... how is it possible for space,
which is utterly empty, to expand?
How can nothing expand?
The answer is: space does not expand.
Cosmologists sometimes talk about expanding space,
but they should know better.
§ 1Warm-up
Problems 2.4.1 – 2.4.9
2.4.1
Problem 2.4.1: A spider on a string
An elastic rubber cord of $1$ meter length $1$ is attached to a wall. A spider sits on it at the junction to the wall, and a man holds the other end. The man starts moving away from the wall with velocity $1\, m/s$, and at the same time the spider starts to run along the cord with velocity $1\, cm/s$. Will the spider come up with the man?
solution
In dimensionless variables
$(x \to x/L,\; t \to t/( L/V),\;v \to v/V)$ the equation for the $x$-coordinate of the spider takes the form
\[\frac{dx}{dt} = v + \frac{x}{1 + t}.\]
Its solution is $x(t) = v(1 + t)\ln (1 + t).$ Equating $x(t)$ to the coordinate of the human at time $t$, which is $t+1$, one finds that $t \approx {e^{1/v}} = e^{100}\,s$. This value considerably exceeds the age of the Universe.
2.4.2
Problem 2.4.2: Law of inertia
Does the law of inertia hold in an expanding Universe?
solution
Let us consider a non-relativistic particle that at time $t$ flies past the first observer with velocity $v_{part}(t)$. Let the second observer be at distance $dR$ from that point. He is receding from the first observer with velocity $dV_{obs}=H(t)dR$. The particle flies past him after the time $dt=dR/V_{part}$ with velocity
\[V_{part}(t+dt)=V_{part}(t)-dV_{obs} =V_{part}(t) - H(t)dR.\]
Then
\[\frac{dV_{part}}{dt} =-H(t)\frac{dR}{dt} =-H(t)V_{part}(t) =-\frac{V_{part}}{a}\frac{da}{dt}.\]
Solution of this equation is
\[V_{part}(t) \propto \frac{1}{a(t)}.\]
Thus in an expanding Universe the velocity of a free particle diminishes with time, and therefore the law of inertia does not hold.
2.4.3
Problem 2.4.3: A particle's momentum in expanding Universe
Suppose a particle's mean free path in an expanding Universe is small enough. Show that its momentum decreases as $^*$ $p(t)\propto a(t)^{-1}$.
$^*$This is a generalization of the previous problem to relativistic case, but still a simplification of the general formulation.
solution
Let us consider the equation for geodesics
\[\frac{du^v}{ds} +
\Gamma _{\alpha \beta }^vu^\alpha u^\beta = 0,\]
where $u^v = dx^v/ds$ is the coordinate $4$-velocity, $x^v$ is the coordinate (comoving) distance, which is related to the proper (physical) one $X^v$ by relation $X^v = a\left( t \right)x^v$. The physical components of the $4$-velocity are equal to $U^v = dX^v/ds$. If we consider propagation of a particle with its mean free path much smaller than the spatial curvature radius (in the case of non-flat Universe), we can neglect curvature and let $k=0$. Then
\[ds^2 = dt^2 - a^2(t)\delta _{ij}dx^{i}dx^{j}.\]
Non-zero Christoffel symbols are:
\[\Gamma _{ij}^0 = a\dot a\delta _{ij},\;
\Gamma _{0j}^i = \frac{\dot a}{a}\delta _j^i.\]
The geodesic equation for the spatial components reduces to
\begin{align*}
0=\frac{du^i}{ds} + \Gamma _{0j}^iu^0u^j +
\Gamma _{j0}^iu^ju^0 =
\frac{du^i}{ds}
+ 2\frac{\dot a}{a}\delta _j^iu^0u^j=
\frac{du^i}{ds}
+ 2\frac{\dot a}{a}\frac{dt}{ds}u^j.
\end{align*}
Taking into account that $u^i = \frac{U^i}{a\left( t \right)}$, one obtains
\[0=\frac{d}{ds}\left(\frac{U^i}{a} \right) +
2\frac{U^i}{a^2}\frac{da}{dt}\frac{dt}{ds} =\frac{1}{a}\frac{dU^i}{ds}
-\frac{U^{i}}{a^2}\frac{da}{ds}
+2\frac{U^i}{a^2}\frac{da}{ds} =\frac{1}{a}\frac{dU^i}{ds}
+\frac{U^{i}}{a^2}\frac{da}{ds},\]
and
\[\frac{dU^i}{U^i} + \frac{da}{a} = 0,\]
thus
\[p^i \sim U^{i} \sim \frac{1}{a(t)}.\]
2.4.4
Problem 2.4.4: Comoving phase volume
Show that the comoving phase volume equals to the physical one.
solution
Let $\left\{ \vec{r},\vec{p} \right\}$ denote the comoving coordinates and momenta, and $\left\{ \vec{R},\vec{P} \right\}$ the proper ones. Then
\[d\vec{R}d\vec{P} =d(a\vec{r})d\left(
\frac{\vec{p}}{a} \right)=d\vec{r}d\vec{p}.\]
2.4.5
Problem 2.4.5: Twin's paraadox
Show that the twins' paradox is resolved in the FLRW Universe in the same way as in the Minkowski spacetime: the twin who experienced non-zero acceleration appears to be younger than his brother.
O.Gron, S. Braeck, arXiv:0909.5364
solution
Let $t_1$ be the time corresponding to "reunion" of the brothers, measured by the clock of the twin at rest. By the clock of the returned twin the corresponding time interval equals to
\[\tau =\int_{0}^{{{t}_{1}}}{\sqrt{1-\frac{{{a}^{2}}(t){{v}^{2}}}{\left( 1-k{{r}^{2}} \right){{c}^{2}}}}dt}.\]
As the integrand is less than unity for all times, $\tau <{{t}_{1}}.$ Note that though Minkowski spacetime is flat and FLRW one is curved, the curvature effect has nothing to do with the paradox resolution.
2.4.6
Problem 2.4.6: Hubble time
Show that the Hubble's time $H_{0}^{-1}$ gives the characteristic time scale for any stage of evolution of the Universe.
2.4.7
Problem 2.4.7: Hubble radius
Show that in a Universe which expands with acceleration the Hubble's radius decreases.
2.4.8
Problem 2.4.8: Velocity of Hubble sphere
Show that the surface of Hubble's sphere recedes with velocity $V=c(1+q)$, where $q=-a\ddot{a}/(\dot{a})^2$ is the deceleration parameter.
2.4.9
Problem 2.4.9: Redshift and Hubble sphere
Show that the standard definition of redshift is valid only inside the Hubble's sphere.
solution
\[z = \sqrt {\frac{c + V}{c - V}} - 1 = \sqrt{\frac{c + HR}{c - HR}}- 1.\]
Objects beyond the Hubble's sphere, with radius $R_H = cH^{-1}$ formally have superluminal velocity of recession from the observer situated in the center, and therefore all quantities described by Special Relativity take on imaginary values.
§ 2The tethered galaxy problem
Problems 2.4.10 – 2.4.19
2.4.10
Problem 2.4.10: Cosmological recession and peculiar velocities
Let us consider radial motion in the uniform and homogeneous Universe. For this case the FLRW metric reduces to
\[ds^2 =c^2 dt^2 -a^{2}(t)d\chi^{2}.\]
Proper (physical) distance is defined as the distance (measured along the constant time section $dt=0$) between an observer and a galaxy with given comoving coordinate. Let us define the total velocity of a test galaxy as the time derivative of the proper distance
\[v_{tot}=\dot{D},\quad
\dot{D}=\dot{a}\chi+a\dot{\chi},\quad
v_{tot}=v_{rec}+v_{pec}.\]
Here $v_{rec}$ is the recession velocity of the test galaxy and $v_{pec}$ is its peculiar velocity. What can be said of the possible values of these velocities?
solution
The peculiar velocity $v_{pec}$ is the velocity with respect to the comoving frame out of which the test galaxy was boosted. It corresponds to our normal, local notion of velocity and must be less than the speed of light. The recession velocity $v_{rec}$ is the velocity of the Hubble flow at the proper distance $D$ and can be arbitrarily large.
2.4.11
Problem 2.4.11: Lightspeed recession
Determine the distance to a galaxy which, due to the Hubble's expansion, recedes from us with the speed of light.
solution
Consider a comoving observer. According to the Hubble's law, a galaxy at distance $r$ recedes along his line of sight with velocity $\vec{v} = H \vec r$. Evidently there is the distance $R_H=cH^{-1}$ where the recession velocity is equal to lightspeed. This distance is called the Hubble's radius.
Richard Feinman wrote$^*$: "This constant ($T=H_{0}^{-1}$) represents a lifetime of the universe; not necessarily that we believe that the universe did begin T years ago, but rather it represents a fundamental dimension of the universe, much in the way that the quantity $e^{2}/mc^2$ represents the "electron radius.""
$^*$Feynman, Morinigo, Wagner. Feynman lectures on gravitation, AW, 1995, p.8.
2.4.12
Problem 2.4.12: Superluminal cosmological velocities
Is it possible for cosmological objects to recede from us with superluminal speeds?
solution
Consider two points in flat FLRW Universe, which were situated at distance $R$ from each other at time $t$. If the points do not change their spatial (comoving) coordinates, so are at rest in this sense, but take part in the general expansion of the Universe (the Hubble's flow), the distance between them increases with velocity $\frac{dR}{dt} = HR$. This means if the distance between them is larger than the Hubble's radius $R=cH^{-1}$, it increases with superluminal velocity. It should be stressed that there is nothing paradoxical here, as one deals with velocity with which the distance between objects increases when they are captured by common cosmological expansion, and it is neither the velocity of signal transmission due to local changes of spatial coordinates of particles nor velocity of their relative motion.
A. D. Linde: "I was told that when Shklovsky was dying he was visited by Rosental, and the latter then told me that Shklovsky was lying in bed and said: "Well, I understand everything ..." -- he was a remarkable astrophysicist, the best astrophysicist ever, -- "... understand everything... but how could they make a theory in which everything expands with superluminal speed?". So do not feel ashame of the lack of understanding… The situation is as follows. There are two different types of expansion. The first is like a wave moves, a signal propagates. The signal cannot go faster that the speed of light. The second type... imagine the Universe as a rubber membrane which is stretched. Let us drive two nails in it. This is the Hubble's type of expansion, which is $a$ dotted and so on, --- it increases the distance between the two nails, the two galaxies. General Relativity can only describe this effect --- the stretching of the membrane. And on the velocity of mutual recession for the two nails there are no limitations."
2.4.13
Problem 2.4.13: Observing superluminal velocities
Is it possible to observe galaxies receding with superluminal speeds?
2.4.14
Problem 2.4.14: Tethered galaxy problem
Imagine that we separate a small test galaxy from the Hubble flow by tethering it to an observer such that the proper distance between them remains constant. We can think of the tethered galaxy as one that has received a peculiar velocity boost toward the observer that exactly matches its recession velocity. We then remove the tether (or turn off the boosting rocket) to establish the initial condition of constant proper distance $\dot{D}_{0}=0$. Determine the future fate of the test galaxy: will it approach the observer, recede from him or remain at constant distance?
solution
The momentum $p$ with respect to the local comoving frame decays as $1/a$ (see this and that problem). This scale factor dependent decrease in momentum is an important basis for many of the results. For nonrelativistic velocities $p=mv_{rec}$, therefore,
\begin{align}
&v_{pec}=\frac{v_{pec\,,0}}{a},\nonumber\\
&a\dot{\chi}=-\frac{\dot{a}_{0}\chi_{0}}{a},\nonumber\\
&\chi=\chi_{0}\Big[1-\dot{a}_{0}
\int\limits_{t_0}^{t}\frac{dt}{a^2}\Big],\nonumber\\
\label{tethered_D}
&D=a\chi_{0}\Big[1-\dot{a}_{0}
\int\limits_{t_0}^{t}\frac{dt}{a^2}\Big].
\end{align}
The integrals can be computed numerically by using $dt=da/\dot{a}$ and $\dot{a}_{0}$, where both are obtained directly from the first Friedman equation. From Eq. (\ref{tethered_D}) we see that the result depends on the explicit function $a(t)$, and thus on the material content of the Universe. It will be obtained for the Big Bang and Standard cosmological models in chapters 3 and 11. Meanwhile, we present the solution for the Milne's Universe, in which $\rho\to 0$. In this case the galaxy experiences no acceleration and stays at a constant proper distance as it joins the Hubble flow.
2.4.15
Problem 2.4.15: Hubble flow
Show that, provided the Universe expands forever, the test galaxy considered in the previous problem asymptotically joins the Hubble flow.
solution
The untethered galaxy asymptotically joins the Hubble flow for all cosmological models that expand forever because
\[\dot D = {v_{res}} + {v_{pec}} = {v_{rec}} + \frac{{{v_{pec,0}}}}{a}.\]
As $a\to\infty$ we have $\dot{D}=v_{rec}=HD$, which is pure Hubble flow. Note that the galaxy joins the Hubble flow solely due to the expansion of the universe ($a$ increasing).
2.4.16
Problem 2.4.16: Hubble's law for acceleration
Obtain the analogue of the Hubble's law for acceleration in presence of radial peculiar velocity.
solution
\begin{align*}
\dot D &= {v_{res}} + {v_{pec}} = \dot a\chi + \frac{{{v_{pec,0}}}}{a};\\
\ddot D &= \left( {\ddot a\chi + \dot a\dot \chi } \right) - \frac{{{v_{pec,0}}}}{a}\frac{\dot{ a}}{a} = \left( {\ddot a\chi + \dot a\dot \chi } \right) - {v_{pec}}\frac{\dot{ a}}{a} = \\
&= \left( {\ddot a\chi + \dot a\dot \chi } \right) - a\dot \chi \frac{\dot{ a}}{a} = \ddot a\chi ;\\
q &= - \frac{\ddot{ a}a}{{{{\dot a}^2}}};\\
\ddot D &= - q{H^2}D
\end{align*}
2.4.17
Problem 2.4.17: Another derivation
Derive the result of the previous problem by direct differentiation of the Hubble's law.
solution
As $\dot{D}=HD$, we have
\[\ddot D = \dot HD + H\dot D = \dot HD + {H^2}D = \left( {\dot H + {H^2}} \right)D =
\frac{\ddot{ a}}{a}D = - q{H^2}D.\]
This method ignores $v_{pec}$ and therefore does not include the explicit cancellation of the two terms in the more general calculation of previous problem. The fact that the results are the same emphasizes that the acceleration of the test galaxy is the same as that of comoving galaxies and there is no additional acceleration on our test galaxy pulling it into the Hubble flow.
2.4.18
Problem 2.4.18: Relative motion and redshifts
In the context of special relativity (Minkowski space), objects at rest with respect to an observer have zero redshift. However, in an expanding universe special relativistic concepts do not generally apply. "At rest" is defined to be "at constant proper distance" ($v_{tot}=\dot{D}= 0$), so our untethered galaxy with $\dot{D}=0$ satisfies the condition
for being at rest. Will it therefore have zero redshift?
That is, are $z_{tot}=0$ and $v_{tot}=0$ equivalent?
2.4.19
Problem 2.4.19: Addition of redshifts
Show that, although radial recession and peculiar velocities add vectorially, their corresponding redshifts combine as
\[1+z_{tot}=(1+z_{rec})(1+z_{pec}).\]
solution
Let $\lambda_{obs}$ be the wavelength we observe, $\lambda_{e}$ be the wavelength measured in the comoving frame of the emitter (the frame with respect to which it has a peculiar velocity $v_{pec}$), and $\lambda_{rest}$ be the wavelength in the rest frame of the emitter. Then
\[1+z_{obs}=\frac{\lambda_{obs}}{\lambda_{rest}} =\frac{\lambda_{obs}}{\lambda_{e}}
\frac{\lambda_{e}}{\lambda_{rest}} =(1+z_{rec})(1+z_{pec}).\]
§ 3Cosmological redshift
Problems 2.4.20 – 2.4.26
Inspired by E. Bunn, D. Hogg. The kinematic origin of the cosmological redshift. Am. J. Phys. 77:688-694, (2009);
arXiv:0808.1081.
2.4.20
Problem 2.4.20: Cosmological redshift as sum of Doppler shifts
Derive the cosmological redshift as the result of addition of infinitesimal Dopper shifts due to relative velocities of galaxies along the worldline of a photon.
solution
Let us consider two adjacent comoving galaxies. Their relative velocities due to the Hubble expansion are small, so we can use the non-relativistic formula for the Doppler effect:
\[\frac{\delta\omega}{\omega}=-\frac{\delta v}{c} =-\frac{H \delta r}{c}=-H\delta t =\frac{\dot{a}}{a}\delta t =-\frac{\delta a}{a}.\]
Then we see that along the trajectory of a photon
\[\omega a=const\quad\Rightarrow\quad
\omega\sim\frac{1}{a}.\]
Thus its redshift by definition is
\[1+z=\frac{\omega_{emit}}{\omega_{obs}} =\frac{a_{obs}}{a_{emit}}.\]
2.4.21
Problem 2.4.21: Relative velocities of distant galaxies
Suppose the source galaxy $A$ and detector galaxy $B$ are moving with the Hubble flow. Imagine a family of comoving observers situated along the trajectory of the photon. Let the observer 1, closest to the source galaxy, measure his velocity $v_1$ relative to the galaxy and send this information along with the photon to the next closest to him observer 2. Observer 2 measures his velocity $u$ relative to observer 1 and calculates his velocity relative to the galaxy $v_2$ according to the special relativistic formula
\[v_{2}=\frac{v_1 +u}{1+v_1 u}.\]
He sends this information along. What will be the velocity $v_{rel}$ of the observers relative to the galaxy, defined this way, in terms of scale factors at the moment of emission and at the moment of detection?
solution
In the limit of continuously distributed observers their relative velocities tend to zero, so if $u$ is such a relative velocity for two adjacent observers, then ($c=1$)
\[v_{2}=\frac{v_{1}+u}{1+v_{1}u}
\approx v_{1}+u(1-v_{1}^{2}),\]
and thus along the photon's worldline ($dR=dt$)
\[dv=du (1-v^2)=H dR (1-v^2)=H dt (1-v^2) =\frac{da}{a}(1-v^2).\]
On integrating with initial conditions $a=a_{emit}$ and $v=0$, we get
\[a\equiv\frac{a_{obs}}{a_{emit}} =\sqrt{\frac{1+v_{rel}}{1-v_{rel}}}
\quad\Leftrightarrow\quad
v_{rel}=\frac{a^{2}-1}{a^{2}+1}.\]
2.4.22
Problem 2.4.22: Cosmological redshift and relative velocity
Show that the registered cosmological redshift corresponds to Doppler effect with this very velocity $v_{rel}$.
solution
If we just plug $v_{rel}$ into the relativistic formula for the Doppler effect, we see that
\[\frac{\omega_{obs}}{\omega_{emit}}|_{v} =\frac{\sqrt{1-v^2}}{1+v} =\sqrt{\frac{1+v}{1-v}} =| v=v_{rel}| =\frac{a_{emit}}{a_{obs}},\]
so the detected cosmological redshift corresponds exactly to the Doppler effect with velocity $v_{rel}$.
2.4.23
Problem 2.4.23: Parellel-transported relative velocity
Find the relative physical velocity of two particles with $4$-velocities $u_{1}^{\mu}$ and $u_{2}^{\mu}$. Let $u_{1}^{\mu}$ be the $4$-velocity of the comoving detector at the moment of detection, and let $u_{2}^{\mu}$ be the $4$-velocity of the source at the moment of emission, parallel transported to the detector along the worldline of the photon$^*$.
$^*$A vector $a$ is parallel transported along a curve with tangent vector $u^{\mu}$, if $u^{\mu}\nabla_{\mu}a^{\nu}=0$.
solution
a) The Lorentz factor of a particle with $4$-velocity $u_{1}^{\mu}$ relative to an observer with $4$-velocity $u_{2}^{\mu}$ is (see the problem on observable invariants)
\[\gamma=u_{1}^{\mu}u_{2\,\mu},\]
therefore their relative physical velocity is
\[v_{12} =\big[1-(u_{1}^{\mu}u_{2\;\mu})^{-2}\big]^{-1/2}.\]
b) Equation of parallel transport along a curve $x^{\mu}(\lambda)$ with tangent vector $u^{\mu}=dx^{\mu}/d\lambda$ reads
\[\frac{dv^{\mu}}{d\lambda} =-\Gamma^{\mu}_{\nu\lambda}v^{\nu}u^{\lambda}.\]
Let us consider its $t$-component. In metric (we are only interested in radial motion, so discard the angular part and work in effectively two-dimensional spacetime)
\[ds^{2}=dt^{2}-a^{2}(t)dr^{2}\]
of all the Christoffel symbols with the first index $t$ the only non-zero one is
\[\Gamma^{t}_{rr}=-\tfrac{1}{2}\partial_{t}g_{rr} =a\dot{a},\]
so the equation is reduced to
\[\frac{dv^{t}}{d\lambda}=-a\dot{a}u^{r}v^{r}.\]
Along the worldline of the photon
\[u^{r}=\frac{dr}{d\lambda} =\frac{a\,dr}{dt}\;\frac{1}{a}\;\frac{dt}{d\lambda} =\frac{1}{a}\;\frac{dt}{d\lambda}.\]
From the normalization condition
\[1=v^{\mu}v_{\mu}=(v^{t})^{2}-a^{2}(v^r)^{2},
\quad\Rightarrow\quad
v^{r}=\frac{1}{a}\sqrt{(v^t)^{2}-1},\]
and the Lorentz factor relative to a comoving observer with $4$-velocity $(u_{stat})^{\mu}=(1,0)$ is
\[\gamma=g_{\mu\nu}v^{\mu}(u_{stat})^{\nu} =v^{t},\]
so
\[v^{t}=\gamma,
\quad v^{r}=\frac{1}{a}\sqrt{\gamma^2 -1}.\]
When we substitute all of this into the equation of parallel transport,
\[\frac{d\gamma}{d\lambda}=
\frac{\sqrt{\gamma^2 -1}}{a}\;
\frac{da}{d\lambda},\quad
\Rightarrow\quad
\frac{d\gamma}{\sqrt{\gamma^2-1}}=\frac{da}{a},\]
and on integration,
\[a=\sqrt{\frac{1+v}{1-v}}.\]
2.4.24
Problem 2.4.24: Killing tensors
A Killing tensor $K_{\mu\nu}$ is a tensor field, which obeys the generalization of the Killing equation
\[\nabla_{(\mu}K_{\nu\lambda)}=0,\]
where parenthesis denote symmetrization over all indices. Prove that the quantity $K_{\mu\nu}u^{\mu}u^{\nu}$ is conserved along a geodesics with tangent vector $u^{\mu}$.
solution
If $\nabla_{(\lambda}K_{\mu\nu)}=0$, then taking into account the geodesic equation $u^{\lambda}\nabla_{\lambda}u^{\mu}=0$, we arrive to
\[ u^{\lambda}\nabla_{\lambda}
\big(K_{\mu\nu}u^{\mu}u^{\nu}\big)=
u^{\lambda}u^{\mu}u^{\nu}
\nabla_{\lambda}K_{\mu\nu}=
u^{\lambda}u^{\mu}u^{\nu}
\nabla_{(\lambda}K_{\mu\nu)}=0.\]
2.4.25
Problem 2.4.25: Killing tensor in FLRW metric
Verify that the tensor
\begin{equation}\label{FLRWKillingTensor}
K_{\mu\nu} =a^{2}\big(u_{\mu}u_{\nu}-g_{\mu\nu}\big),
\end{equation}
where $u^{\mu}$ is the $4$-velocity of a comoving particle, is a Killing tensor for the FLRW metric.
solution
In conformal-comoving frame $(\eta,\chi)$ we have $u_{\mu}=a\delta_{\mu}^{0}$, and thus
\[K_{\mu\nu}=a^{4}\delta_{\mu}^{0}\delta_{\nu}^{0}
-a^{2}g_{\mu\nu}.\]
Then
\begin{align*}
\nabla_{\lambda}K_{\mu\nu}& =a^{4}\nabla_{\lambda}
\big(\delta_{\mu}^{0}\delta_{\nu}^{0}\big)
+\big[2a^{2}\delta_{\mu}^{0}\delta_{\nu}^{0}
-g_{\mu\nu}\big]\cdot 2a\nabla_{\lambda}a=\\
&=-a^{4}
\big(\Gamma^{0}_{\lambda\mu}\delta_{\nu}^{0}
+\Gamma^{0}_{\lambda\nu}\delta_{\mu}^{0}\big)
+h_{\mu\nu}\cdot 2a\partial_{\lambda}a=\\
&=-a^{2}
\big(\Gamma_{0,\lambda\mu}\delta_{\nu}^{0}
+\Gamma_{0,\lambda\nu}\delta_{\mu}^{0}\big)
+2a\dot{a}h_{\mu\nu}\delta_{\lambda}^{0},
\end{align*}
where $h_{\mu\nu}=g_{\mu\nu}
-2a^{2}\delta_{\mu}^{0}\delta_{\nu}^{0}$.
On calculating the Christoffel symbols
\begin{align*}
\Gamma_{0,00}&=\tfrac{1}{2}\partial_{0}g_{00} =a\dot{a}=\frac{\dot{a}}{a}g_{00};\\
\Gamma_{0,0i}&=0,\quad\mbox{for}\; i=1,2,3;\\
\Gamma_{0,ij}&=-\tfrac{1}{2}\partial_{0}g_{ij} =-\frac{\dot{a}}{a}g_{ij},
\quad\mbox{for}\; i,j=1,2,3,
\end{align*}
we see that they can be put down in the form
\[\Gamma_{0,\mu\nu}=\frac{\dot{a}}{a}
\big[2a^2\delta_{\mu}^{0}\delta_{\nu}^{0}
-g_{\mu\nu}\big]=
-\frac{\dot{a}}{a}h_{\mu\nu}.\]
Then
\[\nabla_{\lambda}K_{\mu\nu} =a\dot{a}\big[
2\delta_{\lambda}^{0}h_{\mu\nu}
-\delta_{\mu}^{0}h_{\lambda\nu}
-\delta_{\nu}^{0}h_{\lambda\mu}\big],\]
and on symmetrization we obtain the desired generalization of the Killing equation
\[\nabla_{(\lambda}K_{\mu\nu)}=0.\]
2.4.26
Problem 2.4.26: A particle's momentum in expanding Universe revisited
Show that due to the Killing tensor (\ref{FLRWKillingTensor}), the physical momentum of a particle, measured by comoving observers, changes with time as $p\sim 1/a$.
solution
The Lorentz factor of a particle with $4$-velocity $u^{\mu}$ relative to a particle with $4$-velocity $v^{\mu}$ is (see the problem on observable invariants)
\[\gamma=u^{\mu}v_{\mu}.\]
Then the integral of motion due to the Killing tensor can be written as
\begin{align*}
const&=K_{\mu\nu}v^{\mu}v^{\nu} =a^{2}\big[(u^{\mu}v_{\mu})^{2}-v^{\mu}v_{\mu}\big] =a^{2}(\gamma^{2}-1)=\frac{a^{2}p^{2}}{m^2},
\end{align*}
where $p=mv\gamma$ is the physical momentum of the particle measured by the comoving observer.
Likewise the frequency $\omega$ of a massless particle with wavevector $k^{\mu}$, measured by an observer with $4$-velocity $u^{\mu}$ is
\[\omega=u^{\mu}k_{\mu}.\]
Then for a photon the integral of motion is
\begin{align*}
const&=K_{\mu\nu}k^{\mu}k^{\nu} =a^{2}\big[(u^{\mu}k_{\mu})^{2}-k^{\mu}k_{\mu}\big] =a^{2}(\omega^{2})\sim a^{2}p^{2}.
\end{align*}
Thus, both for massive and massless particles we obtained that
\[p\sim 1/a.\]