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Knot theory studies closed curves in three-dimensional Euclidean space that cannot intersect themselves, and asks which can be deformed into one another without cutting or passing one strand through another. Its basic object is a knot: a smooth embedding of a circle, considered up to ambient isotopy. Several disjoint knots form a link. More generally, knot theory studies such embeddings in other manifolds and in higher dimensions.

The central surprise is that a flexible circle can carry permanent information about how it sits in space. A trefoil knot and an unknot each consist of just one circle, but no ambient isotopy turns one into the other. A knot diagram translates this spatial question into a finite drawing; Reidemeister moves describe the changes of drawing that preserve the knot. Knot invariants translate it into computations: unequal Alexander polynomials of a knot, Jones polynomials, or knot groups certify different knots. Equal values of a particular knot invariant generally leave the question open; even an isomorphism of knot groups alone is not a complete classification.

An early foundation for working with knot diagrams is Kurt Reidemeister's 1927 paper Elementare Begründung der Knotentheorie, published in volume 5 of Abhandlungen aus dem Mathematischen Seminar der Universität Hamburg, pages 24–32.^[ref] Marc Lackenby's authored notes Elementary knot theory, Section 2.4, explain the Reidemeister move theorem and its role in searching for a certificate that two knot diagrams represent the same knot.^[ref]

For $|z|>1$, both logarithms in the display can use their principal logarithms. The real part is $\log (|z+1{|}^2/|z-1{|}^2)$ and the derivative is $2/(z+1)-2/(z-1)$. The two residues cancel on a loop surrounding both points. Equivalently $f=4{\sum }_{m\ge 0}{z}^{-(2m+1)}/(2m+1)$ is globally single-valued outside the unit circle, although individual logarithms of $z+1$ and $z-1$ are not.

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Fix a holomorphic function on a neighbourhood of a closed disk, with $f(0)\ne 0$. On an interval of positive radii whose circles contain no zeros of a function, differentiating Jensen's formula with respect to $x=\log R$ givesEach interior zero of a function contributes slope one for each unit of its multiplicity of a zero. Between consecutive zero moduli, $L_f(e^x)$ is an affine function; passing the modulus of a zero of multiplicity of a zero $m$ increases the adjacent slopes by $m$. The finite sum $\log |f(0)|+{\sum }_j\max (0,x-\log |a_j|)$ defines a continuous convex function of $x$ whose values agree with the circle log-mean of a holomorphic function at every boundary-zero-free radius. This describes the adjacent line segments without asserting a boundary-singularity extension of the integral.

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

A lunar phase is the appearance of the sunlit portion of the Moon as seen from Earth. Outside a lunar eclipse, the distant Sun illuminates approximately one hemisphere of the Moon, but an observer sees a changing overlap of that hemisphere with the hemisphere facing Earth. The lunar phase therefore changes even though synchronous rotation keeps approximately the same near side of the Moon facing Earth. ^[ref].

The cycle runs through new Moon, waxing crescent, first quarter, waxing gibbous, full Moon, waning gibbous, last quarter and waning crescent. “Waxing” means that the lunar illuminated fraction is increasing; “waning” means it is decreasing. A crescent has less than half its apparent disk sunlit, and a gibbous lunar phase has more than half. First and last quarter each show half a disk: “quarter” refers to progress around the cycle, rather than a quarter of the disk being lit. ^[ref].

An instructive circular orbit model puts Earth at the origin, the Sun far away along $+x$, and the Moon at $R(\cos \theta ,\sin \theta )$. The sunward position $x>0$, $\theta =0$, gives new Moon; the opposite position $x<0$, $\theta =\pi$, gives full Moon. Counterclockwise motion gives first quarter at $\theta =\pi /2$ and last quarter at $\theta =3\pi /2$. The Sun's direction is fixed during this snapshot of the cycle. Every small globe in the orbital view is illuminated on its $+x$ half; the adjacent apparent disks are separate views from Earth, rather than globes viewed from above.

The disk views choose north up, with the upward direction perpendicular to the orbital plane. In this convention waxing lunar phases are bright on the right and waning lunar phases on the left. Local horizon orientation and viewing from different locations on Earth can rotate the apparent disk; its lit edge always points toward the projected direction of the Sun. The lunar illuminated fraction is unaffected by this rotation.

The useful angle is the phase angle $\alpha =\angle SME$, with $S$ the Sun, $M$ the Moon and $E$ the observer at Earth's center: its vertex is at the Moon. Full Moon has $\alpha =0$, quarter has $\alpha =\pi /2$, and new Moon has $\alpha =\pi$ in the ideal model. By contrast, the elongation $ϵ=\angle SEM$ has its vertex at Earth. For a distant Sun, its rays are parallel, soHere $k$ is the lunar illuminated fraction: a fraction of projected disk area, not a fraction of the Moon's total surface area or its received radiant flux relative to full Moon. The phase angle and unsigned elongation each lie between $0$ and $\pi$, so either alone gives the illuminated amount without distinguishing waxing from waning. The orbital direction resolves that ambiguity.

The lunar illuminated fraction has an elegant geometric derivation. Model the Moon as a sphere of unit radius and view it along the $z$ direction. Take the Sun's unit direction to be $(\sin \alpha ,0,\cos \alpha )$. A visible surface point above apparent disk coordinates $(x,y)$ has $z=\sqrt{1-x^2-y^2}$. Direct illumination requires the outward normal's dot product with the sunward direction to be positive. The day-night terminator therefore obeys $x\sin \alpha +z\cos \alpha =0$. At a fixed $y$, set $q=\sqrt{1-y^2}$; the visible day-night terminator is $x_t=-q\cos \alpha$, and the illuminated strip extends from $x_t$ to the right edge $q$. Integrating its width givesThe integral is the area of half a unit circle. Thus a three-dimensional visibility problem reduces to a circle's area and a cosine. The derivation assumes parallel sunlight, a distant observer and a sphere, and ignores a lunar eclipse. Real observed brightness additionally depends on surface scattering of light and observing geometry, so $k=1/2$ does not promise half the full-Moon radiant flux.

For a worked example, elongation $ϵ=60^{\circ }$ gives phase angle $\alpha =120^{\circ }$ and $k=(1+\cos 120^{\circ })/2=1/4$: a crescent with $25\%$ of its apparent disk sunlit. At elongation $120^{\circ }$, the phase angle is $60^{\circ }$ and $k=3/4$: a gibbous lunar phase. At $90^{\circ }$, $k=1/2$. In each case the same $k$ occurs once while waxing and once while waning in the ideal cycle.

Ordinary lunar phases are not caused by Earth's shadow. A lunar eclipse occurs only when the Moon actually enters that shadow near full Moon. The orbital inclination of the Moon is about $5^{\circ }$ relative to Earth's orbital plane around the Sun; most full Moons pass above or below the shadow. This is why full Moon does not imply a lunar eclipse every month. ^[ref].

The phase cycle is a synodic month, about $29.5$ days, whereas the Moon's orbital period relative to distant stars is a sidereal month, about $27.3$ days. During one sidereal month, Earth advances around the Sun, so the Moon must move farther to repeat the sunward alignment. For coplanar circular orbits traversed in the same direction, subtract their mean motions: $n_{\mathrm{syn}}=n_{\mathrm{sid}}-n_{\mathrm{year}}$, with $n=2\pi /P$. HenceUsing $P_{\mathrm{sid}}=27.321661\mathrm{days}$ and $P_{\mathrm{year}}=365.256\mathrm{days}$ gives $P_{\mathrm{syn}}=29.530591\mathrm{days}$, consistent with the mean synodic month of $29.53059$ days. After one sidereal month, the sunward direction has advanced about $26.93^{\circ }$; catching up takes about $2.21$ more days. These are mean-period estimates: orbital eccentricity and changing geometry make individual synodic months vary. ^[ref]. ^[ref].

The lunar phase is also a practical clock for observing. Full Moon is opposite the Sun and rises roughly at sunset; first quarter is about a right angle from the Sun and is convenient in the evening, while last quarter is convenient before dawn. These are rough guides, not exact rise-time predictions for every place and season. ^[ref]. Historically, repeated lunar phases provided the month in lunar calendars, while half-lit disks gave Aristarchus of Samos a way to infer astronomical distances. ^[ref].

For a nonempty compact topological manifold $X^d$ without boundary, its suspension is a topological manifold without boundary exactly when $X$ is homeomorphic to $S^d$. The forward implication is a local homology from a link calculation at a suspension vertex: ${\tilde{H}}_j(X)$ is $\mathbb{Z}$ in degree $d$ and zero elsewhere. For $d\ge 2$, put a coordinate ball between two nested cone neighbourhoods. The inclusion of their punctures is a homotopy equivalence but factors through the simply connected punctured ball, forcing ${\pi }_1(X)=0$. Thus $X$ is a homotopy sphere, and the topological generalized Poincare theorem gives the result. The dimensions zero and one follow from the finite-set and circle classifications. Conversely $S(S^d)\cong S^{d+1}$. If manifolds with boundary are allowed, the interval and its disk suspension show why this formulation needs the boundary restriction.

The mapping torus of $f:F\to F$ isIf $f$ is a homeomorphism, the projection to the circle $S^1$ is a fiber bundle with fiber $F$. For a general continuous map, the mapping torus need not be a fiber bundle, but its homology still has the Wang sequence.

For two nonempty planar circles with centers $c_i$ and radii $r_i\ge 0$, the greatest Euclidean distance between their points is the displayed expression. The triangle inequality gives the upper bound. When the centers differ, choose the two boundary points outward along the line of centers to attain it; when they coincide, choose opposite radial directions. For any third nonempty circle and a point $u$ on it, $|z-w|\le |z-u|+|u-w|$ proves $m(C_1,C_2)\le m(C_1,C_3)+m(C_2,C_3)$. This is not a metric, since $m(C,C)=2r$ for a positive-radius circle.

For a closed connected orientable surface of genus $g$, the Künneth theorem gives the Betti numbers of $S^1\times {\Sigma }_g$ as $(1,2g+1,2g+1,1)$. The weak Morse inequalities give the displayed lower bound. A surface Morse function with one minimum, $2g$ saddles and one maximum exists from its standard handle decomposition. Adding a positive multiple of the cosine Morse function on the circle makes a product of Morse functions, whose critical counts are precisely these four Betti numbers. The lower bound is attained.

The Möbius band is the topological surface obtained by gluing the ends of a rectangle with a half-turn. Precisely, it is the quotient spaceThe twist identifies the two apparent long edges into one boundary circle. Following the core circle of a Möbius band reverses local handedness, so this surface is not orientable. An untwisted untwisted band has two boundary circles and is orientable: the distinction survives stretching and bending because orientability is a topological invariant.

To explore the construction with paper, mark the two long edges in different colours, join the short ends after a half-turn, and trace an edge. The trace passes from one marked edge to the other without ever leaving the boundary. Separately, trace the centre line and track a local transverse arrow: it comes back reversed after one circuit. The boundary closes after twice the travel around the centre, while the transverse arrow needs two circuits to return to its original direction.

This original mathematical model has central radius $R=2$ and half-width $a=0.55$:Its seam obeys $F(2\pi ,w)=F(0,-w)$. The dotted circle is the core circle of a Möbius band; the teal curve follows the single boundary through $0\le \theta \le 4\pi$. At the same point on the core, the arrows for $\theta =0$ and $\theta =2\pi$ are opposite. The graph compares a continuously followed unit normal with its starting value in the ambient space; the endpoint reversal, rather than any intermediate zero of the graph, establishes the obstruction. It is not a statement about parallel transport.

The Möbius band has the homotopy type of a circle: a deformation retraction shrinks each width coordinate to zero. Nevertheless its boundary represents twice a generator of its fundamental group. The boundary of a Möbius band winds twice around its core, so with compatible generator choices the boundary generator maps to $c^2$ in ${\pi }_1(M)\cong \mathbb{Z}$.

The historical account by Julyan H. E. Cartwright and Diego L. González places the mathematical discovery by August Möbius and Johann Listing in 1858, while investigating representations of twisted bands predating that work. A familiar-looking picture and a recognition of the underlying topological invariant are different kinds of evidence. ^[ref]

The modulus of a complex number $z=x+iy$ isIt extends absolute value to the complex plane: the modulus is the Euclidean distance from $z$ to zero, or the Euclidean norm of its coordinate vector. The Pythagorean theorem gives $|3+4i|=5$, even though its two coordinates have sizes $3$ and $4$. The modulus is always nonnegative, and is zero exactly at zero.

Two essential rules areThe second is the triangle inequality. For $z=3+4i$ and $w=1+i$, the product has modulus $\sqrt{50}=5\sqrt{2}\approx 7.071068$, while the sum has modulus $\sqrt{41}\approx 6.403124<5+\sqrt{2}\approx 6.414214$. The strict inequality records that the two displacement vectors point in different directions.

The Euclidean distance between two complex numbers $z,w$ is $|z-w|$. Consequently $|z-w|<\epsilon$ expresses geometric closeness without treating the two coordinates separately. This makes the modulus a natural measure of error in complex analysis; the triangle inequality bounds accumulated error, while the product rule tracks how multiplication changes its size. Equal moduli do not imply equal complex numbers: $5$, $-5$, and $3+4i$ all lie on the same circle around zero. The modulus records size and deliberately forgets direction; the argument supplies the missing information.

On the punctured plane, this one-form is closed and locally the differential of the polar angle. Its line integral around a closed curve is $2\pi$ times the winding number. A once-winding circle therefore has nonzero integral, proving that no global single-valued angle potential exists and the form is not an exact differential form.

For a counterclockwise circle of radius $R>0$, substitute $x=R\cos t$, $y=R\sin t$. The pullback of a differential form isThe radius cancels. In coordinates, $P=-y/(x^2+y^2)$ and $Q=x/(x^2+y^2)$ satisfy ${\partial }_{x}Q={\partial }_{y}P$ away from the origin, so $d\omega =0$ there. The nonzero integral is a global obstruction to being an exact differential form, not a nonzero local exterior derivative.

Compare this with the smooth one-form $\eta =(-ydx+xdy)/2$ on the whole plane. Here $d\eta =dx\wedge dy$, and its integral around a radius-$R$ circle is $\pi R^2$, the enclosed area. The Generalized Stokes theorem applies to the full disc for $\eta$. For $\omega$, that disc crosses the excluded origin, where the one-form is undefined. On an annulus the theorem applies, but its induced inner boundary is clockwise: the two integrals $2\pi$ and $-2\pi$ cancel. This explains exactly why closedness and nonzero circulation can coexist.

Model a finite solenoid as a continuum of coaxial turns shaped as circles of radius $a>0$, uniformly distributed over length $\ell$ with solenoid turn density $n=N/\ell$. Put its centre at $z=0$, so its ends are $z=\pm \ell /2$. With the same electric current orientation as in the long-solenoid approximation, its axial magnetic field isThis formula applies only on the axis, where symmetry cancels the transverse magnetic field. It does not describe points off the axis. It is exact for the continuum circular-turn model, which neglects winding thickness, discrete wire spacing, and helical advance. ^[ref].

The derivation is an especially direct use of the superposition principle. The on-axis magnetic field of a circular current loop at source position $z^{\prime }$ is ${\mu }_{0}I{a}^2/[2(a^2+(z-z^{\prime }{)}^2{)}^{3/2}]$. There are $nd{z}^{\prime }$ turns in an axial slice, soSubstitute $u=z-z^{\prime }$ and use the antiderivative $u/[a^2\sqrt{a^2+u^2}]$ to obtain the displayed expression. If the centre is instead at $z=z_0$, replace every $z$ in that expression by $z-z_0$.

At the centre and at either end, respectively,For aspect ratios $\ell /a=2,6,20$, the centre ratios are approximately $0.7071,0.9487,0.9950$. The end ratio tends to $1/2$ as $\ell /a$ becomes large, even though the centre ratio tends to $1$. Thus a long coil can have an excellent central magnetic field while the ends still differ strongly from that value. Convergence to the long-solenoid approximation is not uniform across the end region.

For a concrete geometry, give the earlier coil radius $a=0.025\mathrm{m}$, so $\ell /a=20$. Its central magnetic field is approximately $1.2504\mathrm{mT}$, its end value is $0.6275\mathrm{mT}$, and at $z=\ell$ its axial magnetic field is still $0.00277\mathrm{mT}$. The last point is outside the coil: finite length does not permit the zero-exterior conclusion of the infinite model. Off the axis, the magnetic field lines bend around to form the return field. A central axial value alone also does not determine the total magnetic flux through every turn of a short coil.