- #36
strangerep
Science Advisor
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I guess you meant ##K_x## ?WannabeNewton said:For an elemental pure Lorentz boost in the ##x## direction, for which ##E^{\mu} = (\partial_{x})^{\mu}##, we get ##\xi = x\partial_t+t\partial_x ##. Can an analogous 1-1 correspondence between this and ##K_z##, the generator of Lorentz boosts in the ##x## direction, be made in the same obvious manner? This is just so I can connect the Hermitian generators of the proper orthochronous Lorentz group back to the familiar notion of Minkowski space-time killing fields as the generators of isometric flows in Minkowski space-time.
It's interesting to me to see how you're approaching this in terms of more general techniques usually associated with GR and DG. When I learned this stuff, I never thought about Killing fields, etc, at all.
But anyway, the answer is essentially "yes". A more low-brow way to get the explicit generator is to consider an ordinary 1-parameter transformation of the form ##x^\mu \to x'^{\,\mu}(x^\lambda, p)##, where ##p## is the parameter. The associated generator ##P## can be found via this formula:
$$
\def\Pdrv#1#2{\frac{\partial #1}{\partial #2}}
\def\pdrv#1{\frac{\partial}{\partial#1}}
P ~:=~ \left. \Pdrv{x'^{\,\mu}}{p} \right|_{p = p_0} \pdrv{x^\mu} ~,
$$
where ##p_0## denotes the value of ##p## corresponding to the identity transformation. I.e., the first derivative above is to be evaluated at the identity.
In the case of a coordinate transformation corresponding to a boost in velocity, one gets essentially what you wrote.
Urk, I knew I was just confusing matters by jumping ahead like that.Could you elucidate this point? I understand that when we write ##U(1 + \delta \omega)##, we basically just saying that to each representation of the infinitesimal Lorentz transformation as an operator ##1 + \delta \omega; \delta^{\mu}{}{}_{\nu} + \delta \omega^{\mu}{}{}_{\nu}## on Minkowski space-time there is associated a representation of the same infinitesimal Lorentz transformation as a unitary operator ##I + \frac{i}{2\hbar}\delta \omega_{\mu\nu} M^{\mu\nu}## on Hilbert space, a rep to rep mapping as you put it, but I'm afraid I don't quite understand the quoted statement above.
Rubi has already given a good answer, and I'd probably just add to the confusion by trying to embelish it. But feel free to ask further questions if you can't keep your curiosity under control.