The clock hypothesis says that an ideal clock following a timelike curve measures its proper time: its instantaneous rate depends on velocity through $d\tau =dt\sqrt{1-v^2/c^2}$, with no separate acceleration term. Acceleration still changes the world line and hence the accumulated proper time. A physical clock may suffer disturbances under forces or environmental changes, so experiments must distinguish its mechanism's errors from the geometric prediction.

An atomic clock uses an atomic transition as a frequency reference for an oscillator; counting its calibrated cycles realizes a time interval. In a passive beam implementation, the chemical atoms are repeatedly prepared, interrogated and detected, while feedback control corrects the continuously running oscillator. The Ramsey separated oscillatory fields experiment explains how internal quantum phase becomes a measurable population signal.

The SI second fixes the unperturbed ground state hyperfine structure transition frequency of caesium-133 to exactly $9192631770\mathrm{Hz}$. This is a defining value for the ideal transition, rather than a guarantee that an operating apparatus has no shifts.^[ref]

A compact smooth embedding has a tubular neighborhood identified with its normal bundle. Collapse the complement of its open disk neighborhood to the basepoint and identify the remaining quotient with the Thom space of that normal bundle. Pulling back its Thom class constructs a Gysin class. For a graph embedding into a product with Euclidean space, desuspending the auxiliary coordinates gives the corresponding wrong-way map.

A smooth embedded submanifold has a neighborhood diffeomorphic to a neighborhood of the zero section of its normal bundle, with the zero section identified with the submanifold. One may prescribe the normal-bundle splitting of the derivative along that section. Choose a Riemannian metric making the splitting orthogonal and use its normal exponential map: its derivative there is the required isomorphism, so the inverse function theorem supplies local diffeomorphisms. A sufficiently small, possibly varying, fibre radius makes these fit into a tubular neighborhood.

For $m$ tribes of size $w$ with $m\sim (\log 2)2^w$ and $n=mw$, the success probability under the p-biased product measure is $1-(1-p^w{)}^m$. Its $x$-quantile is $p_x=[1-(1-x{)}^{1/m}{]}^{1/w}$. For fixed $0<x<1$,Thus the fixed-error window has order $1/\log n$. For small fixed error the leading coefficient grows with $\log (1/\epsilon )$, so the general Friedgut-Kalai sharp threshold theorem also has the correct error dependence up to absolute constants.

Choose $w\ge 1$ maximal such that $wm\le n$ for the displayed $m$, and take the OR of $m$ AND blocks of size $w$, ignoring unused coordinates. Its failure probability is $(1-2^{-w}{)}^m$, between one quarter and three quarters. Active coordinates have influence $2^{1-w}(1-2^{-w}{)}^{m-1}$ and unused ones have zero influence. Maximality gives $2^{1-w}<4(\log 2)(w+1)/n$; for $n\ge 2$, $w\le {\log }_{2}n$, so every influence is below $8\log n/n$. The construction is therefore a quite fair Boolean function with small influences for every $n\ge 2$. The logarithmic bound cannot hold at $n=1$, where every quite fair function has influence one.

Forward substitution solves a linear system $Ly=c$ with a lower triangular matrix $L$ and nonzero diagonal, from the first row downward:The first sum is empty, so $y_1=c_1/l_{11}$. Each subsequent row uses only components already found. For a unit triangular matrix $L$, omit the division because $l_{ii}=1$. The complementary procedure, back substitution, solves an upper triangular matrix system from the last row upward.^[ref]

If $np\to 0$, the triangle count in a binomial random graph has expectation at most $(np{)}^3/6\to 0$, so the Markov inequality shows there are no triangles in a graph with probability tending to one. If $np\to \infty$, the triangle variance in a binomial random graph divided by the squared mean is $O((np{)}^{-3})+O((n^{2}p{)}^{-1})\to 0$. The second moment method then gives at least one triangle in a graph with probability tending to one. At the intermediate scale $np\to c\in (0,\infty )$, a nontrivial Poisson limit for triangles in a binomial random graph describes the count.

Choose $n=d^7$ and a binomial random graph of edge probability $d/n$. With high probability its maximum degree is at most $2d$, it has $o(n)$ triangles, and every independent set has size at most $4n\log d/d$. Delete one vertex per triangle. The resulting triangle-free graph has chromatic number at least $d/(8\log d)$. Appending a disjoint star of degree $2d$, if necessary, makes the maximum degree exactly $2d$ without lowering the chromatic number. Thus the order $\Delta /\log \Delta$ is unavoidable.

The pseudoscalar Yukawa interaction with coefficient $-g$ has the displayed Feynman vertex. Its two-fermion tree-level Feynman diagrams are the $t$ and $u$ exchanges; antisymmetry of the final identical fermions requires their relative minus. Fermion-antifermion scattering has $t$ exchange and $s$ annihilation, again with a relative minus. With initial and final states ordered as $b_1^{†}{d}_2^{†}|0⟩$ and $b_3^{†}{d}_4^{†}|0⟩$, the normal-ordered current contributes $-d^{†}d\bar{v}{\gamma }^{5}v$ in the exchange channel but $db$ and $b^{†}{d}^{†}$ in the annihilation channel. These fermionic signs fix the overall sign as well as their interference; changing an external-state phase changes both terms together.

For fixed $k\ge 2$ and $\omega \to \infty$, the expected number of tree components of order $k$ is asymptotic to $k^{k-2}{n}^k{p}^{k-1}{e}^{-knp}/k!$ throughout the displayed window. It tends to infinity: split $np$ at $1$ and $(\log n)/(2k)$ to obtain lower bounds of orders ${\omega }^{k-1}$, $\sqrt{n}$, and $e^{\omega }$. Distinct overlapping vertex sets cannot both be components; disjoint sets have joint-to-product probability ratio $(1-p{)}^{-k^2}=1+o(1)$. Thus the relative variance vanishes, proving a component of order $k$ exists with high probability.

Assume treatment receipt gives event risk $p_1$, nonreceipt gives risk $p_0$, and a fraction $c$ in each equally sized randomized arm crosses to the opposite treatment. The two assignment risks are $(1-c)p_1+c{p}_0$ and $c{p}_1+(1-c)p_0$. Subtracting yields the displayed attenuation, with ${\Delta }_{\mathrm{receipt}}=p_0-p_1$. It depends on the stated equal crossover fractions and unchanged receipt-specific risks; it is not a universal formula for every form of treatment contamination.

For a gravity-current nose of height $h_N$ on a small incline of slope $\theta$, the travelling front of a gravity-driven thin film satisfies $h_y=\theta (1-h_N^2/h^2)$. Close to its dry tip, $h\sim (-3\theta h_N^{2}y{)}^{1/3}$. The slope becomes order one at $h\sim \sqrt{\theta }{h}_N$, giving an $O(\sqrt{\theta }{h}_N)$ region where lubrication theory fails. The main nose length is $h_N/\theta$. Comparing capillary pressure gradients $\gamma h_N/(h_N/\theta {)}^3$ with gravitational forcing $\rho g\theta$ gives the crossover $h_N\sim \theta \sqrt{\gamma /(\rho g)}$. This concerns the main nose length; capillarity may already matter in a much smaller tip region.

On an equal-panel partition of an interval, with $h=(b-a)/n$, sum the trapezoidal rule over adjacent panels:There are $n+1$ distinct quadrature nodes. Each interior sample belongs to two panels and therefore has quadrature weight $h$, while each endpoint has quadrature weight $h/2$. Equivalently, the composite trapezoidal rule integrates the piecewise linear function obtained by linear interpolation exactly.^[ref]

For $f\in C^2[a,b]$ and $K={\max }_{x\in [a,b]}|f^{\prime \prime }(x)|$, summing panel quadrature errors givesfor some $\xi \in (a,b)$. Thus panel quadrature errors of order $O(h^3)$ accumulate to order $O(h^2)$, using big O notation.^[ref] On a nonuniform partition of an interval with widths $h_j$ and largest width $H$, the bound is $K{\sum }_j{h}_j^3/12\le K(b-a)H^2/12$.

For the polynomial $f(x)=x^2$ on $[0,1]$, direct summation yieldsBoth quadrature errors shrink by exactly four when $n$ doubles, with opposite signs. This exact relation belongs to this polynomial example; the trapezoidal rule is not always twice as inaccurate as the midpoint quadrature rule.

The left plot uses exact polynomial sums; the right adds a common sample bias $\delta =0.002$ to the composite trapezoidal rule. That mathematical model converges to $1/3+\delta$, showing why smaller panels alone do not ensure accuracy for noisy samples. It illustrates noise sensitivity of a quadrature rule, rather than measured data.

Periodizing the bulk transverse tangent correlation of a stretched worm-like chain gives $C_L(r)={\sum }_{j\in \mathbb{Z}}C(r+jL)$, where $d_L(r)$ is the distance to zero on the periodic interval. Summing the geometric series yields the displayed result. At zero separation its finite-length factor is $\coth (L/(2\xi ))$; for $L\gg \xi$ the bulk result is recovered. Periodic tangent fluctuations are a mathematical boundary convention, not a claim that a pulled physical chain forms a closed loop.

If $f:M\to N$ satisfies transversality of a map to a submanifold $S$, its nonempty preimage is an embedded submanifold of the displayed codimension. Choose a slice chart for an embedded submanifold around $f(p)$ in which $S$ is the zero set of the first $k$ coordinates. Their composition with $f$ has surjective derivative on its zero set, so the regular level set theorem gives a local embedded slice of dimension $\dim M-k$. These slices give the subspace topology and a compatible smooth atlas on the preimage. Its tangent space is

Smooth maps of finitely many circles into a manifold of dimension at least three can be perturbed within their homotopy classes to disjoint embeddings. The transversality dimension count for two curve branches is $1+1-m<0$, so their images avoid one another after perturbation; one also removes self-intersections and obtains immersions by the jet version of transversality. This allows disjoint representative loops for relators in high-dimensional surgery constructions of prescribed fundamental groups.

This matrix adds $t$ times coordinate $j$ to coordinate $i$. Left multiplication performs the corresponding row addition. Such matrices generate the special linear group over a finite field: row elimination also realizes signed interchanges and compensating diagonal scalings using products of these matrices.

Every nonidentity transvection has an adapted basis in which it is $I+E_{12}$. Thus all such maps are conjugate in the general linear group. In dimension at least three, a scaling of a third basis vector centralizes this matrix and has arbitrary nonzero determinant; adjusting a conjugator by this scaling proves conjugacy in the special linear group as well.

// Target: algebra.bigb

In the Weyl representation of the gamma matrices, $C=i{\gamma }^2{\gamma }^0=\mathrm{diag}(\epsilon ,-\epsilon )$ with $\epsilon =i{\tau }_2$. The antisymmetry of $\epsilon$ gives $C^T=-C$, and direct multiplication gives $C{\gamma }^{\mu T}{C}^{-1}=-{\gamma }^{\mu }$ and $C^{†}C=I$. Any two such intertwining matrices differ by a scalar: their ratio commutes with all gamma matrices, whose products span the full four-by-four matrix algebra. Changing spinor basis sends $C$ to $SC{S}^T$, so transpose antisymmetry is basis-independent. Its sign is therefore fixed, not merely one of the two alternatives supplied by the general transpose symmetry of a charge-conjugation matrix argument.