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Let $\mathbf{u}$ be the boost velocity, and decompose the fields into components parallel and perpendicular to it. The parallel components are unchanged. The perpendicular components obeyAll vectors inside these expressions are perpendicular to $\mathbf{u}$. Thus $|\mathbf{u}\times {\mathbf{B}}_{\perp }{|}^2=u^2{B}_{\perp }^2$, and similarly for ${\mathbf{E}}_{\perp }$. In the difference of squares, the mixed terms cancel becauseIt follows thatAdding the unchanged parallel components provesFor the mixed dot product, expand directly. The terms ${\mathbf{E}}_{\perp }\cdot (\mathbf{u}\times {\mathbf{E}}_{\perp })$ and $(\mathbf{u}\times {\mathbf{B}}_{\perp })\cdot {\mathbf{B}}_{\perp }$ vanish. The remaining cross-product identity isHenceParallel-perpendicular mixed dot products are zero in both frames, soRotations also preserve these two expressions, completing the proof for proper Lorentz transformations. The second expression is a pseudoscalar under spatial parity, but is invariant under the boosts at issue here.

These results also follow from tensor contractions. In the convention used below, $F_{\mu \nu }{F}^{\mu \nu }=2(B^2-E^2/c^2)$ and $F_{\mu \nu }{\tilde{F}}^{\mu \nu }=-4\mathbf{E}\cdot \mathbf{B}/c$. The numerical sign of the latter depends on the chosen orientation of the dual electromagnetic field tensor, while its invariance under proper Lorentz transformations does not.

The three primitive lattice translation vectors join a cube corner to the three adjacent face centres. Thus the Bravais lattice is a face-centered cubic lattice, with conventional cubic side $a$.

Write the given primitive lattice translation vectors as ${\mathbf{a}}_1,{\mathbf{a}}_2,{\mathbf{a}}_3$. Their primitive unit cell volume isThe reciprocal lattice convention ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$ then givesEach reciprocal lattice vector has units ${\mathrm{m}}^{-1}$ when $a$ is in metres. Integer combinations generate a body-centered cubic lattice of conventional cubic side $4\pi /a$; the point $(2\pi /a)(1,1,1)$ is its body centre. Directly taking the nine dot products checks the signs and the $2\pi$ normalization.

The symmetry is electromagnetic reciprocity applied to mutual inductance. For thin closed circuits in vacuum, the magnetic vector potential generated by circuit B isStokes theorem converts the magnetic flux through circuit A into its magnetic vector potential circulation. Dividing by $I_B$ gives Neumann's mutual-inductance formula,The dot product and separation are unchanged by interchanging A and B, soEach has units $\mathrm{Wb}{\mathrm{A}}^{-1}=\mathrm{H}$. The winding orientations must be used consistently to determine the common sign. The same reciprocity holds in ordinary linear reciprocal magnetic media; it is not an unrestricted statement about biased nonreciprocal materials. Equivalently, a reversible magnetic energy function has equal mixed derivatives with respect to the two currents.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

Point-charge field and superposition. Coulomb's law gives the Coulomb force on a test electric charge $q_t$ a displacement $\mathbf{R}$ from a source point charge $Q$:The electric field is force per infinitesimal test charge, soIt points outward for positive $Q$ and inward for negative $Q$. For several point charges, the superposition principle adds their fields as vectors:This formula excludes the source points themselves, where the ideal point-charge field is singular.

Exact field components of the pair. PutThe displacement vectors from the positive and negative charges to the observation point are $(x-a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$ and $(x+a)\hat{\mathbf{i}}+y\hat{\mathbf{j}}$. Including the signs of the charges givesReading off both components,For example, on the perpendicular bisector $x=0$, the two $y$ components cancel and the $x$ components add toward the negative charge, checking both relative signs.

Far-field polar components. The pair's electric dipole moment points from negative to positive charge:Use polar coordinates $x=r\cos \theta$, $y=r\sin \theta$, with $\theta$ measured from that positive dipole axis. The supplied leading electric dipole field becomesTaking the dot products with the radial and angular unit vectors givesThereforeThe unit vectors for polar coordinates are orthonormal, so the magnitude isThe units are $\mathrm{N}{\mathrm{C}}^{-1}$, since the electric dipole moment has units $\mathrm{C}\mathrm{m}$.

Angular sketch and comparison. At fixed large $r$, the magnitude has period $\pi$. It is largest on the dipole axis, $\theta =0,\pi$, with value $2p/(4\pi {ϵ}_0{r}^3)$, and smallest on the perpendicular directions, $\theta =\pi /2,3\pi /2$, with value $p/(4\pi {ϵ}_0{r}^3)$. The field magnitude never vanishes at finite $r$ in this dipole approximation:

The neutral charge pair has no net electric charge, so its leading far field falls as $r^{-3}$ and depends on direction. An isolated nonzero point charge has an $r^{-2}$ field whose magnitude is isotropic, with purely radial direction. The dipole also generally has a nonzero $E_{\theta }$ component.

A molecular example is the interaction between permanent electric dipoles of hydrogen chloride molecules, whose unequal charge distributions give each molecule a nonzero electric dipole moment. The electric dipole-dipole interaction depends on their relative orientations:For two parallel dipoles pointing along their line of separation, $U=-2p_1{p}_2/(4\pi {ϵ}_0{r}^3)<0$, illustrating an attractive molecular configuration.

For an integrable finite-dimensional random vector $Y$, let $L=\{h:h\cdot Y=0\text{ almost surely}{\}}^{\perp }$. The displayed convex cone is open relative to $L$. Given a bounded strictly positive $Z$, the perturbation $Z_{\epsilon }=Z(1+\epsilon \cdot Y/(1+\Vert Y\Vert ))$, $\Vert \epsilon \Vert <1$, changes its weighted expectation by $A_Z\epsilon$, whereFor $0\ne v\in L$, $v^T{A}_{Z}v>0$: otherwise $v\cdot Y=0$ almost surely. Thus $A_Z$ is invertible on $L$, and these perturbations give a neighborhood of every point of the cone. The hyperplane separation theorem consequently gives the alternative: either zero is a strictly positively weighted expectation, or a nonzero direction in $L$ has a nonnegative dot product with $Y$ almost surely and a positive dot product with positive probability. To obtain the latter assertion, test the separating direction with $Z=1_{\{h\cdot Y<0\}}+\epsilon$ and let $\epsilon \downarrow 0$.

Iterate $q^{(k+1)}=A{q}^{(k)}/\Vert A{q}^{(k)}\Vert$. For a self-adjoint operator with a unique dominant eigenvalue separated in modulus from the others, a starting vector that is nonorthogonal to its dominant eigenvector gives convergence of the eigendirection at the spectral-ratio rate. For a general diagonalizable matrix, the condition is a nonzero coefficient in the dominant component of its eigenbasis, detected by the corresponding left eigenvector; an ordinary dot product with the right eigenvector does not generally detect that coefficient.

The power method is power iteration; its convergence depends on the dominant spectral component of the starting vector.

For example, $B=\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has dominant right eigenvector $(1,0{)}^T$ and a dominant left eigenvector $(1,1{)}^T$. The starting vector $q^{(0)}=(-1,1{)}^T/\sqrt{2}$ has nonzero dot product with the right eigenvector, but $B{q}^{(0)}=q^{(0)}$: it stays forever in the eigenspace for the smaller eigenvalue $1$. Its coefficient along the dominant direction is zero, as the left eigenvector detects. An orthonormal eigenbasis makes the two tests agree in the self-adjoint case.

The Poynting theorem is the local energy balance of classical electromagnetism. For vacuum field laws, with current density $\mathbf{J}$, the electromagnetic energy per unit volume isandThus a decrease in local electromagnetic energy can supply outward energy flux or work on electric charge. The term $\mathbf{J}\cdot \mathbf{E}$ is power transferred to matter per unit volume. Its sign matters: positive values remove field energy, while negative values describe matter supplying the field. The magnetic part of the Lorentz force supplies no work term.

For a fixed volume $V$, the divergence theorem gives the directly measurable balanceFor example, $2.0\mathrm{W}$ entering a region with $0.50\mathrm{W}$ leaving and $1.20\mathrm{W}$ transferred to matter means stored electromagnetic energy increases at $0.30\mathrm{W}$. In source-free vacuum only the boundary flux changes the stored energy; a nonzero local Poynting vector can still have zero net closed-surface flux.

A brief proof takes the dot product of the Ampère-Maxwell equation with $\mathbf{E}/{\mu }_0$ and uses Faraday's law with $\mathbf{B}/{\mu }_0$. The identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot (\nabla \times \mathbf{E})-\mathbf{E}\cdot (\nabla \times \mathbf{B})$ turns the remaining terms into the divergence of the Poynting vector. ^[ref].

The projected area of a planar patch viewed from a direction is the area of its perpendicular shadow onto a plane normal to that direction. If $\mathbf{n}$ and $\mathbf{s}$ are its unit normal and viewing unit vector, then $A_{\perp }=A|\mathbf{n}\cdot \mathbf{s}|$ by the dot product. On the outward side, this is $A\cos \theta$. For a rectangle tilted around one edge, the width perpendicular to that edge contracts by $\cos \theta$ and the edge length stays fixed, proving the formula. At grazing view $A_{\perp }=0$; interpret the radiance definition by a limit rather than dividing by zero.

In geometrical optics, suppose radiation has a nonnegative radiance $L(\mathbf{x},\mathbf{s})$ whose directional contributions add as power. Here $\mathbf{s}$ is the actual propagation unit vector, $d\Omega$ is a solid angle element, and the collecting oriented surface $A$ has chosen unit normal $\mathbf{n}(\mathbf{x})$. Assume the following integrals exist. The outward and inward radiant fluxes areBoth quantities are nonnegative; their difference is the signed net radiant flux. The dot product supplies the cosine of the angle with the unit normal, hence the projected area. For incoming radiation the outward unit normal gives a negative dot product; the minus sign makes the collected power positive. Measuring the incoming angle from a source-facing unit normal instead gives the familiar positive $\cos \theta$. The units check is $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\times {\mathrm{m}}^2\times \mathrm{sr}=\mathrm{W}$.

Equivalently, the incoming surface integral is ${\Phi }_{-}={\int }_{A}EdA$ with local irradiance $E$; the emitted part of ${\Phi }_{+}$ is ${\int }_{A}MdA$ with radiant exitance $M$. For a small flat patch and a narrow accepted cone, nearly constant radiance and angle give $\Phi \simeq LA\cos \theta \Delta \Omega$. This approximation requires uniformity across the patch and cone. Direction-independent outgoing radiance over a whole hemisphere gives the different exact result ${\Phi }_{+}=\pi LA$, by Lambert's cosine law and the spherical coordinates calculation in radiant exitance. Replacing the cosine-weighted integral by the hemisphere's $2\pi$ solid angle would double the answer.^[ref]

These radiance formulas presume a description in additive directional power; they do not reconstruct arbitrary coherent wave interference from separate ray contributions. For general electromagnetic fields, the signed instantaneous power crossing an oriented surface is ${\int }_A\mathbf{S}\cdot \mathbf{n}dA$, using the Poynting vector $\mathbf{S}$; optical measurements commonly average over many wave periods. A zero net radiant flux can conceal equal nonzero incoming and outgoing radiant fluxes.

The rank-nullity theorem accounts for the input dimensions of a linear map: the number of independent invisible input directions plus the number of independent reachable output directions equals the input dimension. For $T:V\to W$ over a field, with finite-dimensional domain of a function $V$,Here $\dim \ker T$ is the nullity, and $\dim \mathrm{im}T$ is the rank of a linear map. The kernel lives in the input vector space, while the image of a linear map lives in the output vector space. For an $m$-by-$n$ matrix of matrix rank $r$, the formula is $n=(n-r)+r$: the number of columns counts input coordinates. ^[ref]

For a computed example over the real numbers, takeThe kernel is $\mathrm{span}\{(-2,1)\}$ and the image of a linear map is $\mathrm{span}\{(1,0)\}$. Each has dimension one, so the two input dimensions split as $2=1+1$. The vector $(3,1)$ can be writtenThe second summand is undetectable: $T(3,1)=T(1,2)$. The picture uses the standard dot product to choose a visible direction perpendicular to the kernel; this is a convenient geometric choice, not an extra hypothesis of the rank-nullity theorem.

This explains a practical tradeoff in a linear system. If $T(x_0)=b$, every other input with output $b$ is $x_0+k$ for $k\in \ker T$. The nullity therefore counts independent freedom in an existing solution; the rank counts independent reachable output directions. In the example, $T(x,y)=(3,0)$ has all solutions $(x,y)=(3-2t,t)$, while $T(x,y)=(3,1)$ has none. The cokernel records this failure of reachability.

For an $m$-by-$n$ matrix, injectivity requires matrix rank $n$, and surjectivity requires matrix rank $m$. When $m\ne n$, these conditions differ. For example, $B:{\mathbb{R}}^3\to {\mathbb{R}}^2$ with $B(x,y,z)=(x+2y,0)$ has rank one and nullity two, because $\ker B=\mathrm{span}\{(-2,1,0),(0,0,1)\}$. The input accounting is $3=2+1$, although only one of two output directions is reached.

Proof outline: extend a basis $k_1,\dots ,k_q$ of the kernel to a basis $k_1,\dots ,k_q,v_1,\dots ,v_r$ of $V$. Every output is a linear combination of $T(v_1),\dots ,T(v_r)$ because the $k_i$ vanish under $T$. If a linear combination of the $T(v_j)$ is zero, the corresponding combination of the $v_j$ lies in the kernel; linear independence of the full basis forces all its coefficients to vanish. Thus the added basis vectors give a basis of the image of a linear map. Counting them proves the formula.

A rank-one matrix update changes a matrix $A$ by an outer product $u{v}^T$. For real vectors $u,v\in {\mathbb{R}}^n$, its action on an input vector $x$ isThe dot product $v^{T}x$ reads one scalar from the input, and the correction moves the output along $u$. All correction outputs therefore lie in one linear span. If either vector is zero, the update is zero; otherwise the correction is a rank-one matrix. The full updated matrix need not have rank one.

For example, chooseThe update sends $x=(x_1,x_2{)}^T$ to the correction $(3x_1-x_2)(1,2{)}^T$. Although all four entries change, there is only one direction of correction. Adding it to the identity matrix gives $B=\left(\begin{array}{cc}4& -1\\ 6& -1\end{array}\right)$.

This special structure allows existing work to be reused. The matrix determinant lemma updates the determinant using one scalar, and the Sherman–Morrison formula updates the matrix inverse using an outer product. Such formulas are useful when a linear system changes slightly after a new observation or a parameter adjustment: the unchanged part can retain its LU decomposition. The term rank-one describes the rank of the correction, not the size of the entry changes. ^[ref].

This surface consists of rays from a vertex $\mathbf{a}$ whose directions make a fixed angle $\beta$ with a unit axis vector $\hat{\mathbf{n}}$. The displayed equation follows from the dot product formula for that angle. For $0<\beta <\pi$, $\beta \ne \pi /2$, it gives one circular nappe; $\beta =\pi /2$ gives a plane, and $\beta =0,\pi$ gives a degenerate ray. Squaring the equation produces both nappes, so the unsquared sign distinguishes them. Generalized angular momentum in a radial monopole field confines particle directions to such a surface.

Choose $k,m>0$, $K=(k^2+m^2{)}^{1/2}$ and the positive angular frequency $\omega =Nk/K$. Then the phase velocity is $Nk(k,m)/K^3$ and the group velocity is $N(m^2,-km)/K^3$. Their dot product vanishes and their sum is horizontal with magnitude $N/K$. Draw the group velocity from the tip of the phase velocity: the two form the perpendicular sides of a right triangle. The horizontal hypotenuse is their sum, not the horizontal phase velocity $\omega /k$.

A rotation matrix is an orthogonal matrix with determinant $1$.

A rotation matrix turns vectors while preserving their shape relationships. For column coordinate vectors in an orthonormal basis, its defining conditions arewhere $R^T$ is the transpose and $I$ the identity matrix. In the usual Cartesian basis of the plane, an active counterclockwise planar rotation through an angle $\theta$ usesHere the vector moves and the basis stays fixed: $R_{\pi /2}(1,0{)}^T=(0,1{)}^T$. Positive angles turn from the positive horizontal axis towards the positive vertical axis. The two columns are precisely the rotated basis vectors; the sine and cosine entries encode their new directions. Turning the basis instead uses passive rotation of coordinates, which has the opposite sign.

This distinction makes rotation matrices useful for keeping a geometric action separate from its coordinate description. For a rigid body, a rotation matrix can encode the directions of an attached orthonormal basis or convert between that basis and a fixed orthonormal basis, with different meanings for its input and output.^[ref]

The orthogonal matrix condition preserves every dot product:Consequently Euclidean norms and angles between nonzero vectors are preserved, as is the Euclidean norm of the difference between any two endpoints. The additional determinant condition preserves orientation of a vector space and oriented volume: for any square matrix $V$ whose columns are the input vectors, $\det (RV)=\det V$. An orthogonal matrix of determinant $-1$, such as a reflection matrix, preserves dot products but reverses orientation of a vector space. The matrix inverse is especially simple: $R^{-1}=R^T$.^[ref]

For a rigidly rotating prescribed vacuum path, use spatial coordinates attached to the apparatus and laboratory time $t$. At a path point $\mathbf{r}$, the local apparatus velocity is $\mathbf{v}=\mathbf{\Omega }\times \mathbf{r}$. A light displacement has instantaneous laboratory displacement $d\mathbf{r}+\mathbf{v}dt$, up to the common rotation of the spatial axes. The null condition is thereforeWrite $d\ell =|d\mathbf{r}|$. To first order in $|\mathbf{v}|/c$, the positive root isReversing the path reverses the dot product term but leaves the leading path-length term unchanged. Subtract the two return times and use the scalar triple product identity:This is a line integral version of the Sagnac area law; the vector area identity also follows from Stokes theorem. For a mirror-defined polygon the actual ray paths move slightly as the mirrors rotate; the stationary optical path at zero rotation gives the same first-order result. The prescribed-path calculation is an idealization, not an exact finite-speed formula for every moving-mirror layout.

The dot product explains the geometry: only the projection of the oriented vector area along the angular velocity contributes. A planar loop whose normal vector is perpendicular to $\mathbf{\Omega }$ has zero leading delay. For repeated traversals with the same orientation, the signed vector areas add; oppositely oriented lobes can cancel even when the total unsigned area is large. Equal perimeter alone does not imply equal response. A constant translational velocity contributes nothing to this leading closed-path line integral, since ${\oint }_{C}d\mathbf{r}=0$.