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A first-order irreversible transfer from fluid concentration $C$ to stored concentration $S$, measured per matching fluid storage volume, obeys the displayed balances. A sustained inlet can produce a stationary decaying fluid profile, but the stored amount then grows rather than becoming stationary. A finite pulse instead has $C\to 0$ and a finite final deposit. Decay, desorption or finite adsorption capacity is additional physics, not a consequence of the linear capture term alone.

Define the instantaneous spin magnetization $m=\frac{1}{4}{\sum }_i{\sigma }_i$. Global reversal sends $m$ to $-m$ without changing the zero-field energy. Pairing every microstate with its reversed partner proves $⟨m⟩=0$ at every positive temperature, even when almost all probability lies on the two fully aligned patterns. A vanishing mean does not imply weak alignment: the limiting probability distribution is half at $m=+1$ and half at $m=-1$.

The Ising ring magnetization fluctuations measure this distinction through variance. In the $d=0$ class, both patterns have $m^2=1$. Among the $d=2$ patterns, eight have $m^2=1/4$ and four have $m^2=0$; the two $d=4$ patterns also have $m^2=0$. ConsequentlyIt equals $6/17$ at $K=(\ln 2)/4$, approaches $1/4$ for four independent fair Ising spins at $K\to 0$, and approaches $1$ as $K\to \infty$. Unlike the energy variance, the spin magnetization variance remains large in the low-temperature symmetric mixture.

The same thermal fluctuation controls the field response. Allow $h$ to vary before setting it to zero; differentiating the Boltzmann distribution inserts $\beta {\sum }_i{\sigma }_i=4\beta m$, givingThis is a spin-normalized magnetic susceptibility with respect to the energy-valued parameter $h$; conversion to physical magnetization and magnetic field requires the magnetic moment and sample volume factors.

For finite $N$, $Z$ is a finite positive sum of exponential functions. Its free energy $-k_{B}T\log Z$ is a real analytic function for $T>0$, so the smooth heat capacity peak here is not a critical phase transition. Spontaneous magnetization concerns the order of limits: first a thermodynamic limit, then removal of a symmetry-breaking field. The four-spin Ising ring alone establishes neither that limit nor any two-dimensional Ising model result.^[ref]

An equilibrium statistical ensemble with fixed particle number, pressure and temperature, allowing volume and energy to fluctuate. With a temperature-independent volume measure, its joint weight is proportional to $e^{-\beta (E+PV)}$. Its normalization and any temperature-dependent prefactor must be specified when taking thermodynamic derivatives.

Integrating a canonical partition function over volume gives the displayed continuous-volume normalization. For a temperature-independent measure and Hamiltonian, $-{\partial }_{\beta }\ln \Delta =⟨E+PV⟩$. Multiplying by a volume-independent prefactor leaves normalized volume probabilities unchanged, but a temperature-dependent prefactor changes this derivative.

An isotonic solution produces no sustained change in cellular volume under the stated comparison conditions. Water can still exchange in both directions.

Under uniform compressive pressure $p$, isotropic Hooke's law gives each normal strain as $-p(1-2\nu )/E$. The volume strain is three times this, so the bulk modulus is $B=E/[3(1-2\nu )]$, where $E$ is Young's modulus and $\nu$ is Poisson's ratio. Positivity of $E$ and finite $B$ gives $\nu <1/2$. Positive shear modulus $G_s=E/[2(1+\nu )]$ adds the lower bound $\nu >-1$. Negative Poisson's ratio is allowed; the bulk relation alone does not supply the lower bound.

Here $V_d$ is the volume of the unit ball in $d$-dimensional Euclidean space. Rotational symmetry makes the integral an isotropic second-rank tensor $C{\delta }_{ij}$. Taking its trace gives $dC={\int }_{|x|\le R}|x{|}^2{d}^{d}x=d{V}_d{\int }_0^R{r}^{d+1}dr$. This proves the formula. For a constant matrix $M$, contraction gives $\int x^{T}Mx{d}^{d}x=V_d{R}^{d+2}\mathrm{tr}M/(d+2)$. In three dimensions the coefficient is $4\pi R^5/15$.

During this stage of the cardiac cycle, the ventricular inlet and outlet valves are both closed. Ventricular volume is fixed at its end-filling value $V^{+}$, while increasing ventricular elastance raises pressure as $p_{LV}=E_{LV}(t)(V^{+}-V_0)$. Arterial and venous pressures remain approximately fixed when the stage is short relative to their systemic resistance–compliance times.

A long wire with uniform current density $I/(\pi a^2)$ generates uniform heat per volume $H=I^2/({\pi }^2{a}^4\sigma )$. Radial steady conduction obeys $(r{T}_r{)}_r=-Hr/k$. Regularity at the centre removes a logarithmic solution, leaving the displayed parabolic temperature excess relative to the surface. Conductivity, electrical conductivity and current density are assumed constant.

Physician and natural philosopher who contributed to the conservation-of-energy programme in the 1840s, including an argument for the mechanical equivalent of heat.

A spontaneous symmetry breaking chooses order-parameter directions independently in regions that have not been in causal contact. Mismatches around disconnected components, loops or enclosing spheres can trap domain walls, cosmic strings or magnetic monopoles. The correlation length cannot exceed the causal horizon, so one monopole per horizon volume estimates the smallest causal initial abundance up to geometrical and efficiency factors.

// Target: classical-field-theory-soliton.bigb

For smooth inviscid barotropic fluid motion without gravity or magnetic forces, let $w$ be the specific enthalpy with $\nabla w={\rho }^{-1}\nabla p$, and set $q=\mathbf{u}\cdot \mathbf{\omega }$. ThenThis follows by dotting the barotropic vorticity transport equation with $\mathbf{u}$ and using the divergence of a cross product. A volume integral is constant when the resulting boundary flux vanishes.

The local density of kinetic helicity is $q=\mathbf{u}\cdot \mathbf{\omega }$. Conservation of its volume integral does not imply that $q$ is a materially advected scalar.

The Kirchhoff tensor is the Cauchy stress tensor multiplied by the local volume ratio. Its stress power is per reference volume. Pushing forward the second Piola-Kirchhoff stress tensor gives the displayed formula.

A Lagrangian coordinate labels a material particle and remains attached to it as the continuum moves.

For a three-dimensional isotropic unpolarized spin-one-half Fermi liquid, define $f^s=(f_{\uparrow \uparrow }+f_{\uparrow \downarrow })/2$ and $f^a=(f_{\uparrow \uparrow }-f_{\uparrow \downarrow })/2$. Dimensionless Landau parameters use the spin-summed density of states per volume:$P_{\ell }$ is a Legendre polynomial; $s,a$ label spin symmetry. No extra $(2\ell +1)$ multiplies this expansion.^[ref] The superscript $s$ on $F_0^s$ is unrelated to the dimensionless wave speed $s$ below.

Changing one Landau quasiparticle occupation changes the energies of others. To second order around equilibrium,Here $\mathcal{V}$ is volume, occupations are per momentum and spin state, and $f$ has units of energy times volume.^[ref]

For a toy two-state change at fixed volume, take $\delta n_a=\delta n_b=\eta$, $f_{aa}=f_{bb}=0$, $f_{ab}=f_{ba}=g\mathcal{V}$, and unperturbed energies ${\xi }_a={\xi }_b=e$. The expression gives $2e\eta +g{\eta }^2$. The factor $1/2$ counts the interaction of the same pair once; it does not halve each excitation's energy. This finite-state illustration only explains the bookkeeping of the Landau quasiparticle energy functional.