In 1975, Günter Blobel and Bernhard Dobberstein combined cell-free translation with dog-pancreas microsomes. Density-gradient centrifugation isolated the membranes; removal of resident ribosomes allowed reconstitution with rabbit-reticulocyte ribosome subunits. Added messenger RNA encoded an immunoglobulin light chain; globin provided a comparison. Products made with or without membranes were challenged with proteases and analyzed by SDS-PAGE and autoradiography. Membranes present during synthesis yielded processed, protected light chain; globin remained susceptible. Adding membranes after precursor synthesis failed to reproduce import. These comparisons support selective passage during synthesis in this preparation.^[ref]

The companion paper specifies radioactive amino acid incorporation, ice-cold trypsin plus chymotrypsin treatment, a slab-gel electrophoresis apparatus and dried-gel exposure to X-ray film. Product-size analysis distinguishes a preserved full-size product from residual labeled fragments.^[ref]

Protection was incomplete. The authors considered leaky vesicles, and warned that heat treatment could damage processing as well as ribosome attachment. Their experiments did not identify SRP or the Sec61 complex. Nor does failure of late membrane addition establish that all organelle protein import must occur during synthesis.^[ref]

A translocon is membrane machinery through which a polypeptide chain crosses or enters a membrane. In endoplasmic reticulum protein targeting, the Sec61 complex supplies the central channel.^[ref]

The signal recognition particle is an RNA–protein targeting complex linking a ribosome-associated nascent chain to the SRP receptor.^[ref]

The cosmic microwave background quadrupole is the degree-two spherical harmonic part of the observed cosmic microwave background sky. It has five real coefficients. The axisymmetric example $Q{P}_2(\cos \theta )$ selects only one orientation and shape from this space; $Q$ is its chosen amplitude, not a normalization for every possible cosmic microwave background quadrupole.

The Differential Microwave Radiometer measured differences in microwave power between pairs of sky directions to map Cosmic microwave background anisotropy. Comparing two directions suppresses a common absolute brightness offset; repeated comparisons during scanning are then combined into a sky map.

The COBE instrument used paired horn antennas at three frequencies, with independent channels that tested detector noise and frequency-dependent galactic foreground contamination. A differencing instrument does not by itself determine the absolute monopole temperature. Its measurement therefore complements the spectrum measured by the Far Infrared Absolute Spectrophotometer.^[ref]

The Far Infrared Absolute Spectrophotometer measured the spectrum of the cosmic microwave background by comparing the sky with a calibrated reference. Its differential Fourier transform spectroscopy and blackbody calibration tested whether the radiation followed one Planck function, rather than only measuring differences between sky directions.

The key observable is a spectral residual after fitting a temperature and accounting for instrumental and foreground contributions. A small residual constrains departures from a blackbody spectrum and thus energy exchange in the early universe. This absolute spectral test is distinct from the directional map supplied by the Differential Microwave Radiometer.^[ref]

Factor the sine expansion as $\sin z=z[1-z^2/6+z^4/120+O(z^6)]$. To invert the bracket, put $1+A{z}^2+B{z}^4+O(z^6)$ and equate coefficients after multiplication. The $z^2$ coefficient gives $A=1/6$, while the $z^4$ coefficient gives $B=A/6-1/120=7/360$. Dividing by $z$ proves the displayed Laurent series truncation for $z\to 0$, $z\ne 0$. This reciprocal-series calculation is useful when a small sine argument is itself an inverse power of a large variable; multiply the remainder by any outside prefactor when assigning its final order.

The Technische Universität Darmstadt demonstration uses a steel ball launched outward from the centre of a rotating disc. A fixed lecture-room camera records a straight path, while a camera carried above the disc records a curved path. The instructions specify launching toward the fixed camera so that the laboratory motion is clearly visible. The rotating iPod transmits its view wirelessly through Apple TV; two vertical support rods of length $1\mathrm{m}$ are specified for the slow-motion arrangement. The paired video views are the readout: one physical motion is represented in two reference frames.^[ref]

A useful preparation control appears in the Universität Freiburg instructions: with the disc stationary, check that the ball travels radially toward its edge. A camera is mounted on supports attached to the disc; turn the disc by hand as the ball leaves its centre and observe the camera view.^[ref]

The inference is a dependence of the observed trajectory on the rotating axes; a curved camera trace alone does not isolate the Coriolis force. To compare with the constant-angular velocity free particle in rotating coordinates, measure the disc angular velocity, calibrate lengths in the camera image, and compare launches at rest and with opposite rotation directions. These are checks of the ideal model, not reported measurements from the manuals. A rolling ball can exchange forces with the surface through friction, and hand rotation need not have constant angular velocity; an Euler acceleration then enters too. The ideal free-ball drawing deliberately omits those effects. Neither manual supplies a numerical uncertainty or a precision measurement of the Coriolis acceleration.

The instantaneous power of the Coriolis force, evaluated with the velocity in the same rotating reference frame, vanishes:This follows from the perpendicularity of the cross product and the dot product definition of power. The Coriolis force alone does no work and cannot change the rotating-frame kinetic energy $m|{\mathbf{v}}_{\mathrm{rot}}{|}^2/2$. It can change the direction of the relative velocity. Actual forces and the centrifugal force can still change that kinetic energy, as the free particle in rotating coordinates illustrates. Conservation of an appropriate total energy has the additional conditions stated under energy balance in a uniformly rotating frame.

In a leading meridional-track approximation, an aircraft starts southward from the equator at constant speed $v$, with latitude $\lambda =-vt/R$. If lateral steering applies the Coriolis force rather than its negative, the two eastward effects add: ${\ddot{x}}_E=-4\Omega v\sin \lambda$. With zero initial cross-track velocity and displacement, double integration gives the displayed local-plane offset at southern latitude $-\Lambda$. For small $\Lambda$, it is $2\Omega R^2{\Lambda }^3/(3v)$. This assumes steering is a force transverse to the actual velocity while thrust and drag balance along it. It is not compatible with a literally fixed velocity vector. A spherical calculation also includes the transport of the local east and north basis, changing finite-latitude corrections.

For univariate polynomials $f,g$ over a field, not both zero, there are polynomials $A,B$ such that $Af+Bg=d$, where $d$ is their monic greatest common divisor. In particular, coprime polynomials admit $Af+Bg=1$. The extended Euclidean algorithm constructs the coefficients.

A linear functional cannot exceed its best original value by mixing. For a finite nonempty set $S$ and $x={\sum }_i{\lambda }_i{s}_i$ in its convex hull,Every original point is itself in the convex hull, so equality between the two maxima follows. Moreover, a mixture attains the maximum exactly when all its positive weights are assigned to maximizing original points. Multiple maximizers can therefore produce a whole line segment or a higher-dimensional region of equally good choices.

For the eight samples above, use the linear functional $f(x,y)=x+y$. Its largest sample value is $f(C)=6$, and $C$ is the only maximizing sample. Thus the unique maximum over the whole convex polygon is also at $C$. The line $x+y=6$ is a supporting line: every point of the convex hull lies in the half-space $x+y\le 6$, and $C$ touches its boundary.

The point $q=(3.8,2.5)$ passes the coordinate-wise bounds $0\le x\le 4$, $0\le y\le 3$, but $f(q)=6.3>6$. This single inequality certifies that $q$ is outside the convex hull. A weighted mixture could not have a larger linear functional value than all its ingredients.

This explains a basic feature of linear programming: searching all mixtures of finitely many candidate vectors cannot improve the objective beyond the best candidate, though mixing equally good candidates can give more optimal choices. The conclusion depends on linearity. For the concave function $g(x,y)=-(x-2{)}^2-(y-1{)}^2$, the new point $p=(2,1)$ instead attains zero, strictly above its value at every original sample. A nonlinear objective can favor a mixture.

Take $a>0$ and define $z^s$ using $-\pi /2<\arg z<3\pi /2$, so the branch cut lies on the negative imaginary axis. An indented upper semicircle encloses the simple pole $ia$. Its two straight segments contribute $(1+e^{i\pi s})$ times the desired real integral, while the residue theorem gives $\pi a^{s-1}{e}^{i\pi s/2}$. The large arc vanishes as its radius to the power $s-1$, and the small arc as its radius to the power $s+1$, giving the stated convergence range and result. This contour avoids placing a branch cut on either straight integration segment.

Let $a>0$ and let $R$ be a rational function with $R(z^2)$ finite for real $z$ and $R(u)=O(u^{-1})$ at infinity. Choose the branch $h(z)=\sqrt{z^2-a^2}$ analytic off $[-a,a]$, with $h(z)\sim z$ at infinity. Put $g(z)=zR(z^2)/h(z)$. On the real exterior, $g(-x)=g(x)$; on the upper bank of the cut, $h(x+i0)=i\sqrt{a^2-x^2}$, so $g(x+i0)$ is odd in $x$.

Close a real-axis contour with a large upper semicircle, making clockwise upper indentations at the two branch points. The large arc vanishes because $g=O(z^{-2})$. Each small indentation is $O(\sqrt{\epsilon })$ and vanishes. The integral along the upper bank of $[-a,a]$ is zero by oddness; the two exterior pieces sum to twice the required integral. The residue theorem therefore givesResidues must use the chosen branch, notably $h(iy)=i\sqrt{y^2+a^2}$ for $y>0$. The substitution $u=\sqrt{x^2-a^2}$ provides an independent real-variable check, converting the integral to ${\int }_0^{\infty }R(u^2+a^2)du$.

For real $a>b>0$, put $d=\sqrt{a^2-b^2}$. The change of variables $z=e^{i\theta }$ changes the integral to ${\oint }_{|z|=1}(2/b)dz/[z^2+2i(a/b)z-1]$. The roots are $-i(a-d)/b$ and $-i(a+d)/b$, with only the first inside the circle because $a-d=b^2/(a+d)$. The residue there is $1/(id)$, and the residue theorem proves the displayed result. A shift of the angular variable replaces sine by cosine. More generally for real $a>|b|$, the same answer follows by shifting by $\pi$ when $b<0$, and the case $b=0$ follows directly.

Two distinct normalized plane waves on a periodic one-dimensional ring of length $L$ give the displayed projected contact interaction. Momentum conservation excludes pair conversion between these two modes. Four mixed operator orderings give the coefficient of $n_1{n}_2$. At fixed particle number $N$, the energy is $g[N(N-1)+2n_1(N-n_1)]/(2L)$. For repulsive coupling, its minima are the two endpoint Fock states with all particles in one mode; a one-particle energy difference selects the lower mode. This ground-state statement concerns the two-mode projection.

For distinct box modes ${\phi }_1=\sqrt{2/L}\sin (p\pi x/L)$ and ${\phi }_2=\sqrt{2/L}\sin (q\pi x/L)$, the displayed contact interaction overlaps produce self terms $3g{\sum }_j{n}_j(n_j-1)/(4L)$, a density term $2g{n}_1{n}_2/L$, and pair conversion $g(a_1^{†2}{a}_2^2+a_2^{†2}{a}_1^2)/(2L)$. Pair conversion preserves total particle number but prevents diagonalization by individual occupation numbers. The additional overlaps are $I_{1112}=-{\delta }_{q,3p}/(2L)$ and $I_{1222}=-{\delta }_{p,3q}/(2L)$, following from ${\sin }^{3}u=(3\sin u-\sin 3u)/4$. Third-harmonic mode pairs therefore also permit single-particle transfer assisted by the other particles. For the first two box modes these additional terms vanish.

For a smooth homogeneous solid in thermodynamic equilibrium with a liquid and vapor, Young's equation relates their three interfacial tensions to the contact angle measured through the liquid. A virtual increase $dA$ of wetted area changes the interfacial free energy by $({\gamma }_{SL}-{\gamma }_{SV}+{\gamma }_{LV}\cos \theta )dA$. Setting this to zero proves the equation. Roughness, pinning and chemical heterogeneity can make an observed apparent angle differ from this ideal equilibrium value. The equation does not determine each solid–liquid surface energy from a contact angle alone, but two liquids on the same unchanged dry substrate allow the solid–vapor term to be eliminated by subtraction.

This effective baryon mass model adds a colour-magnetic spin-spin interaction to a sum of constituent quark masses. The common coefficient $A^{\prime }$ absorbs a spatial contact density, colour factors, and the strong coupling. $A^{\prime }$ has dimensions of energy cubed in natural units. A shared coefficient is an approximation because different flavour configurations can have different spatial wavefunctions.

For $A{x}^2+Bxy+C{y}^2+Dx+Ey+F=0$, the homogeneous quadratic coefficient matrix isThe conic is degenerate when $Q$ is a singular matrix. Examples over the real numbers include the point $x^2+y^2=0$, the repeated line $y^2=0$, and the two straight lines in $xy=0$. The real solution set alone need not reveal the algebraic distinction: $x^2+y^2=-1$ has no real points, but its matrix is nonsingular. This determinant test is different from the conic type discriminant, which only uses the quadratic part in $x,y$.