Fluids and Interfaces

These notes are automatically transcribed from my handwritten notes. Expect incomplete sentences and unfinished thoughts.

1. What Is a Fluid?

Two Perspectives

There are two fundamental viewpoints for describing fluid motion:

1. Eulerian — A fixed control volume through which the fluid flows. We account for the flux through the fixed volume.
2. Lagrangian — A frame that moves along with the flow, tracking individual fluid parcels.

Relating the Two Perspectives

Let $F_E = F_E(\mathbf{x}, t)$ be a field quantity in the Eulerian frame, and $F_L = F_L(\mathbf{r}(\mathbf{x}, t), t)$ the same quantity described in the Lagrangian frame (i.e. "talking about the same thing").

Taking the time derivative of the Lagrangian quantity:

\[\partial_t F_L = \partial_t F_E + \frac{\partial F_E}{\partial \mathbf{x}} \cdot \frac{\partial \mathbf{x}}{\partial t} = \left(\partial_t + \mathbf{v} \cdot \nabla\right) F_E\]

(assuming no rotation or scaling of reference frames, i.e. the Jacobian $= \mathbb{1}$).

We define the convective (material) derivative:

\[\boxed{D_t \equiv \partial_t + \mathbf{v} \cdot \nabla}\]

Relative Change of Volume

(Volumetric strain rate, or bulk strain rate)

\[\frac{\partial_t \, \delta V_L}{\delta V_L} = \sum_i \frac{\partial_t \, \delta r_i}{\delta r_i} = \sum_i \frac{\delta v_i}{\delta r_i} \;\longrightarrow\; \nabla \cdot \mathbf{v} \qquad \text{for } \delta V_L \to 0,\]

where $\delta V_L = \delta r_1 \, \delta r_2 \, \delta r_3$.

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2. Conservation of Mass

Starting from the fact that the mass of a Lagrangian fluid parcel is conserved:

\[\delta M_L^{-1} \, \partial_t \, \delta M_L = 0.\]

Write $\delta M_L = \rho_L \, \delta V_L$, so that:

\[\delta M_L^{-1} \, \partial_t \, \delta M_L = \frac{\partial_t \rho_L}{\rho_L} + \frac{\partial_t \, \delta V_L}{\delta V_L} = \frac{1}{\rho}\left(\partial_t + \mathbf{v} \cdot \nabla\right)\rho + \nabla \cdot \mathbf{v} = \frac{1}{\rho}\left(\partial_t \rho + \nabla \cdot \rho\mathbf{v}\right) = 0\]

This gives the continuity equation:

\[\boxed{\partial_t \rho + \nabla \cdot \rho\mathbf{v} = 0}\]

Alternative (Integral Form)

Integrate over a volume and apply the Reynolds transport theorem to the total mass:

\[\frac{dM}{dt} = \frac{d}{dt}\int_V \rho \, dV = \int_V \partial_t \rho \, dV + \oint_{\partial V} \rho\left(\mathbf{v} \cdot d\mathbf{S}\right) = 0\]

Or, from another perspective (fixed volume): the velocity flux is the field of particles through $\partial V$:

\[\frac{dM}{dt} = \frac{d}{dt}\int_V \rho \, dV = -\oint_{\partial V} \rho\left(\mathbf{v} \cdot d\mathbf{S}\right)\]

This is the flux of density out of volume. In any case:

\[\int_V \left(\partial_t \rho + \nabla \cdot \rho\mathbf{v}\right) dV = 0\]

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3. Conservation of Momentum

For point objects we have $\mathbf{F} = \frac{d}{dt}(m\mathbf{v})$. For a fluid volume:

\[\frac{d}{dt}\int_V \rho\mathbf{v}\, dV = \int_V \partial_t(\rho\mathbf{v})\, dV + \oint_{\partial V} \rho\mathbf{v}\left(\mathbf{v} \cdot d\mathbf{S}\right)\]

The surface integral involves a dyadic (tensor) product. Applying the divergence theorem:

\[= \int_V \left[\partial_t(\rho\mathbf{v}) + \nabla \cdot \rho\mathbf{v}\mathbf{v}\right] dV\]

Expanding:

\[= \int_V \left[\partial_t(\rho\mathbf{v}) + \rho\mathbf{v}\cdot\nabla\mathbf{v} + \mathbf{v}\,\nabla\cdot\rho\mathbf{v}\right] dV = \int_V \left[\underbrace{\mathbf{v}\left(\partial_t\rho + \nabla\cdot\rho\mathbf{v}\right)}_{\text{mass creation}} + \rho\underbrace{\left(\partial_t + \mathbf{v}\cdot\nabla\right)}_{D_t}\mathbf{v}\right] dV\]

The first term vanishes by the continuity equation (no mass creation).

Forces on a Fluid Element

This should equal the total force. There are two kinds of forces:

\[\mathbf{F} = \underbrace{\int_V \frac{\partial\mathbf{F}_b}{\partial V}\, dV}_{\text{Bulk forces}} + \underbrace{\oint_{\partial V} \frac{\partial\mathbf{F}_s}{\partial S}\, dS}_{\text{Surface forces}}\]

where $\mathbf{f} \equiv \frac{\partial \mathbf{F}_b}{\partial V}$ is the force density.

Surface Forces and the Stress Tensor

The surface traction $\mathbf{T} \equiv \frac{\partial \mathbf{F}_s}{\partial S}$ should depend on the normal $\hat{n}$ to the surface.

Write (Jacobian of $\mathbf{T}$, local linearization):

\[T_i = (T_0)_i + T_{ij}\, n_j\]

By Newton's third law: $\mathbf{T}(\hat{n}) = -\mathbf{T}(-\hat{n})$, which implies $\mathbf{T}_0 = 0$.

Finally:

\[\frac{\partial \mathbf{F}_s}{\partial S} = \mathbf{T}\, \hat{n} \qquad \text{so that} \qquad \oint_{\partial V} \frac{\partial \mathbf{F}_s}{\partial S}\, dS = \int_V \frac{\partial T_{ij}}{\partial x_j}\, \hat{e}_{(i)}\, dV\]

Asserting independence of control volume $V$:

\[\rho\, D_t \mathbf{v} + \mathbf{v}(\partial_t\rho + \nabla\cdot\rho\mathbf{v}) = \mathbf{f} + \frac{\partial T_{ij}}{\partial x_j}\hat{e}_{(i)}\]

If fixed mass (mass production = 0):

\[\boxed{\rho\, D_t v_i = f_i + \partial_j T_{ij}}\]

This is the Cauchy momentum equation.

Force of Fluid on an Object

No bulk force ($\mathbf{F}_b = 0$) and steady flow ($\partial_t(\rho\mathbf{v}) = 0$), then:

\[\oint_{\partial V}\left(\rho\mathbf{v}\mathbf{v} - \frac{\partial\mathbf{F}_s}{\partial S}\right)\cdot d\mathbf{S} = 0\]

according to the above. Consider an object with boundary $A$ to the fluid. Then the force exerted on the object by the fluid is:

\[\mathbf{F} = -\oint_A \frac{\partial\mathbf{F}_s}{\partial S} \cdot d\mathbf{S} = -\oint_A \left(\rho\mathbf{v}\mathbf{v} - \frac{\partial\mathbf{F}_s}{\partial S}\right)\cdot d\mathbf{S}\]

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4. Stress and Newtonian Fluids

Definition of a Fluid

"A substance that deforms under any form of shear stress."

$\Rightarrow$ $T_{ij} = -p\,\delta_{ij}$ for some $p$ when the fluid is static.

\[T_{ij} = -p\,\delta_{ij} + \underbrace{\sigma_{ij}}_{\substack{\text{deviatoric} \\ \text{stress tensor}}} \qquad (= 0 \text{ at rest})\]

Newtonian Fluids

Because the stress vanishes for static flows, we must have $\sigma_{ij} \sim \partial_\ell v_k$. For Newtonian fluids we assume a linear relation:

\[\boxed{\sigma_{ij} = C_{ijk\ell}\,\frac{\partial v_\ell}{\partial x_k}}\]

The anti-symmetric part $\frac{1}{2}(\partial_j v_i - \partial_i v_j) \sim$ rotations, and not $\sim$ deformations. The traction $\sigma_{ij} n_j$ must be orthogonal to the surface normal $\hat{n}$ if $\sigma_{ij}$ is anti-symmetric.

The symmetric part (for an isotropic medium):

\[C_{ijk\ell} = \lambda\,\delta_{ij}\delta_{k\ell} + \mu(\delta_{ik}\delta_{j\ell} + \delta_{i\ell}\delta_{jk})\]

Full Stress Tensor for Newtonian Fluid

For the strain rate tensor $\dot{e}_{ij} \equiv \frac{1}{2}(\partial_j v_i + \partial_i v_j)$:

\[T_{ij} = -p\,\delta_{ij} + \lambda\,\delta_{ij}\,\nabla\cdot\mathbf{v} + \underbrace{2\mu\dot{e}_{ij}}_{\text{dynamic/shear viscosity}}\]

where $\mu$ is the dynamic (shear) viscosity and $\lambda$ is related to the bulk viscosity.

This can be rewritten as:

\[\boxed{T_{ij} = -\left(p - \zeta\,\partial_k v_k\right)\delta_{ij} + \mu\left(\partial_j v_i + \partial_i v_j - \frac{2}{3}\delta_{ij}\,\partial_k v_k\right)}\]

where $\zeta = \frac{2}{3}\mu + \lambda$ is the coefficient of bulk viscosity (typically set to 0 as it only modifies the pressure). Here $p$ is the thermodynamic pressure (from the equation of state), while $\bar{p} = -T_{ii}/3$ is the mean mechanical pressure; the two differ by $p - \bar{p} = \zeta\,\nabla\cdot\mathbf{v}$.

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5. Navier–Stokes Equations

Consequently, the dynamics is given by ($\zeta = 0$ is called Stokes' assumption):

\[\boxed{\rho\,D_t v_i = -\partial_i\left(p - \zeta\,\partial_k v_k\right) + \partial_j\,\mu\!\left(\partial_j v_i + \partial_i v_j - \frac{2}{3}\delta_{ij}\,\partial_k v_k\right) + f_i}\]

along with:

\[\boxed{\partial_t \rho + \partial_k\,\rho\, v_k = 0}\]

(put mass sources here if needed).

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6. Incompressible Navier–Stokes

Assume $\partial_k v_k = 0$. Then $\partial_t\rho + v_k\,\partial_k\rho = 0$, so $D_t\rho = 0$. With Stokes' assumption, and constant viscosity ($\partial_t\mu = \partial_i\mu = 0$), we get:

\[\rho\,D_t v_i = -\partial_i p + f_i + \mu\,\partial_j^2 v_i\]

Incompressible Newtonian fluids without second viscosity:

\[\boxed{\rho\,D_t\mathbf{v} = -\nabla p + \mathbf{f} + \mu\,\nabla^2\mathbf{v}} \qquad \text{with} \qquad \boxed{(\partial_t + \mathbf{v}\cdot\nabla)\rho = 0}\]

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7. Euler Equation

Inviscid ($\mu = 0$) and incompressible ($\nabla\cdot\mathbf{v} = 0$):

\[\boxed{\begin{aligned} \rho\,D_t\mathbf{v} &= -\nabla p + \mathbf{f} \\ D_t\rho &= 0 \\ \nabla\cdot\mathbf{v} &= 0 \end{aligned}}\]

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8. Bernoulli Equation

For conservative forces $\mathbf{f} = -\nabla\alpha$, the inviscid ($\mu = 0$) Navier–Stokes takes the form:

\[\rho\,\partial_t\mathbf{v} + \rho\,\mathbf{v}\cdot\nabla\mathbf{v} + \nabla(p + \alpha) = 0\]

Now write $\mathbf{v}\cdot\nabla\mathbf{v} = \frac{1}{2}\nabla|\mathbf{v}|^2 - \mathbf{v}\times(\nabla\times\mathbf{v})$. Then:

\[\rho\,\partial_t\mathbf{v} + \frac{1}{2}\rho\,\nabla|\mathbf{v}|^2 + \nabla(p + \alpha) = \rho\,\mathbf{v}\times(\nabla\times\mathbf{v})\] \[\partial_t\mathbf{v} + \nabla\!\left[\frac{1}{2}|\mathbf{v}|^2 + \frac{p}{\rho} + \frac{\alpha}{\rho}\right] + \frac{p + \alpha}{\rho^2}\nabla\rho = \mathbf{v}\times(\nabla\times\mathbf{v})\]

So if $\partial_t\mathbf{v} = 0$ and $\nabla\rho = 0$, then:

\[\boxed{\frac{1}{2}|\mathbf{v}|^2 + \frac{p}{\rho} + \frac{\alpha}{\rho} = \text{const}}\]

This holds everywhere if $\nabla\times\mathbf{v} = 0$ (irrotational). If $\nabla\times\mathbf{v} \neq 0$, then const $= \int \mathbf{v}\times(\nabla\times\mathbf{v})\cdot d\mathbf{\ell}$ (constant along streamlines).

Stream Function

From the continuity equation $\partial_t\rho + \nabla\cdot\rho\mathbf{v} = 0$, for steady incompressible flow $\nabla\cdot\mathbf{v} = 0$.

Now write $\mathbf{v} = \nabla\phi + \nabla\times\mathbf{A}$, so that $\nabla\times\mathbf{v} = \nabla\times(\nabla\times\mathbf{A}) = \nabla(\nabla\cdot\mathbf{A}) - \nabla^2\mathbf{A}$, and $\nabla\cdot\mathbf{v} = \nabla^2\phi = 0$ (i.e. $\nabla\cdot\mathbf{v}_I$).

Streamline: $d\mathbf{x} \times \mathbf{v} = 0$

\[\frac{d\mathbf{x}}{d\lambda} \times \nabla\phi + \frac{d\mathbf{x}}{d\lambda}\times(\nabla\times\mathbf{A})\] \[\left(\frac{d\mathbf{x}}{d\lambda}\cdot\nabla\right)\mathbf{A} - \frac{d\mathbf{x}}{d\lambda}\cdot\nabla\mathbf{A} = \nabla A \cdot \frac{d\mathbf{x}}{d\lambda} \cdot \frac{d\mathbf{x}}{d\lambda} \cdot \nabla A\]

An Important Example

$\mathbf{v} = c\,r^n\,\hat{e}_\theta$:

\[\nabla\times\mathbf{v} = -\hat{e}_r\,\partial_z\,v_\theta + \hat{e}_z\,\frac{1}{r}\,\partial_r(r\,v_\theta)\]

So $\mathbf{v}\times(\nabla\times\mathbf{v}) = -v_\theta(\hat{e}_\theta\times\hat{e}_r)\,\partial_z v_\theta + v_\theta(\hat{e}_\theta\times\hat{e}_z)\,\frac{1}{r}\partial_r(r\,v_\theta)$

$= \hat{e}_z\,v_\theta\,\partial_z v_\theta + \hat{e}_r\,v_\theta\,\frac{1}{r}\partial_r(r\,v_\theta)$

$= \hat{e}_r\,v_\theta\,\partial_r v_\theta + \hat{e}_r\,\frac{1}{r}v_\theta^2$

So $\int \mathbf{v}\times(\nabla\times\mathbf{v})\cdot\hat{e}_r\,dr = \int\left(v_\theta\,\partial_r v_\theta + \frac{1}{r}v_\theta^2\right)dr$

So if $v_\theta = c\,r^n$, then $c^2(n+1)\int r^{2n-1}\,dr$:

\[= c^2(n+1)\begin{cases} \frac{1}{2n}(r^{2n} - R^{2n}) & \text{if } n \neq 0 \\ \ln(r/R) & \text{if } n = 0 \end{cases}\]

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9. Vortex Models

Rankine Vortex

(Convenient phenomenological model)

\[v_r = 0, \quad v_\theta = \begin{cases} \frac{\Gamma}{2\pi}\frac{1}{R}\frac{r}{R} & \text{for } r \leq R \\ \frac{\Gamma}{2\pi}\frac{1}{r} & \text{for } r > R \end{cases}, \quad v_z = 0\]

Vorticity: $\omega_z = \begin{cases} 2\Omega & r \leq R \\ 0 & r > R \end{cases}$, for $\Omega = \frac{\Gamma}{2\pi R^2}$.

Kaufmann (Scully) Vortex

(Differentiable, rank-like)

\[v_\theta(r) = \frac{\Gamma}{2\pi}\frac{r}{r^2 + \ell^2}\]

Much like Rankine, just differentiable.

Lamb–Oseen Vortex

(Vortex decays due to viscosity)

\[v_r = 0, \quad v_z = 0, \quad v_\theta = \frac{\Gamma}{2\pi r}\,g(r,t)\] \[\partial_t g = \nu\left(\partial_r^2 g - \frac{1}{r}\partial_r g\right)\] \[g(r,t) = 1 - e^{-r^2/4\nu t}\]

Non-zero vorticity:

\[\omega_z(r,t) = \frac{\Gamma}{4\pi\nu t}\,e^{-r^2/4\nu t}\]

Pressure: $\partial_r p = \rho\,v_\theta^2 / r$

Burgers Vortex

(Exact solution of Navier–Stokes with viscosity — balance of viscous widening of vortex and converging radial flow, vortex in stationary self-similar flow)

\[\mathbf{v} = -\alpha r\,\hat{e}_r + \frac{\Gamma}{2\pi r}\,g(r)\,\hat{e}_\theta + 2\alpha z\,\hat{e}_z\]

where $\alpha$ is the strain rate. The function $g$ satisfies:

\[r\,g''(r) + \left(\frac{\alpha}{\nu}\,r^2 - 1\right)g'(r) = 0\] \[g(r) = 1 - \exp\!\left(-\frac{\alpha r^2}{2\nu}\right)\] \[\omega_z = \frac{\alpha\Gamma}{2\pi\nu}\exp\!\left(-\frac{\alpha r^2}{2\nu}\right)\]

Sullivan Vortex

(Extension of Burgers)

\[v_r = -\alpha r + \frac{1}{r}f(\eta), \quad v_z = 2\alpha z\left[1 - \frac{1}{2}f'(\eta)\right], \quad v_\theta = \frac{\Gamma}{2\pi}\frac{g(\eta)}{g(\infty)}\]

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10. Draining Tanks

Using the Bernoulli equation between the surface and the bottom hole:

\[\frac{1}{2}v^2 + \frac{p_s}{\rho} + g\,h = \frac{1}{2}v_0^2 + \frac{p_0}{\rho} + g \cdot 0\] \[2g\,h = v_0^2 \quad \Rightarrow \quad v_0 = \sqrt{2g\,h}\]

Lost volume is $\partial_t V = A_h\sqrt{2g\,h}$ (where $A_h$ is the hole area), so:

\[\frac{dh}{dt} = -\frac{A_h}{A}\sqrt{2g\,h}\]

Integrating: $\dfrac{dh}{\sqrt{h}} = -\frac{A_h}{A}\sqrt{2g}\,dt$

\[\Rightarrow \quad 2\sqrt{h(t)} - 2\sqrt{h_0} = -\frac{A_h}{A}\sqrt{2g}\,t\] \[h(t) = \left(\sqrt{h_0} - \frac{A_h}{A}\sqrt{\frac{g}{2}}\,t\right)^2\] \[\boxed{h(t) = h_0\!\left(1 - \frac{A_h}{A}\sqrt{\frac{g}{2h_0}}\,t\right)^2}\]

Assuming flow perpendicular out of the hole and no bulk flow or rotation.

Irrotational when $\nabla\times\mathbf{v} = 0$, i.e. (in cylindrical coordinates) $\tfrac{1}{r}\partial_\theta v_z - \partial_z v_\theta = 0$, $\partial_z v_r - \partial_r v_z = 0$, and $\tfrac{1}{r}\!\left(\partial_r(r v_\theta) - \partial_\theta v_r\right) = 0$.

Vortex Above a Drain (Axially Symmetric)

Now assume vanishing flow. For $\mathbf{v} = v_r\hat{e}_r + v_\theta\hat{e}_\theta + v_z\hat{e}_z$, $v^2 = \frac{C^2 + D^2}{r^2}$.

The Bernoulli integral and detailed computations for $v_r = D/r$ and $v_\theta = C/r$ lead to the constant:

\[\text{const} = (r^2 - R^2)C^2 + G(r, r_0, z, z_0) \quad \text{(vertical)}\]

If $r < R$ (non-vertical).

With Burgers vortex: $\mathbf{v} = -\alpha r\,\hat{e}_r + \frac{\Gamma}{2\pi r}\,g(r)\,\hat{e}_\theta + 2\alpha z\,\hat{e}_z$, detailed calculations involving the exponential integral function $\text{Ei}(x) = \int_{-\infty}^x \frac{e^t}{t}\,dt$ yield expressions for the streamline integrals.

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11. Vorticity

Start with incompressible Navier–Stokes without second viscosity:

\[\rho\,D_t\mathbf{v} = -\nabla p + \mathbf{f} + \mu\,\nabla^2\mathbf{v} \qquad \text{with } (\partial_t + \mathbf{v}\cdot\nabla)\rho = 0\]

Write $\mathbf{v} = \nabla\phi + \nabla\times\mathbf{\xi}$ ($\mathbf{v}_I$ irrotational, $\mathbf{v}_R$ rotational). Then:

\[D_t\mathbf{v} = \partial_t\mathbf{v}_I + \partial_t\mathbf{v}_R + (\mathbf{v}_I + \mathbf{v}_R)\cdot\nabla(\mathbf{v}_I + \mathbf{v}_R)\] \[= D_t^{(I)}\mathbf{v}_I + D_t^{(R)}\mathbf{v}_R + \mathbf{v}_I\cdot\nabla\mathbf{v}_R + \mathbf{v}_R\cdot\nabla\mathbf{v}_I\] \[= (D_t\mathbf{v})_I + (D_t\mathbf{v})_R + \nabla\phi\cdot\nabla(\nabla\times\mathbf{\xi}) + (\nabla\times\mathbf{\xi})\cdot\nabla(\nabla\phi)\]

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12. Parametric Resonance

Consider the damped Mathieu equation:

\[\ddot{\zeta} + 2\gamma\dot{\zeta} + \left(\omega_0^2 + \varepsilon\cos\omega_d t\right)\zeta = 0\]

where $\omega_d$ is the driving frequency. The wave operator is:

\[\hat{D} \equiv \partial_t^2 - 2\delta\partial_t + \omega_0^2 + \varepsilon\,C_\omega(t)\]

Multiscale Analysis

Consider $\zeta = \zeta(\varepsilon, \eta)$, so $\partial_t \mapsto \partial_t + a\,\partial_\eta$ where $a$ is small:

\[\hat{D} \mapsto (\partial_t + a\partial_\eta)^2 + 2\gamma(\partial_t + a\partial_\eta) + \omega_0^2 + \varepsilon\,C_\omega(t)\]

Expanding: $\partial_t^2 + 2a\,\partial_t\partial_\eta + a^2\partial_\eta^2 + 2\gamma\partial_t + 2\gamma a\,\partial_\eta + \omega_0^2 + \varepsilon\,C_\omega$

Then consider an expansion $\zeta = \zeta_0 + a\,\zeta_1 + a^2\,\zeta_2 + \cdots$

0th order:

\[(\partial_t^2 + \omega_0^2)\zeta_0 = 0 \quad \Rightarrow \quad \zeta_0^\pm = C_\pm(\eta)\,e^{\pm i\omega_0 t}\]

or $\zeta_0 = A\sin(\omega_0 t) + B\cos(\omega_0 t)$.

1st order: terms that have $e^{\pm i\omega t}$ on the right hand side result in driven oscillator resonance terms.

Set $\omega_d = \frac{1}{2}\omega$. Using the product-to-sum identities $2\cos\theta\cos\varphi = \cos(\theta - \varphi) + \cos(\theta + \varphi)$, etc., and requiring mode independence (no secular/resonant terms):

\[2\omega_0(\partial_\eta + \gamma)A = -\frac{1}{2}\varepsilon\,B\] \[2\omega_0(\partial_\eta + \gamma)B = -\frac{1}{2}\varepsilon\,A\]

Including $\omega_0 \mapsto \omega_0 + a\,\delta\omega$ from the $\varepsilon$-constant, this gives:

\[\Rightarrow 4\omega_0^2(\partial_\eta + \gamma)^2\,\mathcal{A} = +\frac{\varepsilon^2}{4}\,\mathcal{A} \quad \text{for } \mathcal{A} = A, B\] \[\Rightarrow \lambda \equiv \frac{\varepsilon}{4\omega_0}\]

So $\mathcal{A} \sim \exp\left[(\lambda - \gamma)\eta\right]$, and:

\[(2\omega_0)^2(\partial_\eta + \gamma)^2\,\mathcal{A} = \left[\left(\frac{\varepsilon}{2}\right)^2 - (2\omega_0\delta\omega)^2\right]\mathcal{A}\] \[\Rightarrow A, B \sim e^{\Gamma t} \quad \text{for} \quad \Gamma = \sqrt{\left(\frac{\varepsilon}{4\omega_0}\right)^2 - (\delta\omega)^2} - \gamma\]

So:

\[\ddot{\zeta} + 2\gamma\dot{\zeta} + \tilde{\omega}_0^2\,\zeta + \varepsilon\cos(\omega_d t)\,\zeta = 0\]

is solved by $\zeta_0(t,\eta) = e^{\Gamma t}\left[A\sin(\tfrac{1}{2}\omega_d t) + B\cos(\tfrac{1}{2}\omega_d t)\right]$

for:

\[\boxed{\Gamma = \sqrt{\left(\frac{\varepsilon}{4\omega_0}\right)^2 - (\delta\omega)^2} - \gamma}\]

The system is unstable when $\Gamma > 0$, i.e. the tongue of instability has slope $\frac{1}{2\sqrt{\gamma^2 + 1}}$ at the $2\omega_0$ axis.

Better estimates are found using harmonic balancing.

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13. Lagrangian Surface Waves

(Following the Miles papers)

For each fluid we have:

\[\rho\,\partial_t\Phi + \frac{1}{2}\rho|\nabla\Phi|^2 + p + \rho g z = c \quad \text{in } V\] \[\nabla^2\Phi = 0\] \[\frac{d\Gamma}{dt} = (\partial_t + \nabla\Phi\cdot\nabla)\Gamma = 0 \quad \text{on the surface}\]

Dirichlet Variational Principle

\[\frac{1}{2}\iiint (\nabla\Phi)^2\,dS\,dy - \iint \Phi(\eta)\,\eta_t\,dS \quad \Leftrightarrow \quad \begin{cases} \nabla^2\Phi = 0 & \text{in } V \\ \hat{n}\cdot\nabla\Phi = 0 \\ \partial_y - \nabla_\perp\cdot\nabla\Phi = \eta_t & \text{on } S \end{cases}\]

The kinetic energy is $T = \frac{1}{2}\rho\iiint(\nabla\Phi)^2\,dS\,dz$.

The potential energy is $V = \rho\iint dS\int dz\left[\mathbf{a}_h\cdot\mathbf{x} + (g + \dot{v}_s)\,z\right]$, where $\mathbf{a}_h$ is the horizontal acceleration of the basin and $\dot{v}_s$ is the vertical acceleration.

The interface height is:

\[\eta = \sum_\alpha q_\alpha(t)\,f_\alpha(\mathbf{x}), \quad \Phi = \sum_\alpha \dot{\Phi}_\alpha(t)\,\chi_\alpha(\mathbf{x}^2, z)\]

where $\chi_\alpha = f_\alpha(\mathbf{x})\,\frac{\cosh(k_\alpha(z - z_j))}{\cosh(k_\alpha z_j)}$.

Lagrangian

\[L = \sum_j \iiint_{V_j} \left(\frac{1}{2}\rho_j\dot{v}_j^2 - \rho_j g z\right)\,dV_j - \sigma\iint_S dA\]

with $dA = \sqrt{1 + |\nabla\zeta|^2}\,dS$, $|\nabla\Gamma| = |\nabla\zeta|$, and $1 + \frac{1}{2}|\nabla\zeta|^2 - \frac{1}{8}(|\nabla\zeta|^2)^2 + \cdots$.

Conservation: $\mathbf{v}_j = \nabla\Phi_j$.

Expanding to linear order with the coupling tensors:

\[L_{a,b}^\pm = T_a^{-1}\delta_{ab} - \tilde{L}_{abc}\,\zeta_c + \tilde{L}_{abcd}\,\zeta_c\,\zeta_d + \cdots\] \[D_{a,b}^\pm = \delta_{a,b} \mp \zeta_c\,T_b\,C_{abc} + \zeta_c\,\zeta_d\,k_b^2\,C_{abcd}\]

where $\tilde{L}_{abc} = T_a T_b$, and:

\[\hat{L}_{ab,cd} = \frac{1}{2}\left[C_{dcab}\frac{k_b^2}{T_b} - D_{dcab}\frac{T_s + T_b}{T_a T_b} + 2\frac{D_{abe}}{T_a T_c}\cdot T_c \cdot \left(C_{acka} + \frac{D_{abe}}{k_c T_c}\right)\right]\]

Linear Order

\[L \sim \frac{\Sigma\rho}{2}\sum_\alpha\left[T_\alpha^{-1}\dot{\zeta}_\alpha^2 + \underbrace{\rho'\,g}_{\text{restoring}}\,\zeta_\alpha^2 - \frac{\sigma k^2}{\Sigma\rho}\,\zeta_\alpha^2\right] + \delta L\] \[\frac{\partial L}{\partial\dot{\zeta}_\alpha} = \Sigma\rho\,T_\alpha^{-1}\dot{\zeta}_\alpha + \frac{\partial(\delta L)}{\partial\dot{\zeta}_\alpha}\] \[\frac{\partial L}{\partial\zeta_\alpha} = \Sigma\rho\left(\rho'\,g - \frac{\sigma k^2}{\Sigma\rho}\right)\zeta_\alpha\]

Euler–Lagrange:

\[\frac{d}{dt}\frac{\partial L}{\partial\dot{\zeta}_\alpha} - \frac{\partial L}{\partial\zeta_\alpha} = Q\] \[\Rightarrow \quad \ddot{\zeta}_\alpha + \left(\rho'\,g + \frac{\sigma k_\alpha^2}{\Sigma\rho}\right)T_\alpha\,\zeta_\alpha = \frac{T_\alpha}{\Sigma\rho}\,Q\]

where $Q_j = \frac{d}{dt}\frac{\partial T}{\partial\dot{q}_j} - \frac{\partial T}{\partial q_j}$ is the dissipation function.

\[\frac{T_\alpha}{\Sigma\rho}Q \doteq 2\delta_\alpha\,\dot{\zeta}_\alpha\]

Eigenmodes

\[\Phi_j(t, \mathbf{r}) = \sum_\alpha t_{j,\alpha}(t)\,\eta_{j,\alpha}(\mathbf{x}^2, t) \quad \text{for} \quad \eta_{j,\alpha} = f_\alpha(\mathbf{r})\,\frac{\cosh(k_\alpha(z - z_j))}{\cosh(k_\alpha z_j)}\] \[\zeta(t, \mathbf{r}) = \sum_\alpha \zeta_\alpha(t)\,f_\alpha(\mathbf{x})\]

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14. Important Bulk Forces

Gravitational Force

\[\mathbf{f} = \rho\,\mathbf{g} = -\nabla\,\rho\,g\,z\]

Electromagnetic Force ("Magnetohydrodynamics" / "Electrohydrodynamics")

The full electromagnetic body force on a fluid:

\[\mathbf{f} = (\rho_f - \nabla\cdot\mathbf{P})\mathbf{E} + (\mathbf{J}_f + \nabla\times\mathbf{M} + \partial_t\mathbf{P})\times\mathbf{B}\]

where $\rho_f$ is the free charge density, $\mathbf{P}$ is the polarization density, $\mathbf{J}_f$ is the free current density, $\mathbf{M} = \chi_m\mathbf{H} \approx \chi_m\mathbf{B}/\mu_0$, and $\mathbf{P} = \epsilon_0\chi_e\mathbf{E}$ (equivalently $\mathbf{P} = \frac{\chi_e}{1+\chi_e}\mathbf{D}$).

Maxwell's equations: $\nabla\cdot\mathbf{D} = \rho_f$, $\nabla\cdot\mathbf{B} = 0$, $\nabla\times\mathbf{E} = -\partial_t\mathbf{B}$, $\nabla\times\mathbf{H} = \partial_t\mathbf{D} + \mathbf{J}_f$.

Using: $-\mathbf{B}\times(\nabla\times\mathbf{B}) = (\nabla\times\mathbf{B})\times\mathbf{B} = (\mathbf{B}\cdot\nabla)\mathbf{B} - \frac{1}{2}\nabla|\mathbf{B}|^2$

After simplification:

\[\mathbf{f} = \rho_f\mathbf{E} + \mathbf{J}_f\times\mathbf{B} - \chi_e\left[\mathbf{E}\cdot\nabla\mathbf{D} + \mathbf{B}\times\partial_t\mathbf{D}\right] + \frac{\chi_m}{\mu_0}\left[(\mathbf{B}\cdot\nabla)\mathbf{B} - \frac{1}{2}\nabla|\mathbf{B}|^2\right]\]

Or using $(\nabla\cdot\mathbf{D})\mathbf{E} + (1 - \chi_e)\mathbf{J}_f\times\mathbf{B} - \chi_e\mathbf{E}\cdot\nabla\mathbf{D} + (\chi_e + \chi_m)(\nabla\times\mathbf{H})\times\mathbf{B}$

Possible Force Forms

Maxwell stress tensor: $\frac{1}{4\pi}\left[E_i E_j + H_i H_j - \frac{1}{2}(E^2 + H^2)\delta_{ij}\right]$

Poynting vector: $\mathbf{S} \equiv \mathbf{E}\times\mathbf{H}$

The full body force on a fluid:

\[\boxed{\mathbf{f} = -\nabla\,\rho g z + \rho_f\mathbf{E} + \mathbf{J}_f\times\mathbf{B} + \frac{\chi_m}{2\mu_0}\nabla|\mathbf{B}|^2 - \chi_e\epsilon_0\left[\mathbf{E}\cdot\nabla\mathbf{E} + \mathbf{B}\times\partial_t\mathbf{E}\right]}\]

| Term | Description | |------|-------------| | $-\nabla\rho g z$ | Gravity | | $\rho_f\mathbf{E} + \mathbf{J}_f\times\mathbf{B}$ | Free charges in fluid ("charged fluids") | | $\frac{\chi_m}{2\mu_0}\nabla|\mathbf{B}|^2$ | Magnetic field on fluid ("Magnetization effect"); the Kelvin body force for linear media, pointing toward higher field strength | | $-\chi_e\epsilon_0[\mathbf{E}\cdot\nabla\mathbf{E} + \mathbf{B}\times\partial_t\mathbf{E}]$ | Electric field on fluid ("Polarization effects") |

Clausius–Mossotti relation: $3\frac{\epsilon_r - 1}{\epsilon_r + 2} \approx N\alpha_e$ (electric susceptibility).

Magnetic susceptibility: $\chi_m$.

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15. Sound Waves

Start with Navier–Stokes:

\[\rho\,D_t v_i = -\partial_i p + \partial_j\,\nu\!\left(\partial_j v_i + \partial_i v_j - \frac{2}{3}\delta_{ij}\,\partial_k v_k\right)\] \[\begin{cases} D_t\rho + \rho\,\partial_k v_k = 0 \\ \rho\,\partial_t v_i + \partial_i p = 0 & \text{(small velocity)} \\ \partial_t\rho + v_i\,\partial_i\rho = 0 \end{cases}\]

Take $\partial_t$ of the linearized continuity equation and substitute the momentum equation $\partial_t v_k = -\frac{1}{\rho}\partial_k p$ (treating $\rho$ as constant to leading order), with $p = p(\rho)$:

\[\partial_t^2\rho = -\rho\,\partial_k\,\partial_t v_k = \partial_k^2 p = \frac{\partial p}{\partial\rho}\,\nabla^2\rho\] \[\Rightarrow \quad \boxed{\partial_t^2\rho = c^2\,\nabla^2\rho} \qquad \text{for} \quad c = \sqrt{\frac{\partial p}{\partial\rho}}\]

$c$ is the speed of sound; the related bulk modulus is $K = \rho c^2 = \rho\,\frac{\partial p}{\partial\rho}$.

Ideal Gas

$pV = Nk_BT = \frac{M}{m}k_BT$, so $p = \frac{\rho\,k_BT}{m}$.

More properly: $\frac{\partial p}{\partial\rho} = \gamma\frac{p}{\rho} = \gamma\frac{k_BT}{m}$ (this is not entropic — adiabatic index $\gamma$):

\[\boxed{c^2 = \frac{\gamma k_B T}{m}} \quad \text{for ideal gas with adiabatic index } \gamma\]

$\gamma = 1 + \frac{2}{f}$, where $f$ = degrees of freedom.

Consequences:

• Sound travels slower at higher altitudes
• Lighter molecules → faster speed of sound
• Increasing $T$ by 4× doubles the speed of sound

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16. Hydrostatic Interfaces

Starting from the incompressible Navier–Stokes, we have:

\[\rho(\partial_t + v_j\,\partial_j)\,v_i = -\partial_i p + f_i + \mu\,\partial^2 v_i\] \[(\partial_t + v_i\,\partial_i)\rho = 0, \qquad \partial_i v_i = 0\]

For a static fluid $v_i = 0$ and $\partial_t\rho = 0$, which is only possible for:

\[\partial_i p = f_i\]

For conservative forces $f_i = \partial_i\alpha$, this means:

\[\partial_i[p - \alpha] = 0\]

The most famous example is perhaps $\alpha = -\rho g z$, i.e. bubbles and droplets.

Young's Equation (Wetting)

At a three-phase contact line:

\[\boxed{\sigma_W\cos\theta = \sigma_{SW} - \sigma_{SL}}\]

where $\sigma_W$ = liquid–gas surface tension, $\sigma_{SW}$ = solid–gas, and $\sigma_{SL}$ = solid–liquid.

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17. Droplets and Bubbles

Consider the full Navier–Stokes, only subjected to a gravitational bulk force.

\[\rho\,D_t v_i = -\partial_i p + \rho\,\partial_j\,\nu(\partial_j v_i + \partial_i v_j - \frac{2}{3}\delta_{ij}\,\partial_k v_k) - \partial_i\,\rho g z\]

In the static situation ($v_i = 0$), then:

\[\partial_i(p + \rho g z) = 0\]

So $p = p_0 - \rho g z$.

This holds for both the fluid that is surrounding (1) and the surrounded fluid.

\[\Rightarrow \Delta p = \Delta p_0 - \Delta\rho\,g\,z\]

where $\Delta p_0$ is the interfacial pressure difference given by the Young–Laplace law:

\[\boxed{\Delta p_0 = -\sigma\,\nabla\cdot\hat{n} = -\sigma\frac{\nabla^2 G}{|\nabla G|} + \sigma\frac{(\nabla G)\cdot\nabla|\nabla G|}{|\nabla G|^2}}\]

If the surface is $G(\mathbf{x}) = 0$, then $\nabla G \propto \hat{n}$, so $\hat{n} = \frac{\nabla G}{|\nabla G|}$.

So an overpressure $\Delta p$ in an enclosed fluid has shape defined by $G(\mathbf{x}) = 0$ satisfying:

\[\boxed{\Delta p + \sigma\,\nabla\cdot\frac{\nabla G}{|\nabla G|} + \Delta\rho\,g\,z = 0}\]

Non-dimensionalization

Define $\frac{\sigma}{\Delta\rho\,g} = \lambda^2$ (natural capillary length scale) and $\frac{\Delta p}{\Delta\rho\,g} = d$ (some reference length). Measuring length in $\lambda = \sqrt{\frac{\sigma}{\Delta\rho\,g}}$, this assumes $\Delta\rho > 0$:

\[\boxed{\nabla\cdot\hat{n} = \mathcal{C} - z} \qquad \text{for } \mathcal{C} = -\frac{\Delta p}{\Delta\rho\,g\,\lambda}\]

Curvature Calculations

\[\nabla\cdot\hat{n} = \nabla\cdot\frac{\nabla G}{|\nabla G|} = \frac{\nabla^2 G}{|\nabla G|} - \frac{\nabla G\cdot\nabla|\nabla G|}{|\nabla G|^2}\]

Note that $\nabla|\nabla G|^2 = 2|\nabla G|\,\nabla|\nabla G|$, so $\nabla|\nabla G| = \frac{\nabla|\nabla G|^2}{2|\nabla G|}$.

For $\nabla G = \hat{e}_r\,\partial_r G + \frac{\hat{e}_\theta}{r}\partial_\theta G + \hat{e}_z\,\partial_z G$, and $|\nabla G|^2 = G_r^2 + G_z^2$ (axially symmetric: $G_\theta = 0$):

\[\hat{n} = \frac{\hat{e}_r\,G_r + \hat{e}_z\,G_z}{\sqrt{G_r^2 + G_z^2}}\] \[\nabla\cdot\hat{n} = \frac{1}{r}\partial_r\frac{r\,G_r}{\sqrt{G_r^2 + G_z^2}} + \partial_z\frac{G_z}{\sqrt{G_r^2 + G_z^2}}\]

After expansion (with $S = G_r^2 + G_z^2$):

\[\nabla\cdot\hat{n} = \frac{1}{S\sqrt{S}}\left[G_{rr}\,G_z^2 + G_{zz}(S - G_z^2) - 2G_r G_z\,G_{rz} + \frac{S}{r}\,G_r\right]\]

or equivalently:

\[= \frac{1}{S\sqrt{S}}\left[G_{rr}\,G_z^2 + G_{zz}\,G_r^2 + \frac{1}{r}S\,G_r - 2G_r\,G_z\,G_{rz}\right]\]

For $r = R(z)$ (Axially Symmetric Surface)

Set $G_r = 1$, $G_{rr} = 0$, $G_z = -R'(z)$, $G_{zz} = -R''$, so $S = 1 + (R')^2$.

\[\Rightarrow \nabla\cdot\hat{n} = \frac{1}{S\sqrt{S}}\left[-R''(z) + \frac{1}{r}S\right]\] \[R''(z) = \left[1 + (R')^2\right]\left[\sqrt{1 + (R')^2}\,(z - \mathcal{C}) + \frac{1}{R}\right]\]

For $z = Z(r)$ (Height Profile)

$G_z = -1$, $G_{zz} = 0$, $G_r = Z'$, $G_{rr} = Z''$, so $S = 1 + (Z')^2$.

\[\Rightarrow \nabla\cdot\hat{n} = \frac{1}{S\sqrt{S}}\left[Z'' + \frac{1}{r}S\,Z'\right]\] \[Z''(r) = \left[1 + (Z')^2\right]\left[\sqrt{1 + (Z')^2}\,(\mathcal{C} - Z) - \frac{1}{r}Z'\right]\]

Hanging droplet: $Z'(r) = 0$ at $r = 0$, so $Z = \mathcal{C} - Z''|_\infty$.

Turning Points and Droplet Shape

Consider points where $R'(z) = 0$: there $1 - r R'' = r(\mathcal{C} - z)$, so $z = \mathcal{C} + R'' - \frac{1}{R}$.

Likewise, for $R' \to -\infty$ we get $\mathcal{C} = z$.

$(1 + (R')^2)(1 - rR'') + (R')^2 R'' = R(1 + (R')^2)^{3/2}(\mathcal{C} - z)$ [full curvature equation].

The "height" of a droplet is proportional to the excess pressure inside: $z = \mathcal{C} = -\frac{\Delta p}{\Delta\rho\,g\,\lambda}$.

This is to be compared to the volume $V = \int_{z_0}^{z_1} \pi R(z)^2\,dz$.

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18. Rayleigh–Plateau Instability

Conservative velocity $\mathbf{v} = \nabla\Phi$, Bernoulli gives:

Set $\Phi = A\,e^{i(\mathbf{k}\cdot\mathbf{x} - \omega t)}$, so:

\[\rho\,\partial_t\Phi + \frac{1}{2}\rho|\nabla\Phi|^2 + p + \rho g z = c \quad \Rightarrow \quad -i\omega\Phi - \frac{1}{2}\rho k^2\Phi^2 + p + \rho g z = c\]

$\partial_t G + \nabla\Phi\cdot\nabla G = 0$, and the negligible term is: $= -\sigma\,\nabla\cdot\frac{\nabla G}{|\nabla G|}$.

$R = R_0(1 + \epsilon_0\,e^{i\omega t - ikz})$.

\[\nabla^2\Phi = \frac{1}{r}\partial_r(r\,\partial_r\Phi) + \frac{1}{r^2}\partial_\theta^2\Phi + \partial_z^2\Phi = 0\] \[\Rightarrow \Phi = (C_1\,J_m(ikr) + C_2\,Y_m(ikr))\,e^{im\theta}\,\mathcal{Z}(z)\]

with $\partial_z^2\Phi = -k^2\Phi$, so $C_1\,e^{ikz} + C_2\,e^{-ikz}$.

Requiring regularity at $r = 0$ and matching to force oscillatory behavior in $z$:

\[\Phi(r,\theta,z) = (C_1\,e^{ikz} + C_2\,e^{-ikz})\,J_m(ikr)\,e^{im\theta}\]

Goal: show that $\omega$ is imaginary when $kR < 1$ (the principal radius of curvature).

\[\tilde{\omega}^2 = ikR_0(1 - k^2R_0^2)\frac{J_m'(ikR_0)}{J_m(ikR_0)}\]

which is $\sim ik^2R_0^3$ for small $kR_0$.

$\Rightarrow$ imaginary $\omega$ (i.e. growth) when $kR < 1$, or equivalently $\lambda > 2\pi R$.

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19. The Surface Due to a Drain

\[\rho(\partial_t + v_j\,\partial_j)\,v_i = -\partial_i p + f_i + \mu\,\partial^2 v_i \qquad \text{with } \Delta p = -\sigma\,\nabla\cdot\hat{n} \text{ on interface}\] \[\partial_t\rho + v_i\,\partial_i\rho = 0, \qquad \partial_i v_i = 0, \qquad v_i\,\partial_i\Gamma = 0 \quad \text{(static interface)}\]

For gravity: $\alpha = \rho g z$.

\[\nabla\cdot\frac{\nabla\Gamma}{|\nabla\Gamma|} = \frac{\rho}{2\sigma}\,v^2 - \frac{\rho g}{\sigma}\,z + \frac{\Delta p}{\sigma}\] \[\mathbf{v}\cdot\nabla\Gamma = 0\]

Simplest Case

$v_i = \partial_i\Phi$ and $f_i = \partial_i\alpha$:

\[\frac{1}{2}\rho(\nabla\Phi)^2 = -p + \alpha + \mu\,\nabla^2\Phi\] \[\nabla^2\Phi = 0 \qquad \Rightarrow \qquad \frac{1}{2}\rho(\nabla\Phi)^2 = \sigma\,\nabla\cdot\frac{\nabla\Gamma}{|\nabla\Gamma|} + \alpha + \mu\,\nabla^2\Phi\] \[\sigma\,\nabla\cdot\frac{\nabla\Gamma}{|\nabla\Gamma|} = \frac{1}{2}\rho\,v^2 - \alpha + \Delta p\]

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20. Capillary Action

For a tube of radius $r_0$ with contact angle $\theta$:

\[\Delta p \approx \frac{2\sigma}{r} \quad \text{(radius of curvature)} \qquad \cos\theta = \frac{r_0}{r}\]

Balancing with hydrostatic pressure:

\[\rho g h = \Delta p = \frac{2\sigma}{r_0}\cos\theta\] \[\boxed{\gamma = \frac{\rho g h\,r_0}{2\cos\theta}}\]

Note: if $\theta > \pi/2$ (dry phase), then $h < 0$ (meniscus depression).

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21. Dynamic Fluid Interfaces

Incompressible Navier–Stokes:

\[\rho(\partial_t + v_j\,\partial_j)\,v_i = -\partial_i p + f_i + \mu\,\partial^2 v_i\]

Continuity equation: $\partial_t\rho + v_i\,\partial_i\rho = 0$

Incompressibility: $\partial_i v_i = 0$

Start with the simple case of a container with flat top and bottom, inviscid ($\mu = 0$), conservative vertical force $f_i = -\partial_i\alpha$ for $\alpha = \rho g z$, constant density $\rho$, and irrotational flow, i.e. $v_i = \partial_i\Phi$. Then:

\[\rho\,\partial_t\,\partial_i\Phi + \rho(\partial_j\Phi)\partial_j\,\partial_i\Phi + \partial_i p + \partial_i\alpha = 0\] \[\Rightarrow \partial_i\!\left[\rho\,\partial_t\Phi + \frac{1}{2}\rho|\nabla\Phi|^2 + p + \alpha\right] = 0\] \[\Rightarrow \quad \boxed{\rho\,\partial_t\Phi + \frac{1}{2}\rho|\nabla\Phi|^2 + p + \rho g z = \mathcal{C}} \qquad \text{and} \qquad \nabla^2\Phi = 0\]

The free surface $\Gamma = 0$ moves with the fluid it bounds:

\[\boxed{\frac{d\Gamma}{dt} = \partial_t\Gamma + \nabla\Phi\cdot\nabla\Gamma = 0}\]

The interfacial pressure difference is given by Young–Laplace:

\[\boxed{\Delta p = -\sigma\,\nabla\cdot\frac{\nabla\Gamma}{|\nabla\Gamma|}}\]

Perturbation Around a Stationary Background

\[\Phi = \Phi_0 + \delta\Phi, \quad \rho = \rho_0 + \delta\rho, \quad \alpha = \alpha_0 + \delta\alpha, \quad \Gamma = \Gamma_0 + \delta\Gamma\]

Background (no time dependence, but possibly a velocity flux; $\partial_t\Phi_0 = 0$ and $\nabla\Phi_0 = \mathbf{U}$):

\[\frac{1}{2}\rho\,U^2 + p_0 + \rho g h_0 = \mathcal{C} \quad \text{on } \Gamma_0\] \[\mathbf{U}\cdot(\hat{e}_z - \nabla h_0) = 0\]

Doing the difference at the interface: $\frac{1}{2}\Delta(\rho v^2) + \Delta p_0 + \Delta\rho\,g\,h_0 = \Delta\mathcal{C}$.

Using Young–Laplace on $\Delta p_0 = -\sigma\,\nabla\cdot\frac{\nabla\Gamma_0}{|\nabla\Gamma_0|}$ and setting $\Delta\mathcal{C} = \Delta\rho\,g\,h_0^2$ (reference point where $v^2 = 0$ and surface tension is negligible):

\[\Rightarrow \quad \Delta\rho\,g\,h_0 \approx h_0^2 - \frac{\Delta(\rho v^2\,h_0)}{2\Delta\rho\,g} - \frac{\sigma\,\nabla^2 h_0}{g\,\Delta\rho}\] \[h_0 \approx h_0^2 - \frac{\Delta(\rho v^2) + 2\sigma\,\nabla^2 h_0}{2\Delta\rho\,g}\]

Same reference where $\nabla^2 h_0 = 0$ and $\mathbf{U} = 0$: $\Delta h \sim \frac{v^2}{2g}$.

For irrotational circulation, $\mathbf{v} \approx \frac{C}{r}\hat{e}_\theta$, and negligible medium above (i.e. $\rho_2 \gg \rho_1$) with negligible $\sigma$:

\[h_0^{(\text{vortex})} = h_0^2 - \frac{C^2}{2g\,r^2}\]

Height depression due to rotation of fluid.

Perturbation Equations

\[\rho\,\partial_t\,\delta\Phi + \rho\,\mathbf{U}\cdot\nabla\,\delta\Phi + \delta p + \rho g\,\delta h = 0 \qquad \text{on } \Gamma\] \[\partial_t\,\delta h + \mathbf{U}\cdot\nabla\,\delta h + \nabla\Gamma_0\cdot\nabla\,\delta\Phi = 0, \qquad \nabla^2\delta\Phi = 0\] \[\rho(\partial_t + \mathbf{U}\cdot\nabla)\,\delta\Phi = -\delta p - \rho g\,\delta h\] \[(\partial_t + \mathbf{U}\cdot\nabla)\,\delta h = -\nabla\Gamma_0\cdot\nabla\,\delta\Phi\]

Use $\nabla\Gamma_0 = (\hat{e}_z - \nabla h_0)\sqrt{1 + (\nabla h_0)^2} \approx \hat{e}_z - \nabla h_0$ for $\nabla h_0 \ll 1$. Then introducing $D_t = \partial_t + \mathbf{U}\cdot\nabla$:

\[\rho\,D_t\,\delta\Phi = -\rho g\,\delta h - \delta p\] \[D_t\,\delta h = -\partial_z\,\delta\Phi + \mathbf{U}\cdot\nabla h_0\]

Next, focus on $\nabla^2\delta\Phi = \nabla_\perp^2\delta\Phi + \partial_z^2\delta\Phi = 0$.

Let $\delta\Phi_k$ be an eigenmode of $\nabla_\perp^2$, i.e. $\nabla_\perp^2\delta\Phi_k = -k^2\,\delta\Phi_k$. Then:

\[\partial_z^2\,\delta\Phi_k = k^2\,\delta\Phi_k \quad \Rightarrow \quad \delta\Phi_k = A\,e^{kz} + B\,e^{-kz}\]

Geometry of the tank enters here. Evaluate at impenetrable boundary: $z = z_b$, so $\partial_z\,\delta\Phi_k = kA\,e^{kz_b} - kB\,e^{-kz_b} \leq 0$ (depending on sign convention).

$\Rightarrow B = A\,e^{2kz_b}$

So $\delta\Phi_k = A\,e^{kz_b}(e^{k(z - z_b)} + e^{-k(z - z_b)}) = 2A\,e^{kz_b}\cosh(k(z - z_b))$

\[\Rightarrow \boxed{\partial_z\,\delta\Phi_k = k\,\tanh(k(z - z_b))\,\delta\Phi_k}\]

Matching the Two Fluids at the Interface via $\nabla\Gamma_0\cdot\nabla\,\delta\Phi = 0$

\[\Rightarrow \partial_z\,\delta\Phi_k^{(1)} = k\,\tanh[k(z - z_1)]\,\delta\Phi_k^{(1)} \quad \text{and similarly for fluid 2}\]

So: $\frac{\delta\Phi_k^{(1)}}{\delta\Phi_k^{(2)}} = -\frac{\tanh[k(z_1 - z)]}{\tanh[k(z_2 - z_b)]}$

Define:

\[G_k = \frac{\tanh(k(h_0 - z_b))}{\tanh(k(z_T - h_0))} \qquad \approx \begin{cases} 1 & \text{for } h_0 - z_b = z_T - h_0 \\ \frac{z_T - z_b}{z - z_b} & \text{for } k\,\Delta z \ll 1 \\ -\frac{1}{\tanh(k(z_2 - z_b))} & \text{for infinite top } z_T \to \infty \end{cases}\]

and $\mathcal{G} = \mathcal{G}(k, z_b, z_T)$.

So the interfacial difference gives:

\[\rho_2\,D_t\,\delta\Phi_k^{(2)} - \rho_1\,D_t\,\delta\Phi_k^{(1)} = (\rho_2 + \rho_1\,G_k)\,D_t\,\delta\Phi_k^{(2)}\]

and $\Delta\,\delta p = -\sigma\,\nabla\cdot\frac{\nabla\Gamma_0}{|\nabla\Gamma_0|} \approx \sigma\,\nabla^2\,\delta h$:

\[\boxed{\sum_k(\rho_2 + \rho_1\,G_k)\,D_t\,\delta\Phi_k^{(2)} = -(\rho_2 - \rho_1)\,g\,\delta h + \sigma\,\nabla^2\,\delta h}\]

Two-Fluid Dispersion Relation

If $\rho_2 = \rho \gg \rho_1$, then fluid 1 decouples and we get the familiar:

\[D_t\,\delta\Phi_k = -(g + \sigma k^2)\,\delta h_k\] \[D_t\,\delta h_k = k\,\tanh(k\,h_0)\,\delta\Phi_k + \mathbf{U}\cdot\nabla h_0\]

where $z_b = 0$ has been chosen. Assuming $D_t\,\mathbf{U}\cdot\nabla h_0 \approx 0$:

\[\boxed{D_t^2\,\delta h_k = -(g + \sigma k^2)\,k\,\tanh(k\,h_0)\,\delta h_k} \quad\Longleftrightarrow\quad \omega^2 = \left(g k + \sigma k^3\right)\tanh(k\,h_0)\]

(gravity and surface tension both restoring; $\sigma$ here absorbs $1/\rho$). The interface is stable, with the familiar gravity–capillary dispersion relation.

Rotation of a Bathtub Vortex

For solid body rotation: $v^2 = C^2 r^2$, matching with irrotational part (Rankine), $CR = C/R$, so $C = C/R^2$.

\[h_0 \approx \begin{cases} h_0^2 + \frac{C^2}{2g R^2}\left(\frac{r}{R}\right)^2 - \frac{C^2}{g R^2} & \text{for } r < R \\ h_0^2 - \frac{C^2}{2g\,r^2} & \text{for } r > R \end{cases}\]

$|\mathbf{v}|$ is $\sim r$ for $r < R$ (parabolic surface $\sim 1/r^2$) and $v_{\max} = \frac{C}{R}$ at $r = R$. Thus $\frac{C^2}{gR^2} = \left\{\frac{\text{parabolic}}{\text{inverse square}}\right\} \sim 1/r^2$.

$\Rightarrow$ this can be used to measure vorticity.

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22. Annular Waves

The geometry enters through $\nabla_\perp^2\Phi_k = -k^2\Phi_k$. In polar coordinates, this is:

\[\frac{1}{r}\partial_r(r\,\partial_r\Phi) + \frac{1}{r^2}\partial_\theta^2\Phi = -k^2\Phi\]

First set $\Phi = R(r)\,e^{im\theta}$, so:

\[\frac{1}{r}\partial_r(r\,\partial_r R) - \frac{m^2}{r^2}R + k^2 R = 0\] \[R'' + \frac{1}{r}R' - \frac{m^2}{r^2}R + k^2 R = 0\] \[r^2 R'' + r R' + (k^2 r^2 - m^2)R = 0\]

Define $x \equiv kr$, so $\partial_r = k\,\partial_x$:

\[\boxed{x^2 R''(x) + x R'(x) + (x^2 - m^2)R(x) = 0}\]

This is Bessel's ODE, with solutions $J_m, Y_m$.

\[R = C_1\,J_m(kr) + C_2\,Y_m(kr)\] \[\partial_r R = C_1 k\,\tfrac{1}{2}(J_{m-1}(kr) - J_{m+1}(kr)) + C_2 k\,\tfrac{1}{2}(Y_{m-1}(kr) - Y_{m+1}(kr))\]

So $C_2 = \frac{J_{m+1}(kr_1) - J_{m-1}(kr_1)}{Y_{m-1}(kr_1) - Y_{m+1}(kr_1)}\,C_1$, or $C_2 = -\frac{J_m'(kr_1)}{Y_m'(kr_1)}\,C_1$.

At $r = r_2$ we must have:

\[\boxed{Y_m'(kr_1)\,J_m'(kr_2) - Y_m'(kr_2)\,J_m'(kr_1) = 0}\]

This is the discretization of $k$-space for annular waves.

Bessel function recurrence relations: $J_\alpha\frac{d Y_\alpha}{dx} - \frac{d J_\alpha}{dx}Y_\alpha = \frac{2}{\pi x}$, and $\frac{d\bar{Z}_\alpha}{dx} = \frac{1}{2}(\bar{Z}_{\alpha-1} - \bar{Z}_{\alpha+1})$.

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23. Nonlinear Interface Waves

KdV, Boussinesq, ...

(Section header only — content to be developed)

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24. Energy Density of Fluids

Starting from the Cauchy momentum equation:

\[\rho\,D_t v_i = f_i + \partial_j T_{ij}\]

Inner product with $\mathbf{v}$ gives:

\[(\star) \qquad \rho\,D_t\!\left(\frac{1}{2}v^2\right) = \underbrace{\partial_j(v_i\,T_{ij})}_{\text{total internal work}} - \underbrace{T_{ij}\,\partial_j v_i}_{\text{deformation work}} + \underbrace{v_i\,f_i}_{\text{external work}} = \partial_j(v_i\,T_{ij}) + v_i\,f_i - \Phi\]

where $\Phi$ is the dissipative energy flow.

Using $T_{ij} = -p\,\delta_{ij} + \mu(\partial_i v_j + \partial_j v_i - \frac{2}{3}\delta_{ij}\,\nabla\cdot\mathbf{v})$ (deviatoric stress $\sigma_{ij}$) with Stokes' assumption of vanishing second viscosity:

\[T_{ij}\,\partial_j v_i = -p(\nabla\cdot\mathbf{v}) + \frac{\mu}{2}\left[\partial_i v_j + \partial_j v_i - \frac{2}{3}\delta_{ij}(\nabla\cdot\mathbf{v})\right]^2 \equiv -p(\nabla\cdot\mathbf{v}) + \Phi\]

The dissipative energy flow $\Phi \geq 0$ (values not energy — good!), assuming forces are conservative.

Introduce internal (thermal) energy $\rho\,e$ and heat flux $\mathbf{q}$.

Then the 1st law of thermodynamics states: add to $(\star)$:

\[\rho\,D_t\!\left(e + \frac{1}{2}v^2\right) = \partial_j(v_i\,T_{ij}) + v_i\,f_i - \partial_i q_i\]

where heat equation: $\rho\,D_t e = -\partial_i q_i - p(\partial_k v_k) + \Phi$. The term $-\partial_i q_i$ is heat loss (removed from $(\star)$).

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25. Incompressible Energy Budget

\[T_{ij}\,\partial_j v_i = \frac{1}{2}\mu\left[\partial_i v_j + \partial_j v_i\right]^2 = \Phi\] \[\partial_j(v_i\,T_{ij}) = -v_j\,\partial_j p + \partial_j\,v_i\,\mu(\partial_j v_i + \partial_i v_j)\] \[= \mu(\partial_j v_i)(\partial_j v_i + \partial_i v_j) + \mu\,v_i\,\partial_j^2 v_i + \mu\,\mathbf{v}\cdot\nabla^2\mathbf{v}\] \[\rho\,D_t\!\left(\frac{1}{2}v^2\right) = \mu\left[(\partial_j v_i)(\partial_j v_i + \partial_i v_j) + \mathbf{v}\cdot\nabla\mathbf{v}\mathbf{v}\right] - \frac{1}{2}\mu\left[\partial_i v_j + \partial_j v_i\right]^2 + v_i\,f_i - v_j\,\partial_j p\]

Or:

\[\rho\,D_t\!\left(e + \frac{1}{2}v^2\right) = -v_j\,\partial_j p + \mu\,\partial_j\,v_i(\partial_j v_i + \partial_i v_j) + v_i\,f_i - \partial_i q_i\]

For inviscid and incompressible flow:

\[\rho\,D_t\!\left(e + \frac{1}{2}v^2\right) = -v_j\,\partial_j p + v_i\,f_i - \partial_i q_i\]

And since $D_t\rho = 0$ for incompressible and $\mathbf{f} = -\nabla\alpha$ conservative:

\[D_t\!\left(\rho\,e + \frac{1}{2}\rho\,v^2 + p + \alpha\right) = \partial_t(p + \alpha) - \nabla\cdot\mathbf{q} = 0\]

along with $\rho\,D_t e = -\nabla\cdot\mathbf{q}$:

\[D_t\!\left(\rho\,e + \frac{1}{2}\rho\,v^2 + p + \alpha\right) = \partial_t(p + \alpha) - \nabla\cdot\mathbf{q}\]

Hamiltonian Formulation

Try Hamiltonian density $\mathcal{H} = \frac{1}{2}\rho\,v^2$. So:

\[H = \int_V\!\left(\frac{1}{2}\rho\,v^2 + p + \alpha\right)dV = \int_S\int_0^h\!\left(\frac{1}{2}\rho\,v^2 + p + \alpha\right)dz\,dA\]

Setting $\alpha = \rho g z$:

\[dH = \left(\frac{1}{2}\rho\,v^2 + p + \rho g z\right)dz\,dA\] \[\delta H = \left(\rho\,\mathbf{v}\cdot\delta\mathbf{v} + \delta p + \rho g\,\delta z\right)dz\,dA + \left(\frac{1}{2}\rho\,v^2 + p + \rho g z\right)\delta z\,dA\]

Bernoulli: $\frac{\delta H}{\delta A} = (\rho\,\mathbf{v}\cdot\nabla\,\delta\Phi + \delta p + \rho g\,\delta z)\,dz - \delta z(\rho\,\mathbf{v}\cdot\nabla\,d\Phi + \delta p + \rho g\,\delta z)$

Variations:

• w.r.t. $\delta z$: $\rho\,\mathbf{v}\cdot\nabla\,d\Phi + dp = 0$, or $d(\frac{1}{2}\rho\,v^2) + dp = 0$
• w.r.t. $\delta p$: $dz = 0$
• w.r.t. $\delta\Phi$: $\nabla\cdot\rho\,\mathbf{v}\,dz = 0$ (yes... constant density)

So not $\delta p$, but rather $\frac{\partial p}{\partial z}\,dz$?, then:

\[\rho\,\mathbf{v}\cdot\nabla\,d\Phi + \frac{\partial p}{\partial z}\,dz = 0\]

or $d(\frac{1}{2}\rho v^2) + \partial_z p\,dz = 0$.

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26. Magnetic Distortion of Interfaces (The "Moses Effect")

Consider a Newtonian fluid with magnetic susceptibility $\chi_m$ and permeability $\mu_0$:

\[\rho\,D_t v_i = -\partial_i p + \rho\,\partial_j\,\nu(\partial_j v_i + \partial_i v_j - \frac{2}{3}\delta_{ij}\,\partial_k v_k) - \partial_i\,\rho g z + \frac{\chi_m}{2\mu_0}\partial_i|B|^2\]

The magnetic term is the Kelvin body force $\frac{\chi_m}{2\mu_0}\partial_i|B|^2$ for a linearly magnetizable fluid (pointing toward higher field strength).

For static fluid: $v_i = 0$, with constant density $\rho$:

\[\Rightarrow \partial_i\!\left[p + \rho g z - \frac{\chi_m}{2\mu_0}|B|^2\right] = 0\] \[p + \rho g z - \frac{\chi_m}{2\mu_0}|B|^2 = \text{Constant}\]

Taking the difference at the interface (i.e. $\Delta p = p_2 - p_1$, $\Delta\rho = \rho_2 - \rho_1$, $\Delta\chi = (\chi_m)_2 - (\chi_m)_1$):

\[\Delta p + \Delta\rho\,g\,h = \frac{\Delta\chi}{2\mu_0}|B|^2\]

Young–Laplace: $\Delta p = -\sigma\,\nabla^2 h$, so total:

\[\Delta p = -\Delta\rho\,g\,h_0 - \sigma\,\nabla^2 h\] \[\frac{\Delta p}{\Delta\rho\,g} + h = \frac{\Delta\chi\,|B|^2}{2\mu_0\,\Delta\rho\,g} + h_0\] \[\left(1 - \frac{\sigma}{\Delta\rho\,g}\nabla^2\right)h = \frac{\Delta\chi\,|B|^2}{2\mu_0\,\Delta\rho\,g} + h_0\]

In the absence of surface tension:

\[\boxed{h = h_0 + \frac{\Delta\chi\,B^2}{2\mu_0\,\Delta\rho\,g}}\]

Practical Estimate

As a practical rule, for water and negligible surface tension:

\[h - h_0 = -368.2\;\mu\text{m}\left(\frac{\bar{B}}{1\,\text{T}}\right)^2\!\left(\frac{997\;\text{kg/m}^3}{\Delta\rho}\right)\!\left(\frac{\Delta\chi}{-9.051\times 10^{-6}}\right)\!\left(\frac{9.81\;\text{m/s}^2}{g}\right)\]

The surface is pushed down by a magnet. The squared field strength means that the largest change in the effect can be achieved with changing $\bar{B}$.

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27. Lagrangian Approach

\[dz = d\zeta, \qquad dn_1 = dr + d\zeta\] \[d\hat{n}_1 \wedge d\hat{n}_2 = (da + d\zeta)\wedge(db + d\zeta) = (1 + \partial_a\zeta)(1 + \partial_b\zeta)\,da\wedge db\]

$\nabla\Gamma = \hat{e}_z - \nabla\zeta$, $d\hat{P} = \nabla\Gamma\cdot d\mathbf{r}$, $d\hat{P}_1 = \nabla\Gamma\,dr$, $d\hat{P}_2 = \nabla\Gamma\,d\theta$.

$dA = \sqrt{1 + |\nabla\zeta|^2}\,dS$, $d\mathcal{A} = |\nabla\Gamma|\cdot\hat{n}\,dS = |\nabla\Gamma|\,dS$

where $|\nabla\Gamma| = \sqrt{1 + |\nabla\zeta|^2}$, and if volume is translation-symmetric in $z$: $1 + \frac{1}{2}|\nabla\zeta|^2 - \frac{1}{8}(|\nabla\zeta|^2)^2 + \cdots$

Lagrangian

\[L = \sum_j\iiint_{V_j}\left(\frac{1}{2}\rho_j\,\dot{v}_j^2 - \rho_j\,g\,z\right)dV_j - \sigma\iint_S dA\]

Conservation: $\mathbf{v}_j = \nabla\Phi_j$.

\[= \sum_j \frac{1}{2}\rho_j\sum_{a,b}\left(\frac{1}{2}\iiint_{V_j} dV_j\,\rho_j\,\nabla\Phi_{j,a}\cdot\nabla\Phi_{j,b}\right)\dot{q}_{j,b}\,\dot{q}_{j,a} - \sum_j\frac{1}{2}\rho_j\sum_{a,b}\left(\frac{1}{2}\iint dS\,\rho_j\,f_a\,f_b\right)\] \[- \sigma\iint_S\sqrt{1 + |\nabla\zeta|^2}\,dS\]

With coupling tensors $K_{ab}^j$ and $L_{a,b,c}^E$, $K_{ab}^j\dot{q}_{a,b}^E = \xi_a$:

\[\sum_E L_{a,c}^E\,\dot{\xi}_c = \dot{q}_{z,a}, \qquad K_{ab}\dot{q}_{a,b} = \xi_a\]

Expanding:

\[= \sum_\pm \rho_\pm \frac{1}{2}\sum_{a,b,c} L_{a,b,c}^E\,D_{a,b,c}\,\dot{\xi}_c\,\dot{\xi}_b \pm \sum_\pm \frac{1}{2}\rho_\pm g\,\xi^2 - \frac{1}{2}\rho_\pm g\,h_0^2\] \[\sim D_{cab}\]

Linear Order

\[L \sim \frac{\Sigma\rho}{2}\sum_\alpha\left[T_\alpha^{-1}\,\dot{\zeta}_\alpha^2 + \rho'\,g\,\zeta_\alpha^2 - \frac{\sigma k^2}{\Sigma\rho}\,\zeta_\alpha^2\right] + \delta L\] \[\frac{\partial L}{\partial\dot{\zeta}_\alpha} = \Sigma\rho\,T_\alpha^{-1}\,\dot{\zeta}_\alpha + \frac{\partial(\delta L)}{\partial\dot{\zeta}_\alpha}\] \[\frac{\partial L}{\partial\zeta_\alpha} = \Sigma\rho\!\left(\rho'\,g - \frac{\sigma k^2}{\Sigma\rho}\right)\zeta_\alpha\]

Euler–Lagrange: $\frac{d}{dt}\frac{\partial L}{\partial\dot{\zeta}_\alpha} - \frac{\partial L}{\partial\zeta_\alpha} = Q$

\[\Rightarrow \quad \ddot{\zeta}_\alpha + \left(\rho'\,g + \frac{\sigma k_\alpha^2}{\Sigma\rho}\right)T_\alpha\,\zeta_\alpha = \frac{T_\alpha}{\Sigma\rho}\,Q\] \[\frac{T_\alpha}{\Sigma\rho}\,Q \doteq 2\delta_\alpha\,\dot{\zeta}_\alpha\]

where $Q_j = \frac{d}{dt}\frac{\partial T}{\partial\dot{q}_j} - \frac{\partial T}{\partial q_j}$ is the dissipation function.

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28. Thermality of Interfaces

(Section header — content to be developed)