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I am developing a new derivation of the Lorentz transformation which I think and hope is more attractive to students than those I have seen in currently available texts. I am carefully defining and discussing the important concepts of homogeneity and isotropy of space.

My question is this: am I justified in assuming that a transformation between inertial reference frames cannot reverse handedness? I understand that parity is not conserved in all particle interactions, but my question concerns two relatively moving observers. Can one of them distinguish that the other has adopted the oppositely handed coordinate frame? References would be appreciated.

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  • $\begingroup$ :-DDD the title seems to ask if The Reference Frame is left-winded or right-winded. @Lubos Motl $\endgroup$
    – arivero
    Commented Sep 21, 2014 at 22:59

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[...] my question concerns two relatively moving observers. Can one of them distinguish that the other has adopted the oppositely handed coordinate frame?

What information does observer A have about observer B? If all A knows about B is B's state of motion, and if he assumes that B is going to choose coordinates in which he is at rest at the origin, then there is a multiparameter family of possible coordinate systems that A could impute to B. These could differ by rotation, parity, and time-reversal. If A assumes that the psychological arrow of time is universal, then there's still rotation and parity. What you're asking about is basically the distinction between the Poincaré group and the Lorentz group.

Note that rotation and parity are similar in that we can't fix either without reference to some object. See The Ozma Problem .

I am carefully defining and discussing the important concepts of homogeneity and isotropy of space.

This is tough to do totally rigorously at the freshman level. Einstein's 1905 attempt to formalize it wasn't right. IIRC he says that the transformation has to be linear by the homogeneity of space. But really that's not quite right since you could, e.g., transform into an accelerated frame. That doesn't require GR according to modern definitions.

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  • $\begingroup$ Well, I think perhaps I wasn't very clear. I am trying to carefully state all the assumptions I make which lead to the Lorentz transform. If f maps spacetime bijectively onto itself, may I assume that a reversal of spatial coordinate system parity in the "domain" of the function result in a similar reversal of parity in the spatial subspace of its range? In other words, may I assume that the spatial "component" function belongs only to O(3) or must I assume it is restricted to SO(3)? In either case, why? $\endgroup$
    – user59591
    Commented Sep 22, 2014 at 3:48
  • $\begingroup$ P.S. My interest is not at the freshman level, but upper division. $\endgroup$
    – user59591
    Commented Sep 22, 2014 at 3:49
  • $\begingroup$ I think a number of responses are simply informing me about the definition of parity conservation, so I will be a bit more detailed. The Lorentz transformation has determinant = 1 and the spatial part has a positive determinant (for each t). Thus, it does not reverse frame orientation. But since I am developing the transformation ab initio I want to use this fact in an a priori way and am seeking physical justification for doing so.(Ben, I am still getting acquainted with this site so forgive this add on to my response to you.) $\endgroup$
    – user59591
    Commented Sep 22, 2014 at 14:53
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It is usual to make a distinction between proper rotation as opposed to all rotations (which include those that change handedness). The matrices of proper rotations have determinate +1 while those of improper rotations have determinate -1.

In terms of groups the proper rotations are $\mathrm{SO}(3)$ while all rotations taken together are the orthagonal group $\mathrm{O}(3)$.

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  • $\begingroup$ So you consider a reflection to be a rotation? $\endgroup$
    – Ryan Reich
    Commented Sep 24, 2014 at 18:30
  • $\begingroup$ It's not really me, it's mathematicians. The transformations that include reflections have all the same properties as the pure (proper) rotations, and in 2D there is no distinction at all. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Sep 24, 2014 at 19:04
  • $\begingroup$ I'm a mathematician and I dispute this. In 2D you do have reflections about a line that are not rotations; only reflection through the origin is. In general the definition of a rotation requires it to preserve orientation. It is true that any reflection can be made a rotation if an extra dimension is added, though. $\endgroup$
    – Ryan Reich
    Commented Sep 24, 2014 at 20:48
  • $\begingroup$ Er ... you're right about that. Don't know where that came from. In any case, the notion that there are "proper rotation" and "improper rotations" (which are a proper rotation plus a reflection) is not mine. You find it in various texts on group theory. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Sep 24, 2014 at 20:57

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