Ideal Fluids, Vorticity, and Bernoulli Relations

An ideal fluid has no viscous stress and no conductive heat flux, so its stress is isotropic: $\boldsymbol\sigma=-p\mathbf I$. That is a constitutive approximation, not a consequence of conservation. The thermodynamic closure must still say how pressure responds to density and entropy before Euler’s equations predict a flow.

Required background. Classical and Variational Mechanics supplies force balance and conservative potentials; States, Processes, and the Laws of Thermodynamics supplies internal energy, entropy, enthalpy, and equations of state. Helpful background. Calculus and Linear Algebra for Relativity supplies curls, line integrals, and vector identities; Classical Fields and Their Actions gives practice separating equations from constitutive input; Differential Equations for Relativity supplies linearization and characteristic reasoning.

Closing the Euler equations

Use the material derivative $D/Dt=\partial_t+\mathbf u\cdot\nabla$ introduced in Continuum Kinematics and Balance Laws.

For a conservative body acceleration $\mathbf b=-\nabla\Phi$, mass and momentum balance become

$$\partial_t\rho+\nabla\cdot(\rho\mathbf u)=0,$$ $$\frac{D\mathbf u}{Dt} =-\frac{1}{\rho}\nabla p-\nabla\Phi.$$

These four equations evolve $\rho$ and the three components of $\mathbf u$, but pressure is another unknown. For a simple compressible fluid, take specific internal energy $e=e(\rho,s)$, where $s$ is entropy per unit mass. The Gibbs relation is

$$de=T\,ds+\frac{p}{\rho^2}\,d\rho, \qquad p=\rho^2\left(\frac{\partial e}{\partial\rho}\right)_s.$$

With no viscosity, conduction, or heat source, the internal-energy balance implies

$$\frac{Ds}{Dt}=0.$$

Thus a general adiabatic ideal-fluid model consists of continuity, Euler momentum, entropy advection, and an equation of state such as $p=p(\rho,s)$. A gamma-law gas has

$$p=(\gamma-1)\rho e, \qquad p=K(s)\rho^\gamma$$

along an isentrope. A barotrope, $p=p(\rho)$, is a stronger specialization that closes mass and momentum without evolving an independent entropy field.

The adiabatic sound speed is

$$c_s^2=\left(\frac{\partial p}{\partial\rho}\right)_s.$$

Mechanical stability of the homogeneous branch requires $c_s^2\gt0$; the waves page derives the associated real acoustic speeds. An algebraic equation count can close a model even when its chosen branch is physically unstable, so closure and stability are separate tests.

Incompressibility is also separate. If a parcel keeps constant density, $D\rho/Dt=0$, then continuity gives $\nabla\cdot\mathbf u=0$. A flow can nevertheless carry different parcel densities, and in the strict constant-density incompressible model pressure acts as a Lagrange multiplier enforcing $\nabla\cdot\mathbf u=0$ rather than being supplied by a compressible equation of state. Tong, current notes, §§2.1 and 4.4.1–4.4.2, printed pp. 12–20 and 144–149 develops the Euler and thermodynamic specializations.

For a barotrope define the specific enthalpy function

$$h(\rho)=\int^\rho\frac{dp(\tilde\rho)}{\tilde\rho}, \qquad \nabla h=\frac{1}{\rho}\nabla p.$$

For a general simple fluid, $h=e+p/\rho$ instead obeys

$$dh=T\,ds+\frac{dp}{\rho}.$$

Keeping these two statements distinct prevents an isentropic identity from being applied across entropy gradients or shocks.

Vorticity: advection, stretching, and generation

Define vorticity by

$$\boldsymbol\omega=\nabla\times\mathbf u.$$

Taking the curl of Euler’s equation gives, for smooth flow,

$$\frac{D\boldsymbol\omega}{Dt} =(\boldsymbol\omega\cdot\nabla)\mathbf u -\boldsymbol\omega\,\nabla\cdot\mathbf u +\frac{\nabla\rho\times\nabla p}{\rho^2}.$$

Using continuity packages expansion into

$$\boxed{ \frac{D}{Dt}\left(\frac{\boldsymbol\omega}{\rho}\right) =\left(\frac{\boldsymbol\omega}{\rho}\cdot\nabla\right)\mathbf u +\frac{\nabla\rho\times\nabla p}{\rho^3}. }$$

The first term stretches and tilts vortex lines. The second is baroclinic generation: it is nonzero when pressure and density surfaces are not parallel. It vanishes for a barotrope because $\nabla p$ is parallel to $\nabla\rho$.

In a two-dimensional flow, $\mathbf u=(u_x,u_y,0)$ with no $z$ dependence, the stretching term vanishes because $\boldsymbol\omega=\omega_z\hat{\mathbf z}$. If the flow is also barotropic, the conserved parcel quantity is vorticity divided by density:

$$\frac{D}{Dt}\left(\frac{\omega_z}{\rho}\right)=0, \qquad \frac{D\omega_z}{Dt}=-\omega_z\nabla\cdot\mathbf u.$$

Thus compression can strengthen $\omega_z$ even without three-dimensional vortex stretching. Scalar vorticity itself is materially conserved when expansion also vanishes, as in constant-density two-dimensional flow. Conversely, imposing $\nabla\cdot\mathbf u=0$ alone does not remove baroclinic generation in a variable-density fluid. Tong’s §2.2.1, printed pp. 26–29 derives the constant-density incompressible limit; the displayed equation above retains both compressibility and the baroclinic source.

Kelvin circulation theorem, with its hypotheses visible

For a closed material curve $C(t)$ define circulation

$$\Gamma(t)=\oint_{C(t)}\mathbf u\cdot d\boldsymbol\ell.$$

Differentiate while following the curve. The line element changes with the flow, so

$$\frac{d\Gamma}{dt} =\oint_{C(t)} \left[\frac{D\mathbf u}{Dt}\cdot d\boldsymbol\ell +\mathbf u\cdot d\mathbf u\right].$$

The second term is $\oint d(u^2/2)=0$. For an inviscid barotrope acted on by a conservative body force,

$$\frac{D\mathbf u}{Dt}=-\nabla(h+\Phi),$$

and therefore

$$\boxed{ \frac{d\Gamma}{dt} =-\oint_{C(t)}d(h+\Phi)=0. }$$

This is Kelvin’s theorem. Viscosity, nonconservative forcing, baroclinicity, a curve that is not material, or a discontinuity can invalidate the derivation. Conservation of circulation does not say the magnitude of vorticity is fixed: a vortex tube may become thinner while its vorticity strengthens. Tong, current notes, §2.2.2, Eqs. (2.23)–(2.24), printed pp. 29–31 gives the material-loop argument.

Three Bernoulli statements that should not be conflated

Use the identity

$$(\mathbf u\cdot\nabla)\mathbf u =\nabla\left(\frac{u^2}{2}\right) -\mathbf u\times\boldsymbol\omega.$$

For a steady barotropic flow, Euler’s equation becomes

$$\nabla B=\mathbf u\times\boldsymbol\omega, \qquad B=\frac{u^2}{2}+h+\Phi.$$

This yields three different scopes:

1. Along a steady streamline. Dotting with $\mathbf u$ gives $\mathbf u\cdot\nabla B=0$, so $B$ is constant along each streamline. Different streamlines may have different constants.

2. Throughout a steady irrotational region. If $\boldsymbol\omega=0$, then $\nabla B=0$, so one value of $B$ holds on each connected region.

3. For unsteady potential flow. If a barotropic flow has a single-valued local potential $\mathbf u=\nabla\phi$, Euler’s equation integrates to

   $$\partial_t\phi+\frac{|\nabla\phi|^2}{2}+h+\Phi=C(t).$$

   The time-only function can be absorbed into $\phi$. A global single-valued potential requires the appropriate topology; zero local vorticity alone does not remove circulation around a hole.

For a general steady adiabatic ideal fluid, the thermodynamic identity gives Crocco’s form

$$\nabla B=\mathbf u\times\boldsymbol\omega+T\nabla s, \qquad B=\frac{u^2}{2}+h+\Phi.$$

Because steady entropy advection gives $\mathbf u\cdot\nabla s=0$, $B$ is still constant along a streamline, but entropy gradients obstruct a global constant. The familiar constant-density result is the special case $h=p/\rho$. Tong, current notes, §2.1.4, Eqs. (2.9)–(2.13), printed pp. 18–20 derives that specialization.

Worked example: a narrowing duct

Consider steady horizontal constant-density flow through cross-sectional areas $A_1$ and $A_2=A_1/2$. For a streamtube, mass conservation gives

$$\rho A_1u_1=\rho A_2u_2, \qquad u_2=2u_1.$$

If the two points lie on the same streamline and viscosity, heat input, and shaft work are negligible, Bernoulli gives

$$p_1+\frac12\rho u_1^2 =p_2+\frac12\rho u_2^2.$$

Hence

$$\boxed{p_2=p_1-\frac32\rho u_1^2.}$$

The pressure drop supplies the kinetic-energy increase. The result cannot be applied unchanged across a pump, a viscous loss, or a shock. That would be a change of model, not a failure of energy conservation.

Worked example: baroclinic vorticity at one instant

Suppose initially $\mathbf u=0$ near the origin while

$$\rho=\rho_0(1+\alpha x), \qquad p=p_0+\beta y.$$

At that instant stretching and expansion vanish, but

$$\nabla\rho\times\nabla p =\rho_0\alpha\beta\,\hat{\mathbf z}.$$

Therefore

$$\left.\frac{D}{Dt}\left(\frac{\boldsymbol\omega}{\rho}\right)\right|_{t=0} =\frac{\rho_0\alpha\beta}{\rho^3}\,\hat{\mathbf z}.$$

Misaligned density and pressure gradients generate vorticity even though the instantaneous velocity is divergence free. This is the missing qualification in the slogan “two-dimensional incompressible vorticity is advected.”

Limits and onward use

Euler flow can develop discontinuities from smooth initial data, after which streamline and curl manipulations must be replaced by weak conservation laws. Free surfaces need kinematic and dynamic boundary conditions. Turbulent closure is not supplied by ideal equations. In relativistic flow, particle current and stress–energy replace Newtonian mass and momentum density, but the same equation-of-state and entropy questions remain.

Viscosity, Heat Flow, and Dissipation adds the leading gradient corrections. Fluid Waves, Shocks, and Instabilities returns to $c_s$, characteristics, and the weak-solution boundary.

Exercises

Compression in two dimensions

A small material patch in a smooth two-dimensional barotropic flow shrinks to half its initial area while retaining constant thickness in the third direction. Assume conservative body forces and nonzero initial vorticity. How do its density and vorticity change in the infinitesimal-patch limit? Check the result against circulation conservation.

Solution

Mass conservation gives $\rho_2\mathcal A_2=\rho_1\mathcal A_1$, so $\rho_2=2\rho_1$. Since $\omega_z/\rho$ is conserved along each parcel, $\omega_{z,2}=2\omega_{z,1}$. Stokes’ theorem gives the small-patch circulation $\Gamma\simeq\omega_z\mathcal A$, which remains unchanged: twice the vorticity acts over half the area. This agrees with Kelvin’s theorem. Setting $D\omega_z/Dt=0$ merely because the flow is two-dimensional would fail this check.

A global Bernoulli constant?

A steady barotropic flow has nonzero vorticity but $\mathbf u\times\boldsymbol\omega=0$. What can be concluded about $B$?

Solution

The steady barotropic equation is $\nabla B=\mathbf u\times\boldsymbol\omega$. The given alignment makes the right-hand side zero, so $B$ is constant throughout each connected region even though the flow is not irrotational. Irrotationality is sufficient, not necessary, for a global Bernoulli constant.

Circulation and a nonconservative force

Replace $-\nabla\Phi$ by a body acceleration $\mathbf b$ with nonzero curl. Show the additional circulation source.

Solution

The material-loop derivation now gives

$$\frac{d\Gamma}{dt} =\oint_{C(t)}\mathbf b\cdot d\boldsymbol\ell =\int_{S(t)}(\nabla\times\mathbf b)\cdot\mathbf n\,dA,$$

where the second equality is Stokes’ theorem. A nonzero curl of the body acceleration can therefore create circulation even for an inviscid barotrope.

References

• Tong, David. Fluid Mechanics. University of Cambridge lecture notes, current official PDF, §§2.1–2.2.2, especially Eqs. (2.9)–(2.13) and (2.19)–(2.24), printed pp. 12–31; §§4.4.1–4.4.4, printed pp. 144–153. Official author PDF.