Continuum Kinematics and Balance Laws

A continuum model replaces molecular detail by fields that vary smoothly on a chosen averaging scale. Its power comes from a precise chain of reasoning: state a balance for an arbitrary region, account for boundary motion and fluxes, and then localize the result. Conservation supplies the skeleton of the theory; equations of state and transport laws are additional physical input.

Required background. Classical and Variational Mechanics supplies force, momentum, work, and energy balance; States, Processes, and the Laws of Thermodynamics supplies internal energy, entropy, temperature, pressure, and the distinction between conservation and irreversibility. Helpful background. Calculus and Linear Algebra for Relativity supplies the chain rule, divergence theorem, tensors, and contractions; Classical Fields and Their Actions gives practice reading local conservation laws; Differential Equations for Relativity supplies parcel trajectories and initial-value reasoning.

Parcels, fields, and the material derivative

Label a fluid parcel by its reference position $\mathbf X$ and write its current position as

$$\mathbf x=\boldsymbol\chi(\mathbf X,t), \qquad \mathbf u(\boldsymbol\chi(\mathbf X,t),t)=\partial_t\boldsymbol\chi(\mathbf X,t).$$

The Lagrangian description follows the map $\boldsymbol\chi$ and therefore follows matter. The Eulerian description gives fields such as density $\rho(\mathbf x,t)$ and velocity $\mathbf u(\mathbf x,t)$ at fixed spatial points. Applying the chain rule to a scalar field $a$ along a parcel gives

$$\frac{d}{dt}a(\boldsymbol\chi(\mathbf X,t),t) =\partial_ta+\mathbf u\cdot\nabla a \equiv\frac{Da}{Dt}.$$

Thus $\partial_ta$ asks what a stationary instrument sees, whereas $Da/Dt$ asks what one moving parcel experiences. A temperature field can be time independent and still heat a parcel that moves into a hotter region. The same operator acts componentwise on vectors and tensors. Tong, current notes, §1.1.2, printed pp. 8–9 develops this distinction and its basic examples.

The local velocity gradient separates into expansion, deformation, and rotation:

$$\partial_j u_i =D_{ij}+W_{ij}, \qquad D_{ij}=\frac12(\partial_j u_i+\partial_i u_j), \qquad W_{ij}=\frac12(\partial_j u_i-\partial_i u_j).$$

The scalar $\theta=\nabla\cdot\mathbf u=\operatorname{tr}\mathbf D$ is the fractional rate of volume expansion. The antisymmetric part $\mathbf W$ describes local rigid rotation and cannot by itself deform a small material sphere. These kinematic objects do not yet specify any force.

A moving control volume

Let $V(t)$ have boundary velocity $\mathbf w$, not necessarily equal to the fluid velocity. For any density $q(\mathbf x,t)$, the transport theorem is

$$\frac{d}{dt}\int_{V(t)}q\,dV =\int_{V(t)}\partial_tq\,dV +\oint_{\partial V(t)}q\,\mathbf w\cdot\mathbf n\,dA.$$

Equivalently, after the divergence theorem,

$$\frac{d}{dt}\int_{V(t)}q\,dV =\int_{V(t)}\left[\partial_tq+\nabla\cdot(q\mathbf w)\right]dV.$$

Here $\mathbf n$ points outward. A fixed laboratory control volume has $\mathbf w=0$; a material volume has $\mathbf w=\mathbf u$. Mei 2002, §1.3, Eqs. (1.3.1)–(1.3.2), printed pp. 1–2 derives both cases directly.

For example, mass crosses a moving boundary at the relative rate

$$\rho(\mathbf u-\mathbf w)\cdot\mathbf n.$$

This one expression resolves a common confusion. A fixed boundary sees the usual advective flux $\rho\mathbf u\cdot\mathbf n$; a material boundary has zero relative mass flux even though every point on it moves.

Localizing mass and a general scalar balance

Mass conservation for every material volume states

$$\frac{d}{dt}\int_{V(t)}\rho\,dV=0.$$

Using $\mathbf w=\mathbf u$ and then exploiting the arbitrariness of $V(t)$ gives

$$\boxed{\partial_t\rho+\nabla\cdot(\rho\mathbf u)=0,} \qquad \boxed{\frac{D\rho}{Dt}+\rho\theta=0.}$$

The first form is conservative and remains meaningful for weak solutions. The second is intuitive for smooth flow: a parcel that expands, $\theta\gt0$, becomes less dense. Mei gives the localization in §1.3.1, Eqs. (1.3.3)–(1.3.8), printed p. 3.

Now let $a$ be any amount per unit mass, $\mathbf j_a$ its non-advective outward flux, and $s_a$ its production per unit volume. The balance on a material region is

$$\frac{d}{dt}\int_{V(t)}\rho a\,dV =-\oint_{\partial V(t)}\mathbf j_a\cdot\mathbf n\,dA +\int_{V(t)}s_a\,dV.$$

Transport, the divergence theorem, and localization give

$$\boxed{ \partial_t(\rho a) +\nabla\cdot(\rho a\,\mathbf u+\mathbf j_a)=s_a, }$$

or, after using continuity,

$$\rho\frac{Da}{Dt}=-\nabla\cdot\mathbf j_a+s_a.$$

The local equation follows pointwise only where the fields are regular. Across a shock, the integral statement is primary and the differential equation is interpreted distributionally.

Momentum, stress, and angular momentum

Traction is the force per area exerted across an oriented internal surface. Cauchy’s stress theorem writes it as

$$\mathbf t(\mathbf n)=\boldsymbol\sigma\cdot\mathbf n.$$

With body acceleration $\mathbf b$, linear momentum balance for a material volume is

$$\frac{d}{dt}\int_{V(t)}\rho\mathbf u\,dV =\oint_{\partial V(t)}\boldsymbol\sigma\cdot\mathbf n\,dA +\int_{V(t)}\rho\mathbf b\,dV.$$

Therefore

$$\boxed{ \rho\frac{D\mathbf u}{Dt}=\nabla\cdot\boldsymbol\sigma+\rho\mathbf b, }$$

or, in conservative form,

$$\boxed{ \partial_t(\rho\mathbf u) +\nabla\cdot(\rho\mathbf u\otimes\mathbf u-\boldsymbol\sigma) =\rho\mathbf b. }$$

Both tensors inside the divergence have the units of pressure, equivalently momentum flux. A stress has units of force per area; its divergence has units of force per volume. This catches both a missing derivative and a wrong sign. In a simple fluid one later writes $\boldsymbol\sigma=-p\mathbf I+\boldsymbol\tau$, but that split is constitutive, not part of momentum conservation itself. In the absence of internal couples or spin angular momentum, angular-momentum balance requires $\sigma_{ij}=\sigma_{ji}$. Tong, current notes, §§2.1.1–2.1.2, printed pp. 13–17 gives the pressure-force specialization.

Energy and the second-law boundary

For the following energy balance, take a simple continuum with symmetric Cauchy stress and no independent internal spin or couple stress, as in the ordinary fluids treated in this chapter.

Let $e$ be internal energy per unit mass and

$$E=e+\frac12u^2$$

the total material energy per unit mass. Let $\mathbf q$ be heat flux outward, $r$ heating per unit mass, and retain the stress-work sign fixed above. Total energy balance is

$$\boxed{ \partial_t(\rho E) +\nabla\cdot\left(\rho E\mathbf u-\boldsymbol\sigma\cdot\mathbf u+\mathbf q\right) =\rho\mathbf b\cdot\mathbf u+\rho r. }$$

The term $-\boldsymbol\sigma\cdot\mathbf u$ is inside the outward energy flux because $\boldsymbol\sigma\cdot\mathbf n$ does work on the material inside the surface. Subtracting the kinetic-energy equation obtained from momentum gives

$$\boxed{ \rho\frac{De}{Dt} =\boldsymbol\sigma:\nabla\mathbf u-\nabla\cdot\mathbf q+\rho r. }$$

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

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

Compression has $\theta\lt0$ and therefore raises the internal energy through $-p\theta\gt0$. Viscous stress and heat conduction are still unspecified.

Entropy is different: it obeys an inequality rather than a conservation law. With specific entropy $s$, entropy flux $\mathbf J_s$, and an externally supplied heating rate $r$ at local temperature $T$,

$$\boxed{ \partial_t(\rho s) +\nabla\cdot(\rho s\mathbf u+\mathbf J_s) -\frac{\rho r}{T}\geq0. }$$

A proposed equation of state, stress, heat flux, and entropy flux must make this production nonnegative for every allowed process. The balance laws alone cannot do that job. On the next pages, local equilibrium gives $\mathbf J_s=\mathbf q/T$, Newtonian viscosity makes mechanical production quadratic, and Fourier conduction makes heat-production quadratic.

Worked example: uniform expansion

In $d$ spatial dimensions take

$$\mathbf u=a\mathbf x, \qquad a\gt0,$$

with initially uniform density $\rho_0$. A parcel obeys $\dot{\mathbf x}=a\mathbf x$, so

$$\mathbf x(t)=e^{at}\mathbf X.$$

Since $\theta=da$, continuity along the parcel gives

$$\frac{D\rho}{Dt}=-da\rho, \qquad \rho(t)=\rho_0e^{-dat}.$$

Every material length grows by $e^{at}$ and every material volume by $e^{dat}$. Hence

$$\rho(t)V(t)=\rho_0V_0,$$

so the localized equation exactly reproduces the global conserved mass. Here $\partial_t\rho=D\rho/Dt$ only because the density stays spatially uniform. That coincidence is not a license to omit advection in a general flow.

Optional bridge: relativistic currents

Greek indices run over spacetime, and $\eta^{\mu\nu}=\operatorname{diag}(-1,1,1,1)$ is the inverse Minkowski metric in the site’s $(-,+,+,+)$ convention. Readers new to four-vectors can first review Events, Intervals, and Minkowski Geometry.

The relativistic generalization does not add a separate conservation principle; it repackages the densities and fluxes so that all observers agree on the equations. For a perfect fluid in flat spacetime,

$$N^\mu=nu^\mu, \qquad T^{\mu\nu}=(\epsilon+p)u^\mu u^\nu+p\eta^{\mu\nu},$$

where $n$ is rest-frame particle density, $\epsilon$ is rest-frame total energy density including rest mass, and $u^\mu u_\mu=-1$. The local balances are

$$\partial_\mu N^\mu=0, \qquad \partial_\mu T^{\mu\nu}=0.$$

Let $D=u^\mu\partial_\mu$, $\Theta=\partial_\mu u^\mu$, $a^\mu=Du^\mu$, and $h^{\mu\nu}=\eta^{\mu\nu}+u^\mu u^\nu$. The tensor $h^{\mu\nu}$ projects into the three-dimensional rest space orthogonal to $u^\mu$. Contracting energy–momentum conservation once along $u_\nu$ and once with this projector gives

$$D\epsilon+(\epsilon+p)\Theta=0,$$ $$(\epsilon+p)a^\alpha+h^{\alpha\nu}\partial_\nu p=0.$$

These are the relativistic energy and Euler equations. Their low-speed, weak-pressure limit has $\epsilon+p\simeq\rho$ in units with $c=1$, and the spatial projection reduces to $\rho D\mathbf u/Dt=-\nabla p$. Tong 2019, §4.5.4, Eqs. (4.53)–(4.55), printed pp. 186–187 gives the projection. The dedicated pages Stress–Energy for Particles, Fields, and Fluids and Local Balance Laws and Flux Integrals develop the tensor and spacetime-flux viewpoints. This chapter remains with nonrelativistic constitutive models.

When the continuum description applies

A continuum element must be large compared with microscopic correlation or mean-free-path scales, yet small compared with the scale on which the macroscopic fields change. One useful warning parameter is the Knudsen number $\mathrm{Kn}=\ell_{\rm micro}/L$; ordinary local fluid closure normally requires $\mathrm{Kn}\ll1$. Conservation laws may remain valid when that closure fails, but the stress and heat flux then need kinetic or nonlocal descriptions.

Mixtures add diffusion currents and reactions. Electromagnetic matter requires field momentum and energy as well as the Lorentz force. Curved spacetime replaces ordinary divergence by covariant divergence. None of these changes the organizing pattern: define the transported quantity, specify all boundary fluxes and sources, then localize with the correct geometry.

Ideal Fluids, Vorticity, and Bernoulli Relations now chooses the simplest stress and thermodynamic closure. Stress–Energy for Particles, Fields, and Fluids gives the later relativistic packaging in $T^{\mu\nu}$.

Exercises

A moving control surface

A one-dimensional control interval has endpoints moving at speed $w$ through a uniform fluid of density $\rho$ and velocity $u$. Show that the outward mass flux through its right endpoint is $\rho(u-w)$ and explain the cases $w=0$ and $w=u$.

Solution

In time $dt$, fluid advances by $u\,dt$ while the boundary advances by $w\,dt$. The relative length crossing the right endpoint is $(u-w)dt$, so the transported mass per area is $\rho(u-w)dt$ and the rate is $\rho(u-w)$. For $w=0$ this is the Eulerian flux $\rho u$. For $w=u$ the endpoint is material and no mass crosses it.

Stress-work sign

For a homogeneous isotropic compression with $\boldsymbol\sigma=-p\mathbf I$, no heat flux, and no source, use the internal-energy equation to show that $De/Dt\gt0$ when $p\gt0$ and $\theta\lt0$.

Solution

Because $\boldsymbol\sigma:\nabla\mathbf u=-p\mathbf I:\nabla\mathbf u=-p\theta$,

$$\rho\frac{De}{Dt}=-p\theta.$$

For positive density and pressure, compression means $\theta\lt0$ and therefore $De/Dt\gt0$. The sign agrees with mechanical work being done on the material.

References

• Mei, Chiang C. “Kinematic Transport Theorem.” MIT 1.63J/2.21J Advanced Fluid Dynamics of the Environment (2002), §1.3, Eqs. (1.3.1)–(1.3.10), printed pp. 1–4. MIT OpenCourseWare PDF.
• Tong, David. Fluid Mechanics. University of Cambridge lecture notes, current official PDF, §§1.1.2–1.1.3, printed pp. 8–10; §§2.1.1–2.1.2, printed pp. 13–17; §§3.1.1–3.1.2, printed pp. 60–64. Official author PDF.
• Tong, David. General Relativity. University of Cambridge Part III lecture notes (2019), §4.5.4, Eqs. (4.53)–(4.55), printed pp. 186–187. Official author PDF.