At fixed chain length, each harmonic mode contributes a Bose–Einstein thermal occupation and a temperature-independent zero-point energy. The latter contributes no heat capacity. With $x=\hslash \omega /(k_{B}T)$,At low temperature only the approximately linear part of the acoustic dispersion is excited, so $g(\omega )\simeq 1/(\pi v_s)$. Using ${\int }_0^{\infty }{x}^2{e}^x/(e^x-1{)}^{2}dx={\pi }^2/3$ givesIt is linear in temperature, rather than the three-dimensional Debye $T^3$ law. At high temperature every scalar oscillator tends to $k_B$, henceThe cutoff temperature is ${\Theta }_m=\hslash {\omega }_m/k_B=305.5$ K. The curve rises smoothly from zero toward this plateau; ${\Theta }_m$ sets a crossover scale, not a sharp transition. In normalized variables its initial slope is $C_V/(N{k}_B)\simeq (2\pi /3)T/{\Theta }_m$. The low-temperature continuum limit also requires temperature above the finite-size level spacing; uniform center-of-mass translation is treated separately. Below the lowest nonzero mode energy a finite chain does not retain a continuum linear law down to arbitrary temperature.

For a long periodic chain of length $\mathcal{L}=Na$, neighboring allowed signed wavevectors differ by $2\pi /\mathcal{L}$. Counting both signs once gives a state density per length $dk/\pi$ on the positive half-zone. Therefore the density per angular-frequency interval and per length isIt is zero outside this band, starts at $g(0)=1/(\pi v_s)=7.958\times 10^{-5}$ s/m and has an integrable inverse-square-root Van Hove singularity as $\omega \to {\omega }_m^{-}$. The state-count normalization isThus the scalar chain has one vibrational mode per atom, with no extra factor of two for the already counted propagation directions. If the horizontal variable were ordinary frequency $\nu$, the density would be $g_{\nu }(\nu )=2\pi g_{\omega }(2\pi \nu )$. A finite chain has discrete lines instead of an actual infinite peak; the smooth divergent curve is the large-chain continuum description.

Differentiate the positive-frequency branch on either side of the origin:It tends to $+4.0$ km/s as $k\to 0^{+}$ and to $-4.0$ km/s as $k\to 0^{-}$, and vanishes at either zone boundary. A positive-$k$ wavepacket therefore transports energy to the right, while a negative-$k$ one transports it to the left. At the zone edge, neighboring atoms oscillate out of phase and the dispersion is flat, giving zero group velocity even though the frequency is nonzero.

The graph has separate one-sided limits at zero because $\omega (k)$ has a cusp there. Assigning one differentiable slope at exactly $k=0$ would erase the distinction between the two propagation directions; a precisely uniform translation has no preferred propagation direction. Plotting only $0\le k\le \pi /a$ is also legitimate if the missing negative branch is stated. Group velocity and phase velocity $\omega /k$ are different except in the long-wavelength linear limit.

The supplied sound speed determines the spring-to-mass ratio without needing either parameter separately:ConsequentlyThe curve is even in $k$, rises linearly from the origin with one-sided slopes $\pm 4.0$ km/s, and flattens at both zone edges. Its maximum is ${\omega }_m=4.0\times 10^{13}$ rad/s, or ordinary frequency ${\nu }_m={\omega }_m/(2\pi )=6.366$ THz. Keeping angular frequency distinct from cycles per second avoids a factor-of-$2\pi$ error. Repeating the curve outside the first zone is a representation of equivalent crystal momenta, not the discovery of additional modes.

For $B=5$ T, the same spin-only two-state probabilities apply, but the low-temperature argument is no longer small. At 300 K, $x=0.0111952$; at 2 K, $x=1.67928$. HenceThe 2 K sample is about $93.3\%$ of its saturation magnetization, with about $96.6\%$ of the moments in the aligned state. The two percentages describe different observables. A linear extrapolation would predict $M/M_s\simeq x=1.679$, exceeding saturation and therefore contradicting the bounded two-state population.

For completeness, at this 2 K, 5 T bias the dimensionless secant susceptibility is approximately ${\chi }_0\tanh x/x=0.01199$, while the differential susceptibility is ${\chi }_0{\mathrm{sech}}^{2}x=0.002805$ under the same prescribed-field approximation. Neither equals the zero-field Curie slope $0.02159$. Thus saturation reduces the incremental response even while the magnetization itself is large.

For $B=0.05$ T, use the two spin-only moment states. Their probabilities and normalized magnetization areAt 300 K, $x=1.11952\times 10^{-4}$; at 2 K, $x=0.0167928$. ThereforeThe phrase “fraction aligned” is ambiguous if it is used to mean net alignment. Half the moments are already in the parallel state at zero field, but the net magnetization is then zero because the antiparallel half cancels them. The table reports both the literal parallel-state population and the imbalance that produces magnetization. At this weak field even the 2 K sample is far from saturation, so the Curie approximation is accurate within this independent-spin model.

For one spin-degenerate parabolic two-dimensional band, $k_F=\sqrt{2\pi n}=1.3729\times 10^8$ m${}^{-1}$. The free-electron model with $m=m_e$ givesThe given $10^3\Omega$ is a sheet resistivity, not a bulk resistivity in ohm-metres. The two-dimensional Drude sheet conductance is $n{e}^2\tau /m_e$, soThus $v_F=\hslash k_F/m_e=1.589\times 10^4$ m/s andThe sample temperature is not given. This is specifically the Fermi-surface mean free path, appropriate in the degenerate regime $T\ll 8.3$ K for the bare-mass model. At higher temperatures the characteristic thermal speed should replace $v_F$ for a thermal transport distance. If a real two-dimensional band has mass $m^{\ast }$, the stated results scale as $E_F,T_F\propto 1/m^{\ast }$, $\tau \propto m^{\ast }$ and $v_F\propto 1/m^{\ast }$; the inferred ${\ell }_F=\hslash k_F/(n{e}^2{\rho }_{\square })$ is mass independent under these same Drude assumptions. No semiconductor effective mass or valley degeneracy is silently introduced.

A conventional face-centered cubic cell contains four atoms. With three conduction electrons per atom,Using the bare electron mass,The Drude transport relation is $\rho =m_e/(n{e}^2\tau )$, givingFor a degenerate electron gas, transport samples electrons near the Fermi surface, so use $v_F=\hslash k_F/m_e=2.025\times 10^6$ m/s:This is a transport relaxation time and associated transport mean free path inferred from the specified resistivity. It need not equal a single-particle quantum lifetime when scattering is strongly angle dependent. The free-electron valence count also neglects the detailed band structure of aluminium, as the stipulated model requires. A classical thermal speed is not the appropriate velocity scale for this strongly degenerate metal.

Use the integrated constant-compressibility model from c, with $\kappa p=0.0059\times 100=0.59$:ThereforeCompression reduces the spacing and moves the same allowed reflection to larger deflection, as the positive pressure derivative predicts. The input compressibility corresponds to a constant bulk modulus $1/\kappa \simeq 169.5$ GPa in this ideal model.

This is the stipulated unchanged cubic-phase, constant-compressibility extrapolation, rather than a reliable prediction for real Fe at 100 GPa. Real Fe exhibits a pressure-induced body-centered-cubic to hexagonal-close-packed transformation, and compressibility can change with pressure.^[ref] A real high-pressure diffraction interpretation would require its actual phase and equation of state. The source gives neither, so its numerical exercise is answered within its explicitly stated idealization.

The 110 plane spacing of the conventional cubic cell isFor the first-order Bragg condition,The quoted experimental deflection is twice the angle between the beam and the reflecting planes; reporting only $22.298^{\circ }$ would report the Bragg angle instead of the requested scattering angle. The 110 reflection is allowed for the identical-Fe body-centered lattice because $1+1+0$ is even. Wavelength and spacing have been expressed in the same units, so their ratio is dimensionless. Higher Bragg orders are conventionally represented by the corresponding higher-index reflection, rather than inserted as an unannounced extra order here.

Removing an electron of charge $-e$ leaves charge $+e$ relative to the filled-band reference. This also reproduces the current. If the missing electron state had velocity ${\mathbf{v}}_e(\mathbf{k})$, subtracting its contribution from the zero current of a filled band givesThe hole dispersion with ${\mathbf{k}}_h=-\mathbf{k}$ gives ${\mathbf{v}}_h({\mathbf{k}}_h)={\mathbf{v}}_e(\mathbf{k})$, soLikewise $\hslash \dot{\mathbf{k}}=-e\mathbf{E}$ becomes $\hslash {\dot{\mathbf{k}}}_h=+e\mathbf{E}$. Positive hole charge and positive hole effective mass consistently describe drift along the electric field, while the valence electrons themselves drift through an almost full band. A hole is a quasiparticle, not a proton moving through the crystal.

An electron hole is a missing electron in an otherwise almost filled valence band. A completely filled band has no net transport current: its velocities cancel over the Brillouin zone. Describing a few missing states as mobile excitations is therefore simpler than following almost every occupied electron state.

Near an isotropic valence maximum, write the electron dispersion as $E_v(\mathbf{k})=E_v(0)-{\hslash }^2{k}^2/(2m_h)$ with $m_h>0$. The electron curvature mass is negative there. Removing that electron creates an excitation with momentum $\hslash {\mathbf{k}}_h=-\hslash \mathbf{k}$ and energy measured upward from the filled reference,ThusThe sign refers to the positive excitation energy of a missing electron near a maximum, not to negative vacuum mass. A general maximum gives the corresponding positive hole-curvature tensor.^[ref]

Fix $r<1$ and construct $g_n$ from the boundary values of $u_n$ on the circle $|z|=r$ as in ii. By the assumed locally uniform convergence, those boundary values converge uniformly to $u$. For each $s<r$, the kernel is bounded on $|z|\le s$, so the same integral shows that $g_n$ converges uniformly there toThe Weierstrass convergence theorem makes $g$ holomorphic. Since $\mathrm{Re}{g}_n=u_n$, passing to the limit gives $u=\mathrm{Re}g$ on $|z|<r$. The real part of a holomorphic function is harmonic. As $r<1$ was arbitrary, $u$ is harmonic throughout the unit disc. This proves the locally uniform limit of harmonic functions theorem without presuming convergence of the derivatives of $u_n$.

For $|z|<r$, the Schwarz integral on the unit disk, rescaled to radius $r$, isOn each smaller closed disk $|z|\le s<r$, its kernel and all its derivatives in $z$ are uniformly bounded. Since the boundary values of the real harmonic function $u$ are continuous on this circle, differentiation under the integral sign proves that $g$ is holomorphic. Taking the real part gives the Poisson integral of those boundary values. It agrees with $u$ by uniqueness of the Dirichlet problem on the radius-$r$ disk. Hence $g$ is holomorphic and $\mathrm{Re}g=u$ on $|z|<r$. Its imaginary part is the harmonic conjugate normalized by $\mathrm{Im}g(0)=0$.

For $0<|z|<R$, set $w=R^2/\bar{z}$, which lies outside the closed disk. On $|\zeta |=R$, direct algebra gives the Poisson kernel identityUsing $d\theta =d\zeta /(i\zeta )$, the proposed Poisson formula for a holomorphic function becomesThe first term is $f(z)$ by the Cauchy integral formula; the second vanishes by the Cauchy integral theorem, since its pole lies outside the disk. At $z=0$ the kernel is one, and the same identity is the mean value property for holomorphic functions. Therefore

The concentric normalization of an eccentric annulus here usesThis Möbius transformation preserves the unit circle and sends the inner circle's real endpoints $0,4/5$ to $-1/2,1/2$. Its image circle is symmetric about the real axis and hence is $|w|=1/2$. Equivalently, expanding $|T(z){|}^2=1/4$ gives $|z{|}^2=(4/5)\mathrm{Re}z$, precisely the inner-circle equation. Thus $T$ sends the domain to $1/2<|w|<1$. Conformal invariance of harmonic measure and i giveThe zero of $T$ lies in the removed inner disk, so this logarithm is harmonic throughout $\Omega$.

The radial harmonic function with value zero at $|z|=r$ and one at $|z|=1$ isIndeed $\log |z|$ is harmonic away from zero, and the displayed expression has the required boundary values. Uniqueness of the Dirichlet problem identifies it as the harmonic measure of an annulus.

Use $h(\sqrt{r}{e}^{i\theta }{)}^2=\varphi (r{e}^{2i\theta })$. As $\theta$ runs around the circle, the doubled angle goes around it twice, so iv givesThe Cauchy integral formula for the Taylor coefficients now impliesFor $n\ge 2$, choosing $r=(n-1)/n$ gives the linear coefficient bound for normalized univalent functions,The last inequality follows from $\log (1+x)\le x$ for $x>-1$.

Because $|h^2|=|h{|}^2$, the Parseval identity for the Taylor series givesDifferentiate this power series for $r<1$ and use iii:Since $M(0)=0$, integration yields the integral mean bound for an odd square-root transform,

For an injective holomorphic function, the real Jacobian determinant is $|h^{\prime }{|}^2$. The area formula for a univalent power series, obtained by change of variables, is thereforeIndeed, insert $h^{\prime }(\rho e^{i\theta })={\sum }_{n\ge 1}n{c}_n{\rho }^{n-1}{e}^{i(n-1)\theta }$. Orthogonality eliminates the cross terms in the angular integral, and radial integration gives $2\pi n^2{\int }_0^r{\rho }^{2n-1}d\rho =\pi n{r}^{2n}$. The power series converges uniformly on each smaller closed disk, justifying the calculation for every $r<1$.

Since $|h(z){|}^2=|\varphi (z^2)|$, the Koebe growth theorem gives $|h(z)|\le |z|/(1-|z{|}^2)$. Hence $h(r\mathbb{D})$ lies in the disk of radius $r/(1-r^2)$, so