What does it mean when a star is in hydrostatic equilibrium?

In this series we are exploring the weird and wonderful world of astronomy jargon! You’ll feel balanced with today’s topic: hydrostatic equilibrium!

Table of Contents Show

Derivation from force summation
Derivation from Navier–Stokes equations
Derivation from general relativity
Astrophysics
Planetary geology
Atmospheric modeling

Hydrostatic equilibrium is a state of balance between gravity, which wants to pull things together, and pressure, which wants to blow it apart. It appears a lot in astrophysics, from the Earth’s own atmosphere to gigantic clusters of galaxies.

The word “hydro” appears because the concept was first studied in the context of fluids, but it can apply to concepts like the composition of rocky planets. The word “static” implies that once this state is achieved, it’s going to hang out like that for a long time.

The Earth’s atmosphere is in a state of hydrostatic equilibrium. The gravity of the Earth is constantly pulling the atmosphere down. If nothing could resist it, all the air around the Earth would compress down into a thin shell. On the other hand, the atmosphere is warm, and that warmth has a pressure associated with it. With only the pressure, our atmosphere would blow off into space.

The inward pull of gravity balances the outward pressure, and the two are locked. In the end, the atmosphere stays nice and thick…and exactly where it is.

The same thing happens to clouds of interstellar gas and dust. These nebulae can maintain hydrostatic equilibrium for millions of years, but if they become disturbed or acquire too much mass, gravity can win and they begin to collapse, forming stars in the process.

On the very largest scales, entire galaxy clusters maintain hydrostatic equilibrium. The plasma that fills the space between galaxies, called the intracluster medium, is incredibly hot. It would all collapse into the center of the cluster if it wasn’t able to support itself from its own pressure.

The concept of hydrostatic equilibrium figures prominently in the definition of a planet. Objects that are too small don’t have enough gravity to pull themselves into a spherical shape. However, larger objects, like the Earth, can achieve it.

Lastly, stars themselves are in hydrostatic equilibrium, which allows them to shine for billions of years or longer. The outward pressure generated by the energy of the fusion reactions balances the pull of gravity. If the star were to begin fusing too much hydrogen, the extra energy would puff the star out, reducing the amount of fuel and lowering the fusion rate, returning the star to equilibrium.

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In fluid mechanics, hydrostatic equilibrium (hydrostatic balance, hydrostasy) is the condition of a fluid or plastic solid at rest, which occurs when external forces, such as gravity, are balanced by a pressure-gradient force.[1] In the planetary physics of Earth, the pressure-gradient force prevents gravity from collapsing the planetary atmosphere into a thin, dense shell, whereas gravity prevents the pressure-gradient force from diffusing the atmosphere into outer space.[2][3]

Diagram of a newly formed planet in a state of hydrostatic equilibrium.

Hydrostatic equilibrium is the distinguishing criterion between dwarf planets and small solar system bodies, and features in astrophysics and planetary geology. Said qualification of equilibrium indicates that the shape of the object is symmetrically ellipsoid, where any irregular surface features are consequent to a relatively thin solid crust. In addition to the Sun, there are a dozen or so equilibrium objects confirmed to exist in the Solar System.

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:

d P = − ρ ( P ) ⋅ g ( h ) ⋅ d h {\displaystyle dP=-\rho (P)\cdot g(h)\cdot dh}

Derivation from force summation

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 3 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,

F t o p = − P t o p ⋅ A {\displaystyle F_{top}=-P_{top}\cdot A}

Similarly, the force on the volume element from the pressure of the fluid below pushing upwards is

F b o t t o m = P b o t t o m ⋅ A {\displaystyle F_{bottom}=P_{bottom}\cdot A}

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:

F w e i g h t = − ρ ⋅ g ⋅ V {\displaystyle F_{weight}=-\rho \cdot g\cdot V}

The volume of this cuboid is equal to the area of the top or bottom, times the height — the formula for finding the volume of a cube.

F w e i g h t = − ρ ⋅ g ⋅ A ⋅ h {\displaystyle F_{weight}=-\rho \cdot g\cdot A\cdot h}

By balancing these forces, the total force on the fluid is

∑ F = F b o t t o m + F t o p + F w e i g h t = P b o t t o m ⋅ A − P t o p ⋅ A − ρ ⋅ g ⋅ A ⋅ h {\displaystyle \sum F=F_{bottom}+F_{top}+F_{weight}=P_{bottom}\cdot A-P_{top}\cdot A-\rho \cdot g\cdot A\cdot h}

This sum equals zero if the fluid's velocity is constant. Dividing by A,

0 = P b o t t o m − P t o p − ρ ⋅ g ⋅ h {\displaystyle 0=P_{bottom}-P_{top}-\rho \cdot g\cdot h}

Or,

P t o p − P b o t t o m = − ρ ⋅ g ⋅ h {\displaystyle P_{top}-P_{bottom}=-\rho \cdot g\cdot h}

Ptop − Pbottom 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 infinitesimally small, the equation can be written in differential form.

d P = − ρ ⋅ g ⋅ d h {\displaystyle dP=-\rho \cdot g\cdot dh}

Density changes with pressure, and gravity changes with height, so the equation would be:

d P = − ρ ( P ) ⋅ g ( h ) ⋅ d h {\displaystyle dP=-\rho (P)\cdot g(h)\cdot dh}

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

u = v = ∂ p ∂ x = ∂ p ∂ y = 0 {\displaystyle u=v={\frac {\partial p}{\partial x}}={\frac {\partial p}{\partial y}}=0}

Then the only non-trivial equation is the $z {\displaystyle z}$ -equation, which now reads

∂ p ∂ z + ρ g = 0 {\displaystyle {\frac {\partial p}{\partial z}}+\rho g=0}

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

T μ ν = ( ρ c − 2 + P ) u μ u ν + P g μ ν {\displaystyle T^{\mu \nu }=(\rho c^{-2}+P)u^{\mu }u^{\nu }+Pg^{\mu \nu }}

into the Einstein field equations

R μ ν = 8 π G c 4 ( T μ ν − 1 2 g μ ν T ) {\displaystyle R_{\mu \nu }={\frac {8\pi G}{c^{4}}}(T_{\mu \nu }-{\frac {1}{2}}g_{\mu \nu }T)}

and using the conservation condition

∇ μ T μ ν = 0 {\displaystyle \nabla _{\mu }T^{\mu \nu }=0}

one can derive the Tolman–Oppenheimer–Volkoff equation for the structure of a static, spherically symmetric relativistic star in isotropic coordinates:

d P d r = − G M ( r ) ρ ( r ) r 2 ( 1 + P ( r ) ρ ( r ) c 2 ) ( 1 + 4 π r 3 P ( r ) M ( r ) c 2 ) ( 1 − 2 G M ( r ) r c 2 ) − 1 {\displaystyle {\frac {dP}{dr}}=-{\frac {GM(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)}{rc^{2}}}\right)^{-1}}

In practice, Ρ and ρ are related by an equation of state of the form f(Ρ,ρ)=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:

M ( r ) = 4 π ∫ 0 r d r ′ r ′ 2 ρ ( r ′ ) . {\displaystyle M(r)=4\pi \int _{0}^{r}dr'r'^{2}\rho (r').}

Per standard procedure in taking the nonrelativistic limit, we let c→∞, so that the factor

( 1 + P ( r ) ρ ( r ) c 2 ) ( 1 + 4 π r 3 P ( r ) M ( r ) c 2 ) ( 1 − 2 G M ( r ) r c 2 ) − 1 → 1 {\displaystyle \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)}{rc^{2}}}\right)^{-1}\rightarrow 1}

Therefore, in the nonrelativistic limit the Tolman–Oppenheimer–Volkoff equation reduces to Newton's hydrostatic equilibrium:

d P d r = − G M ( r ) ρ ( r ) r 2 = − g ( r ) ρ ( r ) ⟶ d P = − ρ ( h ) g ( h ) d h {\displaystyle {\frac {dP}{dr}}=-{\frac {GM(r)\rho (r)}{r^{2}}}=-g(r)\,\rho (r)\longrightarrow dP=-\rho (h)\,g(h)\,dh}

(we have made the trivial notation change h=r and have used f(Ρ,ρ)=0 to express ρ in terms of P).[4] A similar equation can be computed for rotating, axially symmetric stars, which in its gauge independent form reads:

∂ i P P + ρ − ∂ i ln ⁡ u t + u t u ϕ ∂ i u ϕ u t = 0 {\displaystyle {\frac {\partial _{i}P}{P+\rho }}-\partial _{i}\ln u^{t}+u_{t}u^{\phi }\partial _{i}{\frac {u_{\phi }}{u_{t}}}=0}

Unlike the TOV equilibrium equation, these are two equations (for instance, if as usual when treating stars, one chooses spherical coordinates as basis coordinates $( t , r , θ , ϕ ) {\displaystyle (t,r,\theta ,\phi )}$ , the index i runs for the coordinates r and $θ {\displaystyle \theta }$ ).

The hydrostatic equilibrium pertains to hydrostatics and the principles of equilibrium of fluids. A hydrostatic balance is a particular balance for weighing substances in water. Hydrostatic balance allows the discovery of their specific gravities. This equilibrium is strictly applicable when an ideal fluid is in steady horizontal laminar flow, and when any fluid is at rest or in vertical motion at constant speed. It can also be a satisfactory approximation when flow speeds are low enough that acceleration is negligible.

Astrophysics

In any given layer of a star, there is a hydrostatic equilibrium between the outward thermal pressure from below and the weight of the material above pressing inward. The isotropic gravitational field compresses the star into the most compact shape possible. A rotating star in hydrostatic equilibrium is an oblate spheroid up to a certain (critical) angular velocity. An extreme example of this phenomenon is the star Vega, which has a rotation period of 12.5 hours. Consequently, Vega is about 20% larger at the equator than at the poles. A star with an angular velocity above the critical angular velocity becomes a Jacobi (scalene) ellipsoid, and at still faster rotation it is no longer ellipsoidal but piriform or oviform, with yet other shapes beyond that, though shapes beyond scalene are not stable.[5]

If the star has a massive nearby companion object then tidal forces come into play as well, distorting the star into a scalene shape when rotation alone would make it a spheroid. An example of this is Beta Lyrae.

Hydrostatic equilibrium is also important for the intracluster medium, where it restricts the amount of fluid that can be present in the core of a cluster of galaxies.

We can also use the principle of hydrostatic equilibrium to estimate the velocity dispersion of dark matter in clusters of galaxies. Only baryonic matter (or, rather, the collisions thereof) emits X-ray radiation. The absolute X-ray luminosity per unit volume takes the form $L X = Λ ( T B ) ρ B 2 {\displaystyle {\mathcal {L}}_{X}=\Lambda (T_{B})\rho _{B}^{2}}$ where $T B {\displaystyle T_{B}}$ and $ρ B {\displaystyle \rho _{B}}$ are the temperature and density of the baryonic matter, and $Λ ( T ) {\displaystyle \Lambda (T)}$ is some function of temperature and fundamental constants. The baryonic density satisfies the above equation $d P = − ρ g d r {\displaystyle dP=-\rho gdr}$ :

p B ( r + d r ) − p B ( r ) = − d r ρ B ( r ) G r 2 ∫ 0 r 4 π r 2 ρ M ( r ) d r . {\displaystyle p_{B}(r+dr)-p_{B}(r)=-dr{\frac {\rho _{B}(r)G}{r^{2}}}\int _{0}^{r}4\pi r^{2}\,\rho _{M}(r)\,dr.}

The integral is a measure of the total mass of the cluster, with $r {\displaystyle r}$ being the proper distance to the center of the cluster. Using the ideal gas law $p B = k T B ρ B / m B {\displaystyle p_{B}=kT_{B}\rho _{B}/m_{B}}$ ( $k {\displaystyle k}$ is Boltzmann's constant and $m B {\displaystyle m_{B}}$ is a characteristic mass of the baryonic gas particles) and rearranging, we arrive at

d d r ( k T B ( r ) ρ B ( r ) m B ) = − ρ B ( r ) G r 2 ∫ 0 r 4 π r 2 ρ M ( r ) d r . {\displaystyle {\frac {d}{dr}}\left({\frac {kT_{B}(r)\rho _{B}(r)}{m_{B}}}\right)=-{\frac {\rho _{B}(r)G}{r^{2}}}\int _{0}^{r}4\pi r^{2}\,\rho _{M}(r)\,dr.}

Multiplying by $r 2 / ρ B ( r ) {\displaystyle r^{2}/\rho _{B}(r)}$ and differentiating with respect to $r {\displaystyle r}$ yields

d d r [ r 2 ρ B ( r ) d d r ( k T B ( r ) ρ B ( r ) m B ) ] = − 4 π G r 2 ρ M ( r ) . {\displaystyle {\frac {d}{dr}}\left[{\frac {r^{2}}{\rho _{B}(r)}}{\frac {d}{dr}}\left({\frac {kT_{B}(r)\rho _{B}(r)}{m_{B}}}\right)\right]=-4\pi Gr^{2}\rho _{M}(r).}

If we make the assumption that cold dark matter particles have an isotropic velocity distribution, then the same derivation applies to these particles, and their density $ρ D = ρ M − ρ B {\displaystyle \rho _{D}=\rho _{M}-\rho _{B}}$ satisfies the non-linear differential equation

d d r [ r 2 ρ D ( r ) d d r ( k T D ( r ) ρ D ( r ) m D ) ] = − 4 π G r 2 ρ M ( r ) . {\displaystyle {\frac {d}{dr}}\left[{\frac {r^{2}}{\rho _{D}(r)}}{\frac {d}{dr}}\left({\frac {kT_{D}(r)\rho _{D}(r)}{m_{D}}}\right)\right]=-4\pi Gr^{2}\rho _{M}(r).}

With perfect X-ray and distance data, we could calculate the baryon density at each point in the cluster and thus the dark matter density. We could then calculate the velocity dispersion $σ D 2 {\displaystyle \sigma _{D}^{2}}$ of the dark matter, which is given by

σ D 2 = k T D m D . {\displaystyle \sigma _{D}^{2}={\frac {kT_{D}}{m_{D}}}.}

The central density ratio $ρ B ( 0 ) / ρ M ( 0 ) {\displaystyle \rho _{B}(0)/\rho _{M}(0)}$ is dependent on the redshift $z {\displaystyle z}$ of the cluster and is given by

ρ B ( 0 ) / ρ M ( 0 ) ∝ ( 1 + z ) 2 ( θ s ) 3 / 2 {\displaystyle \rho _{B}(0)/\rho _{M}(0)\propto (1+z)^{2}\left({\frac {\theta }{s}}\right)^{3/2}}

where $θ {\displaystyle \theta }$ is the angular width of the cluster and $s {\displaystyle s}$ the proper distance to the cluster. Values for the ratio range from .11 to .14 for various surveys.[6]

Planetary geology

The concept of hydrostatic equilibrium has also become important in determining whether an astronomical object is a planet, dwarf planet, or small Solar System body. According to the definition of planet adopted by the International Astronomical Union in 2006, one defining characteristic of planets and dwarf planets is that they are objects that have sufficient gravity to overcome their own rigidity and assume hydrostatic equilibrium. Such a body will often have the differentiated interior and geology of a world (a planemo), though near-hydrostatic or formerly hydrostatic bodies such as the proto-planet 4 Vesta may also be differentiated and some hydrostatic bodies (notably Callisto) have not thoroughly differentiated since their formation. Often the equilibrium shape is an oblate spheroid, as is the case with Earth. However, in the cases of moons in synchronous orbit, nearly unidirectional tidal forces create a scalene ellipsoid. Also, the purported dwarf planet Haumea is scalene due to its rapid rotation, though it may not currently be in equilibrium.

Icy objects were previously believed to need less mass to attain hydrostatic equilibrium than rocky objects. The smallest object that appears to have an equilibrium shape is the icy moon Mimas at 396 km, whereas the largest icy object known to have an obviously non-equilibrium shape is the icy moon Proteus at 420 km, and the largest rocky bodies in an obviously non-equilibrium shape are the asteroids Pallas and Vesta at about 520 km. However, Mimas is not actually in hydrostatic equilibrium for its current rotation. The smallest body confirmed to be in hydrostatic equilibrium is the dwarf planet Ceres, which is icy, at 945 km, whereas the largest known body to have a noticeable deviation from hydrostatic equilibrium is Iapetus being made of mostly permeable ice and almost no rock.[7] At 1,469 km this moon is neither spherical nor ellipsoid. Instead, it is rather in a strange walnut-like shape due to its unique equatorial ridge.[8] Some icy bodies may be in equilibrium at least partly due to a subsurface ocean, which is not the definition of equilibrium used by the IAU (gravity overcoming internal rigid-body forces). Even larger bodies deviate from hydrostatic equilibrium, although they are ellipsoidal: examples are Earth's Moon at 3,474 km (mostly rock),[9] and the planet Mercury at 4,880 km (mostly metal).[10]

Solid bodies have irregular surfaces, but local irregularities may be consistent with global equilibrium. For example, the massive base of the tallest mountain on Earth, Mauna Kea, has deformed and depressed the level of the surrounding crust, so that the overall distribution of mass approaches equilibrium.

Atmospheric modeling

In the atmosphere, the pressure of the air decreases with increasing altitude. This pressure difference causes an upward force called the pressure-gradient force. The force of gravity balances this out, keeping the atmosphere bound to Earth and maintaining pressure differences with altitude.

Gemology

Gemologists use hydrostatic balances to determine the specific gravity of gemstones. A gemologist may compare the specific gravity they observe with a hydrostatic balance with a standardized catalogue of information for gemstones, helping them to narrow down the identity or type of gemstone under examination.

List of gravitationally rounded objects of the Solar System; a list of objects that have a rounded, ellipsoidal shape due to their own gravity (but are not necessarily in hydrostatic equilibrium)
Statics
Two-balloon experiment

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