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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 ${\oint }_{C}d\mathbf{r}=0$.

A curved feasible set can have the same certificate structure. ConsiderWrite $x=(t,z{)}^T$ and use the second-order cone $K=\{(s_0,s_1,s_2):s_0\ge \sqrt{s_1^2+s_2^2}\}$. In this coordinate ordering,Self-duality of a second-order cone makes $K^{\ast }=K$. The equality $A^{T}y=c$ forces $y_0=1$ and $y_2=0$, leaving $-1\le y_1\le 1$. Its dual objective function is $b^{T}y=-y_1$, maximized at $y=(1,-1,0{)}^T$. Thus every feasible point satisfies $t\ge 1$, and $(t,z)=(1,0)$ attains that bound: $\boxed{p_{\ast }=d_{\ast }=1}$.

The zero dot product is $(1,1,0)\cdot (1,-1,0)=0$. Individual coordinate products need not be nonnegative here: the cancellation $1-1=0$ occurs across coordinates. The valid complementary slackness condition is cone-level orthogonality, rather than the coordinatewise interpretation of the nonnegative orthant.

A linear code is self-dual when it equals its dual code under the standard dot product. For a length-$2k$ code with a rank-$k$ generator matrix $G$, this is equivalent to $G{G}^T=0$: orthogonality gives inclusion in the dual code, and equal dimensions give equality. Over ${\mathbb{F}}_2$, a generator matrix $(I\mid A)$ therefore generates a self-dual code exactly when $A{A}^T=I$.

A signed spatial filter model of a simple cell assigns a weight $K_{ij}$ to each sampled location and takes its dot product with the visual contrast image $C_{ij}$. Positive weights favor bright features; negative weights favor dark features. The signed driveis a linear function of the image, while the rectified linear unit output $R$ is nonlinear. $D$ and $R$ below are arbitrary model units. To obtain a predicted firing rate one would need a calibrated gain, background response and possibly a different output function. The model does not generate a spike train; a linear-nonlinear-Poisson cascade model adds a stochastic event-generation assumption.

For the first diagram, every one of five rows has weightsA bright central column, with $C_{i3}=1$ and zero elsewhere, gives $D=5/5=1$. A bright middle row gives $D=-1/20-1/20+1/5-1/20-1/20=0$; a uniformly bright patch also gives $D=0$, since every row sums to zero. Move the bright column one site right: $D=5(-1/20)=-1/4$, so $R=0$. Make that shifted column dark instead: $D=1/4$ and $R=1/4$. Thus an OFF-like weight can increase output for darkness. A negative weight is a descriptive sensitivity, not evidence for a particular inhibitory synapse.

For a smooth example use the real, even Gabor filterThe Gaussian function localizes the weights; the cosine supplies alternating polarity. Choose ${\sigma }_x=1/2$, ${\sigma }_y=1$ and $q=\pi$ in arbitrary spatial units. The stripes' spatial frequency is $q/(2\pi )=1/2$ cycles per unit. Their axis makes angle $\theta$ with horizontal, while brightness changes across the perpendicular axis. $\phi$ is their spatial phase. Normalize the full-plane integral of $K{C}_{\theta ,\phi }$ by its value at $\theta =\pi /2$, $\phi =0$. The Gaussian integral and the product identity for cosines givewhereAt $\phi =0$, horizontal stripes give $R\approx 0.004159$, $60^{\circ }$ stripes give $R\approx 0.286741$, and vertical stripes give $R=1$. At the same vertical orientation, $\phi =\pi /2$ gives $D=R=0$, while reversing the polarity with $\phi =\pi$ gives $D=-1$ and $R=0$. The preferred axis persists while the response changes dramatically with position. Shifting the vertical stripes by a quarter of their wavelength changes phase by $\pi /2$ up to sign, without rotating them.

These are two different toy filters, not discretizations of the same fitted receptive field. The discrete filter exactly rejects a uniform contrast patch; this Gabor filter need not reject uniform contrast because its weight integral need not vanish. Neither toy includes stimulus history, recurrent interactions, saturation or a measured response baseline. Their value is a transparent prediction: orientation selectivity can emerge from spatial summation, and it does not by itself imply tolerance to phase.

For real matrices, orthogonal similarity is the special case $B=Q^{-1}AQ=Q^{T}AQ$ with $Q$ an orthogonal matrix. Then a symmetric matrix remains symmetric, because $B^T=Q^T{A}^{T}Q=B$. The expression $QA{Q}^T$ describes the reverse conversion and has the same property. An orthonormal basis retains the ordinary dot product formula for its coordinate vectors.

An arbitrary change of basis need not preserve entrywise symmetry. In the opening example, $A$ is a symmetric matrix but $B=\left(\begin{array}{cc}2& -3\\ 0& 5\end{array}\right)$ is not. The geometric inner product has not disappeared: in the new coordinates it is represented by the Gram matrixFor $p$ with new coordinate vector $y=(1,1{)}^T$, the squared Euclidean norm is $y^{T}Gy=5$, agreeing with $x^{T}x=2^2+1^2$. The naive coordinate sum $y^{T}y=2$ would use the wrong inner product.

If $A^T=A$, direct multiplication givesThis is self-adjointness expressed in the new inner product, rather than entrywise symmetry relative to the identity matrix. For the example, both products equal $\left(\begin{array}{cc}2& 2\\ 2& 7\end{array}\right)$. Thus arbitrary matrix similarity preserves the linear operator, while its geometric interpretation requires transforming the Gram matrix as well. Matrix congruence supplies the corresponding rule for a bilinear form and should be distinguished from matrix similarity.

A simple cell is a visual cortical neuron whose receptive field has spatially separable ON and OFF receptive-field subregions. Elongated neighboring subregions give a preferred orientation and a sensitivity to the location and polarity of visual contrast. “Simple” describes the response organization; it does not describe a simpler cell body or a particular shape of neuron.

Hubel and Wiesel's 1962 classification emphasized four observations: distinguishable ON and OFF receptive-field subregions, summation within each region, antagonism between regions, and qualitative prediction of responses to differently shaped stimuli from those regions' arrangement. A slit fitting an elongated center was more effective than a smaller spot; a larger patch could invade opposing flanks and lose effectiveness. An edge separating bright and dark regions could be effective when its boundary matched the boundary between subregions. These were observations about spike trains, rather than an assertion of exact linearity. ^[ref]

The following deliberately coarse signed spatial filter model of a simple cell makes the geometry easy to count. A centered bright bar contributes positively at five sites. Rotating it preserves its area but mixes positive and negative weights; shifting it preserves its orientation but changes which weights it samples. The model is useful because it predicts distinct effects of these three manipulations instead of treating any bright stimulus as equivalent.

A simple cell is not an ideal linear sensor. Its spike train is nonnegative in event count, and neural adaptation can change responsiveness. The signed spatial filter model of a simple cell supplies a useful approximation to spatial drive and an explicit output rule. Comparing predictions with unseen stimuli is more informative than assuming the rule from the name “simple”. In contrast, the existing complex cell models explore tolerance to phase and local displacement.

At a smooth mirror with unit normal vector $n$, an incoming unit direction $i$ has outgoing direction $i-2(i\cdot n)n$. Thus the incident and reflected rays make equal angles with the normal.

More precisely, specular reflection redirects each incident light ray into the direction prescribed by the law of reflection, rather than spreading it over many outgoing directions. The incident light ray, reflected light ray, and local normal vector lie in one plane. The surface must be sufficiently smooth on the scale of the wavelength for this model to describe its reflected beam; a visibly flat surface can still scatter light if its microscopic roughness is appreciable. ^[ref].

The displayed formula uses unit vectors $\mathbf{i},\mathbf{n},\mathbf{r}$ and the dot product. It preserves the component parallel to the mirror and reverses the normal component. For a mirror along $y=0$, take $\mathbf{n}=(0,1)$ and $\mathbf{i}=(3/5,-4/5)$. Then $\mathbf{i}\cdot \mathbf{n}=-4/5$ and $\mathbf{r}=(3/5,4/5)$: both directions have vector norm one, and both make an angle $\arccos (4/5)\approx 36.87^{\circ }$ with the relevant normal direction. Replacing $\mathbf{n}$ by $-\mathbf{n}$ leaves the result unchanged.

This same local rule explains a flat mirror's virtual image. For a source at $(0,a)$ above $y=0$, backward extensions of reflected light rays intersect at $(0,-a)$. The light remains above the mirror; its apparent origin lies below it. A curved mirror changes its normal vector from point to point and can instead make reflected light rays converge. The geometry of image formation therefore starts with one simple direction rule. ^[ref].

// Target: optics.bigb

Spherical geometry studies lengths and angles measured along the surface of a sphere, rather than through its interior. It is a useful model whenever travel is constrained to a round surface. Its straightest paths are great circle arcs: they are geodesics, even though they appear curved in the surrounding Euclidean space. Two distinct great circles meet at an antipodal pair, so the parallel-line picture of Euclidean geometry does not apply.

For a sphere of radius $R>0$, write two positions as $Ru,Rv$, with $u,v$ unit vectors. Their shorter surface route has great-circle distancewhere the dot product gives the central angle in radians. The straight chord through the surrounding Euclidean space instead has length $2R\sin (d/(2R))$. For a quarter-circle separation these are $\pi R/2\approx 1.571R$ and $\sqrt{2}R\approx 1.414R$: specifying which distance we mean has a measurable effect. On the unit sphere, $R=1$ and great-circle distance equals the central angle.

Consider the spherical triangle with vertices $(R,0,0)$, $(0,R,0)$ and $(0,0,R)$, joined by the shorter great circle arcs. Each side has length $\pi R/2$. At each vertex the two outgoing tangent vectors point along the other coordinate axes and are orthogonal, so all three interior angles are $\pi /2$. This is the positive coordinate octant of the sphere, not a flat triangle drawn inside it.

For a convex minor spherical triangle with interior angles $\alpha ,\beta ,\gamma$ in radians, the spherical excess formula statesOur octant therefore has area $R^2(3\pi /2-\pi )=\pi R^2/2$. Independently, eight congruent octants partition the sphere, giving $4\pi R^2/8$ again. Thus angles alone determine area when $R$ is fixed, unlike Euclidean geometry, where similar triangles can have arbitrarily different areas. The Gauss-Bonnet theorem explains the connection: the sphere's constant Gaussian curvature $1/R^2$ times the area equals this excess. For spherical triangles whose area is small compared with $R^2$, the excess is small, recovering the familiar flat picture.

For a nondegenerate convex minor spherical triangle, the spherical law of cosines relates two angular sides $b,c$, their included interior angle $A$, and the opposite angular side $a$:Relabelling the vertices gives the corresponding formulas for $b,B$ and $c,C$. As with the spherical law of sines, physical side lengths must first be divided by the sphere's radius. The spherical law of cosines determines $a\in (0,\pi )$ uniquely from $b,c,A$ when the data describe such a spherical triangle. It also recovers each interior angle from three sides, for exampleUsing the cosine and the range $0<A<\pi$ selects a unique angle, avoiding the supplementary-angle ambiguity of an inverse sine calculation.

To derive the spherical law of cosines, let $u,v,w$ be the vertex unit vectors at $A,B,C$, so $u\cdot v=\cos c$, $u\cdot w=\cos b$ and $v\cdot w=\cos a$. The outgoing unit tangent vectors at $u$ are $t_B=(v-\cos cu)/\sin c$ and $t_C=(w-\cos bu)/\sin b$. Their dot product is both $\cos A$ and $(\cos a-\cos b\cos c)/(\sin b\sin c)$, proving the identity. This gives a geometric explanation: spherical trigonometry compares angles in the tangent plane using vectors in the surrounding Euclidean space.

For a worked example on the unit sphere, take the verticesAt $A$, the outgoing unit tangent vectors are $(1,0,0)$ and $(0,1,0)$, so the interior angle is a right angle. The vertex dot products give $b=c=\pi /3=60^{\circ }$. The spherical law of cosines then reduces to the right-angle identity $\cos a=\cos b\cos c$, givingTo find $B$, use the relabelled spherical law of cosines:The physical side lengths on a sphere of radius $R$ are ${\ell }_a\approx 1.318116R$ and ${\ell }_b={\ell }_c=\pi R/3\approx 1.047198R$. For example, on a toy sphere with $R=10$ metres they are approximately $13.18116$, $10.47198$ and $10.47198$ metres; the interior angles stay the same. Applying the planar Pythagorean theorem directly to the angular sides would instead give $\sqrt{60^2+60^2}\approx 84.852814^{\circ }$. That flat right triangle calculation does not give the great-circle distance between $B$ and $C$.

The familiar law of cosines reappears as a local approximation. With $a,b,c$ all small in radians, substitute the small-angle approximations $\sin x\approx x$ and $\cos x\approx 1-x^2/2$ into the spherical law of cosines and keep terms through second order:Multiplication by $R^2$ gives the planar law of cosines for the physical side lengths. For a fixed small surface triangle on a progressively larger sphere, the angular sides shrink and this approximation improves. Thus the curved and flat formulas fit together: the Pythagorean theorem is recovered locally when $A$ is a right angle.

Spherical trigonometry calculates the sides and interior angles of spherical triangles using trigonometric functions. Its sides must be shorter great circle arcs, which are geodesics on a round sphere. A curve at constant nonzero latitude does not qualify. The spherical law of sines and spherical law of cosines play the roles of the familiar flat-triangle rules, but account for the sphere's curvature.

Use uppercase $A,B,C$ both for the vertices and for their interior angles, and lowercase $a,b,c$ for the opposite angular sides. For example, $a$ is the central angle subtended by the arc from $B$ to $C$. On a sphere of radius $R$, its physical arc length is ${\ell }_a=Ra$ when $a$ is in radians; similarly ${\ell }_b=Rb$ and ${\ell }_c=Rc$. Only on the unit sphere do the angular and physical side lengths have the same numerical values. See J. B. Tatum, Celestial Mechanics, chapter 3, section 3.5: great-circle sides and angular units.

One useful application of spherical trigonometry is finding great-circle distance from spherical coordinates. Write a unit vector as $u(\varphi ,\lambda )=(\cos \varphi \cos \lambda ,\cos \varphi \sin \lambda ,\sin \varphi )$, where $\varphi$ is latitude and $\lambda$ is the equatorial azimuth, or longitude, measured from a chosen reference direction. The dot product of two such unit vectors givesHere $\delta \in [0,\pi ]$ is the central angle and $d$ is the great-circle distance. The same identity follows from the spherical law of cosines by adding the north pole as the third vertex, with angular sides $\pi /2-{\varphi }_1$ and $\pi /2-{\varphi }_2$. This connects spherical geometry to position calculations in astronomy and to navigation on a spherical model of Earth. Choosing the surface model matters: Paula Neeley and Anthony Narkawicz, Map Projection Induced Variations in Locations of Polygon Geofence Edges, NASA/TM-2017-219675 (2017), sections 2-3 distinguishes spherical and ellipsoidal geodesics and gives this coordinate identity.

For a toy example, put two points at latitude $30^{\circ }$, with equatorial azimuths $-60^{\circ }$ and $60^{\circ }$. The great-circle distance and the arc length along constant latitude areThe great circle takes a shorter route by bending towards the pole in these spherical coordinates. To locate its highest latitude, the endpoint unit vectors are $p=(\sqrt{3}/4,-3/4,1/2)$ and $q=(\sqrt{3}/4,3/4,1/2)$. Their great circle lies in the plane through the origin with $z=2x/\sqrt{3}$. On its shorter arc, the maximum latitude occurs at $y=0$ and is $\arctan (2/\sqrt{3})\approx 49.106605^{\circ }$. Thus a geodesic can change latitude substantially even when its endpoints have equal latitude. These are lengths on an ideal sphere; the example illustrates the geometry rather than specifying an actual flight route.

Historically, spherical trigonometry was prominent in teaching as well as in astronomy. John Conway and Alex Ryba note its presence in early twentieth-century school textbooks in Remembering spherical trigonometry, The Mathematical Gazette 100(547), March 2016, pages 1-8, DOI 10.1017/mag.2016.3. Their introductory observation documents that teaching history; it does not identify a single inventor of the spherical law of sines or spherical law of cosines.

A spin-spin interaction couples two spin angular momenta. In an isotropic model it is proportional to their dot product, whose expectation in a total-pair-spin state follows from $2{\mathbf{S}}_i\cdot {\mathbf{S}}_j=S_{ij}^2-S_i^2-S_j^2$. Magnetic interactions generate examples in atoms and effective colour-magnetic interactions in hadron models.

Static friction is the tangential contact force that maintains zero contact slip. “Static” describes the contact state: both surfaces can move together in the chosen inertial frame. In the ideal dry-contact model, the actual static friction adjusts to the equations of motion, subject to a bound, rather than always taking its maximum value.^[ref]Here $N$ is the magnitude of the normal force and ${\mu }_s$ the coefficient of static friction. Equality means limiting static friction; it is a threshold, not an instruction to replace every unknown contact force by ${\mu }_{s}N$.

For a stipulated example, place a nonrotating $2\mathrm{kg}$ block initially at rest on a stationary horizontal surface. Take gravitational acceleration $g=9.8m/{\mathrm{s}}^2$ and coefficient of static friction ${\mu }_s=0.40$, with no other vertical force. The vertical force balance gives $N=mg=19.6\mathrm{N}$, so the largest allowed static friction magnitude is $7.84\mathrm{N}$. Choose right as positive and apply a horizontal force $F$. Remaining at rest requires Newton's second law to giveAt $F=2\mathrm{N}$, the static friction is $-2\mathrm{N}$; at $F=6\mathrm{N}$, it is $-6\mathrm{N}$. Both states have zero acceleration. At $F=10\mathrm{N}$, balancing would require $f_s=-10\mathrm{N}$, outside the bound, so this initially resting block cannot remain at rest. Its subsequent acceleration requires a kinetic friction law; setting $f_s=-7.84\mathrm{N}$ after contact slip begins would still use the wrong contact state. At exactly $F=7.84\mathrm{N}$, the ideal inequality still permits rest at the threshold. These are model calculations, not measured material properties.

The toy model used in kinetic friction makes the contact-state distinction explicit: $m=1\mathrm{kg}$, $N=10\mathrm{N}$, ${\mu }_s=0.5$ and coefficient of kinetic friction ${\mu }_k=0.3$. On stationary ground, an initially resting box pushed right with $4\mathrm{N}$ has $f_s=-4\mathrm{N}$. At $6\mathrm{N}$ it cannot remain at rest; once it slides right, $f_k=-3\mathrm{N}$ and $a=3m/{\mathrm{s}}^2$. At $5\mathrm{N}$, rest is allowed with $f_s=-5\mathrm{N}$, whereas an already right-sliding box has $f_k=-3\mathrm{N}$ and $a=2m/{\mathrm{s}}^2$. The applied force alone does not specify the friction branch: contact slip and motion history matter.

For a flat contact with two tangential directions, the bound applies to a vector norm, not separately to each component:For the stipulated $7.84\mathrm{N}$ bound, balancing horizontal components $4\mathrm{N}$ and $4\mathrm{N}$ requires a force magnitude $\sqrt{32}\mathrm{N}\simeq 5.657\mathrm{N}$, which is allowed. Components $6\mathrm{N}$ and $6\mathrm{N}$ require $\sqrt{72}\mathrm{N}\simeq 8.485\mathrm{N}$, which is forbidden even though each component separately is below $7.84\mathrm{N}$. This circular bound assumes the same coefficient of static friction in every tangential direction. Near impending contact slip, the limiting static friction opposes the relative contact motion that would otherwise begin. Once actual contact slip exists, kinetic friction instead opposes its tangential relative velocity, which need not have the direction of the body's center of mass motion.

The normal force must also come from the equations of motion. On horizontal ground, an upward pull of magnitude $P$ at angle $\alpha$ above horizontal gives, for a block remaining at rest with no other forces,The sine component unloads the contact and the cosine component demands tangential support. These equations require $N>0$; if the supporting contact opens, this dry-contact model cannot supply a negative normal force. For the stipulated $2\mathrm{kg}$ block and $\alpha =30^{\circ }$, the positive threshold is $P={\mu }_{s}mg/(\cos \alpha +{\mu }_s\sin \alpha )\simeq 7.3544\mathrm{N}$, with $N\simeq 15.9228\mathrm{N}$ and $|f_s|\simeq 6.3691\mathrm{N}$. A downward pull increases $N$ instead. An inclined plane gives another reason why $N$ is not always $mg$.

To solve a new problem, impose zero contact slip, find the required static friction and normal force from the equations of motion, then check the bound. A demand inside it permits that contact state; a demand outside it rules the state out. For an ideal driven tire undergoing rolling without slipping on stationary ground, a forward static friction can accelerate the vehicle. Its role is to prevent contact slip, rather than to oppose all motion of the whole body.

Nor does the word “static” by itself make friction do zero work. The contact power delivered to a body is $P_s={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, a dot product with its material contact-point velocity in the chosen inertial frame. At an ideal contact undergoing rolling without slipping on stationary ground, ${\mathbf{v}}_c=0$, so this power is zero. For a rigid body, adding the translational and rotational contributions gives ${\mathbf{f}}_s\cdot {\mathbf{V}}_{\mathrm{CM}}+{\mathbf{\tau }}_{\mathrm{CM}}\cdot \mathbf{\omega }={\mathbf{f}}_s\cdot {\mathbf{v}}_c$, where ${\mathbf{\tau }}_{\mathrm{CM}}$ is the contact torque about the center of mass and $\mathbf{\omega }$ its angular velocity. Counting only the first term omits the rotational contribution.

At a moving contact with zero contact slip, the two material contact points share a velocity ${\mathbf{v}}_c$ that can be nonzero. For instance, a nonrotating block moving with an accelerating belt receives positive contact power $f_{s}U$ when the belt's rightward velocity $U$ and rightward static friction $f_s$ are positive. The belt receives $-f_{s}U$ by Newton's third law. These contact contributions sum to zero in the ideal no-slip model, while the block's kinetic energy can increase. Zero contact slip and zero work on each body are different conditions.

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.

In tetrahedral molecular geometry, four neighbouring chemical atoms surround a central chemical atom in the directions of the vertices of a tetrahedron. In the regular idealization, the central atomic nucleus lies at the centre of a regular tetrahedron, and every pair of chemical bonds makes the same bond angle. Methane, ${\mathrm{CH}}_4$, provides the standard example: its carbon has four bonding electron domains and no lone pair. See Chemistry 2e, §7.6: tetrahedral methane.

The exact ideal bond angle follows from four directions proportional to $(1,1,1)$, $(1,-1,-1)$, $(-1,1,-1)$, and $(-1,-1,1)$. Each has squared Euclidean norm $3$, and each distinct pair has dot product $-1$. HenceThis is a statement about a regular tetrahedron, rather than a universal exact value for every molecule labelled tetrahedral. It also explains why a cross-shaped flat drawing with $90^{\circ }$ gaps is misleading: four equal directions can spread out more evenly in three dimensions.

The same vectors sum to zero. Consequently, four equal bond-dipole contributions along those directions cancel. The symmetric methane molecule is nonpolar, but a tetrahedral shape alone does not imply a nonpolar molecule: replacing one neighbour can make the electronic distribution and bond contributions unequal. Chemical polarity depends on both molecular geometry and electron density. See McMurry, Organic Chemistry, §2.2: molecular dipoles and methane.

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.

Choose positions in one inertial frame and let $M=m_1+m_2$. The centre of mass and two-particle relative coordinate areThe lighter body is farther from the centre of mass. Reconstructing both positions from these equations is part of solving the two-body problem; $\mathbf{r}$ alone describes their separation, not either body's laboratory position.^[ref]

Substitute the reconstructed velocities into the kinetic energy. The result is the exact two-particle instance of kinetic-energy decomposition about the centre of mass:The mixed dot product has coefficient $-m_1{m}_2/M+m_2{m}_1/M=0$; the coefficient of $|\dot{\mathbf{r}}{|}^2$ is $(m_1{m}_2^2+m_2{m}_1^2)/M^2=\mu$. This algebra needs constant positive masses but no particular force law. In the stationary centre of mass frame, the fraction of internal kinetic energy carried by body 1 is $m_2/M$, and by body 2 is $m_1/M$: the lighter body carries more.

For a closed pair with no external forces, write $\mathbf{F}$ for the force on body 2 and $-\mathbf{F}$ for that on body 1, as required by Newton's third law. Adding and subtracting their Newton's second law equations yieldsThus the centre of mass moves uniformly while the two-particle relative coordinate behaves like one body of reduced mass. Equal-opposite internal forces suffice for these equations. For an internal central force, $\mathbf{F}=f(r)\hat{\mathbf{r}}$ with $r=|\mathbf{r}|>0$, the internal torque vanishes and conservation of angular momentum gives constant ${\mathbf{L}}_{\mathrm{rel}}=\mu \mathbf{r}\times \dot{\mathbf{r}}$. If the central force comes from a time-independent potential energy $U(r)$, then $f(r)=-dU/dr$ and the internal mechanical energy $E_{\mathrm{rel}}=\mu |\dot{\mathbf{r}}{|}^2/2+U(r)$ is conserved. The kinetic energy split alone does not establish either conservation law.

A unit normal is a normal vector of Euclidean norm one. Given a nonzero normal vector $n_0$, normalization gives $n=n_0/\Vert n_0\Vert$. This makes a dot product $w\cdot n$ measure the signed length of the normal component, without an extra factor from the length of $n_0$. On a hypersurface, the two opposite choices $n$ and $-n$ correspond to opposite normal orientations; a continuous choice on an orientable surface fixes its orientation of a surface.

The VSEPR model predicts approximate molecular geometry by arranging the electron domains around a central chemical atom to reduce their mutual repulsion. Start with a Lewis structure, count the domains, and then distinguish the resulting electron pair geometry from the positions of the chemical atoms themselves. A lone pair counts as a domain even though it is not another chemical atom.^[ref]

Three worked counts have the same total of eight valence electrons. In methane, carbon's four plus the contributions of four hydrogen chemical atoms give $4+4=8$: four bond pairs, no lone pairs, and tetrahedral molecular geometry. In ammonia, nitrogen's five plus three from hydrogen give $5+3=8$: three bond pairs and one lone pair, producing trigonal pyramidal molecular geometry. In water, oxygen's six plus two from hydrogen give $6+2=8$: two bond pairs and two lone pairs, producing bent molecular geometry. All three have four electron domains; their molecular geometries differ.

The regular tetrahedral directions can be represented by the unit vectorsEach pair has dot product $-1/3$, so the ideal bond angle is $\theta =\arccos (-1/3)\approx 109.47^{\circ }$. This is a geometric calculation, rather than a prediction that every four-domain molecule has that exact bond angle. Lone pairs and different kinds of chemical bonds distort the ideal arrangement; water's H–O–H angle is about $104.5^{\circ }$.^[ref]

The model is attractive because small-number bookkeeping produces a spatial prediction without first solving for every molecular orbital. Shape then helps explain chemical polarity and which hydrogen bond donor or hydrogen bond acceptor sites are accessible. It is a qualitative model, not a full calculation of the electronic energy or a universal solution for every chemical compound. Gillespie and Nyholm's 1957 review ^[ref] is a historical anchor for its development.