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The two real branches start at $\omega =\pm {\omega }_{\mathrm{p}e}$, have zero initial slope, and approach the lines $\omega =\pm k{V}_e$ from outside the populated parallel-velocity region. The original plot solves the degenerate-plasma dielectric response numerically; the shaded wedge shows where particles with resonant parallel velocities exist.

The distinction from a nondegenerate gas with a Maxwellian Maxwell-Boltzmann velocity distribution is its tail. A Maxwellian has resonant particles at every finite phase velocity. Its long-wavelength Langmuir waves have exponentially weak Landau damping, but at shorter wavelengths that damping becomes strong. The ideal degenerate cutoff instead permits this undamped branch to persist continuously into degenerate-plasma zero-point sound. The coefficient $3V_e^2/5$ in c is also different from the classical thermal dispersive coefficient.

A natural additional branch is a degenerate ion-acoustic wave. For singly charged ions with equal electron and ion number densities, degenerate electron pressure supplies a restoring force while ion mass supplies inertia. The electron equation of state has $d{P}_e/d{n}_e=m_e{V}_e^2/3$, suggesting, at long wavelengths,Both species are degenerate here, so $V_i=(m_e/m_i)V_e$. The hierarchy $V_i\ll c_s\ll V_e$ explains the damping: there are no resonant ions, but there are resonant electrons inside their parallel marginal distribution. That electron marginal distribution has a small nonzero slope at $u=c_s$, even though the full three-dimensional occupation is constant inside the Fermi sphere. One therefore expects weak electron Landau damping, of relative order $\sqrt{m_e/m_i}$, rather than exact absence of damping. Ion degeneracy pressure adds a relatively small correction to $c_s^2$. This reasoning concerns an electrostatic ion-acoustic wave; transverse or magnetized branches require additional field dynamics or a background magnetic field.

Since $L_z=J_{\varphi }$, Hamilton's equations giveFor the spherical unperturbed Hamiltonian, $H_0=H_0(J_r,L)$; in these particular action-angle variables the third component ${\Omega }_{\varphi }=0$ describes the fixed orientation of the orbital plane. It should not be confused with the physical azimuthal angular velocity of a particle around its orbit.

The conserved normalization of a probability distribution $\int f{d}^3\theta d^{3}J=1$ makes the expected value of specific angular momentum equal to $\int L_{z}f{d}^3\theta d^{3}J$. Its time derivative is the torque per unit halo mass:This follows either by averaging the single-particle equation or by integrating the Collisionless Boltzmann equation against $L_z$, with vanishing boundary fluxes.

The term $f_0{\partial }_{{\theta }_{\varphi }}\delta \Phi$ integrates to zero over the periodic angle variable. Thus the first nonzero torque is quadratic in the perturbation. Use the Fourier series expansions of b; the source's reference here to part (a) is a numbering slip. Integrating the product over all three angle variables yieldsReality of the potential energy per unit mass gives $\delta {\Phi }_{-\mathbf{k}}=\delta {\Phi }_{\mathbf{k}}^{\ast }$; likewise $\delta f_{-\mathbf{k}}=\delta f_{\mathbf{k}}^{\ast }$. Conjugate harmonics make the total torque real. The sum is over all integer vectors, so no extra factor of two should be inserted.

Apply the chain rule to $f_0=f_0(E)$ with $E=H_0(\mathbf{J})$:On the resonance selected by the Dirac delta function, $\mathbf{k}\cdot \mathbf{\Omega }=k_{\varphi }{\Omega }_{\mathrm{p}}$. ThusEvery weight is nonnegative, and $-f_0^{\prime }(E)>0$. Strict positivity additionally requires a populated nonaxisymmetric resonance with nonzero coupling on a set of positive surface measure. The unconditional statement is nonnegativity, rather than positive definiteness. As a concrete counterexample to strict positivity of this formula, retain one nonzero conjugate harmonic pair with $k_{\varphi }\ne 0$ and choose its smooth coefficient ${\Psi }_{\mathbf{k}}$ supported in an open region $|{\Delta }_{\mathbf{k}}|>d>0$. The perturbation is nonzero but has zero resonant coupling, so $\mathcal{T}=0$ even when $f_0^{\prime }(E)<0$. The even simpler zero-perturbation case also gives zero. Whether a particular physical stellar bar has nonzero resonant overlap is additional information.

For a coupled stellar bar and halo, the halo gains positive angular momentum and exerts the opposite torque on the stellar bar. If the halo mass is $M_h$ and the rigid stellar bar has constant positive moment of inertia $I_b$, thenThis bar-halo angular-momentum transfer is a form of dynamical friction: a resonantly coupled stellar bar tends to slow down while the dark matter halo spins up. Actual long-term evolution changes both the stellar bar pattern and the halo distribution function, so the fixed-pattern response must then be updated. If the stellar bar changes its moment of inertia, the general relation is $d(I_b{\Omega }_{\mathrm{p}})/dt=-M_h\mathcal{T}$; positive moment of inertia alone would not fix the sign of $d{\Omega }_{\mathrm{p}}/dt$ without that rigidity assumption.

Put $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$ and $d{z}_i=d^3{x}_i{d}^3{v}_i$. Normalize the symmetric $N$-particle phase-space distribution function $D_N$ to one. Use factorial normalization for its one-particle marginal and two-particle distribution function:Thus $\int fdz=N$ and $\int f_2(z_1,z_2)d{z}_2=(N-1)f(z_1)$. For unit particle mass, $f$ also carries mass density. These conventions explain why no extra factor $N-1$ occurs in the result.

Separate the Hamiltonian as $H=H_1+H_{\mathrm{rest}}+H_{\mathrm{int}}$, whereand $H_{\mathrm{rest}}$ contains all terms involving only particles $2,\dots ,N$. The stated Poisson bracket gives Hamilton's equationsConservation of probability along the Hamiltonian flow gives the Hamiltonian Liouville equationAssume decay of the required velocity moments and either decay or periodic boundary conditions in position. Integrating over $z_2,\dots ,z_N$ eliminates the $H_{\mathrm{rest}}$ terms by integration by parts. The terms in $H_{\mathrm{int}}$ differentiating an integrated velocity also vanish. Indistinguishability makes the remaining $N-1$ terms equal, soRelabeling the two particles gives the first BBGKY hierarchy equation:This is exact for a differentiable pair potential; no independent-particle closure has been made.

Spatial homogeneity makes $\rho$ independent of position and makes the two-particle number density depend only on the separation. Isotropy removes its directional dependence. ConsequentlyHere $g_2$ is a radial distribution function, with the finite-$N$ normalization inherited from a; statistical independence is unnecessary. In particular, the translated pair in c has separation $\mathbf{R}$ for every $s$, so its $s$ integral is simply ${\rho }^2{g}_2(R,t)$.

Isotropy of the random velocity $\mathbf{c}=\mathbf{v}-\mathbf{u}$ makes its second velocity moment proportional to the identity. Taking its trace fixes the proportionality constant:Thus $\theta$ is the one-component random velocity variance; for unit mass in thermal equilibrium, $\theta =k_{B}T$. A Maxwell-Boltzmann velocity distribution is not required for the pressure tensor to be isotropic.

For the interaction contribution, the angular integral is $\int d\Omega {\hat{R}}_i{\hat{R}}_j=(4\pi /3){\delta }_{ij}$, by isotropy and its trace. With $d^{3}R=R^{2}dRd\Omega$, c becomesCombining the two pressure tensors gives the virial pressure from pair correlations:For a repulsive pair potential, the outward radial force is $-{\varphi }^{\prime }(R)\ge 0$. Since $g_2\ge 0$, the second term in $p$ is nonnegative, and positive when occupied separations feel nonzero repulsion. It expresses additional momentum transfer through interactions. The displayed integral assumes a differentiable pair potential and convergence; a hard core requires a contact formula or a limiting argument.

For an integrable or approximately integrable background Hamiltonian, action-angle variables turn an otherwise complicated stellar orbit into $\dot{\mathbf{J}}=0$ and $\dot{\mathbf{\theta }}=\mathbf{\Omega }(\mathbf{J})$. This has two useful consequences. First, a phase-mixed equilibrium can be represented directly as $f_0(\mathbf{J})$: the action variables label the unperturbed orbits, while angle variables describe position along them. Second, the angles are periodic, so Fourier series reduce unperturbed transport to multiplication by $i\mathbf{n}\cdot \mathbf{\Omega }$ and make orbital resonances explicit. The action variables also separate rapid orbital motion from slow secular changes. These advantages apply on regular orbital regions; a globally chaotic background need not admit global action-angle variables.

The transformation is canonical, preserving the phase space measure and the stated total mass $M=\int f{d}^3\theta d^{3}J$. The collisionless Vlasov equation isHere $H$ is the single-star specific Hamiltonian; the normalization of $f$ by mass changes neither the Poisson bracket nor the transport equation.

To obtain the Boltzmann equation, take a dilute gas with a short-range pair potential. Its interaction range is small compared with the mean free path and the spatial scale on which the phase-space distribution function changes. The binary collision approximation treats encounters as isolated elastic collisions; simultaneous encounters involving three or more particles can be neglected. The encounter duration is short compared with the collision relaxation time and macroscopic evolution time, and the external force changes negligibly during an encounter. This permits local, effectively instantaneous scattering at one position $\mathbf{x}$.

The essential statistical closure is molecular chaos for the Boltzmann equation: the distribution of incoming pairs factorizes as $f{f}_{\star }$. Outgoing pairs need not be independent. Follow each incoming pair through its two-body scattering, and use conservation of momentum and kinetic energy, together with time reversal and the invariant scattering measure, to express both gain and loss in the same variables. This givesHere $d\Omega =\sin \theta d\theta d\phi$; when the measure is written as $d\theta d\phi$, its Jacobian is included in the Boltzmann scattering kernel. In a scattering-angle convention, that kernel contains relative speed times differential scattering cross-section. The hemisphere prevents double counting equivalent outcomes for indistinguishable particles; the precise factor is included in $B$. The primed velocities are connected to the unprimed ones by the reversible elastic collision map. This closure is appropriate for a dilute short-range gas, rather than a generic long-range interacting system with an important collective mean-field theory contribution.

The further replacement by the BGK model is a phenomenological relaxation closure. Replace the full nonlinear Boltzmann collision operator by relaxation towards a local Maxwellian, with a single, velocity-independent collision relaxation time $\tau >0$:It models relaxation towards a local Maxwellian without resolving the detailed scattering. Treating $\tau$ as constant is an additional simplification; it is of the order of a mean time between effective collisions. Near a local Maxwellian, departures in the nonconserved velocity moments decay on this timescale.

Choose $f^{(0)}$ to match the collision invariants $1$, $\mathbf{v}$ and $|\mathbf{v}{|}^2$ of $f$. For unit particle mass, this meansHere $\rho$ is the mass density, $\mathbf{u}$ is the mean velocity, and $\Theta$ is the temperature in squared-velocity units, $\Theta =k_{B}T/m$ before setting $m=1$. In particular,Thus relaxation preserves local mass, momentum and kinetic energy rather than relaxing them to externally prescribed values.

Write $c_i=v_i-u_i$ and introduce the raw third velocity moment $M_{ijk}$ and the third central velocity moment $Q_{ijk}$:The expansion of their product givesThe heat flux is $q_i=Q_{ijj}/2$ for unit mass.

Multiply the BGK model by $v_i{v}_j$. The external force term isTherefore the raw second velocity moment satisfiesThe matching local Maxwellian has ${\Pi }_{ij}^{(0)}=\rho u_i{u}_j+\rho \Theta {\delta }_{ij}$, soTo remove the bulk motion, let $D_t={\partial }_t+u_k{\partial }_k$ be the material derivative. The continuity equation and the momentum conservation equation implyand henceSubtract this identity from the raw second velocity moment equation and insert the expansion of $M_{ijk}$. The terms $u_i{\partial }_k{P}_{jk}$ and $u_j{\partial }_k{P}_{ik}$ cancel by the product rule. The external force terms also cancel, since a common acceleration changes the local mean velocity without directly changing the relative velocities. What remains isThe trace of the relaxation term is zero because $P_{ii}=3\rho \Theta$. Thus the BGK model relaxes anisotropy of the pressure tensor while preserving the local random kinetic energy. The evolution is still a velocity-moment hierarchy, since $Q_{ijk}$ requires further information about $f$.

The horizontal axis is copper mass percentage. The two liquidus branches meet at $(28\%,779^{\circ }\mathrm{C})$ and connect to pure silver at $960^{\circ }\mathrm{C}$ and pure copper at $1083^{\circ }\mathrm{C}$. The terminal solid-solution boundaries at the eutectic temperature are $9\%$ Cu for $\alpha$ and $92\%$ Cu for $\beta$. Label the single-phase fields $L,\alpha ,\beta$, the two liquid-plus-solid fields $L+\alpha ,L+\beta$, and the lower two-solid field $\alpha +\beta$.

The solidus boundaries link the pure-component melting points to their terminal solubility limits; the eutectic isotherm spans the three-phase tie line. Below it, the solvus branches approach pure Ag and pure Cu at room temperature. The supplied data determine this topology and the marked points, not the exact shapes of every boundary. The drawing therefore does not invent a numerical liquidus temperature for an intermediate composition.

The room-temperature endpoints are pure Ag and pure Cu. Conservation of the total copper mass therefore givesThe answer is 72% silver phase and 28% copper phase by mass. The change from the high-temperature fractions is caused by decreasing solid solubility: the phases reject the opposite component during equilibrium cooling. Simply retaining the eutectic-temperature lever-rule fractions would violate the stipulated room-temperature equilibrium compositions.

Apply the lever rule on the $779^{\circ }\mathrm{C}$ solid tie line. All compositions are copper mass percentages:Thus the equilibrium material is 77.1% $\alpha$ and 22.9% $\beta$ by mass just below the eutectic temperature. These are phase fractions; converting them to volume fractions would require densities.

The cooling curve first falls in the entirely liquid field. On crossing the liquidus, primary Ag-rich $\alpha$ begins to grow. Latent heat release makes the descent shallower; the remaining liquid becomes richer in copper and approaches the eutectic composition.

At $779^{\circ }\mathrm{C}$, the remaining liquid solidifies as $\alpha +\beta$, giving an isothermal arrest under the ideal equilibrium, heat-extraction model. After the liquid is exhausted, temperature falls through the two-solid field. Continued loss of solid solubility causes additional redistribution and precipitation; it does not require another invariant liquid-freezing plateau.

Immediately before the eutectic reaction, the remaining liquid fraction isThe arrest represents freezing of roughly $58\%$ of the sample by mass. Its duration cannot be calculated without the heat-extraction rate and latent heat, and the onset temperature cannot be calculated exactly from the supplied phase-diagram points alone.

At height $z$ above the free lower end, a uniform vertical filament carries the weight of length $z$. With cross-sectional area $A$,The maximum stress is at the top and is independent of diameter. The diameter cancels because both mass and load-bearing area scale with $A$. Neglecting self-heating gradients, evaporation and initial fabrication stress, this top point sets the first creep-strain limit.

For a clearly labelled external reference estimate, take $\rho =1.93\times 10^4\mathrm{kg}{\mathrm{m}}^{-3}$. Then ${\sigma }_{\max }=1.89\times 10^4\mathrm{Pa}=0.0189\mathrm{MPa}$; the area is $3.14\times 10^{-8}{\mathrm{m}}^2$ and the total suspended mass is about $6.06\times 10^{-5}\mathrm{kg}$. The density is independently tabulated, not invented as a supplied 2016 value.^[ref]

Use the original map's top temperature axis at $2500^{\circ }\mathrm{C}$ and its right-hand tensile-stress axis, so no uncertain shear-stress conversion or additional shear modulus is required. The point lies in C, between the $10^{-9}$ and $10^{-8}{\mathrm{s}}^{-1}$ contours, nearer $10^{-8}$. A logarithmic interpolation of the scanned contours gives about $6\times 10^{-9}{\mathrm{s}}^{-1}$, with only graphical precision. Thusor of order three months in this steady-rate, reference-density map estimate. Reading only the bracketing contours gives the appropriately broad interval $5\times 10^6$ to $5\times 10^7\mathrm{s}$; the graph does not support an exact service-life number.

For the exact prescribed input, the answer remainsThis estimate uses the cited reference density and the grain-size assumptions of the supplied map. Thermal expansion lowers density, and an evolved area or tertiary material creep changes the rate; a steady initial-rate estimate is not a guaranteed lifetime. In particular, a different filament grain size could require a different map. Other failure mechanisms could shorten useful life.

With $w$ fixed, bending strength requires $h=\sqrt{6FL/(w{\sigma }_f)}$. Substitution into the cantilever mass givesThe material performance index to maximize is thereforeChanging the free geometric variable changes the strength exponent. A stiffness-limited problem would instead involve Young's modulus; the supplied modulus does not enter this strength-only index.

The rectangular second moment of area is $I=w{h}^3/12$. The maximum bending stress in the cantilever is $FL(h/2)/I=6FL/(w{h}^2)$. If $h$ is fixed, choose the minimum allowed width $w=6FL/({\sigma }_f{h}^2)$. Its mass is thenThus the material performance index to maximize isFor unequal tensile and compressive strengths, use the smaller permissible stress in this symmetric section. The derivation assumes bending-strength control, not a separate deflection, shear, buckling or existing-crack constraint.

At identical applied load and fixed geometric constraints, minimum mass is inversely proportional to the appropriate material performance index. ThereforeThe same percentages apply to weight under the same gravitational acceleration. It would be incorrect to compare densities alone, since the required beam dimensions also change.

Check stiffness and tip deflection using Young's modulus, failure from existing cracks using fracture toughness or the supplied critical energy-release rate, local crushing at supports, shear failure, buckling, fatigue and joint design. In a simple linear-elastic crack model, $K_c^2\simeq E^{\prime }{G}_c$, so toughness cannot be inferred from strength alone. Wood also has grain-direction dependence, moisture sensitivity, natural defects, property scatter and time-dependent creep; the design must control grain alignment and compression-side stability. Metals raise issues of corrosion, mass, joining, machining, residual stress and heat-treatment state. Cost, availability, reproducibility, service temperature and end-of-life treatment can decide between otherwise feasible designs. A high merit index is a screening result, not a complete safety specification.

The characteristic equation of the damped harmonic oscillator is $r^2+\gamma r+{\omega }_0^2=0$. In the underdamped case its roots are $-\gamma /2\pm i{\omega }_d$, where ${\omega }_d=\sqrt{{\omega }_0^2-{\gamma }^2/4}$. Thus the exact real solution is $y=A{e}^{-\gamma t/2}\cos ({\omega }_{d}t+\alpha )$. For $\gamma \ll {\omega }_0$, ${\omega }_d={\omega }_0+O({\gamma }^2/{\omega }_0)$, giving the requested leading approximationFor mass $m$ and spring constant $m{\omega }_0^2$, the mechanical energy is $E=\frac{1}{2}m({\dot{y}}^2+{\omega }_0^2{y}^2)$. Differentiating and using the differential equation gives $\dot{E}=-m\gamma {\dot{y}}^2$. In particular, energy decay of a viscously damped harmonic oscillator is not an exact pointwise exponential decay: $\dot{E}=0$ at a turning point. To leading order in $\gamma /{\omega }_0$, the period average isThe quality factor is $Q=2\pi \overline{E}/(\text{energy lost in one cycle})$. Using the period $T\simeq 2\pi /{\omega }_0$ and the small fractional loss $\gamma T$ gives $\boxed{Q\simeq {\omega }_0/\gamma }$. Equivalently, the energy-decay definition $Q={\omega }_0\overline{E}/{\overline{P}}_{\mathrm{loss}}$ yields the same leading result. Replacing ${\omega }_d$ by ${\omega }_0$ in the oscillatory phase is not uniform over arbitrarily long times; its phase error is $O({\gamma }^{2}t/{\omega }_0)$.

Assume a physical mass $m>0$, spring constant $k>0$, and nonnegative viscous damping parameter $\gamma$. The homogeneous damped harmonic oscillator has characteristic equationIts roots are $r=-\gamma /2\pm \sqrt{{\gamma }^2/4-{\omega }_0^2}$. They have nonzero imaginary parts precisely when ${\gamma }^2<4k/m$. Defining $\Omega =\sqrt{k/m-{\gamma }^2/4}$, the real oscillatory solutions areFor strictly decaying oscillations also require $\gamma >0$. The limiting $\gamma =0$ case is an undamped harmonic oscillator. Equality gives critical damping with $z_h=(C+Dt)e^{-\gamma t/2}$, and a larger positive $\gamma$ gives two real decaying exponentials rather than oscillations.

For a short string element $dx$, small slope makes its tension magnitude approximately constant and its vertical component approximately $T{y}_x$. The net vertical force is $T[y_x(x+dx,t)-y_x(x,t)]\simeq T{y}_{xx}dx$. Its mass is $\rho dx$. Newton's second law givesThis assumes uniform equilibrium tension, small slopes, negligible bending stiffness and dissipation. The symbol $\rho$ is linear mass density here, not volume density.