What is the physical meaning of the derivative of the norm of a Killing vector?
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Hey mate ^_^
It might be simpler to ask this question in two dimensions, getting rid of the θ and t coordinates entirely....
The regularity condition you get at the end would reduce to the statement that the metric, near the origin, is a rescaling of the usual flat metric in polar coordinates...
#Be Brainly❤️
It might be simpler to ask this question in two dimensions, getting rid of the θ and t coordinates entirely....
The regularity condition you get at the end would reduce to the statement that the metric, near the origin, is a rescaling of the usual flat metric in polar coordinates...
#Be Brainly❤️
Answered by
2
Hello mate here is your answer.
It might be simpler to ask this question in two dimensions, getting rid of the θθ and tt coordinates entirely. The regularity condition you get at the end would reduce to the statement that the metric, near the origin, is a rescaling of the usual flat metric in polar coordinates, ∼a(r,ϕ)2(dr2+r2dϕ2)∼a(r,ϕ)2(dr2+r2dϕ2). I've always found physical meanings a bit elusive in GR, but you may be able to rephrase (4) in freshman physics language and centripetal acceleration. Using the definition of a Killing vector, ∂μX=−2ην∇νημ∂μX=−2ην∇νημ.
Hope it helps you.
It might be simpler to ask this question in two dimensions, getting rid of the θθ and tt coordinates entirely. The regularity condition you get at the end would reduce to the statement that the metric, near the origin, is a rescaling of the usual flat metric in polar coordinates, ∼a(r,ϕ)2(dr2+r2dϕ2)∼a(r,ϕ)2(dr2+r2dϕ2). I've always found physical meanings a bit elusive in GR, but you may be able to rephrase (4) in freshman physics language and centripetal acceleration. Using the definition of a Killing vector, ∂μX=−2ην∇νημ∂μX=−2ην∇νημ.
Hope it helps you.
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