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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.

Qmechanic
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2 Answers2

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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.

Community
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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 $\mathrm{O}(3)$.

dmckee --- ex-moderator kitten
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