 Solution

The sign of $T_{ab}{X}^a{X}^b$ is unchanged by a nonzero rescaling of the timelike vector $X$. Normalize it to a four-velocity $W$ and, after reversing its direction if necessary, write $W=\gamma (u+w)$, where $u\cdot w=0$, $w\cdot w=s\in [0,1)$ and $\gamma =(1-s{)}^{-1/2}$ is the Lorentz factor. For the perfect fluid in general relativity,Taking $W=u$ requires $\rho \ge 0$. Allowing the Lorentz factor to become arbitrarily large requires $\rho +p\ge 0$; otherwise sufficiently fast observers measure a negative value. Conversely, these two inequalities make both terms in the last expression nonnegative. Therefore the weak energy condition for a perfect fluid is exactlyProperty (1) is the weak energy condition: every physical observer measures nonnegative local energy density, namely $T_{ab}{W}^a{W}^b$ for their normalized four-velocity $W$. It allows negative pressure down to $p=-\rho$.