- #1
binbagsss
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Homework Statement
Question attached.
Homework Equations
3. The Attempt at a Solution [/B]I'm not really sure how to work with what is given in the question without introducing my knowledge on lie derivatives.
We have: ##(L_ug)_{uv} = U^{\alpha}\nabla_{\alpha}g_{uv}+g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}##
Where ##L_u## denotes the lie derivative in the direction of ##u##
Where the first term vanishes on the assumption of a levi-civita connection by the fundamental theorem of riemmanian geometry. ##U^{\alpha}\nabla_{\alpha}g_{uv}= U^{\alpha}(\partial_{\alpha}g_{uv} + \Gamma^{c}_{\alpha v} g_{uc} + \Gamma^{c}_{\alpha u} g_{vc}) ##
And so the fact that ##w^u \partial_u g_{ab} =0 \implies \partial_t g_{ab} =0 ## vanishes here, implies that the connection terms vanish (here I need an argument that they vanish individually, and that the sum can not vanish) I can then conclude that this implies we are working in Minkowski space-time.
And I can then use my the following knowledge on Lie derivatives to obtain an answer:
##w^u \partial_u g_{ab} =0 \implies \partial_t g_{ab} =0 ## ; i.e. the metric has no ##t## dependence so I know that this means that the Lie derivative (1) vanishes:
And then I have
##(L_ug)_{uv} =0= g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}=\nabla_vU_u+\nabla_u U_v##
##\implies k=-1 ##
However the question makes no reference to the requirement of needing Lie derivative, so I'm not too sure about what I've done here,
Can I start from first principles more? Any hint appreciated.
Thanks in advance.