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Two routes to electric polarization are displacement and reorientation. An electric field can shift the centres of positive and negative electric charge in a neutral chemical atom or molecule, inducing an electric dipole moment. A polar molecule already has an electric dipole moment; a field biases its orientation. At finite temperature, random molecular motion opposes complete alignment, so a nonzero average need not mean that every electric dipole points the same way.^[ref]

For a single species with number per unit volume $n$, the electric polarization isThe brackets denote the local average electric dipole moment, rather than its magnitude. Opposite orientations can cancel while individual electric dipole moments remain nonzero.^[ref] This averaging explains how an initially unpolarized material can respond strongly without acquiring net electric charge.

A real scalar field is minimally coupled to a metric tensor when its kinetic term uses the metric contraction of first derivatives and its scalar potential depends only on the field, without an explicit scalar curvature coupling. For metric signature $(-,+,+,+)$ the displayed Lagrangian and volume density $\sqrt{-g}$ give $\square \varphi -V^{\prime }(\varphi )=0$. Its stress-energy tensor is $T_{ab}={\nabla }_a\varphi {\nabla }_b\varphi -g_{ab}[\frac{1}{2}(\nabla \varphi {)}^2+V]$. The scalar stress-energy divergence identity explains its conservation on the field equation.

A historical unit of volume; in the British apothecaries' system, 480 minims make one fluid ounce. It is not generally interchangeable with an uncalibrated drop.

Let $M$ be molar mass in $\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$, $\bar{v}$ partial specific volume, ${\rho }_s$ solvent density, and $N_A$ the Avogadro constant. The buoyant molecular mass is $m_b=M(1-\bar{v}{\rho }_s)/N_A$. Combining sedimentation coefficient $s=m_b/f_h$ and the thermal diffusion-friction relation $D=k_{B}T/f_h$ eliminates the unknown friction and proves the displayed formula, with $R_{\mathrm{gas}}=N_A{k}_B$. Once friction is known, $R_h=f_h/(6\pi \eta )$ defines a hydrodynamic radius using Stokes's law. Mass and radius are different inferences: an extended molecule need not occupy the volume of that friction-equivalent sphere.

The single-molecule state sum $q={\sum }_s{e}^{-\beta E_s}$ for specified energy references and external constraints. In a dilute classical ideal gas, indistinguishability gives $Q_N=q^N/N!$; translation supplies the factor proportional to volume.

For a free molecule in a uniform ideal gas, $q_i$ is proportional to volume. The ratio $f_i=q_i/V$ is independent of the container volume at fixed temperature and has units of inverse volume.

For an inviscid material volume element, neglecting body forces, Newton's second law gives $D(\rho \mathbf{u}\delta V)/Dt=-\nabla p\delta V$. Since mass conservation keeps $\rho \delta V$ constant, the result is $\rho D\mathbf{u}/Dt=-\nabla p$. Cancelling $\delta V$ before differentiating would incorrectly omit its changing volume. A fixed control volume instead requires the advected momentum flux, yielding ${\partial }_t(\rho \mathbf{u})+\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=-\nabla p$.

For a periodic box of volume $V$, expanding $\psi (\mathbf{r})=V^{-1/2}{\sum }_{\mathbf{k}}{e}^{i\mathbf{k}\cdot \mathbf{r}}{a}_{\mathbf{k}}$ in $H_{\mathrm{int}}=\frac{u}{2}\int d^{3}r{\psi }^{†}{\psi }^{†}\psi \psi$ gives the displayed normal ordering. The spatial integral is $V{\delta }_{{\mathbf{k}}_1+{\mathbf{k}}_2,{\mathbf{k}}_3+{\mathbf{k}}_4}$ by Fourier orthogonality, enforcing momentum conservation and leaving the factor $V^{-1}$. The coupling $u$ in three dimensions has units of energy times volume.

A high-pressure apparatus in which several anvils compress a relatively large specimen. Compared with a diamond anvil cell, it typically offers greater sample volume for synthesis and bulk measurements, with less optical access and a lower ultimate pressure range.

Multivariable calculus extends differentiation and integration to functions of several variables.

Murray's law balances viscous pumping power against the metabolic cost of maintaining blood volume. For cylindrical vessels it gives $Q\propto R^3$ and hence $R_0^3=R_1^3+R_2^3$ at an ideal bifurcation.

Diffusional material creep dominated by transport through grain interiors. For an equiaxed-grain model, $\dot{\epsilon }\sim C{D}_V\Omega \sigma /(k_{B}T{d}^2)$, with geometric factor $C$, atomic volume $\Omega$ and grain size $d$. It is linear in stress in this ideal limit.

Ordinary solid sodium chloride has the rock salt structure. Choose a conventional cube of side $a$ with chlorine anions at the corners and face centers, and sodium cations at the edge centers and cube center. Corner sites are shared by eight cubes, face sites by two and edge sites by four. The unit cell counts are thereforeThere are four formula units per conventional unit cell, rather than four separate NaCl molecules.^[ref]

The Bravais lattice is a face-centered cubic lattice. One choice of primitive lattice translation vectors and atomic basis isThe determinant gives a primitive unit cell volume of $a^3/4$; that cell represents one cation and one anion. Four such volumes make the conventional cube. The union of all ion positions resembles a primitive cubic lattice of step $a/2$ only if the chemical labels are discarded: translation by $a(1,0,0)/2$ exchanges unlike ions and fails to preserve the chemical structure.

Each ion has six nearest unlike neighbors at distance $a/2$, along the positive and negative directions of the three cube axes. Thus the coordination number is six. Within either same-species face-centered cubic lattice, the nearest same-ion neighbors instead number twelve: their displacements are permutations of $(\pm a/2,\pm a/2,0)$, all of length $a/\sqrt{2}$. Since $a/\sqrt{2}>a/2$, those twelve are a more distant shell in the complete sodium chloride structure. The number twelve describes the same-ion FCC lattice neighborhood; six describes the nearest unlike-ion neighborhood.

A particle with dimensions on the nanometre scale.

Engineering of structures, materials and devices with important nanoscale dimensions.

Removing a central cylindrical hole from a sphere leaves a solid whose volume depends only on its remaining height $h$. With $a=h/2$, sphere radius $R$ and hole radius $r_0$, the rim gives $R^2-r_0^2=a^2$. Every horizontal slice is an annulus of area $\pi (a^2-z^2)$, so integrating from $-a$ to $a$ gives $V=4\pi a^3/3=\pi h^3/6$. This slice equality explains why changing $R$ while retaining $h$ leaves the volume unchanged.

For observations $X_1,\dots ,X_n$, the nearest neighbour distance of order $k$ at $x$ is the $k$th order statistic of the distances $|X_i-x|$. An atomless distance distribution function ensures ties occur with probability zero.

Localization of tensile deformation into a narrowing region of a specimen. The onset depends on the competition between declining cross-sectional area and work hardening.

A decrease in equilibrium temperature when the total energy increases along a specified family of equilibria. A virialized nonrelativistic self-gravitating ideal gas has $E=-U$ and $U$ proportional to its particle-weighted mean temperature, producing this behavior. It does not mean the local gas parcel has negative heat capacity at fixed volume. It occurs naturally for isolated gravitational systems rather than stable ordinary canonical equilibrium with an infinite heat bath.

Suppose a solid is above the elliptic paraboloid $z=x^2+y^2$ and below a plane whose vertical separation from it is $A-|\mathbf{x}-\mathbf{p}{|}^2$, where $A>0$. A second lower plane exceeds the elliptic paraboloid by $B-|\mathbf{x}-\mathbf{q}{|}^2$, with $B>0$. If $|\mathbf{p}-\mathbf{q}|+\sqrt{B}\le \sqrt{A}$, the disk where the lower plane matters lies within the upper projection. Raising the bottom to this lower plane therefore removes exactly its excess volume, without changing the outer projection.

Translate polar coordinates separately to the two centres. The double integral over a disk of squared radius $A$ isSubtracting the analogous integral with $B$ proves the displayed volume formula. The containment condition is essential: if the two disks overlap only partly, subtract only over their intersection.