at the origin) on $\gamma$ only, but I'm not sure of that yet.

In any case, I'm going to put the subject aside for the moment: I think it be proper for me to deepen my knowledge in group theory now, in order to be at least aware of the correct concepts and terminology. For what regards precausality, I'm still convinced that it is not needed in the end.

 

[vanhees71]

Indeed, only the subgroup that is continuously (or here in the case of Lie groups even smoothly) connected with the identity is necessarily a symmetry group for any closed system, if the physical laws should be consistent with the underlying spacetime structure. The "disconnected parts" can still be symmetries. In Nature that's only partially the case since the weak interaction violates parity and time reversal.

In the quantum realm, i.e., in local relativistic QFT you have in addition also charge conjugation, i.e., that for each particle there's also its antiparticle (in the case of strictly neutral ones both are the same). From the symmetry under $\mathrm{ISO}(1,3)^{\uparrow}$, i.e., the proper orthochronous Poincare group and locality (microcausality) it follows that CPT must also be a symmetry. The weak interaction fufills this symmetry since it's described by such a local QFT, but it breaks all the other "discrete" symmetries, i.e., P, T, CP, and CT. All this symmetry breakings are experimentally verified.