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The dimensionless bacterial growth and substrate depletion system $b_t=b_{xx}+gb$, $g_t=-gb$ has travelling-wave equations $b\prime \prime +cb\prime +gb=0$, $cg\prime =gb$ in coordinate $y=x-ct$. For a front with $(b,g)\to (b_0,0)$ behind and $(b,g)\to (0,1)$ ahead, its first integral is $b\prime +c(b+g)=c$, forcing $b_0=1$. Positivity of the leading edge requires $c\ge 2$.

Kepler's laws describe three connected features of an elliptic orbit: its shape, how fast it is traversed, and how its size determines its orbital period. Kepler's first law gives an ellipse with the attracting centre at one focus of a conic; Kepler's second law gives equal swept areas in equal times; Kepler's third law relates the squared orbital period to the cubed semi-major axis. Together they turn a drawing of a Kepler orbit into a prediction of where the bodies will be and when they will return. ^[ref]

Here the mathematical model is an isolated two-body problem in Newtonian gravity, with point masses $M,m>0$ and a bound, noncolliding elliptic orbit. Let $\mathbf{r}={\mathbf{r}}_m-{\mathbf{r}}_M$ be the relative displacement, $r=|\mathbf{r}|$ the separation, and $\mu =G(M+m)$ the gravitational parameter, where $G$ is the gravitational constant. Newton's law of universal gravitation then givesThe relative Kepler orbit has semi-major axis $a$ and orbital eccentricity $0\le e<1$. Both bodies move around their shared astronomical barycenter, the system's centre of mass. In its rest frame their displacements are ${\mathbf{r}}_m=M\mathbf{r}/(M+m)$ and ${\mathbf{r}}_M=-m\mathbf{r}/(M+m)$. Their individual semi-major axes therefore add to $a$; the relative ellipse's attracting focus of a conic represents zero separation, rather than a body fixed in space.

One example will connect all three laws. After nondimensionalization, choose $a=1$, $e=0.6$ and $\mu =1$. The semi-minor axis is $b=a\sqrt{1-e^2}=0.8$, while the periapsis distance and apoapsis distance areThus $a=(0.4+1.6)/2=1$ measures the ellipse's size even though the separation changes fourfold. The specific angular momentum will be $0.8$, the two endpoint speeds $2$ and $0.5$, and the orbital period $2\pi$. These are values in the chosen scaled units, not measurements of a particular planet.

Johannes Kepler published Kepler's first law and Kepler's second law in Astronomia Nova in 1609 and Kepler's third law in Harmonices Mundi in 1619. These publication dates are documented by ^[ref]. Their appeal is that three short statements organize both geometry and timing; Newtonian gravity supplies a common dynamical explanation.

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

The Newtonian gravitational field is the gravitational force per unit mass of a test particle. For a point mass $M$ at the origin,The test particle's mass cancels: different small masses have the same gravitational acceleration at the same position. In SI, the Newtonian gravitational field has units $\mathrm{N}\mathrm{k}{\mathrm{g}}^{-1}=\mathrm{m}{\mathrm{s}}^{-2}$. The same exterior expression holds for a spherically symmetric source, by the spherical shell theorem.

Choose a reference radius $r_0>0$, set $g_0=GM/r_0^2$, and use nondimensionalization with $z=r/r_0$. Then $g/g_0=z^{-2}$: at $z=1,2,4$, the magnitudes are $1,1/4,1/16$. This makes the spatial variation visible without specifying a planet. ^[ref]

Use $K$ as the population scale, $N_0$ as the nutrient concentration scale, $r^{-1}$ as the time scale, and $L$ as the length scale. The nondimensionalization isFor example, the chemotaxis term becomesDividing the population equation by $rK$ and the nutrient equation by $r{N}_0$ gives the requested system with the following six dimensionless quantities:The rescaled domain is $0<x<1$, $t>0$. Its boundary conditions and initial conditions are

Use the three standard reduction assumptions: $m$ relaxes much faster than $V,n,h$ and can be replaced by $m_{\infty }(V)$; the two slower gates have the same time constant, ${\tau }_n(V)={\tau }_h(V)=\tau (V)$; and their limiting open fractions have a voltage-independent sum, $n_{\infty }(V)+h_{\infty }(V)=K$. The constant-sum assumption is essential: equality of two time constants by itself does not preserve a moving voltage-dependent sum.

For $q=n+h$, addition of the two gate ordinary differential equations givesThe specified initial condition is $q(0)=K$. Uniqueness for this first-order linear differential equation gives $q(t)=K$, hence $h=K-n$ for all time. The remaining two ordinary differential equations areThese assumptions give a planar reduction of the Hodgkin-Huxley model. The FitzHugh-Nagumo model keeps its fast activation and slow recovery variable geometry, approximating the voltage response by a cubic polynomial and the recovery response by a linear function after nondimensionalization. The polynomial equations in part (c) are a further model simplification; they are not an exact algebraic consequence of arbitrary original gate functions.

Choose positive reference concentrations $C,C_s$ and a reference flux $J_0$, and defineOverdots now mean derivatives with respect to the dimensionless time $\tau$. The store balance becomesThe total balance becomes $\dot{u}+(C_s/C)\dot{v}=r/(kC)-u$. ThereforeandThe reference quantities are scale choices, not additional kinetic variables; typically the response functions supply natural saturation concentrations and a maximal uptake flux. The identity $\delta /\epsilon =k_s/k$ checks the nondimensionalization. A small $\epsilon$ means store exchange is fast compared with external cytosolic removal.

The conserved total is $e_0=[E](0)$ under the chosen initial conditions. Substitute $E=e_0-X-Y-E_i$ into the equations from the previous part. This closes the subsystem for $S,X,Y,E_i$; product can be recovered afterward by integrating $\dot{P}=k_{3}Y$.

Let $k_{\Sigma }=k_{-1}+k_2$ and $K_m=k_{\Sigma }/k_1$. The required nondimensionalization is completed byIndeed $X=e_0\sigma x/(1+\sigma )$ and the free-enzyme fraction is $1-\sigma x/(1+\sigma )-y-e_i$. Substitution and division by the indicated time scale giveFor example $k_2/k_{\Sigma }=1/(1+\rho )$ explains the coefficient in the third equation. The small multiplier on the intermediate-species derivatives identifies their potentially fast adjustment relative to substrate depletion.

Choose the nondimensionalizationThe chain rule gives ${\partial }_{\tau }={\beta }_u{\partial }_t$ and ${\partial }_x^2=({\beta }_u/D_u){\partial }_s^2$. Dividing the first equation by ${\beta }_u$ and the second by ${\beta }_v$ gives the required scaled reaction–diffusion system. Positive degradation and diffusion constants are assumed.

Applying the law of mass action to the sequential two-site enzyme reaction network givesAdding the appropriate equations gives the conservation laws $e+c_1+c_2=e_0$ and $s+c_1+2c_2+p=s_0$. In particular $e=e_0(1-v_1-v_2)$ after nondimensionalization.

Define the dimensionless reaction rates $a_1=k_{-1}/(k_1{s}_0)$, $a_3=k_{-3}/(k_1{s}_0)$, $b_2=k_2/(k_1{s}_0)$, $b_4=k_4/(k_1{s}_0)$ and $r_3=k_3/k_1$. Substitution, with primes denoting $d/d\tau$, yieldsThese right sides are $f,g_1,g_2$, respectively, with $u(0)=1$ and $v_1(0)=v_2(0)=0$. When $ϵ\ll 1$, the form of these ordinary differential equations displays the faster relaxation of the complexes, which motivates a subsequent quasi-steady-state approximation.

Substitute $gb=c{g}^{\prime }$ into the bacterial profile equation. This giveswhich integrates once toAt the asymptotic constant states the profile derivative tends to zero. The state ahead of the travelling wave, where $b\to 0$ and $g\to 1$, therefore gives $C=c$. The state behind it, where $g\to 0$ and $b\to b_0$, gives $C=c{b}_0$. Since $c>0$,This first integral expresses the matching between biomass produced and biological growth factor consumed after the chosen nondimensionalization.

Define the aspect ratio and the thickness-based Reynolds number byUse the nondimensionalizationThe small transverse velocity scale follows from incompressibility. Dropping stars, the base fields become $\bar{u}=2z-z^2$ and $\bar{p}=(1-z)\cot \alpha$. Before subtracting the base state, the dimensionless Navier-Stokes equation readsFor example, division of the downslope momentum equation by $\mu U_0/h_0^2$ makes the inertial coefficient $\rho U_0{h}_0^2/(\mu L)=\mathrm{Re}\epsilon$ and the pressure gradient coefficient $2\epsilon$. The transverse equation has been divided instead by $\mu U_0/h_0^2$ after using its smaller velocity scale; this gives the displayed $\mathrm{Re}{\epsilon }^2$ coefficient.

Now substitute $u=\bar{u}+\hat{u}$, $w=\hat{w}$, $p=\bar{p}+\hat{p}$. Since ${\bar{u}}_{zz}+2=0$, ${\bar{p}}_z=-\cot \alpha$, and the base fields have no $x$ or $t$ dependence, subtraction givesThese are nonlinear equations for the deviations; small amplitude has not been assumed. The later asymptotic expansion takes $\mathrm{Re}=O(1)$ as $\epsilon \to 0$, or more generally requires the retained inertial correction to remain small.

Introduce the positive outlet constantFor a draining similarity profile, $J$ is finite and positive. A regular linear profile would have $f{f}^{\prime }\to 0$ and $(f{f}^{\prime }{)}^{\prime }\ge 0$ at the origin, incompatible with the rain-fed aquifer similarity solution $(f{f}^{\prime }{)}^{\prime }(0)=-1$. Instead there is a square-root drainage boundary. At leading order, $(f^2{)}^{\prime }=2f{f}^{\prime }\to 2J$, givingMore explicitly, the rain-fed aquifer similarity solution implies $(f{f}^{\prime }{)}^{\prime }=-1+O(\sqrt{\eta })$, hence $f{f}^{\prime }=J-\eta +O({\eta }^{3/2})$ and $f^2=2J\eta -{\eta }^2+O({\eta }^{5/2})$. The height vanishes but its derivative diverges; their product remains finite.

Use $t_d,x_d,h_d$ temporarily for dimensional quantities. The nondimensionalization in part (a) givesTaking the drainage flux to be positive out of the left-hand end, $Q_0=-q(0,t_d)$, givesThe printed expression in terms of $f(0)$ and $f^{\prime }(0)$ must be understood as this limit; a literal product of their separate endpoint values is undefined. This is discharge per unit transverse width. Very close to the river, the divergent slope marks the limit of the Dupuit approximation, while the outer solution still supplies the leading discharge.

Physics investigates matter, motion, and interactions by building mathematical models and testing their consequences against observations. A physical law supplies a relation; a model specifies the objects, conditions, and approximations used to apply it. Classical mechanics, electromagnetism, thermodynamics, and quantum mechanics organize different aspects of these investigations. Their usefulness comes from turning measurements into predictions that can be checked and refined.

Consider a simple pendulum: how does its oscillation period change when its length changes? For small amplitude, the ideal model gives $T_0=2\pi \sqrt{\ell /g}$, where $g$ is gravitational acceleration. Taking $g=9.81m/{\mathrm{s}}^2$ as the model parameter, a one-metre pendulum has predicted period $2.0061\mathrm{s}$. Increasing its length to four metres gives $4.0121\mathrm{s}$: quadrupling the length doubles the period. The point mass cancels from the equation, so changing it alone does not change this model's answer.

The prediction identifies a mechanism and a way to challenge it. The small-angle approximation replaces $\sin \theta$ by $\theta$; the full ideal equation predicts a longer period at large amplitude. At $60^{\circ }$, its period is about $1.07318T_0$, already a $7.3\%$ change. Measurements at such amplitudes can reveal the limit of the approximation, while appreciable drag requires a damped pendulum. These distinctions are developed in simple pendulum and dimensional analysis. The small-amplitude model is also derived in Ling, Sanny and Moebs, University Physics, pendulums.

Several different experiments can become one calculation. Nondimensionalization rewrites the pendulum equation using $\tau =t\sqrt{g/\ell }$, leaving $d^2\theta /d{\tau }^2+\sin \theta =0$. All lengths share that dimensionless equation; the amplitude remains a parameter. Similarly, conservation laws and symmetry reveal features that many models share. This is one source of the subject's beauty: a compact relation can organize a large range of phenomena.

The connection to evidence remains essential when a model is elegant. In his introductory lectures, Richard Feynman emphasizes testing proposed relations by experiment and using idealizations whose accuracy depends on the conditions. ^[ref] A discrepancy can motivate a better measurement, a better approximation, or a different physical model; the mathematical calculation and the empirical assessment are both part of the explanation.

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.