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The integrand defining $\epsilon (t)$ is nonnegative by the dominant energy condition, henceApply the spacetime divergence theorem to $J^a=T^{ab}{V}_b$ on the slab $S$. The flux through the cylinder at $r\to \infty$ vanishes by the stated decay. With the Lorentzian boundary orientation, the two spacelike boundary contributions givePart (ii) makes the right-hand side nonnegative, so for $t_1\ge t_0$,Therefore $\epsilon$ is a nonnegative monotone decreasing energy functional.

The Roche lobe of star 1 is the volume around it bounded by the critical closed equipotential passing through the inner Lagrange point $L_1$. Material on a lower potential surface remains confined to star 1; at the critical surface a path opens toward star 2.

Let $\mathbf{g}=-\nabla \Psi$ and take $d\mathbf{S}$ outward. For a nearly spherical interior surface, the divergence theorem givesThe self-gravity term contributes $-4\pi G{M}_1$. The companion lies outside $S$, so its potential is harmonic inside and contributes zero net flux. For the centrifugal term, ${\nabla }^2(-{\Omega }^2{s}^2/2)=-2{\Omega }^2$, so its acceleration has divergence $2{\Omega }^2$ and contributes $2{\Omega }^{2}V$. Using $V=4\pi r^3/3$ and $S=4\pi r^2$,The sign is the outward-normal component; the dominant self-gravity is inward and therefore negative.

For two body-force-free Stokes flows $({\mathbf{u}}^{(1)},{\mathbf{\sigma }}^{(1)})$ and $({\mathbf{u}}^{(2)},{\mathbf{\sigma }}^{(2)})$ in the same domain, the Lorentz reciprocal theorem for Stokes flow isIndeed, the difference of the two volume integrands isbecause both flows are incompressible and the Newtonian fluid stress tensor is symmetric. The divergence theorem proves the boundary identity.

On a rigid body, ${\mathbf{u}}^{(r)}={\mathbf{U}}^{(r)}+{\mathbf{\Omega }}^{(r)}\times \mathbf{x}$. The reciprocal theorem becomesWriting $(\mathbf{F},\mathbf{G}{)}^T=\mathsf{R}(\mathbf{U},\mathbf{\Omega }{)}^T$ and choosing arbitrary pairs of rigid velocities shows thatThus the hydrodynamic resistance matrix is symmetric.

The divergence theorem is ${\iint }_{\partial V}F\cdot ndS={\iiint }_V\nabla \cdot FdV$. Here $\nabla \cdot F=3z+2yz$. The $y$ term integrates to zero by symmetry, while the ellipse has area $\pi /\sqrt{ab}$. Hence the flux is

Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives

Letbe the scalar moment of inertia and total kinetic energy. Sincesubstitution of the Euler momentum equation reduces the stress contribution, by the divergence theorem and isotropic pressure $\mathbf{P}=P1$, toFor self-gravity, $\int \rho \mathbf{r}\cdot \mathbf{F}dV=\Omega$, where $\Omega <0$ is the gravitational potential energy. Thus the stellar virial theorem is

Apply hydrostatic equilibrium to the isothermal core, taking its boundary pressure to be $P_c$. Its ideal-gas pressure integral scales as $\int P,dV\propto M_c{T}_c$, its volume as $R_c^3$, and its gravitational energy as $-G{M}_c^2/R_c$. The virial theorem therefore givesor, after absorbing fixed dimensional and structural factors into positive constants,Hydrogen-burning reactions are extremely temperature-sensitive, so expansion cools and suppresses burning while contraction heats and enhances it. This stellar thermostat keeps the shell and adjoining isothermal core near an approximately fixed $T_c$.

At fixed $M_c,T_c$, differentiating $P_c(R_c)$ gives

For a homologous envelope, hydrostatic balance gives $P_c\propto G{M}^2/R^4$ and the ideal-gas temperature scale gives $T_c\propto GM/R$. Eliminating $R$ yieldsup to composition and gravitational constants common to the sequence. A matching core exists only if this required pressure does not exceed $P_{c,\max }\propto T_c^4/M_c^2$. ThereforeThis maximum fractional isothermal-core mass is the mechanism behind the Schönberg-Chandrasekhar limit.

Divergence theorem says ${\int }_V\nabla \cdot FdV={\int }_{\partial V}F\cdot ndS$ for the outward normal. Here $\nabla \cdot (x{e}^{-r^2})=(3-2r^2)e^{-r^2}$, so radial integration gives $4\pi a^3{e}^{-a^2}$. On the sphere $F\cdot n=a{e}^{-a^2}$, whose integral is the same $4\pi a^3{e}^{-a^2}$.

Differentiate the ball mean-value formula and use the divergence theorem:ConsequentlyThus one may take $C(n)=n$.

Integrating the Euler-Lagrange equation and applying the divergence theorem on the compact boundaryless manifold givesEquivalently, if this integral were nonzero, replacing $u$ by $u+C$ would leave the gradient term unchanged and make the energy unbounded below in one direction, so no minimizer could exist.

Let ${\Sigma }_t$ be a Cauchy hypersurface with induced metric $h_{ij}$, lapse $N$, shift $N^i$, and future unit normal $n^a$. For the normalization of the action in the question, the canonical momentum density isThe equal-time canonical commutation relations areandWith the conventional extra factor $1/2$ in the action, $\Pi$ loses the factor two.

For complex classical solutions, the Klein-Gordon inner product isThe integrand is a conserved current because both fields obey the Klein-Gordon equation. Applying the divergence theorem between two Cauchy hypersurfaces shows that the value is independent of the foliation, provided there is no boundary flux.

A total divergence has the form $\nabla \cdot \mathbf{F}$. The divergence theorem converts its volume integral into a boundary flux, which vanishes when the field has suitable decay or boundary conditions.