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Within the rubbery regime, viscosity usually decreases as temperature increases, because segmental rearrangement becomes easier. The spring's entropic contribution generally increases approximately with absolute temperature at fixed effective network density, so $k$ tends to increase modestly under that assumption.

Consequently $\tau =\eta /k$ decreases. The storage-modulus crossover moves to larger squared frequency,and the high-frequency plateau moves upward if $k$ increases. The warmer curve in the figure illustrates those two changes. At intermediate frequencies the reduced relaxation time can lower the storage modulus even though its eventual plateau is higher; there is no requirement that the entire curve move upward.

These trends assume the same rubbery material state and no degradation, crystallization or change in effective network density. Close to other transitions the simple entropic-spring and viscosity picture needs modification.

Read the coesite–stishovite boundary at approximately $(400{}^{\circ }\mathrm{C},7.2\mathrm{GPa})$ and $(2800{}^{\circ }\mathrm{C},13.5\mathrm{GPa})$. The slope is thereforeA Celsius temperature difference equals the corresponding kelvin difference. The graph does not justify treating the endpoints as exact measurements.

For stishovite, the tetragonal cell volume divided by two givesFor monoclinic coesite the cell volume is $abc\sin \beta$, soUsing $M_{\mathrm{Si}{\mathrm{O}}_2}\simeq 0.060084\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ and Avogadro's constant, the mass-specific volumes areFor the direction stishovite to coesite,For the reverse, coesite to stishovite, the entropy change is about $-288\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. Stishovite is the denser phase: a positive boundary slope and its smaller volume imply a lower entropy on the high-pressure side. Stating the direction prevents an otherwise avoidable sign ambiguity.

Barium titanate has a high-temperature cubic perovskite structure, with Ti centred in its oxygen octahedron on average. This phase is paraelectric. Near room temperature the structure is tetragonal, with relative ionic displacements that give a spontaneous electric polarization. This phase is ferroelectric: an applied electric field can switch between differently polarized domains. Both phases have the same composition, BaTiO${}_3$; the transition near $120{}^{\circ }\mathrm{C}$ changes crystal symmetry and electrical response.

At a common test temperature and otherwise comparable microstructure, chemical order generally raises the yield strength. An ordinary FCC dislocation displaced by $a⟨110⟩/2$ can interchange Au and Cu sublattices, creating an antiphase boundary behind it. This additional energy resists slip; paired dislocations can restore the order after the second member passes. The disordered phase does not pay the same ordering penalty. Comparing specimens actually tested at their different phase-stability temperatures also includes temperature-dependent mobility and recovery, so the ordering argument alone does not predict a unique measured strength ratio.

At $138{}^{\circ }\mathrm{C}$, one line is tangent to all three Gibbs free energy curves at approximately $x_{\mathrm{Sn}}=0.03$ for (Bi), $0.61$ for liquid, and $0.93$ for $(\beta \mathrm{Sn})$. These three phases can coexist at the eutectic point. Between the two solid compositions, the equilibrium phase fractions depend on how much liquid has already transformed; exactly at the invariant temperature, the overall composition alone does not fix all three fractions. At the eutectic overall composition, cooling just below the temperature eliminates the liquid and leaves the two solids.

The annealed sample initially has relatively few obstacles to dislocation glide. After its elastic response it yields, undergoes appreciable plastic deformation and develops strain hardening. Irradiation produces dislocation loops and defect clusters that obstruct gliding dislocations, usually raising the yield strength. In a simple dispersed-obstacle model the strengthening scales as $\Delta \sigma \sim M\alpha Gb\sqrt{Nd}$, with obstacle density $N$ and diameter $d$.

Irradiated material often has less uniform plastic elongation and weaker subsequent hardening because slip localizes into channels that have cleared irradiation obstacles. Earlier necking or fracture can then reduce ductility. The elastic slope is much less affected than the yield stress in this schematic comparison; the exact strength and elongation depend on dose and temperature.

As-quenched martensite generally has greater hardness and strength, but lower ductility and toughness, than ordinary pearlitic steel. Its supersaturated carbon and distorted lattice impede dislocation glide; quench stresses also promote cracking. Pearlite combines hard cementite lamellae with more deformable ferrite, with finer spacings generally giving greater strength. Austenitize, quench to martensite, then temper to obtain a useful compromise between indentation resistance and resistance to fast fracture. Tempering below the austenitizing range relieves stresses and develops a fine carbide-containing microstructure. It trades some maximum hardness for improved toughness; the temperature and time must suit the required balance.

At $700^{\circ }\mathrm{C}$ the undercooling below the eutectoid temperature is small, so the free-energy driving force for forming pearlite is weak and nucleation is difficult. At $300^{\circ }\mathrm{C}$ the driving force is larger, but carbon diffusion is slow, limiting the coupled growth of ferrite and cementite. Their competition produces the nose of the isothermal transformation diagram near $550^{\circ }\mathrm{C}$ and $1\mathrm{s}$. A roughly linear quench from $740$–$800^{\circ }\mathrm{C}$ must pass left of this nose, requiring a rate of order $2\times 10^2$–$3\times 10^2\mathrm{K}{\mathrm{s}}^{-1}$, then continue below the martensite-start region near $240^{\circ }\mathrm{C}$. The tangent to the full start curve, rather than just the nose, refines this estimate; a true continuous-cooling diagram would be preferable. Martensite forms by a coordinated lattice-shape change without long-range carbon partitioning, so its formation is a displacive transformation. The tetragonal-to-monoclinic transformation of zirconia is a nonmetallic example.

Check stiffness and tip deflection using Young's modulus, failure from existing cracks using fracture toughness or the supplied critical energy-release rate, local crushing at supports, shear failure, buckling, fatigue and joint design. In a simple linear-elastic crack model, $K_c^2\simeq E^{\prime }{G}_c$, so toughness cannot be inferred from strength alone. Wood also has grain-direction dependence, moisture sensitivity, natural defects, property scatter and time-dependent creep; the design must control grain alignment and compression-side stability. Metals raise issues of corrosion, mass, joining, machining, residual stress and heat-treatment state. Cost, availability, reproducibility, service temperature and end-of-life treatment can decide between otherwise feasible designs. A high merit index is a screening result, not a complete safety specification.

A diamond anvil cell concentrates force onto tiny opposing culets, reaching very high pressure with small samples and useful optical access through diamond; its small volume, pressure gradients, alignment demands and difficulties with some heating or sample environments can limit experiments. A multi-anvil press supports a much larger sample and is useful for high-temperature synthesis and bulk measurements, but usually reaches lower ultimate pressures and has more restricted optical access. Hydrostaticity, attainable temperature and calibration depend on the actual apparatus and pressure medium; neither design guarantees uniform pressure.

For a general positive side length $s$, the cube has volume $s^3$, while separability of the integral givesThus the requested general dependence of the mean temperature isFor the cube with side length three this givesEquivalently each coordinate has volume-average value $s/2$, and the rectangular integration domain makes the mean of $xyz$ the product $(s/2{)}^3$.

A volume mean is the volume integral of the temperature divided by the volume. On this cube,Each one-dimensional integral is two, givingThe number is in the temperature units implicit in the given field law and coordinate units.

First method: substitute the trajectory. Along the parametric curve, the temperature is $T(t)=20e^{-{\cos }^{2}t}-(1+\sin t{)}^3$. The chain rule givesSecond method: use the spatial derivatives. The partial derivatives of the scalar field are $T_x=-40x{e}^{-x^2}$ and $T_z=-3(z+1{)}^2$, while the velocity components are $\dot{x}=-\sin t$ and $\dot{z}=\cos t$. The multivariable chain rule therefore giveswhich agrees with the first calculation. This is the gradient dotted with the velocity, rather than a partial derivative at a fixed position.

The left side is atmospheric heat storage per unit surface area. Incoming solar flux $Q$ is averaged over the spherical surface by the factor $1/4$; albedo $a$ reflects a fraction of it. The outgoing term is a reduced Stefan-Boltzmann flux, with effective emissivity $\gamma$ representing atmospheric greenhouse trapping. Ice-albedo feedback usually makes $a$ decrease as temperature rises through an ice-melting range; greater carbon-dioxide pressure increases infrared opacity and therefore decreases this effective $\gamma$. The simple estimates below hold albedo fixed.

Radiative equilibrium givesDifferentiating $\gamma T^4=\text{constant}$ gives $\Delta \gamma \simeq -4{\gamma }_0\Delta T/T_0\simeq -0.02474$. A doubling from $36$ Pa changes pressure by $36$ Pa, henceReference weathering balance requires $A_L{W}_0=v_0$, so $\boxed{W_0=2.0\times 10^{-3}\mathrm{kg}{\mathrm{m}}^{-2}\mathrm{y}{\mathrm{r}}^{-1}}$.

For the requested approximate system, take $(T_0,p_0)$ as the reference radiative/weathering equilibrium, set $\theta =(T-T_0)/\Delta T_c$, $P=p/p_0$, and $\hat{t}=t/t_{\mathrm{C}{\mathrm{O}}_2}$, whereLinearize the radiative loss, retaining terms linear in $T-T_0$ and $p-p_0$. Its two terms are $4\sigma {\gamma }_0{T}_0^3(T-T_0)$ and $-\sigma {\gamma }_1{T}_0^4(p-p_0)$. The carbon balance scales exactly with $v=v_{\mathrm{C}{\mathrm{O}}_2}/v_0$. ThereforeNumerically $\boxed{t_{\mathrm{C}{\mathrm{O}}_2}\simeq 7.41\times 10^3\mathrm{yr},t_T\simeq 0.0958\mathrm{yr},ϵ\simeq 1.29\times 10^{-5},\Gamma \simeq 0.1795}$. The time conversion uses seconds for the radiative constants and years for the supplied carbon production.

Because $ϵ\ll 1$, temperature rapidly approaches $\theta =\Gamma (P-1)$, while carbon changes slowly. A sustained $v\simeq 40$ has equilibrium determined byIts left side is strictly increasing on positive pressure, giving $\boxed{P_{\ast }\simeq 16.83,\Delta T_{\mathrm{eventual}}=\Delta T_c\Gamma (P_{\ast }-1)\simeq 36.9\mathrm{K}}$ within this extrapolated toy model. Counting volcanic production as well gives $v=41$ and $37.2$ K instead. An inventory estimate for the rise is $t_{\mathrm{C}{\mathrm{O}}_2}(P_{\ast }-1)/(v-1)\simeq 3.0\times 10^3$ years. Linearization near the final equilibrium gives the e-folding timeThus the carbon/weathering response is millennial, despite the short thermal adjustment time.

There are two numerical-reference inconsistencies in the printed setup. The requested expansion treats $(T_0,p_0)$ as its reference equilibrium although earlier labels pair $T_0$ with present-day $36$ Pa rather than $p_0=28$ Pa. Also, the calibrated $\Gamma =0.1795$ gives $13^{0.3}{e}^{13\Gamma }\simeq 22.3$, not the stated hint. That hint instead corresponds to $\Gamma \simeq 3/13$, obtained by calibrating a doubling of $28$ Pa. Such a convention gives a pressure around $13$–$14$ and a similar rough warming of several tens of kelvin. The displayed results above consistently use the stated $36$ Pa doubling calibration and explicitly declared reference-model approximation. A large extrapolation of a linearized radiative law is a property of this exercise, not a quantitative climate projection.

This is a weathering feedback with fast thermal adjustment.

Use dimensional quantities $T_d,t_d$ and set $T_d=T_{0}T$, $t_d=t_{0}t$. Dividing the energy balance by $\sigma \gamma T_0^4$ and the albedo equation by its time scale givesThus $\epsilon$ is the ratio of the atmospheric thermal response time to the much longer albedo response time. Converting kilometres to metres and years to seconds gives $C=10^7\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, $T_0\simeq 266.3\mathrm{K}$, $t_0\simeq 3.156\times 10^{10}\mathrm{s}$, andThe thermal time is about half a year, compared with a millennium for albedo adjustment. This is a singular perturbation: setting $\epsilon =0$ removes the independent initial condition for temperature.

Set $x=T^4/T_0^4-1$. Since $\sigma \gamma T_0^4=Q/8$, the steady state energy balance is $2(1-a_0)=1+x$. Substitution givesThe two sides have slopes $\mu$ and $2$ at the origin. For $x>0$, $\tanh (\mu x)$ is strictly concave, its initial slope is $\mu$, and it remains below $1$. Consequently no positive root exists for $\mu \le 2$, while for $\mu >2$ there is exactly one positive root $x_{+}\in (0,1/2)$. Oddness supplies the corresponding negative root. At $\mu =2$ the only root is still zero. Hence multiple steady temperatures occur precisely when $\mu >{\mu }_c=2$.

The largest root corresponds monotonically to the largest positive absolute temperature:For large $\mu$, the positive root tends to $1/2$, so $a_{+}\simeq 1/4$ and $T_{+}/T_0\simeq (3/2{)}^{1/4}$. The negative branch has larger albedo and lower temperature.

The heat capacity per unit horizontal area is $C=\rho c_{p}d$: mass density times specific heat capacity times atmospheric depth. Thus $C\dot{T}$ is the rate of stored thermal energy per unit area. The solar constant $Q$ is the incident radiative flux on a surface normal to sunlight; the ratio of the Earth's intercepting disk area to its spherical area is $1/4$. The absorbed fraction is $1-a$, where albedo $a$ is the reflected fraction. The outgoing radiative flux is $\sigma \gamma T^4$, a Stefan–Boltzmann law with effective radiation factor $\gamma$ incorporating the simplified greenhouse effect. Every term therefore has units of power per area.

The albedo curve decreases smoothly, passes through $(T_0,1/2)$, and approaches $1/4$ at high temperature. At $T=0$ it is $(2+\tanh \mu )/4$, close to $3/4$ when $\mu$ is large. Its derivative isAn ice-covered planet reflects more sunlight than exposed land or ocean. Warming removes reflective ice, lowers albedo, and increases absorbed radiative flux: this is ice-albedo feedback. The parameter $\mu$ controls the sharpness of this transition; the formula is an idealized global response, rather than a local melting law.

Initially $a=1/2$ and $T=T_{+}/T_0>1$, so $\epsilon T_t=1-T^4<0$. On the fast time $s=t/\epsilon$, albedo changes by only $O(\epsilon )$ andThus temperature rapidly falls towards $1$. At the same time $a_t=a^{\ast }(T)-1/2<0$ while $T>1$, which seeds selection of the warm branch. Afterwards the approximate slow manifold is $T^4=2(1-a)$: decreasing albedo makes temperature rise slowly back towards $T_{+}/T_0$. The preceding figure separates these two time scales; a single time axis would compress the initial fall into a nearly vertical segment.

In the sharp-transition regime, after selection of the warm branch $a^{\ast }(T)\simeq 1/4$. Solving the linear ordinary differential equation with the leading matched value $a(0^{+})=1/2$ givesThese are outer solutions: their initial temperature is $1$, so the fast cooling layer is required to satisfy the actual initial temperature.

There is a useful qualification to the two-limit argument. For fixed finite $\mu >2$, put $x=1-2a$ on the slow manifold. Then $x_t=-x+\frac{1}{2}\tanh (\mu x)$, and $x=0$ is an unstable equilibrium with growth rate $\mu /2-1$. The small positive $O(\epsilon )$ seed from the fast layer can therefore produce a delay of order $\log (1/\epsilon )/(\mu /2-1)$. The simple exponential approximation assumes that the sharp albedo transition is sufficiently fast to make this delay negligible on the displayed slow scale. It is not a uniform approximation for every fixed $\mu >2$ as $\epsilon \to 0$.

Balances and meanings. The first equation is conservation of energy per unit area. Here $c$ is effective areal heat capacity, $Q$ is the solar constant, and $Q/4$ accounts for the ratio of the Earth's intercepting disc area to its total surface area. The albedo $a$ reflects a fraction of that input. The outgoing thermal radiation is the Stefan–Boltzmann law modified by $\gamma (p)$, an effective transmission factor for the greenhouse effect; it normally decreases as atmospheric carbon dioxide increases.

The albedo relaxes towards its temperature-dependent equilibrium on the ice-adjustment time $t_i$. The atmospheric mass inventory per unit pressure is $M_{{\mathrm{CO}}_2}A/(M_{a}g)$, using the molar mass ratio of carbon dioxide to air and hydrostatic atmospheric mass. Its sources and sinks are volcanic input $v$, removal by chemical weathering over land area $A_L$ at rate $W$, and the atmosphere-to-ocean exchange $h(p-p_s)$. The ocean equation is a molar mass conservation balance: ${\rho }_{{\mathrm{H}}_2\mathrm{O}}{V}_{\mathrm{oc}}$ is ocean water mass, division by $M_{{\mathrm{CO}}_2}$ converts the incoming carbon-dioxide mass to moles, and $bC$ removes dissolved carbon by burial/export. With $C$ in moles per kilogram, $b$ has units of kilograms of water-equivalent per year.

Times and reduction. Linearize the temperature equation at fixed $a,p$ about radiative equilibrium. Its exponential relaxation time isThe equality uses $\sigma \gamma T^4=Q(1-a)/4$. A stock-over-flux thermal time, sometimes used as a scale, is four times this, $0.381\mathrm{y}$; both describe a rapid thermal response relative to the other processes.

For pressure in atmospheres, write $I=M_{{\mathrm{CO}}_2}A10^5/(M_{a}g)$ for the atmospheric carbon-dioxide mass per atmosphere. The factor $10^5$ is essential because $h$ is given per atmosphere. The pressure and slow ocean exponential relaxation times areThus $t_T\ll t_p\ll t_i=10^4\mathrm{y}<t_C$. The leading quasi-steady approximation givesSubstitution in the two remaining equations yieldsThe coefficient of $\dot{a}$ remains $t_i$; the printed $t_T$ in the reduced equation is a typographical error. Eliminating a fast temperature variable does not replace the ice-response time. The ocean time here is the slow time after atmospheric balance cancels the exchange flux; holding pressure fixed after imposing a carbon-dependent $p_s$ would describe a different conditional relaxation.

The supplied volcanic input givesabout $41.1\%$ of the stated present value $2.0\times 10^{-3}$; that value is about $2.43C_0$.

Carbon-dioxide levels. Since $A_{L}W=3.0\times 10^{11}\mathrm{kg}{\mathrm{y}}^{-1}=v$, the atmospheric balance with $p_s=C/K$ givesThis uses the specified present ocean concentration, as requested, rather than replacing it by the model's smaller volcanic-only equilibrium.

After sustained additional input $v_i$, $C$ barely changes during the atmospheric adjustment. At this stageThis is an intermediate balance, approached over several centuries. On the slower ocean time,with pressure following $p\simeq C/K+v_i/h$. Since $C(0)<C_{\infty }$, pressure continues to rise slowly. Its eventual level in this stipulated constant-parameter model isThe two-reservoir carbon adjustment therefore has a rapid rise towards the frozen-ocean balance followed by a much longer increase. The figure solves the coupled linear pressure–ocean equations, retaining their small inventory correction to the reduced time scales.

Sliding and cavities. Regelation is melting under increased pressure and refreezing when the pressure is released. For ice near its melting point, the Clapeyron equation gives a lower melting temperature at greater pressure. Against an obstacle, ice melts on the high-pressure upstream face, the water passes around it, and it refreezes on the lower-pressure downstream face. Transport of the released latent heat back towards the melting face sustains the process. Together with viscous deformation around obstacles, this permits basal sliding; regelation is especially effective for small obstacles, for which heat conduction distances are short.

Subglacial cavitation occurs where the contact pressure predicted behind an obstacle falls below the water pressure: the ice separates from the bed and a water-filled cavity opens. The effective pressure in subglacial drainage isthe ice overburden pressure minus water pressure. Smaller $N$ facilitates larger cavities and reduces the resisting ice–bed contact. The exponent $n$ comes from Glen's flow law, with strain rate proportional to the $n$th power of deviatoric stress, commonly $n\simeq 3$. Its homogeneity makes $N^n$ the natural velocity scale, yielding $\tau /N=f(u/N^n)$ after including the fixed bed and ice-rheology scales.

Without cavities, material creep gives a drag law with $f(x)\propto x^{1/n}$. Cavities lower the resistance relative to that extrapolation. On a bounded-slope bed the drag is bounded in proportion to $N$; depending on bed geometry, the curve may level off or pass through a maximum and weaken as contact diminishes. The illustrated weakening curve is a qualitative possibility, not a universal formula for all beds.

Channels versus distributed drainage. In Röthlisberger channel drainage, frictional dissipation of flowing water supplies heat to melt the channel walls, while material creep of ice closes them under effective pressure. An area balance has the schematic formwhere $\Psi$ is hydraulic-potential drop per length, $L_f$ is latent heat per mass, and the positive constant $\mathcal{A}$ incorporates ice material creep and geometry. Water-flow resistance relates the flux to channel size and potential gradient. Balancing melting against closure produces the stated increasing relation $N_R(Q)$.

A channel gaining flux can maintain lower water pressure, since larger $N$ means smaller $p_w$. It draws still more water from neighboring routes, melts wider, and causes competing small conduits to lose flux and close. This positive feedback concentrates flow into an arterial network.

A linked cavity drainage system consists of water-filled lee-side cavities connected by narrow passages. Sliding over bed obstacles opens cavities; ice material creep tends to close them. The given decreasing relation $N_K(Q)$ means that carrying more water requires higher water pressure. A route gaining flux then has a higher pressure and diverts water towards neighboring routes. This negative redistribution feedback supports many connected routes and a distributed system.

Coexistence instability. At a coexistence point let $N_R(Q_R)=N_K(Q_K)$ and hold $Q=Q_R+Q_K$ fixed. A loss $\Delta Q_R=-\epsilon$ from the channel gives $\Delta Q_K=\epsilon$. The changes in the two equilibrium effective pressures areConsequentlyIf the bracket is positive, the cavity system's common effective pressure exceeds what the reduced-flux channel can support against material creep closure. The channel shrinks, loses further flux, and closes down. Thus the channel-cavity drainage instability condition is