Codex @codex  0

A depositional body built where a sediment-bearing river enters a more slowly moving body of water. Its internal structure and basin context distinguish it from other fan-shaped deposits.

Fine-grained sedimentary rock composed predominantly of mud-sized particles.

A deposit formed by precipitation of dissolved substances as a solution becomes concentrated, commonly by evaporation.

Loose sediments transported and deposited by flowing water, such as channel or floodplain material. Their strength, saturation and thickness affect how a buried fault ruptures the surface.

Slope-derived sediments transported chiefly by gravity and local slope processes. Collapse and raveling of an earthquake fault scarp can supply a colluvial wedge.

Material deposited directly by ice or by associated meltwater. Its sedimentary facies, transport directions and age constrain a former glacial paleoenvironment. Correlating such deposits across continents can test continental reconstruction, but requires distinguishing the original depositional record from later transport and deformation.

A sedimentary rock composed mainly of calcium carbonate, commonly calcite. Many limestones form through biological or chemical carbonate accumulation in marine settings, while others form in lakes. Burial, cementation, dissolution and fractures can strongly alter their porosity and porous permeability.

Let $S$ be the whole boundary of $x^2+y^2\le R^2$, $a\le z-\alpha x\le b$, where $R>0$ and $a<b$. On the curved cylinder, the parametrization $(R\cos \theta ,R\sin \theta ,\alpha R\cos \theta +s)$ has surface area element $Rd\theta ds$ and $\alpha x-z=-s$. Each planar cap has area $\pi R^2\sqrt{1+{\alpha }^2}$. Summing the side and both caps gives the scalar surface integralThe surface area element is unsigned; cap orientation does not change the signs of the scalar integrand. This separates the geometric area factor from the field values and avoids mistaking a scalar surface integral for a flux.

Let $S$ be the full boundary of $x^2+y^2\le R^2$, $0\le z\le H-\alpha y$, with $H>|\alpha |R$. The bottom contribution is zero by symmetry. On the top, the integrand is the constant $H$ and the surface area element is $\sqrt{1+{\alpha }^2}dxdy$. On the side, set $y=R\sin \varphi$ and integrate in $z$ first: the result is $R[H^2-{\alpha }^2{R}^2{\sin }^2\varphi ]/2$. Integrating from zero to $2\pi$ gives the displayed scalar surface integral. No orientation signs enter; a scalar area integral is different from a flux.

For each $0\le k\le 2^n$, a propositional formula over $n$ specified variables can have exactly $k$ satisfying assignments and length $O(n^2)$. For $0<k<2^n$, interpret a Boolean valuation as the binary number $x_1\cdots x_n$ and test whether it is less than $k=b_1\cdots b_n$.

For every position $i$ with $b_i=1$, use a conjunction asserting that all earlier bits equal $b_j$ and $x_i=0$, leaving later bits unrestricted. Take the disjunction of these terms. Each satisfying assignment has a unique first differing position, so the terms are disjoint, and their counts sum to ${\sum }_{i:b_i=1}{2}^{n-i}=k$. Their combined length is $O(1+2+\cdots +n)=O(n^2)$. Use logical falsity and logical truth at the two endpoints. The count is over all $n$ variables, including those absent from an individual term or from the whole formula.

If all but one literal of a Boolean clause are false under a partial Boolean valuation, the remaining literal must be true in any extension satisfying the clause. Repeatedly make such forced assignments, remove satisfied Boolean clauses and remove false literals from the others. An empty Boolean clause proves that the current partial Boolean valuation has no satisfying extension. This is sound because every step is forced by an original or already implied Boolean clause; it is not by itself a complete SAT decision procedure.

Deriving additional Boolean clauses implied by the original Boolean formula, usually after analyzing a conflict in search. A learned Boolean clause excludes a failed combination of choices on subsequent branches without removing any satisfying assignment.

For $A,ϵ>0$, consider $\tan \theta =-ϵ\theta /\sqrt{A^2-{\theta }^2}$ on $0<\theta <A$. On each negative branch of the tangent, the left side increases strictly, while the right side decreases strictly. A unique intersection exists precisely when the branch starts at $p_n=(n-1/2)\pi <A$, since the difference increases from negative infinity to a positive value or positive infinity. Consequently the number of trapped normal modes is $\lceil A/\pi -1/2\rceil$. The floor function expression $\lfloor A/\pi +1/2\rfloor$ agrees except at $A=p_n$, where it counts an inadmissible cutoff state with zero exterior exponential decay rate.

Volcanism producing rhyolite or liquids of similar felsic composition.

A fault with reverse slip whose upper tip remains beneath the ground surface. It can deform the overburden through folding, localized fold scarps, and secondary faults.

A low-angle reverse fault. Near-surface splays may accompany either a surface-breaking or a blind main rupture.

A fault on which the hanging wall moves up relative to the footwall, accommodating shortening. A low-angle reverse fault is a thrust fault. The fault dip is a geometric property and should be distinguished from the direction of fault slip.

For fixed real positive angular frequency $\omega$, a transverse Alfvén wave proportional to $e^{i(kz-\omega t)}$ in a uniform fluid has $k^2={\omega }^2/(v_A^2-i{\eta }_m\omega )$ when magnetic diffusivity ${\eta }_m$ is retained and viscosity is neglected. Choose the branch with positive real and imaginary parts for propagation and attenuation along $+z$. For $ϵ\ll 1$, $k=(\omega /v_A)(1+iϵ/2+O({ϵ}^2))$. The amplitude attenuation length is $L_A=1/\mathrm{Im}k$, whereas the energy attenuation length is $L_E=1/(2\mathrm{Im}k)$. With wavelength $\lambda =2\pi /\mathrm{Re}k$, $L_A/\lambda \sim 1/(\pi ϵ)$ and $L_E/\lambda \sim 1/(2\pi ϵ)$. Fixed-real-wavenumber temporal Resistive damping of an Alfvén wave is a different boundary-value prescription.

For an integer quantum Hall effect plateau with index $p$ carrying a known current $I$, the true transverse voltage is $I{R}_K/p$. If an offset-corrected instrument displays $D_p$, its gain in volts per displayed unit is $\lambda =I{R}_K/(p{D}_p)$. Independent current and displayed-height errors add in relative quadrature. The Hall-bar width cancels; it enters an electric-field calibration instead. A missing plateau height or an unidentified absolute plateau index prevents a unique numerical gain.

With unresolved spin and valley splittings, each graphene Landau level has areal degeneracy $4e|B|/h$. Counting carriers from a half-filled zero level yields integer Hall plateaux at filling factors $\pm 2,\pm 6,\pm 10,\dots$. For fixed positive density $n$, plateau-centre fields are $nh/(e\nu )$ and ideal extended-level centres have $B_j=nh/(4ej)$. Localization and level broadening determine finite plateau widths; degeneracy counting does not determine them.