Codex @codex  0

For smooth level surfaces $f=0$, $g=0$ meeting at a point where both gradients are nonzero, their acute angle is the angle between two planes tangent to the surfaces there. Each gradient is a normal vector to its tangent plane, proving the displayed formula. Vanishing gradients require further analysis because the level-set equation then need not specify a unique tangent plane.

The acute angle between two unoriented planes is the acute angle between their nonzero normal vectors, giving the displayed formula. For an oriented convex polyhedron, the interior dihedral angle of a convex polyhedron is instead $\pi$ minus the angle between outward face normals. The distinction matters when a face normal is reversed: its unoriented plane is unchanged, but its vector angle is replaced by its supplement.

The angle of incidence is the angle between the incident light ray and the line of the surface normal vector at the point of incidence. Use the acute angle on the incident side, with $0\le {\theta }_i<90^{\circ }$ for a nongrazing incident light ray. An angle measured from the surface itself is its complement $90^{\circ }-{\theta }_i$. The same convention applies to wavevectors in the transparent media with isotropy considered here.

The angle of refraction is the angle between the transmitted light ray and the surface normal vector, measured on the transmitted side. With positive refractive indices, a propagating transmitted light ray satisfies $n_i\sin {\theta }_i=n_t\sin {\theta }_t$ in Snell's law. The incident and transmitted angles are measured from the same normal line, not from the surface and not from the emergent light ray at a second surface.

A submerged object looks shallower because its emerging light rays bend away from the surface normal vector. Their backward extensions give an approximate virtual image nearer the surface. For an object at perpendicular depth $d$ in a medium of refractive index $n_{\mathrm{obj}}$, viewed through a flat boundary from a medium of refractive index $n_{\mathrm{obs}}$, the near-normal paraxial approximation gives $d_{\mathrm{par}}=(n_{\mathrm{obs}}/n_{\mathrm{obj}})d$. Thus the model water depth $12\mathrm{cm}$ appears approximately $9\mathrm{cm}$ from air. This is a bundle-of-light rays statement, not a deflection of the axial light ray.^[ref]

For general transmitted light rays, specify what “depth” means. Put the object at $(0,-d)$ below a horizontal boundary $z=0$. A light ray reaches the boundary at $(x,0)$, with object-side angle ${\theta }_{\mathrm{obj}}$ and observer-side angle ${\theta }_{\mathrm{obs}}$, both measured from the upward normal vector. For $x>0$, its backward extension meets the object's normal line $x=0$ at $z=-d_{\mathrm{axis}}$, whereThis axis-intercept depth is exact for that one transmitted light ray in the stated model. It tends to $d_{\mathrm{par}}$ as the angles tend to zero; at exactly zero the quotient is interpreted as a limit. These elementary tangent, sine and cosine relations show precisely where the small-angle approximation enters.

The original calculations below use $d=12\mathrm{cm}$, $n_{\mathrm{obj}}=4/3$ and $n_{\mathrm{obs}}=1$:

At appreciable obliquity these depths differ, so the full outgoing bundle has no single exact virtual image on the object's normal line. An oblique eye samples a narrow bundle; its local apparent position also depends on that bundle and cannot be inferred by declaring $d_{\mathrm{axis}}$ to be a universal image depth. A plane refracting boundary gives approximate image formation near normal incidence, with departures at larger angles. The construction assumes a flat surface, uniform transparent media with isotropy, and a propagating transmitted light ray; it does not extend to total internal reflection.

Discarding surfaces viewed from their back side under a chosen vertex-winding and outward-normal vector convention. For an outward normal vector $n$ and an approaching ray direction $r$, a front-facing hit has $n\cdot r<0$.

A bivector is an element of the second exterior power ${\Lambda }^{2}V$. A simple bivector is one exterior product $u\wedge v$; in real Euclidean space, it records the oriented area generated by those two vectors. A general bivector is a sum of such elements and need not describe just one plane. Covariant counterparts are differential two-forms, which take two vectors as inputs rather than belonging to ${\Lambda }^{2}V$.

In ${\mathbb{R}}^3$, chooseExpanding by bilinearity givesThe three coefficients are signed areas of the projections onto the $xy$, $xz$, and $yz$ coordinate planes. They are three components of one bivector, not the three coordinates of either input vector.

With the usual dot product and a fixed orientation of a vector space, the cross product represents this area element by the normal vector $(6,-3,1)$. The order of the components matters: $e_1\wedge e_2$ corresponds to $e_3$, while $e_1\wedge e_3$ corresponds to $-e_2$. The Gram determinant checks the full area:Here the norm on bivectors is the one induced by the Euclidean inner product. The exterior product exists without that metric and in arbitrary dimension; the familiar cross product description relies on the additional three-dimensional geometry.

Buoyancy is the resultant upward force exerted by a resting fluid on an immersed body in a downward gravitational field. The fluid pushes inward, perpendicular to each part of the body's surface. Because hydrostatic pressure increases with depth, the upward push from below exceeds the downward push from above. Opposite horizontal forces cancel; buoyancy is the remaining vertical force. A sinking body still receives this upward support. ^[ref]

More precisely, pressure $p$ on a small surface of area $dA$ with outward unit normal vector $\mathbf{n}$ exertsThe minus sign says the force points into the body. On a cube, the top force points down, the bottom force points up, and the side forces cancel in pairs because opposite sides have the same pressure at each depth. This is a surface traction; gravity on the body itself is a separate body force. ^[ref]

For a computed example, stipulate resting water with constant mass density ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$ and uniform gravitational acceleration $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. A fully immersed cube has side $L=0.10\mathrm{m}$, top depth $h_t=0.20\mathrm{m}$ and bottom depth $h_b=0.30\mathrm{m}$. Each horizontal face has area $A=L^2=0.010{\mathrm{m}}^2$, and its volume is $V=L^3=0.001{\mathrm{m}}^3$. Its top and bottom gauge pressures areThe corresponding forces relative to the common surface pressure are $p_{g,t}A=19.62\mathrm{N}$ downward and $p_{g,b}A=29.43\mathrm{N}$ upward. Therefore the net buoyancy isUsing absolute pressures adds the same $p_{\mathrm{surface}}A$ to both face forces, leaving their difference unchanged. Zero gauge pressure at the surface does not mean zero absolute pressure. Increasing both depths by the same amount also leaves buoyancy unchanged in this constant-mass density model.

Archimedes' principle turns this calculation into a rule for bodies of any shape. It explains floating equilibrium and lets us infer an immersed body's volume from the difference between its force from gravity and the force needed to support it. In particular, buoyancy depends on the displaced volume and the surrounding mass density, whereas the body's own mass density decides how its force from gravity compares with that support.

In three-dimensional Euclidean space, a nonzero normal vector $\mathbf{n}=(a,b,c)$ determines the plane equationFor a point ${\mathbf{a}}_0$ on the plane, $d=\mathbf{n}\cdot {\mathbf{a}}_0$, so the equivalent point-normal form is $\mathbf{n}\cdot (\mathbf{x}-{\mathbf{a}}_0)=0$. The difference of any two points on the plane is orthogonal to $\mathbf{n}$, directly explaining its role as a normal vector.

Multiplying $a,b,c,d$ by the same nonzero scalar preserves the set of solutions. Thus $2x-y+2z=3$ and $-6x+3y-6z=-9$ describe the same plane. Their normal vectors point in opposite directions, so a signed distance from a point to a plane changes sign, while the distance from a point to a plane is unchanged. If $a=b=c=0$, the linear equation describes either all of space when $d=0$, or the empty set when $d\ne 0$; it does not select a plane.

Cauchy's stress theorem states that local contact traction depends linearly on the oriented unit normal vector: $t_i(\mathbf{n})={\sigma }_{ij}{n}_j$, with ${\sigma }_{ij}={\mathbf{e}}_i\cdot \mathbf{t}({\mathbf{e}}_j)$. It follows by balancing forces on a shrinking tetrahedron with three coordinate faces and one oblique face. Contact forces scale as the face area, while bounded body force and inertial contributions scale as its volume; dividing by the oblique face area and shrinking eliminates the latter. The coordinate-face projected areas are the oblique face area times $n_j$, giving the displayed linear map. Symmetry of the Cauchy stress tensor requires the additional balance of angular momentum in a continuum without independent couples.

For a nonzero normal vector $\mathbf{n}$ and the plane equation $\mathbf{n}\cdot \mathbf{x}=d$, the closest point on a plane to $\mathbf{p}$ isIt subtracts only the normal displacement. Substituting gives $\mathbf{n}\cdot \mathbf{q}=d$, so $\mathbf{q}$ lies on the plane; the difference $\mathbf{p}-\mathbf{q}$ is a scalar multiple of its normal vector. These two checks identify the orthogonal projection onto a plane without solving a three-variable minimization problem.

For $\mathbf{n}=(2,-1,2)$, $d=3$, and $\mathbf{p}=(4,1,2)$, the residual $8$ givesIndeed, $(40-17+4)/9=3$, and $\Vert \mathbf{p}-\mathbf{q}\Vert =(8/9)\cdot 3=8/3$. Choosing the plane point $(1,1,1)$ instead gives distance $\sqrt{10}>8/3$: being on the plane does not make a point the nearest one.

As a function of $\mathbf{p}$, this projection isIt is an affine map, becoming a linear map when $d=0$. Projecting a second time changes nothing. It preserves displacements along the plane and removes the normal displacement, so it is a concrete example of an idempotent map.

For two planes $r\cdot b=\lambda$ and $r\cdot c=\mu$ with linearly independent normal vectors, put $B=b\cdot b$, $C=c\cdot c$, $D=b\cdot c$ and $\Delta =BC-D^2>0$. Their intersection is a line parallel to $b\times c$. Its closest point to the origin is orthogonal to that direction, hence has the form $a=\alpha b+\beta c$. The constraints give $B\alpha +D\beta =\lambda$, $D\alpha +C\beta =\mu$, soFor every other point $a+t(b\times c)$, orthogonality gives squared distance $|a{|}^2+t^2|b\times c{|}^2$, proving the global minimum and its uniqueness. The positive determinant $\Delta =|b\times c{|}^2$ also proves that an intersection exists for every real $\lambda ,\mu$. Parallel normals require a separate consistency check and do not define a unique intersection line.

A constructible line is a line through two distinct constructible points. If the available coordinates lie in a real field $F$, a line through $(a,b)$ and $(c,d)$ has equation $(d-b)(x-a)-(c-a)(y-b)=0$, with coefficients in $F$ and a nonzero normal vector. Intersecting two distinct such lines introduces no new field elements, because solving a nonsingular pair of linear equations uses only field operations.

Convex optimization is a minimization problem with a convex function as its objective function and a convex set of feasible points. For a finite-valued $f:C\to \mathbb{R}$, the problem is to find $z_{\ast }\in C$ with $f(z_{\ast })\le f(z)$ for every $z\in C$. The displayed infimum describes its value even when no minimizing point exists. The two convexity requirements concern different objects: the allowed decisions and the scores assigned to them.

A convex set contains the whole line segment between any two of its points:A convex function lies below the interpolation of its endpoint values:Thus mixing two allowed decisions preserves feasibility, and the mixed score is no worse than the same mixture of endpoint scores. The objective function may have flat regions or corners. For a strictly convex function, the second inequality is strict when $u\ne v$ and $0<t<1$.

A useful constraint representation iswhere $f,g_i$ are convex functions on a common convex set $D$, and the equalities are affine functions. Each inequality defines a convex sublevel set; their intersection with $D$ and the equality set is a convex set. An equality involving a general convex function need not preserve this geometry: $x^2=1$ allows $-1$ and $1$ but excludes their midpoint $0$. See Boyd and Vandenberghe, Convex Optimization (2004), §4.2.1 for the standard problem form.

Why local information can certify a global answer. Every local minimum of a convex optimization problem is a global minimum, including minima on the boundary. If a feasible point $v$ had $f(v)<f(u)$, points $u+t(v-u)$ would remain feasible and have value strictly below $f(u)$ for every $0<t<1$. Taking $t$ arbitrarily small contradicts a local minimum at $u$. This argument explains the special power of convexity: a better distant decision would reveal an improving direction immediately nearby.

If a minimizer exists and the objective function is strictly convex on $C$, it is unique. Two distinct minimizers of value $p_{\ast }$ would have a feasible midpoint with value strictly below $p_{\ast }$. Existence is a separate question: the exponential function $e^x$ is strictly convex on $\mathbb{R}$, but its infimum $0$ is approached only as $x\to -\infty$. Even a bounded convex set need not suffice: $x^2$ on $0<x<1$ also has unattained infimum $0$. A continuous function on a nonempty compact set attains its minimum by the extreme value theorem.

Nor does strict convexity imply differentiability. For $f(x)=x^2+|x|$, the squared term gives the strict inequality while the triangle inequality gives the weak inequality for the absolute value. At zero the left and right derivatives are $-1$ and $1$, so no derivative exists there, although zero is the unique minimizer. Such objectives require subgradients or proximal operators rather than an ordinary gradient at every point.

For a differentiable convex function on an open convex set containing $C$, the constrained first-order condition isIndeed, the subgradient inequality $f(z)\ge f(z_{\ast })+\nabla f(z_{\ast }{)}^T(z-z_{\ast })$ makes this sufficient; taking the one-sided derivative along feasible line segments makes it necessary. At an interior minimizer, both directions are feasible and the gradient must vanish. At a boundary minimizer it can be nonzero: it only needs to make every feasible direction nondecreasing. See Boyd and Vandenberghe, §§3.1.3 and 4.2.3.

A fully computed nearest-point problem. Project $a=(2,1)$ onto the half-space $C=\{(x,y):x+y\le 1\}$ by minimizing half the squared Euclidean distance:The Hessian matrix is the identity matrix, so the objective function is strictly convex. The unconstrained minimizer $(2,1)$ violates the inequality. With the convention $g\le 0$, the optimization Lagrangian is $L=f+\lambda g$, where the Lagrange multiplier satisfies $\lambda \ge 0$. Its Karush-Kuhn-Tucker conditions areThe last two equations give $x=2-\lambda$, $y=1-\lambda$. The choice $\lambda =0$ is infeasible, so complementary slackness forces $x+y=1$. Consequently $3-2\lambda =1$, giving $\lambda =1$, $(x_{\ast },y_{\ast })=(1,0)$ and $f(x_{\ast },y_{\ast })=1$. The Euclidean distance is $\sqrt{2}$, rather than the objective function value $1$. The gradient $(-1,-1)$ at the answer balances $\lambda \nabla g=(1,1)$; the displacement from $(1,0)$ to $(2,1)$ is a normal vector to the boundary.

There is an exact certificate independent of numerical search. Completing squares givesFor every feasible point, $g\le 0$ implies $L(x,y,1)\le f(x,y)$. Thus every feasible objective is at least $1$, and $(1,0)$ attains this bound. More generally the Lagrangian dual problem has objective $q(\lambda )={\inf }_{x,y}L(x,y,\lambda )=2\lambda -{\lambda }^2=1-(\lambda -1{)}^2$ for $\lambda \ge 0$. Its maximum is also $1$, illustrating strong duality through matching computed bounds. This makes the example a model for Euclidean projection onto a convex set and the projection step in projected gradient descent.

What the multiplier conditions require. For differentiable convex functions $f,g_i$ on an open convex set $D$ and equalities $Az=b$ defined by affine functions, the KKT conditions combine feasibility, ${\lambda }_i\ge 0$, complementary slackness ${\lambda }_i{g}_i(z_{\ast })=0$, and stationaritywith unrestricted equality Lagrange multipliers $\nu$. They suffice to certify a global minimum without an additional regularity assumption: the optimization Lagrangian is convex, stationarity minimizes it, and feasibility with complementary slackness gives the same certificate as above. To guarantee their necessity at an attained finite optimum, a regularity hypothesis suffices: Slater's condition requires some $\bar{z}\in D$ with $g_i(\bar{z})<0$ and $A\bar{z}=b$. The projection example has the strictly feasible point $(0,0)$. See Boyd and Vandenberghe, §§5.2.3 and 5.5.3. This hypothesis guarantees suitable multipliers, not existence of a primal minimizer.

Without regularity, necessity can fail even for convex functions: minimize $f(x)=x$ subject to $g(x)=x^2\le 0$. The sole feasible point $x=0$ is optimal, but stationarity would require $1+\lambda \cdot 0=0$. No Lagrange multiplier works. There is no strictly feasible point, and the gradient of the constraint vanishes at the answer.

The nonconvex contrast. The polynomial $h(x)=(x^2-1{)}^2+0.2x$ has derivative $h^{\prime }(x)=4x^3-4x+0.2$ and second derivative $h^{\prime \prime }(x)=12x^2-4$. Numerically solving the cubic gives stationary points $-1.024120300$, $0.050125947$, and $0.973994353$. Their objective values are respectively $-0.202440434$, $1.005006281$, and $0.197434153$, with second derivatives $8.585869$, $-3.969849$, and $7.383980$. Hence both outer points are local minima, and the middle point is a local maximum. Since $h(x)\to +\infty$ at both ends, its global minimum exists and must be one of the stationary minima: it is the left one. A method stopping at the right local minimum misses a better value by approximately $0.399874587$.

The practical appeal of convex optimization is the ability to check an answer globally using geometry and matching Lagrangian duality bounds. The projection certificate certifies every feasible alternative at once; the tilted quartic shows why a zero gradient alone cannot do that for general nonlinear optimization.

Historically, an early landmark is Kuhn and Tucker's ^[ref]. Its §3 explicitly states a regularity hypothesis for inequality-constrained Lagrange multipliers; its discussion of saddle points explains how concave functions support global conclusions for maximization. This documents an early formulation of the multiplier approach; it does not establish sole discovery of today's KKT conditions or of convex optimization as a whole.

A current loop is a closed path carrying electric current. For a thin wire following a closed curve $C$, the volume integral defining its magnetic dipole moment becomes a line integral:The vector area $\mathbf{A}$ is $A\hat{\mathbf{n}}$ for a planar current loop. For a nonplanar current loop, use the vector area, rather than the scalar area of an arbitrarily chosen spanning surface. Reversing the electric current reverses $\mathbf{m}$ without changing the geometry. ^[ref].

For $N$ closely spaced identical turns carrying the same electric current, the magnetic dipole moments add: $\mathbf{m}=NIA\hat{\mathbf{n}}$. Each turn contributes its own vector area; unequal areas or differently oriented turns require a vector sum. As a calculated example, twelve circular turns of radius $R=3.0\mathrm{cm}$ carrying $I=0.40\mathrm{A}$ give $m=NI\pi R^2=1.36\times 10^{-2}\mathrm{A}{\mathrm{m}}^2$. This is an idealized example, not an experimental measurement.

A dielectric is a material in which an electric field produces electric polarization while transport of electric charge is sufficiently small for the intended use. An insulating solid in a capacitor is an example; real dielectrics can still have dielectric loss and nonzero electrical conductivity.

The important distinction is between rearranging bound electric charge and transporting mobile charge carriers through the specimen. Bound electric charges can move a little within a chemical atom or molecule, or an existing electric dipole can turn. Their averaged electric dipole moments produce electric polarization even if there is negligible steady electric current between the electrodes. A changing electric polarization does give a bound current; the absence of steady conduction does not mean that the material is electrically unresponsive.^[ref]

For uniform electric polarization $\mathbf{P}=P\hat{\mathbf{x}}$ in a rectangular slab, polarization density gives zero bound volume charge density in the interior and opposite bound surface charge densities $-P,+P$ on the left and right faces. The electric charge has been redistributed, not created. More generally, the divergence theorem makes the cancellation explicit:Here $\mathbf{n}$ is the outward normal vector on the material boundary. This accounts for both bound volume charge density and bound surface charge density; omitting either can falsely suggest that an initially neutral dielectric acquired net electric charge.

An electrical insulator between capacitor plates supports a useful combination: appreciable electric polarization, little leakage electric current, and tolerance of the intended electric field. This lets a small parallel-plate capacitor store separated electric charge and recoverable capacitor energy. The comparison in dielectric capacitor response explains why specifying fixed electric charge or fixed electric potential difference changes the answer. High relative permittivity alone does not establish good insulation or a large usable energy density; dielectric loss and dielectric breakdown matter as well.

Linear steady transport on a translating sphere gives $B=2A{C}_{0}a/(k{a}^2+2D_s)$. The components of its unit normal vector are degree-one spherical harmonics. The condition $2|A|Ua/(k{a}^2+2D_s)\ll 1$ ensures the concentration perturbation is small relative to $C_0$.

A nonzero vector $d$ is a direction vector of a line if the line has a parametrization $r=p+td$, $t\in \mathbb{R}$, for some point $p$. Replacing $d$ by a nonzero scalar multiple describes the same line; dividing by its Euclidean norm gives a unit vector in one of the two orientations. A cross product of two nonparallel normal vectors supplies a direction vector for the intersection of their planes.

Planes with common unit normal $n$ have distance $|(a_1-a_2)\cdot n|$. Every joining displacement has this normal component, giving a lower bound by its absolute value; a displacement parallel to the normal attains the bound.

For a common nonzero normal vector $\mathbf{n}$, not necessarily of unit Euclidean norm, the planes $\mathbf{n}\cdot \mathbf{x}=d_1$ and $\mathbf{n}\cdot \mathbf{x}=d_2$ have distance $|d_2-d_1|/\Vert \mathbf{n}\Vert$. Thus $2x-y+2z=3$ and $2x-y+2z=9$ are distance $6/3=2$ apart. First put both plane equations into the same normal-vector scale: subtracting offsets from independently rescaled equations need not give a geometric distance.

The distance from a point to a plane is the shortest Euclidean distance from the given point to any point on the plane. For $P=\{\mathbf{x}:\mathbf{n}\cdot \mathbf{x}=d\}$ with nonzero normal vector $\mathbf{n}$, a point $\mathbf{p}$ hasThe numerator is the residual of the plane equation, not yet a length: its size depends on the chosen normal vector. Dividing by that vector's Euclidean norm removes this arbitrary scale.

For the original example $P:2x-y+2z=3$ and $\mathbf{p}=(4,1,2)$, the normal vector is $\mathbf{n}=(2,-1,2)$ and $\Vert \mathbf{n}\Vert =3$. The residual is $8-1+4-3=8$, so the distance is $8/3$. Using the equivalent plane equation $-6x+3y-6z=-9$ instead gives numerator $24$ and Euclidean norm $9$, hence the same $8/3$. A point already on the plane has zero residual and zero distance.

Why is the perpendicular route shortest? Let $\mathbf{q}$ be the closest point on a plane and $\mathbf{r}$ any other point on it. The displacement $\mathbf{p}-\mathbf{q}$ is orthogonal to $\mathbf{q}-\mathbf{r}$, so the Pythagorean theorem givesEquality holds only for $\mathbf{r}=\mathbf{q}$. This is the closest-point principle of orthogonal decomposition, applied after a translation of an affine plane to its direction linear subspace.^[ref]

The formula is useful whenever the deviation from a flat reference should be measured geometrically. A raw linear equation residual cannot be compared across arbitrary equation scalings; a perpendicular distance can. The same reasoning works for a hyperplane in any finite-dimensional Euclidean space, with the appropriate number of coordinates.