Let $(M, g_{ab})$ be a spacetime and define a new metric, $\tilde{g}_{ab}$, on $M$ by $\tilde{g}_{ab} = \Omega^2 g_{ab}$, where $\Omega$ is a smooth, positive function. Let $\nabla_a$ denote the derivative operator associated with $g_{ab}$ and let $\tilde{\nabla}_a$ denote the derivative operator associated with $\tilde{g}_{ab}$. Let $v^a$ be an arbitrary smooth vector field on $M$. Do we necessarily have the following identity:$$\tilde{\nabla}_a v^b = \nabla_a v^b = {\delta^b}_a v^c \nabla_c \text{ln}\,\Omega - v^b \nabla_a \text{ln}\,\Omega - g_{ac} v^c g^{bd} \nabla_d \text{ln}\,\Omega?$$

  • $\begingroup$ Maybe you already know, but formula D.5 on pag. 446 of Wald's General Relativity looks quite similar to this identity. $\endgroup$ – Giuseppe Negro Apr 6 '16 at 21:00

It seems you have some misprints in your formula. Start by calculating the Cristoffell symbols for the new metric

$$\tilde\Gamma_{ab}^c = \Gamma_{ab}^c + g^{cd}(g_{db}\nabla_a \ln\Omega + g_{ad}\nabla_b \ln\Omega- g_{ab}\nabla_d \ln\Omega)$$ or $$\tilde\Gamma_{ab}^c = \Gamma_{ab}^c + (\delta_b^c\nabla_a \ln\Omega + \delta_a^c\nabla_b \ln\Omega- g_{ab}g^{cd}\nabla_d \ln\Omega)$$.

Next step is to use $\tilde\nabla_a v^b = \partial_a v^b + \tilde\Gamma_{ac}^b v^c$. Combine this formula with previous one yields

$$\tilde\nabla_a v^b = \nabla_a v^b + v^b \nabla_a\ln \Omega + \delta_a^b v^c\nabla_c \ln\Omega - g_{ac}v^c g^{bd}\nabla_d \ln\Omega$$.

| cite | improve this answer | |
  • $\begingroup$ I agree. Corrected the answer accordingly. $\endgroup$ – user26977 Apr 10 '16 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy