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The Fenchel–Young inequality says that every input and every slope obeywhere $f$ is any proper extended-real function and $f^{\ast }$ is its convex conjugate. No convexity assumption on $f$ is needed for this inequality: the defining supremum is at least the particular value $p\cdot x-f(x)$. If either term on the left is $+\infty$, the inequality still holds, but cannot be an equality because the dot product on the right is finite.

For a proper convex function, equality is equivalent to the subgradient conditionTo see this, with $f(x)$ finite, equality says that $x$ attains the supremum defining $f^{\ast }(p)$, so $p\cdot z-f(z)\le p\cdot x-f(x)$ for every $z$. Rearranging gives the subgradient inequality $f(z)\ge f(x)+p\cdot (z-x)$. Geometrically, the affine minorant ${\ell }_p(z)=p\cdot z-f^{\ast }(p)$ touches $f$ at $x$.

If the proper convex function is also closed, meaning its epigraph is a closed set, or equivalently it is lower semicontinuous, thenThe forward implication holds without closedness; the reverse one requires the additional hypothesis in general. Applying the equality criterion to $f^{\ast }$ relates $x\in \partial f^{\ast }(p)$ to $f^{\ast \ast }(x)+f^{\ast }(p)=p\cdot x$, and the Fenchel-Moreau theorem supplies $f^{\ast \ast }=f$ when $f$ is closed. See Bertsekas, Convex Optimization Theory: A Summary, §5.4.1, Proposition 5.4.3 for precisely this distinction.

A concrete failure of the reverse implication occurs for the indicator functional of a constraint set $f$ of the open interval $(-1,1)$. Its convex conjugate is $f^{\ast }(p)=|p|$: the endpoint gives the supremum, even though it is excluded from the interval. At $(x,p)=(1,1)$, $x\in \partial f^{\ast }(p)=\{1\}$, but $f(1)=+\infty$, so equality fails and $\partial f(1)$ is empty. Closing the interval changes the endpoint value and restores equality there.

For the convex quadratic function $f(x)=x^2/2$, the Fenchel–Young inequality is $x^2/2+p^2/2\ge px$, and equality holds exactly when $x=p$. For the absolute value $f(x)=|x|$, finite equality requires $p=\mathrm{sgn}(x)$ at $x\ne 0$, where $\mathrm{sgn}$ is the sign function; at $x=0$ every $p\in [-1,1]$ gives equality. These are exactly the subgradients of the absolute value, including its entire interval of slopes at the corner.

Every self-adjoint operator on a finite-dimensional real or complex inner product space has an orthonormal eigenbasis, and all its eigenvalues are real. In orthonormal coordinates, a real symmetric matrix has the form $A=Q\Lambda Q^T$, with $Q$ an orthogonal matrix and $\Lambda$ a real diagonal matrix. A complex Hermitian matrix instead has $A=U\Lambda U^{\ast }$, with $U$ a unitary matrix and $U^{\ast }$ its conjugate transpose. These are the self-adjoint cases of the spectral theorem; Axler gives the real and complex finite-dimensional statements in Section 7B.^[ref]

An orthonormal eigenbasis turns a seemingly coupled linear map into independent scalar multiplications while preserving lengths and angles. If $x={\sum }_j{c}_j{e}_j$ and $A{e}_j={\lambda }_j{e}_j$, thenThe last identity removes every cross term from the quadratic form. The eigenvalues tell us whether it is positive, negative, or mixed in sign, and the eigenvectors identify the corresponding directions. This connects positive-definite matrices, quadratic forms, and the Hermitian Rayleigh quotient range.

For a fully computed example, takeDirect multiplication gives $A{e}_{+}=3e_{+}$ and $A{e}_{-}=e_{-}$. Their dot product is zero and each has norm one. ThusWriting $x=c{e}_{+}+d{e}_{-}$ changes $x^{T}Ax=2x_1^2+2x_1{x}_2+2x_2^2$ into $3c^2+d^2$. Its level set $x^{T}Ax=1$ is therefore an ellipse with semiaxis lengths $1/\sqrt{3}$ and $1$, along $e_{+}$ and $e_{-}$. The coordinates change, but the ellipse does not: orthogonal diagonalization reveals its existing geometry.

For distinct eigenvalues, orthogonality follows at once from self-adjointness: $\lambda ⟨u,v⟩=⟨Au,v⟩=⟨u,Av⟩=\mu ⟨u,v⟩$. Within a repeated eigenspace, choose an orthonormal basis; arbitrarily chosen eigenvectors of the same eigenvalue need not be orthogonal. One outline of the full argument finds an eigenvector, observes that its orthogonal complement is invariant, and repeats in that smaller space. Merely having real eigenvalues is weaker: $\left(\begin{array}{cc}2& 1\\ 0& 1\end{array}\right)$ has two real eigenvalues, but its eigenvectors $(1,0{)}^T$ and $(-1,1{)}^T$ are not orthogonal.

A Gabor filter has a kernel formed by multiplying a Gaussian function by a sinusoid or a complex oscillation. In two-dimensional image processing, its orientation and spatial frequency select local image structure; evaluating the kernel's translated dot product with an image is spatial filtering. The real even example under signed spatial filter model of a simple cell uses a cosine.

There is experimental motivation for this family without making every simple cell an exact Gabor filter. Jones and Palmer fitted two-dimensional Gabor filters to measured spatial and spectral response profiles of simple cells in cat cortex and compared independently estimated parameters as a test of spatial linearity. Their reported residuals supported the approximation in that dataset. The chosen parameters and orientation tuning curve in this article are not extracted from their measurements. ^[ref]

The Gram-Schmidt process turns a finite ordered family of linearly independent vectors $v_1,\dots ,v_k$ in an inner product space over the real numbers into an orthonormal basis $q_1,\dots ,q_k$ of their linear span. It keeps every successive linear span, so the first $j$ new vectors describe exactly the same directions as the first $j$ inputs. Its idea is geometric: remove the orthogonal projection onto directions already found, then divide the remaining vector by its norm. This makes a complicated basis easier to use without changing the vector subspace it describes.^[ref]

Start with $u_1=v_1$ and $q_1=v_1/\Vert v_1\Vert$. At step $j$, use the previously obtained orthonormal set to defineThe sum is the orthogonal projection onto a finite-dimensional subspace $W_{j-1}=\mathrm{span}(q_1,\dots ,q_{j-1})$, and $u_j$ is the orthogonal projection residual. Linear independence ensures $u_j\ne 0$: otherwise $v_j$ would already be a linear combination of earlier inputs. If the input family is linearly dependent, use only the retained orthonormal vectors in the sum and omit an input whenever its remainder is the zero vector; do not divide by zero. The retained outputs still form an orthonormal basis of the original linear span, but their number can be smaller than $k$.

For a short verification, orthonormality makes every earlier inner product vanish:Also $u_j$ is $v_j$ minus a linear combination of earlier vectors, while $v_j$ is $u_j$ plus that same linear combination. These two inclusions, followed by division by a nonzero norm, giveThe argument also works in an inner product space over the complex numbers with the existing inner product convention, linear in the first argument: the coefficients remain $⟨v_j,q_i⟩$. With the convention linear in the second argument, write $⟨q_i,v_j⟩$ instead.

For a first picture, take $v_1=(1,1{)}^T$ and $v_2=(1,0{)}^T$ in ${\mathbb{R}}^2$ with the ordinary dot product. ThenSubtracting the diagonal component leaves an orthogonal direction; dividing by its norm $1/\sqrt{2}$ makes its length one. If the second input were $(2,2{)}^T$ instead, its orthogonal projection would be the entire input and $u_2=0$, revealing linear dependence.

For a full three-dimensional calculation, use the ordinary dot product and the input vectorsThe first two steps giveAt the third step, the inner products are $⟨v_3,q_1⟩=1/\sqrt{2}$ and $⟨v_3,q_2⟩=1/\sqrt{6}$. ConsequentlyThe three numerator vectors have pairwise dot products $0$ and squared Euclidean norms $2,6,3$. Thus the displayed outputs are indeed an orthonormal basis, and the three nonzero remainders also confirm linear independence of the inputs. Keeping the removed coefficients produces the QR decomposition below.

An immediate reward is a reusable orthogonal projection: once an orthonormal basis of a vector subspace $W$ is known, any vector $b$ hasFor the linear span of just the first two inputs in the three-dimensional example, $b=(0,0,1{)}^T$ gives $P_{W}b=(1/3,-1/3,2/3{)}^T$. Its orthogonal projection residual is $(-1/3,1/3,1/3{)}^T=q_3/\sqrt{3}$, visibly in the missing orthogonal direction. This is the geometry behind the linear least-squares problem.

The formulas above describe exact arithmetic. With floating-point arithmetic, subtracting an orthogonal projection nearly equal to the input can leave a small remainder; dividing by its small norm can magnify existing roundoff error. The Classical Gram-Schmidt and Modified Gram-Schmidt arrangements agree in exact arithmetic but can give different rounded results. Exact linear dependence and a small computed remainder therefore require different judgments in a numerical implementation.^[ref]

Choose a fixed support point $O$ and nonrotating axes in an inertial frame. All angular momenta, torques, and moments of inertia below refer to $O$. Let $\hat{\mathbf{e}}$ point along the axle from $O$ toward the centre of mass, and let $\hat{\mathbf{z}}$ point vertically upward. Write $\theta$ for their angle. For a rapidly spinning symmetric top, the dominant angular momentum is ${\mathbf{L}}_O\simeq L_s\hat{\mathbf{e}}$, where $L_s=I_3{\omega }_3$ is signed and $I_3$ is the axial principal moment of inertia. The sign of the axial angular velocity ${\omega }_3$ follows the right-hand rule.

With mass $M$, centre of mass distance $d$, and constant gravitational acceleration $g$, the gravitational force and its torque areAn ideal support applies a force at $O$, so its lever arm and torque about $O$ vanish. At the instant the axle lies in the $xz$ plane with a positive $x$ component, the gravitational torque points along $+y$. It adds $d{\mathbf{L}}_O={\mathbf{\tau }}_{O}dt$ toward $+y$, rather than along the downward force. The dot product identity $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$ shows why perpendicular torque changes direction without instantly changing magnitude. This vector balance and the slow-turning relation ${\mathbf{\tau }}_O=\mathbf{\Omega }\times {\mathbf{L}}_O$ are developed in Chapter 20: Rotation in space (sections 20-2 and 20-3, equations 20.13–20.15).

For steady precession about the vertical, $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$ and $d\hat{\mathbf{e}}/dt=\Omega \hat{\mathbf{z}}\times \hat{\mathbf{e}}$. Equating the two cross products givesThe angular momentum tip traces a circle of radius $|L_s|\sin \theta$: its sideways speed is $|\Omega L_s|\sin \theta$. Both that radius and the gravitational lever arm contain $\sin \theta$, explaining its cancellation for $0<\theta <\pi$. Exactly vertical alignment has zero gravitational torque and no distinguishable azimuthal precession to infer by this cancellation. Positive $L_s$ gives positive precession, counterclockwise viewed from above; reversing the spin reverses the slow precession. ^[ref]

The normalized Hadamard transform maps a vector indexed by binary strings through entries $2^{-n/2}(-1{)}^{x\cdot y}$, where the dot product is evaluated modulo two. It is its own inverse and an orthogonal transformation, hence also a unitary operator on complex input vectors. The Hadamard gate is the $n=1$ quantum case, and the Hadamard basis is the image of the computational basis. The Walsh-Hadamard transform gives the same tensor product construction.

For two qubits, applying one Hadamard gate to each prepares the uniform quantum superpositionEach probability amplitude is $1/2$, so each computational-basis measurement outcome has probability $1/4$. Applying the Hadamard transform again reconstructs $|00⟩$: signed contributions cancel at the other three outputs. Its usefulness lies in combining simple local quantum gates into an operation on exponentially many amplitudes. Extracting useful information still requires an appropriate subsequent transformation and measurement.

A model evolving the entire instantaneous forward rate curve. In the displayed one-factor risk-neutral measure dynamics the drift restriction makes every suitably integrable discounted zero-coupon bond price a martingale. The multi-factor form replaces the product of volatilities by their dot product.

A Hermitian inner product is a Hermitian form $H$ on a complex vector space which has positive definiteness: $H(v,v)>0$ for every nonzero vector $v$. It is precisely a complex inner product. Only with this definiteness condition doesdefine a norm rather than a seminorm or an expression that may not be real. The standard Hermitian inner product on ${\mathbb{C}}^n$ is $H(z,w)={\sum }_j{z}_j\overline{w_j}$; positive weights $c_j$ give ${\sum }_j{c}_j{z}_j\overline{w_j}$, with squared norm ${\sum }_j{c}_j|z_j{|}^2$.^[ref]

A Hermitian inner product need not have a real matrix. For instance, with the linear-first coordinate convention,The last expression proves positive definiteness without a lengthy calculation of eigenvalues. For $z=(1,i)$ it gives $6$, while for $z=(1,-i)$ it gives $2$. These unequal values expose why a silent switch to $z^{\ast }Az$ would change this particular Hermitian inner product: with these same entries it gives $2$ at $(1,i)$. All entries must keep their roles from the matrix of a Hermitian form.

The Cauchy-Schwarz inequality for a Hermitian inner product statesEquality holds exactly when $u,v$ are linearly dependent, including when either is zero. For a nonzero $u$, outline the proof by setting $a=H(v,u)/H(u,u)$ and $r=v-au$. Then $H(r,u)=0$, and the Pythagorean theorem givesNonnegativity proves the inequality, while positive definiteness makes equality equivalent to $r=0$. The scalar $a$ has no extra conjugate because the first argument is linear.^[ref]

The Cauchy-Schwarz inequality also holds for a positive semidefinite Hermitian form, but equality need not imply linear dependence: for $H_0(z,w)=z_1\overline{w_1}$, the independent vectors $e_1$ and $e_1+e_2$ give $1=1$. Equality then describes dependence after passing to the quotient vector space by the radical of a Hermitian form. The indefinite Hermitian form in Hermitian form shows that conjugate symmetry alone does not even guarantee the inequality.

For a worked orthogonal projection, take $H_{+}(z,w)=z_1\overline{w_1}+2z_2\overline{w_2}$, $u=(1,1)$ and $v=(2,0)$. Then $a=2/3$, $au=(2/3,2/3)$ and $r=(4/3,-2/3)$, with $H_{+}(r,u)=0$ and $4=4/3+8/3$ for the squared norms. The residual is perpendicular in this Hermitian inner product even though its ordinary real dot product with $u$ is $2/3$. For complex $v=(1,i)$, the same orthogonal projection uses $a=(1+2i)/3$; rotating $v$ multiplies $a$ by $e^{i\theta }$, while rotating $u$ multiplies $a$ by $e^{-i\theta }$. This combines geometric decomposition with sensitivity to the relative complex arguments of the multipliers.

A Hermitian inner product therefore supplies more than a positive diagonal: orthogonality, orthogonal projection onto a span, and the triangle inequality for its norm make approximation and distance meaningful. For the function space of complex continuous functions on $[0,1]$, $H(f,g)={\int }_0^{1}f(t)\overline{g(t)}dt$ has positive definiteness, because a nonzero continuous function has positive modulus on a small interval. The same construction turns overlap under integration into lengths and best approximation in finite-dimensional vector subspaces.

A horizontal coordinate system describes a viewing direction using altitude $\gamma$ above the observer's astronomical horizon and azimuth $\eta$ around it. Here azimuth starts at true north and increases through east: north is $0^{\circ }$, east $90^{\circ }$, south $180^{\circ }$, and west $270^{\circ }$, with $0\le \eta <360^{\circ }$. True north is the northward intersection of the celestial meridian with the astronomical horizon, rather than an uncorrected magnetic-compass reading. Other origins and directions are legitimate, but their formulae must be converted consistently. ^[ref].

Construct a local orthonormal basis of east ${\mathbf{e}}_E$, north ${\mathbf{e}}_N$, and upward ${\mathbf{e}}_U$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. The upward unit vector points to the zenith; the horizontal plane is perpendicular to it. For a ground instrument, a freely hanging weight establishes the local vertical through gravity. In an elementary spherical-Earth model, this vertical is radial; precision work distinguishes it from the normal to a reference surface. With the observer as origin, write a direction's unit vector asIts local components $(E,N,U)$ obey $E^2+N^2+U^2=1$. Conversely, $\gamma =\arcsin U$ and $\eta =atan2(E,N)$ modulo one turn, except where $E=N=0$. The quadrant-aware arctangent takes east as its first argument and north as its second. A synthetic direction with $\eta =60^{\circ }$ and $\gamma =30^{\circ }$ has $(E,N,U)=(3/4,\sqrt{3}/4,1/2)$. Thus its horizontal component has norm $\sqrt{3}/2$; its altitude is not the angle of the projected arrow in the horizontal plane.

A horizontal coordinate system depends on the observer's site and the observation time. The Earth's rotation changes the local axes relative to a fixed distant direction, while changing the site changes the zenith and the astronomical horizon. This makes the horizontal coordinate system directly useful for pointing an optical telescope and checking whether terrain blocks a target. A pair of angles without a site, time, azimuth convention, and geometric or refracted interpretation is an incomplete pointing instruction.

For a check against an equatorial coordinate system, let $\theta$ be the observer's geographical latitude, $\delta$ the direction's declination, and $h$ its west-positive hour angle, obtained from local sidereal time minus right ascension. In the spherical-Earth, unrefracted fixed-direction model, projection onto the local orthonormal basis givesTaking the dot product with the upward unit vector gives the last formula; the other two follow by the same projection. Recover altitude and azimuth from $(E,N,U)$ as above, using radians inside software trigonometric functions. The minus sign in $E$ is essential: a negative hour angle puts this synthetic direction east of the celestial meridian. ^[ref].

For illustrative geographical latitude $\theta =45^{\circ }$ and invented declination $\delta =15^{\circ }$, hour angle $h=-60^{\circ }$ gives $\gamma \simeq 31.636^{\circ }$ and $\eta \simeq 100.729^{\circ }$. At $h=+60^{\circ }$, the same altitude accompanies $\eta \simeq 259.271^{\circ }$: the direction is now westward. At $h=0$, it is due south with altitude $60^{\circ }$, agreeing with $90^{\circ }-|\theta -\delta |$ at upper meridian transit. These are analytical checks, not present positions of any star. For nearby objects, the direction must also be referred to the observer; simply rotating a geocentric direction omits the change of viewing position. Catalogue-to-observation refinements are described under equatorial coordinate system.

A line integral can be defined as an improper integral of its parametrized integrand when that integrand is undefined or unbounded at isolated parameter values. At an interior singular parameter, split the integral and require both one-sided limits to exist separately; their cancellation in a symmetric Cauchy principal value is a different convention. For a scalar arc length integral, use $f(\mathbf{r}(t))|{\mathbf{r}}^{\prime }(t)|$ in place of the vector-field dot product.

If the restricted integrand has a removable singularity, integrating its continuous extension gives the same value. This does not require the original vector field to extend continuously to a full neighbourhood of the curve. Likewise, a differential identity may hold only on the regular segments: a zero integrand there has improper integral zero, although its printed classical expression may not be defined at the omitted points.

The inertia tensor packages the moments of inertia about every axis through a specified point $O$ into one symmetric matrix. An axis still needs a direction: if $\mathbf{n}$ is a unit vector, its scalar moment of inertia is $I_{O,\mathbf{n}}={\mathbf{n}}^T{\mathbf{I}}_O\mathbf{n}$. The same mass layout can therefore be easy to spin about one direction and harder about another. The inertia tensor also determines rotational kinetic energy and relates angular velocity to angular momentum in the appropriate rigid body model.

Take position vectors $\mathbf{r}=(x,y,z)$ relative to $O$, expressed in an orthonormal basis. For point masses $m_a$, or a continuous distribution with mass density $\rho$ and $dm=\rho dV$,Here ${\delta }_{ij}$ is the Kronecker delta. Equivalently, using the identity matrix $1$ and an outer product,Every entry has SI units $\mathrm{kg}{\mathrm{m}}^2$. The diagonal entries are axial moments of inertia; the off-diagonal entries carry a minus sign in this convention. Some texts instead define the integrals without the prefixed minus sign $\int xydm$, $\int xzdm$, $\int yzdm$ as their named off-diagonal quantities and put the minus signs into the displayed matrix. Check that convention before combining formulas. ^{[ref] [ref]}

The symmetric matrix is positive semidefinite, because its quadratic form isThe dot product and cross product identify this as a sum of squared distances when $\mathbf{u}$ is a unit vector. A zero moment of inertia occurs precisely when all the mass lies on that axis, up to sets of zero mass. A genuine three-dimensional body with positive mass density in a volume has strictly positive axial moments of inertia. Do not confuse this inertia tensor with the mass second-moment tensor $\mathbf{S}=\int \mathbf{r}{\mathbf{r}}^{T}dm$: they obey ${\mathbf{I}}_O=\mathrm{tr}(\mathbf{S})1-\mathbf{S}$, where $\mathrm{tr}$ denotes the matrix trace.

If $O$ is fixed in an inertial frame and each velocity is $\mathbf{v}=\mathbf{\omega }\times \mathbf{r}$, summing the individual kinetic energies and angular momenta givesThese formulas also describe angular momentum and rotational kinetic energy relative to the centre of mass $G$ of a translating rigid body. In that case the total kinetic energy in the inertial frame is $K_{\mathrm{total}}=\frac{1}{2}M|{\mathbf{V}}_G{|}^2+\frac{1}{2}{\mathbf{\omega }}^T{\mathbf{I}}_G\mathbf{\omega }$. For a general moving point $O$, the body's actual velocities include ${\mathbf{V}}_O$, so ${\mathbf{I}}_O\mathbf{\omega }$ alone need not be its angular momentum about $O$. ^{[ref] [ref]}

Changing the orthonormal basis changes components, rather than the physical distribution: if the columns of an orthogonal matrix $Q$ are the new basis vectors expressed in the old basis, then ${\mathbf{I}}_O^{\prime }=Q^T{\mathbf{I}}_{O}Q$. Changing the reference point is a different operation, governed by the tensor parallel-axis theorem. Both principal axes and principal moments of inertia must therefore be quoted with their reference point.

Wave interference is the reinforcement or cancellation that occurs when overlapping wave amplitudes add by the superposition principle. For two monochromatic waves of the same frequency, mutual optical coherence means that their relative wave phase is stable during the observation. Equal wave phases give constructive interference; a half-cycle phase difference gives the destructive interference intensity minimum. Wave amplitudes add before their squares determine wave intensity. ^[ref].

Consider two overlapping beams in a homogeneous medium, with the same polarization of an electromagnetic wave and individual local time-averaged light intensities $I_1,I_2$. Their phase difference $\delta$ is measured in radians. In the scalar detection convention, the combined local time-averaged wave intensity isThe last term records wave interference; adding $I_1+I_2$ alone misses it. The formula describes irradiance when both fields propagate along the same direction, or when the same detector geometry factor is included in all three intensities. For crossing beams it describes the squared electric field response, proportional to scalar wave intensity; a specified component of the Poynting vector also requires the propagation geometry. The ideal crossing plane waves below make that distinction explicit.

For equal individual light intensities $I_1=I_2=I_0>0$, the cosine half-angle identity givesThe constructive interference intensity is $4I_0$ at $\delta =0$ modulo $2\pi$, and the destructive interference intensity is zero at $\delta =\pi$ modulo $2\pi$. At phase quadrature, $\delta =\pi /2$, it is $2I_0$. With $I_1=1$ and $I_2=0.25$ in the same arbitrary wave intensity unit, the maximum is $2.25$, the minimum is $0.25$, and the phase quadrature value is $1.25$: unequal wave amplitudes cannot cancel completely.

Both curves use the same wave intensity unit and a horizontal angle in radians. The dashed levels are the averages over a full phase difference cycle. Equal beams have interferometric visibility one; the unequal example has interferometric visibility $0.8$.

A spatially varying phase difference turns the same law into interference fringes. Take two ideal plane electromagnetic waves with common wavelength $\lambda$, equal individual light intensities $I_0$, and propagation directions at angles $+\theta$ and $-\theta$ from the $z$ axis in the $xz$ plane, with $0<\theta <\pi /2$. Both have linear polarization along $y$, perpendicular to both propagation directions. Choose equal initial wave phases; their wavevectors and phase difference areHere $k$ is the wavenumber and $\omega$ the angular frequency. The common $z$ and time terms cancel, so the interference fringes stay at fixed $x$ for every $z$. Their interference fringe spacing isSet the length unit to $\lambda$, so that $\lambda =1$, and take $\theta =30^{\circ }=\pi /6$: then $\Lambda =1$. The constructive interference intensity bands lie at integer $x/\lambda$; the destructive interference intensity bands lie at half-integer $x/\lambda$.

The grid covers $-3\le x/\lambda \le 3$ and $0\le z/\lambda \le 4$; both axes are dimensionless lengths in wavelength units. The colour scale is $I/I_0$ from zero to four, using the same single-beam normalization as the equal-beam curve above. This is an ideal plane wave superposition, with no slits or point sources and no near-field approximation. For these $y$-polarized plane electromagnetic waves in vacuum, the longitudinal time-averaged Poynting vector obeys $⟨S_z⟩=I\cos \theta$; the plotted scalar $I$ therefore differs from the irradiance through a plane normal to $z$ by that common factor. This relation follows by adding the magnetic fields as well as the electric fields.

The bright peaks do not create energy. Averaging over one interference fringe spacing, or equivalently over one full phase difference cycle, givesbecause the integral of the cosine is zero, with sine as its antiderivative. Equal beams thus average to $2I_0$, despite bright bands at $4I_0$. For the crossing-wave geometry, the average longitudinal energy flux is correspondingly $2I_0\cos \theta$. Bright and dark regions redistribute the incoming energy; conservation of energy concerns the full energy flux, rather than the height of one peak.

Wave interference is useful because phase differences convert small changes in optical path length into measurable changes of wave intensity, as in a Mach-Zehnder interferometer. Its beauty is that the same superposition principle produces both complete cancellation and fourfold brightening; adding more coherent contributions also produces the narrow peaks of a diffraction grating. Feynman's treatment connects this many-source pattern directly to addition of complex amplitudes. ^[ref].

A short proof outline uses local real fields $E_j(t)=A_j\cos (\omega t+{\varphi }_j)$ with parallel linear polarizations. Write $I=C⟨(E_1+E_2{)}^2{⟩}_t$ and $I_j=C{A}_j^2/2$ for the same positive conversion factor $C$. Averaging over many wave periods while the phase difference stays fixed uses $⟨{\cos }^2(\omega t+{\varphi }_j){⟩}_t=1/2$ and $⟨\cos (\omega t+{\varphi }_1)\cos (\omega t+{\varphi }_2){⟩}_t=\cos \delta /2$, giving the stated law. Equivalently, add the complex amplitudes and take the squared complex modulus. ^[ref].

For different fixed linear polarizations in a common transverse plane, the cross term acquires the dot product of their real unit vectors, equal to $\cos \alpha$ for relative angle $\alpha$. Orthogonal linear polarizations remove the cross term in polarization-insensitive scalar detection; different orientations generally only reduce it. A common polarizer can select overlapping components and recover wave interference. Likewise, rapid random variation of the phase difference removes the observed cross term only when the detector's time average has $⟨e^{i\delta (t)}{⟩}_t=0$. Slowly varying wave phases can instead move resolved interference fringes, and biased or incompletely sampled wave phases can leave residual contrast. Finite optical coherence therefore depends on the observation time as well as the fields.

The interplanar spacing $d$ is the perpendicular distance between parallel planes in a specified family. Measuring along an oblique line gives a larger distance, so the direction of measurement matters.

There is a short reciprocal lattice derivation. With the angular convention $a_i\cdot b_j=2\pi {\delta }_{ij}$, put $G=h{b}_1+k{b}_2+l{b}_3$. Then $G\cdot r=2\pi m$ is the same plane equation as $hx+ky+lz=m$. Thus $G$ is a normal vector. Moving between consecutive sheets along this normal changes the dot product by $2\pi$, givingFor perpendicular cubic axes of length $a$, this becomesFor example, choose a primitive cubic lattice with $a=0.600\mathrm{nm}$. Its $(210)$ lattice planes have $d_{210}=0.600/\sqrt{5}\simeq 0.268328\mathrm{nm}$. This is a chosen geometry, not an experimental value. The distance from a point to a plane gives the same result: the planes $2X+Y=0$ and $2X+Y=a$ in ordinary Cartesian length coordinates are separated by $a/\sqrt{5}$.

Specify whether $d_{hkl}$ denotes spacing between successive occupied lattice planes or the formal spacing assigned to a reflection label. In a primitive cubic lattice, $(200)$ has formal spacing $a/2$ but every other sheet contains no lattice points; its reflection is the second diffraction order of $(100)$. In a face-centered cubic lattice with conventional cubic axes, the $(200)$ sheets at half-integer $X/a$ do contain lattice points. Here $a/2$ is the actual spacing of the complete family perpendicular to that cube edge. The Primitive cubic Bragg peak positions example explains how to avoid counting a reflection twice.

A collection of holdings in traded assets. Its value is the dot product of the holdings with the asset prices. A self-financing portfolio is an investment portfolio whose trading requires no external funding.

Kinetic energy is the energy associated with motion. For a nonrelativistic point mass of mass $m>0$, measured in a specified inertial frame,Here $\mathbf{v}$ is velocity and $v$ is speed; $T$ is another common symbol for kinetic energy. The dot product makes $K$ nonnegative regardless of the direction of velocity. At fixed mass, doubling speed quadruples kinetic energy: a $2\mathrm{kg}$ point mass at $2m/s$ has $4\mathrm{J}$, whereas at $4m/s$ it has $16\mathrm{J}$. The SI units are $\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2=\mathrm{J}$. Different reference frames can assign different velocities and therefore different kinetic energies. ^[ref]

A short outline explains why kinetic energy has this formula. In a nonrelativistic inertial frame, constant mass and Newton's second law give, using the chain rule,Integrating over time proves the work-energy theorem, $W_{\mathrm{net}}=K_f-K_i$. The net force is the sum of all forces, and $W_{\mathrm{net}}$ is their total work, not merely the work of a chosen driving force. This route to final speed is useful when the forces as functions of position are known but the full motion as a function of time would be harder to find. ^[ref]

In the running example, a $2\mathrm{kg}$ point mass starts at rest and receives a constant net $3\mathrm{N}$ force through $4\mathrm{m}$. Its net work is $12\mathrm{J}$, soAt any intermediate distance $x\ge 0$, $K(x)=(3\mathrm{N})x$ and $v(x)=\sqrt{2Fx/m}$: kinetic energy is a linear function of distance, while speed is proportional to the square root of distance. If a net opposing $3\mathrm{N}$ braking force subsequently acts for $4\mathrm{m}$, it does $-12\mathrm{J}$ of work and brings this point mass to rest. Negative net work decreases kinetic energy; it does not make kinetic energy negative. Beyond the stopping point the motion must be reconsidered, since the point mass cannot keep moving forward with a negative $K$.

Kinetic energy also makes the exchange with potential energy intuitive: a falling body's loss of gravitational potential energy can appear as increased kinetic energy. The formula above applies when speed is small compared with the speed of light; at higher speeds, special relativity changes the expression. ^[ref]

The electric field does work on matter at local power per volumeHere $\mathbf{J}$ is current density and the dot product includes the relative direction. Positive $\mathbf{J}\cdot \mathbf{E}$ means energy passes from the field to matter; negative means matter supplies energy to the field. In general this is a work rate, not automatically an irreversible heating rate.^[ref]

For a stationary isotropic Ohmic conductor with real scalar electrical conductivity $\sigma >0$, local Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Thus the local Joule heating rate isThe electrical resistivity ${\rho }_{\mathrm{el}}$ has units $\Omega \mathrm{m}$, and electrical conductivity has units $(\Omega \mathrm{m}{)}^{-1}$. These formulas use the material's rest frame and dissipative conduction current density. Moving media, complex conductivity, and additional storage or transport processes require their appropriate energy balance rather than this scalar substitution.

For uniform current in a straight homogeneous wire of length $L$, wire cross-sectional area $A$ and constant electrical conductivity, $E=V/L$, $J=I/A$ and $R=L/(\sigma A)$. The volume integral recovers terminal power:As a numerical check, the wire already used in electrical resistance has $L=1\mathrm{m}$, $A=10^{-6}{\mathrm{m}}^2$, $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$ and $V=5\mathrm{V}$. It gives $J=2.5\times 10^{6}A/{\mathrm{m}}^2$, $p_{\mathrm{vol}}=1.25\times 10^{7}W/{\mathrm{m}}^3$ and $P=12.5\mathrm{W}$, agreeing with $I=2.5\mathrm{A}$ and $R=2\Omega$. Concentrating the same electric current in a smaller wire cross-sectional area therefore raises the local heating rate if electrical resistivity stays fixed.

For a locally oscillatory wave written $u(\mathbf{x},t)=A(\mathbf{x},t)\cos \varphi (\mathbf{x},t)$, the local wavevector is ${\mathbf{k}}_{\mathrm{loc}}=\nabla \varphi$, where $\nabla$ is the spatial gradient. The local wavelength $2\pi /|{\mathbf{k}}_{\mathrm{loc}}|$ measures local wave phase spacing normal to the wavefront. Interpreting it as a crest separation requires the amplitude and local wavevector to vary little over one wavelength, as in the WKB approximation. Along a unit direction $\mathbf{e}$ the corresponding local repeat distance is $2\pi /|{\mathbf{k}}_{\mathrm{loc}}\cdot \mathbf{e}|$, if that dot product is nonzero; oblique sampling measures this projected spacing.

A real snapshot alone does not uniquely specify $A$ and $\varphi$: their meaning requires a chosen oscillatory component or physical approximation. Strong interference, zeros of amplitude, and rapidly changing media can defeat a single-carrier description. If $\nabla \varphi =0$, the local reciprocal formula gives no finite wavelength; it does not establish an exact global spatial period.

The Lorentz force is the force exerted by an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units its form in nonrelativistic classical mechanics, using mass $m$ and Newton's second law, isBoth fields are evaluated at the body's position at the current time. The electric field contributes $q\mathbf{E}$ even at rest, whereas the magnetic force depends on velocity through a cross product. ^[ref]; ^[ref].

For a concrete charged particle in a uniform magnetic field, use nondimensionalization to stipulate $m=q=B=1$, magnetic field $\mathbf{B}=(0,0,1)$, zero electric field, initial position $(0,0,0)$, and initial velocity $(1,0,0)$. These are rescaled quantities, not an assertion that the mass, electric charge, and magnetic field have the same physical units. At the start, $(1,0,0)\times (0,0,1)=(0,-1,0)$, so the magnetic acceleration points toward $-y$. The resulting circular motion isAt $t=\pi /2$ the position is $(1,-1,0)$ and the velocity is $(0,-1,0)$; at $t=\pi$ the position is $(0,-2,0)$ and the velocity is $(-1,0,0)$. Thus the circle has center $(0,-1,0)$ and radius one. Viewed from $+z$ toward the $xy$ plane, the motion is clockwise. Changing only the electric charge to $q=-1$ gives position $(\sin t,1-\cos t,0)$ and reverses the bending to counterclockwise.

Adding initial velocity $v_z=0.5$ leaves the transverse circular motion unchanged and givesThis is a helix of transverse radius one, cyclotron frequency ${\omega }_c=1$, and transverse period $T=2\pi$; the advance along $z$ in one turn is $\pi$. The parallel component of velocity stays constant because the magnetic force has no parallel component: there is no magnetic acceleration along the magnetic field. If the initial velocity were entirely parallel to the magnetic field, the magnetic force would vanish and the path would be straight. ^[ref].

More generally, with zero electric field, no other forces, and a uniform constant magnetic field of magnitude $B>0$, decompose velocity into perpendicular and parallel components. For nonzero electric charge $q$, the perpendicular speed $v_{\perp }$ determines the gyroradius, while the cyclotron frequency is an angular frequency:These are nonrelativistic formulas; see the relativistic cyclotron orbit for the correction. The beauty is that increasing perpendicular speed increases the radius in exactly the proportion needed to leave the period unchanged. Conversely, known magnetic field, perpendicular speed, and measured radius determine the magnitude of the electric charge divided by mass. This makes the same geometry useful for comparing bodies with different electric charge and mass.

Two short checks explain the result. First, the cross product gives $\mathbf{v}\times \mathbf{B}=(-\sin t,-\cos t,0)$ for both stipulated positive-electric charge paths, exactly their acceleration. Second, the transverse magnetic force supplies the centripetal acceleration, so $m{v}_{\perp }^2/r=|q|v_{\perp }B$; for $v_{\perp }>0$ this gives the gyroradius, and ${\omega }_c=v_{\perp }/r$ gives the cyclotron frequency.

Magnetic forces do no work: the instantaneous power of the magnetic part of the Lorentz force vanishes because a cross product is perpendicular to its first argument. The work-energy theorem therefore givesThe dot product $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})$ is zero at every instant, rather than only after a full turn. With zero electric field and no other forces, the kinetic energy and speed remain constant; the electric field can increase or decrease kinetic energy according to the sign of $q\mathbf{E}\cdot \mathbf{v}$. In the two stipulated examples the constant kinetic energy is respectively $K=1/2$ and $K=5/8$ in rescaled units. The extra parallel velocity raises kinetic energy without changing the gyroradius or the cyclotron frequency.

For an electric example in rescaled units, stipulate $m=q=1$, zero magnetic field, constant electric field $\mathbf{E}=(2,0,0)$, no other forces, and rest at $t=0$. The Lorentz force gives acceleration $(2,0,0)$ and velocity $(2t,0,0)$, hence kinetic energy $K=2t^2$ and power $dK/dt=4t=q\mathbf{E}\cdot \mathbf{v}$. At $t=1$, the kinetic energy has risen from zero to two: the electric field has done positive work.

The existing electromagnetic field tensor formulation expresses the same interaction covariantly. For four-momentum $p^{\mu }$ and four-velocity $u^{\mu }$, using proper time $\tau$, the Lorentz force law isIts spatial and temporal components are