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Use the Gibbs-Duhem equation for a homogeneous, extensive, single-component thermodynamic phase:At fixed temperature, this becomes $dp=nd\mu$, with number density $n=N/V$. Separately, at fixed particle number,The chain rule along the same isothermal process in the equation of state now givesConsequentlyThis is isothermal compressibility from chemical-potential response. Since chemical potential has units of energy per particle, the right-hand side has units of volume per energy, or $\mathrm{P}{\mathrm{a}}^{-1}$, as required. The derivation assumes a smooth bulk equation of state; changing the fractions of coexisting thermodynamic phases at fixed pressure needs separate treatment.

Measure single-particle energy from the lowest state and write ${\epsilon }_k=c{k}^{\alpha }$, with $c>0$. In a homogeneous system, the number of wavevector states in a thin spherical shell is proportional to $k^{d-1}dk$. Changing variables from $k$ to $\epsilon$ gives the density of states per unit spatial volume,where $S_{d-1}$ is the area of the unit sphere in $d$ dimensions; an additional internal state degeneracy would multiply this expression by a constant.

The Bose-Einstein distribution has $\mu \le 0$. At saturation, $\mu \to 0^{-}$, the maximum excited-state number density isOnly the lower endpoint can cause divergence. There the integrand is proportional toThe improper integral converges precisely when its power exceeds $-1$. ThusWhen $d=\alpha$, the divergence is logarithmic; when $d<\alpha$, it is a power divergence. In either case the excited states can accommodate any fixed finite density at every positive temperature, so there is no bulk Bose-Einstein condensation at nonzero temperature.

For completeness, putting $s=d/\alpha >1$ and using the Gamma function and Riemann zeta function givesAt fixed density $n$, the right-hand side reaches $n$ at $T_c$; below it the excess particles occupy the lowest state, with condensate fraction $1-(T/T_c{)}^s$. This is Bose condensation with power-law dispersion in the homogeneous thermodynamic limit. A finite vessel or a trapping potential changes the low-energy state count and must not be silently substituted for that setting.

Canonical probabilities. A canonical ensemble describes a system of fixed particle number and fixed external parameters such as volume, weakly coupled to a much larger thermal reservoir. They exchange energy and reach thermal equilibrium, while the reservoir's temperature stays effectively constant. The interaction energy is negligible in the state counting. The combined isolated system is described by equal probabilities for its accessible microstates.

If its total energy is $E_{\mathrm{tot}}$, the probability of a particular system microstate $i$ is proportional to the reservoir multiplicity at $E_{\mathrm{tot}}-E_i$. Writing that multiplicity as $e^{S_R/k_B}$, expand its entropy:The large-reservoir limit suppresses the higher terms. Normalization therefore givesThe sum counts individual microstates; when grouping them by energy, their state degeneracies must be included. This $Z$ is the canonical partition function.

Classical orientational heat capacity. Hold the magnetic field fixed. Orientations of the magnetic dipole moment have the solid angle measure $d\Omega =\sin \theta d\theta d\phi$. With $x=\beta mB$, the angular partition function isChoosing a normalized angular measure would remove $4\pi$, with no effect on the energy derivatives. Differentiating at fixed $B$,The angular average $⟨\cos \theta ⟩=\coth x-1/x$ is the Langevin function. Since $dx/dT=-x/T$,This is the heat capacity of a classical magnetic dipole, in $\mathrm{J}{\mathrm{K}}^{-1}$ for one magnetic moment. The word specific here refers to that single moment; $N$ independent magnetic moments have $N$ times this heat capacity. At $B=0$ the energy is independent of orientation and $C_B=0$, obtained by the continuous $x\to 0$ limit. At fixed nonzero $B$, the classical low-temperature limit instead gives $C_B\to k_B$.

Equipartition and its proof. The equipartition theorem states that an independent quadratic term $a{q}^2/2$ in a classical Hamiltonian, with $a>0$, contributes $k_{B}T/2$ to its mean energy. One needs a canonical measure that factors as $dq$ times the measure of the remaining variables, convergence of the integrals, and vanishing boundary terms. The coordinate must not be restricted by a thermally important hard boundary or a coupled constraint.

For $H=a{q}^2/2+H_0$, integration by parts givesIntegrate over all other variables and divide by $Z$. It follows thatThe same reasoning applies to quadratic momenta among the canonical variables. It is the number of quadratic energy contributions that is counted, not an unrestricted count of coordinate labels.

Two soft orientational directions. When $x\gg 1$, the moment is concentrated near alignment. Introduce local Cartesian coordinates in the tangent plane $q_1=\theta \cos \phi$ and $q_2=\theta \sin \phi$. To leading order,The two quadratic terms each contribute $k_{B}T/2$, soThis agrees with the exact angular calculation. Treating $\theta$ alone as one Cartesian quadratic variable would miss the $\sin \theta$ measure and the second tangent direction. There are no extra rotational kinetic terms in the stipulated Hamiltonian; adding such terms would be a different model.

Quantum spin one. The spin projections are $m_s=-1,0,1$, with $S_z=\hslash m_s$. Put $\Delta =\hslash |\gamma |B>0$, where $\gamma$ is the gyromagnetic ratio. The three energies are $-\Delta ,0,+\Delta$, irrespective of the sign of $\gamma$. Relative to the ground state they are $0,\Delta ,2\Delta$. Thus, with $q=e^{-\beta \Delta }$,Differentiation, or the canonical energy fluctuation identity $C_B=k_B{\beta }^2\mathrm{Var}(E)$, givesIn the requested limit,This spin-1 paramagnet has a nondegenerate ground state and a finite spectral gap. Excitations are exponentially rare, so its heat capacity tends to zero. Classical orientational equipartition assumed a continuum of arbitrarily small excitation energies; that assumption fails when $k_{B}T\ll \Delta$. This is a change in the accessible state spectrum, not a violation of the equipartition theorem within its classical domain.

Occupation and the two-dimensional Fermi energy. For an independent single-particle state, the Fermi-Dirac distribution isThe chemical potential enforces the chosen mean particle number through $N={\sum }_{j}n({\epsilon }_j)$, including spin states in the sum. It is common to all states in particle-exchange equilibrium and must generally change with temperature or field when the total mean number is held fixed. At zero temperature, the distribution becomes a Heaviside step function at the Fermi energy.

Let the sample area be $\mathcal{S}$ and let $g_s$ be its internal spin spin degeneracy. In the continuum limit, filling the disk $k<k_F$ givesThereforeFor spin-half fermions with both spin states available, $g_s=2$ and ${\epsilon }_F=\pi {\hslash }^{2}N/(m\mathcal{S})$; for a spin-polarized gas $g_s=1$. The question does not specify the area or spin degeneracy, so these conventions must be stated rather than extracting an energy from $N$ alone. Although the original text calls $\mathbf{k}$ momentum, the supplied ${\hslash }^2{k}^2/(2m)$ dispersion makes it a wave vector; physical momentum is $\mathbf{p}=\hslash \mathbf{k}$. The result agrees with the constant two-dimensional free-electron density of states. Discrete level spacings in a finite sample would require a specified boundary spectrum.

One level occupied on average. Write $\delta ={\mu }_{B}B$ and $z=e^{\beta (\epsilon -\mu )}>0$. The two mean spin occupations areImposing $N_{+}+N_{-}=1$ gives $2+z(e^{\beta \delta }+e^{-\beta \delta })=1+z(e^{\beta \delta }+e^{-\beta \delta })+z^2$, hence $z^2=1$. Thus, for every positive temperature,The total magnetic moment is $\mathcal{M}=-{\mu }_B(N_{+}-N_{-})$. ConsequentlyThis is the grand-canonical spin susceptibility of a half-filled single level. Empty, singly occupied and doubly occupied orbital states are allowed in the grand canonical ensemble. At zero field, the singly occupied configurations together have probability $1/2$, explaining the factor of two relative to one permanent spin-half moment. If exactly one electron were required at all times, a canonical two-state partition function would instead give $\mathcal{M}={\mu }_B\tanh (\beta {\mu }_{B}B)$ and $\chi ={\mu }_B^2/(k_{B}T)$. That different answer does not implement the stated average-number constraint.

Density-of-states formula for the susceptibility. Here $g$ is the zero-field single-spin density of states. Regard its argument $E$ as the full spin-shifted energy. The shifted functions are $g_{\pm }(E)=g(E\mp {\mu }_{B}B)$, soThe total number is even in $B$; thus $\mu (B)=\mu (0)+O(B^2)$ at fixed total number on a smooth branch. More directly, terms from ${\mu }^{\prime }(0)$ cancel in the spin difference because the two densities coincide at zero field. To first order,It follows thatFor a physical energy band, take $g$ to vanish below its lower edge. Provided the endpoint term vanishes, integration by parts also givesThis form displays the positive response and the narrow thermal window around $\mu$. A sharp band edge is included by the corresponding distributional derivative in the first expression. The susceptibility defined here is the derivative of total magnetic moment with respect to $B$, with units $\mathrm{J}{\mathrm{T}}^{-2}$; it is not automatically a dimensionless SI volume susceptibility. A density-of-states convention per unit volume likewise gives a response per unit volume. These are the field conventions for magnetic susceptibility.

Chemical potential at fixed number. Define the integrated single-spin state countThe probability kernel $K_T$ has unit integral, mean $\mu$, and variance $a(k_{B}T{)}^2$. Using integration by parts,Expand $G$ about $\mu$. The linear term integrates to zero, while $G^{\prime \prime }=g^{\prime }$ givesThis is the number form of the Sommerfeld expansion. Since $N/2=G({\epsilon }_F)$ at zero temperature, set $\mu ={\epsilon }_F+\delta \mu$ with $\delta \mu =O(T^2)$:ThereforeThe coefficient can be obtained directly. With $y=(E-\mu )/(k_{B}T)$,For example, the sum follows by expanding $e^{-y}/(1+e^{-y}{)}^2$ on $y>0$, using the evenness of the kernel and integrating each exponential. Thus $a/2={\pi }^2/6$. A density of states increasing with energy gives a downward quadratic chemical-potential shift. For the constant two-dimensional density of states that shift vanishes to this order; a finite lower band edge still allows exponentially small corrections.

Low-temperature susceptibility at fixed number. Apply the same kernel expansion to $g$ itself:The shift of $\mu$ must be retained in the first term. Writing $g_F=g({\epsilon }_F)$ and similarly for its derivatives,ConsequentlyThis is the fixed-number low-temperature Pauli susceptibility. The $g_F^{\prime \prime }$ contribution is thermal broadening at fixed chemical potential; the $-g_F^{\prime 2}/g_F$ contribution enforces fixed total number. Both are needed. If the density of states were instead defined with both spin directions included, the prefactor $2$ would be absorbed into that definition. Only the spin response has been calculated; no orbital magnetic response is included.

The expansion assumes $g_F>0$, sufficient smoothness near the Fermi energy, and a temperature small compared with the energy scale over which $g$ changes or encounters a band edge. It does not apply unmodified at a singular density of states. For the ideal two-dimensional parabolic band, $g_F^{\prime }=g_F^{\prime \prime }=0$, so the leading Pauli response is temperature independent, consistently with the earlier state count.

An additive-looking energy expression does not by itself establish two noninteracting subsystems. Their allowed microstates must also form a Cartesian product of independently available states. Here the second label's allowed values depend on the first. The four allowed pairs and their energies areThe canonical partition function, with $\beta =(k_{B}T{)}^{-1}$, is thereforeThe conditional sum over $m_J$ is $2\cosh (\beta ϵJ)$ and depends on $J$; it cannot be pulled out as the partition function of an independent second subsystem. Summing independently over $J\in \{1,2\}$ and all four values $m\in \{-2,-1,1,2\}$ would invent four forbidden pairs and give eight states instead of four.

One can make the state labels independent by writing $s=m_J/J=\pm 1$, but then $E(J,s)=ϵJ(1+s)$ contains a coupling between them. For genuinely additive independent energies, the difference between the $s=+1$ and $s=-1$ states would be independent of $J$; here it is $2ϵJ$. Hence the proposed $J$ and $m_J$ subsystems cannot be treated as noninteracting independent subsystems. The literal two-value reading of the PDF's endpoint notation has been used. If that notation were instead intended as shorthand for full angular-momentum multiplets, including every integer between the endpoints, the eight allowed states would giveThe conclusion about the failure of independent-subsystem factorization is unchanged. The two state-counting conventions should not be mixed.

A nonzero superconducting energy gap suppresses the population of thermally excited quasiparticles. When $k_{B}T\ll \Delta$, their Boltzmann factor produces an approximately activated electronic heat capacity,where the prefactor $A(T)$ varies algebraically rather than exponentially. For the rough estimate requested, neglect its change between the two quoted temperatures. Taking a ratio eliminates its unknown magnitude:ThereforeThe corresponding excitation energy is $\Delta \simeq 2.19\times 10^{-22}\mathrm{J}\simeq 1.36\mathrm{meV}$. The units $\mathrm{mJ}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ of the two heat-capacity readings cancel in their ratio.

This is a heat-capacity activation estimate, not an exact determination of a gap from two data points. More generally, if $C_e=A{T}^p{e}^{-\Delta /(k_{B}T)}$, the same measurements giveFor example, the low-temperature BCS theory form for a fully gapped isotropic superconductor has $p=-3/2$ and would give about $19.5\mathrm{K}$. The stated approximate data and constant-gap assumption support the simpler $16\mathrm{K}$ estimate; they do not specify a detailed density of states or heat-capacity prefactor. Here $\Delta$ means the minimum quasiparticle excitation energy, rather than automatically twice that value.

Put $\beta =1/(k_{B}T)$ and define the thermal de Broglie wavelength of species $\alpha$ byIgnoring the internal degrees of freedom as stipulated, each molecule's translational canonical partition function is $V/{\lambda }_{\alpha }^3$. The two species do not interact, so their partition functions multiply; the treatment of indistinguishable particles requires a separate factorial for each species, not a factorial of the total particle number. ThusUsing $F=-k_{B}T\ln Z$ and Stirling's approximation gives the extensive Helmholtz free energyThese are ideal gas expressions, valid for $n_{\alpha }{\lambda }_{\alpha }^3\ll 1$, with $n_{\alpha }=N_{\alpha }/V$. The energy zero has been chosen so that separated CO and oxygen molecules have zero internal reference energy.

The given binding energy $ϵ$ belongs to the formation of two carbon-dioxide molecules. Relative to the reference used in (a), the internal energy offset of one carbon-dioxide molecule is therefore $-ϵ/2$, not $-ϵ$. Its mass is $m_c=m_{\mathrm{CO}}+m_{{\mathrm{O}}_2}/2$ and its thermal de Broglie wavelength is ${\lambda }_c=h/\sqrt{2\pi m_c{k}_{B}T}$, where the subscript $c$ denotes $\mathrm{C}{\mathrm{O}}_2$.

The single-molecule canonical partition function is $(V/{\lambda }_c^3)e^{\beta ϵ/2}$. For $N_c$ indistinguishable noninteracting molecules,The corresponding Helmholtz free energy, using Stirling's approximation, isThe negative sign in the last term expresses the stabilizing binding energy and ensures that low temperature favors association.

At fixed temperature, volume and the numbers of all other species, the chemical potential is ${\mu }_{\alpha }=(\partial F/\partial N_{\alpha }{)}_{T,V,N_{{\alpha }^{\prime }\ne \alpha }}$. For the combined noninteracting mixture, add the free energies from (a) and (b). Since the thermal de Broglie wavelengths are independent of particle number at fixed temperature, differentiation givesEach logarithm has a dimensionless argument, and each chemical potential has units of energy per molecule. The binding offset must be kept consistently when comparing species.

Introduce an extent of reaction $\xi$ increasing in the dissociation direction. The allowed particle-number changes conserve both carbon and oxygen atoms:At fixed temperature and volume, the resulting Helmholtz free energy change isAt an interior chemical equilibrium, both forward and backward changes are allowed, so the coefficient must vanish:Equality of the three individual chemical potentials would be wrong: a reaction exchanges species in its stoichiometric proportions rather than molecule for molecule. Stability is also explicit here, sincewhen all three species are present. Thus the equilibrium composition minimizes the free energy along the conserved-composition constraint.

A phase transition is a qualitative change of thermodynamic phase as control parameters such as temperature, pressure or chemical potential are varied. A genuine equilibrium singularity is a nonanalyticity of the appropriate thermodynamic potential in the thermodynamic limit; finite systems generally show rounded crossovers instead.

An order parameter is a macroscopic quantity that distinguishes the phases. It often vanishes in a symmetric disordered phase and becomes nonzero after spontaneous symmetry breaking, although density-driven transitions need not have that simple symmetry interpretation. It can be a scalar, vector or tensor. At a first-order phase transition an appropriate first derivative of the free energy is discontinuous, and the order parameter commonly jumps; at a continuous phase transition the order parameter can approach zero continuously while a susceptibility or a higher derivative becomes singular.

Three examples are:

These examples also illustrate why an order parameter must reflect the physical distinction between phases rather than merely be any quantity that changes with temperature.

Write $M=\mathcal{N}=V/b$ for the number of cells; this is fixed even though the particle number $N$ fluctuates. Approximate each position integral by the cell sum $\int d^3{r}_i\simeq b{\sum }_{{\alpha }_i=1}^M$, evaluating the slowly varying intercell pair potential at cell centres. With the assumed single-occupancy restriction, the fixed-$N$ contribution becomesAn occupation configuration $\{n_{\alpha }\}$ with $N={\sum }_{\alpha }{n}_{\alpha }$ selects an unordered set of $N$ occupied cells. There are exactly $N!$ ways to assign the temporary particle labels to that set. They all have the same interaction energy, cancelling the prefactor $1/N!$.

Summing over all particle numbers is therefore equivalent to summing each cell occupation independently over $0,1$:The activity per cell, $e^{\beta \mu }b/{\lambda }^3$, is dimensionless. This is a lattice gas, with each cell either empty or occupied.

The exclusion of multiple occupation is an additional physical modeling assumption, explicitly imposed by the question. Making a cell small does not by itself forbid two point particles from entering it; a repulsive core or a dilute-occupancy approximation would be needed to justify that restriction starting from a general continuum pair potential. In the remaining parts it is part of the defined cell model.

In the one-mole Van der Waals equation, $b>0$ represents excluded volume: the volume available to molecular motion is $V-b$, rather than $V$. Subtracting $b$ in the denominator therefore increases the repulsive contribution to pressure. The $a>0$ term represents intermolecular attraction, which reduces the outward momentum flux at a wall and hence appears as the negative correction $-a/V^2$. The two minus signs do not mean that both physical effects lower pressure.

For a reversible isothermal process along a homogeneous gas branch with $V_2>V_1>b$, the work done by the gas isTo find the internal energy change without guessing a caloric law, combine $dU=TdS-pdV$ with the Helmholtz free energy Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$. This givesThereforeThe attractive contribution $-a/V$ becomes less negative as the gas expands: separating attracting particles requires energy. This differs from the ideal gas result $\Delta U=0$ at fixed temperature. The required heat is $Q=\Delta U+W_{\mathrm{by}}=RT\ln [(V_2-b)/(V_1-b)]$. Below the critical temperature, an equilibrium path crossing a liquid–gas coexistence region needs the equilibrium coexistence construction rather than the unstable homogeneous Van der Waals isotherm.

The equilibria are $\boxed{(0,0),(\pm \sqrt{2/3},0)}$. The Jacobian is $J=\left(\begin{array}{cc}0& 1\\ 2-9x^2& -1\end{array}\right)$. At the origin it has eigenvalues $1,-2$ and tangent eigenvectors $(1,1)$ and $(1,-2)$, so the origin is a saddle equilibrium. At either nonzero point the eigenvalues are $\boxed{(-1\pm i\sqrt{15})/2}$, giving stable foci.

The energy-like function $H=y^2/2-x^2+3x^4/4$ has $\dot{H}=-y^2\le 0$. It is coercive at large distance, so trajectories remain bounded; there are no periodic orbits because their energy could not return to its starting value unless $y\equiv 0$. The stable manifold of the saddle equilibrium separates the two attracting basins. Its unstable branches leave the origin and approach the right and left foci, respectively. Their negative energy near the origin prevents crossing back through $x=0$.

For either invariant manifold write $y=h(x)=sx+c{x}^3+O(x^5)$. Odd symmetry rules out even powers. Tangency to the vector field requires $h^{\prime }(x)h(x)=2x-h(x)-3x^3$. Matching coefficients yields $s^2+s-2=0$ and $c(4s+1)=-3$. HenceThese are local graphs, not globally exact cubic curves.

For $\nu =1$, the equilibria are $(0,0)$ and $(\pm 1,0)$. At the origin the eigenvalues are $1,-1$, so it is a saddle for every $\kappa$.

At either nonzero point the Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -2& \kappa \end{array}\right)$, with eigenvaluesBoth points are attracting for $\kappa <0$ and repelling for $\kappa >0$. For $-2\sqrt{2}<\kappa <0$ they are stable foci; for $\kappa <-2\sqrt{2}$ stable nodes. For $0<\kappa <2\sqrt{2}$ they are unstable foci; for $\kappa >2\sqrt{2}$ unstable nodes. At the two repeated-root boundaries they are degenerate nodes. At $\kappa =0$, linearization has purely imaginary eigenvalues; the conserved energy in the next part shows that they are actual nonlinear centres. Thus the zero-real-part case is resolved by an additional first integral, rather than by the linear test alone.

An equilibrium is asymptotically stable when it is Lyapunov stable and attracts every initial state in some neighbourhood. Lyapunov stability means that sufficiently small initial displacement stays small for all future times; attraction additionally requires convergence to the equilibrium. A Lyapunov function is a differentiable positive-definite scalar function near the equilibrium whose derivative along solutions is nonpositive. Strict negativity, or an appropriate invariant-set argument for a non-strict derivative, gives asymptotic stability.

For $\nu =-1$, equilibria satisfy $y=0$ and $x+x^3=0$, so only $(0,0)$ is real. Its Jacobian matrix is $\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, with eigenvalues $\pm i$ for every $\kappa$. It is a linear centre equilibrium, so linearization alone cannot decide nonlinear attraction.

Use the positive proper energyFor $\kappa <0$, energy sublevel sets are compact and invariant. The zero-derivative set is the union of the axes, but its only invariant subset is the origin: on $x=0,y\ne 0$ one immediately leaves that axis, and on $y=0,x\ne 0$ one has $\dot{y}=-x-x^3\ne 0$. LaSalle's invariance principle therefore proves global asymptotic stability. At $\kappa =0$ the conserved energy levels are closed periodic orbits, giving a nonlinear centre equilibrium. For $\kappa >0$, time reversal together with $y\mapsto -y$ converts the system to the attracting negative-$\kappa$ case, so the origin is a nonlinear repeller and is unstable. Nonzero forward trajectories cannot remain bounded; this does not assert existence for all time if finite-time escape occurs.

Orient the flow so $q>0$, and let subscripts 1 and 2 denote upstream and downstream. The mechanical energy flux is $\rho qB$, where $B=u^2/2+gh$. Inserting the relation for $q^2$ in i gives the hydraulic jump head loss,An admissible jump dissipates mechanical energy, so $\rho q(B_1-B_2)>0$. The denominator is positive, henceThe zero depth-change case is continuous flow. A depth-decreasing discontinuity would require an energy input at the jump; it is not the dissipative hydraulic jump selected by this model.

The star is near the southwestern end of the main fault slip patch, so a much longer part of the finite-source rupture propagates northeastward. For comparable distances, ground conditions and buildings, the northeastern site receives the stronger, shorter direct pulse; the southwestern site receives the weaker, longer direct shaking. This is rupture directivity.^[ref]

For a segment of length $\ell$, rupture velocity $v_r$ and seismic wave speed $c>v_r$, successive arrival times span approximatelyForward arrivals overlap and concentrate energy into a shorter pulse. Backward arrivals spread out. The supplied finite-source rupture is not perfectly unilateral, so the prediction concerns the dominant contribution. Seismic radiation patterns, seismic attenuation, sediment amplification and building response can alter the ranking of actual damage or total coda duration.

Specify the amplitude convention. For incident stress/pressure amplitude one, the normal-incidence seismic reflection coefficient and stress seismic transmission coefficient areAt limestone–sandstone, $R_P=-3/19=-0.1579$ and $T_{\sigma P}=16/19=0.8421$. At sandstone–salt, $R_P=0$ and $T_{\sigma P}=1$. The first return reverses the chosen stress-pulse polarity; the second interface produces no normal-incidence P return.

For a particle-velocity amplitude convention, $R_v=-R$ and $T_v=1-R$, giving $(R_v,T_v)=(0.1579,1.1579)$ and $(0,1)$ respectively. These are amplitude coefficients, not fractions of energy. The lossless reflected and transmitted energy fractions are $\mathcal{R}=R^2$ and $\mathcal{T}=4Z_a{Z}_b/(Z_a+Z_b{)}^2=1-R^2$, which sum to one.

The normal-incidence acoustic impedance is $Z=\rho V_p$, with units $\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. ThusDefine the pressure-amplitude reflection coefficient and pressure-amplitude transmission coefficient for incidence downwards from layer $i$ into layer $j$ byHere $p$ denotes compressional normal-stress amplitude; in the water it is the acoustic pressure. Continuity of normal traction and normal particle velocity gives these expressions. They describe each interface separately, before any layer layer reverberation.

For a lossless normal-incidence plane wave, intensity is proportional to $p^2/Z$, soA transmitted pressure amplitude exceeding unity is therefore compatible with conservation of energy. If instead the coefficient is defined for particle velocity measured along the same downward coordinate, $r_v=-r_p$ and $t_v=2Z_i/(Z_i+Z_j)=1-r_p$: the two transmitted velocity coefficients are $0.8000$ and $18/29=0.62069$. Upward incidence reverses the sign of $r_p$, interchanges the transmitted pressure and velocity coefficients, and leaves $R_E,T_E$ unchanged. At normal incidence between these ideal isotropic layers there is no P-to-S seismic mode conversion.