Riemann's differential equation is a second-order ordinary differential equation with three regular singular points, normalized to zero, one and infinity by a Möbius transformation. Its local characteristic exponents are encoded by the Papperitz symbol and constrained by the Fuchs relation.

A mesa is a localized broad plateau joined to its surrounding state by two separated fronts. Its leading spatial average equals the plateau fraction times the plateau value; the transition layers give corrections proportional to their width divided by the domain length. A conserved mean can determine the fraction occupied by the plateau, as in the coexistence fraction of a conserved-field amplitude mesa.

A stationary front solution of $u_t=u_{xx}+f(u)$ connecting equilibria $u_{-}$ and $u_{+}$ requires ${\int }_{u_{-}}^{u_{+}}f(s)ds=0$. Multiply the stationary equation by $u_x$ and integrate: the derivative term has zero endpoint contribution. This equal-potential condition is the Maxwell balance. For $f(u)=\mu u+\alpha u^3-u^5$, a front from zero to a nonzero constant requires $u_{+}^2=3\alpha /4$ and $\mu =-3{\alpha }^2/16$.

For a FRW metric with positive scale factor, nonnegative dust and radiation, and negative vacuum density, the Friedmann acceleration equation gives $a_{tt}\le -{\omega }^{2}a$, with ${\omega }^2=8\pi G|{\rho }_{\Lambda }|/3$. At maximum expansion, $a(0)=a_{\max }$ and $a_t(0)=0$. The forced-oscillator integral formula gives $a(t)\le a_{\max }\cos (\omega t)$ until the first zero, since its forcing is nonpositive and its sine kernel is nonnegative for $0\le t\le \pi /(2\omega )$. Thus each half of the expansion-recollapse history lasts at most $\pi /(2\omega )$. Nonzero dust or radiation makes the endpoint a physical density singularity; the matter-free limit instead bounds the open anti-de Sitter coordinate patch.

In the ideal kinetic friction model, the loss of combined mechanical energy of the two contacting bodies becomes thermal internal energy at rateThis follows by adding the contact powers $f_{k}v$ and $-f_{k}U$ from Newton's third law. It depends on contact slip, whereas the contact power received by either body separately depends on its velocity in the chosen inertial frame. Thus positive frictional heating is compatible with increasing kinetic energy of one body, as the moving-belt example in kinetic friction shows. Heating need not cause a temperature rise if thermal energy escapes by heat transfer.

Contact slip is the tangential relative velocity between the two material points instantaneously in contact. For a nonrotating box on a belt it is $w=v-U$; for a rigid body, use the velocity of its contact point, including the contribution from angular velocity. Zero contact slip defines rolling without slipping at a rolling contact, even when the center of mass moves.

Kinetic friction is friction during nonzero contact slip. For one-dimensional motion along a surface, choose right as positive and write $v$ for the body's contact-point velocity and $U$ for the surface's contact-point velocity, both measured in the same inertial frame. The signed relative velocity is $w=v-U$. The ideal Coulomb law for the tangential force on the body isHere $N\ge 0$ is the magnitude of the normal force, ${\mu }_k\ge 0$ is the coefficient of kinetic friction, and $\mathrm{sgn}$ is the sign function. Thus kinetic friction opposes contact slip, which can differ from the body's direction of motion. In several tangential directions the same law reads ${\mathbf{f}}_k=-{\mu }_{k}N\mathbf{w}/|\mathbf{w}|$ for a nonzero tangential relative velocity $\mathbf{w}$. The magnitude law is documented in Ling, Sanny and Moebs, OpenStax University Physics 1, §6.2: static and kinetic friction.

For a stipulated toy model, take a nonrotating box with mass $m=1\mathrm{kg}$, normal force $N=10\mathrm{N}$, coefficient of static friction ${\mu }_s=0.5$, and coefficient of kinetic friction ${\mu }_k=0.3$. These are assumed model parameters, rather than measured properties of any particular pair of materials; the prescribed normal force is consistent with horizontal support under $g=10m/{\mathrm{s}}^2$ and no other vertical forces. On stationary ground, a box already sliding right has $f_k=-3\mathrm{N}$. With an applied horizontal force $F=6\mathrm{N}$ to the right, Newton's second law gives $a=(F+f_k)/m=3m/{\mathrm{s}}^2$, for as long as rightward contact slip persists.

Direction is clearer by comparing two instantaneous states of this same model. If the box has $v=1m/s$ on stationary ground, $U=0$ and $w=+1m/s$, so kinetic friction acts left with magnitude $3\mathrm{N}$. If the box still has $v=1m/s$ but a belt beneath it has $U=2m/s$, then $w=-1m/s$: kinetic friction acts right with magnitude $3\mathrm{N}$, even while the box moves right.

In the belt case, the power delivered to the box is $P_{\mathrm{box}}=f_{k}v=+3\mathrm{W}$. With no other horizontal force, its kinetic energy therefore increases by the work-energy theorem. By Newton's third law, the box exerts $-f_k$ on the belt, so the contact power delivered to the belt is $P_{\mathrm{belt}}=-f_{k}U=-6\mathrm{W}$. Their sum is $-3\mathrm{W}$: frictional heating produces thermal internal energy at $3\mathrm{W}$. An ideal motor maintaining the prescribed belt velocity supplies $+6\mathrm{W}$ against this contact load, neglecting other losses, so its input divides into $3\mathrm{W}$ of box kinetic energy and $3\mathrm{W}$ of frictional heating. These are instantaneous rates, since the box's velocity subsequently changes. This balance follows directly from power as force times velocity and conservation of energy; the power law is given in Ling, Sanny and Moebs, OpenStax University Physics 1, §7.4: power.

When contact slip vanishes, including a transition to rolling without slipping, the nonzero-slip law no longer applies. Determine the required static friction from the no-slip condition and the equations of motion, then test its bound. Substituting $\mathrm{sgn}(0)=0$ into the sliding law would incorrectly exclude a nonzero static friction.

In the unscaled convention, the finite Fresnel integrals are $C(\lambda )={\int }_0^{\lambda }\cos (t^2)dt$ and $S(\lambda )={\int }_0^{\lambda }\sin (t^2)dt$. Their Taylor series follow by termwise integration. At infinity both tend to $\sqrt{\pi /8}$, and their oscillatory tails have asymptotic expansions obtained by integration by parts. This convention differs from the convention with phase $\pi t^2/2$.

This empirical sea ice growth relation gives thickness $H$ in metres from accumulated freezing degree-days $N$. Its coefficient includes the specified units and calibration. It is not a complete energy-balance model: snow, ocean heat and changing forcing can alter its applicability. Inversion gives $N=(H/0.0133{)}^{1/0.58}$.

Suppose a group homomorphism $\pi :G\to Q$ is injective on a subgroup freely generated by $x_1,\dots ,x_n$. If $y_i$ have $\pi (y_i)=\pi (x_i)$, then the $y_i$ freely generate their subgroup: a relation among them projects to a relation among the free basis images, so the corresponding abstract reduced word must be empty. In particular, this applies to $y_i=x_{i}g$ when the normal closure of $g$ intersects $⟨x_1,\dots ,x_n⟩$ trivially.

If $a,b$ freely generate, then $a^{i}b{a}^{-i}$ for any distinct integer indices freely generate their subgroup. Group a reduced word in these new generators into blocks with nonzero exponents and different successive indices. Expansion produces alternating nonzero powers of $b$ and interior powers $a^{i_{j+1}-i_j}$, so it cannot freely reduce to the empty word. Consequently every finite-rank free group embeds in a rank-two free group; free rank is not monotone under subgroup inclusion.

In $F(a,b)$, any set of distinct conjugates $b^{-i}a{b}^i$ freely generates its subgroup. Combine adjacent powers of the same conjugate in a reduced word. After expansion, distinct successive indices leave nonzero intervening powers of $b$, and the intervening nonzero powers of $a$ cannot cancel in the free product $⟨a⟩\ast ⟨b⟩$. The same argument works with $a,b$ interchanged and for finite or countably infinite sets of indices.

An endomorphism of $F_n$ has image generated by at most $n$ elements. If this image has finite index $d$, the Nielsen–Schreier formula gives rank $1+d(n-1)$. Its abelianization requires at least that many generators. For $n\ge 2$, comparison with $n$ forces $d=1$.

// Target: geometry-and-topology.bigb

Let a test particle start at rest at radius $r_0$ outside a fixed spherically symmetric source. The spherical shell theorem gives the exterior Newtonian gravitational potential $\Phi (r)=-\mu /r$, where $\mu =GM$ is the source's standard gravitational parameter. Zero initial transverse velocity makes the motion radial. Conservation of mechanical energy per unit mass givesThe minus sign selects inward motion; the positive square root is its speed. For a fall from $r_0=2R$ to the surface $r=R$, that speed is $\sqrt{\mu /R}$. In the limit of release from rest infinitely far away, the familiar free-fall speed is $\sqrt{2\mu /r}$, equal in magnitude to the local escape velocity. A finite starting radius gives a smaller speed.

The arrival time follows from the same first integral. Set $r=r_0{\cos }^2\eta$. ThenFor arrival at $R>0$ this is finite, and the field formula is used only outside the source. For $r_0=2R$, $\eta =\pi /4$ givesKeeping the smaller initial gravitational magnitude $g(r_0)=\mu /(4R^2)$ constant would predict $\sqrt{8}\sqrt{R^3/\mu }$, about $10.0\%$ longer. The actual acceleration increases during the fall.

The curves compare a shallow fall, $h/R=0.02$, with a fall from $r_0=2R$, so $h/R=1$. Each comparison holds the release height and initial gravitational magnitude fixed. The constant-gravity model is close for the shallow fall and increasingly inaccurate for the larger one. This is also a practical use of nondimensionalization: the normalized trajectories depend on $h/R$, while $\sqrt{R^3/\mu }$ supplies the dimensional time scale.

Take downward displacement $s$ as positive, assume a constant gravitational acceleration $g>0$, and release a point mass from rest at $s=0$. Integrating $\ddot{s}=g$ twice givesFor a downward distance $h>0$, the arrival time and speed areFor the model values $h=20\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, these are approximately $2.0193\mathrm{s}$ and $19.8091\mathrm{m}{\mathrm{s}}^{-1}$. Doubling the height multiplies both quantities by $\sqrt{2}$; it does not double them. The same speed follows from the work-energy theorem, $mgh=m{v}^2/2$.

For intervals of length $\Delta t$, the successive downward displacements areThus equal spacing in time produces increasingly separated positions. This is the odd-number pattern in Galileo's result. It assumes constant acceleration and release from rest; a nonzero initial velocity adds a linear term to $s(t)$.

The approximation can be estimated for a spherical source of radius $R$ and standard gravitational parameter $\mu$. Writing $g_0=\mu /R^2$, the gravitational magnitude at height $h$ isFor $h/R\ll 1$, a Taylor expansion gives $g(R+h)/g_0=1-2h/R+O((h/R{)}^2)$. Small height relative to the source's radius makes nearly constant gravity plausible; negligible drag force remains a separate requirement.

For the insulated gas-only free expansion of an ideal gas, the equilibrium endpoints are $(T,V_1)$ and $(T,V_2)$ with $V_2>V_1$. Because entropy is a state function, calculate their difference along a hypothetical reversible thermodynamic process: an isothermal expansion using a frictionless piston and a thermal reservoir at $T$. This comparison process receives heat and delivers work, even though the actual free expansion does neither. Its reversible heat increment is $\delta Q_{\mathrm{rev}}=pdV=nRTdV/V$, soHere $R$ is the molar gas constant, and the logarithm acts on a dimensionless ratio. The reversible reference gives the correct endpoint entropy difference because entropy is a state function; putting the actual $Q=0$ into $\Delta S=Q/T$ would misuse a relation requiring reversible heat transfer. For the actual isolated gas, all this increase is entropy production. Thus an adiabatic process can increase entropy and need not be an isentropic process. ^[ref]

In the figure's volume doubling, the ideal gas law gives $p_1=99.768\mathrm{kPa}$ and $p_2=49.884\mathrm{kPa}$, while $\Delta S=R\ln 2\simeq 5.763\mathrm{J}{\mathrm{K}}^{-1}$. The reference process receives and delivers $nRT\ln 2\simeq 1728.85\mathrm{J}$ as heat and work; its thermal reservoir loses the same entropy that the gas gains. The actual free expansion has $Q_{\mathrm{in}}=W_{\mathrm{out}}=0$ and produces that entropy internally. Reversibly compressing the gas back at $300\mathrm{K}$ would require $1728.85\mathrm{J}$ of work input and rejection of equal heat. Restoring the gas alone therefore changes the surroundings: the actual release is irreversible.

For a real gas the final temperature may differ, so the ideal-gas logarithm formula cannot simply be copied. Evaluate $S(T_2,V_2)-S(T_1,V_1)$ using that gas's equilibrium properties or a suitable reversible comparison path. If the walls and water exchange heat with it, the Second law of thermodynamics constrains the total entropy of the isolated apparatus; the gas alone is then not an isolated system.

In his paper published in 1845, Joule connected two copper receivers with stopcocks, filled one with compressed dry air, and evacuated the other with an air pump. He immersed the connected receivers in water in an insulated arrangement, stirred the water, and read a sensitive thermometer. He then opened the stopcocks, let the air redistribute, stirred again, and compared the water's temperature with its initial value. Alternating control trials accounted for effects such as stirring and evaporation. The observation was no detectable net cooling of the common bath, rather than a record of the gas temperature at every instant. ^{[ref] [ref]}

Joule also immersed the receivers and their connecting piece in separate water vessels. The source receiver cooled while the receiving side warmed, with nearly cancelling thermal effects. ^[ref]

This distinguishes local transient changes from the final observation for the combined apparatus. Insulating its outer boundary suppresses external heat exchange; it does not prevent heat exchange between the gas, the receivers and the water inside that boundary.

An idealized modern version can use two rigid chambers joined by a valve, with thermometers in the chambers and on the walls. Prepare one chamber with gas and the other evacuated, disconnect the pump, close the apparatus to matter exchange, and let the initial temperature settle. Record initial and final equilibrium pressure and temperature, along with the wall response. Repeated closed-valve runs can reveal drift and valve-actuation effects; independent heating calibrations can estimate the heat capacity of the participating apparatus. This is a proposed measurement protocol, not a claim of a particular measured result. Fast local readings can characterize the release, but they cannot replace the final equilibrium readings in the endpoint internal energy balance.

A subgroup $L\subseteq M$ of a finitely generated free abelian group is saturated if $am\in L$ for a nonzero integer $a$ implies $m\in L$. Equivalently $M/L$ is torsion-free. It is then free abelian and the quotient map splits, so $L$ is a direct summand. Intersecting $M$ with a real linear subspace always gives a saturated sublattice.

// Target: algebra.bigb

Single-slit diffraction is the spreading of light transmitted by one slit. Consider an ideal rectangular slit of width $a$, illuminated normally by a uniform plane wave with a single wavelength $\lambda$ in the surrounding medium. This monochromatic wave has uniform wave amplitude and equal incident wave phases across the slit, providing the required optical coherence. In the scalar Fraunhofer diffraction model, the normalized angular wave intensity isHere $\theta$ is measured from the forward normal in the plane across the slit width. The sinc function uses the unnormalized convention $\mathrm{sinc}u=\sin u/u$. The limit makes the central value one. This ideal envelope neglects additional direction-dependent propagation and detector factors; near the forward direction it describes the normalized screen irradiance. ^[ref].

The noncentral zeros satisfyThe restriction follows from the range of the sine of a real angle. The first minima have $\sin \theta =\pm \lambda /a$ when $\lambda /a\le 1$. If $a<\lambda$, this model has no noncentral zero in the forward hemisphere; equality puts the first zeros at grazing directions, outside the small-angle approximation and off any finite perpendicular screen. ^[ref].

For $0<\lambda /a<1$, define the central width as the separation between the two first minima. Its angular width and its width on a perpendicular screen areThe angular approximation uses radians and requires $\lambda /a\ll 1$; the screen approximation uses $y\approx L\theta$. The tangent relation is exact straight-line geometry, even though the underlying Fraunhofer diffraction field is approximate. This width is distinct from the full width at half maximum.

For $\lambda =500\mathrm{nm}$, $a=0.1\mathrm{mm}$ and $L=1\mathrm{m}$, the positive first minimum isThe dimensionless far-field parameter is $a^2/(\lambda L)=0.02$. Doubling the slit width to $a=0.2\mathrm{mm}$ gives $W\approx 5.00002\mathrm{mm}$ and parameter $0.08$; both curves below use the Fraunhofer diffraction approximation. Choosing $L=2\mathrm{m}$ halves those parameters and doubles both screen widths.

Each curve is normalized to its own $I_0$; the arrows measure between first minima. The calculation uses $\theta =\arctan (y/L)$ before evaluating the sinc function, rather than replacing $\sin \theta$ by $y/L$. The lower diagram shows the exact screen geometry with the vertical scale enlarged. Normalization removes the change in peak brightness when the slit width changes.

The wave interference explanation is especially simple: divide the slit into equal narrow strips and add their complex amplitudes, rather than their wave intensities. With wavenumber $k=2\pi /\lambda$, the continuous slit givesIntegrating the complex exponential gives the endpoint difference; its continuous value at zero is one. Taking the squared complex modulus gives the stated wave intensity. This is the Fourier transform of an interval indicator expressed in angular coordinates. Feynman's continuous-source argument obtains the same pattern by letting the number of sources tend to infinity. ^[ref].

At the positive first minimum, the wave phases span exactly $2\pi$ across the slit. Every strip in one half has a partner $a/2$ away, with equal wave amplitude and phase difference $\pi$, so the pairs cancel. At $\theta =0$, all the wave phases align instead. ^[ref].

The drawing uses $N=8$ equally weighted midpoint samples $x_j/a=-1/2+(j+1/2)/8$, with $j=0,\dots ,7$. At the first minimum their complex amplitudes are a common rotation of the eight roots of unity, so their head-to-tail polygon closes exactly. This finite sum illustrates cancellation; the continuous integral, rather than eight point sources, defines the slit pattern at general angles. The circle shows the continuous range of wave phases; coloured arrows show the eight samples.

A narrower aperture requires a larger angle to accumulate the cancelling phase difference, so narrowing the slit widens the beam. This is a useful limit on geometrical optics: confining light more tightly at an opening increases its subsequent angular spread. At fixed incident irradiance and slit height, reducing $a$ also reduces transmitted power, exposing a practical balance between diffraction-limited resolution and light throughput. In a diffraction grating, this single-slit diffraction envelope modulates the wave interference between different slits.

Let $m=p^a$ and let an $r$-uniform set family have no distinct pair whose intersection is congruent to $r$ modulo $m$. Its size is bounded as displayed; the customary nontrivial form $\left(\genfrac{}{}{0ex}{}{n}{m-1}\right)$ requires $m-1\le r$. The Lucas theorem makes $\left(\genfrac{}{}{0ex}{}{z-r+m-1}{m-1}\right)$ a residue detector modulo $p$, equal to one exactly when $z\equiv r(\mathrm{mod}m)$. Its Boolean evaluation functions have degree at most $m-1$ and a diagonal evaluation matrix on the family. The low-degree evaluation rank on a uniform layer supplies the sharp dimension bound. Without the size qualification, three singletons with $n=3,r=1,m=4$ contradict the customary printed form, since $3>\left(\genfrac{}{}{0ex}{}{3}{3}\right)$.