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Use Cartesian coordinates $x^a=(t,x,y,z)$ with $c=1$ and ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$. Spatial dot and cross products below use the Euclidean three-metric. The alternating tensor is given the orientation ${ϵ}^{0123}=+1$.

Vector and covector transformation. The vector $A=A^b{\partial }_b$ is independent of coordinates, and the chain rule gives ${\partial }_b=(\partial x^{\prime a}/\partial x^b){\partial }_a^{\prime }$. Likewise the one-form $A_{b}d{x}^b$ is independent of coordinates. Comparing coefficients provesTheir contraction $A_a{B}^a$ is unchanged because the forward and inverse Jacobians cancel. Here lowering an index uses the metric, not an identification of the numerical upper and lower components.

Why antisymmetrization matters. Write $J^c{}_a=\partial x^c/\partial x^{\prime a}$. Differentiating the covector law givesThe last term obstructs the general tensor transformation law. It is symmetric in $a,b$, so it cancels from ${\partial }_a^{\prime }{A}_b^{\prime }-{\partial }_b^{\prime }{A}_a^{\prime }$. ThereforeEquivalently $F=dA$ is the exterior derivative of the electromagnetic four-potential one-form. The statement that ${\partial }_a{A}_b$ is not a tensor needs a qualification: under affine Cartesian Lorentz transformations it does transform tensorially, because the second derivatives vanish. The failure concerns arbitrary nonlinear coordinate changes, including curvilinear coordinates in flat spacetime. This distinction is precisely the difference between general covariance and covariance under the restricted inertial-frame transformations.^[ref]

Electric and magnetic components. Choose $A^a=(\Phi ,\mathbf{A})$, so $A_a=(\Phi ,-\mathbf{A})$, $\mathbf{E}=-\nabla \Phi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$. ThenThus the lower-index field tensor isThis is a derived component representation of the field, with the source's sign convention.

Raised indices and the first invariant. Raising both indices changes the time-space signs and leaves the spatial block unchanged:In the scalar contraction, each independent antisymmetric pair is counted twice, givingUnder a Lorentz transformation each inverse transformation from a lower index cancels its matching transformation from an upper index, leaving the same scalar. In particular a plane electromagnetic wave has $E=B$ in these units and makes this invariant vanish, while a pure electric field makes it negative.

The printed dual convention. The PDF explicitly defines ${}^{\ast }{F}^{ab}={ϵ}^{abcd}{F}_{cd}$, with no $1/2$. Each independent pair therefore contributes twice: for example ${}^{\ast }{F}^{01}=F_{23}-F_{32}=-2B_x$. The complete result isThe time-space pairs contribute $-4\mathbf{E}\cdot \mathbf{B}$ and the spatial pairs contribute another $-4\mathbf{E}\cdot \mathbf{B}$. The conventional Hodge dual includes $1/2$ and would instead give the contraction $-4\mathbf{E}\cdot \mathbf{B}$; it must not silently replace this paper's definition. With fixed orientation the latter contraction is a pseudoscalar: invariant under proper Lorentz transformations but odd under orientation reversal if compared using a fixed alternating symbol.

Energy density and flux. Substitution into the given mixed stress-energy tensor yieldsFor the first formula, $F^{0c}{F}_{0c}=-E^2$ and $F_{cd}{F}^{cd}/4=(B^2-E^2)/2$. For the second, $F^{0j}{F}_{ij}=(\mathbf{E}\times \mathbf{B}{)}_i$. The minus sign on $T^0{}_i$ arises from its lowered spatial index; it does not reverse the physical energy flow. Raising that index gives $T^{0i}=S_i$, and symmetry gives $T^{i0}=S_i$, with $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$ the Poynting vector in $c=1$ units. Thus $T^{00}$ is energy density and $T^{i0}$ represents energy flux. In SI coordinates $x^0=ct$, restore $T^{0i}=S_i/c$ and $T^{00}=({ϵ}_0{E}_{\mathrm{SI}}^2+B^2/{\mu }_0)/2$, with ${ϵ}_0{\mu }_0{c}^2=1$. This keeps the mixed-index signs and the units distinct.

For representatives $p=(a,b,c)$ and $p^{\prime }=(a^{\prime },b^{\prime },c^{\prime })$, a point $[x:y:z]$ belongs to their projective line exactly when these three vectors are linearly dependent. The determinant criterion givesExpanding along the first row givesThe coefficient vector is the cross product $p\times p^{\prime }$. It is nonzero because the projective points are distinct. Rescaling either representative merely rescales all three coefficients, leaving the projective line unchanged. This is the join formula in projective joins and meets by cross products.

Apply projective joins and meets by cross products to each corresponding pair. The cross products of their representatives are proportional to $(0,1,0)$, $(0,0,1)$, and $(0,1,-1)$, respectively. HenceTheir common point isIt is a valid projective point, because its representative is nonzero. The three equations force $y=z=0$, so this common projective point is unique. Thus the two projective triangles are in perspectivity from a point $O$.

For $v=|\mathbf{v}|>0$, decompose the velocity as $\mathbf{u}={\mathbf{u}}_{\parallel }+{\mathbf{u}}_{\perp }$, parallel and perpendicular to $\mathbf{v}$. A rotation-free Lorentz boost gives the velocity-addition formulaThese two components are orthogonal, soThe last equality is Lagrange's identity for the cross product. For physical subluminal speeds, $D>0$, givingThe $\mathbf{v}=0$ case follows directly or by continuity. An independent form useful for checks is $1-|\mathbf{w}{|}^2/c^2=(1-u^2/c^2)(1-v^2/c^2)/D^2$, obtained from the transformed four-velocity normalization.

A monochromatic vacuum plane electromagnetic wave has phase $\mathbf{k}\cdot \mathbf{r}-\omega t$, where $\mathbf{k}$ is the wavevector, $\omega >0$ the angular frequency, and $\omega =c|\mathbf{k}|$. Its field amplitudes obey $\mathbf{k}\cdot {\mathbf{E}}_0=\mathbf{k}\cdot {\mathbf{B}}_0=0$ and ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$.

For the linear polarization and orientation in the visualization, choose a right-handed Cartesian coordinate system and putConstant phase means $z=ct+\text{constant}$, so propagation is along $+z$. The cross product $\hat{\mathbf{x}}\times \hat{\mathbf{y}}=\hat{\mathbf{z}}$ fixes the positive Poynting vector. Both fields reverse together every half-period and are in phase. Reversing the propagation direction with the same electric orientation reverses the magnetic orientation.

These fields are spatially uniform across every transverse plane. An infinite plane electromagnetic wave is an idealization with infinite total electromagnetic energy; it is useful as a local approximation and as a building block for superposition. For frequency $f=100\mathrm{MHz}$, $\omega =2\pi f$ and wavelength $\lambda =c/f\simeq 3.00\mathrm{m}$, independently of $E_0$.

A small longitudinal electric field drives conductor current in a lossy coaxial line. Its cross product with the azimuthal magnetic field sends energy radially into both conductors, where it becomes heat; forward power decreases with distance.

In the real projective plane, represent a projective point by a nonzero vector and a projective line by its nonzero coefficient covector, identified with a vector in chosen coordinates. The join of distinct projective points represented by $p,q$ has coefficient vector $p\times q$, using the cross product. It is nonzero because the representatives are linearly independent, and its dot product with each representative vanishes. Dually, two distinct projective lines with coefficients $\ell ,m$ meet in the projective point represented by $\ell \times m$. Scaling an input scales the output and leaves the projective object unchanged. This makes joins, intersections, concurrency, and collinearity exact calculations with homogeneous coordinates.

The right-hand rule fixes the orientation of a cross product and relates the circulation direction of an electric current to the direction of its magnetic field.

For unstressed central Hookean springs, the quadratic elastic energy is $\frac{1}{2}{\sum }_{ij}{k}_{ij}[{\mathbf{n}}_{ij}\cdot ({\mathbf{u}}_i-{\mathbf{u}}_j){]}^2$, with unit edge direction ${\mathbf{n}}_{ij}$. A common translation gives zero displacement difference. An infinitesimal rotation ${\mathbf{u}}_i=\mathbf{\Omega }\times {\mathbf{r}}_i$ also gives zero extension because a vector is perpendicular to its cross product with $\mathbf{\Omega }$. Thus a noncollinear three-dimensional free framework has three translational and three rotational zero modes. Extra mechanisms can give additional zero modes; six is a lower bound until rigidity is established.

Differentiating $R^{T}R=I$ gives $\mathfrak{so}(3)=\{A\in M_3(\mathbb{R}):A^T=-A\}$. The generators $L_{a}v=e_a\times v$ satisfy $[L_a,L_b]={ϵ}_{abc}{L}_c$ by the cross product identity. This identifies it with the SU(2) Lie algebra by matching a skew-Hermitian Pauli basis.

Let $a,b,c\in {\mathbb{R}}^n$, with $c\ne 0$. The equation $x=a+(b\cdot x)c$ is the linear system $(I-c{b}^T)x=a$, a rank-one matrix update of the identity matrix. Every solution has $x=a+\lambda c$. Substitution reduces the entire problem toIf $b\cdot c\ne 1$, there is one solution, $x=a+c(b\cdot a)/(1-b\cdot c)$. If $b\cdot c=1$ and $b\cdot a=0$, every $\lambda \in \mathbb{R}$ works and the solution set is the straight line $a+\mathbb{R}c$. If $b\cdot c=1$ and $b\cdot a\ne 0$, there is no solution.

The first reduction follows directly from the original equation in any dimension. In three dimensions it can also be obtained by taking the cross product with $c$, giving $(x-a)\times c=0$. The singular denominator therefore represents two different possibilities, inconsistency or a family of solutions, and must not simply be discarded as an undefined formula.

For a positively oriented basis $\mathbf{a},\mathbf{b},\mathbf{c}$, put $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The center of an interior sphere at distance $r$ from each of the three tetrahedron face planes through the origin has the displayed expression. The three inward unit vectors are the normalized cross products $\mathbf{b}\times \mathbf{c}$, $\mathbf{c}\times \mathbf{a}$, $\mathbf{a}\times \mathbf{b}$. Taking inner products with these vectors solves three independent signed-distance equations. Reversing some signs instead produces centers on exterior sides. Existence of a sphere contained in the full tetrahedron also requires clearance from the fourth face.

For a nondegenerate convex minor spherical triangle, with angular sides $a,b,c$ opposite interior angles $A,B,C$, the spherical law of sines statesThe sines of sides refer to central angles, so a physical side length ${\ell }_a$ on a sphere of radius $R$ must enter as $\sin ({\ell }_a/R)$. All six angles lie in $(0,\pi )$, and the denominators are positive. The reciprocal form, $\sin a/\sin A=\sin b/\sin B=\sin c/\sin C$, is equivalent. This is Tatum's sine formula, equation 3.5.1, in Celestial Mechanics, section 3.5: spherical sine and cosine formulas.

A short proof explains the symmetry of the spherical law of sines. Let $u,v,w$ be the vertex unit vectors at $A,B,C$. At $u$, the outgoing unit tangent vectors towards $v,w$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their angle is $A$, so the magnitude of their cross product is $\sin A$. Since both tangent vectors are orthogonal to $u$, the scalar triple product obeysDividing by $\sin a\sin b\sin c$ proves the spherical law of sines. Its beauty is that the same three-dimensional scalar triple product can be calculated at any vertex, producing all three ratios at once.

For the spherical triangle worked out under the spherical law of cosines, $A=90^{\circ }$, $b=c=60^{\circ }$ and $\cos a=1/4$. Hence $\sin a=\sqrt{15}/4$, and the spherical law of sines givesThe inverse sine gives $B\approx 63.434949^{\circ }$, but the supplementary angle $116.565051^{\circ }$ has the same sine. The spherical law of sines alone therefore leaves two candidates for this angle. The spherical law of cosines, using the known three sides, gives $\cos B=1/\sqrt{5}>0$ and selects $63.434949^{\circ }$; likewise $C=B$. When only two sides and an opposite angle are given, two distinct valid spherical triangles can remain. A cosine relation resolves an inverse sine branch only when enough additional data are known; it does not make every ambiguous data set unique.

For a regular parametrization $\mathbf{X}(u,v)$ of a surface in three-dimensional Euclidean space, the scalar surface area element is the area of the infinitesimal parameter parallelogram after mapping. Its value is the displayed cross product magnitude. The vector surface area element additionally records the chosen orientation.

For the radial profile $(\rho (t),z(t))$, with $\rho >0$, use the parametrized surface $\mathbf{R}(t,\theta )=(\rho \cos \theta ,\rho \sin \theta ,z)$, $0\le \theta <2\pi$. The two tangent vectors are orthogonal and have Euclidean norms $\sqrt{{\rho }^{\prime 2}+z^{\prime 2}}$ and $\rho$. Their cross product gives the displayed surface area. Geometric surface area assumes a single covering of the radial profile. A general three-dimensional generating curve must first be replaced by its actual radius $\rho (t)=\sqrt{x(t{)}^2+y(t{)}^2}$; its spatial arc length cannot simply replace the radial profile's arc length.

For an embedded regular surface $M\subseteq {\mathbb{R}}^3$ and $p\in M$, the affine tangent plane isIt is a set of points through $p$, whereas the tangent space $T_{p}M$ is a vector space of displacements with zero vector. Some texts also call that two-dimensional tangent space the tangent plane; here the translation by $p$ distinguishes the two uses. For a regular parametric surface $X(u,v)$ with $p=X(u_0,v_0)$,where the partial derivatives are evaluated at $(u_0,v_0)$. Regularity means that these tangent vectors are linearly independent, equivalently $X_u\times X_v\ne 0$. Their cross product is a normal vector. ^[ref]

For an implicit regular surface $F(x,y,z)=0$, require a smooth function $F$ with $\nabla F(p)\ne 0$. ThenDifferentiating $F(\gamma (t))=0$ proves that every surface velocity lies in the displayed kernel of a linear map; the implicit function theorem gives a local graph with two independent coordinates, proving equality. The gradient is the normal vector with respect to the ambient dot product. ^[ref]

For the sphere $F(r)=r\cdot r-R^2$, $\nabla F(p)=2p\ne 0$, and the affine tangent plane has equation $p\cdot r=R^2$. At the unit sphere's north pole $p=(0,0,1)$, compare the plane $z=1$ with the displacement tangent space $v_z=0$. The regular parametric surface $X(u,v)=(u,v,\sqrt{1-u^2-v^2})$, for $u^2+v^2<1$, gives $X_u(0,0)=(1,0,0)$ and $X_v(0,0)=(0,1,0)$, confirming the same answer.

This analytical illustration shows a unit sphere and a vertical cross-section. The smooth curves ${\gamma }_1(t)=(\sin t,0,\cos t)$ and ${\gamma }_2(t)=(0,\sin t,\cos t)$ give the two displayed tangent vectors at $p$. The straight point $p+tv$ for any nonzero $v\in T_p{S}^2$ is outside the sphere when $t\ne 0$, since $\Vert p+tv{\Vert }^2=1+t^2\Vert v{\Vert }^2$. A tangent plane therefore captures first-order motion, not an exact patch of the regular surface.

The same idea makes local approximation useful. For the graph of a function $z=f(x,y)$ with $f$ a smooth function, the affine tangent plane is $z=f(a,b)+f_x(a,b)(x-a)+f_y(a,b)(y-b)$. ^[ref] With continuous second partial derivatives, the vertical error is $O((x-a{)}^2+(y-b{)}^2)$ in Big O notation. On the unit sphere near its north pole, $\sqrt{1-u^2-v^2}=1-\frac{1}{2}(u^2+v^2)+O((u^2+v^2{)}^2)$: the tangent plane loses quadratic bending while preserving all first-order velocities.

A failed regularity test needs interpretation. Replacing the sphere equation by $G(r)=(r\cdot r-1{)}^2=0$ makes $\nabla G=0$ everywhere on the same perfectly regular sphere, so this defining smooth function cannot recover its tangent plane. Similarly, the usual spherical parametric surface loses matrix rank at the poles, while the graph parametrization above remains regular there. A vanishing gradient or degenerate parametrization invalidates that calculation; it alone does not prove that the underlying geometry has no tangent plane.

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

At regular transverse intersections, the displayed map has a rank-three $3\times 4$ Jacobian matrix. Its one-dimensional kernel is the parameter tangent, while the physical tangent is parallel to the cross product of the surface normals. A continuation method predicts along that tangent and corrects $F=0$ together with a transverse section equation using Newton method. Patch bounds and component isolation supply seeds; one seed alone cannot discover every disconnected curve or closed loop. Tangency or coincident surfaces violate the regular-curve assumptions and require separate analysis.

For planar vectors $u=(u_x,u_y)$ and $v=(v_x,v_y)$, the scalar $u\times v=u_x{v}_y-u_y{v}_x$ is the determinant of their coordinate columns. It equals the perpendicular component of the three-dimensional cross product after embedding the plane in three dimensions. Its sign records orientation, and its absolute value is twice the area of the triangle spanned by $u,v$.

Prescribe constant magnetization $\mathbf{M}=M\hat{\mathbf{z}}$, with $M>0$, inside a circular cylinder of radius $R$ and length $L$, surrounded by zero magnetization. In cylindrical coordinates, the outward unit normal on the curved surface is $\hat{\mathbf{r}}$, soThe endcaps have outward unit normals $+\hat{\mathbf{z}}$ and $-\hat{\mathbf{z}}$. Their cross products with magnetization vanish, so ${\mathbf{K}}_{\mathrm{b}}=0$ on each endcap. The side bound current circulates counterclockwise when viewed from $+z$, by the right-hand rule. Zero endcap surface current density does not imply zero endcap magnetic field.

For an original example, let $M=800A/m$, $R=0.020\mathrm{m}$, and $L=0.10\mathrm{m}$. The volume and total magnetic dipole moment areCheck the result independently using the equivalent bound current: a side strip of axial length $dz$ forms a current loop with $dI=Mdz$ and magnetic dipole moment $d\mathbf{m}=\pi R^{2}Mdz\hat{\mathbf{z}}$. Its integral over $L$ gives the same $M\pi R^{2}L\hat{\mathbf{z}}$. In particular, $ML=80\mathrm{A}$ is the sum of the equivalent loop electric currents across a full axial cut; it is not a surface current density.

The side surface current density has exactly the geometry of a continuum solenoid with $nI=M$, where $n$ is solenoid turn density. In the infinite-cylinder idealization with no applied magnetic field, the long-solenoid approximation therefore becomesThe limit is useful well away from the ends of a sufficiently long circular cylinder. A finite circular cylinder instead has fringing magnetic field and a demagnetizing field. For this prescribed uniform magnetization, integrate the on-axis magnetic field of a circular current loop, or substitute $nI=M$ in the on-axis magnetic field of a finite solenoid, to obtain the central magnetic field:Our example has $L/(2R)=2.5$ and $B_z(0)/({\mu }_{0}M)=5/\sqrt{29}\approx 0.92848$, rather than unity. With the rounded vacuum permeability ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$, this is approximately $0.9334\mathrm{mT}$, compared with the infinite-cylinder value $1.0053\mathrm{mT}$. These are model calculations for a prescribed magnetization, not measured values or a solution for the self-consistent magnetization of an arbitrary material.