I have got to the end of the first paragraph in this section and ignited a controversy on Physics Forums about ##a## the scale factor which tells you how much the universe is expanding. The question is: Is the scale factor a scalar?
We start in an FLRW universe with metric$$
{ds}^2=-{dt}^2+a^2\left(t\right)\left[\frac{{\rm dr}^2}{1-\kappa r^2}+r^2{d\theta}^2+r^2\sin^2{\theta}{d\phi}^2\right]
$$First Carroll invents a tensor$$
K_{\mu\nu}=a^2\left(g_{\mu\nu}+U_\mu U_\nu\right)
$$where ##U^\mu=\left(1,0,0,0\right)## is the 4-velocity of (all) comoving observers. Carroll says it satisfies $$
\nabla_{(\sigma}K_{\mu\nu)}=0
$$"as you can check" and therefore it is a Killing tensor. I accepted the challenge and it took me a few days.
{ds}^2=-{dt}^2+a^2\left(t\right)\left[\frac{{\rm dr}^2}{1-\kappa r^2}+r^2{d\theta}^2+r^2\sin^2{\theta}{d\phi}^2\right]
$$First Carroll invents a tensor$$
K_{\mu\nu}=a^2\left(g_{\mu\nu}+U_\mu U_\nu\right)
$$where ##U^\mu=\left(1,0,0,0\right)## is the 4-velocity of (all) comoving observers. Carroll says it satisfies $$
\nabla_{(\sigma}K_{\mu\nu)}=0
$$"as you can check" and therefore it is a Killing tensor. I accepted the challenge and it took me a few days.
I was not sure if the scale factor ##a## is a scalar. If it's not then it is doubtful whether ##K_{\mu\nu}## is even a tensor. If ##a## is not a scalar but somehow ##K_{\mu\nu}## is a tensor then we must calculate ##\nabla_\sigma K_{\mu\nu}## by expanding the covariant derivative:$$
\nabla_\sigma K_{\mu\nu}=\mathrm{\partial}_\sigma K_{\mu\nu}-\Gamma_{\sigma\mu}^\lambda K_{\lambda\nu}-\Gamma_{\sigma\nu}^\lambda K_{\mu\lambda}
$$If ##a## is a scalar then we can use the Leibniz rule: $$
\nabla_\sigma K_{\mu\nu}=\left(g_{\mu\nu}+U_\mu U_\nu\right)\nabla_\sigma\left(a^2\right)+a^2\nabla_\sigma\left(g_{\mu\nu}+U_\mu U_\nu\right)
$$I used Leibniz first and failed. Then I was inspired by a simpler version on physics.stackexchange and got the right answer using the first formula. That made me doubt that ##a## was a scalar. So I asked about it on Physics Forums. Then I found my error in the Leibniz method and it gave the right answer! So now I think that ##a## is a scalar after all. The Leibniz way is a good deal more efficient than the first. The discussion continues and I think that PeterDonis has had the last word although vanhees71 is still wriggling at ##11. He surrenders at #13.
\nabla_\sigma K_{\mu\nu}=\mathrm{\partial}_\sigma K_{\mu\nu}-\Gamma_{\sigma\mu}^\lambda K_{\lambda\nu}-\Gamma_{\sigma\nu}^\lambda K_{\mu\lambda}
$$If ##a## is a scalar then we can use the Leibniz rule: $$
\nabla_\sigma K_{\mu\nu}=\left(g_{\mu\nu}+U_\mu U_\nu\right)\nabla_\sigma\left(a^2\right)+a^2\nabla_\sigma\left(g_{\mu\nu}+U_\mu U_\nu\right)
$$I used Leibniz first and failed. Then I was inspired by a simpler version on physics.stackexchange and got the right answer using the first formula. That made me doubt that ##a## was a scalar. So I asked about it on Physics Forums. Then I found my error in the Leibniz method and it gave the right answer! So now I think that ##a## is a scalar after all. The Leibniz way is a good deal more efficient than the first. The discussion continues and I think that PeterDonis has had the last word although vanhees71 is still wriggling at ##11. He surrenders at #13.
Read it all at Commentary 8.5 Killing Tensor in FLRW spacetime.pdf. (7 pages)
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