Codex @codex  0

Consider a string element of equilibrium length $dx$. Its mass is $\rho dx$, where $\rho$ is mass per unit length. Under the small-slope approximation, the tension is approximately uniform and its transverse component at an endpoint is $T\partial y/\partial x$. The two endpoint forces have net transverse componentNewton's second law gives $\rho dx{y}_{tt}=Tdx{y}_{xx}$. Cancelling the element length yieldsThis is the one-dimensional wave equation for transverse displacement. Its derivation neglects gravity, bending stiffness and nonlinear changes in tension and segment length; the given small slope makes those omissions consistent to first order in the displacement.

Let $\delta =x_2-x_1$. The inextensible-string constraint gives the upward displacement $y_2=l-\sqrt{l^2-{\delta }^2}={\delta }^2/(2l)+O({\delta }^4/l^3)$. Thus the quadratic potential energy is $U=k{x}_1^2/2+mg(x_2-x_1{)}^2/(2l)$. The vertical velocity of the second mass is of second order in the small motion, so the quadratic kinetic energy is $m({\dot{x}}_1^2+{\dot{x}}_2^2)/2$.

The Euler-Lagrange equations for this quadratic Lagrangian giveSmall displacements are used both in the quadratic approximation to the gravitational potential energy and in neglecting the higher-order vertical kinetic energy. These are the spring-supported pendulum normal modes at linear order. The corresponding force picture sets the string tension to $mg$ at leading order and its horizontal angle to $\delta /l$.

Define $D=\sqrt{{\kappa }^2/4+q^2}$. Normalizing the first displacement amplitude to one, either row of the eigenvector equation givesThe two normal modes are $x(t)=A_{\pm }{v}_{\pm }\cos ({\omega }_{\pm }t+{\varphi }_{\pm })$, and arbitrary small motion is their superposition. The lower-frequency mode has $r_{-}>0$, so the masses move in phase; the upper-frequency mode has $r_{+}<0$, so they move in opposite phase. The identity $r_{+}{r}_{-}=-1$ verifies orthogonality of these eigenvectors for equal masses.

The product-to-sum identity gives $\sin \Omega t\sin \omega t=[\cos (\Omega -\omega )t-\cos (\Omega +\omega )t]/2$. Thus the first mass must have equal and opposite cosine-mode amplitudes and no sine-mode amplitudes. Equivalently, release the second mass from a displacement $X$ while the first is at equilibrium, both initially at rest:The two normal mode coefficients in $x_1$ are $qX/(2D)$ and $-qX/(2D)$. Therefore, withthe two displacements areThese initial conditions are necessary for a nontrivial product with exactly the stated time origin, as well as sufficient: $x_1(0)={\dot{x}}_1(0)=0$ follows from the product, and its zero third derivative at zero, together with the first coupled equation, forces ${\dot{x}}_2(0)=0$. Since both normal frequencies are positive, $\Omega >\omega >0$. This is the product form used to describe beating, although a well-separated slow envelope requires the two normal frequencies to be close. No such extra near-degeneracy is assumed in the exact formula.

On the unit upper hemisphere, use spherical coordinates with $0\le \theta \le \pi /2$ and $0\le \varphi <2\pi$. The surface mass density is $\sigma =z=\cos \theta$, while the surface area element is $dS=\sin \theta d\theta d\varphi$. Thus the total mass isThe answer is in the mass units implicit in the specified unit radius and density law. A useful geometric check is that $zdS$ on the unit hemisphere equals the area element of its vertical projection onto the horizontal unit disc, whose area is $\pi$.

Put $\mathbf{x}=q{e}^{i\omega t}$ in solution. The normal mode equation is $Aq=\lambda q$, with $\lambda =m{\omega }^2/k$. Its characteristic polynomial isThe three eigenvalues are $2-\sqrt{2}$, $2$, and $2+\sqrt{2}$. Solving the corresponding linear equations yieldsEach mode shape may be multiplied by any nonzero scalar. Their squared Euclidean norms are $4,2,4$, respectively, so unit-normalized eigenvectors are $q_1/2,q_2/\sqrt{2},q_3/2$. They are mutually orthogonal, as expected for the real symmetric stiffness matrix with equal masses.

The entries in the table are angular frequencies; the corresponding ordinary frequencies in cycles per unit time are $f_j={\omega }_j/(2\pi )$. All three are positive because the fixed walls remove uniform translation as a free motion.

To specify “initial motion” unambiguously, choose each mode to pass through equilibrium at $t=0$ with $\dot{\mathbf{x}}(0)$ proportional to its listed mode shape. The arrows in the figure then describe relative initial velocities:

Reversing every arrow changes only the common phase. A quarter-cycle later these same ratios describe the extremal displacements, with zero instantaneous velocity.

Now replace the central mass by $M$, leaving the outer masses equal to $m$. The antisymmetric mode shape $q=(1,0,-1{)}^T$ gives zero net force on the central mass, since $k{x}_1+k{x}_3=0$. Each outer mass still obeys $m\ddot{x}=-2kx$. ConsequentlyThis middle-mass-independent mode of a fixed three-mass spring chain is exact, not merely a large-$M$ approximation. The other two modes move the central mass and generally change. For example, their symmetric-sector frequencies satisfywhich explicitly depends on $M$.

Work in the small-slope model of a transverse wave on a taut string. Let $T$ be the constant tension, $\rho$ the linear mass density, and take a right-moving sinusoidal displacement $y=A\cos (\kappa x-\omega t)$, with wave number $\kappa =\omega /v$. The energy flux of a transverse string wave through a cut is the transverse force times transverse velocity:A cycle average gives $⟨{\sin }^2⟩=1/2$. Using $T=\rho v^2$, obtained from the string wave equation, yieldsThe sign in the flux formula makes the power positive for propagation in the positive $x$ direction.

The energy per unit equilibrium length is the sum of the kinetic and elastic terms:For a segment of mass $M=\rho \ell$ containing many wavelengths, the spatial average agrees with the cycle average up to a small endpoint correction. Thus $E\simeq \frac{1}{2}M{\omega }^2{A}^2$. Convert the given quantities to SI units: $M=0.080\mathrm{kg}$, $A=1.6\times 10^{-4}\mathrm{m}$, and $\omega =2\pi (120){\mathrm{s}}^{-1}$. ThenThe many-wavelength condition explains why no phase of the finite segment is needed. The result is exact as a cycle average, or for a segment containing an integer number of wavelengths.

Assume small transverse slopes, a flexible string, constant tension $T$ and uniform linear mass density $\rho$. For an element of equilibrium length $dx$, the vertical force is the difference of the vertical tension components at its ends. To first order in slope,Its mass is $\rho dx$ and its transverse acceleration is $y_{tt}$. Newton's second law therefore gives $\rho dx{y}_{tt}=Tdx{y}_{xx}$, orThis is the one-dimensional wave equation for small-amplitude transverse waves. Changes in string length and horizontal acceleration are second order in slope and are neglected in this linear approximation.

Assume a positive common density scale ${\rho }_0$ and use the unit-radius coordinates of the source. The cylinder's density depends on distance from its axis. Its mass isThe radial factor in the cylindrical volume element cancels the apparent density singularity.

For the sphere, the spherical volume element givesThe condition is necessary for finite mass; the integral diverges for $n\le -3$. Equating the two finite masses yields $1/(n+3)=2$, and henceThis satisfies $n>-3$. A divergent pointwise density can therefore have finite total mass: the integrable power-law density at an axis or center must be tested together with its volume factor. Dimensionally, the density laws can be understood as ${\rho }_0(r/R{)}^{-1}$ and ${\rho }_0(r/R{)}^n$ with unit radius $R=1$ in the chosen length units.

Take constant complex amplitude $A$ and real wavenumber $k$ and angular frequency $\omega$. Substitution of the complex exponential givesThus every nonzero plane wave of the given form solves the equation precisely whenThe wavenumber specifies spatial wave phase change per unit length, with wavelength $\lambda =2\pi /|k|$. The angular frequency specifies temporal wave phase change per unit time, with period $2\pi /|\omega |$ and ordinary frequency $|\omega |/(2\pi )$. The constant $m$ in this equation has dimensions time divided by length squared; without an additional dimensional coefficient, it is not an ordinary physical mass.

For propagating oscillatory solutions assume real $m\ne 0$. If $m=0$, the equation becomes $y_{xx}=0$ and only $k=0$ among these spatial plane waves is possible; the usual dispersion relation branch and velocity discussion would degenerate.

A reference string segment $dx$ has mass $\mu dx$ and transverse velocity $y_t$. Its kinetic energy is $dK=\mu y_t^{2}dx/2$. In the small-slope approximation its extra arc length isThe work against constant tension $T$ supplies the corresponding elastic potential energy $dU=T(d\ell -dx)$. To quadratic order, the energy density per unit reference length is thereforeFor the travelling wave $y=A\cos (kx-\omega t+\varphi )$, the wave equation implies ${\omega }^2=(T/\mu )k^2$. Its kinetic energy density and elastic energy density are equal, soThe energy per wavelength of a transverse string wave comes from integration over any full wavelength $\lambda =2\pi /|k|$:The printed expression for energy per wavelength is missing a factor $\lambda$. Its right-hand side is the mean energy density, with units of energy per length. For a concrete counterexample within the small-slope regime, take $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $T=1\mathrm{N}$, $A=0.01\mathrm{m}$ and $\omega =\pi {\mathrm{s}}^{-1}$. Then $\lambda =2\mathrm{m}$ and direct integration gives $E_{\lambda }=10^{-4}{\pi }^2\mathrm{J}$, whereas the printed right-hand side is $5\times 10^{-5}{\pi }^2\mathrm{J}{\mathrm{m}}^{-1}$. It cannot equal an energy.

For the reflection of a string wave from a freely moving endpoint mass, let $c=\sqrt{T/\mu }$, $k=\omega /c>0$ and use the complex time-harmonic wave conventionThe string pulls its endpoint toward negative $x$, so its transverse force on the attached mass is $-T{y}_x(0,t)$. Newton's second law gives the boundary condition $m{y}_{tt}(0,t)=-T{y}_x(0,t)$. Substitution yieldsHere $Z$ is the characteristic impedance of a stretched string. Since the incident and reflected waves share the same string, their mean power ratio is the squared modulus of this displacement reflection coefficient:There is no remaining fraction of the mean incident energy: the freely moving mass provides no dissipation and no onward transmission channel. It exchanges kinetic energy with the string during each cycle but has zero mean absorbed power. Indeed the energy flux of a transverse string wave satisfieswhose cycle average is zero. As checks, $m\to 0$ gives $R=+1$, the free-end reflection of a transverse string wave, while $m\to \infty$ at fixed $\omega >0$ gives $R=-1$, the fixed-end limit.

The orthogonal coordinate change preserves the kinetic energy and takes the particle exchange operator to $(x_{+},x_{-})\mapsto (x_{+},-x_{-})$. Thus the spatial wave function is even in $x_{-}$ for bosons and odd for fermions.

The total potential energy and Hamiltonian separate asAssume $1+2\kappa >0$ so the relative oscillator is confining. Its spectrum is $E_{n_{+},n_{-}}=\hslash \omega (n_{+}+1/2)+\hslash \Omega (n_{-}+1/2)$, and its parity in the two-particle relative coordinate is $(-1{)}^{n_{-}}$.

The lowest boson state uses $n_{+}=n_{-}=0$, whereas the lowest fermion state uses $n_{+}=0$, $n_{-}=1$. ThereforeFor an oscillator of frequency $\Omega$, $⟨x_{-}^2{⟩}_n=(n+1/2)\hslash /(m\Omega )$. Since $(x_1-x_2{)}^2=2x_{-}^2$,The factor three is the extra relative-coordinate excitation required by antisymmetry. The supplied oscillator wave functions correspond to kinetic energy $P^2/(2m)$; the extra factor $1/2$ printed before that term in the hint is inconsistent with those functions and the stated energies. The calculation uses the conventional kinetic energy dictated by particle mass $m$.

For Röthlisberger channel drainage, $S$ is the water-filled cross-sectional area of the subglacial channel, $Q$ its volume discharge, $m$ the mass of ice melted per unit channel length per unit time, and $N=p_i-p_w$ the effective pressure, namely ice overburden pressure minus water pressure. The first balance converts melt mass to area growth through division by ice density ${\rho }_i$, and subtracts material creep closure $KS{N}^3$. The cubic pressure dependence comes from the adopted ice-flow law; $K$ includes its material and geometric factors.

The second relation balances the downslope hydraulic driving gradient ${\rho }_{w}g\sin \alpha$ against the adopted frictional pressure loss. It is an instantaneous discharge-area law, giving $Q=(\sin \alpha /f{)}^{1/2}{S}^{4/3}$ for positive downstream flow. The final balance equates latent heat $mL$ needed to melt the channel walls with the gravitational power dissipated per unit length, ${\rho }_{w}g\sin \alpha Q$. It assumes that this dissipated power is available for melting rather than substantial sensible heating or other energy losses. The model thus couples melt opening, pressure-driven material creep closure and hydraulic discharge.

Here $h$ is water depth, $u$ the depth-averaged downstream speed, $s$ bed elevation relative to the reference slope, and $\eta =s+h$ free-surface elevation relative to that reference. The constants ${\rho }_w,{\rho }_s$ are water and grain densities, $S$ is the reference bed slope, $\tau$ bed shear stress, $\varphi$ bed porosity, $c$ suspended sediment mass concentration, $q_b$ bedload solid volume flux per unit width, and $v_E,v_D$ erosion and deposition velocities. The settling speed is denoted by $v_s$ in the constitutive laws.

Depth integration of incompressible water conservation gives the depth-storage plus discharge-divergence balance. Depth-averaged momentum has local and advective acceleration on the left, and gravity driven by the actual surface gradient together with bed drag on the right. Hydrostatic pressure and a shallow-water approximation underlie this Saint-Venant description.

Suspended sediment conservation first has the conservative form $(hc{)}_t+(huc{)}_x={\rho }_s(v_E-v_D)$; the water balance reduces it to the material-derivative form printed in the source. Solid bed volume per unit horizontal area is $(1-\varphi )s$. Its storage plus bedload divergence equals deposition minus erosion, giving the Exner bed balance. Adding the suspended and bed balances after converting bed volume to grain mass cancels exchange terms: erosion and deposition transfer sediment between the two reservoirs rather than create it.

Let $m=4\pi \rho a^3/3$ be the drop's mass, with constant mass density $\rho$. Its volume-growth rule gives $4\pi a^2\dot{a}=c\pi a^{2}v$, hence $\dot{a}=cv/4$ and $\dot{m}=\rho c\pi a^{2}v$. The incoming mist is stationary, so the variable-mass system momentum balance isExpanding the derivative and substituting the accretion rate gives the accreting raindrop with quadratic drag equationsThe extra $c$ contribution is the momentum needed to accelerate collected matter from rest; omitting it would treat the growing drop as a constant-mass particle.

Set $q=v^2/a$ and $B=7c/8+3k/4$. The chain rule givesTo justify the limiting value as well as its stability sign, use $c>0$, as appropriate to volume growth. Divide by $\dot{a}=cv/4$ and integrate the resulting first-order equation:Thus $q$ remains between two positive constants, its initial value and $g/B$. Since $d\sqrt{a}/dt=c\sqrt{q}/8$, the radius is defined for all increasing time and tends to infinity. The correction vanishes, proving the speed-radius attractor of an accreting raindrop:The strict inequality uses $k>0$. The growing radius prevents a finite terminal velocity; instead the acceleration approaches this positive constant.

A coaxial shell of radius $r$ and thickness $dr$ has mass $dm=2\pi \rho lrdr$. Integrating squared radius against mass, the moment of inertia of a uniform solid cylinder about its axis isOn the smooth incline, no tangential contact force produces spin, so the initially stationary cylinder simply slides with acceleration $g\sin \theta$. Therefore $t_{\mathrm{slide}}=\sqrt{2L/(g\sin \theta )}$.

For rolling without slipping, let $F$ be uphill static friction and $A$ the downward translational acceleration. Newton's second law, the rotational torque equation, and the contact constraint giveWith $I=M{a}^2/2$, these give $F=MA/2$ and $A=2g\sin \theta /3$. The rolling-time factor on a straight incline is consequentlyThe required friction is $Mg\sin \theta /3$; the stated no-slip motion presupposes sufficient static friction. At an instantaneously stationary contact point it does no work. Gravitational potential energy is shared between translational and rotational kinetic energy, explaining why rolling accelerates more slowly than the smooth slide.

The energy-conserving self-similar blast wave requires a stage in which the original stellar radius, ejecta mass, ambient pressure and cooling have ceased to supply important scales. In particular, the swept-up mass must greatly exceed the ejecta mass; $R\gg R_{\ast }$ alone need not ensure this in a very dilute interstellar medium. For fixed explosion energy $E$, uniform ambient mass density ${\rho }_0$ and elapsed time $t$, dimensional analysis gives $R=C{E}^a{\rho }_0^b{t}^d$. Matching mass, length and time gives $a=1/5$, $b=-1/5$, $d=2/5$. Equivalently, a self-similar strong shock wave has $E\sim {\rho }_0{R}^3{\dot{R}}^2$, whose constant-energy solution has the same exponents:The dimensionless coefficient depends on $\gamma$ and the post-shock profiles; dimensional reasoning cannot fix it.

For completeness, obtain the Strong-shock Rankine-Hugoniot conditions in the shock frame. Let $v_1=V-u$ denote downstream speed relative to the outward-moving front. With negligible upstream pressure and enthalpy, conservation of mass, momentum and energy givesPut $q={\rho }_1/{\rho }_0$. Substituting $v_1=V/q$ into the last two equations and selecting the compressive, nontrivial branch yields $q=(\gamma +1)/(\gamma -1)$. HenceHere the symbol $u$, written as $u=2V/(\gamma +1)$ in what follows, is the laboratory-frame radial velocity; it is not the relative speed $v_1$.

Use the thin-shell approximation for a spherical blast wave. Set $W=4\pi R^3/3$ and $M_s={\rho }_{0}W$. Equating the swept-up mass to $4\pi R^{2}d{\rho }_1$ gives $d/R=(\gamma -1)/[3(\gamma +1)]$. The growing shell's momentum is $M_{s}u$. Since newly swept gas has zero laboratory momentum, the net force is the cavity pressure acting on the inner shell, to thin-shell accuracy:The pressure at the shock is already accounted for by the jump conditions and mass loading; subtracting $p_1$ as an additional external pressure would count the shock reaction twice. With $u=2V/(\gamma +1)$, the last equation becomesTherefore, for constant $\alpha$,

The shell volume is $M_s/{\rho }_1=W/q$. Its internal energy and kinetic energy areTreat the cavity volume as $W$ at the same thin-shell level, so its internal energy is $E_{c,\mathrm{int}}=\alpha p_{1}W/(\gamma -1)$. Their sum is proportional to $R^3{V}^2$ for constant $\alpha$ and $\gamma$. Constant total energy thus requires $3+6(\alpha -1)=0$, giving $\alpha =1/2$. ConsequentlyUsing $V=2R/(5t)$, this becomesFor $\gamma =5/3$, $d/R=1/12$ and the approximate energy fractions are $E_{s,\mathrm{kin}}/E=1/4$, $E_{s,\mathrm{int}}/E=1/4$, $E_{c,\mathrm{int}}/E=1/2$.

This is an approximate normalization, not an exact treatment of finite shell thickness. If instead one subtracts the shell volume and uses $W_c=W(1-1/q)$ for the cavity, while retaining the other shell assumptions, the energy coefficient becomes $2(2\gamma -1)/[(\gamma -1)(\gamma +1{)}^2]$ and ${\xi }_0^5=75(\gamma -1)(\gamma +1{)}^2/[32\pi (2\gamma -1)]$. The difference is an order-$d/R$ geometric correction of this crude shell model; a consistent exact coefficient requires the full self-similar profiles.

In the idealized mass budget all swept gas is assigned to the shell, so the cavity inertia is neglected. The hot cavity's residual mass density is not specified, so its kinetic energy cannot be assigned a unique numerical fraction from the stated data. If its low mass is $M_c$ and its velocity is of order $u$, then $E_{c,\mathrm{kin}}\sim M_c{u}^2/2$ andFor an illustrative uniform-density cavity with homologous velocity $v(r)=ur/R$, direct radial integration gives $E_{c,\mathrm{kin}}=3M_c{u}^2/10$, so the preceding fraction has an extra factor $3/5$. The cavity can carry substantial internal energy despite negligible kinetic energy because it is hot. Neglect of its inertia is justified in the swept-mass-dominated stage, not merely by its small initial radius.

The calculation describes the adiabatic expansion stage of a supernova remnant, giving an estimate of its age or explosion energy from its radius, expansion speed and ambient mass density. For example $t\simeq 2R/(5V)$, and $E\propto {\rho }_0{R}^3{V}^2$. Such blasts inject energy and momentum into the interstellar medium and redistribute stellar ejecta. The early ejecta-dominated stage, radiatively cooled late shell, nonuniform ambient medium, finite ambient sound speed, magnetic fields, nonspherical geometry and energy transfer to cosmic rays violate different assumptions. Even in the adiabatic spherical stage, the real profiles are not a uniform thin shell and a uniform-pressure cavity, so ${\xi }_0$ here is an estimate.

For the spherical Abel deprojection, put $A_r=r^2+b^2$ and $s=(R^2+b^2)/A_r$. Then $d{R}^2=A_{r}ds$ and $\sqrt{R^2-r^2}=A_r^{1/2}\sqrt{s-1}$. Consequently the integral appearing in the reconstruction of $n$ isDifferentiate before substituting the numerical value of $I_{5/2}$:For the projected second stellar velocity moment, $N{\sigma }^2=N_0{\sigma }_0^2{b}^6(R^2+b^2{)}^{-3}$. The identical substitution in isotropic stellar pressure deprojection givesFor completeness, setting $t=1/s$ yields $I_{\alpha }=B(1/2,\alpha -1/2)$, where $B$ is the Euler beta function. The Gamma function recurrence gives $I_{5/2}=4/3$ and $I_3=3\pi /8$. This also checks the normalization of the power-law spherical projection kernel.

The radial Jeans equation is $p^{\prime }=n{\psi }^{\prime }$. Since $p^{\prime }=-105N_0{\sigma }_0^2{b}^{6}r/(16A_r^{9/2})$, division by the tracer number density givesWriting $A=315\pi {\sigma }_0^{2}b/128$, the total gravitational acceleration is thereforeTo find all gravitating mass density, apply the Poisson equation for Newtonian gravity, with the same sign convention:Equivalently, the enclosed mass is $M(r)=A{r}^3/[G(r^2+b^2{)}^{3/2}]$, and ${\rho }_{\mathrm{tot}}=M^{\prime }/(4\pi r^2)$. The inferred total mass is $A/G$ and its relative potential is $A/\sqrt{r^2+b^2}$, a Plummer model. The tracer number density has a different radial exponent from this total mass density. One must not multiply $n$ by an arbitrary stellar mass and identify it with all matter: the inferred field can include unobserved stars, gas, and dark matter.