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Hey guys! Problem #15 of Chapter 3 in Carroll's text says: Use Raychaudhuri's equation to show that if a fluid is flowing on geodesics through space - time with zero shear and expansion, then space - time must have a time - like killing vector.
As background: Raychaudhuri's equation and all the things formulated below apply to a time - like geodesic congruence. The equation itself is ##\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = -\frac{1}{3}\theta^2 - \sigma_{\mu\nu}\sigma^{\mu\nu} + \omega_{\mu\nu}\omega^{\mu\nu} - R_{\mu\nu}U^{\mu}U^{\nu}## where ##\omega_{\mu\nu}## is the rotation of the 4 - velocity field ##U^{\mu}## of the congruence, ##\sigma_{\mu\nu}## its shear, and ##\theta## its expansion. Given the tensor ##B_{\mu\nu} = \triangledown _{v}U_{\mu}## (this is its definition - its physical significance is that it measures the failure of separation vectors between neighboring geodesics in the congruence to be parallel transported), the rotation,shear, and expansion are defined as ##\theta = \triangledown _{\mu}U^{\mu},\sigma_{\mu\nu} = B_{(\mu\nu)} - \frac{1}{3}\theta P_{\mu\nu}, \omega_{\mu\nu} = B_{[\mu\nu]}## where ##P_{\mu\nu}## is just the projection tensor onto the subspace orthogonal to the 4- velocity field at each point.
Here we have a fluid traveling on geodesics which is exactly a time - like geodesic congruence so the above formulations can be used. Now the problem says that both the shear and expansion are zero which tells us immediately that ##B_{(\mu\nu)} = \triangledown _{(\nu}U_{\mu)} = 0## thus ##U^{\mu}## solves killing's equation and is time - like since it is the 4 - velocity field so it gives us a time - like killing vector field. I don't even see why Carroll mentioned explicitly using Raychaudhuri's equation because the result the problem wanted is so immediate from the definitions above when given the zero shear and expansion that it makes me question why this problem was even put in...did I over look something?
As background: Raychaudhuri's equation and all the things formulated below apply to a time - like geodesic congruence. The equation itself is ##\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = -\frac{1}{3}\theta^2 - \sigma_{\mu\nu}\sigma^{\mu\nu} + \omega_{\mu\nu}\omega^{\mu\nu} - R_{\mu\nu}U^{\mu}U^{\nu}## where ##\omega_{\mu\nu}## is the rotation of the 4 - velocity field ##U^{\mu}## of the congruence, ##\sigma_{\mu\nu}## its shear, and ##\theta## its expansion. Given the tensor ##B_{\mu\nu} = \triangledown _{v}U_{\mu}## (this is its definition - its physical significance is that it measures the failure of separation vectors between neighboring geodesics in the congruence to be parallel transported), the rotation,shear, and expansion are defined as ##\theta = \triangledown _{\mu}U^{\mu},\sigma_{\mu\nu} = B_{(\mu\nu)} - \frac{1}{3}\theta P_{\mu\nu}, \omega_{\mu\nu} = B_{[\mu\nu]}## where ##P_{\mu\nu}## is just the projection tensor onto the subspace orthogonal to the 4- velocity field at each point.
Here we have a fluid traveling on geodesics which is exactly a time - like geodesic congruence so the above formulations can be used. Now the problem says that both the shear and expansion are zero which tells us immediately that ##B_{(\mu\nu)} = \triangledown _{(\nu}U_{\mu)} = 0## thus ##U^{\mu}## solves killing's equation and is time - like since it is the 4 - velocity field so it gives us a time - like killing vector field. I don't even see why Carroll mentioned explicitly using Raychaudhuri's equation because the result the problem wanted is so immediate from the definitions above when given the zero shear and expansion that it makes me question why this problem was even put in...did I over look something?
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