Problem 3.7.1: Energy conditions for perfect fluid
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
\[\rho\geq0,\quad \rho+p\geq 0.\]
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 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
\[|p|\leq \rho,\]
which also implies $\rho\geq 0$.