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The Anderson positron track experiment identified a light positively charged particle, now called the Positron, by combining magnetic bending, loss of kinetic energy in lead, and particle range. Each observation answered a different question: bending constrained momentum, the lead supplied a direction of travel, and particle range tested the proposed mass.

Carl David Anderson's The Positive Electron (1933), pp. 491–493, describes the 1932 discovery in a vertical cloud chamber with a $15,000\mathrm{gauss}$ magnetic field. A $6\mathrm{mm}$ lead plate separated two photographed track segments. The effective viewing depth was $1\mathrm{cm}$ along the magnetic field, with $14\mathrm{cm}$ transverse diameter: this restricted out-of-plane motion. The photographic readout supplied curvature on both sides. Comparisons with electron tracks under the same conditions constrained ionization; specific ionization was not directly measured for these events. He tested accidental alignment of unrelated tracks and rejected a reversed electron interpretation requiring an increase of kinetic energy in lead. Assuming unit electric charge, the mass bound was about twenty electron masses, rather than an exact mass determination. The paper credits Robert Andrews Millikan with the design and Seth Henry Neddermeyer with assistance.^[ref]

Preparation used rapid piston expansion and a water–alcohol mixture in the cloud chamber; cooling permitted droplets to form along ionization trails. A camera photographed these droplets. Initially, expansions occurred at random, so many photographs contained no tracks. Later counter triggering improved the useful yield.^[ref]

In The Production and Properties of Positrons, his 1936 lecture, Carl David Anderson describes the chamber inside an electromagnet and explains why the lead plate was decisive: after losing kinetic energy, a particle bent more sharply. He describes a proton hypothesis predicting only $5\mathrm{mm}$ of particle range, against an observed path longer than $50\mathrm{mm}$. These are historical values from the lecture, not the stipulated example below. The point was the incompatibility of particle range with the momentum inferred from curvature, rather than positive electric charge alone.^[ref]

At radius $a$ in a ring rotating at rate $\omega$, a walker with signed tangential speed $u$ relative to the ring has inertial speed $a\omega +u$. The normal reaction per mass is $(a\omega +u{)}^2/a$; the curvature of the relative path matters as well as the centrifugal force and the Coriolis force.

In the weak-field, short-wavelength averaging regime, a transverse-traceless tensor strain $H_{ij}$ carries the displayed effective stress-energy tensor in units $c=1$. The average spans several wave periods while remaining small compared with the background curvature scale. The stress is second order in amplitude, so it is absent from a first-order vacuum Linearized Einstein equations calculation. For a plane wave, summing the two equal transverse diagonal contributions gives flux $(16\pi G{)}^{-1}⟨{\dot{H}}_{+}^2+{\dot{H}}_{\times }^2⟩$. This effective averaged stress is not an arbitrary pointwise gravitational energy tensor.

A circular elastic plate with bending stiffness $B$, constant excess pressure $p$ and clamped boundary conditions at radius $R$ obeys $B{\nabla }_r^{4}h=p$, where ${\nabla }_r^2=r^{-1}{\partial }_r(r{\partial }_r)$. Its regular solution is $h=p(R^2-r^2{)}^2/(64B)$. Consequently its enclosed volume is $V=\pi p{R}^6/(192B)$, and its edge curvature is $h_{rr}(R)=p{R}^2/(8B)=24V/(\pi R^4)$.

A bending moment is an internal torque transmitted across a cut through a beam or elastic plate. For a slender linear elastic beam, $M=EI\kappa$ up to sign convention, where $E$ is Young's modulus, $I$ the second moment of area, and $\kappa$ the centreline curvature. The corresponding bending response of an elastic plate is described by its bending stiffness and curvature tensor.

The Brans-Dicke theory adds a dynamical scalar field $\Phi$ to a metric tensor theory of gravity, with constant dimensionless coupling $\omega$. In a $(+---)$ signature and the reversed-curvature convention used here, the displayed Lagrangian is supplemented by matter and integrated against the metric tensor volume density with prefactor $1/(16\pi G)$. Its scalar multiplier of the Ricci scalar lets the curvature coupling vary in spacetime. Both the metric tensor and scalar are varied; the scalar equation of Brans-Dicke theory follows by eliminating curvature between the scalar equation and the trace of the metric tensor equation. The original construction is described in Brans and Dicke's 1961 article.

A thin liquid lining inside a rigid cylindrical tube has pressure perturbation $p_1=-\sigma (a^{-2}{h}_1+h_{1,xx})$ at leading order in $h_0/a$. Lubrication theory gives flux $q=-h_0^3{p}_x/(3\mu )$, so mass conservation gives the displayed growth rate. Long waves with $0<k^2{a}^2<1$ grow, while short waves are stabilized by curvature. The maximum occurs at $k^2{a}^2=1/2$.

For a fixed spherical gravitating source, place a test particle at $(a,0)$ with initial velocity $(0,\sqrt{\mu /a})$, where $a$ is outside the source. Its exact circular orbit isDifferentiating twice gives $\ddot{\mathbf{r}}=-{\omega }^2\mathbf{r}=-\mu \mathbf{r}/a^3$, precisely the Newtonian gravitational field at each point of that circle. There is no outward physical force to cancel gravity in this inertial frame. Gravity supplies the inward centripetal acceleration that continually changes the velocity's direction.

Over a short interval, the displacement from the straight tangent path begins inward with magnitude $\frac{1}{2}(\mu /a^2)t^2$. The curvature of the orbit matters at the same order: the particle moves sideways while continually falling away from its instantaneous tangent. The full circular solution remains at radius $a$ rather than striking the source.

Its speed is $\sqrt{\mu /a}$, its oscillation period is $2\pi \sqrt{a^3/\mu }$, and its specific orbital energy is $-\mu /(2a)$. Removing gravity while keeping the initial velocity would give the straight path $(a,a\omega t)$, rather than a stationary particle. This contrast explains why orbital weightlessness is compatible with a substantial gravitational force: a freely coasting spacecraft and its contents fall together. ^[ref]

For a twice continuously differentiable regular curve $r(t)$ in Euclidean space, its unsigned curvature measures how quickly its unit tangent vector changes per unit arc length. WriteThe curvature iswhere the norm is the Euclidean norm. Differentiating with respect to arc length removes the effect of traversal speed; differentiating $T$ with respect to $t$ alone generally does not. See Ghomi's university lecture notes, section 1.7.

For a circle of radius $R>0$, use $r(t)=(R\cos t,R\sin t)$. Then $v=R$, $T=(-\sin t,\cos t)$ and $\Vert dT/dt\Vert =1$, giving $\kappa =1/R$. A straight line $r(t)=p+ta$ with $a\ne 0$ has constant unit tangent vector, so $\kappa =0$.

The same circle traversed as $q(u)=(R\cos (3u),R\sin (3u))$ has speed $3R$ and $\Vert dT/du\Vert =3$: its curvature remains $1/R$. By contrast, enlarging the circle to radius $2R$ changes its geometry and gives curvature $1/(2R)$. More generally, a dilation $r\mapsto ar$ with $a>0$ divides curvature by $a$, because it multiplies arc length by $a$ while leaving unit tangent vectors unchanged. Thus curvature has units of inverse length, and a reparametrization changes neither its value at a corresponding point nor the shape.

Differentiating $T\cdot T=1$, using the dot product, shows $T\cdot (dT/ds)=0$: the change is perpendicular to the unit tangent vector. Wherever $\kappa >0$, the principal normal vector $N_p=(dT/ds)/\kappa$ points toward the bend. For a plane curve, the osculating circle at such a point has radius of curvature $\rho =1/\kappa$ and center $r+\rho N_p$. It matches the point, tangent line, and curvature, providing a local circular approximation. At $\kappa =0$ there is no finite osculating circle defined by this formula. See Ghomi's university lecture notes, section 1.13.

This makes curvature useful for comparing bends without confusing them with speed. If $t$ denotes time, differentiating $r^{\prime }=vT$ gives acceleration $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$, whose component perpendicular to the unit tangent vector has magnitude $v^2\kappa$. The curvature of a plane curve example below computes the bend and osculating circle of a parabola explicitly.

For smooth surfaces, bending has more than one direction. The principal curvatures $k_1,k_2$ instead define Gaussian curvature $K=k_1{k}_2$ and mean curvature $H=(k_1+k_2)/2$. These are surface quantities, distinct from the curvature of an individual curve. In particular, the circle value $1/R$ must not be confused with the Gaussian curvature $1/R^2$ of a sphere of radius $R$. See Ritter's Oxford lecture notes, section 12.2.

For a twice continuously differentiable regular curve $r(t)=(x(t),y(t))$ in the plane, unsigned curvature is the absolute value of its signed curvature:The nonvanishing velocity vector makes the denominator positive. For a graph of a function $(t,f(t))$, this reduces to $|f^{\prime \prime }(t)|/(1+f^{\prime }(t{)}^2{)}^{3/2}$; omitting the absolute value gives the signed curvature of a plane graph for increasing $t$.

For the parabola $r(t)=(t,t^2)$, the derivatives are $r^{\prime }=(1,2t)$ and $r^{\prime \prime }=(0,2)$. ThereforeThe curvature is greatest at the vertex $t=0$, where $\kappa =2$ and the radius of curvature is $1/2$. At $t=\pm 1/2$ the curvature is $1/\sqrt{2}$, so the parabola bends less sharply despite its larger speed $\sqrt{2}$.

Its unit tangent vector and left unit normal areSince $k>0$, $N_L$ is also its principal normal vector. At $t=0$ the osculating circle consequently has center $(0,0)+(1/2)(0,1)=(0,1/2)$ and equationThe lower branch $y=1/2-\sqrt{1/4-x^2}$ has value $0$, derivative $0$, and second derivative $2$ at $x=0$, matching the parabola $y=x^2$ through second order. Away from the vertex the two curves separate: an osculating circle describes a local bend, not the whole curve.

Vanishing curvature at one point does not imply a straight segment. The graph of a function $y=x^3$ has signed curvature $6x/(1+9x^4{)}^{3/2}$, which vanishes at its inflection point $x=0$ and changes sign across it. By contrast, if curvature vanishes throughout a connected regular interval, $dT/ds=0$ throughout that interval: the constant unit tangent vector forces the image to lie on a straight line.

Definitions of the Riemann curvature tensor can differ by an overall sign, independent of the metric signature. Reversing that definition reverses the Ricci tensor, Ricci scalar and Einstein tensor, as well as the curvature side of the geodesic deviation identity and the Palatini identity. A derivation must choose the contraction and derivative ordering explicitly before combining formulas from different conventions. Changing a metric signature alone is a separate operation.

Differential geometry uses calculus to study smooth curves, smooth surfaces, and smooth manifolds. A tangent space describes possible instantaneous motions; a Riemannian metric turns those motions into measurable lengths and angles; curvature records how the resulting geometry differs from Euclidean geometry. Local calculations with derivatives can therefore answer geometric questions about whole shapes.

An especially revealing experiment is rolling a flat sheet into a circular cylinder without stretching it. The sheet visibly bends, yet a small creature measuring only arc lengths, angles, and surface areas detects the same geometry. This separates geometry measured within a surface from its bending in the surrounding Euclidean space. The first fundamental form encodes the measurements; the second fundamental form encodes the bending. The Theorema Egregium explains why a sphere behaves differently, while geodesics on a circular cylinder turn a route-planning problem into drawing a straight line.

The differential geometry of smooth surfaces studies their induced Riemannian metrics and intrinsic or extrinsic curvature. The first fundamental form measures lengths, while the second fundamental form and shape operator describe bending in an ambient Euclidean space. A regular parametric surface $X(u,v)$ must have linearly independent tangent vectors $X_u,X_v$: otherwise its parameters cannot provide a manifold chart with two independent directions.

For a sphere of radius $R>0$, use colatitude $0<\theta <\pi$ and longitude $\varphi$:Thus a change $d\varphi$ has arc length $R\sin \theta |d\varphi |$: equally spaced longitudes draw closer together toward a pole. The coordinate degeneracy at the poles belongs to this parametrization, not to the smooth surface.

Both principal curvatures of the sphere have magnitude $1/R$ and the same sign, giving Gaussian curvature $K=1/R^2$. A plane has $K=0$, and a circular cylinder also has $K=0$ because one principal curvature vanishes. These three familiar shapes show why visible bending alone cannot measure intrinsic Gaussian curvature.

With signature $(-,+,+,+)$ and vanishing cosmological constant, the Einstein field equations areIn four spacetime dimensions, contraction with $g^{\mu \nu }$ gives the useful trace relation $R=-\kappa T$.

In the usual commutator convention for the Riemann curvature tensor, $\kappa =8\pi G/c^4$: $G$ is the gravitational constant, whereas $G_{\mu \nu }$ is the Einstein tensor. The Ricci tensor $R_{\mu \nu }$ and scalar curvature $R$ encode contractions of curvature. The stress-energy tensor $T_{\mu \nu }$ includes energy density, momentum flow and stresses, so pressure also gravitates. A cosmological constant adds $\Lambda g_{\mu \nu }$ to the left-hand side. See Tong's Einstein equations and matter coupling.

For a concrete consequence, set $T_{\mu \nu }=0$ and $\Lambda =0$. The trace gives $R=0$, and substitution gives $R_{\mu \nu }=0$. Yet the exterior Schwarzschild spacetime still has the tidal forces computed under geodesic deviation: vanishing Ricci tensor does not imply vanishing Riemann curvature tensor. Geometry carries gravitational effects through empty space. Einstein presented the final matter equations on 25 November 1915, including the essential trace correction, in Die Feldgleichungen der Gravitation, pages 844-847, especially equation (2a). His symbols and sign conventions differ from modern ones; the displayed formula uses modern notation.

The planar equilibrium of an inextensible slender beam can have large tangent angle while retaining small elastic strain across its thickness. For constant horizontal compressive force $F$, moment balance and curvature ${\theta }^{\prime }=d\theta /ds$ give the displayed nonlinear equation. Multiplication by ${\theta }^{\prime }$ gives the first integral $EI{\theta }^{\prime 2}/2-F\cos \theta =C$. A tensile load has $F<0$ in this convention.

The Euler spiral is the planar curve parametrized by the Fresnel integrals. In the optical convention,The tangent has unit length and argument $\pi u^2/2$, so its signed curvature is $\pi u$, proportional to signed arc length. In Fresnel diffraction, an interval integral is the directed chord joining its endpoint parameters on the Cornu spiral. Chord length gives the interval factor's modulus, while chord direction gives its complex argument.

If physical ordinate is $Y=M^2$ and abscissa $X=H/M$, slope $S=dY/dX$ and horizontal intercept $X_0$ give $b={\mu }_0/S$ and $a={\mu }_0{X}_0$. Any displayed squared-magnetisation scale must be converted before using S. For example, an ordinate unit $(10^4\mathrm{A}{\mathrm{m}}^{-1}{)}^2$ equals $10^8{\mathrm{A}}^2{\mathrm{m}}^{-2}$, not $10^4$. With a free-energy density, the resulting units are $[a]=\mathrm{H}{\mathrm{m}}^{-1}$ and $[b]=\mathrm{J}\mathrm{m}{\mathrm{A}}^{-4}$. Curvature or differing isotherm slopes limits the accuracy of a constant-coefficient fit.

For positive slope parameter $A$, the exponential-potential field increases during cosmic inflation. The quantity $aH$ grows by $e^{(1-{ϵ}_H)N}$ with constant ${ϵ}_H=A^2/(16\pi )$ and $N=8\pi ({\varphi }_R-{\varphi }_i)/(A{m}_{\mathrm{pl}})$. Starting from order-one curvature and demanding order-one or smaller present curvature gives the displayed upper bound on the initial field. A specified tighter final curvature tolerance adds one-half its logarithm to the numerator. Thus sufficient cosmic inflation sets an upper, not a lower, initial-field bound for this sign of the potential slope. Instantaneous reheating and subsequent entropy conservation determine $a_R{H}_R/a_0{H}_0$.

Fracture toughness measures a material's resistance to crack advance. In a thin elastic plate model, the threshold for advance can be represented by a prescribed front curvature ${\kappa }_f$. Such a curvature parameter is not numerically interchangeable with a bulk fracture-toughness parameter without specifying the fracture model.

Rolling a sheet into a circular cylinder bends it without changing its internal lengths and angles. Fitting that same sheet smoothly onto a sphere requires distortion of its first fundamental form. Gaussian curvature distinguishes these situations: it measures the intrinsic curvature of a smooth surface, even though its simplest definition uses how the surface bends in Euclidean space.

The Gaussian curvature at a point is the determinant of the shape operator, or the product of the two principal curvatures:Here $I=Ed{u}^2+2Fdudv+Gd{v}^2$ is the first fundamental form and $II=ed{u}^2+2fdudv+gd{v}^2$ is the second fundamental form, with $S=-dN$. Regularity makes $EG-F^2>0$. Reversing the unit normal negates $e,f,g$ and both principal curvatures, leaving Gaussian curvature unchanged. ^[ref]

The sign describes bending at the point: positive Gaussian curvature means the principal curvatures have the same nonzero sign, as at an elliptic point; negative Gaussian curvature means opposite signs, as on a saddle; zero Gaussian curvature means at least one principal curvature vanishes. Zero Gaussian curvature at a single point does not imply that a neighborhood is a plane. ^[ref]

Four worked models make the signs concrete:

The patches use a common length unit, with $R=1$ for the sphere and circular cylinder. The marked red and blue curves have principal directions at the black point; the arrow is the chosen unit normal. For physical lengths, principal curvature and mean curvature have units of inverse length, while Gaussian curvature has units of inverse squared length. Rescaling a surface by $\lambda >0$ divides its principal curvatures by $\lambda$ and its Gaussian curvature by ${\lambda }^2$. Writing the saddle as $h=(x^2-y^2)/(2L)$ restores an explicit length scale: $K(0)=-1/L^2$.

The Theorema Egregium says that Gaussian curvature depends only on the first fundamental form and its first two coordinate derivatives. A local isometry therefore preserves Gaussian curvature, even if the shape operator and the principal curvatures change. ^[ref]

Here is the outline, rather than a full proof. The Gauss formula splits second derivatives of a parametric surface into tangential Christoffel symbols and the second fundamental form. The Christoffel symbols are determined by the first fundamental form and its first derivatives. Commuting the next partial derivatives gives the Gauss equation, which identifies $\det S$ with the intrinsic sectional curvature computed from the Riemann curvature tensor. Thus the apparently extrinsic product $k_1{k}_2$ is determined by the Riemannian metric. ^[ref]

One can see the distinction directly by rolling a plane strip with coordinates $(u,v)$ into the circular cylinder $X(u,v)=(R\cos (u/R),R\sin (u/R),v)$. Both have $I=d{u}^2+d{v}^2$, so the map is a local isometry, and is one-to-one on an open strip of width less than $2\pi R$. Extending $u$ over all real numbers wraps around periodically; it does not give a global isometry between the entire plane and the entire circular cylinder. A nonempty open patch of a sphere cannot be flattened by a local isometry into a plane, because its Gaussian curvature $1/R^2$ cannot become zero. The obstruction concerns preservation of the first fundamental form, rather than a particular drawing or choice of seam.