A spatial image after a chosen dark interval is a different readout from continuously recording arrivals at a fixed probe. Here is an original idealised fluorescence imaging model: release a dilute cloud, let it move freely, then apply a short uniform probe pulse. Neglect collisions, residual forces, attenuation and motion during the pulse. Use each image's cloud centre to remove mean drift. Uniform gravity shifts that centre without changing the centred variance.

For each atom let $x(t)=x_0+v_{0}t$ along a horizontal imaging axis. Expanding the square about the expected values gives exactlyNo Gaussian distribution is needed for this variance identity. A jointly Gaussian distribution of initial positions and velocities does make the later density Gaussian. For a separable three-dimensional cloud viewed along $z$, integration along the viewing direction gives the column densityHere $N$ is atom number. Integrating this image along $y$ leaves a one-dimensional Gaussian distribution with standard deviation ${\sigma }_x(t)$. This is a projected density measurement, not a three-dimensional radius or a direct velocity histogram. Under the uniform, optically thin probe assumptions, expected fluorescence is proportional to this column density. A known, independent point spread function with finite variance adds its own variance to the measured width; probe duration and intensity gradients need separate treatment.

For a Maxwell-Boltzmann distribution, $T_x=m{\sigma }_v^2/k_B$, where $m$ is atomic mass and $k_B$ the Boltzmann constant. For a nonthermal distribution this defines a directional second-moment temperature, which need not describe every feature of the distribution. A covariance between position and velocity can initially narrow the cloud even though ${\sigma }_v^2$ is positive. The physical covariance matrix requires $C_{xv}^2\le {\sigma }_{x0}^2{\sigma }_v^2$.

Choose ${\sigma }_{x0}=0.60\mathrm{mm}$, ${\sigma }_v=0.060\mathrm{m}{\mathrm{s}}^{-1}$ and $C_{xv}=-0.018\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-1}$, giving correlation coefficient $-1/2$. With $t$ in milliseconds, the numerical variance in square millimetres is $0.36-0.036t+0.0036t^2$. Its minimum occurs at $5\mathrm{ms}$, with standard deviation $\sqrt{0.27}\mathrm{mm}$. These chosen values satisfy the covariance matrix constraint; early contraction is compatible with free motion.

Generate thirteen synthetic variance observations at equally spaced times from $0$ to $30\mathrm{ms}$, adding independent Gaussian noise of standard deviation $0.012\mathrm{m}{\mathrm{m}}^2$. Linear least squares fits all three coefficients in $a+bt+c{t}^2$. The example yields $a\simeq 0.37026\mathrm{m}{\mathrm{m}}^2$, $b\simeq -0.036413\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-1}$ and $c\simeq 0.0035820\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-2}$. With the chosen approximate sodium mass $m=3.82\times 10^{-26}\mathrm{kg}$, the true $T_x$ is $9.96\mu \mathrm{K}$ and the fitted value is $9.91\mu \mathrm{K}$. This is a fit to generated width data, not historical observations.

If the same data are fitted to $a+c{t}^2$, forcing $C_{xv}=0$, the inferred $T_x$ falls to $6.79\mu \mathrm{K}$. The omitted term is partly absorbed into both fitted coefficients. In noiseless linear least squares with a free intercept, its precise effect on the quadratic coefficient iswhere the subscript denotes equal-weight moments over the chosen observation times, using one consistent unit system. Since the ratio is positive for distinct nonnegative times, negative $C_{xv}$ biases this estimate downwards. An incorrect release-time origin can also generate a linear term, so fitting it does not by itself identify a physical initial covariance. Timing calibration, spatial resolution, residual magnetic fields and probe-induced recoil energy remain separate controls. Neither a cold-looking image nor a single long expansion time identifies all three coefficients.

The 1988 experiment loaded sodium optical molasses from a slowed atomic beam. After stopping loading, a shutter extinguished the six cooling beams. Released atoms crossed a lower probe; a photomultiplier recorded fluorescence versus arrival time. Calculations integrated the measured initial cloud distribution and an assumed three-dimensional Maxwell-Boltzmann distribution, including gravity and detection geometry. The minimum was $43\pm 20\mu \mathrm{K}$, versus the $240\mu \mathrm{K}$ Doppler cooling limit.^[ref]

Probe separation, probe intensity, cloud density and shutter timing were varied. Fountain-height, horizontal-spreading and release/recapture measurements also gave below-limit results. Geometry remained an uncertainty; agreement among complementary thermometers made a detector artifact less plausible. These measured released atoms, rather than an in-situ velocity distribution.^[ref]

Steven Chu's account connects the explanation to optical pumping, Stark effect shifts and spatially changing electromagnetic polarization; Claude Cohen-Tannoudji's lecture separates this mechanism from subrecoil schemes. Their lectures and William D. Phillips's describe complementary experimental and theoretical routes.^[ref][ref]

For the radial Laplace operator $L=-r^{-2}(r^2{u}^{\prime }{)}^{\prime }$ on $0<r<R$, require a finite value at the origin and $u(R)=0$. The map $u\mapsto v=ru$ preserves the norm from $L^2((0,R),r^{2}dr)$ to $L^2(0,R)$ and sends $L$ to $-d^2/d{r}^2$. The regular endpoint condition becomes $v(0)=0$. The self-adjoint realization therefore corresponds to the ordinary Dirichlet boundary conditions for the one-dimensional second derivative. Its eigenvalues and eigenfunctions are displayed above, with $u_n(0)=n\pi /R$ understood by a limit. Their squared weighted norm is $R/2$. Completeness and orthogonality follow from the Fourier sine series on $(0,R)$.

For $\ell >0$ and $k>0$, the Sturm-Liouville problem with one Dirichlet boundary condition and one Robin boundary condition has positive eigenvalues. Indeed, integration by parts givesfor every nonzero eigenfunction. Let $z=\ell \sqrt{\lambda }$ and $\kappa =k\ell$. The eigenfunctions are proportional to $\sin (zx/\ell )$, withFor each $n\ge 1$, there is exactly one root $z_n$ in $((n-1/2)\pi ,n\pi )$: on that interval $\tan z+z/\kappa$ increases strictly from $-\infty$ to a positive value. There are no other positive roots. The apparent root $z=0$ gives only the zero sine function and is not an eigenvalue.

For fixed $k,\ell$, write $z_n=(n-1/2)\pi +{\epsilon }_n$. The root equation becomes $\kappa \cot {\epsilon }_n=(n-1/2)\pi +{\epsilon }_n$, henceAn optional unit normalization uses ${\int }_0^{\ell }{\sin }^2(\sqrt{{\lambda }_n}x)dx=\ell /2+k/[2(k^2+{\lambda }_n)]$. The positive sign of the Robin boundary term matters: reversing it can introduce nonpositive modes, as in nonpositive modes in a Dirichlet-Robin interval expansion.

Suppose $p(x)=c(x-a)+O((x-a{)}^2)$ with $c>0$, and the other coefficients in $(p{y}^{\prime }{)}^{\prime }+(\lambda w-q)y=0$ extend continuously to $a$, with $w(a)>0$. A bounded eigenfunction has $(p{y}^{\prime }{)}^{\prime }=O(1)$, hence $p{y}^{\prime }=C+O(x-a)$. A nonzero $C$ would give $y^{\prime }\sim C/[c(x-a)]$ and an unbounded logarithmic singularity. Thus $C=0$, and the regular branch has bounded first derivative. The second local branch carries a logarithm. The title condition excludes that branch.

For two regular eigenfunctions, $p(\overline{u}{v}^{\prime }-\overline{u^{\prime }}v)\to 0$. This removes the singular-end contribution to the Lagrange identity for a Sturm-Liouville operator. Together with a separated self-adjoint condition at the other endpoint, it selects the regular self-adjoint differential operator realization. The precise operator domain uses the maximal differential-operator domain and the no-log endpoint condition; boundedness of classical eigenfunctions is its convenient local expression, not a replacement for specifying an operator domain.

Engineering stress is axial force divided by the initial cross-sectional area. The initial area remains the denominator throughout a tensile modulus experiment, even if the specimen later changes shape.

Engineering strain is the change in length divided by that segment's initial length. For an extensometer, use its own initial gauge length and extension.

Engineering shear strain is the first-order change of the right angle between material lines initially directed along two perpendicular axes. For a displacement $\mathbf{u}$, ${\gamma }_{xy}={\partial }_y{u}_x+{\partial }_x{u}_y=2{\epsilon }_{xy}$, where ${\epsilon }_{xy}$ is the tensor strain component. Slender Euler-Bernoulli beam equation models neglect the transverse shear strain contribution to deflection.

Let $x_n$ be a centered scalar beam-deviation coordinate in a deliberately controlled toy model. Let the readout be $s_n=x_n+{\eta }_n$, where the centered measurement noise ${\eta }_n$ has variance $N$ and is independent of $x_n$ and of earlier noise. Stipulate a correctly timed correction of strength $g$: $x_{n+1}=x_n-g{s}_n$. This scalar observation is a teaching surrogate, not a claim that real cooling measures each particle. From variance additivity for independent random variables,For fixed current $V>0$ and $N\ge 0$, differentiate the next-step variance with respect to feedback gain:The unique one-step minimizing feedback gain and resulting variance areThe noisier the readout relative to the current spread, the smaller the useful correction. The exact one-step cooling condition follows by subtraction:Thus excessive positive feedback gain heats this model. The threshold depends on the current variance: a gain that initially cools can later heat if the beam has become narrower or the noise has grown.

For chosen dimensionless $V_0=1$ and $N=1/4$, $g_{\ast }=0.8$ gives $V_1=0.20$. Taking $g=1.8$ instead gives $V_1=1.45$, larger than the initial spread. Holding $g=0.8$ fixed givesso the variance approaches the nonzero noise floor $1/6$. More generally, for $0<g<2$ the fixed-gain limit is $V_{\infty }=gN/(2-g)$. This long-run stability condition differs from the condition for cooling the current beam on the very next step.

If the toy feedback gain is recomputed after every step and $N>0$, the reciprocal recurrence is even simpler:With the chosen inputs, $V_5=1/21\simeq 0.047619$. This ideal model has no additional diffusion, finite observation time or lost correlation; its indefinitely falling spread is not a prediction for an operating storage ring.

To expose the missing physics mathematically, replace $s=x+\eta$ by an arbitrary centered sample signal $s$. Defining $C=\mathrm{Cov}(x,s)$ and $S=\mathrm{Var}(s)>0$ givesUnwanted contributions from other particles or electronics can enlarge $S$ without increasing useful $C$; poor timing or lost correlation can reduce $C$ or even reverse its sign. This is why real stochastic cooling needs measured beam fluctuations, appropriate beam mixing and correctly phased feedback control, rather than an imagined catalogue of individual particle positions.

Beam mixing changes sample membership between turns through revolution-time differences. Repeatedly correlated samples and thermal noise limit stochastic cooling.^[ref]

Fresh samples help across turns; losing pick-up–kicker correlation hurts. Useful feedback gain therefore depends on bandwidth, delay, beam dynamics and signal-to-noise ratio.

The time after which a response remains within a specified tolerance of its final value. State both the tolerance and its reference scale.

For a positive step potential and left incidence, take $E\to V_0^{+}$ at fixed incident probability amplitude $A$. The transmitted wavenumber tends to zero while the incident wavenumber tends to $k=\sqrt{2m{V}_0}/\hslash$. Wavefunction matching at a finite potential step gives reflected amplitude $A$ and transmitted amplitude $2A$, producing the spatial wavefunction in the title. The left probability density is $4|A{|}^2{\cos }^{2}kx$ and the right density $4|A{|}^2$. The transmitted probability current still tends to zero because its velocity tends to zero. Over adjacent intervals of length $2\pi /k$, the right integrated density is twice the left. These ratios use finite local integrals of a non-normalizable stationary scattering state.

For $V_0<0$, $k=\sqrt{2mE}/\hslash$ tends to zero as $E\to 0^{+}$ while $q=\sqrt{2m(E+|V_0|)}/\hslash$ stays finite. Matching gives probability amplitudes $r=(k-q)/(k+q)$ and $t=2k/(k+q)$, so quantum reflection probability $R=r^2\to 1$ and transmission probability $T=(q/k)|t{|}^2\sim 4k/q\to 0$. The final region is classically allowed; reflection results from the abrupt spatial change in the wavefunction's wavenumber. These are limits of positive-energy incoming states, not a flux ratio evaluated at exactly zero incident current.

William Alfred Fowler's 1983 lecture connects laboratory nuclear reaction cross-sections to stellar nucleosynthesis and credits Fred Hoyle's stellar synthesis framework.^[ref] This complements Bethe stellar energy-production theory: a fuel reservoir alone cannot predict surviving stellar chemical abundances.

Transient beryllium-8 must capture an alpha particle before dissociation; subsequent electromagnetic decay of the Hoyle state can stabilize carbon-12 despite dominant breakup.

Why can a star shine for so long, and why do hotter stellar cores favor different nuclear reactions? A useful theory must supply both an adequate energy reservoir and the right stellar energy-generation rate. Nuclear binding energy supplies the reservoir; the thermonuclear reaction rate, stellar composition and stellar structure determine how quickly it is spent. The proton–proton chain and CNO cycle are two routes for hydrogen burning with very different temperature sensitivities.

In his 1967 lecture, Hans Albrecht Bethe contrasted the short Kelvin-Helmholtz cooling time with long ages inferred from terrestrial and stellar evidence.^[ref]

For an original solar-scale estimate, choose mass $M=1.99\times 10^{30}\mathrm{kg}$, radius $R=6.96\times 10^8\mathrm{m}$ and luminosity $L=3.83\times 10^{26}\mathrm{W}$. The characteristic Kelvin-Helmholtz cooling time $G{M}^2/(RL)$ is about $31\mathrm{Myr}$; the precise gravitational reservoir has a structure-dependent coefficient. If a fraction $f=0.10$ of the mass can burn its initial hydrogen mass fraction $X=0.70$, with a chosen rest-energy conversion fraction $\eta =0.007$, the fuel estimate isThis assumes fixed luminosity and neglects neutrino losses. It demonstrates the reservoir advantage, rather than predicting the Sun's age or its full stellar evolution. Having enough fuel still does not identify a reaction: the fuel must release energy at a rate compatible with hydrostatic equilibrium and radiative transfer.

Hans Albrecht Bethe and Charles Louis Critchfield calculated deuterium formation from two protons in 1938, combining quantum tunnelling with a weak conversion of a proton into a neutron. Their calculated nuclear reaction cross-section was a theoretical input to hydrogen burning.^[ref] In 1939, Hans Albrecht Bethe analysed the catalytic carbon–nitrogen route and compared required stellar core temperatures with stellar structure estimates in Table VIII. His reported solar agreement used historical abundance and structure assumptions; it should not be read as a modern demonstration that the CNO cycle dominates the Sun.^[ref]

Hans Albrecht Bethe's 1967 lecture credits Carl Friedrich von Weizsäcker's independent 1938 carbon–nitrogen cycle discovery, distinguishing it from calculation of rates and temperature dependence.^[ref] The 1967 Nobel Prize in Physics recognized Hans Albrecht Bethe's contributions to nuclear reaction theory, especially stellar energy production.^[ref]

Laboratory nuclear reaction cross-sections were inputs, not measurements of an artificial star. The 1939 paper converted thick-target yields to nuclear reaction cross-sections using a range–energy relation and tabulated capture data.^[ref] A concrete apparatus example is the Cockcroft-Walton lithium disintegration experiment, with an accelerated proton beam, lithium target, zinc sulfide screen and beam-blocking control. Its nuclear reaction differs from both stellar burning routes. The 1967 lecture stresses extrapolation below laboratory access and tests for nearby nuclear levels.^[ref]

Suppose $h$ is differentiable and $F^{\prime }>0$ on its range. The multivariable chain rule shows that $F\circ h$ has exactly the same stationary points as $h$. Since $F$ is strictly increasing, the type of every local maximum, local minimum, and saddle point is preserved. If both functions are twice differentiable, then at a stationary pointso the Hessian matrix signature is unchanged. The hypothesis $F^{\prime }>0$ matters for the converse about stationary points: a strictly increasing function such as $F(s)=s^3$ can have a zero derivative and create new stationary points in a composition.

The physical radius assigned to a star, with a specified surface convention.