Past exam of the general relativity II course of the University of Oxford / 2021 / 1

Write $D=u^a{\nabla }_a$, $\theta ={\nabla }_a{u}^a$ for the expansion scalar, $a_b=u^a{\nabla }_a{u}_b$ for the four-acceleration, and $P_b{}^c={\delta }_b{}^c+u_b{u}^c$ for the spatial projector of a timelike congruence. The normalization of the four-velocity gives $u^b{a}_b=0$ and $u^b{\nabla }_a{u}_b=0$. The product rule and metric compatibility give the covariant derivative of the perfect fluid in general relativity stress-energy tensor:Its tensor contraction with the four-velocity isand its projection with the spatial projector of a timelike congruence isThus stress-energy conservation implies the relativistic perfect-fluid energy equation and the relativistic Euler equation:Conversely, the two displayed equations set both projections of $F_b$ to zero. Since $F_b=P_b{}^c{F}_c-u_b{u}^c{F}_c$, they imply $F_b=0$. This establishes equivalence, including points where $\rho +p=0$, without dividing by that quantity.

Use stress-energy conservation and symmetry of the stress-energy tensor. The Killing equation for the Killing vector field $V$ then gives the stress-energy current from a Killing vector:Hence $\boxed{{\nabla }^a{J}_a=0}$. The assumption of stress-energy conservation, supplied by the equations of motion in part (a), is essential; the Killing equation alone only eliminates the second term.

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 $p=-\rho$.

For relativistic dust, $p=0$, the relativistic Euler equation gives $\rho a_b=0$. The nowhere-vanishing density therefore makes the four-acceleration zero: $u$ is a timelike geodesic vector field. Let $\alpha =u^{♭}$ be the differential one-form obtained by the musical isomorphism. Its exterior derivative satisfiesHere ${\iota }_u$ is the interior product. By hypersurface orthogonality and the Frobenius theorem, $\alpha \wedge d\alpha =0$. Taking the interior product and using $\alpha (u)=-1$ givesThus $\boxed{d{u}^{♭}=0}$. By the Poincaré lemma, locally $\alpha =-dt$, and then $u(t)=-\alpha (u)=1$. Choose spatial coordinate functions $x^i$ on one level set of $t$ and extend them constantly along the integral curves of a vector field $u$, using its smooth flow. This constructs $u={\partial }_t$. Since $dt({\partial }_i)=0$,The metric on the orthogonal complement of a timelike vector is positive-definite, so the synchronous coordinates for irrotational dust haveThis construction is local; crossing geodesics can prevent these synchronous coordinates from covering a larger region.

In the prescribed synchronous coordinates, the Christoffel symbols satisfyConsequently the four-acceleration of $u={\partial }_t$ vanishes. For a spatial index $i$, $u_i=0$, so the spatial component of the relativistic Euler equation becomesOn a connected coordinate neighbourhood,A nonzero pressure is allowed, but its spatial gradient must vanish if the fluid's four-velocity follows these freely falling coordinate lines. No division by $p$ or $\rho +p$ is needed.