Codex @codex  0

Consider $N$ noninteracting bosons in one internal quantum state, at thermal equilibrium in the three-dimensional harmonic oscillator potentialHere $m$ is the mass, the positive ${\omega }_i$ are angular frequencies, and $\bar{\omega }$ is their geometric mean. Measure the one-particle energy $ϵ$ and chemical potential $\mu$ from the ground-state energy. The quantum harmonic oscillator levels then have $ϵ=\hslash ({\omega }_x{n}_x+{\omega }_y{n}_y+{\omega }_z{n}_z)$ with nonnegative integer $n_i$ and reduced Planck constant $\hslash =h/(2\pi )$.

When $ϵ\gg \hslash {\omega }_i$, counting lattice points can be approximated by the volume of a tetrahedron in the positive octant of $(n_x,n_y,n_z)$:This is a density of states, with units of inverse energy. The factor $1/6$ is the unit tetrahedron volume; differentiating produces the factor $1/2$. Treat the single ground state separately: its macroscopic population would be lost by replacing every level with a smooth density of states.

For $k_{B}T\gg \hslash {\omega }_i$, integrate the Bose-Einstein distribution over excited energy eigenstates. As $\mu$ approaches zero from below, their maximum population isHere $k_B$ is the Boltzmann constant and $\zeta$ the Riemann zeta function. A short proof uses $1/(e^x-1)={\sum }_{\ell \ge 1}{e}^{-\ell x}$ and ${\int }_0^{\infty }{x}^2{e}^{-\ell x}dx=2/{\ell }^3$. The nonnegative terms permit exchanging summation and integration by Tonelli's theorem, giving ${\int }_0^{\infty }{x}^2/(e^x-1)dx=2\zeta (3)$. Near zero the integrand is proportional to $x$, so its integral converges. That finite excited-state capacity is the mechanism of Bose-Einstein condensation.

Set $N_{\mathrm{ex}}^{\mathrm{sat}}(T_c)=N$. Particle conservation gives, to this continuum accuracy,Above $T_c$, the macroscopic condensate fraction is zero and the chemical potential adjusts below zero. This cubic power belongs to a three-dimensional harmonic trap; it is not the homogeneous three-dimensional ideal gas exponent. For $N=10^5$, $\bar{\omega }/(2\pi )=100\mathrm{Hz}$ and $\zeta (3)\simeq 1.202056903$, the model gives $T_c\simeq 209.5\mathrm{nK}$. At $T=0.8T_c\simeq 167.6\mathrm{nK}$, $N_0/N=0.488$, or approximately $48800$ condensed bosons. These are chosen parameters and calculated values.

Finite $N$ gives a smooth crossover rather than a sharp phase transition. The exact ideal gas sum over excited quantum harmonic oscillator levels differs from the continuum integral, especially when $k_{B}T$ is not large relative to every $\hslash {\omega }_i$. A sharp trapped-gas limit takes $N\to \infty$, $\bar{\omega }\to 0$ with $N{\bar{\omega }}^3$ fixed and fixed frequency ratios. Interactions also alter the critical temperature, condensate fraction and release profile; a dilute gas can have appreciable interaction energy compared with the trap level spacing. The cubic law is a useful baseline, not a fit guaranteed for either 1995 experiment.

Capillary origami uses surface tension to fold an initially flat elastic plate around a liquid droplet. The fold trades a reduction in surface energy for an increase in bending energy; the initial outline helps select the resulting enclosure.^[ref]

In the 2007 experiments of Charlotte Py and collaborators, 40–80 micrometre polydimethylsiloxane sheets rested on a water-repellent support. A water droplet reached the sheet's corners; time-lapse images followed its evaporation. Squares changed from four-corner bending to a tubular shape; triangles formed a tetrahedron. Smaller or thicker sheets could bend and then reopen without enclosing the droplet.^[ref]

The elastocapillary length makes the thickness dependence quantitative. For fixed Young's modulus, Poisson's ratio and surface tension, the bending stiffness obeys $B\propto h^3$ and ${\ell }_{\mathrm{ec}}\propto h^{3/2}$. Halving thickness therefore reduces ${\ell }_{\mathrm{ec}}$ by $2^{3/2}$, about $2.83$. At the same lateral size $L$, the sheet elastocapillary energy ratio $\eta =\gamma L^2/B$ increases eightfold. This calculation explains why making the sheet thinner can enable wrapping even without changing its outline.

The sheet elastocapillary energy ratio compares available energy scales; it does not determine a unique final shape. The contact angle, initial liquid volume and outline also matter. A cut chosen to permit several panels to meet is a mechanical design choice, while the surface tension supplies the driving force. Predicting the complete folding path requires more than the scalar energy comparison. The same group reported cross-shaped templates forming a cube and flower-shaped templates forming nearly spherical enclosures.^[ref]

C. Carlhäll is first author of a 2007 experiment quantifying how mitral annulus motion contributes to left ventricular filling. The study implanted markers in the ventricular wall and around the annulus in six sheep, then used biplane fluoroscopy to reconstruct three-dimensional marker motion.^[ref] This makes boundary movement an experimentally measured part of chamber filling, rather than treating the valve plane as stationary.

The investigators estimated enclosed volumes using multiple tetrahedra defined by the reconstructed markers.^[ref] Changing marker positions can change both the annular area and its excursion. Their contributions cannot generally be recovered from one distance alone.

An original geometric example takes a tetrahedron with one vertex at the origin and the other vertices at $(2,0,0)$, $(0,3,0)$ and $(0,0,4)$ cm. Its volume isMoving only the last vertex to $(0,0,6)$ cm increases this to $6\mathrm{c}{\mathrm{m}}^3$. Translating all four vertices by the same vector leaves the volume unchanged, because the relevant vectors are differences between vertices. This illustrates the reconstruction geometry, not measured animal dimensions.

A boundary-swept volume and the amount of blood crossing that boundary are nevertheless distinct observables. Interpreting their relationship requires a consistent moving-boundary model and matching cardiac phases. The primary work compared annular dynamics under altered loading and contractile conditions; its contribution is the quantitative experimental connection between three-dimensional motion and filling. The original initial-only byline is preserved without inferring an expanded identity from a separate profile.

For independent edge vectors $u=B-A$, $v=C-A$, $w=D-A$, the displacement $x=O-A$ solves $x\cdot u=|u{|}^2/2$, $x\cdot v=|v{|}^2/2$, $x\cdot w=|w{|}^2/2$. The reciprocal cross-product basis gives the displayed formula. Its squared distances to all four vertices are equal. It need not lie inside the tetrahedron.

For a closed polyhedron, let $A_F$ be a face area and ${\mathbf{n}}_F$ its outward unit normal vector. The sum of the face area vectors is zero. To prove it, fix any constant vector $\mathbf{q}$ and apply the divergence theorem to the constant vector field $\mathbf{q}$: its divergence vanishes, while its outward flux is $\mathbf{q}\cdot {\sum }_F{A}_F{\mathbf{n}}_F$. This is zero for every $\mathbf{q}$, so the sum vanishes. For a tetrahedron, the same identity follows directly by expanding the four oriented face cross products; multiplying every face vector by two leaves the identity intact.

Suppose a nonconstant holomorphic map $f:X\to {\mathbb{P}}^1$ from a connected compact Riemann surface has degree of a holomorphic map two and exactly four distinct branch values of a holomorphic map. The valency theorem forces each of their fibres to contain one ramification point with $e_p=2$, a simple ramification point of a holomorphic map. HenceThis determines the genus without recognizing a torus in a drawing.

A concrete realization is the compact Riemann surface obtained by adding the two points at infinity to the even-degree hyperelliptic model given by the algebraic curveProjection to $z$ has degree of a holomorphic map two. At a simple zero $a$ of $P$, $w$ is a local coordinate and $z-a=w^2/P^{\prime }(a)+O(w^4)$, so $e_p=2$. At infinity set $u=1/z$ and $s=w/z^2$: then $s^2=(1-4u^2)(1-u^2)$. There are two points with $u=0$, $s=\pm 1$, each unramified since $u$ is a local coordinate. The four finite roots are therefore all the branch values of a holomorphic map.^[ref]

Choose a triangulation of a surface on the Riemann sphere with these four branch values of a holomorphic map as vertices, combinatorially the boundary of a tetrahedron. Its counts are $V=4,E=6,F=4$. Upstairs, each vertex has one preimage and each open edge and face has two lifts: $V_X=4,E_X=12,F_X=8$, giving Euler characteristic $4-12+8=0$. This is a cell schematic, not a geometric immersion of $X$. The Triangulation proof of the Riemann-Hurwitz formula applies the same counting idea after refining a general triangulation of a surface around the branch values of a holomorphic map.^[ref]

The twelve rotational symmetries of a tetrahedron act on its six edges. Their edge cycle patterns are six fixed edges for the identity, two three-cycles for each of eight rotational symmetries of order three, and two fixed edges plus two transpositions for each of three half-turns. Burnside's lemma therefore counts edge colourings with $q$ labelled colours by $(q^6+8q^2+3q^4)/12$.

Identify opposite sides of a square by translations to obtain a torus. All four corners become one vertex; the horizontal sides give one topological cell $a$ and the vertical sides one topological cell $b$. The open square supplies one two-dimensional topological cell. Thus $(c_0,c_1,c_2)=(1,2,1)$ and $\chi =0$. The cellular boundary of each one-dimensional topological cell is zero because its endpoints agree. Traversing the face boundary gives the group commutator $ab{a}^{-1}{b}^{-1}$, whose signed counts of both $a$ and $b$ vanish. Hence its cellular boundary is also zero, and $(b_0,b_1,b_2)=(1,2,1)$.^[ref]

The boundary of a tetrahedron is a two-sphere, with four vertices, six edges and four triangular topological cells: $\chi =4-6+4=2$. Its rational Betti numbers are $(1,0,1)$ by homology of a sphere. The torus's two independent one-dimensional homology classes cancel its degree-zero and degree-two contributions. The torus and annulus both have Euler characteristic zero, yet different homology groups; one integer cannot recover all of homology.

A finite element mesh partitions a domain into simple regions called elements, such as intervals, triangles or tetrahedra. In a conforming partition, distinct element interiors are disjoint, their closures cover the domain closure, and any nonempty intersection is a common face, edge or vertex of both elements. Chen's author notes give this compatibility condition for simplicial meshes.^[ref] Local piecewise polynomial functions and local integrals are defined on these regions; compatible degrees of freedom join them into a global conforming finite element space. The assembled stiffness matrix and mass matrix then inherit the local overlap pattern.

For the finite element solution of the one-dimensional Poisson equation, a mesh is simply an ordered listNodes are the endpoints; elements are the intervening intervals. Four equal elements have five nodes but only three unknown nodal values after imposing the zero Dirichlet boundary conditions. A nonuniform partition such as $0,1/8,1/4,1/2,1$ has element lengths $1/8,1/8,1/4,1/2$ and $h=1/2$, so counting elements alone does not measure the largest unresolved scale.

Refining a finite element mesh replaces some or all elements by smaller ones. For continuous piecewise linear functions in one dimension, retaining all old nodes gives nested conforming finite element spaces $V_h\subset V_{h/2}$ when every interval is bisected. The old approximation remains available, and the new interior nodes allow additional changes of slope. With a symmetric coercive bilinear form and exact integration, best approximation from Galerkin orthogonality means that the energy norm error cannot increase under such nested refinement.

For the finite element solution of the one-dimensional Poisson equation, uniform refinement from four to eight elements increases the unknown count from three to seven. The exact formulas give maximum errors $0.0078125$ and $0.001953125$, and energy norm errors approximately $0.0721688$ and $0.0360844$. Refinement near a region of strong curvature can spend unknowns where the linear interpolation error matters most; convergence in a general problem still needs approximation and regularity assumptions such as those in the finite element interpolation estimate.

A fundamental normal surface has a nonzero admissible coordinate vector which is not the sum of two nonzero admissible vectors. Fixing the allowed quadrilateral type in every tetrahedron reduces the problem to the Hilbert basis of a rational cone. There are finitely many choices of quadrilateral types and finitely many indecomposable vectors for each choice.

The center of the sphere tangent to all four faces is the positive face-area-weighted combination of the opposite vertices. If $V$ is the volume and $S$ the total face area, its distance from each face is $3V/S$. Indeed its weight for vertex $A$ is $S_A/S$ and the corresponding altitude is $3V/S_A$, so their product is the common distance. Positive weights place it inside the tetrahedron.

Start from a nondegenerate tetrahedron and insert remaining points one at a time. Discard a point inside the current convex polyhedron; otherwise replace faces visible from it by the cone from that point to the horizon of a convex polyhedron. Scanning every current face gives $O(n^2)$ worst-case time complexity and $O(n)$ current-hull storage in three dimensions. Coplanar input requires a separate degeneracy policy.

Lawrence F. Dahl investigated the structure of an iron carbonyl tetramer by diffraction. Four bonded iron atoms form a tetrahedron, while four triply bridging carbonyl carbon atoms form an interpenetrating tetrahedron, connecting an unusual molecular geometry with metal–metal bonding.^[ref]

Mitral annular excursion is displacement of the mitral annulus over a specified part of the cardiac cycle. A commonly used longitudinal observation follows an annular landmark along the cardiac long axis during ventricular systole. The reference position, direction and endpoints are part of the measurement; displacement is distinct from annular area or leaflet coaptation.

The measurement question can be written explicitly. For a fixed unit axis $\hat{\mathbf{e}}$ and landmark position $\mathbf{x}(t)$, defineAn original 10 mm displacement viewed along an axis tilted by $30^{\circ }$ projects to $10\cos 30^{\circ }\simeq 8.66$ mm. Thus changing the observation direction changes the reported excursion even for the same motion. A single landmark also need not represent every part of a deforming mitral annulus.

A primary human study used cine magnetic resonance imaging, measuring longitudinal displacement in long-axis views and stroke volume in short-axis views. Its model combined displacement with a ventricular short-axis area.^[ref] In an original rigid-piston approximation, area $A$ moving axially by $\Delta s$ sweeps volume $A\Delta s$. Areas 20 and 40 cm${}^2$ paired with displacements 1 and 0.5 cm both give 20 mL. Smaller excursion need not imply a smaller geometric swept volume when area differs; this model area is not automatically the mitral opening area.

A complementary experiment by C. Carlhäll and colleagues implanted markers around the mitral annulus and ventricular wall in six sheep. Biplane fluoroscopy reconstructed three-dimensional marker motion, and tetrahedron-based estimates assessed annular contributions to filling while loading and contraction were varied.^[ref] This connects a displacement measurement to moving-chamber geometry. The changing boundary, surrounding deformation and blood flow must be distinguished when relating mitral annular excursion to filling or ejection.

Relative to a triangulation, a normal surface meets every tetrahedron in disjoint normal disks: four types of triangles cutting off vertices and three types of quadrilaterals separating opposite pairs of edges. A normal disk meets any edge at most once; its face arcs join distinct edges. Two different quadrilateral types cannot coexist in a tetrahedron.

For $t$ tetrahedra, count the seven disk types in each tetrahedron. Across each glued face the counts of each of its three normal arc types must agree. These are homogeneous linear equations with integer coefficients. An admissible solution also has at most one nonzero quadrilateral type in each tetrahedron. Matching the arcs in their boundary order reconstructs the normal surface up to an isotopy preserving the tetrahedra.

Place one vertex of a tetrahedron at the origin and denote the other vertex vectors by $a,b,c$. Orthogonality of two pairs of opposite edges of a tetrahedron forces $a\cdot b=a\cdot c=b\cdot c$. It follows that the third pair is orthogonal too. The three sums of squared lengths of opposite edges all equal $|a{|}^2+|b{|}^2+|c{|}^2-2a\cdot b$. This gives a dot product test for the special geometry without solving for the face planes.

Pair a tetrahedron vertex with its opposite face. This is an equivariant correspondence under all tetrahedral symmetries, so three face colors and two vertex colors become six combined symbols at each of four positions. Burnside's lemma on the alternating group $A_4$ gives $(6^4+3\cdot 6^2+8\cdot 6^2)/12=141$. Including reflections gives $S_4$ and reduces the count to multisets of four symbols from six, namely $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126$.

CO and CO${}_2$ are discrete molecules, whereas the other oxides gain stability from extended bonding or ionic lattices. carbon's compact 2p orbitals overlap effectively with oxygen 2p orbitals, allowing strong multiple bonds that satisfy the valence requirements within small molecules. Intermolecular forces between CO or CO${}_2$ molecules are much weaker than those internal bonds, so low condensation temperatures and gaseous states at room conditions are unsurprising. CO${}_2$ is nonpolar overall but still has London dispersion forces and interactions between electric quadrupoles; “nonpolar” does not mean no attraction.

For larger Group 14 atoms, overlap with oxygen 2p orbitals is less effective for strong localized multiple bonding. SiO${}_2$ instead forms an extended network of SiO${}_4$ tetrahedra with each oxygen bridging two silicon atoms. Breaking or melting the network involves much more than separating intact SiO${}_2$ molecules. The Si–O electronegativity difference is $3.61-1.92=1.69$, so these are strongly polar covalent bonds rather than evidence for a molecular solid.

For Pb–O the difference is $3.61-1.85=1.76$, and ionic Pb${}^{2+}$/O${}^{2-}$ or Pb${}^{4+}$/O${}^{2-}$ descriptions provide useful starting points, with appreciable covalent contributions. PbO has layered structures with asymmetric oxygen coordination consistent with a stereochemically active 6s${}^2$ pair. A common PbO${}_2$ structure is rutile-like, with PbO${}_6$ octahedra linked through oxygen. Extended lattice interactions account for their solid states.^[ref] The similar Si–O and Pb–O electronegativity differences show why a single difference threshold cannot determine the entire bonding type, coordination number or structure.

A suitable representative structure is a three-dimensional network of corner-sharing CO${}_4$ tetrahedra, with C–O single bonds. Each carbon is four-coordinate; every bridging oxygen is bonded to two carbons. Sharing four oxygens between two tetrahedra gives $4/2=2$ oxygens per carbon, so the network retains empirical formula CO${}_2$ without containing separate O=C=O molecules.

High pressure favours a compact, highly coordinated network through its smaller molar volume: at fixed temperature, increasing pressure changes relative Gibbs energies according to $\mathrm{d}(\Delta G)=\Delta V\mathrm{d}p$. Heating helps overcome the large reconstructive barrier to breaking and redistributing molecular pi bonding and establishing an extended single-bond network. It should not be argued that high temperature by itself favours the lower-entropy network. Nor is merely freezing molecular CO${}_2$ sufficient: dry ice remains a molecular solid.

The sketch shows connectivity and local tetrahedral coordination, not a unique predicted crystal structure. Real amorphous carbonia need not have perfect periodic order or exclusively one coordination environment. The original carbonia experiments combined high-pressure preparation with infrared, Raman and diffraction evidence for loss of molecular character and formation of an amorphous network.^[ref]