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An ER signal sequence is polypeptide chain targeting information recognized by machinery delivering a protein to the endoplasmic reticulum. Many soluble cargoes have a removable N-terminus signal; membrane insertion can use internal signals. An ER signal sequence alone does not specify the final destination.^[ref]

For an original counting example, stipulate an unbranched precursor polypeptide chain of $180$ amino acid residues whose first $20$ amino acid residues form a cleaved ER signal sequence. With no further processing, the remaining chain has $180-20=160$ amino acid residues and $159$ peptide bonds. The initial chain had $179$ peptide bonds; the detached signal has $19$, so $159+19=178$ remain after one bond is cut. The check counts adjacent positions: an unbranched chain of $n$ amino acid residues has $n-1$ peptide bonds. These stipulated lengths are not a natural sequence, a measured size or evidence that a particular sequence functions as an ER signal sequence.

A real rotation matrix is a unitary matrix because it is an orthogonal matrix. It is a Hermitian matrix precisely when it is symmetric, equivalently when its square is the identity. Its eigenvalues are then $\pm 1$. In three dimensions a proper rotation is therefore Hermitian only for the identity or a half-turn. A half-turn about a unit vector $n$ has matrix $2n{n}^T-I$: it fixes $n$ and reverses the perpendicular plane. For every real rotation $R$, the difference $R-R^T$ is anti-Hermitian; it is also Hermitian only in the zero case.

Attach a spring of spring constant $k$ to the axle of a round body of mass $m$, radius $R$ and axial moment of inertia $I$. Assume rolling without slipping, a horizontal drag force $-b\dot{x}$ applied through the axle, and no additional resisting torque. The translational force equation is $m\ddot{x}=-kx-b\dot{x}+f$, where $f$ is signed static friction. The rotational equation and rolling constraint give $f=-I\ddot{x}/R^2$, yielding the displayed equation. Equivalently, the kinetic energy is $(m+I/R^2){\dot{x}}^2/2$. This added effective inertia lowers the undamped angular frequency to $\sqrt{k/(m+I/R^2)}$. A distributed viscous torque would add a separate damping term and cannot be inferred from the drag force alone.

For self-adjoint operators and an admissible pure state, the quantum covariance is the real, symmetrized second moment of two centered observables. The displayed expression uses their anticommutator; equivalently $C_{AB}=\mathrm{Re}⟨A^{\prime }\psi |B^{\prime }\psi ⟩$, which requires only the individual finite second moments. It reduces to ordinary covariance for commuting observables. For noncommuting observables, its definition does not assert that both have simultaneous sharp values drawn from a joint classical probability distribution.

For $d{s}^2=(\alpha x{)}^{2}d{t}^2-d{x}^2$ in a patch with $x\ne 0$, the null paths solve $dx/dt=\pm \alpha x$, giving $x=C{e}^{\pm \alpha t}$ with $C\ne 0$. They are geodesics after choosing an affine parameter. The Rindler coordinates $X=x\cosh (\alpha t)$, $cT=x\sinh (\alpha t)$ give $d{s}^2=c^{2}d{T}^2-d{X}^2$ and $X\pm cT=x{e}^{\pm \alpha t}$. Thus each exponential path is a straight inertial null line. A literal $x=0$ curve is outside this degenerate coordinate chart, not an extra null geodesic within it.

The line element of a Riemannian metric is its local expression for squared infinitesimal length. Along a parametrized curve $x(t)$, the arc length is $\int \sqrt{g_{ij}(x(t)){\dot{x}}^i{\dot{x}}^j}dt$. For a circular cylinder of radius $R$ parametrized by angle $\theta$ and axial position $z$, the induced metric gives $d{s}^2=R^{2}d{\theta }^2+d{z}^2$. Unrolling with $u=R\theta$ yields the Euclidean form $d{u}^2+d{z}^2$, explaining the straight unrolled descriptions of geodesics on a circular cylinder.

Consider a planar surface with outward normal along $z$. For an auxiliary equilibrium electron gas just outside it, let $n_v$ be the total number density, counting both signs of $v_z$. Its Maxwell-Boltzmann velocity distribution has normal-component densityAn electron with $v_z>0$ crosses more area per unit time when it moves faster. The outgoing number flux therefore uses $v_z{f}_z$, rather than just the probability of $v_z>0$:This integral contributes only $T^{1/2}$. Holding a classical bulk number density fixed and imposing a normal kinetic energy threshold $W$ would instead give $\Gamma =n\sqrt{k_{B}T/(2\pi m_e)}{e}^{-W/(k_{B}T)}$; that argument alone does not produce $T^2$.

For the ideal surface model, measure external electron energies from the metal's chemical potential $\mu$. A freely propagating external state has $E-\mu =W+p^2/(2m_e)$. When $W\gg k_{B}T$, its Fermi-Dirac distribution reduces to the Maxwell–Boltzmann statistics tail. Counting two spin states per phase space cell of volume $h^3$ givesEach of the three Gaussian integrals supplies $T^{1/2}$. Combining this $T^{3/2}$ state count with the $T^{1/2}$ crossing factor givesThe auxiliary atmosphere supplies an equilibrium state-counting argument; an operating collector removes the outgoing electrons. It need not maintain a real equilibrium gas throughout the gap. Unit transmission and the ideal free electron state count are substantive assumptions, so this calculation does not determine an arbitrary real surface's measured $A$.

In particular, electrons inside a metal generally form a degenerate Fermi gas. A classical distribution for the escaping tail does not imply a classical distribution for the whole interior. Richardson himself emphasized this distinction in his lecture.^[ref]

Choose a fixed positive reference $A_{\mathrm{ref}}$ with the same units as $A$. Taking the natural logarithm of a dimensionless ratio givesThus the Richardson plot is a straight line when $A$ and $W$ are constant, with slope $-W/k_B$ and intercept $\ln (A/A_{\mathrm{ref}})$. The familiar notation $\ln (J/T^2)$ silently chooses units; the reference makes that choice explicit. Changing $A_{\mathrm{ref}}$ shifts the intercept and leaves the inferred work function unchanged. If the horizontal axis instead uses $1000\mathrm{K}/T$, its slope is $-W/(1000\mathrm{K}{k}_B)$.

For the chosen $2.0\mathrm{eV}$ example, choose $A_{\mathrm{ref}}=120\mathrm{A}\mathrm{c}{\mathrm{m}}^{-2}{\mathrm{K}}^{-2}$. Then $y(1000\mathrm{K})=-23.2090$ and $y(1200\mathrm{K})=-19.3409$. Their slope is approximately $-23209\mathrm{K}$, giving $W=2.0\mathrm{eV}$. Two temperatures determine a line algebraically; several temperatures allow curvature and drift to be detected. These values are calculated, rather than experimental observations.

A straight trace alone does not establish a unique surface work function. Adsorption, changing surface coverage, or a mixture of regions with different barriers can change the inferred slope. For independent emitting patches with area fractions $f_i$, constant $A_i$ and barriers $W_i$,Lower barriers contribute disproportionately. Even with fixed patches, the slope of the Richardson plot gives a temperature-dependent weighted mean of the $W_i$, rather than their simple area average. If $A$ or $W$ themselves vary with temperature, the apparent slope-derived barrier becomesCheck collection, temperature calibration, and surface stability before assigning curvature to a new emission mechanism.

The coefficient $A$ has dimensions of electric current density divided by squared temperature. For ideal nonrelativistic particles with the free electron mass, two spin states, a planar uniform surface, and unit escape transmission, the thermionic tail-flux derivation giveswhere $e$ is the positive elementary charge and $h$ is the Planck constant. The two area units differ by a factor of $10^4$. A fitted $A$ need not equal $A_0$: reflection at the surface, the electronic band structure, and uncertain emitting area affect the inferred coefficient. A constant escape transmission $\tau$ would give $A=\tau A_0$ in this simple model; an energy-dependent transmission generally requires an integral instead.

The name joins Owen Willans Richardson with Saul Dushman. Dushman's 1923 paper develops a thermodynamic treatment of electron emission analogous to evaporation and evaluates the prefactor using quantum ideas.^[ref]

Consider the original pair of polynomial equationsTreat them as polynomials in $y$ with coefficients in $\mathbb{R}[x]$. Their Sylvester matrix and resultant areEvery common solution must therefore have $x=-1$ or $x=1$. Back-substitution into $F=0$ gives $y=x^2=1$, and both pairs also satisfy $G=0$. The complete real solution set isHere elimination and direct substitution agree: subtracting $F$ from $G$ gives $2x^2-2$, exactly the same condition.

In general, first solve the one-variable resultant equation, then recover and check the other coordinate in both original polynomial equations. A real coordinate candidate can correspond only to nonreal values of the eliminated variable. Also, vanishing leading coefficients require the separate check in degree drop in a specialized resultant. The general elimination-and-back-substitution method is described in the lecture's §3.1.^[ref]

Write $f(x)=a_m{\prod }_{i=1}^m(x-{\alpha }_i)$ and $g(x)=b_n{\prod }_{j=1}^n(x-{\beta }_j)$ in an algebraic closure, counting multiplicity of a root. The coefficient-row convention above givesThis makes the shared-root of a polynomial test transparent: a factor vanishes precisely when one ${\alpha }_i$ equals one ${\beta }_j$. Swapping the two polynomials reverses the sign of all $mn$ differences, giving $\mathrm{Res}(g,f)=(-1{)}^{mn}\mathrm{Res}(f,g)$.^[ref]

The value measures an algebraic product, not a normalized geometric distance. Replacing $f$ by $cf$ for nonzero $c\in K$ leaves its roots of a polynomial unchanged but multiplies the resultant by $c^n$. Therefore a small numerical resultant alone does not quantify how close two sets of roots of a polynomial are. Exact zero testing and numerical sensitivity are different questions.

A multiple root of a polynomial is also a root of a polynomial of its formal derivative. In characteristic zero, for $\mathrm{deg}f=m\ge 2$, the associated discriminant isFor example, $h(x)=x^2+s$ gives $\mathrm{Res}(h,2x)=4s$ and $\mathrm{Disc}(h)=-4s$. The two roots of a polynomial merge precisely at $s=0$.^[ref]

The respiratory control ratio divides the oxygen consumption rate during ADP-stimulated respiration by that after supplied ADP has been used. It compares reaction rates, not ATP formed per oxygen.^[ref]

A large ratio means strong stimulation under the comparison's conditions; it is not a universal measure of all mitochondrial functions.

For a constructed example, choose constant suspension volume $V=2\mathrm{mL}$ and oxygen concentration declines of $0.030$ and $0.006\mathrm{mmol}{\mathrm{L}}^{-1}\mathrm{mi}{\mathrm{n}}^{-1}$ during the two comparison windows. Assume no gas exchange or dilution and a correctly calibrated readout. A derivative converts each concentration slope into an amount of substance consumption rate:The common volume cancels in the ratio, so dividing the positive slope magnitudes gives the same answer. These are chosen inputs, not the 1955 observations. If an uncorrected influx of $3\mathrm{nmol}\mathrm{mi}{\mathrm{n}}^{-1}$ of oxygen occurred in both windows, the apparent rates would instead be $57$ and $9$, giving $19/3\approx 6.33$. The same absolute error can inflate a ratio because it removes a larger fraction of its smaller denominator.