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Write physical position as $\mathbf{r}=a(t)\mathbf{x}$ and physical velocity asSubtract the homogeneous expanding background from the pressureless Euler equation. To linear order in peculiar velocity and density contrast, the convective term is negligible andorHere $\nabla$ differentiates with respect to comoving position and $\Phi$ is the peculiar gravitational potential.

The linear continuity and Poisson equations areFor the growing mode $\delta (\mathbf{x},t)=D(a){\delta }_i(\mathbf{x})$, matter conservation gives ${\bar{\rho }}_m\propto a^{-3}$, so the potential scales asafter absorbing the fiducial normalization into ${\Phi }_i$. Integrating the linear Euler equation with a negligible decaying mode gives

Taking the divergence of Euler and using continuity and Poisson yields the linear growth equationIn the fiducial normalization used in the question this can be rearranged asThe cosmological Lagrangian displacement obeys $\dot{\mathbf{\psi }}=\mathbf{v}/a$. Combining the last two relations and choosing the growing displacement gives the Zeldovich approximation

DefineThen $\mathbf{x}={\mathbf{x}}_i-b\nabla {\Phi }_i$ and $\mathbf{v}=-a\dot{b}\nabla {\Phi }_i$. Keeping the lowest nonvanishing order in displacement in the halo angular momentum givesThe angular momentum vanishes for a spherical Lagrangian patch, and also whenever the patch's inertia principal axes align with the local tidal-field principal axes. It likewise vanishes in a locally isotropic tidal field.

Taylor-expandThe constant-gradient term vanishes by the barycentre definition. Withone obtains the tidal torque theory result$I$ is the protohalo inertia tensor and $T$ is the local tidal, or gravitational Hessian, tensor. Their eigenframe misalignment produces the torque.

In an Einstein-de Sitter universe, $D\propto a\propto t^{2/3}$. Hence $\dot{b}\propto \dot{a}\propto t^{-1/3}$ andThis linear estimate is normally stopped near turnaround. Simulated haloes subsequently gain angular momentum through nonlinear torques, anisotropic accretion, and mergers, including the orbital angular momentum of infalling subhaloes, so the early linear estimate underpredicts the final value.

In the homogeneous background, $\mathbf{v}=0$ and the physical velocity is the Hubble flowThe comoving coordinate $\mathbf{x}$ is held fixed by the time derivative in the fluid equation, soMoreover, spatial homogeneity gives $\nabla \bar{P}=0$. The Euler equation therefore reduces toTaking the divergence in three spatial dimensions givesConsistency with the Poisson equation,then requiresThis is the pressureless, zero-cosmological-constant Raychaudhuri equation, obtained by the Newtonian fluid derivation of the Raychaudhuri equation.

The elementary Newtonian derivation of the Friedmann equation follows a test particle on the boundary of a finite expanding sphere. It therefore introduces a distinguished centre and can misleadingly suggest that all matter expands away from one origin.

The fluid derivation is local: the Euler equation and Poisson equation must be compatible in every comoving neighbourhood. It consequently describes a genuinely homogeneous infinite universe, in which every comoving observer sees the same Hubble flow, without selecting a physical centre. This is the improvement captured by the Newtonian fluid derivation of the Raychaudhuri equation.

Project the steady Euler equations for an inviscid fluid along a streamline. With no gravity,For an isentropic fluid, $dp=c_s^{2}d\rho$, soThe scalar mass flux is $j=\rho u$, and henceIt increases with speed in subsonic flow and decreases with speed in supersonic flow.

Write proper position as $\mathbf{r}=a(t)\mathbf{x}$ and proper velocity asSubstitute this decomposition into the proper-coordinate Euler equations for an inviscid fluid, use ${\nabla }_r=a^{-1}{\nabla }_x$, and subtract the homogeneous-background acceleration. With $\rho =\bar{\rho }(1+\delta )$, one obtains the comoving peculiar-velocity equationThe peculiar gravitational potential $\Phi$ is the total Newtonian potential with the potential of the exactly homogeneous expanding background subtracted. Its gradient therefore generates only accelerations relative to the Hubble flow.

For a pressureless fluid, linearization discards the quadratic advection term, leavingIn the early Einstein-de Sitter universe, the growing mode has $\Phi (\mathbf{x},t)=[D(a)/a]{\Phi }_i(\mathbf{x})$. Integration gives

The linear growth factor obeysSince $\bar{\rho }={\bar{\rho }}_i/a^3$ after choosing $a_i=1$, this is equivalent toIt follows that $a\mathbf{v}=-a^2\dot{D}\nabla {\Phi }_i/(4\pi G{\bar{\rho }}_i)$. Since $\dot{\mathbf{x}}=\mathbf{v}/a$, one final integration gives the Zeldovich approximationwhere an additive initial displacement has been absorbed into ${\mathbf{x}}_i$.

For an inviscid fluid of constant mass density $\rho$ with no body force, the Euler equations for an inviscid fluid areThe vector identityand irrotational flow give $(u\cdot \nabla )u=\nabla (|u{|}^2/2)$. Writing $u=\nabla \varphi$ in terms of a velocity potential, we obtainThus the bracket depends only on time:Since adding a function of time to $\varphi$ leaves $u=\nabla \varphi$ unchanged, that function may be chosen to absorb $C(t)$. This is the Unsteady Bernoulli equation.

Steady spherical mass conservation gives $4\pi r^2\rho u=\dot{M}$. For the polytropic equation of state, $c_s^2=dp/d\rho$, soSubstitution into the radial Euler equations for an inviscid fluid gives the Parker wind equationAt a sonic point, the coefficient of $du/dr$ vanishes. A smooth transonic branch can pass it only if the right-hand side vanishes simultaneously. Hence

For $\gamma =5/3$, the shock data give ${\rho }_{\mathrm{ps}}=4{\rho }_0$ and $u_{\mathrm{ps}}=3V_s/4$. Since $V_s=R/(3t)$, the prescribed homologous spherical flow isPut $\xi =r/R(t)$ and $\rho ={\rho }_{0}D(\xi )$. The spherical continuity equation becomesso $D\propto {\xi }^9$. Matching the post-shock density givesThe material acceleration isThe radial Euler equations for an inviscid fluid thus give $dp/dr=3\rho r/(16t^2)$. Integrating from the centre to the shock,Since $p_{\mathrm{ps}}={\rho }_0{R}^2/(12t^2)$,

The steady Euler equations for an inviscid fluid areTaking the scalar product with $u$ and usinggivesThusThe Bernoulli function $H$ is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.

Let $y(t)$ be the downward displacement of the free surface from its initial level, and let $U$ be the speed in the tube. Conservation of volume givesThe free surface and the outlet are both at atmospheric pressure, and their vertical separation is $H+h_0-y$. Applying the Bernoulli equation between them, while retaining the small free-surface speed, givesThereforeThe surface reaches the upper tube end when $y=h_0$. Hence the draining time of a uniform tank through a siphon is

The steady radial equation first gives the base-pressure gradientFor the stated normal mode, putRetaining terms linear in the disturbance in the Euler equations for an inviscid fluid givestogether with incompressible flowThe $-2\Omega v^{\prime }$ and $(2\Omega +r{\Omega }^{\prime })u^{\prime }$ terms are the linearized centrifugal and angular-momentum couplings of the swirling base flow.

Yes. Synchronous gauge in cosmology uses freely falling time lines and sets the lapse and shift perturbations to zero, so no gravitational-potential force appears explicitly in this component of momentum conservation. Metric perturbations still affect the other field equations, the evolution of matter variables, and the relation between coordinate-dependent variables and gauge-invariant observables; synchronous gauge also retains residual gauge modes.

In Newtonian gauge in cosmology, the time-time metric perturbation is a Newtonian gravitational potential. Its spatial gradient therefore appears explicitly as a force term in the Euler equation. The difference is a coordinate representation of the same covariant conservation law.

For a fluid in an expanding universe, the peculiar-velocity equation is the Euler equations for an inviscid fluid after decomposing physical velocity into Hubble flow and peculiar velocity. It contains the Hubble-drag term $H\mathbf{v}$ and force $-\nabla \Phi /a$ from the peculiar gravitational potential.

For steady flow of constant-density fluid in a uniformly rotating frame with conservative body forces, the displayed quantity is constant along each relative streamline. The centrifugal potential contributes the negative quadratic term, and the Coriolis force contributes no work because it is perpendicular to the relative velocity. The result follows by dotting the rotating Euler equations for an inviscid fluid with that velocity.

Consider steady isentropic flow toward a Newtonian gravitational potential $-GM/r$ through a tube of cross-sectional area $A(r)=C{r}^n$. Write $v>0$ for the inward speed and use a polytropic equation of state with specific-heat ratio $\gamma >1$. Mass conservation and the Euler equations for an inviscid fluid givewhere $c_s$ is the adiabatic sound speed. A regular sonic point therefore has $v_s=c_s$ and $r_s=GM/(n{c}_s^2)$. The Bernoulli equation, matched to a nearly stationary reservoir with sound speed $c_0$, givesFor $n>1/2$, a finite positive sonic point requires $1<\gamma <(2n+1)/(2n-1)$. Differentiating the flow equation at that point, with $x=r_s{v}_s^{\prime }/c_s$, givesIts discriminant is $4n[(2n+1)-(2n-1)\gamma ]$, so the same bound permits real regular slopes. The branch on which the Mach number rises inward selects the negative sign inFor a spherical tube $n=2$, this recovers the $5/3$ threshold of Bondi accretion; for a dipolar flux-tube area, $n=3$ gives $7/5$. The tube approximation must remain valid between the reservoir matching region and the accretor.

Circular streamlines and constant vorticity imply $u_{\varphi }=\omega r/2+C/r$. Zero velocity at $2a$ and positive azimuthal velocity $V$ at $a$ giveThe radial Euler equations for an inviscid fluid give $p_r=\rho u_{\varphi }^2/r$, yielding outer-minus-inner pressure $(15-16\log 2)\rho V^2/18$. The pressure difference is independent of circulation orientation.

For time-dependent irrotational flow with $u=\nabla \varphi$, constant mass density, and no body force, the Euler equations for an inviscid fluid integrate toA time-dependent change of gauge in $\varphi$ may set $C(t)=0$.