# Difference between revisions of "Energy conditions and the Raychaudhuri equation"

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− | <p style="text-align: left;">The first Friedman equation implies that under the imposed conditions of | + | <p style="text-align: left;">The first Friedman equation implies that under the imposed conditions of "unexpected-ness" density must be finite. However, $p$ can diverge along with $\ddot{a}$ and $\dot{\rho}$: |

\[\frac{\ddot{a}}{a}\sim 4\pi k\,p, | \[\frac{\ddot{a}}{a}\sim 4\pi k\,p, | ||

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=== Problem 14. === | === Problem 14. === | ||

Consider a solution of Friedman equations of the form | Consider a solution of Friedman equations of the form |

## Revision as of 22:22, 23 July 2012

## Energy conditions

S. Carroll writes$^{*}$: * "Sometimes it is useful to think about Einstein's equation without specifying the theory of matter from which $T^{\mu\nu}$ is derived. This leaves us with a great deal of arbitrariness; consider for example the question, What metrics obey Einstein's equation? In the absence of some constraints on $T^{\mu\nu}$, the answer is any metric at all; simply take the metric of your choice, compute the Einstein tensor $G^{\mu\nu}$ for this metric, and then demand that $T^{\mu\nu}$ be equal to $G^{\mu\nu}$. It will automatically be conserved, by the Bianchi identity. Our real concern is with the existence of solutions to Einstein's equation in the presence of "realistic" sources of energy and momentum, whatever that means. One strategy is to consider specific kinds of sources, such as scalar fields, dust, or electromagnetic fields. However, we occasionally wish to understand properties of Einstein's equations that hold for a variety of different sources. In this circumstance it is convenient to impose energy conditions that limit the arbitrariness of $T^{\mu\nu}$." *

The energy conditions are formulated in coordinate-independent way, but in the context of cosmology they are most useful in application to the energy-momentum tensor of a perfect fluid$^*$: \[\begin{array}{lcc} \text{Name}&\text{Statement}&\text{For perfect fluid}\\ \text{Weak}\phantom{\Big|}& T_{\mu\nu}v^{\mu}v^{\nu}\geq0& \rho\geq 0,\quad \rho+p>0;\\[0.2cm] \text{Null}& T_{\mu\nu}k^{\mu}k^{\nu}\geq 0& \rho+p\geq 0;\\[0.2cm] \text{Strong}& (T_{\mu\nu}-\tfrac{1}{2}Tg_{\mu\nu}) v^{\nu}v^{\nu}\geq 0\quad& \quad\rho+p\geq 0,\quad \rho+3p\geq 0;\\[0.2cm] \text{Dominant}& \quad T^{\mu}_{\nu}v^{\nu}\; \mbox{is non-spacelike and future-directed}\quad & \rho\geq |p\,|. \end{array}\] The conditions are assumed to hold for arbitrary timelike vectors $v^{\mu}$ and arbitrary null vectors $k^{\mu}$.

$^*$For more detailed discussion see textbooks: Carroll S. *Spacetime and geometry: an introduction to General Relativity*. AW, 2003; ISBN 0805387323, 525p., and Poisson E. *A relativist's toolkit*. CUP, 2004; ISBN 0521830915, 248p. (ch 2)

### Problem 1.

Derive the energy conditions for the perfect fluid, shown in the last column, from the coordinate-independent formulations.

Let us use the locally Euclidean frame (such that $g_{\mu\nu}=\eta_{\mu\nu}$) comoving with the fluid, so that $u^{\mu}$ in
\begin{equation}\label{EnCond_PFl}
T^{\mu\nu}=\rho u^{\mu}u^{\nu}-pg^{\mu\nu}
\end{equation}
has the coordinates $u^{\mu}=(1,0,0,0)$. Next we choose the $x$-coordinate to be in the direction of the arbitrarily chosen timelike $v^\mu$ or null $k^\mu$ vector. Then those can be parametrized as
\begin{equation}\label{EnCond_VKparametrization}
v^{\mu}=\gamma(1,v,0,0),\qquad k^{\mu}=(1,1,0,0),
\end{equation}
where $v<1$ is the velocity of the particle in the comoving frame, and $\gamma=(1-v^2)^{-1/2}$ its Lorentz factor. We remember that normalization of a null vector is arbitrary.

**1)** *Weak energy condition (WEC).*

Plugging in (\ref{EnCond_VKparametrization}), and taking into account that $v^{\mu}v_{\mu}=1$, we obtain
\[ T_{\mu\nu}v^{\mu}v^{\nu}
=\gamma^{2}(\rho+p)-p
=\gamma^{2}(\rho+v^{2}p)\geq0.\]
Since this must hold for any $v\in[0,1)$, it follows that
\[\rho\geq0,\quad \rho+p> 0.\]

**2)** *Null energy condition (NEC).*

In the same way using $k^{\mu}$ from (\ref{EnCond_VKparametrization}) with $k_{\mu}k^{\mu}=0$, we arrive to
\[T_{\mu\nu}k^{\mu}k^{\nu}
=\rho+p\geq 0,\]
which is already what we wanted to prove. We can also obtain this from the weak energy condition by removing $\gamma$ (normalization is not fixed) and formally replacing $v\to 1$.

**3)** *Strong energy condition (SEC):*
\begin{align*}
&\big[T_{\mu\nu}-\tfrac{1}{2}Tg_{\mu\nu}\big]
v^{\mu}v^{\nu}=
\big[(\rho+p)u_{\mu}u_{\nu}
-\tfrac{1}{2}(\rho-p) g_{\mu\nu}\big]
v^{\mu}v^{\nu}=\\
&\quad=\gamma^{2}(\rho+p)-\tfrac{1}{2}(\rho-p)
=\tfrac{1}{2}\gamma^{2}
\big\{(\rho+3p)+v^{2}(\rho-p)\big\}\geq 0.
\end{align*}
As this must hold for $v\in[0,1)$, we get the conditions
\[\rho+3p\geq 0,\qquad \rho+p\geq 0.\]
Note that in the limit $v=1-0$ ($v$ tends to $1$ from below) the SEC is rewritten as
\[\rho(1-0)+p(1+0)\geq 0,\]
so, as opposed to the WEC, the non-strict inequality sign remains in $\rho+p\geq 0$.
The strong energy condition is in the most simple way formulated for the Ricci tensor:
\[R_{\mu\nu}v^{\mu}v^{\nu}\geq 0,\]
This is the reason it is used in the focusing theorem discussed below.

**4)** *Dominant energy condition (DEC).*

The vector $w^{\mu}=T^{\mu}_{\nu}v^{\nu}$ is the momentum density as measured by the observer with 4-velocity $v^{\mu}$. In the chosen coordinate frame
\[w^{\mu}=\gamma (\rho+p)u^{\mu}-pv^{\mu}
=\gamma (\rho,-pv).\]
As this vector has to be future-directed, we have $\rho\geq 0$ and by demanding for it to be non-spacelike, we arrive to
\[w^{\mu}w_{\mu}
=\gamma^{2}\big(\rho^{2}-v^{2}p^{2})\geq 0.\]
Thus the DEC for a perfect fluid can be written as
\[|p|<\rho,\]
which also implies $\rho\geq 0$.

### Problem 2.

Does the weak energy condition follow from the strong one? Which of the energy conditions imply the others?

No. Small negative $\rho$ with large positive $p$ obey the SEC and NEC but violate WEC and DEC.

Despite the naming, which is a bit confusing here, the *null* energy condition is the weakest, *not* the weak energy condition$^*$. The weak, strong and dominant conditions all imply NEC, but of all three only the dominant energy condition implies the weak one.

All other variants of matter, obeying one condition but violating another, are hypothetically possible.

$^*$Strictly speaking, the weak energy condition allows $\rho+p=0$, which is prohibited by the null condition, but from here on we will not usually distinguish the strict and non-strict equalities. However, this makes a difference when considering the cosmological constant (see chapter on dark energy).

### Problem 3.

Express the energy conditions in terms of scale factor and its derivatives.

With the help of Friedman equations we can express density and pressure through the scale factor and its derivatives \[\rho = \frac{3}{8\pi G} \left[\Big(\frac{\dot a}{a}\Big)^{2} + \frac{k}{a^2}\right],\quad p = - \frac{1}{8\pi G}\left[2\frac{\ddot a}{a} +\Big(\frac{\dot a}{a}\Big)^{2}+\frac{k}{a^2} \right],\] and then rewrite the energy conditions in the form \[\begin{array}{ll} \text{Null:}&\displaystyle \phantom{\text{NEC}} - \frac{\ddot a}{a} +\Big(\frac{\dot a}{a}\Big)^{2} + \frac{k}{a^2} \geq 0;\\[0.4cm] \text{Weak:}&\displaystyle \text{NEC and}\quad \Big(\frac{\dot a}{a}\Big)^{2} +\frac{k}{a^2} \geq 0;\\[0.4cm] \text{Strong:}&\displaystyle\text{NEC and}\quad \frac{\ddot a}{a} \leq 0. \end{array}\] The strong energy condition follows immediately from the second Friedman the expansion of the universe is decelerating equation. We should stress, that the SEC implies that the Unverse is decelerating and this conclusion holds regardless to whether the Universe is open, closed or flat. For the dominant EC we have two variants: $\rho\geq p\geq 0$ and $-p\geq \rho\geq 0$, which lead to \[\text{Dominant}:\quad\left\{ \begin{array}{l}\displaystyle \frac{\ddot a}{a}+\frac{1}{2} \left[\Big(\frac{\dot a}{a}\Big)^{2} +\frac{k}{a^2}\right]\leq 0,\\[0.3cm] \displaystyle \frac{\ddot a}{a}+2\left[ \Big(\frac{\dot a}{a}\Big)^{2}+\frac{k}{a^2}\right]\geq0, \end{array}\right. \quad\text{or}\quad\left\{ \begin{array}{l}\displaystyle \frac{\ddot a}{a}+\tfrac{1}{2} \left[\Big(\frac{\dot a}{a}\Big)^{2} +\frac{k}{a^2}\right]\geq0,\\[0.3cm] \displaystyle \frac{\ddot a}{a} -\Big(\frac{\dot a}{a}\Big)^{2}\geq0, \end{array}\right..\]

### Problem 4.

Express the null, weak and strong energy conditions in terms of the Hubble parameter and redshift.

Using the definitions $H=\dot{a}/a$ and relation to the redshift $a_{0}/a=(z+1)$, we get \[\begin{array}{ll} \text{Null:}&\displaystyle \phantom{\text{NEC and}} \frac{\partial H^2}{\partial z} \ge -\frac{2k\left(1 + z\right)}{a_0^2};\\[0.4cm] \text{Weak:}&\displaystyle\text{NEC and}\quad \frac{k\left(1+z\right)^2}{a_0^2H^2}\geq -1; \\[0.4cm] \text{Strong:}&\displaystyle\text{NEC and}\quad \frac{\partial \ln H}{\partial z}\geq \frac{1}{1+z}. \end{array}\]

### Problem 5.

Find the restrictions that the energy conditions impose on the deceleration parameter in a flat Universe with $\rho>0$.

Dividing one Friedman equation by the other, we get \[q=-\frac{\ddot{a}/a}{H^2} =\frac{1}{2}\big(1+3p/\rho\big)=\frac{1}{2}(1+3w).\] Then from the energy conditions, if we assume $\rho>0$, we straightforwardly obtain \begin{eqnarray*} \text{NEC or WEP:} & 1+w\geq 0 & \Rightarrow\quad q \geq - 1; \\ \text{SEC:} &1+3w\geq 0 & \Rightarrow\quad q \geq 0;\\ \text{DEC:} & |w|<1& \Rightarrow\; -1\leq q \leq 2\end{eqnarray*}

## Raychaudhuri equation

### Problem 6.

Consider a timelike curve $x^{\mu}(\tau)$ and find the projection operators on its tangent vector and on its orthogonal complement.

The decomposition of unit operator into projectors is \[\delta^{\mu}_{\nu}=u^{\mu}u_{\nu}+h^{\mu}_{\nu},\] or in terms of metric \[g_{\mu\nu}=u_{\mu\nu}+h_{\mu\nu},\quad \mbox{where}\quad h_{\mu\nu}=g_{\mu\nu}-u_{\mu}u_{\nu}.\] One can see that $u^{\mu}u_{\nu}$ and $h^{\mu}_{\nu}$ are indeed projection operators from their action on $u^{\mu}$ ($u^{\mu}u_{\mu}=1$): \[(u^{\mu}u_{\nu})u^{\nu}=u^{\mu},\quad h^{\mu}_{\nu}u^{\nu}=0,\] and on any vector $e^{\mu}$, orthogonal to $u^{\mu}$ ($u^{\mu}e_{\mu}=0$): \[(u^{\mu}u_{\nu})e^{\nu}=0,\quad h^{\mu}_{\nu}e^{\nu}=e^{\mu}.\] At the same time the quantity $h_{\mu\nu}$ acts as the metric in the orthogonal complement to linear span of $u^{\mu}$. \[h_{\mu\nu}e^{\mu}f^{\nu} =g_{\mu\nu}e^{\mu}e^{\nu}.\]

### Problem 7.

A *congruence* is a set of curves having the property that each point in a given region belongs to one and only one curve of the set. Consider a congruence of timelike geodesics. Let us mark two infinitely close geodesics and look at their relative evolution along their length. Let $\xi^{\mu}$ be the infinitesimal $4$-vector that is directed normal to one of the curves towards the other. Show that
\[\frac{d\xi_{\nu}}{d\tau}
=B_{\nu\mu}\xi^{\mu},\]
where
\[B_{\nu\mu}=u_{\nu;\mu},\]
is a three-dimensional spacelike tensor orthogonal to $u^{\mu}$, and $\tau$ is the parameter along the geodesic.

Let $\lambda$ be the parameter along a curve with tangent vector $\xi^{\mu}$. Then \[B^{\nu\mu}\xi_{\mu} =\xi^{\mu}\nabla_{\mu}u^{\nu} =\frac{du^{\nu}}{d\lambda} =\frac{dx^{\nu}}{d\tau d\lambda} =\frac{dx^{\nu}}{d\lambda d\tau} =\frac{d\xi^{\nu}}{d\tau}\] and \[\frac{d\xi_{nu}}{d\tau}=B_{\nu\mu}\xi^{\mu}.\] From the geodesic equation and normalization condition \[\left\{\begin{array}{l} B_{\mu\nu}u^{\nu}=u^{\nu}\nabla_{\nu}u_{\mu}=0,\\ B_{\mu\nu}u^{\mu}=u^{\mu}\nabla_{\nu}u_{\mu} =\tfrac{1}{2}\nabla_{\nu}(u^{\mu}u_{\mu}) \sim\nabla_{\nu} (1)=0. \end{array}\right.\] On the other hand, we can always choose a coordinate frame in such a way that locally $g_{\mu\nu}=\eta_{\mu\nu}$, and by additional Lorentz boost we can direct the time along $u^{\mu}$. If ${e^{\mu}}_{i}$ ($i,j=1,2,3$) are the orthonormal unit vectors of spatial directions, such that $h_{\mu\nu}{e^{\mu}}_{i}{e^{\nu}}_{j}=\eta_{ij}=-\delta_{ij}$, then due to orthogonality $B$ is expressible as \[B^{\mu\nu}=B_{ij}{e^{\mu}}_{i}{e^{\nu}}_{j},\] so in particular we obtain $B_{\mu\nu}B^{\mu\nu}=B_{ij}B^{ij}\geq 0$.

### Problem 8.

Show that any tensor field of second rank defined on an $n-$dimensional Riemannian manifold with positive definite metric $g_{\mu\nu}$ can be uniquely decomposed into \begin{equation}\label{TensorDecomposition} B_{\mu\nu}=\frac{1}{n}\Theta g_{\mu\nu} +\sigma_{\mu\nu}+\omega_{\mu\nu}, \end{equation} where $\Theta={B^{\mu}}_{\mu}$, $\sigma_{\mu\nu}$ is the symmetric traceless part of $B_{\mu\nu}$, and $\omega_{\mu\nu}$ is the antisymmetric part of $B_{\mu\nu}$.

Let us decompose $B_{\mu\nu}$ into the symmetric and anti-symmetric parts (symmetrization is denoted by parenthesis $()$ and anti-symmetrization by brackets $[]$): \[B_{\mu\nu} =\frac{B_{\mu\nu}+B_{\nu\mu}}{2} +\frac{B_{\mu\nu}-B_{\nu\mu}}{2} =B_{(\mu\nu)}+B_{[\mu\nu]}.\] Then search for the representation of the symmetrical part in the form \[B_{(\mu\nu)}=\alpha g_{\mu\nu}+\sigma_{\mu\nu}, \quad\mbox{where}\quad \sigma^{\mu}_{\mu}=0.\] The contraction is $\Theta=B^{\mu}_{\mu}=\alpha \delta^{\mu}_{\mu}=\alpha n$, so we obtain $\alpha=\Theta/n$ and thus derive the representation of $B_{\mu\nu}$ we had sought for.

### Problem 9.

Let there be a congruence in a three-dimensional Riemannian manifold. What is the geometric meaning of $\Theta$, $\sigma$ and $\omega$ for $B_{\mu\nu}=u_{\nu;\mu}$?

Let the vector $\xi^{\mu}$ be a two-dimensional vector that lies in the space $S$ orthogonal to the tangent vector, and let us consider the pipe of curves from the congruence in the neighborhood of the given curve and the evolution of its form with the parameter $\tau$. Each of the generating lines of the pipe is parametrized by the corresponding vector $\xi^{\mu}$.

a) Let $B_{\mu\nu}=\frac{\Theta}{2}g_{\mu\nu}$. Then \[B^{\mu}_{\nu}=\frac{\Theta}{2} \begin{pmatrix} 1&0\\0&1\end{pmatrix} \quad\mbox{and}\quad \frac{d\xi^{\mu}}{d\tau} =\frac{\Theta}{2}\xi^{\mu},\] so the pipe's section is merely stretched $(1+\Theta/2)$ times. The area $S$ of the pipe's section changes as $dS/S=d(|\xi|^2)/|\xi|^2 = \Theta d\tau$, so \[\Theta=\frac{1}{S}\frac{dS}{d\tau}.\] b) The matrix of the anti-symmetric part is \[{\omega^{\mu}}_{\nu}=\begin{pmatrix} 0&\omega\\-\omega&0 \end{pmatrix},\] and it describes the torsion of the pipe, without changing its section area. c) The matrix of the symmetric traceless part can be presented in the form \[\sigma^{\mu}_{\nu}= \begin{pmatrix}\sigma_{+}&\sigma_{\times}\\ \sigma_{\times}&-\sigma_{+}\end{pmatrix} =\sigma_{+} \begin{pmatrix}1&0\\ 0&-1\end{pmatrix} +\sigma_{\times} \begin{pmatrix}1&0\\ 0&-1\end{pmatrix} \begin{pmatrix}0&-1\\ -1&0\end{pmatrix}.\] The first term describes expansion in one direction in $(1+\sigma_{+})$ times, and contraction by the same factor in the perpendicular direction. The second term is the superposition of the same deformation without change of volume and a rotation. So the general deformation is a shear in arbitrary direction, without change of section area.

### Problem 10.

Derive the Raychaudhuri equation for a congruence of timelike geodesics in spacetime \begin{equation}\label{Raychaudhuri} \frac{d\Theta}{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^{\mu}. \end{equation} Here $\Theta$, $\sigma_{\mu\nu}$ and $\omega_{\mu\nu}$ are the components (\ref{TensorDecomposition}) of decomposition of $B_{\mu\nu}=u_{\mu;\nu}$.

Let us first consider the evolution of tensor $B_{\mu\nu}$ along the geodesic: changing the order of the covariant derivatives and using one of the definitions of the curvature tensor, we get \begin{align*} \frac{dB_{\mu\nu}}{d\tau}& =u^{\lambda}\nabla_{\lambda}\nabla_{\nu}u_{\mu} =\\& =u^{\lambda}\nabla_{\nu}\nabla_{\lambda}u_{\mu} +u^{\lambda}{R^{\sigma}}_{\mu\nu\lambda}u_{\sigma} =\\& =\nabla_{\nu}(u^{\lambda}\nabla_{\lambda}u_{\mu}) -(\nabla_{\nu}u^{\lambda})(\nabla_{\lambda}u_{\mu}) -R_{\rho\mu\sigma\lambda}u^{\rho}u^{\sigma} =\\& =-{B_{\nu}}^{\lambda}B_{\lambda\mu} -R_{\mu\rho\lambda\sigma}u^{\rho}u^{\sigma}. \end{align*} On contraction we obtain \[\frac{d\Theta}{d\tau}=-B^{\mu\nu}B_{\nu\mu} -R_{\mu\nu}u^{\mu}u^{\nu}.\] Now rewrite the contraction $B_{\mu\nu}B_{\nu\mu}$ in terms of components of $B$'s decomposition and take into account that pairwise contractions are zero due to the symmetries and due to $h^{\mu\nu}\sigma_{\mu\nu}=g^{\mu\nu}\sigma_{\mu\nu}=0$: \[B^{\mu\nu}B_{\nu\mu} =\frac{\Theta^2}{9}h^{\mu\nu}h_{\mu\nu} +\sigma^{\mu\nu}\sigma_{\mu\nu} -\omega^{\mu\nu}\omega_{\mu\nu}\] (the last term goes with the opposit sign due to anti-symmetry of $\omega^{\mu\nu}\omega_{\nu\mu}=-\omega^{\mu\nu}\omega_{\mu\nu}$), so taking into account that $h^{\mu\nu}h_{\mu\nu}=3$, we derive the Raychaudhuri equation (\ref{Raychaudhuri}).

### Problem 11.

Show that for a congruence of geodesics orthogonal to a family of hypersurfaces $\omega_{\mu\nu}=0$. Prove further, that in case the strong energy condition $R_{\mu\nu}u^{\mu}u^{\nu}\geq0$ holds (see problem and the following), then the following (the focusing theorem) also is true: if $\Theta=\Theta_{0}<0$ at some initial moment, then in a finite period of proper time $\Theta$ diverges and tends to $-\infty$.

Orthogonality of a congruence to a family of hypersurfaces means that there is a scalar field $\phi(x)$, the level lines of which $\phi(x)=const$ define this family and $u_{\mu}=\alpha \phi_{,\nu}$. Then \[\omega_{\mu\nu}=B_{[\mu\nu]}=u_{[\mu;\nu]} =\alpha\phi_{[;\mu;\nu]}+\alpha_{[,\nu}\phi_{,\mu]} =\alpha_{[,\nu}\phi_{,\mu]} \sim \alpha_{[,\nu}u_{\mu]} =\alpha_{,\nu}u_{\mu}-\alpha_{,\mu}u_{\nu}.\] But on the other hand \[0=\omega_{\mu\nu}u^{\nu} =\alpha_{,\nu}u^{\nu}u_{\mu}-\alpha_{,\mu},\] so $\alpha_{,\mu}\sim u_{\mu}$. But then $\omega_{\mu\nu}\sim u_{[\mu}u_{\nu]}=0$, and thus torsion turns to zero. Then in case the strong energy condition $R_{\mu\nu}u^{\mu}u^{\nu}$ holds, there are only non-positive terms left in the right hand part of the Raychaudhuri equation and \[\frac{d\Theta}{d\tau}\leq -\frac{1}{3}\Theta^{2}.\] On integration \[\frac{1}{\Theta}-\frac{1}{\Theta_0} \geq \frac{\Delta\tau}{3}\quad\Leftrightarrow\quad \Theta\leq \frac{\Theta_{0}} {1+\Delta\tau \Theta_{0}/3},\] so if at some moment $\Theta_{0}<0$, then for $\Delta\tau\to |\Theta_{0}|/3$ we obtain from the inequality that $\Theta\to -\infty$, i.e. the geodesics of the congruence are focused into a point. The focusing theorem implies that the geodesics form a caustics, which in general does not necessarily mean a singularity. Some further elaboration is needed, making use of the focusing theorem and usually some energy conditions, in order to prove that a singularity actually takes place$^{*}$

$^{*}$ For more on this better see textbooks Hawking S.W., Ellis G.F.R. *The large scale structure of space-time*, CUP, 1973 (ISBN 0521099064) and Wald R.M, *General relativity.* U. Chicago, 1984, 505p (ISBN 0226870332).

### Problem 12.

Write out the Raychaudhuri equation for the geodesics of comoving matter in the FLRW Universe and show that it is reduced to the second Friedman equation.

Let the metric have the form \[ds^{2}=dt^{2} -a^{2}(t)h_{ij}dx^{i}dx^{j}.\] Consider the congruence of geodesics orthogonal to the hypersurfaces of constant cosmic time $t$, so that torsion $\omega_{\mu\nu}$ is zero and $u_{\mu}=\partial_{\mu}t=\delta_{\mu}^{t}$. Then \[B_{\mu\nu}=u_{\nu;\mu}=u_{\nu,\mu} +{\Gamma^{\lambda}}_{\mu\nu}u_{\lambda} ={\Gamma^{0}}_{\mu\nu}=\Gamma_{0,\mu\nu} =-\tfrac{1}{2}\partial_{t}g_{\mu\nu}\] and for the spatial components, which are the only ones different from zero, \[B_{ij} =-\tfrac{1}{2}\frac{\partial_{t}a^2}{a^2}(-h_{ij}) =\frac{\dot{a}}{a}h_{ij}.\] We see from the explicit notation (as $t$ is also the natural parameter along the geodesics) that \[\Theta=3\frac{\dot{a}}{a},\quad \sigma_{\mu\nu}=0,\quad \omega_{\mu\nu}=0.\] and the Raychaudhuri equation (\ref{Raychaudhuri}) is rewritten as \[\frac{d}{dt}\Big(3\frac{\dot{a}}{a}\Big) =-3\Big(\frac{\dot{a}}{a}\Big)^{2}-R_{00},\] thus \[R_{00}=-\dot{\Theta}-\Theta^{2}/3 =-3\Big[\frac{d}{dt}\Big(\frac{\dot{a}}{a}\Big) +\Big(\frac{\dot{a}}{a}\Big)^{2}\Big] =-3\frac{\ddot{a}}{a}.\] On substitution of this into the $\binom{0}{0}$ component of the Einstein's equation \[R_{00} =\frac{8\pi G}{c^4}\big(T_{00}-\tfrac{1}{2}Tg_{00}) =\frac{4\pi G}{c^4}\big(\varepsilon +3p\big),\] we obtain the second Friedman equation.

We see now that the focusing theorem in the cosmological context implies that the geodesics of comoving matter must converge at some time in the past, as is already known to be a major feature of the solutions of the Friedman equations. Though in general a caustic of geodesics does not necessarily mean a singularity, in this case the considered geodesics are actually the geodesics of all the comoving matter in the Universe, so their focusing actually implies the Big Band singularity. The theorem becomes inapplicable, however, at small times, when particles' interaction has to be taken into account. This is where the more general singularity theorems work$^{*}$

$^{*}$ For more on this better see textbooks Hawking S.W., Ellis G.F.R. *The large scale structure of space-time*, CUP, 1973 (ISBN 0521099064) and Wald R.M, *General relativity.* U. Chicago, 1984, 505p (ISBN 0226870332).

## Sudden Future Singularities

The following problems are composed in the spirit of (John D. Barrow, Sudden Future Singularities, arXiv:0403084v3).

### Problem 13.

Let us consider the possibility of \textit{sudden future singularities}. The ``suddenness* implies that they occur at some time in the future, while both the scale factor and the Hubble constant remain bounded and separated from zero:*
\[a\to a_{s}\neq 0,\infty,
\qquad H\to H_{s}\neq 0,\infty.\]
What scalars can in principle become unbounded in this scenario?

The first Friedman equation implies that under the imposed conditions of "unexpected-ness" density must be finite. However, $p$ can diverge along with $\ddot{a}$ and $\dot{\rho}$: \[\frac{\ddot{a}}{a}\sim 4\pi k\,p, \qquad \dot{\rho}\sim -3Hp.\]

### Problem 14.

Consider a solution of Friedman equations of the form \[a(t)=A+Bt^{q}+C(t_{s}-t)^{n},\] where $A,B,q,n>0$ and $C$ are some free constants. What values of $q$ and $n$ are compatible with the sudden singularity of the previous problem?

Fixing the zero time by condition $a(0)=0$ and using the freedom of rescaling the scale factor, we can rewrite the solution as \[a(t)=1+(a_{s}-1)\Big(\frac{t}{t_s}\Big)^{q} -\Big(1-\frac{t}{t_s}\Big)^{n},\] where $a_{s}=a(t_{s})$. Then as $t\to t_{s}-0$ we have \begin{equation}\label{sing_eq} \ddot{a}=\frac{q(q-1)}{t_s^q}(a_{s}-1)t^{q-2} -\frac{n(n-1)}{t_s^2}\Big(1-\frac{t}{t_s}\Big)^{n-2} \to -\infty. \end{equation} The solution with singularity exists on the interval $0<t<t_s$ for \[1<n<2 \quad\text{and}\quad 0<q\leq 1.\] Note that $n>1$ is needed for $\dot{a}$ to stay finite and $n<2$ for $\ddot{a}$ to diverge. For $2<n<3$ the values of $\ddot{a}$ and $p$ would remain bounded, while $\dddot{a}$ and $\dot{p}$ diverge.

### Problem 15.

Is any energy condition violated by the solutions with the sudden future singularity? What physical constraint on matter can be introduced that would prevent it?

Due to the second Friedman equation, we see that as (\ref{sing_eq}), both $\rho$ and $(\rho+3p)$ remain positive, so all common energy conditions are satisfied. The considered singularity is possible, however, only if pressure is allowed to be unbounded at finite values of density. If we demand that $p<C\rho$ for some $C>0$, for example, along with the common conditions $\rho>0$ and $\rho+3p>0$, then the sudden future singularity is eliminated.