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For a nondegenerate two-dimensional electron gas, the two quadratic momentum terms supply $k_{B}T/2$ each by the equipartition theorem. There is no lattice restoring potential energy for a free electron. ThusOn cooling below the Fermi temperature, the Pauli exclusion principle restricts the thermally active electrons to within approximately $k_{B}T$ of the Fermi energy. The linear low-temperature heat capacity of a two-dimensional Fermi gas isThe heat capacity therefore decreases from its classical value as quantum degeneracy becomes important. The particle number of electrons and sample area are held fixed when differentiating the internal energy.

Write $U(x)=-e\varphi (x)$ for the potential energy of an electron in an electrostatic potential $\varphi$, where $e>0$. With $G=2\pi /a$, the coupling in the nearly-free electron model is the Fourier coefficientIf a symbol $V(x)$ already denotes the electron potential energy, as in part (c), it replaces $U(x)$ without another factor $-e$.

At the Brillouin zone boundary the two resonant plane waves have unperturbed energy $E_{\mathrm{ZB}}$. Their Hamiltonian operator in the two-state basis has diagonal entries $E_{\mathrm{ZB}}$ and off-diagonal entries $V_G,V_G^{\ast }$; its eigenvalues are $E_{\mathrm{ZB}}\pm |V_G|$. ThusThis is the local band gap in the two-wave nearly-free electron model.

The boundary of the first Brillouin zone is $|k|=\pi /a$. The unperturbed free electron model assigns the kinetic energywhere $m_e$ is the bare electron mass and $\hslash$ is the reduced Planck constant. This is the free kinetic energy at the Brillouin zone boundary, not the already split energy band edge. A constant mean potential energy may be absorbed into the choice of zero of energy.

Regard $V(x)$ as an electron potential energy, and take $G=2\pi /a$. Each cosine supplies half its coefficient to each of its two Fourier coefficients. The nonzero coefficients areThe Fourier coefficient connecting the two plane waves at $k=\pm \pi /a$ is $V_G$, so the two-wave band gap isAt $\delta =0$ the actual periodic potential has period $a/2$ and no Fourier harmonic at $G$. The crossing at the artificially halved Brillouin zone boundary then has zero band gap. Higher-order weak-potential corrections can change the proportionality coefficient at nonzero $\delta$; the displayed coefficient belongs to the given two-wave approximation.

There is also a small-displacement qualification in the source: the claimed positions of the potential energy minima are correct to first order in $\delta$, not exactly. Put $t=2\pi x/a$. ThenAt $\delta =0.01$ and $x/a=1/4-\delta =0.24$, this equals approximately $-2.802\times 10^{-3}$, rather than zero. For $V_0>0$, expanding about $x/a=1/4$ and $3/4$ gives minima at $1/4-\delta +O({\delta }^3)$ and $3/4+\delta +O({\delta }^3)$. This does not alter the leading Fourier coefficient or the linear opening of the band gap.

The small-slope excess length is $\int (w^{\prime }{)}^2/2$, so the first term is the work needed against uniform tension. The second is the potential energy of the downward distributed load. Their sum is total potential energy subject to the obstacle constraint.

Expand the difference exactly:The first two integrals together are nonnegative by part (b), and the square is nonnegative. ThusFor $T>0$ this also proves uniqueness: equality for two minimizers forces identical derivatives, and the fixed endpoints remove their additive constant. The convex admissible set and strictly convex energy make the variational inequality a global minimization condition.

Make the Wick rotation $t^{\prime }=-i\tau$, and write $T_E=\beta \hslash$ for the final imaginary time, where $\beta$ is inverse temperature and $\hslash$ is the reduced Planck constant. Then $d{t}^{\prime }=-id\tau$ and $dx/d{t}^{\prime }=idx/d\tau$. Thus the continued action isBoth the kinetic energy and the potential energy enter this Euclidean action with positive signs relative to their real-time definitions. Also $e^{-iH(-i\beta \hslash )/\hslash }=e^{-\beta H}$, so the Euclidean path integral isIts boundary conditions fix the two endpoint positions. They need not coincide: identifying and integrating over them would instead give the thermal trace.

The normalization of $\mathcal{D}x$ can be fixed without continuing a formal measure. Apply the Trotter product formula to $H={\hat{p}}^2/(2m)+V(\hat{x})$, divide $T_E$ into $M$ intervals of size $ϵ=T_E/M$, and integrate the intermediate positions. The resulting time-sliced configuration-space path integral isThis also explains the upper limit $\beta \hslash$ rather than $\beta$ when the reduced Planck constant has not been set to one.

virial pressure-radius curve with uniform-sphere binding energy. Put $c_s^2=k_{B}T/\mu$. Isothermality makes $\int (p/\rho )dM=c_s^2{M}_0$, so the virial theorem givesThe curve crosses zero at $r_z=B/A=G{M}_0/(5c_s^2)$, rises to its maximum atthen decreases toward zero at large radius. Since $p_0^{\prime }=(4B-3A{r}_0)/r_0^5$, a cloud displaced outward at fixed external pressure has a restoring pressure deficit only on the descending branch. Thus the homologous, isothermal radial mode is stable for $r_0>r_c$, unstable for $r_z<r_0<r_c$, and marginal at $r_c$. The negative-pressure branch is not a physical equilibrium with a positive confining pressure. This conclusion uses the stipulated gravitational-energy approximation, not the exact isothermal density profile.

critical mass from a virial pressure turning point. At the turning point, $B=3A{r}_c/4$; henceUsing the cloud's mean density $\bar{\rho }=3M_0/(4\pi r_0^3)$, the condition $r_0>4G{M}_0/(15c_s^2)$ becomesUsing the boundary density instead changes the numerical coefficient, not this scaling. The coefficient should not be identified with the exact Bonnor-Ebert sphere threshold because the assumed potential energy is different.

mass and luminosity weighting of a stellar initial mass function. Impose finite birth-mass cutoffs $m_l,m_u$. The stellar birth mass and main-sequence luminosity scale asFor a broad mass interval, the low-mass end controls the first integral and the high-mass end the second. Per logarithmic mass interval the weights are respectively $m^2\xi \propto m^{-0.35}$ and $mL\xi \propto m^{2.65}$. Finite lifetimes remove massive stars from an aging population, so the luminosity conclusion applies while those stars remain on the main sequence.

exponential closed-box gas consumption. The gross formation rate is $\Psi =ϵM_g$, but the permanent locking rate is $\alpha \Psi$. The instantaneous recycling approximation and mass conservation give ${\dot{M}}_g=-\alpha ϵM_g$, ${\dot{M}}_{\ast }=\alpha ϵM_g$. ThereforeHere $M_{\ast }$ means surviving locked stellar mass in the galactic gas fraction, not the cumulative gross mass that has passed through star formation.

locked-mass and returned-mass yield conventions. Pre-existing metals are locked at rate $\alpha Z\Psi$; the stipulated newly produced metal source is $b(1-\alpha )\Psi$. ThusFor $\alpha >0$ and initially pristine gas,The effective yield $y$ is per unit permanently locked mass; it is not the source parameter $b$. For $\alpha =0$, $M_g=M_0$ and the direct time equation gives $Z=bϵt$. For $\alpha =1$, the explicitly stipulated metal source vanishes, so this particular enrichment model gives $Z=0$. Constant yield is a trace-enrichment approximation; the formal logarithmic divergence as gas is exhausted must not be interpreted as a physical metal fraction exceeding one.

backward extrapolation of exponential closed-box star formation. Denote the present age of the universe by $t_{\mathrm{now}}$. With the last paragraph's $\alpha =1$,Adopting a present age of the universe $t_{\mathrm{now}}=13.8\mathrm{Gyr}$ for the estimate gives about $365M_{\odot }\mathrm{y}{\mathrm{r}}^{-1}$, requiring $7.3\times 10^{11}{M}_{\odot }$ of gas at that epoch in the same model. The boxed expression makes the dependence on the adopted present age explicit. This is an extreme history for a present Milky-Way-like system, although intense high-redshift star formation by itself is not forbidden.

The extrapolation forces a fixed gas reservoir to decline exponentially over many gas-depletion times. Gas accretion, outflows, evolving gas density and star-formation efficiency, delayed recycling, and galaxy assembly break that assumption. The empirical disk-averaged relation is not a universal constant-efficiency law for all gas: the original measurements combined Balmer series H-alpha, atomic gas, carbon monoxide and infrared data, with tracer calibration and dust corrections entering the inference.^[ref] Cosmological calculations explicitly support continuing cold-gas supply to rapidly forming high-redshift galaxies, providing a physical alternative to this closed-box backward extrapolation.^[ref]

For $E>0$, put $k=\sqrt{2mE}$ and $q=k/\hslash >0$. Outside the support of the potential energy, a generalized eigenfunction of the Hamiltonian operator has the plane wave formHere $R_0$ denotes the support radius, to distinguish it from the quantum reflection probability $R$. The momenta are $\pm k$ and the velocities are $\pm k/m$. Thus $A,D$ describe incoming scattering channels from the left and right, and $B,C$ describe outgoing scattering channels towards the left and right. These energy eigenfunctions are not square-integrable functions; normalizable incoming wave packets are formed by superposition over energy, and their passage through the interaction region gives the physical quantum scattering experiment.

For incidence from the left, normalize $A=1,D=0$ and write $B=r_L,C=t_L$. The quantum reflection amplitude is $r_L$ and the transmission amplitude is $t_L$. The ratios of outgoing to incident probability currents defineNo extra velocity ratio occurs because both exterior regions have the same potential energy and mass. The scattering matrix, with a fixed choice of channel order, isIts first and second columns are the responses to unit incidence from the left and right, respectively. For a reflection-symmetric potential energy, these conventions give $r_L=r_R=r$ and $t_L=t_R=t$.

For a real potential energy and a stationary state, the probability current isThe time-independent Schrödinger equation gives ${\psi }^{\prime \prime }=2m(V-E)\psi /{\hslash }^2$, so $j^{\prime }=\hslash \mathrm{Im}({\psi }^{\ast }{\psi }^{\prime \prime })/m=0$. In an exterior plane wave superposition, interference terms contribute no current andConsequently $|A{|}^2+|D{|}^2=|B{|}^2+|C{|}^2$ for every choice of incoming amplitudes. If $u=(A,D{)}^T$, this says $u^{†}(I-S^{†}S)u=0$ for every complex vector $u$. Since $I-S^{†}S$ is Hermitian, vanishing of its quadratic form implies that the matrix itself vanishes. ThereforeThe second identity follows by choosing $u=(1,0{)}^T$. Since $S$ is a square matrix, the first also implies $S{S}^{†}=I$, proving unitarity. A real delta potential preserves this proof: continuity of $\psi$ and the real derivative jump make the current continuous across the interaction point.

Beyond the outer classical turning point, let $\chi (x)=\sqrt{2m[V(x)-E]}$ and $U(x)={\hslash }^{-1}{\int }_{x_2}^x\chi (y)dy$. The bound state condition removes the growing branch at positive infinity, leavingHere $F$ is an arbitrary common amplitude pending the matching and wavefunction normalization. For the confining potential energy in the native diagram, the decaying branch gives the required exterior square-integrable function.

The WKB approximation fails at a simple classical turning point $x_t$, because $p(x_t)=0$ makes its amplitude singular and violates the WKB validity condition. Locally replace $V(x)-E$ by $V^{\prime }(x_t)(x-x_t)$. For a forbidden region on the right, $V^{\prime }(x_t)>0$ and the rescaled coordinate $z=[2m{V}^{\prime }(x_t)/{\hslash }^2{]}^{1/3}(x-x_t)$ changes the time-independent Schrödinger equation into the Airy ordinary differential equation ${\psi }_{zz}-z\psi =0$. Reversing the coordinate handles a forbidden region on the left.

The two independent Airy functions have asymptotic forms, for $z,w>0$,Match these to the WKB approximation in an overlap region where the linearization of the potential energy and the large-argument asymptotic expansions both apply. If $K_t={\hslash }^{-1}|{\int }_{x_t}^x\chi dx|$ on the forbidden side and $s_t={\hslash }^{-1}|{\int }_{x_t}^{x}pdx|$ on the allowed side, the Airy turning-point connection formula isThus a decaying exponential acquires a factor two in its allowed-side amplitude, whereas a growing exponential does not. The phase convention uses positive integrals measured away from the classical turning point, so it applies at both ends of the well without an extra orientation sign. These are leading semiclassical matching rules for simple, separated classical turning points.

A tall, broad central potential barrier has $\varphi \gg 1$. Let $\sigma =+1$ denote even parity and $\sigma =-1$ odd parity, and write $\theta =(n+\frac{1}{2})\pi +{\delta }_{\sigma }$, where $n=0,1,\dots$. Since $\tan [(n+\frac{1}{2})\pi +\delta ]=-\cot \delta$, the condition in part (d) givesThus the leading correction to the isolated-well Bohr-Sommerfeld quantization condition isThis is an implicit condition, with all action integrals evaluated at the corrected energy. To convert it into an explicit leading energy correction, let $E_n^{(0)}$ solve $\theta (E_n^{(0)})=(n+\frac{1}{2})\pi$ and define the full classical oscillation period in one well by $\mathcal{T}(E)=2m{\int }_{x_1}^{x_2}dx/p(x)$. Differentiation of the action integral has no endpoint terms because $p(x_1)=p(x_2)=0$, and givesConsequently, to leading exponentially small order within the WKB approximation,The even member lies below its odd partner. For an approximately harmonic oscillator well, $\mathcal{T}\simeq 2\pi /\omega$, giving the familiar double-well tunneling splitting $\Delta E\simeq (\hslash \omega /\pi )e^{-\varphi }$. The explicit energy formula retains only its leading exponential correction; varying $\varphi$ and ${\theta }^{\prime }$ with the shifted energy also contributes at higher orders.

The ground state has an even, everywhere same-sign wavefunction with a peak in each well and a small nonzero value beneath the central potential barrier. Its derivative vanishes at zero. The first excited state is odd, has opposite-sign lobes in the two wells, and has exactly one interior node of a wavefunction, at zero. Both wavefunctions decay outside the wells. In terms of localized normalized wavefunctions ${\phi }_L,{\phi }_R$, with overlap $s=⟨{\phi }_L|{\phi }_R⟩$, their qualitative shapes are the lowest parity states of a symmetric double well:The overall sign of the odd wavefunction is arbitrary. The sketch uses localized Gaussian profiles to show the parity and the central node of a wavefunction, rather than claiming exact energy eigenfunctions for an unspecified potential energy.

Choose $y$ northward and $z$ upward in the Northern Hemisphere. In a locally linear Boussinesq approximation, write the background buoyancy as $\bar{b}=b_{\ast }+By+N^{2}z$, with horizontal buoyancy gradient $B<0$ and positive squared buoyancy frequency $N^2$. The stable density stratification resists a purely vertical displacement, but the isopycnals slope upward poleward:The geopotential surfaces are horizontal in this local approximation. Their inclination differs from that of the isopycnals, which is the relevant baroclinicity.

Exchange two equal small fluid elements, each of volume $V$, between positions separated by $(\delta y,\delta z)$. The adiabatic parcel exchange has no thermal diffusion, so each parcel retains its buoyancy. The background buoyancy difference is $\Delta \bar{b}=B\delta y+N^2\delta z$. Because the variable-density contribution to gravitational potential energy is $-{\rho }_{0}Vbz$, the change in the pair's gravitational potential energy isFor a poleward-rising exchange, set $\delta z=s\delta y$ with $s>0$. ThenThe trajectory lies between a geopotential surface and an isopycnal. The poleward-rising parcel starts more buoyant than the parcel it replaces, and the equatorward-sinking parcel starts less buoyant: the buoyancy does positive work on both. This releases available potential energy into kinetic energy, allowing eddies to amplify if the full rotating dynamics supports the displacement and dissipation does not remove the gain.

A trajectory steeper than the isopycnals, $s>s_{\rho }$, instead costs gravitational potential energy; an equatorward-rising trajectory also costs it. Motion exactly along a geopotential surface or an isopycnal gives zero change in this ideal exchange. At fixed $\delta y$, the largest release occurs at $s=s_{\rho }/2$, since $N^{2}s(s-s_{\rho })$ is minimized there.

This is an energetic possibility argument for baroclinic instability, not a sufficient normal mode instability criterion. In particular, unrestricted parcel motion must not be assumed in a rotating fluid: conservation of angular momentum can impose an additional restoring effect. Symmetric instability has a separate Ertel potential vorticity criterion; ordinary Eady instability can occur without violating that criterion.

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.

Normalize the symmetric $N$-particle phase-space distribution function $D_N$ to one and write $z_i=({\mathbf{x}}_i,{\mathbf{v}}_i)$, $d{z}_i=d^3{x}_i{d}^3{v}_i$. The number-normalized one- and two-particle distribution functions areThus $\int fdz=N$ and $\int f_2(z,z_{\star })d{z}_{\star }=(N-1)f(z)$. These conventions absorb the otherwise explicit factor $N-1$ in the first BBGKY hierarchy equation.

Separate the Hamiltonian into kinetic energy, external potential energy, and pair potential terms. Their Hamilton's equations giveThe sign follows directly from the stated Poisson bracket: ${\dot{v}}_{i\alpha }=\{v_{i\alpha },H\}=-{\partial }_{x_{i\alpha }}H$. The Hamiltonian Liouville equation is consequentlyIntegrate over $z_2,\dots ,z_N$ and multiply by $N$. Assume sufficient decay in velocity space and either decay or periodic boundary conditions in position, so that integration by parts removes all transport terms with $i\ge 2$. The surviving interaction term iswhere indistinguishability makes all $N-1$ terms equal. Relabeling $(z_1,z_2)$ as $(z,z_{\star })$ yields the exact resultNo assumption of molecular chaos for the Boltzmann equation has entered: the dependence on the two-particle distribution function is still exact.

The total kinetic energy and small-slope elastic potential energy areOrthogonality removes cross terms between the two distinct sine or cosine modes. A mode of amplitude $A_n$ contributes $K_n=\mu L{A}_n^2{\omega }_n^2{\sin }^2({\omega }_{n}t)/4$ and $V_n=TL{A}_n^2{k}_n^2{\cos }^2({\omega }_{n}t)/4$. For amplitudes $1,1/2$ and $n=2,4$, the prefactors agree. Writing $\Omega =2\pi v/L$,Thusindependent of time. This is another instance of equal modal energies for inverse-harmonic string amplitudes. An explicit initial amplitude $A$ multiplies every energy by $A^2$.

In the small-slope string model, the additional elastic potential energy is $V=(T/2){\int }_0^L{y}_x^{2}dx$. For the fundamental mode, $y_x=A_1{k}_1\cos (k_{1}x)\cos ({\omega }_{1}t)$. Since ${\int }_0^L{\cos }^2(\pi x/L)dx=L/2$,Using $T{k}_1^2=\mu {\omega }_1^2$ from the dispersion relation givesThe expression vanishes as the string passes through its straight equilibrium shape and is maximal at the turning shapes, when the corresponding kinetic energy is zero.

A symmetric quadratic form has twice its off-diagonal entry as the coefficient of the mixed product. Comparing with the preceding potential energy and kinetic energy expressions givesHere $\mathbf{V}$ is the stiffness matrix and $\mathbf{T}$ the mass matrix in angular coordinates; their units differ accordingly. The dimensionless mass-matrix determinant is $8-4=4>0$ and its first diagonal entry is positive, so the kinetic energy is positive for every nonzero angular velocity.

For a single normal mode $y_n=A_n\sin (k_{n}x)\cos ({\omega }_{n}t+{\varphi }_n)$, the kinetic energy is the integral of $\frac{1}{2}\rho y_t^2$. Using ${\int }_0^l{\sin }^2(k_{n}x)dx=l/2$ givesThe extra string length is ${\int }_0^l[\sqrt{1+y_x^2}-1]dx\simeq \frac{1}{2}{\int }_0^l{y}_x^{2}dx$. Multiplying by tension gives the quadratic potential energy, so ${\int }_0^l{\cos }^2(k_{n}x)dx=l/2$ impliesSince $\rho {\omega }_n^2=T{k}_n^2$, the sum is time independent:For the released shape, $A_1=A$ and $A_2=A/2$, soThe second mode's smaller displacement is offset by its doubled spatial wavenumber and angular frequency. orthogonality of distinct sine and cosine spatial modes eliminates cross terms in the integrated energies, so the total is their sum. Initially the kinetic energy is zero and all the energy is stored in the string's deformation; during the motion the total remains constant.

For small transverse displacements, the string's kinetic energy and extra elastic potential energy areThe orthogonality of trigonometric functions removes cross terms between the two normal modes. For a mode $B_n\sin (k_{n}x)\sin ({\omega }_{n}t)$, use $T{k}_n^2=\rho {\omega }_n^2$ and ${\int }_0^L{\sin }^2(k_{n}x)dx={\int }_0^L{\cos }^2(k_{n}x)dx=L/2$ to obtainConsequently the modal energy distribution isThe second mode contributes more total energy at every time, despite its smaller displacement amplitude. The total energy is $13\rho L{V}_0^2/64$, with fractions $4/13$ and $9/13$ in the first and second modes. Each mode exchanges kinetic energy and potential energy, while its sum is conserved; the comparison is of these sums, not of instantaneous kinetic energies alone.