The discussion revolves around the Killing Equation in the context of differential geometry, specifically addressing the term ##g_{\mu \nu, \rho} V^\rho## where ##V## is the Killing Vector. Participants explore the reasoning behind setting this term to zero and the implications of such a decision in the derivation process.
Participants express differing views on the necessity and justification for setting the term to zero, indicating that the discussion remains unresolved with multiple competing perspectives on the derivation process.
There are indications of missing assumptions and incomplete derivations, particularly regarding the handling of the metric and the Killing Vector, which contribute to the ongoing debate.
What kind of details? I tried to put everything into the opening post.martinbn said:Can you give more details?
You did not though. You said ”This term is usually set to zero. Why?” without any reference to a text actually doing so or reciting an actual derivation doing so.kent davidge said:What kind of details? I tried to put everything into the opening post.
It's me. I'm deriving it. However I get ##\mathcal L g _{\sigma \rho} = \nabla_\sigma V_\rho + \nabla_\rho V_\sigma + 2 V_\mu \Gamma^\mu{}_{\sigma \rho} + g_{\sigma \rho, \kappa} V^\kappa##.stevendaryl said:Are you sure that they didn't write ##g_{\mu \nu;\rho} V^\rho##?
So why are you withholding the rest of your derivation?kent davidge said:It's me. I'm deriving it. However I get ##\mathcal L g _{\sigma \rho} = \nabla_\sigma V_\rho + \nabla_\rho V_\sigma + 2 V_\mu \Gamma^\mu{}_{\sigma \rho} + g_{\sigma \rho, \kappa} V^\kappa##.
because I was not wanting to bore you with it. but never mind, i found my fault. i was equating ##g_{\mu \nu} V^\nu{}_{, \ \rho} = V_{\mu, \ \rho}## right away.Orodruin said:So why are you withholding the rest of your derivation?
kent davidge said:because I was not wanting to bore you with it