Eliminating the shear stress from the two linear equations gives the Maxwell shear-wave dispersion equation in physical space,where $\nu$ is the kinematic viscosity. Substitution of a Fourier mode $e^{iky+\sigma t}$ givesThese are temporal exponents: the oscillation angular frequency is the imaginary part of $\sigma$, with the sign determined by the selected time convention. For $4\tau \nu k^2<1$ both exponents are real and nonpositive. At equality the repeated exponent is $-1/(2\tau )$; generic solutions include $t{e}^{-t/(2\tau )}$. Above the threshold the modes are damped propagating shear disturbances:At large $|k|$ their phase velocity tends to $\sqrt{\nu /\tau }=\sqrt{G_{\mathrm{el}}/{\rho }_0}$.

For the limit $\tau \to 0$ at fixed $\mu ,{\rho }_0$ and $k$, expansion of the square root givesAfter a short initial relaxation, the surviving branch obeys the diffusion equation $u_t=\nu u_{yy}$ of a Newtonian fluid. The fast branch disappears on the vanishing viscoelastic relaxation time; at $k=0$ the two exponents are exactly $0$ and $-1/\tau$. This limit is not uniform in $k$: modes with $k$ of order $(\tau \nu {)}^{-1/2}$ still sample the oscillatory transition. If instead $G_{\mathrm{el}}=\mu /\tau$ is held fixed as $\tau \to 0$, then $\mu \to 0$ and the fixed-$k$ slow exponent tends to zero. The ordinary Newtonian fluid limit therefore specifies which material parameter is held fixed.

Linearize about a relaxed state $\mathsf{T}=0$ with constant mass density ${\rho }_0$. In the unforced momentum conservation equation ${\rho }_0{D}_t\mathbf{u}=-\nabla p+\nabla \cdot \mathsf{T}$, the $x$ component of the pressure gradient is zero for a freely evolving unidirectional flow independent of $x$. A prescribed external pressure gradient would add a forcing term.

For $\mathbf{u}=u(y,t)\hat{\mathbf{x}}$, incompressibility is automatic and the only nonzero entries of the rate-of-strain tensor are $E_{xy}=E_{yx}=\frac{1}{2}{\partial }_{y}u$. Products of stress and velocity gradient are quadratic in the small amplitude. The constitutive equation consequently becomes $\mathsf{T}+\tau {\partial }_t\mathsf{T}=2\mu \mathsf{E}$ to first order. With $T=T_{xy}=T_{yx}$, its shear component and the $x$ component of momentum conservation areThe diagonal stress components have only homogeneous relaxation at this order and vanish if initially relaxed. In particular, $\alpha$ does not enter this linear response.

When deformation is stopped, the rate-of-strain tensor and velocity gradient vanish, leaving $\mathsf{T}+\tau {\partial }_t\mathsf{T}=0$. HenceThe positive constant $\tau$ is the viscoelastic relaxation time. For slow weak deformation, all the transport and quadratic terms are smaller than $\mathsf{T}$, so $\mathsf{T}\simeq 2\mu \mathsf{E}$. Thus $\mu >0$ is the zero-rate dynamic viscosity, measured in stress times time. The corresponding short-time shear modulus of the Maxwell fluid response is $G_{\mathrm{el}}=\mu /\tau$. The source's $\alpha$ has units of time, since $\alpha \mathsf{TE}$ must have units of stress.

The Adjoint representation of G2 has one weight at each of its twelve roots and two at zero. Its six long roots form the A2 root system; together with the two zero weights they give the Adjoint representation of SU(3). The remaining six short roots form the weight systems of its defining representation and dual representation. Hence restriction to the long-root SU(3) gives $14=8\oplus 3\oplus \overline{3}$. This is a branching of a representation; the complement is invariant under the subgroup, not under the full G2 action.

The weight system of a finite-dimensional Lie algebra representation is its collection of weights with their weight multiplicities. Restriction to a Lie subalgebra projects each weight to the smaller Cartan subalgebra, retaining and combining weight multiplicities. Grouping the resulting weights into irreducible weight systems computes the branching rule.

Suppose two dissolution reactions to a reference mineral $C$ have been normalized to the same amount of a conserved reference component. After division by the same positive energy factor, let their Gibbs energies of reaction be $y-{\ell }_1(x)$ and $y-{\ell }_2(x)$. Their relative thermodynamic potentials can then be written as displayed. The stable mineral has the smallest potential.

It follows that $C$ is stable below $y=\min ({\ell }_1,{\ell }_2)$, mineral 1 is stable where ${\ell }_1<{\ell }_2$ and $y>{\ell }_1$, and mineral 2 is stable where ${\ell }_2<{\ell }_1$ and $y>{\ell }_2$. The two nonreference minerals coexist stably where ${\ell }_1={\ell }_2<y$. Their equality below that height is a metastable continuation. This comparison proves the stable lower envelope and selects which portions of the pairwise mass-action lines belong on the mineral activity diagram.

For the B2 root system with long simple root first, the four-dimensional spinor representation has highest weight $(0,1)$. Its symmetric square has highest weight $(0,2)$ and dimension ten, so it is the Adjoint representation. The exterior square has the four nonzero defining weights and two zero weights; it splits into the five-dimensional vector representation of highest weight $(1,0)$ and a trivial representation. Thus the tensor square decomposes as displayed. The invariant alternating bilinear form reflects the pseudoreal representation structure. The spinor representation is a group representation of the Spin group $\mathrm{Spin}(5)$, while its tensor square descends to $SO(5)$.

Choose five Hermitian four-by-four matrices obeying the Euclidean Clifford algebra relations ${\gamma }_a{\gamma }_b+{\gamma }_b{\gamma }_a=2{\delta }_{ab}I$. For $M_{ab}=E_{ab}-E_{ba}$, the displayed anti-Hermitian matrices satisfy the Special orthogonal Lie algebra Lie bracket relations. A direct proof first establishes $[\rho (M_{ab}),{\gamma }_c]={\delta }_{bc}{\gamma }_a-{\delta }_{ac}{\gamma }_b$, then applies the commutator product rule to $[{\gamma }_c,{\gamma }_d]$. One explicit choice is ${\gamma }_1={\sigma }_1\otimes I$, ${\gamma }_2={\sigma }_2\otimes I$, ${\gamma }_3={\sigma }_3\otimes {\sigma }_1$, ${\gamma }_4={\sigma }_3\otimes {\sigma }_2$, ${\gamma }_5={\sigma }_3\otimes {\sigma }_3$, with the Pauli matrices. The Cartan subalgebra operators $-i\rho (M_{12})$ and $-i\rho (M_{34})$ have simultaneous eigenvalues $(\pm 1/2,\pm 1/2)$, identifying the four-dimensional spinor representation. A full $2\pi$ rotation acts as $-I_4$, proving it does not descend to $SO(5)$.

The Spin group $\mathrm{Spin}(7)$ double covers SO(7). Its central kernel element $-1$ acts trivially in the vector representation and as $-I$ in the spinor representation. Thus a spinor Lie algebra representation does not define an ordinary SO(7) group representation. Under $\mathrm{Spin}(6)\cong SU(4)$ the same obstruction is visible because $-I_4$ acts trivially on ${\Lambda }^{2}4$ but negatively on $4\oplus \overline{4}$.

For a spin-degenerate parabolic two-dimensional band, the Drude model gives $\tau =m^{\ast }/(n{e}^2{\rho }_{\square })$ and $v_F=\hslash k_F/m^{\ast }$. Their product is the displayed mass-independent mean free path. A Fermi-surface transport interpretation requires a sufficiently degenerate electron gas; a quantum lifetime need not equal the transport relaxation time.

For the Driven two-level Rabi model at fixed resonant pulse duration $T_{\pi }=\pi /\Omega$, put $x=\delta /\Omega$. Its transfer probability isThe sine-squared numerator differs from one only at order $x^4$, while the denominator gives $1-x^2+O(x^4)$. Thus the leading population-transfer error is $x^2$. For a target transfer error $ϵ$, the frequency mismatch in cycles per second should satisfy $|\delta \nu |\lesssim \Omega \sqrt{ϵ}/(2\pi )$ in this small-detuning regime. This is not a complete characterization of every coherent gate error.

The fugacity coefficient gives $f_{{\mathrm{H}}_2\mathrm{O}}=\varphi P=0.82\times 4500=3690\mathrm{bar}$. Keep the solid pressure correction at the actual $4500\mathrm{bar}$; only the water activity uses the fugacity. ThusThe decrease from part (c) follows because $\varphi <1$ lowers the chemical potential of the water product. All logarithms in these thermodynamic formulae are natural logarithms.

The compositions alone do not specify a unique thermodynamic activity model. Adopt the usual ideal, independent mixing on crystallographic sites approximation: three Mg–Fe positions per formula unit, with the other sites fixed at their stated endmember compositions. Then the mineral endmember activities areThe exponent counts the three mixing cations; treating each whole mineral formula as a single mixing site would be a different model. In particular, the magnesian biotite and garnet components here are phlogopite and pyrope.

For the written reaction, the reaction quotient iswith the other solid activities equal to one. The Gibbs energy of reaction balance therefore givesThe smaller product-endmember activity lowers the reaction chemical potential and lowers the equilibrium temperature in this model. Nonideal activity coefficients or unequal occupancies of crystallographic sites would require further information rather than a revised universal numerical answer.^[ref]

Use the pressure correction to a mineral reaction for the solids, with $P_0=1\mathrm{bar}$. The water contribution must instead use its chemical fugacity. For pure solid phases and water vapour treated as an ideal gas, $f_{{\mathrm{H}}_2\mathrm{O}}=P$, givingThe last term already accounts for the gas-pressure contribution; adding another gas-volume correction would double count it. The pressure ratio inside the logarithm is dimensionless.

With pressure expressed in bars, the solid correction is $-2.28(4500-1)=-10257.72\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}$. ThusUsing $4500$ rather than $4499$ for the solid pressure increment would silently shift the specified one-bar reference.

At the one-bar standard state, pure mineral endmembers and water vapour treated as an ideal gas all have unit thermodynamic activity. The Gibbs energy of reaction is ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$. Setting it to zero givesThe quantities are taken per mole of the reaction as written. The simplified result uses the stipulated temperature-independent enthalpy and entropy differences.

At $t=0$, the displacement expansion is $f(x)={\sum }_n{A}_n\sin (n\pi x/L)$. Fourier orthogonality givesDifferentiating the modal solution at zero gives $g(x)={\sum }_n{\omega }_n{B}_n\sin (n\pi x/L)$, soThe factor $1/{\omega }_n$ is required because the derivative of $\sin ({\omega }_{n}t)$ at zero is ${\omega }_n$. The dimensions agree: $A_n$ and $B_n$ are displacements, while the sine coefficients of $g$ are velocities. These projections also establish uniqueness within the finite-energy solution class.

The wave equation and its homogeneous Dirichlet boundary conditions are linear, so any finite linear combination of the separated normal modes is another solution. More strongly, the Fourier sine basis is complete in L2 space on $(0,L)$. Expanding the initial fields in this basis yieldsCompleteness of the Fourier sine basis is the reason the expansion covers general data; linearity alone would not prove that every solution has this form. The convergence and differentiation are understood in a function space appropriate to the data. For finite wave energy, one can take $f\in H_0^1(0,L)$ and $g\in L^2(0,L)$, obtaining an energy solution; stronger smoothness and endpoint compatibility give a classical solution.

For each spatial normal mode, the temporal equation is $T_n^{\prime \prime }+{\omega }_n^2{T}_n=0$, with angular frequency ${\omega }_n=c{k}_n=cn\pi /L$. HenceThe ordinary frequency is ${\omega }_n/(2\pi )=cn/(2L)$; it should not be confused with the angular frequency used in the sine and cosine arguments.

For $k>0$, the harmonic oscillator equation gives $X=C\sin (kx)+D\cos (kx)$. The condition at zero requires $D=0$, and the condition at $L$ requires $C\sin (kL)=0$. A nonzero normal mode therefore hasA multiplicative normalization has been absorbed into the temporal amplitude. The $k=0$ candidate is affine and the two zero endpoint values force it to vanish. A positive separation constant gives hyperbolic functions, whose two zero endpoint values likewise give only the trivial solution. Negative $n$ adds no new eigenfunctions because it only reverses their signs.

Use separation of variables with a nontrivial product $\eta =XT$. Substitution into the wave equation on a string gives ${\tau }_0{X}^{\prime \prime }T={\lambda }_{0}X{T}^{\prime \prime }$. Wherever the factors are nonzero,The left side depends only on $x$ and the right only on $t$, so both are a constant. For nontrivial fixed-end solutions this constant is negative: multiplying $X^{\prime \prime }=\sigma X$ by $X$ and integrating, using the homogeneous Dirichlet boundary conditions, givesWrite it as $-k^2$. The separated equations are thereforeThese equations extend across the zeros of $X$ and $T$; the division is a derivation device, not a restriction excluding their nodes.