Osmotic pressure is the excess pressure needed on a chemical solution to maintain osmotic thermodynamic equilibrium with pure solvent across a semipermeable membrane. For a nearly incompressible solvent, equality of chemical potential gives $\Pi \simeq -(RT/{\bar{V}}_w)\ln a_w$. It is an equilibrium property; an observed initial swelling rate is a different measurement.^[ref]

For a dilute ideal chemical solution of retained particles,Here $n$ is the amount of substance of independently dissolved solute particles, measured in moles; $V$ is the final volume of chemical solution, not the starting volume of pure solvent. Count the resulting ions separately if a solute dissociates; do not multiply twice if $n$ already counts them. With $c_{\mathrm{osm}}$ in $\mathrm{mol}{\mathrm{m}}^{-3}$ and $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, $RT{c}_{\mathrm{osm}}$ has units $\mathrm{J}{\mathrm{m}}^{-3}=\mathrm{Pa}$. This resembles the ideal gas law, although the particles here are dissolved in a liquid.

The assumptions are uniform temperature, a dilute ideal chemical solution, a nearly incompressible solvent, and a semipermeable membrane that retains the counted solutes while passing solvent. Association, incomplete dissociation and interactions change the relation between added material and independent particles; nonideal thermodynamic activities invalidate simple particle counting. The formula does not describe a concentrated brine merely because the apparatus contains a membrane.

A short derivation explains why particles are counted. In an ideal mixture, $a_w=x_w=n_w/(n_w+n)$, with $n_w$ the amount of substance of water. If $n\ll n_w$, then $-\ln x_w=\ln (1+n/n_w)\simeq n/n_w$ and $V\simeq n_w{\bar{V}}_w$. Substitution gives $\Pi \simeq RTn/V$. This is a dilute-limit approximation, not a claim that water molecules stop moving or that dissolved particles are a gas.

Water potential expresses the chemical potential of water relative to a stated pure-water reference in units of pressure. For a nearly incompressible liquid at fixed temperature, use a constant reference molar volume ${\bar{V}}_w$ and writeThe reference is pure water at the same temperature and a chosen pressure $P_0$. Both sides must use the same reference. Neglecting height differences and interactions with solid surfaces, the approximation isHere $a_w$ is the dimensionless thermodynamic activity of water, $R$ the molar gas constant, and $T$ the absolute temperature. The natural logarithm is therefore applied to a dimensionless quantity. The pressure contribution ${\Psi }_p=P-P_0$ increases with pressure; the solute potential ${\Psi }_s$ is negative when $a_w<1$. Across an available passive water route, net water transfer favors higher to lower water potential. The familiar plant description adds effects of height and interactions with solid surfaces when they matter; they are omitted in the two liquid compartments here.^[ref]

Adding solute and raising pressure can thus cancel in water potential while each contribution remains nonzero. Equality of water potential does not imply equality of pressure or of every solute concentration. In a supported plant cell, the cell wall permits a positive turgor pressure to balance a negative solute potential; plasmolysis illustrates the opposite loss of support after water leaves.

Let $U=|\mathbf{U}|$, $\nu =\mu /\rho$, and $r=|\mathbf{x}|$. A decaying outer disturbance matched to the leading translating sphere in Stokes flow isHere $\Delta \varphi =0$ and $(\nu \Delta -\mathbf{U}\cdot \nabla )\chi =0$ away from zero. The substitution $\chi =h{e}^{\mathbf{U}\cdot \mathbf{x}/(2\nu )}$ reduces the latter equation to $(\Delta -U^2/(4{\nu }^2))h=0$, whose decaying radial solution is proportional to $e^{-Ur/(2\nu )}/r$. In the inner overlap, cancellation of the potential monopole gives $\mathbf{u}\sim -3a[\mathbf{U}+\mathbf{n}(\mathbf{U}\cdot \mathbf{n})]/(4r)$. The potential source's volume flux is $6\pi \nu a$, and its mass flux is $6\pi \mu a$. Downstream, the fluid wake width grows as $\sqrt{\nu x/U}$; its mass deficit balances this source, while its momentum deficit is $6\pi \mu aU$, the Stokes drag law.

For an orthonormal sequence $(e_n)$ in a Hilbert space,

An orthogonal projection residual is the difference $r=x-P_{W}x$ between a vector and its orthogonal projection onto a linear subspace $W$. It is orthogonal to every vector in $W$, so $⟨r,w⟩=0$ for all $w\in W$. The Gram-Schmidt process uses such residuals to remove directions already represented by earlier basis vectors.

This is an optimality condition as well as a geometric description. No permitted change within $W$ can remove a further orthogonal component of $r$. In the preceding dot product example, $r=(2,-1{)}^T$ is perpendicular to the line spanned by $(1,2{)}^T$.

The selected inner product matters. Keep $x=(3,1{)}^T$ and the same line, but use the weighted Euclidean inner product with $M=\mathrm{diag}(4,1)$. The projection coefficient and residual becomeNow $4(r_M{)}_1{e}_1+(r_M{)}_2{e}_2=5-5=0$, although $r_M\cdot e=-15/4\ne 0$. The orthogonality is with respect to $M$, so it need not look perpendicular in the ordinary drawing. The squared weighted error isThe weighted optimum reduces the first-coordinate error from $2$ to $5/4$, at the cost of increasing the magnitude of the second-coordinate error from $1$ to $5/2$. The weights express which errors are more costly.

In linear least squares, residual orthogonality to the column space is expressed by the normal equations for linear least squares. With a weighted Euclidean inner product, the corresponding real-matrix equation is $A^{T}M(x-Ac)=0$. This makes the choice of error measurement part of the mathematical model, rather than merely a plotting convention.

At $(p,o)$ in the orientation double cover, use the derivative of the covering projection to transport the chosen orientation $o$ to the tangent space of the cover. In local oriented sheets this prescription is smooth, so it defines a global orientation. The involution $(p,o)\mapsto (p,-o)$ reverses it.

For a connected smooth manifold, every component of the orientation double cover maps onto the base by the path lifting theorem. There can therefore be at most two components. A global orientation splits the cover into two copies of the base. Conversely, if there are two components, each meets every fibre once, so its inverse projection gives a smooth global choice of orientation.

On the smooth boundary of an orientable surface, let $n$ be the chosen unit normal vector and $t$ the positively oriented unit tangent vector used by Stokes theorem. The outward conormal is $u=t\times n$. It is a unit vector, tangent to the surface and orthogonal to its boundary. This convention turns $(n\times v)\cdot d\mathbf{r}$ into $u\cdot vds$ by cyclic invariance of the scalar triple product.

An oriented atlas is a smooth atlas whose smooth transition maps have positive Jacobian determinants on every overlap. The coordinate bases then give compatible choices of orientation of a vector space on all tangent spaces. A nowhere-vanishing top-degree differential form gives such an atlas: restrict each chart until its coefficient has constant sign, reverse one coordinate if necessary to make that sign positive, and compare the positive coefficients on overlaps. Conversely, a partition of unity glues positive local coordinate top forms to a global volume form.

An orientation of a smooth manifold is a continuous choice of orientation of a vector space in every tangent space, equivalently an orientation of a vector bundle on its tangent bundle. It can be specified by an atlas with positive Jacobian determinants on all overlaps or by a nowhere-vanishing top-degree differential form. On a connected orientable manifold there are two choices, interchanged by reversing every tangent-space orientation.

// Target: differential-geometry.bigb

For a multiplicative set $S$ of regular elements satisfying the left Ore condition, represent a fraction by $(s,a)$. Pairs $(s,a)$ and $(t,b)$ agree when there are $u,v$ with $us=vt\in S$ and $ua=vb$. Addition uses this common denominator. For multiplication choose $v\in S,c\in R$ with $va=ct$, then $(s^{-1}a)(t^{-1}b)=(vs{)}^{-1}cb$. The Ore condition gives common multiples for successive operations; regularity supplies cancellation. In a domain with $S$ all nonzero elements, every nonzero fraction has inverse $a^{-1}s$.

// Destination: associative-algebra.bigb

If two nonzero principal right ideals in a right Noetherian domain had zero intersection, the sums $bR\oplus abR\oplus \cdots \oplus a^{m}bR$ would be strictly increasing. Directness follows by splitting the lowest term into $bR$ and the other terms into $aR$, then cancelling $a$ and repeating. This contradicts the ascending chain condition. A nonzero common element $au=bv$ gives the right Ore condition. The left version follows from the opposite ring. Consequently a noncommutative domain Noetherian on both sides has a single division ring of fractions with both left and right fraction presentations.

This is the localization of a ring at all its regular elements of a ring, with left fractions. It exists precisely when those denominators satisfy the left Ore condition. If $R$ embeds in a ring $Q$, all regular elements become units in $Q$, and every element of $Q$ is such a fraction, the Ore condition follows by writing $r{c}^{-1}=s^{-1}b$, which gives $sr=bc$.

In the ordinary topology on infinite subsets, the family $D$ of cofinite sets is a countable set, a dense and a meagre set. Countability follows by indexing its members by their finite complements. Each singleton is closed with empty interior, while every basic cylinder with finite initial segment $s$ contains $s\cup \{n:n>\max s\}$. Consequently $D$ is not a nowhere dense set. This example uses the ordinary cylinder topology, rather than the Ramsey cone topology.

For ordinals $\alpha$ and $\delta >0$, there are unique ordinals $q,r$ with the displayed relation. The function $q\mapsto \delta q$ is strictly increasing, continuous at limit ordinals and unbounded. Its first value above $\alpha$ occurs at a successor, giving the largest $q$ with $\delta q\le \alpha$. The ordered tail of $\alpha$ after that initial segment gives $r$, and maximality gives $r<\delta$.

A fast-growing hierarchy indexes number-theoretic functions by an effective system of ordinal notations. Successor stages iterate the preceding function, while limit stages diagonalize along a chosen fundamental sequence of a limit ordinal. These choices are part of the definition. With the standard Cantor normal form sequences below epsilon zero, each fixed $F_{\alpha }$ for $\alpha <{\epsilon }_0$ is provably total in Peano arithmetic, but the uniform endpoint $F_{{\epsilon }_0}$ is not. Every function provably total in Peano arithmetic is eventually bounded by some lower level. These classification results are proved in epub.ub.uni-muenchen.de/3843/1/3843.pdf .

For a countable limit ordinal, a fundamental sequence is an increasing cofinal sequence of smaller ordinals. Effective choices allow a fast-growing hierarchy to turn limit recursion into recursion at a finite input. Below epsilon zero, write the final term as $\gamma +{\omega }^{\beta }$. For successor $\beta =\delta +1$, take $\gamma +{\omega }^{\delta }(n+1)$; for nonzero limit $\beta$, take $\gamma +{\omega }^{\beta [n]}$. Earlier terms in the Cantor normal form remain fixed.

For $C=B^{-1}$, ordered Grassmann Gaussian contractions give $⟨\bar{\theta }M\theta ⟩=-\mathrm{tr}(MC)$ and $⟨(\bar{\theta }M\theta )(\bar{\theta }N\theta )⟩=\mathrm{tr}(MC)\mathrm{tr}(NC)-\mathrm{tr}(MCNC)$. The disconnected contribution has two fermion loops; the connected contribution has one, giving its minus sign. This covariance is an algebraic normalized Berezin integral, rather than a positivity statement about a probability measure.

For independent and identically distributed random variables with a continuous probability distribution, let $N\ge 2$ be the index of the first increase. The event $N\ge n$ requires the first $n-1$ observations to be decreasing; symmetry among their $(n-1)!$ possible strict orders gives probability $1/(n-1)!$ for $n\ge 2$. Subtracting successive tails gives the displayed mass function for $r\ge 2$. The tails tend to zero, so $N$ is finite almost surely. The tail-sum formula for expectation gives $\mathbb{E}N=1+{\sum }_{j\ge 1}1/j!=e$, independently of the continuous sampling law.

In a second-order reaction $\mathrm{A}+\mathrm{B}⟶\mathrm{P}$ with rate law $-d[\mathrm{A}]/dt=k_2[\mathrm{A}][\mathrm{B}]$, a large excess of B can keep $[\mathrm{B}]\approx [\mathrm{B}{]}_0$ during the observation window. ThenThis is a pseudo-first-order reaction: the approximate first-order reaction describes that window, while changing the excess B concentration changes $k_1$. Its units are ${\mathrm{s}}^{-1}$, distinct from the second-order reaction rate constant $k_2$. For one-to-one stoichiometry, the maximum fractional depletion of B is $[\mathrm{A}{]}_0/[\mathrm{B}{]}_0$; an original $100:1$ initial excess bounds its change by $1\%$. A twofold excess permits $50\%$ depletion and does not provide the same control.^[ref]