Editing Hydrostatic equilibrium (section)

== Mathematical consideration ==
[[File:Hydrostatic equilibrium.svg|thumb|right|If the highlighted volume of fluid is not accelerating, the forces on it upwards must equal the forces downwards.]]

For a hydrostatic fluid on Earth:
<math display="block">dP = - \rho(P) \, g(h) \, dh</math>

=== Derivation from force summation ===
{{further|Mechanical equilibrium}}
[[Newton's laws of motion]] state that a volume of a fluid that is not in motion or that is in a state of constant velocity must have zero net force on it. This means the sum of the forces in a given direction must be opposed by an equal sum of forces in the opposite direction. This force balance is called a hydrostatic equilibrium.

The fluid can be split into a large number of [[cuboid]] volume elements; by considering a single element, the action of the fluid can be derived.

There are three forces: the force downwards onto the top of the cuboid from the pressure, ''P'', of the fluid above it is, from the definition of [[pressure]],
<math display="block">F_\text{top} = - P_\text{top} A</math>
Similarly, the force on the volume element from the pressure of the fluid below pushing upwards is
<math display="block">F_\text{bottom} = P_\text{bottom} A</math>

Finally, the [[weight]] of the volume element causes a force downwards.  If the [[density]] is ''ρ'', the volume is ''V'' and ''g'' the [[standard gravity]], then:
<math display="block">F_\text{weight} = -\rho g V</math>
The volume of this cuboid is equal to the area of the top or bottom, times the height&nbsp;– the formula for finding the volume of a cube.
<math display="block">F_\text{weight} = -\rho g A h</math>

By balancing these forces, the total force on the fluid is
<math display="block">\sum F = F_\text{bottom} + F_\text{top} + F_\text{weight} = P_\text{bottom} A - P_\text{top} A - \rho g A h</math>
This sum equals zero if the fluid's velocity is constant.  Dividing by A,
<math display="block">0 = P_\text{bottom} - P_\text{top} - \rho g h</math>
Or,
<math display="block">P_\text{top} - P_\text{bottom} = - \rho g h</math>
''P''<sub>top</sub> − ''P''<sub>bottom</sub> is a change in pressure, and ''h'' is the height of the volume element—a change in the distance above the ground.  By saying these changes are [[infinitesimal]]ly small, the equation can be written in [[differential equation|differential]] form.
<math display="block">dP = - \rho g \, dh</math>
Density changes with pressure, and gravity changes with height, so the equation would be:
<math display="block">dP = - \rho(P) \, g(h) \, dh</math>

=== Derivation from Navier–Stokes equations ===
Note finally that this last equation can be derived by solving the three-dimensional [[Navier–Stokes equations]] for the equilibrium situation where
<math display="block">u = v = \frac{\partial p}{\partial x} = \frac{\partial p}{\partial y} = 0</math>
Then the only non-trivial equation is the <math>z</math>-equation, which now reads
<math display="block">\frac{\partial p}{\partial z} + \rho g = 0</math>
Thus, hydrostatic balance can be regarded as a particularly simple equilibrium solution of the Navier–Stokes equations.

=== Derivation from general relativity ===
By plugging the [[energy–momentum tensor]] for a [[perfect fluid]]
<math display="block">T^{\mu\nu} = \left(\rho c^{2} + P\right) u^\mu u^\nu + P g^{\mu\nu}</math>
into the [[Einstein field equations]]
<math display="block">R_{\mu\nu} = \frac{8\pi G}{c^4} \left(T_{\mu\nu} - \frac{1}{2} g_{\mu\nu} T\right)</math>
and using the conservation condition
<math display="block">\nabla_\mu T^{\mu\nu} = 0</math>
one can derive the [[Tolman–Oppenheimer–Volkoff equation]] for the structure of a static, spherically symmetric relativistic star in isotropic coordinates:
<math display="block">\frac{dP}{dr} = -\frac{G M(r)\rho(r)}{r^2} \left(1+\frac{P(r)}{\rho(r)c^2}\right) \left(1+\frac{4\pi r^3 P(r)}{M(r) c^2}\right) \left(1 - \frac{2GM(r)}{r c^2}\right)^{-1}</math>
In practice, ''Ρ'' and ''ρ'' are related by an equation of state of the form ''f''(''Ρ'',''ρ'')&nbsp;=&nbsp;0, with ''f'' specific to makeup of the star. ''M''(''r'') is a foliation of spheres weighted by the mass density ''ρ''(''r''), with the largest sphere having radius ''r'':
<math display="block">M(r) = 4\pi \int_0^r dr' \, r'^2 \rho(r').</math>
Per standard procedure in taking the nonrelativistic limit, we let {{nowrap|''c'' → ∞}}, so that the factor
<math display="block">\left(1+\frac{P(r)}{\rho(r)c^2}\right) \left(1+\frac{4\pi r^3P(r)}{M(r)c^2}\right) \left(1-\frac{2GM(r)}{r c^2} \right)^{-1} \rightarrow 1</math>
Therefore, in the nonrelativistic limit the Tolman–Oppenheimer–Volkoff equation reduces to Newton's hydrostatic equilibrium:
<math display="block">\frac{dP}{dr} = -\frac{GM(r)\rho(r)}{r^2} = -g(r)\,\rho(r)\longrightarrow dP = - \rho(h)\,g(h)\, dh</math>
(we have made the trivial notation change ''h''&nbsp;=&nbsp;''r'' and have used ''f''(''Ρ'',''ρ'')&nbsp;=&nbsp;0 to express ''ρ'' in terms of ''P'').<ref>{{cite book|last1=Zee|first1=A.|title=Einstein gravity in a nutshell|date=2013|publisher=Princeton University Press | location=Princeton | isbn=9780691145587|pages=451–454}}</ref> A similar equation can be computed for rotating, axially symmetric stars, which in its gauge independent form reads:
<math display="block">\frac{\partial_i P}{P+\rho} - \partial_i \ln u^t + u_t u^\varphi\partial_i\frac{u_\varphi}{u_t}=0</math> 
Unlike the TOV equilibrium equation, these are two equations (for instance, if as usual when treating stars, one chooses spherical coordinates as basis coordinates <math>(t,r,\theta,\varphi)</math>, the index ''i'' runs for the coordinates ''r'' and <math>\theta</math>).