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A question from the book "Quantum Field Theory and the Standard Model" by Matthew D. Schwartz.

Is the transformation $Y: (t,x,y,z)\to (t,x,-y,z)$ a Lorentz transformation? If so, why is it not considered with $P$ and $T$ as a discrete Lorentz transformation? If not, why not?

I think it is a Lorentz transformation because it preserve the Minkowski metric. But I don't know why is it not considered as a discrete Lorentz transformation?

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  • $\begingroup$ You are very close to answering the question yourself. Consider a rotation by $\pi$ in the x-z plane followed by P. $\endgroup$ – user178876 Dec 18 '17 at 4:13
  • $\begingroup$ Actually it's an improper Lorentz transformation: see en.wikipedia.org/wiki/… $\endgroup$ – ZeroTheHero Dec 18 '17 at 4:53
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As you noted, this transformation preserves the Minkowski metric and is hence a Lorentz transformation. Further, a rotation by $\pi=180^\circ$ in the $x$-$z$ plane is described by the Lorentz transformation $\Lambda=\text{diag}(1,-1,1,-1)$, and $Y=\text{diag}(1,1,-1,1)$. Thus, $Y=P\cdot\Lambda$.

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To add to Marmot's answer that the transformation conserves the Minkowski inner product and is therefore a Lorentz transformation: your particular transformation is an improper Lorentz transformation, meaning that, although it conserves the Minkowski inner product and thus belongs to the group $O(1,\,3)$, it does not belong to the identity connected component $SO^+(1,\,3)$ of proper (of unity determinant), orthochronous (conserving direction of the time component, i.e. "causal") Lorentz transformations. This last group $SO^+(1,\,3)$ has particular physical significance in that its members connect all inertial frames that can be reached from one another by finite sequences of boosts and rotations. The rest frames of any two spaceships in our universe that can contact one another can be transformed into one another by a unique member of this identity connected component $SO^+(1,\,3)$, modulo a translation. The technical jargon for this state of affairs is that $SO^+(1,\,3)$ acts transitively (any two frames can be linked) and freely (connecting transformations are unique) (aka "sharply transitively") on the set of inertial frames with common origins in Minkowski spacetime.

I describe this state of affairs in more detail here. $O(1,\,3)$ splits into the semidirect product of $SO^+(1,\,3)$ and four discrete cosets. Your transformation $Y$ belongs to the same coset as the parity flipper $P=\mathrm{diag}(1,\,-1,\,-1,\,-1)$.

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protected by Qmechanic Dec 18 '17 at 5:55

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