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In the small-amplitude approximation, $\dot{x}$ is the speed along the track, and $y=B{x}^2$ supplies the gravitational potential. The conserved energy isDifferentiating and extending through turning points by continuity givesThis is simple harmonic motion with ${\omega }_0^2=4gB/\{(r_2/r_1{)}^2+2\}$. HenceThe frequency is lower than for a sliding point mass because energy is also stored in rotation.

Prefer molecular rotational spectroscopy or rovibrational spectroscopy, because nuclear masses enter their reduced mass directly. For an isotopologue with nuclear reduced mass ${\mu }_N$,Isotope substitution therefore produces appreciable shifts and distinct sets of lines, assuming the same electronic potential and approximately unchanged bond length and force constant. A suitable heteronuclear molecule can provide electric-dipole rotational/vibrational spectra; other molecules may require Raman spectroscopy.

Atomic electronic energies instead depend on the electron-nucleus reduced mass $m_{e}M/(m_e+M)$, whose fractional change is suppressed by $m_e/M$. High-resolution atomic spectroscopy can still detect isotope shifts, including nuclear-size and many-electron mass effects; it is not intrinsically incapable of identifying isotopes. Molecular spectra generally offer the stronger mass sensitivity for this purpose.

A normal mode is a small motion in which all coordinates oscillate at one common angular frequency with fixed relative amplitudes. For conservative linearized motion, these are eigenvectors of the mass-weighted stiffness matrix.

All three springs are initially stretched from $l$ to $2l$, giving equal tension $T_0=kl$ and zero resultant force on each mass. This prestress is essential for the transverse modes found below.

For the final mixed-axis trajectory of a prestressed spring pair, the initial displacements excite only the symmetric mode in each direction. Therefore, to first order in $ϵ/l$,The second mass has the same displacements about $4l$. Over the first transverse period $T_y=2\pi \sqrt{2m/k}$, write $s=t/T_y$:The trajectory is not closed after $T_y$, because the frequency ratio is $\sqrt{2}$. It ends at approximately $(-0.858,1)$ in these scaled coordinates rather than $(1,1)$.

For photons emitted back to back in the measurement frame, their total invariant mass obeys $m^2{c}^4=(E_1+E_2{)}^2-(E_1-E_2{)}^2=4E_1{E}_2$. ThusThe parent's total laboratory energy is consequently about $160.2\mathrm{GeV}$, exceeding its rest energy. The measured-frame opposite-directions condition is essential: opposite directions only in the parent's rest frame would hold automatically and would not determine the second laboratory energy by itself.

One conduction electron per atom gives number density $n=\rho N_A/M\simeq 8.50\times 10^{28}{\mathrm{m}}^{-3}$. Filling two spin states per momentum state in the three-dimensional Fermi gas gives $k_F=(3{\pi }^{2}n{)}^{1/3}$ andThe Fermi temperature greatly exceeds ordinary temperatures, explaining why these conduction electrons are degenerate rather than a classical Maxwell gas. The numerical result uses the stipulated free-electron band and bare electron mass, rather than a detailed copper band structure.

For equally nondegenerate levels, the partition function is $Z=e^{-\beta ϵ}(1+e^{-\beta \Delta })$. The mean energy per particle is $U=ϵ+\Delta /(1+e^{\beta \Delta })$, so the Schottky anomaly has heat capacityAt low temperature, $C\sim k_B(\Delta /k_{B}T{)}^2{e}^{-\Delta /(k_{B}T)}$ because excitation is suppressed. At high temperature, $C\sim {\Delta }^2/(4k_B{T}^2)$ because the two populations approach equality and the energy saturates. The peak is about $0.439k_B$ at $k_{B}T\simeq 0.417\Delta$. The common offset $ϵ$ does not affect heat capacity. Multiply by particle number for the total heat capacity, or divide by the corresponding mass for a mass-specific quantity.

Let $a$ be the cube's side and $z$ a small upward displacement. Its displaced water volume decreases by $a^{2}z$, so the change in buoyancy is $-{\rho }_{w}g{a}^{2}z$. With cube mass $m=0.9{\rho }_w{a}^3$,The displacement must be small enough that the upper face remains above water and the bottom remains submerged. This buoyancy estimate neglects fluid added mass, damping and melting.

A normal mode is a motion in which all generalized coordinates oscillate at one frequency with fixed amplitude ratios. Small oscillations about stable equilibrium in a time-independent conservative system give $\mathsf{M}\ddot{\mathbf{\theta }}+\mathsf{K}\mathbf{\theta }=0$, with positive mass matrix and symmetric stiffness matrix, allowing a generalized normal-mode eigenproblem.

For the three-to-one mass double-pendulum modes, the angles are absolute displacements from vertical, as printed. To quadratic order,For $\mathbf{\theta }=\mathbf{A}\cos \omega t$ and $q={\omega }^2\ell /g$, the determinant isThusThe low mode has the strings leaning together, with the lower angle twice as large; the high mode has opposite angular signs. Both reverse together after half their respective periods.

For the specified release, decompose $(\alpha ,\alpha )=(3\alpha /4)(1,2)+(\alpha /4)(1,-2)$. Thereforeand the lower mass's horizontal displacement isThis is the supremum of two incommensurate harmonic modes, conventionally described as the maximum-amplitude estimate. Strictly, ${\omega }_{+}/{\omega }_{-}=\sqrt{3}$ is irrational, so the opposite cosine extrema are approached arbitrarily closely but never coincide at a finite time. The quoted value is a supremum of the ideal linearized motion.

A spin-half gas has two spin states. Filling the Fermi sphere at zero temperature gives $n=k_F^3/(3{\pi }^2)$, soWith $m_3\simeq 5.008\times 10^{-27}\mathrm{kg}$, the number density is $1.62\times 10^{28}{\mathrm{m}}^{-3}$ and $k_F\simeq 7.82\times 10^9{\mathrm{m}}^{-1}$. ThusThe associated Fermi temperature isFermionic degeneracy matters on cooling to a few kelvin or below in this model, and is pronounced for $T\ll T_F$. Liquid helium-3 is interacting; its quasiparticle effective mass and thermodynamics need not equal the bare-mass free-gas predictions. The requested calculation is the ideal Fermi-gas scale.

For a finite step in a one-dimensional potential and constant mass, the stationary Schrödinger equation requires continuity of the wavefunction and its first derivative. A discontinuous derivative would introduce a delta-function term in its second derivative, unsupported by a merely finite step. A discontinuous wavefunction is still more singular. A delta-function potential has a different derivative jump condition; an infinite wall instead imposes zero wavefunction and does not require matching to an exterior zero derivative.

For a bound state of the half-well, $-V_0<E<0$. The wall at zero excludes a cosine term, and square integrability excludes exponential growth outside. WriteContinuity at $a$ gives $A\sin ka=B$ and $Ak\cos ka=-qB$. HencePut $z=ka$, $w=qa$, and $\nu =a\sqrt{2m{V}_0}/\hslash$. Matching becomes $z\cot z=-w$ with $z^2+w^2={\nu }^2$, $w>0$. The negative branches of $z\cot z$ occur on $((n-1/2)\pi ,n\pi )$. There is one bound-state intersection in the $n$th interval exactly whenAt equality the new level has $q=0$, so its outside tail is constant and not normalizable: it is a threshold state, not a bound state. Thus no bound state exists for $\nu \le \pi /2$, including the strict inequality asked in the question.

Exactly two negative-energy bound states require the second threshold to have been crossed and the third not crossed:The lower endpoint is excluded and the upper endpoint included precisely because a zero-energy threshold state does not count as bound. This half-line finite-well bound-state count differs from an unrestricted symmetric one-dimensional well, which has no hard wall at its centre.

Newton's law of universal gravitation gives the force $-GmM\hat{\mathbf{r}}/r^2$ on a test mass a distance $r$ from a point source. Gauss's law for gravity statesThe inverse-square point field has the stated flux through a surrounding sphere. Superposition extends it to distributed sources; the divergence theorem relates the integral and local forms. Symmetry is still needed to turn a flux into a pointwise field value.

For the gravitational field of an infinite uniform cylinder, a coaxial cylindrical Gaussian surface has no end-cap flux and constant inward field on its curved surface. The enclosed mass per unit length is $\pi \rho r^2$ inside and $\pi \rho R^2$ outside. HenceIts magnitude rises linearly from zero, reaches $2\pi G\rho R$ at the surface, then decreases as $1/r$. The field is continuous at $R$, while its radial derivative changes because the density has a sharp edge.

Use signed distance $y$ along the diameter hole. A rod displaced by $x$ occupies $y\in [x-R,x+R]$, so its force is $F(x)=\lambda {\int }_{x-R}^{x+R}{g}_y(y)dy$. At equilibrium symmetry makes $F(0)=0$. Differentiating the moving endpoints givesThus the leading restoring force and oscillation frequency areFor an explicit positive-displacement check, put $C=2\pi G\rho$ and integrate the inside and outside pieces for $0<x<2R$:Its first-order term is $-2C\lambda Rx$, agreeing with the endpoint argument. The result is the diameter-rod oscillation in cylindrical gravity; it assumes a sufficiently long source cylinder and a narrow hole whose removed mass can be neglected.

For an ideal gas of fixed molecular composition, translational energy is supplemented by rotational, vibrational and eventually electronic excitations as their level spacings become thermally accessible. Activating more internal degrees of freedom generally increases the heat capacity. In a harmonic vibrational mode the contribution rises from nearly zero when $k_{B}T\ll \hslash \omega$ to its classical limit when $k_{B}T\gg \hslash \omega$; summing modes with different frequencies gives a smooth increase over an intermediate temperature range.

A monatomic gas has no molecular rotations or vibrations, so its mass-specific $c_p$ is nearly constant until electronic excitation or ionization becomes important: $B$ is then small. A simple diatomic gas can also have a small slope when its rotations are already active but its vibrational modes remain frozen. Large flexible polyatomic molecules have many internal modes, often with relatively low excitation energies, and can give a larger slope over their activation range.

The linear law is a local approximation, not an asymptotic result. A finite set of modes approaches finite classical heat-capacity contributions, whereas $A+BT$ grows without bound. At still higher temperatures dissociation and ionization change the composition and gas constant, invalidating the fixed-molecule ideal-gas model as well. The constants have units $[A]=\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $[B]=\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-2}$.

Use downward optical depth $t$ measured from the top, with total $D={\tau }_{\infty }$. This is the effective optical-depth/two-stream convention that gives an attenuation factor $e^{-t}$. Constant absorption per unit mass and constant gravity give $t=Dp/p_s$. Write the locally emitted flux source as $S(t)=\pi B(t)=S_0+bt$.

The atmospheric contribution to upward outgoing longwave flux isBecause the source is linear, the effective emission pressure makes this exactly $(1-e^{-D})S(t_{\mathrm{eff}})$, whereThus for $D\gg 1$, $t_{\mathrm{eff}}\to 1$ and $p_{\mathrm{rad}}\simeq p_s/D$. For $D\ll 1$, expansion gives $t_{\mathrm{eff}}=D/2-D^2/12+O(D^4)$, so $p_{\mathrm{rad}}\simeq p_s/2$. In the thin limit $F_{\mathrm{atm}}\simeq D\pi B(p_s/2)$: it is the small atmospheric contribution, not the whole outgoing flux, which this level characterizes. Surface radiation transmitted through the column must be added separately.

Angular convention: these numerical coefficients use an effective optical depth. If $t$ instead means literal vertical monochromatic optical depth and the angular integral is retained exactly, the flux weight is $2E_2(t)$, not $e^{-t}$. Its thick-limit mean depth is $2/3$, giving $p_{\mathrm{rad}}\simeq 2p_s/(3D)$ for a linear source. The coefficient one in the requested expression therefore specifies a two-stream/diffusivity convention; it is not an angle-independent identity. The thin-limit half-column result survives to leading order.

Assume the added gas mixes throughout the convective column and retain constant mass absorption coefficient for the original gas. The original absorber column mass is unchanged, so its total effective optical depth $D={\tau }_{\infty }$ is unchanged. Under transparent-gas addition at fixed absorber mass, the absorber mass fraction becomes $q=p_s/p_a$, and consequently $t=Dp/p_a$. The new radiating pressure is $p_{\mathrm{rad},a}\simeq p_a/D$, even though the ratio of surface to radiating pressure stays $D$.

The mixture's mass-specific constants areThese pressure weights represent column mass fractions, not mole fractions. Repeating the adiabatic argument givesAt fixed $D>1$, this is warmer than the original surface precisely ifThe equivalence follows by cross multiplication of the positive heat capacities; it holds for any positive added mass. If the two ratios agree, adding a transparent gas leaves this model's surface temperature unchanged. Increased total pressure alone is not a new opacity in a model that fixes absorber mass and mass absorption coefficient. Pressure broadening or collision-induced absorption could change the answer in a more realistic atmosphere, but neither is part of the stated constant-opacity model.

$S=e/e_s^l$ is the ambient saturation ratio over a flat liquid-water surface; $S-1$ is the liquid atmospheric supersaturation. Curvature increases the surface equilibrium vapour pressure, producing the Kelvin penalty $A/r$, while dissolved solute lowers it and gives the opposing Raoult contribution $B/r^3$. Thus the dilute Köhler equilibrium curve is $S_{\mathrm{eq}}-1=A/r-B/r^3$, and the numerator is the atmospheric supersaturation available to drive growth beyond surface equilibrium.

$F_d^l$ describes the resistance of vapour diffusion through the surrounding air. $F_k^l$ describes the thermal-conduction resistance to removing latent heat released by condensation: a warmed droplet surface has a higher equilibrium vapour pressure and hence a weaker mass-transfer drive. Common continuum approximations are $F_d^l\simeq R_{v}T/(D_v{e}_s^l)$ and $F_k^l\simeq L_v[k_{\mathrm{air}}T{]}^{-1}[L_v/(R_{v}T)-1]$, showing the roles of diffusivity, conductivity, saturation pressure and latent heat. Both resistances have the supplied units $\mathrm{s}\mathrm{m}\mathrm{k}{\mathrm{g}}^{-1}$ and add for this coupled transport problem.

For a spherical droplet of constant density, $m=4\pi \rho r^3/3$. Multiplying the radius equation by $4\pi \rho r$ givesAt sufficiently large radius, curvature and solute corrections are negligible, soThis limit also requires sustained positive atmospheric supersaturation. It need not describe the earliest submicrometre stage or an unactivated haze particle.

The target liquid-water mass isThe thin-plate ice-growth time uses the prescribed ice mass law, which therefore requires $r_{i\ast }=(m_{\ast }/\alpha {)}^{1/3}=45.31\mu \mathrm{m}$, not the same radius as the water droplet. With $C=2r_i/\pi$ and $F_i=F_k^i+F_d^i$,Hence, for negligible seed mass and constant conditions,This is about $8.24$ times faster than the liquid estimate to the same mass. Both laws have $dm/dt\propto m^{1/3}$ under the given shape assumptions. Ice benefits here from five times the atmospheric supersaturation and a larger geometrical capacitance for the same mass, partly offset by its slightly greater transport resistance. The mass coefficient $\alpha$ encodes the assumed plate shape; replacing it by bulk ice density would change the supplied model.

In a mixed-phase cloud, saturation vapour pressure over ice is lower than over supercooled liquid water. Air can therefore be subsaturated with respect to liquid yet supersaturated with respect to ice: droplets evaporate while ice crystals grow, transferring water through the vapour phase. This is the Wegener–Bergeron–Findeisen mechanism. Subsequent aggregation, riming and melting can produce atmospheric precipitation much faster than condensation alone.

The large difference is not a universal fixed ice/liquid time ratio. Atmospheric supersaturation evolves, latent heating increases thermal resistance, crystal habits and capacitance change, and surface attachment or ventilation may matter. Ice nucleation and the availability of freezing nuclei also precede the calculated growth. The boxed time describes the specified already-present crystals under maintained conditions, not a complete prediction of the time until rain begins.

Let $R_A,R_B$ be the specific specific gas constants and $c_{pA},c_{pB}$ the specific heat capacities of the noncondensible gas and vapour. Write $e_s(T)$ for the saturation vapor pressure, $r$ for vapour mass per unit dry-gas mass, and $ϵ=R_A/R_B$. For a dilute ideal gas mixture,The second equation is the Clausius-Clapeyron equation, neglecting condensed-phase volume and taking constant latent heat $L$. The first law of thermodynamics for saturated ascent, with condensate removed and negligible condensate heat capacity, is to leading orderSince $dr=r[LdT/(R_B{T}^2)-d\ln p]$, the resulting moist adiabat has logarithmic slopeConsequently a sufficient dry adiabat criterion isThe first two conditions justify neglecting the vapour's effects on the gas law and sensible heat capacity; the last two suppress latent heating. With $x=R_A/c_{pA}$ and $\beta =L/(R_{B}T)$, the latent conditions are $\beta e_s/p\ll 1$ and $x{\beta }^2{e}_s/p\ll 1$. Usually $x\beta \gg 1$, so the squared-latent term gives the stronger condition. At sufficiently low surface temperature, the exponential smallness of $e_s(T_s)$ makes these conditions hold andThese are local conditions along the column. They can be checked at the surface when $x{\beta }_s>1+2x$: on this dry profile the strongest latent parameter is proportional to $\exp [(1+2x)\zeta -{\beta }_s(e^{x\zeta }-1)]$ and decreases upwards. Without that restriction a surface-only test need not bound every altitude. If suspended condensate is retained, its own heat capacity and loading must also be negligible; saturation alone does not specify those quantities.

Interpret the requested mass path as the vapour column; no condensed-water loading is prescribed. Hydrostatic equilibrium gives $dp=-\rho gdz$, so if $q_B={\rho }_B/\rho$ is vapour mass fraction,For the mixture $q_B=ϵe_s/[p+(ϵ-1)e_s]$. Under the dilute conditions of part (a), $q_B\simeq ϵe_s/p$. Introduce $\zeta =-\ln (p/p_s)$, so $T=T_s{e}^{-x\zeta }$ and $-dp=pd\zeta$. Integrating the constant-$L$ Clausius-Clapeyron equation relative to the surface givesThereforeThe explicit $p_s$ cancels: decreasing vapour mass fraction in a denser dry column compensates its greater total mass. This cancellation belongs to the dilute, dry-adiabatic approximation, and is not a general property of moist atmospheres. With suspended condensate, an additional and unspecified condensate mass path would have to be added.

The first term describes collision-coalescence: a collector sweeps area $\pi R^2$, overtakes small droplets at relative terminal speed $u_R-u_r$, encounters liquid mass per volume $L$, and retains a fraction $E$ of that intercepted liquid. The second term describes diffusion-limited condensational growth, with vapour supersaturation $S-1$ and combined thermal and vapour-diffusion resistances $F_k+F_d$. Condensation is positive for $S>1$ and reverses to evaporation for $S<1$. The liquid-water content $L$ here is a local mass-per-volume quantity, not a vertically integrated liquid-water path.

Neglect condensation and the smaller droplets' fall speed, while retaining their specified liquid-water content. For exponential cloud collector growth, the collection law and spherical collector mass giveThus $R(t)=R_0{e}^{{\alpha }_{R}t}$. The growing terminal speed eventually exceeds the updraft, andwhere the last expression uses $t$ in seconds and $U=1\mathrm{m}{\mathrm{s}}^{-1}$. Numerically,The trajectory reaches maximum height when $k_{R}R=U$, at $t=1145\mathrm{s}$, $R=125\mu \mathrm{m}$ and $h_{\max }=620\mathrm{m}$. It returns to cloud base near $t=1860\mathrm{s}$, with radius about $392\mu \mathrm{m}$. The left-to-right sketch in the preceding figure shows this rise and subsequent fall.

Once the drop leaves the cloud, the prescribed in-cloud $L$, $U$ and collection law need not persist. Consequently the negative $2000$-second value describes the requested constant-coefficient model's extension, not a prediction that collection continues below cloud base. Its actual subsequent height needs an outside-cloud model. Within the cloud, the exponential feedback between increasing radius and increasing swept volume explains how a minority of collector drops can escape the condensation bottleneck and initiate warm rain; depletion, breakup and changes in fall law eventually limit indefinite exponential growth.