Viscosity, Heat Flow, and Dissipation

Mass, momentum, and total energy balances constrain every continuum model, but they do not determine how unresolved microscopic motion transports momentum or heat. Viscosity and thermal conductivity enter through constitutive laws: symmetry-constrained, experimentally measurable approximations valid in a declared long-wavelength and slow-variation regime.

Required background. Classical and Variational Mechanics supplies force, work, kinetic energy, and dimensional analysis; States, Processes, and the Laws of Thermodynamics supplies the Gibbs relation and the entropy inequality. Helpful background. Calculus and Linear Algebra for Relativity supplies symmetric tensors, traces, and contractions; Classical Fields and Their Actions gives practice distinguishing equations of motion from material response; Differential Equations for Relativity supplies diffusion modes and Fourier analysis.

From balances to a Newtonian constitutive law

Use the material derivative $D/Dt=\partial_t+\mathbf u\cdot\nabla$ and the stress/work sign convention introduced in Continuum Kinematics and Balance Laws. Here $\mathbf b$ is body acceleration, $\mathbf q$ is outward heat flux, and $r$ is heating per unit mass. With these symbols, split the stress as

$$\boldsymbol\sigma=-p\mathbf I+\boldsymbol\tau,$$

where $\boldsymbol\tau$ is the viscous stress. Momentum and internal energy are then

$$\rho\frac{D\mathbf u}{Dt} =-\nabla p+\nabla\cdot\boldsymbol\tau+\rho\mathbf b,$$ $$\rho\frac{De}{Dt} =-p\theta+\boldsymbol\tau:\nabla\mathbf u -\nabla\cdot\mathbf q+\rho r, \qquad \theta=\nabla\cdot\mathbf u.$$

The balances do not yet determine $\boldsymbol\tau$ or the heat flux $\mathbf q$. For a three-dimensional isotropic fluid, decompose the symmetric velocity gradient as

$$D_{ij}=\frac12(\partial_i u_j+\partial_j u_i), \qquad \mathbf D_0=\mathbf D-\frac{\theta}{3}\mathbf I.$$

The leading local response that is linear in first derivatives is

$$\boxed{ \boldsymbol\tau=2\eta\mathbf D_0+\zeta\theta\mathbf I, \qquad \mathbf q=-\kappa\nabla T. }$$

Here $\eta$ is shear viscosity, $\zeta$ bulk viscosity, and $\kappa$ thermal conductivity. In SI units, $\eta$ and $\zeta$ have units Pa s, while $\kappa$ has units W m$^{-1}$ K$^{-1}$. The antisymmetric velocity gradient does not appear because rigid rotation produces no strain in an isotropic Newtonian fluid.

For constant coefficients,

$$\nabla\cdot\boldsymbol\tau =\eta\nabla^2\mathbf u +\left(\zeta+\frac{\eta}{3}\right)\nabla\theta.$$

This gives the compressible Navier–Stokes equation. In incompressible flow, $\theta=0$, leaving $\eta\nabla^2\mathbf u$. An alternative notation writes $\boldsymbol\tau=2\eta\mathbf D+\lambda\theta\mathbf I$, in which $\zeta=\lambda+2\eta/3$. “Stokes’ hypothesis” sets $\zeta=0$ in the traceless convention; it is an additional approximation, not a conservation law. Tong, current notes, §3.1.1, Eqs. (3.4)–(3.7), printed pp. 58–62 derives the Newtonian shear response and its incompressible limit.

Magnetized plasmas can have direction-dependent viscosity and conduction. Polymer solutions can remember deformation history. Rarefied gases can be nonlocal. None of these failures alters momentum or energy conservation; they alter the constitutive closure.

Deriving entropy production step by step

For a simple fluid, the Gibbs identity is

$$T\,ds=de+p\,d(1/\rho).$$

Continuity gives

$$\frac{D}{Dt}\left(\frac1\rho\right)=\frac{\theta}{\rho}.$$

Multiply the Gibbs identity by $\rho$ and follow a parcel:

$$\rho T\frac{Ds}{Dt} =\rho\frac{De}{Dt}+p\theta.$$

Substituting the internal-energy balance cancels the reversible compression work:

$$\rho T\frac{Ds}{Dt} =\boldsymbol\tau:\nabla\mathbf u -\nabla\cdot\mathbf q+\rho r.$$

Because $\boldsymbol\tau$ is symmetric, only $\mathbf D$ contributes. Also,

$$\boldsymbol\tau:\mathbf D =2\eta\,\mathbf D_0:\mathbf D_0+\zeta\theta^2.$$

Use the product rule to move $-\nabla\cdot\mathbf q$ into the entropy flux $\mathbf q/T$. The result is

$$\partial_t(\rho s) +\nabla\cdot\left(\rho s\mathbf u+\frac{\mathbf q}{T}\right) =\frac{\rho r}{T}+\sigma_s,$$

with

$$\boxed{ \sigma_s= \frac{2\eta\,\mathbf D_0:\mathbf D_0+\zeta\theta^2}{T} -\frac{\mathbf q\cdot\nabla T}{T^2}. }$$

Fourier’s law then gives

$$\boxed{ \sigma_s= \frac{2\eta\,\mathbf D_0:\mathbf D_0+\zeta\theta^2}{T} +\frac{\kappa|\nabla T|^2}{T^2}\geq0 }$$

provided

$$\eta\geq0, \qquad \zeta\geq0, \qquad \kappa\geq0, \qquad T\gt0.$$

The three quadratic forms correspond to independent shear, compression, and temperature-gradient processes, so their signs are not decorative. Viscosity converts organized kinetic energy into internal energy; conduction transports heat from hot to cold. Tong’s §3.1.2, Eqs. (3.8)–(3.9), printed pp. 62–64 supplies the matching mechanical-energy loss.

Diffusive modes and dimensionless controls

A transverse velocity perturbation of a uniform fluid decouples at linear order:

$$\partial_t\delta\mathbf u_\perp =\nu\nabla^2\delta\mathbf u_\perp, \qquad \nu=\frac{\eta}{\rho}.$$

For a Fourier mode $e^{i\mathbf k\cdot\mathbf x}$,

$$\delta\mathbf u_\perp(t) =\delta\mathbf u_\perp(0)e^{-\nu k^2t}.$$

Negative shear viscosity would amplify arbitrarily short wavelengths. In a stationary constant-density medium with specific heat $c_v$, thermal conduction similarly gives

$$\partial_t\delta T=\chi\nabla^2\delta T, \qquad \chi=\frac{\kappa}{\rho c_v}.$$

For characteristic speed $U$ and length $L$,

$$\mathrm{Re}=\frac{UL}{\nu}, \qquad \mathrm{Pe}=\frac{UL}{\chi}, \qquad \mathrm{Pr}=\frac{\nu}{\chi}.$$

The Reynolds number compares advection with momentum diffusion, the Péclet number compares advection with heat diffusion, and the Prandtl number compares momentum with heat diffusion. They do not establish local thermodynamic equilibrium. That separate requirement is measured by microscopic ratios such as

$$\mathrm{Kn}=\frac{\ell_{\rm mfp}}{L}\ll1, \qquad \omega\tau_{\rm coll}\ll1.$$

Tong discusses Reynolds scaling in §3.3.1, printed pp. 73–75. The kinetic page explains why small Knudsen number is the more fundamental closure condition.

Worked example: planar Couette flow

Two parallel plates are separated by $H$. The lower plate is fixed, the upper moves at speed $U$, and a steady constant-viscosity incompressible flow has no pressure gradient. With no-slip boundary conditions,

$$0=\eta\frac{d^2u_x}{dy^2}, \qquad u_x(0)=0, \qquad u_x(H)=U.$$

Therefore

$$u_x(y)=\frac{U}{H}y, \qquad \tau_{xy}=\eta\frac{U}{H}.$$

The viscous heating per volume is

$$\boldsymbol\tau:\nabla\mathbf u =\eta\left(\frac{U}{H}\right)^2.$$

If the temperature variation is negligible in estimating the local production,

$$\sigma_s=\frac{\eta}{T}\left(\frac{U}{H}\right)^2.$$

The upper plate supplies power per area

$$\tau_{xy}U=\frac{\eta U^2}{H},$$

which exactly equals the volume heating integrated across the gap. This checks the sign and the factor of two in the stress convention. The no-slip condition is empirical boundary input; it does not follow from the bulk Navier–Stokes equation.

An exactly steady thermal state needs a path for removing this heat. For constant $\kappa$ and both plates held at the same temperature $T_w$, with no bulk heat sink, the energy equation gives

$$\kappa T''+\eta\left(\frac UH\right)^2=0, \qquad T(y)=T_w+\frac{\eta U^2}{2\kappa H^2}y(H-y).$$

The interior is warmer than the walls. Each wall removes power $\eta U^2/(2H)$ per area, adding up to the mechanical input. Uniform temperature is therefore an approximation when $\eta U^2/(8\kappa)\ll T_w$, or requires a separate distributed heat sink; wall thermostats alone do not make it exact.

Worked example: entropy made by conduction

A slab $0\leq x\leq L$ has fixed temperatures $T_h\gt T_c\gt0$, constant $\kappa$, no flow, and no internal source. The steady temperature is

$$T(x)=T_h-\frac{T_h-T_c}{L}x,$$

and heat flows to the right with

$$q_x=\kappa\frac{T_h-T_c}{L}\gt0.$$

The total entropy production per unit area is

$$\int_0^L\frac{\kappa(T')^2}{T^2}\,dx =q_x\left(\frac1{T_c}-\frac1{T_h}\right)\gt0.$$

This is exactly the entropy delivered to the cold reservoir minus that removed from the hot reservoir. Energy transfer is steady, but entropy is still produced.

Relativistic boundary of the model

Newtonian viscosity and Fourier conduction are first-order gradient laws. Their diffusion equations have instantaneous mathematical tails, so they cannot simply be promoted to covariant relativistic laws and assumed causal. Hiscock and Lindblom find short-wavelength instabilities in the traditional Eckart and Landau–Lifshitz first-order classes under their stated assumptions in Hiscock and Lindblom 1985, §§II–IV, printed pp. 726–733.

That historical result is not a theorem that every first-order hydrodynamic frame is unstable. Kovtun exhibits stable parameter regions in a more general first-order frame family in Kovtun 2019, §§2–3, author-manuscript pp. 4–15. Other causal formulations introduce relaxation variables and additional modes. For a relativistic application, the actual equations, coefficients, characteristic speeds, and initial-value problem must be checked; the label “first order” or “second order” is not enough.

Distribution Functions and the Boltzmann Equation now shows how collision dynamics, local equilibrium, and controlled gradient corrections generate fluid stresses and heat fluxes.

Exercises

Bulk viscosity in a pure expansion

Take $\mathbf u=a\mathbf x$ in three dimensions. Find $\mathbf D_0$ and the viscous entropy production.

Solution

Here $D_{ij}=a\delta_{ij}$ and $\theta=3a$, so $\mathbf D_0=\mathbf D-(\theta/3)\mathbf I=0$. The shear contribution vanishes and

$$\sigma_s=\frac{\zeta(3a)^2}{T}=\frac{9\zeta a^2}{T}.$$

Pure isotropic expansion probes bulk viscosity, not shear viscosity.

Why solid rotation does not dissipate

For $\mathbf u=\boldsymbol\Omega\times\mathbf x$ with constant $\boldsymbol\Omega$, show that the Newtonian viscous stress vanishes.

Solution

The velocity gradient is antisymmetric: $\partial_j u_i=\epsilon_{ikj}\Omega_k$. Hence $\mathbf D=0$ and $\theta=0$, so $\boldsymbol\tau=0$. A rigidly rotating fluid has vorticity but no local strain and therefore no Newtonian viscous dissipation in its interior.

References

• Hiscock, William A., and Lee Lindblom. “Generic Instabilities in First-Order Dissipative Relativistic Fluid Theories.” Physical Review D 31 (1985): 725–733, §§II–IV. Author-hosted journal PDF; DOI.
• Kovtun, Pavel. “First-Order Relativistic Hydrodynamics Is Stable.” Journal of High Energy Physics 2019, 34, §§2–3. arXiv:1907.08191; DOI.
• Tong, David. Fluid Mechanics. University of Cambridge lecture notes, current official PDF, §§3.1–3.3.2, especially Eqs. (3.4)–(3.9), printed pp. 58–76. Official author PDF.