Symbol Z

The passage around this formula

Check it against the elevator directly: $\phi$ = $\ln(1+gZ/c^2)$ gives $\phi$' = (g/$c^2$)/U and $\phi$'' = -(g/$c^2$)^2/$U^2$ = -($\phi$')^2 exactly, for every Z , not merely at leading order. The bracket $\phi_{,ij}$+$\phi_{,i}$$\phi_{,j}$ vanishes identically, and so does $R^i${}_{0j0} — confirming, by direct calculation rather than appeal to intuition, that a uniformly accelerated frame carries exactly zero curvature no matter how large its slope g becomes. The naive second difference of $\ln$ U alone, without the quadratic correction, does not vanish in this case — a fact worth stating plainly, because it is exactly the trap a network audit could fall into if it used $\phi_{,ij}$ by itself as its curvature signal.

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