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Hello. I was going through the problems in the angular momentum chapter of the text "Problem Book in Relativity and Gravitation"-Lightman et al in preparation for a project and I came upon part (c) of problem 11.10. It basically starts with a family of observers with 4-velocity ##u^{\mu} = (-\xi_{\gamma}\xi^{\gamma})^{-1/2}\xi^{\mu}##, where ##\xi^{\mu}## is a time-like Killing field, and a Lorentz frame ##\{e_a \}## Lie transported along ##\xi^{\mu}##, and asks to show that the angular velocity of comoving gyroscopes relative to this frame is given by ##\omega_{\text{gyro}}^{\alpha} = -\frac{1}{2}(\xi_{\gamma}\xi^{\gamma})^{-1}\epsilon^{\alpha\beta\mu\nu}\xi_{\beta}\nabla_{\mu}\xi_{\nu}##.
It's the requirement that ##\xi## be a Killing field that I would like to discuss. Assume that Lie transport ##\mathcal{L}_{\xi}e_a =0## holds (which is physically equivalent to the condition that spatial axes constructed by any observer in the family remain fixed to the same neighboring observers in the family for all of his proper time); denote by ##e^{\mu}{}{}_a## the coordinate components of ##e_a##. Under what conditions will ##\xi^{\delta}\nabla_{\delta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = 0## hold? Well ##\xi^{\delta}\nabla_{\delta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = \mathcal{L}_{\xi}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = e^{\mu}{}{}_a e^{\nu}{}{}_b\mathcal{L}_{\xi}g_{\mu\nu}## so ##\xi^{\delta}\nabla_{\delta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = 0## if and only if ##\xi^{\mu}## is a Killing field given that ##\mathcal{L}_{\xi}e_a = 0##. I bring this up because for this physical setting we require ##\mathcal{L}_{\xi}e_a = 0## so on top of this if we also want ##\{e_a \}## to be a Lorentz frame then the Lie transport along ##\xi^{\mu}## has to preserve the condition ##g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b = \eta_{ab}## and as we've seen, under said Lie transport, the necessary and sufficient condition for this is that ##\xi^{\mu}## be a Killing field.
So we've established that we need ##\xi^{\mu}## to be a Killing field at minimum in order for the desired physical setup to remain consistent. But is that all we need ##\xi^{\mu}## to be a Killing field for? The reason I ask is, Lightman et al provide their own (rather long) calculation in the solutions showing that ##\omega_{\text{gyro}}^{\alpha} = -\frac{1}{2}(\xi_{\gamma}\xi^{\gamma})^{-1}\epsilon^{\alpha\beta\mu\nu}\xi_{\beta}\nabla_{\mu}\xi_{\nu}## wherein, on top of making use of a handful of various relations from relativistic kinematics, they also explicitly make use of the fact that ##\xi^{\mu}## is a Killing field.
To me however this seems completely unnecessary so I was hoping someone could check my (rather short) calculation, which makes no use of the fact that ##\xi^{\mu}## is a Killing field, and see if it has any fundamental errors in it:
We start with the relation ##u^{\beta}\nabla_{\beta}e^{\alpha}{}{}_a = -\Omega^{\alpha}{}{}_{\beta}e^{\beta}{}{}_a## where ##\Omega_{\alpha\beta} = a_{\alpha}u_{\beta} - u_{\alpha}a_{\beta} + \epsilon_{\alpha\beta\mu\nu}u^{\mu}\omega^{\nu}## (see MTW exercise 6.8); here ##\omega^{\alpha}## is the precession angular velocity of ##\{e_i \}## relative to comoving gyroscopes, which is (minus) the quantity we're after.
Using Lie transport ##\xi^{\beta}\nabla_{\beta}e^{\alpha}{}{}_a =e^{\beta}{}{}_a \nabla_{\beta}\xi^{\alpha}##, and defining ##\varphi \equiv (-\xi_{\gamma}\xi^{\gamma})^{-1/2}## we then have ##\varphi e^{\beta}{}{}_a\nabla_{\beta}\xi^{\alpha} = -\Omega^{\alpha}{}{}_{\beta}e^{\beta}{}{}_a##; contracting both sides with ##e_{\gamma}{}{}^a## and using ##e_{\gamma}{}{}^a e^{\beta}{}{}_a = \delta_{\gamma}^{\beta}## (see Carroll appendix J) we get [tex]\varphi \nabla_{\beta}\xi_{\alpha} = - \Omega_{\alpha\beta} = -a_{\alpha}u_{\beta} + u_{\alpha}a_{\beta} - \epsilon_{\alpha\beta\mu\nu}u^{\mu}\omega^{\nu}[/tex] hence [tex]\varphi \epsilon^{\alpha\beta\delta \lambda}\nabla_{\beta}\xi_{\alpha} = -\varphi\epsilon^{\delta\lambda \alpha\beta}\nabla_{\alpha}\xi_{\beta} = 2\epsilon^{\alpha\beta\delta\lambda}u_{\alpha}a_{\beta} + 2u^{\delta}\omega^{\lambda} - 2u^{\lambda}\omega^{\delta}[/tex] where I've used ##\epsilon^{\alpha\beta\delta \lambda}\epsilon_{\alpha\beta\mu\nu} = -4\delta^{[\delta}_{\mu}\delta^{\lambda]}_{\nu}##.
Contracting both sides with ##u_{\lambda}## yields ##-\varphi^2\epsilon^{\delta\lambda \alpha\beta}\xi_{\lambda}\nabla_{\alpha}\xi_{\beta} = 2\omega^{\delta}## since ##\epsilon^{\alpha\beta\delta\lambda}u_{\lambda}u_{\alpha}a_{\beta} = \omega^{\lambda}u_{\lambda} = 0## and ##u^{\lambda}u_{\lambda} = -1##.
This yields ##\omega^{\delta} = \frac{1}{2}(\xi^{\gamma}\xi_{\gamma})^{-1}\epsilon^{\alpha\beta\mu\nu}\xi_{\beta}\nabla_{\mu}\xi_{\nu}## as desired.
Are there any flaws in the solution? Thanks in advance.
It's the requirement that ##\xi## be a Killing field that I would like to discuss. Assume that Lie transport ##\mathcal{L}_{\xi}e_a =0## holds (which is physically equivalent to the condition that spatial axes constructed by any observer in the family remain fixed to the same neighboring observers in the family for all of his proper time); denote by ##e^{\mu}{}{}_a## the coordinate components of ##e_a##. Under what conditions will ##\xi^{\delta}\nabla_{\delta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = 0## hold? Well ##\xi^{\delta}\nabla_{\delta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = \mathcal{L}_{\xi}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = e^{\mu}{}{}_a e^{\nu}{}{}_b\mathcal{L}_{\xi}g_{\mu\nu}## so ##\xi^{\delta}\nabla_{\delta}(g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b) = 0## if and only if ##\xi^{\mu}## is a Killing field given that ##\mathcal{L}_{\xi}e_a = 0##. I bring this up because for this physical setting we require ##\mathcal{L}_{\xi}e_a = 0## so on top of this if we also want ##\{e_a \}## to be a Lorentz frame then the Lie transport along ##\xi^{\mu}## has to preserve the condition ##g_{\mu\nu}e^{\mu}{}{}_a e^{\nu}{}{}_b = \eta_{ab}## and as we've seen, under said Lie transport, the necessary and sufficient condition for this is that ##\xi^{\mu}## be a Killing field.
So we've established that we need ##\xi^{\mu}## to be a Killing field at minimum in order for the desired physical setup to remain consistent. But is that all we need ##\xi^{\mu}## to be a Killing field for? The reason I ask is, Lightman et al provide their own (rather long) calculation in the solutions showing that ##\omega_{\text{gyro}}^{\alpha} = -\frac{1}{2}(\xi_{\gamma}\xi^{\gamma})^{-1}\epsilon^{\alpha\beta\mu\nu}\xi_{\beta}\nabla_{\mu}\xi_{\nu}## wherein, on top of making use of a handful of various relations from relativistic kinematics, they also explicitly make use of the fact that ##\xi^{\mu}## is a Killing field.
To me however this seems completely unnecessary so I was hoping someone could check my (rather short) calculation, which makes no use of the fact that ##\xi^{\mu}## is a Killing field, and see if it has any fundamental errors in it:
We start with the relation ##u^{\beta}\nabla_{\beta}e^{\alpha}{}{}_a = -\Omega^{\alpha}{}{}_{\beta}e^{\beta}{}{}_a## where ##\Omega_{\alpha\beta} = a_{\alpha}u_{\beta} - u_{\alpha}a_{\beta} + \epsilon_{\alpha\beta\mu\nu}u^{\mu}\omega^{\nu}## (see MTW exercise 6.8); here ##\omega^{\alpha}## is the precession angular velocity of ##\{e_i \}## relative to comoving gyroscopes, which is (minus) the quantity we're after.
Using Lie transport ##\xi^{\beta}\nabla_{\beta}e^{\alpha}{}{}_a =e^{\beta}{}{}_a \nabla_{\beta}\xi^{\alpha}##, and defining ##\varphi \equiv (-\xi_{\gamma}\xi^{\gamma})^{-1/2}## we then have ##\varphi e^{\beta}{}{}_a\nabla_{\beta}\xi^{\alpha} = -\Omega^{\alpha}{}{}_{\beta}e^{\beta}{}{}_a##; contracting both sides with ##e_{\gamma}{}{}^a## and using ##e_{\gamma}{}{}^a e^{\beta}{}{}_a = \delta_{\gamma}^{\beta}## (see Carroll appendix J) we get [tex]\varphi \nabla_{\beta}\xi_{\alpha} = - \Omega_{\alpha\beta} = -a_{\alpha}u_{\beta} + u_{\alpha}a_{\beta} - \epsilon_{\alpha\beta\mu\nu}u^{\mu}\omega^{\nu}[/tex] hence [tex]\varphi \epsilon^{\alpha\beta\delta \lambda}\nabla_{\beta}\xi_{\alpha} = -\varphi\epsilon^{\delta\lambda \alpha\beta}\nabla_{\alpha}\xi_{\beta} = 2\epsilon^{\alpha\beta\delta\lambda}u_{\alpha}a_{\beta} + 2u^{\delta}\omega^{\lambda} - 2u^{\lambda}\omega^{\delta}[/tex] where I've used ##\epsilon^{\alpha\beta\delta \lambda}\epsilon_{\alpha\beta\mu\nu} = -4\delta^{[\delta}_{\mu}\delta^{\lambda]}_{\nu}##.
Contracting both sides with ##u_{\lambda}## yields ##-\varphi^2\epsilon^{\delta\lambda \alpha\beta}\xi_{\lambda}\nabla_{\alpha}\xi_{\beta} = 2\omega^{\delta}## since ##\epsilon^{\alpha\beta\delta\lambda}u_{\lambda}u_{\alpha}a_{\beta} = \omega^{\lambda}u_{\lambda} = 0## and ##u^{\lambda}u_{\lambda} = -1##.
This yields ##\omega^{\delta} = \frac{1}{2}(\xi^{\gamma}\xi_{\gamma})^{-1}\epsilon^{\alpha\beta\mu\nu}\xi_{\beta}\nabla_{\mu}\xi_{\nu}## as desired.
Are there any flaws in the solution? Thanks in advance.
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