3
$\begingroup$

Why only the subgroup of orthochronous and proper Lorentz transformations (i.e. those simultaneously satisfying $\Lambda_{\;0}^{0}> 0$ and $\text{det}\,\Lambda =1$, respectively) are considered to be physically realisable transformations?

Is it simply because by Lorentz symmetry, all Lorentz transformations must be orthochronous, since if they were not, then one could distinguish between two different inertial frames (time would be running in the opposite sense in one frame relative to another)?

Assuming the above statement is true, then, since all orthochronous Lorentz transformations are continuously connected to the identity, it follows that the physically realizable Lorentz transformations must also be proper, since improper, orthochronous Lorentz transformations cannot be continuously connected to the identity.

$\endgroup$
  • $\begingroup$ $\lim_{\epsilon \to 0} 1+\epsilon = 1$, whilst $\lim_{\epsilon \to 0} -1+\epsilon \neq 1$. That's how they're continuously connected to the identity. $\endgroup$ – QuantumBrick Oct 10 '16 at 11:44
  • $\begingroup$ @QuantumBrick Is $\epsilon$ an infinitesimal change in the time coordinate? Also, is the reasoning I gave for why we only consider proper, orthochronus Lorentz transformations correct? $\endgroup$ – user35305 Oct 10 '16 at 11:50
  • $\begingroup$ $\epsilon$ is just what you've said. The answer given clarifies why proper ortochronous transformations are not the only physically interesting ones. $\endgroup$ – QuantumBrick Oct 10 '16 at 12:16
  • $\begingroup$ Hi user35305: I removed your last subquestion, cf. this meta post. $\endgroup$ – Qmechanic Oct 10 '16 at 13:07
1
$\begingroup$

The full Lorentz group features multiple branches, between which no continuous transformations can interpolate. They are related through discrete transformations though. The restriction to proper orthochronous Lorentz transformations ensures that the transformations we deal with are a continuous branch of a Lie group, for which the representation theory is easier.

Note that the transformations used to restrict the Lorentz group are not always realised in the system. Time reversal and parity are violated in the Standard Model, for example. But if one keeps these subtleties in mind, the restriction is w.l.o.g. (with which I mean that the objects in our theory have to be (not necessarily trivial) representations under T and P).

$\endgroup$
  • $\begingroup$ Is there any physical motivation for why we stick to proper orthochronus Lorentz transformations? $\endgroup$ – user35305 Oct 10 '16 at 12:29
  • $\begingroup$ @user35303 Yes. This is the brach that includes the identity, and can therefore be treated as a Lie group. $\endgroup$ – Neuneck Oct 11 '16 at 10:04
  • $\begingroup$ Ah ok, so is it the case that if we want a continuous symmetry connected to the identity (such the trivial case in which the Lorentz transformation maps back to the same inertial frame is included) and therefore a Lie group, we must consider the branch that is continuously connected to the the identity, i.e. the proper orthochronus branch?! $\endgroup$ – user35305 Oct 11 '16 at 10:53
  • $\begingroup$ @user35305 exactly! $\endgroup$ – Neuneck Oct 11 '16 at 13:41
0
$\begingroup$

I wouldn't say that orthochronous proper Lorentz transformations are the only physically realizable. The rest of the components of the Lorentz group can be reached via parity or/and time reversal operating on the orthochronus proper Lorentz subgroup, and we know that these symmeties exist in our universe.

A different thing is that because of this relation between the different components of the Lorentz group, we can limit the study of the Lorentz group to the Lorentz orthochronous proper subgroup.

$\endgroup$
  • $\begingroup$ Why is it so often the case that analysis is restricted to the proper orthochronus subgroup? $\endgroup$ – user35305 Oct 10 '16 at 12:31
  • $\begingroup$ It's what I say in the last paragraph. The different components of the Lorentz group are the same up to a discrete transformaton ($P$ or $T$), so one only needs to study one of them to know the properties of the other ones. $\endgroup$ – dpravos Oct 10 '16 at 12:35
  • $\begingroup$ Ah ok. The reason why I argued as I did in my OP for why one considers proper orthochronus Lorentz transformations was due to reading this: books.google.co.uk/…. It seems like a reasonable physical requirement from observations... $\endgroup$ – user35305 Oct 10 '16 at 13:37

protected by Qmechanic Oct 10 '16 at 13:10

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.