Let $S_g$ be the fraction of the pore space occupied by gas, and let $\varphi$ be the porosity. At fixed porosity and unchanged rock framework, the bulk density isReplacing water with much less dense gas therefore decreases the bulk density approximately linearly, as in Figure 1. The liquid and gas densities here are those at reservoir conditions.

A pore fluid has negligible static shear modulus, so the rock framework chiefly controls the effective shear modulus $\mu$. If $\mu$ changes little, $V_s=\sqrt{\mu /\rho }$ increases gently as the density falls. This does not imply that S waves travel through the pore gas itself: their restoring force comes from the connected solid framework.

The P wave speed depends on both bulk modulus and shear modulus:Gas is much more compressible than water. Even a small connected gas fraction can strongly reduce the effective fluid bulk modulus, approximately through the compliance average ${\kappa }_f^{-1}\simeq (1-S_g)/{\kappa }_w+S_g/{\kappa }_g$ for a pressure-equilibrated mixture. The saturated rock's effective bulk modulus consequently falls rapidly at first, overwhelming the smaller density reduction and causing the sharp initial fall in $V_p$. Once the fluid is already very compliant, further gas replacement changes the stiffness less; the continuing density decrease can then make $V_p$ rise again. Figure 1 assumes a particular saturation response: patchy fluids, changing effective pressure, and frequency-dependent fluid motion can alter it. The plotted $V_s$ is only nearly constant, so the plot should not be interpreted as an exact fixed-$\mu$ constitutive model.

For fugacity coefficient $\varphi =0.75$, use $f=\varphi P=3375\mathrm{bar}$ for the water activity. The solid pressure correction still uses the actual pressure $4500\mathrm{bar}$. ThusLowering the water chemical potential favours the product side, which lowers the equilibrium temperature relative to part (c) under the stated constant-property model.

Adopt ideal independent mixing on crystallographic sites, with Mg and Fe sharing three positions per formula unit and the other endmember sites fixed. This gives the phlogopite and pyrope activitiesThe reaction quotient contains their ratio, since pyrope is a product and phlogopite a reactant. ConsequentlyThe site-mixing assumption is part of this numerical answer: compositions alone do not determine nonideal thermodynamic activities. More detailed activity coefficients or unequal site occupancies would need an additional model. The cube counts the three magnesium-bearing positions rather than three molecules of each mineral.

Apply the pressure correction to a mineral reaction to the solids and use the chemical fugacity of the water product. For pure solids and water vapour treated as an ideal gas,The gas contribution is already included by the logarithm and must not be counted again as an independent volume correction. With pressures in bars,

At the one-bar standard state, pure phases have unit thermodynamic activities. The Gibbs energy of reaction balance is ${\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }=0$, givingThis uses the stipulated temperature-independent reaction enthalpy and entropy differences, taken per mole of the written reaction.

Write ${\delta }_j=\sqrt{2K/{\omega }_j}$ for each component's thermal penetration depth. Its contribution at depth isThus its amplitude is multiplied by $e^{-x/{\delta }_j}$ and its wave phase lags the surface by $x/{\delta }_j$ radians. The corresponding unwrapped time lag is $x/({\delta }_j{\omega }_j)$; a phase measured from a single periodic signal is observable only modulo $2\pi$.

Higher-frequency variations have smaller penetration depths. In particular,The daily oscillation is therefore strongly attenuated at depths where a substantial annual oscillation survives. This is a thermal diffusion wave, whose phase lag is not the arrival time of a sharp thermal front. The displayed solution is the established periodic response; an additional transient would depend on initial subsurface data not needed for this interpretation.

Apply a Fourier transform or discrete Fourier transform, retain the desired low-frequency coefficients and attenuate those in the high-frequency noise band with a low-pass filter, then invert. For a real signal, apply matching gains at positive and negative frequencies to preserve Hermitian symmetry.

Parseval identity for Fourier series shows what an ideal coefficient cutoff does: the output mean-square signal power is the sum of the retained squared coefficients, and cannot exceed that of the unfiltered input. Much of the high-frequency noise can be removed, but a finite cutoff also removes genuine square wave harmonics. Its edges become less sharp, and a sharp spectral cutoff can exhibit the Gibbs phenomenon. There is no claim of exact recovery if the noise overlaps the retained band. Noise already folded into low frequencies by aliasing cannot in general be removed merely by discarding the high-frequency bins afterward.

Most of the mean-square signal power lies at low frequencies. Each nonzero coefficient has magnitude $2/(\pi |r|)\mathrm{V}$, so its power contribution decreases as $r^{-2}$. The pair at the fundamental frequencies $\pm 0.25\mathrm{Hz}$ alone containsof the total. The first three positive odd harmonics, together with their negative-frequency partners, contain about $93.3\%$. The harmonic power of a square wave still has a nonzero tail: higher harmonics are needed to reproduce sharp edges even though their combined power is small.

Use Parseval identity for Fourier series with the coefficient normalization above. Only odd nonzero indices contribute; summing the positive and negative indices givesThusThe factor of two from the negative frequencies is essential. The discontinuities do not obstruct Parseval's identity, which is a statement about L2 space.

The squared signal equals $1{\mathrm{V}}^2$ except at irrelevant isolated jump times. Its mean-square signal power is thereforeHere “power” has the signal-processing definition specified in the question. To convert this voltage mean square into electrical power in watts, a load electrical resistance would also be required: $P_{\mathrm{electrical}}=⟨g^2⟩/R$.

The zero Fourier coefficient is $u_0=0$. For a nonzero integer $r$, integrate the two constant halves:ConsequentlyThe complex Fourier series is $g(t)\sim {\sum }_{r\in \mathbb{Z}}{u}_r{e}^{ir\pi t/(2\mathrm{s})}$. The negative-index coefficients are the complex conjugates of the positive-index ones, as required for a real signal. Combining each pair gives the equivalent real expansionAt a jump, pointwise convergence of a piecewise smooth Fourier series gives zero, the average of the two one-sided limits. The value assigned to $g$ at isolated jumps does not affect its Fourier coefficients or mean-square signal power.