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.

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.

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 \[p^i \sim U^{i} \sim \frac{1}{a(t)}.\]

Let $\left\{ \vec{r},\vec{p} \right\}$ denote the comoving coordinates and momenta, and $\left\{ \vec{R},\vec{P} \right\}$ the proper ones. Then \[d\vec{R}d\vec{P} =d(a\vec{r})d\left( \frac{\vec{p}}{a} \right)=d\vec{r}d\vec{p}.\]

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.

\[z = \sqrt {\frac{c + V}{c - V}} - 1 = \sqrt{\frac{c + HR}{c - HR}}- 1.\] 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.

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.

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.

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 Shklovkiy was dying he was visited by Rosental, and the latter then told me that Shklovkiy was lying in bed and said: "Well, I understand everything ..." -- he was a remarkable astrophysicist, the best astrophysicist ever, -- "... understand everything\ldots but how could they make a theory in which everything expands with superluminal speed?". So do not feel ashame of the lack of understanding\ldots 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."

The momentum $p$ with respect to the local comoving frame decays as $1/a$ (see problems \ref{equGeo2} and \ref{equ_exp2}). 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.

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).

\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[[:Template:\dot a]]{a} = \left( {\ddot a\chi + \dot a\dot \chi } \right) - {v_{pec}}\frac[[:Template:\dot a]]{a} = \\ &= \left( {\ddot a\chi + \dot a\dot \chi } \right) - a\dot \chi \frac[[:Template:\dot a]]{a} = \ddot a\chi ;\\ q &= - \frac[[:Template:\ddot aa]]{{{{\dot a}^2}}};\\ \ddot D &= - q{H^2}D \end{align*}

As $\dot{D}=HD$, we have \[\ddot D = \dot HD + H\dot D = \dot HD + {H^2}D = \left( {\dot H + {H^2}} \right)D = \frac[[:Template:\ddot a]]{a}D = - q{H^2}D.\] 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.

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 \[1+z_{obs}=\frac{\lambda_{obs}}{\lambda_{rest}} =\frac{\lambda_{obs}}{\lambda_{e}} \frac{\lambda_{e}}{\lambda_{rest}} =(1+z_{rec})(1+z_{pec}).\]

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 (see problem \ref{equ70}) is \[1+z=\frac{\omega_{emit}}{\omega_{obs}} =\frac{a_{obs}}{a_{emit}}.\]

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 \[a\equiv\frac{a_{obs}}{a_{emit}} =\sqrt{\frac{1+v_{rel}}{1-v_{rel}}} \quad\Leftrightarrow\quad v_{rel}=\frac{a^{2}-1}{a^{2}+1}.\]

If we just plug $v_{rel}$ into the relativistic formula for the Doppler effect, we see that \[\frac{\omega_{obs}}{\omega_{emit}}\Bigg|_{v} =\frac{\sqrt{1-v^2}}{1+v} =\sqrt{\frac{1+v}{1-v}} =\Big\lwavy v=v_{rel}\Big\rwavy =\frac{a_{emit}}{a_{obs}},\] so the detected cosmological redshift corresponds exactly to the Doppler effect with velocity $v_{rel}$.

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 problem \ref{equ_oto1a}) \[\gamma=u_{1}^{\mu}u_{2\,\mu},\] 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, \[a=\sqrt{\frac{1+v}{1-v}}.\]

If $\nabla_{(\lambda}K_{\mu\nu)}=0$, then taking into account the geodesic equation $u^{\lambda}\nabla_{\lambda}u^{\mu}=0$, we arrive to \[ u^{\lambda}\nabla_{\lambda} \big(K_{\mu\nu}u^{\mu}u^{\nu}\big)= u^{\lambda}u^{\mu}u^{\nu} \nabla_{\lambda}K_{\mu\nu}= u^{\lambda}u^{\mu}u^{\nu} \nabla_{(\lambda}K_{\mu\nu)}=0.\]

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,i=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 \[\nabla_{(\lambda}K_{\mu\nu)}=0.\]

The Lorentz factor of a particle with $4$-velocity $u^{\mu}$ relative to a particle with $4$-velocity $v^{\mu}$ is (see problem \ref{equ_oto1a}) \[\gamma=u^{\mu}v_{\mu}.\] 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 \[p\sim 1/a.\]