Hydrostatic equilibrium

4.1 Hydrostatic equilibrium

For a neutron star to be in true equilibrium, the spacetime must be stationary as discussed in Section 2.4. This means that the spacetime possesses both “temporal” and “angular” Killing vectors (cf. Ref. [35

]). If the matter is also to be in equilibrium, then the 4-velocity of the matter $uμ$ must be a linear combination of these two Killing vectors. If we use coordinates as defined in (53

) with the angular Killing vector in the $ϕ$ direction, then

Here, $t u$ and $Ω$ are functions of $r$ and $𝜃$ only. $Ω$ is the angular velocity of the matter as measured at infinity.

It is common to define $v$ as the relative velocity between the matter and a normal observer (often called a zero angular momentum observer) so that

The velocity $v$ is then fixed by the normalization condition $uμu μ = − 1$ .

If we assume that the matter source is a perfect fluid, then the stress-energy tensor is given by

where $𝜀$ and $P$ are the total energy density and pressure, respectively, as measured in the rest frame of the fluid. The vanishing of the divergence of the stress-energy tensor yields the equation of hydrostatic equilibrium (often referred to as the relativistic Bernoulli equation). In differential form, this is

If the fluid is barytropic¹² , then we can define the relativistic enthalpy as

and rewrite the relativistic Bernoulli equation as

The constants $h0$ , $t u 0$ , and $Ω0$ are the values their respective quantities have at some reference point, often taken to be the surface of the neutron star at the axis of rotation. When uniform rotation is assumed ( $dΩ = 0$ ), Eq. (118

) is rather easy to solve. The case of differential rotation is somewhat more complicated. An integrability condition of (116

) requires that $utuϕ$ be expressible as a function of $Ω$ , so

$F (Ω)$ is a specifiable function of $Ω$ which determines the rotation law that the neutron star must obey [35