For an exactly soluble example, set $L_1=L_2=L$, $R_1=R_2=R$, and $0<M<L$. Start with zero electric currents, and apply a constant electric potential difference $V_0>0$ at $t=0$. The sums $s=i_1+i_2$ and $d=i_1-i_2$ obey separate first-order linear differential equations:Their time constants are ${\tau }_{+}=(L+M)/R$ and ${\tau }_{-}=(L-M)/R$. Thus, for $t\ge 0$,Since ${\tau }_{-}<{\tau }_{+}$, $i_2(t)<0$ for $t>0$, but $i_2(t)\to 0$ as $t\to \infty$. The primary approaches $V_0/R$. The secondary response opposes the rise of linked magnetic flux, in accordance with Lenz's law. Removing mutual inductance by setting $M=0$ makes the two exponentials equal and gives zero secondary electric current.

For stipulated $L=0.1\mathrm{H}$, $M=0.06\mathrm{H}$, $R=10\Omega$ and $V_0=1\mathrm{V}$, the two time constants are $16\mathrm{ms}$ and $4\mathrm{ms}$. Setting ${\dot{i}}_2=0$ gives its largest negative magnitude atThe continuing primary electric current eventually reaches $100\mathrm{mA}$ while the secondary returns to zero. These numbers belong to the stated model, not the 1831 apparatus.

Victor Franz Hess's 1912 series used two sealed, 3-mm-walled ionization chambers with Theodor Wulf's fiber electrometers. Karl Bergwitz suggested internal plating; Günther & Tegetmeyer improved focusing. Basket-mounted ground comparisons, insulation-loss measurements, simultaneous detectors and barograph-derived heights supplied checks. The highest flight reached 5350 m; its comparisons came from Vienna and after landing, without a launch-site baseline.^[ref]

A third, thin-walled chamber admitted ambient air. Hess multiplied its readings by $750/b$, with pressure $b$ in millimeters of mercury, warning that vessel-wall contributions might violate this scaling. He recorded outside temperature but applied no absolute-temperature correction. Night flights, an eclipse and weather observations constrained alternatives to radiation from above; the eclipse argument addressed directly propagating solar radiation.^[ref]

The electrometers read gradual electric potential loss caused by continuing air ionization. The low-altitude dip and strong high-altitude rise favored an extraterrestrial contribution, without measuring proton composition.^[ref] An ionization chamber integrates electric charge: one energetic particle may create many ion pairs, several particles may contribute to one reading, and a photon can create charged particles in the gas or walls. A voltage-loss reading is therefore neither a count of primary cosmic rays nor a chemical analysis of them.

Domenico Pacini supplied independent evidence: in June 1911 he compared a copper-box-enclosed electroscope at the sea surface and 3 m underwater, more than 300 m offshore above an 8-m-deep seabed. Reduced underwater ionization, stable electrical loss between runs, and a lake comparison supported an external contribution independent of soil radioactive decay. Water provided shielding; this established a different constraint from Hess's altitude comparison.^[ref]

An air shower is a cascade of secondary particles produced by an energetic particle interacting in the atmosphere. Both a charged particle and a gamma ray can start one.^[ref]

A secondary cosmic ray here is a particle produced by an atmospheric interaction of a primary cosmic ray or its descendants. A muon at ground level need not have traveled from the original astronomical source.

Here a primary cosmic ray means a particle arriving from space before its first atmospheric collision.

Let $X$ have the Standard Cauchy distribution and let $Y$, independently, have the uniform distribution on $[-h,h]$, where $h>0$. The convolution of probability densities gives the probability density of $Z=X+Y$ asThis is an even function. Its derivative isso its unique mode is zero. Its tails satisfy $p_h(z)\sim 1/(\pi z^2)$ as $|z|\to \infty$: multiplying the integral by $z^2$ makes the integrand converge uniformly to $1$ on the bounded interval $[-h,h]$. Consequently both positive and negative parts of the first moment diverge, so its expected value does not exist. The symmetric Cauchy principal value of that moment is zero, which is a different assertion. Adding a bounded independent displacement smooths the probability density without removing the divergent first moment of the Cauchy distribution.

Use $N_0={\nu }_s$ for the normal-state single-spin density of states per unit energy and volume at the Fermi surface. Write the attraction as $-V/\mathcal{V}$, with $V>0$, and define $\lambda =N_{0}V$.^[ref]

Since $V/\mathcal{V}$ has units of energy, $V$ has units of energy times volume, making $\lambda$ dimensionless. Set ${\Omega }_D=\hslash {\omega }_D$, an energy cutoff using the Debye frequency ${\omega }_D$, rather than a frequency itself. These conventions also fix the meaning of the earlier constant-shell BCS gap equation.

Here is the reduction of the BCS gap equation in these conventions. Substituting the zero-temperature BCS anomalous average givesThe $k$ sum counts one spin species. For $\Delta >0$, cancelling $\Delta$ and replacing this sum by an integral with constant $N_0$ givesThe factor two from the symmetric integral cancels the $1/2$ in the BCS anomalous average; it is not an extra spin degeneracy. Substituting $\xi =\Delta \sinh t$ evaluates the integral as $\mathrm{arsinh}({\Omega }_D/\Delta )$. Solving givesThe first equality is exact within the constant-density of states, sharp-cutoff mean-field approximation; the last is the weak coupling asymptote. Extending the first equality to large $\lambda$ does not make this simplified model an accurate description of a strongly coupled material.

For an original numerical example, choose $\hslash {\omega }_D=20\mathrm{meV}$ and $\lambda =0.25$. The exact constant-shell BCS gap equation gives $\Delta (0)=0.732871\mathrm{meV}$; the weak coupling formula gives $0.732626\mathrm{meV}$. Its fractional underestimate, measured against the exact result, is exactly $e^{-2/\lambda }=e^{-8}\simeq 0.03355\%$. At $\lambda =1$ that error is $e^{-2}\simeq 13.53\%$, so the two formulas are no longer interchangeable. No material parameters have been fitted here.

At nonzero temperature, substitute the finite-temperature BCS anomalous average instead. The same cancellation yieldsAt the superconducting transition temperature take the continuous $\Delta \to 0$ limit. The integrand at $\xi =0$ then has the finite limiting value $1/(2k_B{T}_c)$, so no divergent endpoint needs to be inserted into a numerical quadrature. For $k_B{T}_c\ll {\Omega }_D$, the resulting integral gives $k_B{T}_c\simeq 1.134{\Omega }_D{e}^{-1/\lambda }$ and $2\Delta (0)/(k_B{T}_c)\simeq 3.528$. These ratios concern the isotropic weak coupling model.^[ref]

A nonzero pivot element can still be a poor numerical choice. Partial pivoting bounds the absolute values of the elimination multipliers by one in exact arithmetic, because each numerator is no larger than the selected denominator. This usually improves numerical stability, but does not prevent every possible growth of intermediate entries or guarantee accurate answers for every linear system.^[ref]

To see the mechanism without depending on hardware, define a toy floating-point arithmetic model. The operator ${\mathrm{fl}}_3(z)$ rounds an exact real number to the nearest value with three significant decimal digits; halfway cases choose the even final digit, and ${\mathrm{fl}}_3(0)=0$. Allow an unbounded exponent range. Round the inputs and the result of every addition, subtraction, multiplication and division separately; a product followed by a subtraction rounds twice. This is a prescribed arithmetic model, not a measurement of a physical computer. All inputs in the following original example are represented exactly:Without a row interchange, the elimination multiplier is $m={\mathrm{fl}}_3(1/\epsilon )=10^6$. The last coefficient and right-hand side becomeTwo distinct exact values now have the same rounded representation. Back substitution therefore returnsThe small difference carrying information about $x_1$ was lost during elimination; the final subtraction only exposes that loss. Dividing by the tiny pivot element amplifies sensitivity to the lost information.

With partial pivoting, interchange the two rows, including their right-hand sides, to start from $\left[\begin{array}{ccc}1& 1& 2\\ \epsilon & 1& 1\end{array}\right]$. Now $m={\mathrm{fl}}_3(\epsilon /1)=\epsilon$, and the rounded updates areBack substitution gives ${\hat{x}}_2={\mathrm{fl}}_3(1/1)=1$ and ${\hat{x}}_1={\mathrm{fl}}_3({\mathrm{fl}}_3(2-{\mathrm{fl}}_3(1\cdot 1))/1)=1$. Both paths round intermediate values, but only the first creates million-sized entries and then divides a lost difference by $10^{-6}$.

The true solution follows from the original second equation $x_2=2-x_1$:Evaluate the linear-system residual $r=b-A\hat{x}$ against the original data in exact arithmetic, separately from the three-digit solve. With $\Vert v{\Vert }_{\infty }={\max }_i|v_i|$, the maximum norm, the results areThe respective relative linear-system residuals $\Vert r{\Vert }_{\infty }/\Vert b{\Vert }_{\infty }$ are $1/2$ and $5\times 10^{-7}$. Computing the diagnostic in higher precision matters: a check made only against the rounded reduced equations would miss the original error.

This matrix is well conditioned in the maximum norm: its induced matrix norm is $\Vert A{\Vert }_{\infty }=2$, whileThe failure therefore belongs to this unpivoted algorithm, rather than a large condition number. In general, however, partial pivoting cannot remove sensitivity inherent in the data. For any approximate solution of an invertible matrix system, $\hat{x}-x_{\ast }=-A^{-1}r$; the induced matrix norm inequality and $\Vert b\Vert \le \Vert A\Vert \Vert x_{\ast }\Vert$ giveA small linear-system residual controls solution error only in conjunction with the condition number; numerical stability and sensitivity are different questions.^[ref]

Substituting $u=-2\nu {\partial }_x\log \Psi$ into the viscous Burgers equation gives ${\partial }_x[({\Psi }_t-\nu {\Psi }_{xx})/\Psi ]=0$. Integration permits the arbitrary function $C(t)$ in the title. Multiplication of $\Psi$ by $e^{-\int C(t)dt}$ preserves $u$ and reduces its equation to the heat equation. Thus the heat-equation representation uses a permissible time normalization rather than a condition uniquely imposed on every representative $\Psi$.

For a Planck function expressed as spectral radiance per frequency interval,$h$ is the Planck constant, $c$ the speed of light and $k_B$ the Boltzmann constant. Differentiating at fixed frequency givesThe proof uses ${\partial }_{T}x=-x/T$ and the chain rule. At low frequency, the multiplier tends to one, recovering the Rayleigh-Jeans law. At high frequency it grows approximately as $x$: the same fractional temperature change has a larger fractional brightness effect on the declining spectral tail.

For an original example choose $T=2.725\mathrm{K}$ and $\delta T=1\mathrm{mK}$. At $100\mathrm{GHz}$, $x=1.76119$, so the first-order fractional residual is $7.80415\times 10^{-4}$. At $300\mathrm{GHz}$, $x=5.28357$, it is $1.94881\times 10^{-3}$. Both spectra are exact blackbodies: this residual is a temperature mismatch, not evidence of a nonthermal spectrum. Fitting temperature before assessing remaining spectral residuals therefore removes a physically distinct degree of freedom. The figure divides by a fixed peak brightness rather than by the rapidly vanishing high-frequency tail.

An ideal differential measurement constrains two sky temperatures through $d_{ij}=T_i-T_j$. Represent every observed pair by a row of a matrix $A$, with $+1$ in column $i$, $-1$ in column $j$ and zeros elsewhere. Then $d=AT$ and $A1=0$: adding any constant to the entire map changes no measurement. This null space is the angular angular monopole ambiguity, explaining why a directional comparison alone cannot measure the absolute blackbody temperature established by FIRAS.

For an original three-pixel example choose $T=(30,-10,-20)\mu \mathrm{K}$ relative to an unspecified mean. Measurements $d_{12}=40\mu \mathrm{K}$ and $d_{23}=10\mu \mathrm{K}$ determine the third difference $d_{31}=-50\mu \mathrm{K}$. The zero-mean condition recovers the chosen map, but adding $100\mu \mathrm{K}$ to every pixel gives identical differences. The closure sum $d_{12}+d_{23}+d_{31}=0$ checks consistency without establishing the mean.

More generally, the graph joining measured pixel pairs must be a connected graph. If $AT=0$, each edge forces equal endpoint temperatures; a path propagates that equality throughout each connected component. Thus a connected observation graph leaves exactly one undetermined constant, whereas disconnected coverage leaves one per component. Fixing the mean makes an ideal connected reconstruction unique; real noise calls for least-squares solution fitting and uncertainty analysis.

The figure chooses a cosmic microwave background quadrupole sky $T(\theta )=Q{P}_2(\cos \theta )$, with $Q=30\mu \mathrm{K}$ and Legendre polynomial $P_2(u)=(3u^2-1)/2$. On a great-circle scan, with $\varphi$ measured from the chosen symmetry axis and ideal beams separated by $\alpha =60^{\circ }$,Expanding ${\cos }^2\varphi =(1+\cos 2\varphi )/2$ proves the formula; its peak magnitude is $45\sqrt{3}/2\simeq 38.9711\mu \mathrm{K}$. A constant sky gives zero. This deliberately restricted scan also misses some angular patterns, so rejecting the angular monopole is not a proof of complete sky reconstruction. Broad, overlapping coverage supplies the missing constraints; finite antenna beam patterns additionally smooth small angular features.

A hotter stellar core can favor a higher-barrier route despite fewer catalyst nuclei. To isolate this competition, prescribe a local hydrogen burning toy model at fixed mass density $\rho$, with reference $T_0=15\mathrm{MK}$ andChoose $E_{G,pp}=493\mathrm{keV}$ and $E_{G,\mathrm{CNO}}=45000\mathrm{keV}$, rounded barrier parameters for a proton pair and proton capture on nitrogen-14. These prescribe the temperature dependence, while the following freely chosen heating normalization incorporates effective nuclear factors and deposited energy:Here $X$ is hydrogen mass fraction, $X_{\mathrm{CNO}}$ is CNO mass fraction, and ${ϵ}_{\ast }$ is an arbitrary positive unit of stellar energy-generation rate. For $X=0.70$ and $X_{\mathrm{CNO}}=0.01$, the reference ratio is $0.01$ by construction. It is not a measured solar ratio. The CNO cycle bottleneck is represented by one effective rate, without evolving the existing proton-proton chain branches or catalyst network.

At $T_0$, the logarithmic temperature slopes are $({\tau }_{pp,0}-2)/3=3.90$ and $({\tau }_{\mathrm{CNO},0}-2)/3=19.90$. The larger Coulomb barrier gives a steeper fractional response. Because these slopes change with temperature, this example retains the exponential rather than extending two fixed power laws. Writing $R_0=0.01(X_{\mathrm{CNO}}/0.01)(0.70/X)$ and $D={\tau }_{\mathrm{CNO},0}-{\tau }_{pp,0}$ givesprovided the bracket is positive. With the chosen composition, $T_{\ast }=20.30\mathrm{MK}$. Reducing $X_{\mathrm{CNO}}$ to $0.001$ at unchanged $X$ shifts it to $23.91\mathrm{MK}$. For $X_{\mathrm{CNO}}=0$, this catalytic route supplies no heat in this model. These are local toy crossovers: a whole star's luminosity requires integrating its stellar energy-generation rate through its stellar composition profile and stellar structure. Nuclear features, plasma screening and changes of the dominant bottleneck limit the model's extrapolation.

At fixed atom number, the entropy of the classical ideal gas in a spherical linear trap depends on temperature and strength through $Z_1\propto T^{9/2}{a}^{-3}$. Reversible adiabatic change therefore holds this product constant, giving the displayed relation. Reducing trap strength lowers the temperature while increasing cloud size. Its peak classical phase-space occupation $N/Z_1$ remains constant, so this cooling alone does not make the gas more quantum degenerate. Heat exchange, particle loss or irreversible entropy production would change this relation.

For a dilute classical trapped gas, $\mu \simeq k_{B}T\ln (N/Z_1)$ with energy zero at the potential minimum. Thus $z=e^{\mu /(k_{B}T)}=N/Z_1$. Quantum occupation effects begin when $z$ is no longer small, giving the displayed order-of-magnitude particle threshold. Equivalently, the central density $n(0)=N/[8\pi (k_{B}T/a{)}^3]$ satisfies $n(0){\lambda }_T^3\sim 1$. This is a degeneracy crossover criterion rather than an exact Bose condensation threshold. In the continuum approximation, spinless bosons have excited-state saturation $N_c=\zeta (9/2)Z_1$, from the trap density of states proportional to $E^{7/2}$. A quantum ground energy should be subtracted consistently when discreteness matters.

The clamped-beam Green function is symmetric. Its explicit formula depends only on $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. More generally, two integrations by parts give $\int u{v}^{\prime \prime \prime \prime }dx=\int u^{\prime \prime }{v}^{\prime \prime }dx=\int u^{\prime \prime \prime \prime }vdx$ when $u=u^{\prime }=v=v^{\prime }=0$ at both endpoints. Applying this identity to two point-load responses shows that the displacement at one point caused by a unit force at a second equals the displacement at the second caused by a unit force at the first.

For a positive-energy Dirac spinor normalized to unit probability and with positive helicity, the left-chiral component has squared weight $(1-\beta )/2$, where $\beta =|\mathbf{p}|/E$ in natural units. Its right-chiral weight is $(1+\beta )/2$; for negative helicity the two are exchanged. In an ultrarelativistic positive-helicity particle, the left-chiral amplitude is suppressed by a mass-to-energy ratio.

A charge carrier population per unit volume or, for a sheet, per unit area. The dimensional convention and whether different carrier species are counted separately must be specified.

X-ray fluorescence is Characteristic X-ray emission following excitation by incident photons. In a simple initially unexcited chemical atom, photoionization can eject an inner electron if the incident photon supplies its binding energy. An outer electron can then fill the vacancy, releasing an emitted photon with the difference of shell binding energies. Nonradiative relaxation competes with this route, so one absorbed incident photon need not yield one escaping characteristic photon.

Choose a separate stipulated electron shell model with values $B_K=10.0$, $B_L=1.5$ and $B_M=0.3$, all in $\mathrm{keV}$. Ignore subshell splitting, recoil and changes of binding energy during relaxation. The possible shell differences areThese labels denote simplified shell groups, not resolved subshell electronic transitions. With $hc=1.23984\mathrm{keV}\mathrm{nm}$, their calculated wavelengths are respectively $0.145864$, $0.127819$ and $1.03320\mathrm{nm}$. An incident $9.8\mathrm{keV}$ photon exceeds both K-line energies but cannot directly create the K vacancy in this initially unexcited model. A $12.0\mathrm{keV}$ photon can, leaving approximately $2.0\mathrm{keV}$ of photoelectron kinetic energy. This comparison fixes neither the excitation probability nor the radiative branching probability.

X-ray emission spectroscopy studies the frequency spectrum of emitted X-rays. The spectral line positions identify electronic energy differences; their strengths also depend on excitation, photon absorption inside the sample, geometry and detector response. An X-ray tube may excite a sample with electrons, while X-ray fluorescence uses incident photons. The distinction matters when choosing what the experiment measures: identifying a chemical element from a line position is different from determining its abundance from a line strength.

For two ideal strings under a common tension, the positive mechanical impedances satisfy $Z_j=\sqrt{T{\mu }_j}$, where ${\mu }_j$ are their linear mass densities. The reflected fraction of incident power is $\mathcal{R}=[(Z_1-Z_2)/(Z_1+Z_2){]}^2$. Solving for the positive ratio $z=Z_2/Z_1$ gives $z=(1+\sqrt{\mathcal{R}})/(1-\sqrt{\mathcal{R}})$ or its reciprocal. Squaring yields the displayed linear mass density ratios. For $\mathcal{R}=1/2$, the heavier string has $17+12\sqrt{2}\simeq 33.97$ times the lighter string's linear mass density; the impedance ratio is only $3+2\sqrt{2}$.