In a region of spacetime that does not wrap all the way round the -axis

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Differential geometry of curved spacetime · Differential geometry of curved spacetime

The Geometry of Gravity, Ch Symmetry: Lie derivatives, Killing vectors and conserved quantities

Problem

In a region of spacetime that does not wrap all the way round the zz-axis, so that ϕ\phi need not be treated as periodic, the line element in coordinates t,r,ϕ,zt,r,\phi,z is (with kk a constant length and no coefficient depending on tt) ds2=−c2dt2+2kc dt dϕ+dr2+r2dϕ2+dz2.ds^2 = -c^2dt^2 + 2kc\,dt\,d\phi + dr^2 + r^2d\phi^2 + dz^2. Someone argues: "No coefficient depends on tt, so ∂t\partial_t is a Killing vector and the spacetime is stationary. But no rotation ϕ=ϕ′+ωt\phi = \phi' + \omega t with constant ω\omega removes the dt dϕ′dt\,d\phi' cross term at every rr at once, so the spacetime is stationary and not static." ((a)) Confirm that ∂t\partial_t is a timelike Killing vector of this metric. ((b)) Check the claim about rotations: find the ω\omega that removes the cross term at a given rr. ((c)) Now shift the time coordinate instead, t=T+kϕ/ct = T + k\phi/c. Write the metric in (T,r,ϕ,z)(T,r,\phi,z), decide whether the spacetime is static in this region, and state exactly where the quoted argument goes wrong.

Browse the problem library