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There exists an earlier question about how to verify that, if $U^\mu = (1,0,0,0)^\mu$ is the four-velocity of comoving observers, then

$$K_{\mu\nu} = a^2(g_{\mu\nu} + U_\mu U_\nu)\tag{1}$$

is a Killing tensor, $\nabla_{(\sigma} K_{\mu\nu)} = 0$, for the FRW metric

$$ds^2 = -dt^2 + a(t)^2 \left[\frac{dr^2}{1 - \kappa r^2} + r^2 d\Omega^2 \right], \quad d\Omega^2 = d\theta^2 + \sin^2\theta \, d\phi^2.\tag{2}$$

However, they only considered the flat case, $\kappa = 0$, but Sean Carroll claims that this is a Killing tensor without assuming flatness (see section 8.5, page 344, of his textbook, or chapter 8, page 228, in his lecure notes). I have tried for a while now to verify this claim, but I have not managed to do so and would therefore appreciate your help.

Partial results

By using $\nabla_\sigma g_{\mu\nu} = 0$, $\nabla_\sigma a^2 = \partial_\sigma a^2 = \delta_\sigma^0 2 a \dot a$, and the definition of $U^\mu$, I have found that

$$\nabla_\sigma K_{\mu\nu} = (g_{\mu\nu} + \delta_\mu^0 \delta_\nu^0)\delta_\sigma^0 2 a \dot a,\tag{3}$$

where $\dot a \equiv \partial_0 a$. This is diagonal in $\mu$ and $\nu$, and zero for $\sigma \ne 0$. I tried to use this to show that $\nabla_{(\sigma} K_{\mu\nu)} = 0$, but to no avail. Of course, I might have made a computation error somewhere along the way.

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    $\begingroup$ I'll point out that your condition can't even be true for the flat case, because $\kappa$ is not the reason that the permutations don't cancel. I'd try a more general form for your $K$ tensor, like $K = \alpha g_{ab} + \beta u_{a}u_{b}$, and see if you can find a functional form for $\alpha$ and $\beta$ that make your condition work out. $\endgroup$ Dec 13, 2021 at 22:26
  • $\begingroup$ @JerrySchirmer From my own calculations, yes, I would agree that setting $\kappa = 0$ does not seem to fix the problem. However, in the linked question they do show it for the flat case (albeit with a different parametrization), and I cannot spot any obvious errors in their computations. Are you saying they are wrong? $\endgroup$
    – ummg
    Dec 13, 2021 at 22:43
  • $\begingroup$ I'm saying that it's easy to make little mistakes in copying over the ansatz, even with simple things like missing that the author is using the (+, -, -, -) signature or something $\endgroup$ Dec 13, 2021 at 23:18
  • $\begingroup$ (and that the way to catch something like that is to add a little wiggle room in your expressions, and see if you can get a cancellation) I think there's a 90% chance you're missing a sign or have a parenthesis in the wrong place. Another thought is maybe the time coordinate is not the comiving one, but one that has a $r$ term that is proportional to $k$ or something. $\endgroup$ Dec 14, 2021 at 16:10

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I worked this problem in detail, starting with the ansatz:

$u_{a} = (A, 0, 0, 0)$

$K_{ab} = \alpha g_{ab} + \beta u_{a}u_{b}$

With some laborious algebra, you find that the $ttt$ component of $\nabla_{c}K_{ab}$ vanishes if $\alpha = A^{2}\beta$, and for simplicity, assuming that there is not $r$ dependence in any of our unsolved functions. Then, letting $q_{ab} = {\rm diag}(0, 0, 1, \sin^{2}\theta)$

$$\nabla_{c}K_{ab} = \delta_{a}^{r}\delta_{b}^{r}\delta_{c}^{t}\frac{a^{2}{\dot \alpha}}{1-kr^{2}} + a^{2}r^{2}{\dot \alpha}\delta_{c}^{t}q_{ab} - 2\frac{a{\dot a}\alpha}{1-kr^{2}}\delta_{(a}^{t}\delta_{b)}^{r}\delta_{c}^{r} - 2a {\dot a}r^{2}\alpha\delta_{(a}^{t}q_{b)c}$$

The RHS is manifestly symmetric in a and b, so, if permuting a and c is antisymmetric, then symmetrizing over abc will give zero. Solving for this, you get the condition $2\frac{\dot a}{a} = \frac{\dot \alpha}{\alpha}$, which, choosing $A = -1$ (future-pointing lowered index), and $\alpha = a^{2}$ we get:

$u_{a} = (-1, 0, 0, 0)$

$K_{ab} = a^{2}(g_{ab} + u_{a}u_{b})$

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  • $\begingroup$ Looking at your partial results section above, I suspect that you are missing some Christoffel symbols in your expression for $\nabla_{a}u_{b}$ $\endgroup$ Dec 14, 2021 at 21:27
  • $\begingroup$ Thank you for taking the time to go through these laborious computations. Is it a typo when you write $u_a=(-1,0,0,0)$, because just above it you wrote $A=1$? Also, the sign of $A$ should not really matter should it? $\endgroup$
    – ummg
    Dec 15, 2021 at 20:50
  • $\begingroup$ I did not use any Christoffel symbols to get my partial result above. Instead I used something like $U_\mu U_\nu = (-\delta_\mu^0) (-\delta_\nu^0) = \delta_\mu^0 \delta_\nu^0$, and hence $\nabla_\sigma (U_\mu U_\nu) = 0$, say, by metric compatibility. Is this not correct? $\endgroup$
    – ummg
    Dec 15, 2021 at 21:03
  • $\begingroup$ @ummg: that is not correct. $\nabla_{a}u_{b} = \partial_{a}u_{b} - \Gamma_{ab}{}^{c}u_{c} = \Gamma_{ab}{}^{t} =a {\dot a}h_{ab}$, where $h_{ab}$ is the 3-metric of the relevant slice $\endgroup$ Dec 15, 2021 at 22:21
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    $\begingroup$ @ummg the order in which you set the index to zero is important. if you think of $\nabla_{a}g_{b0}$ treating that 0 as a tensor index, then the expression is: $\partial_{a}g_{b0} - \Gamma_{ab}{}^{c}g_{c0} - \Gamma_{a0}{}^{c}g_{bc}$, but if you interpret $g_{b0}$ as simply a vector (which is what you do when writing $u_{b}$ down), then you do not have that second Christoffel symbol, and the "metric compatibility" argument no longer works. $\endgroup$ Dec 16, 2021 at 22:24

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