Killing Equation: Why Set g_μν,ρ V^ρ = 0?

  • Context: Undergrad 
  • Thread starter Thread starter kent davidge
  • Start date Start date
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
8 replies · 2K views
kent davidge
Messages
931
Reaction score
56
When deriving the Killing Equation, one ends up with a term ##g_{\mu \nu, \rho} V^\rho## where ##V## is the Killing Vector. This terms is often set to zero. Why?
I see no motivation for demanding that the derivative of the metric coefficients vanish.
 
Physics news on Phys.org
martinbn said:
Can you give more details?
What kind of details? I tried to put everything into the opening post.
 
stevendaryl said:
Are you sure that they didn't write ##g_{\mu \nu;\rho} V^\rho##?
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##.
 
Orodruin said:
So why are you withholding the rest of your derivation?
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.
 
kent davidge said:
because I was not wanting to bore you with it

As you see, this is not a good strategy. When you ask a question about something you did, you need to include all of what you did. You can't assume that just the part you think you need to include is relevant, because if you already knew what was relevant and what was not, you wouldn't have made a mistake and you wouldn't have needed to post a question in the first place. Please keep that in mind for the future.

Since you say you have found your mistake, this thread is closed.
 
  • Like
Likes   Reactions: jim mcnamara and Orodruin