The equilibrium calculation determines where the two temperatures end, without determining how long they take. Stipulate a constant thermal conductance $G>0$ and the linear heat transfer law underlying Newton's law of cooling. With $Q$ positive from body 1 to bath 2, the ordinary differential equations areThe dot denotes a derivative with respect to time. This model assumes internal equilibration is fast compared with exchange between the bodies. Adding the two energy rates gives zero; subtracting the temperature rates givesThe temperature difference has exponential decay with time constant $\tau$. For the finite heat bath example and the stipulated $G=2\mathrm{W}{\mathrm{K}}^{-1}=2\mathrm{J}{\mathrm{s}}^{-1}{\mathrm{K}}^{-1}$, $\tau =45\mathrm{s}$, and solving the ordinary differential equations givesAt one time constant, $T_1\simeq 315.865\mathrm{K}$, $T_2\simeq 293.793\mathrm{K}$ and $Q\simeq 3413.451\mathrm{J}$. Equal temperatures are approached as a limit at infinite time. The finite bath causes the temperature difference to shrink faster than the fixed-bath model, because both bodies change temperature.

For an ideal heat bath with the same $G$, set $T_2=T_b=290\mathrm{K}$ rather than evolving its temperature. Newton's law of cooling then yieldsChanging $G$ changes these time constants, while the final temperature is fixed by the capacities and initial conditions. The dynamical model independently checks the Second law of thermodynamics:Its integral over the relaxation reproduces the endpoint entropy change. The useful separation is between capacity, which governs how much the bath temperature shifts, and thermal conductance, which governs the exchange rate. Neither alone establishes a reliable heat bath.

Differentiate the finite geometric series ${\sum }_{n=0}^N{r}^n=(1-r^{N+1})/(1-r)$ with respect to $r$, then multiply by $r$. This proves the title identity for $r\ne 1$ without any infinite-series differentiation. At $r=1$, the sum is $N(N+1)/2$. For $|r|<1$, both $N{r}^N$ and $N{r}^{N+1}$ tend to zero, yielding ${\sum }_{n=1}^{\infty }n{r}^n=r/(1-r{)}^2$.

The cyclic group generated by $r$ with $r^4=e$ and $r^2\ne e$ has elements $e,r,r^2,r^3$. It is realized by quarter-turn rotations of a plane. The cyclic group of order four and Klein four-group have the same order but different element orders: the latter has no element of order four.

In 2001, Andrew G. Truscott, Randall G. Hulet, and collaborators observed degeneracy pressure in trapped lithium. A Zeeman slower fed a dual magneto-optical trap; optical pumping prepared atoms for a cloverleaf apparatus for magnetic trapping of neutral atoms. Microwave evaporation removed energetic lithium-7 bosons; elastic collisions cooled lithium-6 fermions by sympathetic cooling.^[ref]

Sequential absorption imaging recorded both trapped clouds on a camera. Fits to axial profiles used the boson cloud as a thermometer. At common temperature, the nearly equal magnetic trapping potentials provided a comparison: the clouds had similar sizes near $T/T_F=1$, but the fermion cloud remained broader as cooling reached $T/T_F=0.25$. Figure 3 compared its radius with ideal Fermi gas and classical predictions.^[ref]

Varying the wait before imaging tested thermal equilibrium. Finite imaging resolution restricted the analysis to the axial direction; fits neglected interaction-induced density shifts. Cooling eventually suffered from the dwindling boson heat capacity. The result supports quantum occupation effects, without directly resolving individual occupied states or reaching zero temperature.^[ref]

These were neutral alkali atoms in a spatially varying trap. The Fermi energy of a model electron gas instead uses uniform electrons, two equally populated spin states, and the electron mass. The measured trapped-atom radii are not measurements of that electron model's $p_F$ or $E_F$; comparing their numerical Fermi energies requires the appropriate particle mass, internal-state populations, and confining potential.

Choose $n=10^{28}{\mathrm{m}}^{-3}$ for a uniform noninteracting electron model with two equally populated spin states. This is an original worked example, not a measured number density assigned to an actual material. Using $m_e\simeq 9.10938\times 10^{-31}\mathrm{kg}$, $\hslash \simeq 1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$, and $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ givesThe conversion uses $1\mathrm{eV}=1.602176634\times 10^{-19}\mathrm{J}$: an electronvolt is an energy unit, while a joule is its SI counterpart. The dimensions provide a quick check: $n^{1/3}$ is inverse length, $\hslash n^{1/3}$ has units of momentum, $p_F^2/m$ has units of energy, and $E_F/k_B$ has units of temperature. The ratio $p_F/(m_{e}c)\simeq 0.002574$ verifies the nonrelativistic limit for this model, where $c$ is the speed of light.

Increasing this chosen number density eightfold doubles the Fermi momentum and quadruples the Fermi energy, giving approximately $6.7701\mathrm{eV}$. The exponent $2/3$ has a geometric origin: the number of occupied states grows as the cube of a momentum radius, but kinetic energy of nonrelativistic particles grows as its square. This makes Fermi energy a useful bridge between a count of particles and a mechanical energy scale.

The Fermi temperature expresses a Fermi energy as a temperature scale, $T_F=E_F/k_B$, where $k_B$ is the Boltzmann constant. In the convention $E_F>0$ measured from the relevant band minimum, $T\ll T_F$ is the regime of a strongly degenerate Fermi gas. For an interacting Fermi liquid, an effective Fermi scale can be introduced through its quasiparticle dispersion; its convention should be specified.

A William & Mary teaching experiment places a $10\mathrm{cm}$ glass rod inside a solenoid. Align a polarized helium-neon laser through the rod's centre onto a photodiode after a rotatable polarimeter analyzer. First scan the analyzer from maximum to minimum transmission in $5^{\circ }$ steps and record detector voltage, establishing the angular response. A function generator and electronic amplifier then drive sinusoidal electric current through the solenoid; a multimeter measures the electric current and an oscilloscope reads the photodiode. Trigger the oscilloscope from the generator, remove the large constant detector component by AC coupling, and average repeated traces. Convert the small synchronized voltage modulation to rotation using the calibrated angular slope. The manual fits rotation against magnetic flux density, using its stated coil calibration and converting measured root mean square current to peak current. This is a specified laboratory procedure, rather than measured results reproduced here.^[ref]

To test the inference, add matched zero-electric current readings before and after each reversal, hold the analyzer fixed during paired readings, and compare equal positive and negative electric currents. A blocked beam tests electrical pickup in the detector channel. These are proposed controls. Joule heating scales as current squared, so reversing electric current at equal magnitude helps separate an odd rotation from even heating effects; it does not remove drift between readings. Record temperature and allow repeatable settling. Changes in transmission, mechanical motion, stressed-glass birefringence and detector pickup can imitate a signal. Synchronized averaging suppresses unsynchronized noise, but cannot reject an artifact synchronized to the drive.

Quantitative extraction of the Verdet constant also needs the integral of magnetic flux density through the rod: a finite solenoid can have a nonuniform axial magnetic field. Use its calibrated path integral, or justify a uniform-field approximation; do not silently multiply a central field by the whole rod length. A signed reversal or phase-sensitive readout tests the part odd in applied magnetic field. Its agreement with the simple rotation law supports the model under these controls, rather than proving that every change of transmitted light intensity is Faraday rotation.

Consider an original model at fixed vacuum wavelength $600\mathrm{nm}$ and temperature $20^{\circ }\mathrm{C}$. Stipulate a transparent sample with Verdet constant $V=20\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$, length $L=0.050\mathrm{m}$, uniform adjustable $B_z$, and a field-independent analyzer-zero offset ${\theta }_0=0.020\mathrm{rad}$. These are chosen values, not measured glass properties. On a continuously tracked small-angle branch,Taking half the difference gives ${\theta }_{\mathrm{odd}}=0.100\mathrm{rad}$; half the sum recovers ${\theta }_0=0.020\mathrm{rad}$. Hence $V={\theta }_{\mathrm{odd}}/(LB)=20\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$. A field-reversal difference of $0.200\mathrm{rad}$ is twice the single-field rotation, not that rotation itself. More generally this subtraction removes every contribution that is even under $B\mapsto -B$, provided the sample and apparatus remain otherwise matched.

An ideal polarimeter analyzer at laboratory angle $a$ reads normalized light intensity $I(a,B)/I_0={\cos }^2[a-{\theta }_{\mathrm{obs}}(B)]$. Here $I_0$ is the light intensity immediately before the analyzer, separately calibrated for each condition. A scan locates the maximum at $a={\theta }_{\mathrm{obs}}$ modulo $\pi$. At $a=\pi /4$, expansion about zero gives $I/I_0=1/2+{\theta }_{\mathrm{obs}}+O({\theta }_{\mathrm{obs}}^3)$, with angles in radians. At crossed extinction, the response instead begins as ${\theta }_{\mathrm{obs}}^2$ and loses the sign near zero. The favorable $45^{\circ }$ slope follows directly by differentiating Malus's law; its sign reverses if the angular convention is reversed.

Faraday rotation exhibits optical nonreciprocity: with the bias magnetic field held fixed, a retracing beam gains another rotation in the same laboratory sense. Natural optical activity in a reciprocal medium instead unwinds on retracing the path.^[ref]

To reconcile this with the signed $B_{\parallel }$ formula, retain the laboratory $x,y,z$ axes for comparing physical lines. For the backward ray choose a right-handed ray basis $x^{\prime }=x$, $y^{\prime }=-y$, $z^{\prime }=-z$. Its magnetic flux density component is $B_{\parallel }^{\prime }=-B_z$, so its local rotation increment is $-\theta$. But positive $y^{\prime }$ is negative laboratory $y$: the backward increment is therefore $+\theta$ in the laboratory orientation plot. Reversing the ray reverses both the longitudinal component and the basis used for its signed angle. Reversing only one creates the false prediction of cancellation.

In fixed laboratory transverse components, define the rotation matrixFor an ideal mirror at normal incidence that contributes the same complex amplitude factor $r$ to both transverse components, the round-trip map is $rR(\theta )R(\theta )=rR(2\theta )$. The common mirror factor does not change the oscillation line. A reciprocal natural rotator instead gives $rR(-\alpha )R(\alpha )=rI$. This comparison assumes a retraced straight path and no additional polarization-changing elements.

For example, a chosen single-pass $30^{\circ }$ Faraday rotation produces a returned line at $60^{\circ }$ to the original laboratory $x$ axis. A chosen reciprocal $30^{\circ }$ natural rotator returns it to $0^{\circ }$. The same addition makes an ideal optical isolator possible: polarizers at $0^{\circ }$ and $45^{\circ }$ surround a $45^{\circ }$ Faraday rotator. Forward light follows $0^{\circ }\to 45^{\circ }$ and passes. Returning light selected at $45^{\circ }$ reaches $90^{\circ }$ at the entrance polarizer and is rejected by Malus's law. Real transmission loss and incomplete extinction limit the isolation.

The factor of one half follows from superposition, without assigning ambiguous right/left names to the circular polarizations. Use the complex amplitude convention $\mathrm{Re}[\mathbf{E}(z)e^{-i\omega t}]$ and fixed transverse unit vectorsFor equal transmitted wave amplitudes, let the accumulated wave phases be ${\varphi }_{+}$ and ${\varphi }_{-}$. Their sum isThe common wave phase $\bar{\varphi }$ sets the timing of the electric field oscillation; the real direction in parentheses sets its linear polarization. Opposite circular polarizations have changed their relative wave phase, yet their sum still oscillates along a line. Unequal transmitted wave amplitudes would generally give elliptical polarization instead.

In a uniform sample, with refractive indices $n_{+},n_{-}$ and vacuum wavenumber $k_0=2\pi /{\lambda }_0$, the phase difference is $(k_{+}-k_{-})L=k_0(n_{+}-n_{-})L$. ThusThe second relation identifies the small signed circular birefringence responsible for the rotation in this basis. Unlike a waveplate, whose separated modes have orthogonal linear polarizations, this ideal rotator has modes with opposite circular polarizations.

The Verdet constant is the coefficient relating Faraday rotation to the magnetic flux density path integral. Its SI units are $\mathrm{rad}{\mathrm{T}}^{-1}{\mathrm{m}}^{-1}$. It depends on material, wavelength and temperature; a sign requires the stated viewing convention.^[ref]

The name commemorates Émile Verdet. For nonuniform material or temperature, use $\theta =\int V(\ell )B_{\parallel }(\ell )d\ell$ when the local scalar description remains valid. A single central magnetic field reading cannot generally replace this weighted integral. Strong additional linear birefringence or appreciable circular dichroism requires following the full polarization of an electromagnetic wave, rather than treating the output as merely a rotated line.

For a positive integer $k$, multiply the exponential series by $t^{k-1}$ and integrate term by term from zero to $x$. Its uniform convergence on every compact interval justifies the operation and proves the title identity. The resulting power series converges absolutely for every real or complex $x$, by the ratio test. Repeated integration by parts evaluates the right side as an exponential times a polynomial, with its constant chosen to make the integral vanish at zero.

For an action $S=\int \mathcal{L}(\varphi ,\partial \varphi ,{\partial }^2\varphi )d^{d+1}x$, vary the field with compact support, or fix both $\delta \varphi$ and its first derivatives on the boundary. One integration by parts on the first-derivative term and two on the second-derivative term give the displayed Euler-Lagrange field equation. With ordered repeated indices, define the second-derivative coefficient symmetrically so that $\delta \mathcal{L}={\mathcal{L}}_{\varphi }\delta \varphi +{\mathcal{L}}_{{\varphi }_{,\mu }}{\partial }_{\mu }\delta \varphi +{\mathcal{L}}_{{\varphi }_{,\mu \nu }}{\partial }_{\mu }{\partial }_{\nu }\delta \varphi$. This convention avoids an unintended factor of two in mixed derivative terms. Higher spatial derivatives alone do not require higher time derivatives.

Adding $({\varphi }_1{\varphi }_{2,t}-{\varphi }_2{\varphi }_{1,t})/(2\tau )$ to the cross-coupled scalar Lagrangian density produces independent equations ${\varphi }_{1,tt}-c^2{\varphi }_{1,xx}+{\varphi }_{1,t}/\tau =0$ and ${\varphi }_{2,tt}-c^2{\varphi }_{2,xx}-{\varphi }_{2,t}/\tau =0$. For the convention $e^{i(kx-\omega t)}$, their frequencies are respectively $-i/(2\tau )\pm \sqrt{c^2{k}^2-1/(4{\tau }^2)}$ and $+i/(2\tau )\pm \sqrt{c^2{k}^2-1/(4{\tau }^2)}$. Above the threshold they oscillate with decaying or growing envelopes; below it they have nonoscillatory real temporal exponents. At equality the repeated temporal root supplies $(A+Bt)e^{\mp t/(2\tau )}$. At $k=0$ each equation also has a constant solution. The full system is generically unstable because of the amplified second field, even though it arises from an autonomous variational system with conserved, indefinite total energy.

Leonhard Euler's Elementa doctrinae solidorum, or Elements of the doctrine of solids, appeared in 1758 in volume 4 of Novi Commentarii academiae scientiarum Petropolitanae, pp. 109–140. The repository record dates its composition to 1750. Composition and publication are distinct dates here.^[ref]

The modern cell-count formula for Euler characteristic puts polyhedral counting into an arbitrary-dimensional CW complex. It also identifies what is being counted: the Euler characteristic of a cube boundary and its interior distinguishes a polyhedral surface from the filled solid using the missing three-dimensional term. The modern homology formulation explains why this alternating arithmetic persists across different decompositions.^[ref]

For a concrete example of cellular homology of real projective space, the real projective plane ${\mathbb{RP}}^2$ has one topological cell in each of dimensions zero, one and two. Its one-dimensional topological cell is a loop, and the face attaches around that loop twice. Choose orientations so that the cellular chain complex over the integers iswith nonzero chain groups in degrees two, one and zero. The twice-around attachment is why the first boundary operator shown is multiplication by two.^[ref]

Over the rational numbers, multiplication by two is invertible, so $H_2$ and $H_1$ vanish, while $H_0\cong \mathbb{Q}$. Over the finite field ${\mathbb{F}}_2$, two equals zero. Both boundary operators then vanish, giving one-dimensional homology groups in all three degrees. Their respective Betti numbers areNevertheless,The direct cell count is also $1-1+1=1$.

This calculation makes the strength and loss of information visible. The Euler characteristic survives a change of the field used for coefficients, while individual homology groups can change. Over the integers, the same chain complex has $H_1\cong \mathbb{Z}/2\mathbb{Z}$ and $H_2=0$: a torsion subgroup that rational Betti numbers do not record. The rank cancellation in the Euler-Poincare formula explains why the alternating total still agrees with the cell count.

The boundary of a cube has eight vertices, twelve edges and six square faces. Treating each open face as a two-dimensional topological cell givesThis agrees with its being a two-sphere: stretching the hollow polyhedral surface changes its geometry but not its topology. The interior is excluded from this space.

Now include the solid interior. The cube has the same lower-dimensional topological cells, together with one open three-dimensional topological cell. The cell-count formula for Euler characteristic becomesThe filled cube is contractible, so its homology agrees with that of a point. The term $-1$ is essential: $V-E+F$ by itself is the full count only for a two-dimensional CW complex.

Adding a diagonal within one square face replaces that face by two triangular faces and introduces one edge. Thus the Euler characteristic changes by $-1+1=0$, as in subdivision invariance of Euler characteristic. Including an interior is a different operation: it adds points to the space, rather than giving the same space a finer description. This distinction matters whenever a geometric model could mean either a surface or the solid it encloses.

If the variable-site model counts every $n$ sector with weight $\left(\genfrac{}{}{0ex}{}{N+n}{n}\right)q^n$, its exact relative partition sum is $(1-q{)}^{-N-1}$. Differentiating its logarithm gives the displayed exact mean. Minimizing the leading extensive Gibbs free energy instead gives $Nq/(1-q)$, the same result to relative order $1/N$. Its exact relative equilibrium free energy is $(N+1)k_{B}T\ln (1-q)$. This correction distinguishes a thermodynamic saddle from an exact finite-system average; changing the microscopic site-creation convention can also change subextensive factors.

Fix a failure probability $0<\alpha <1$ and assume an independent and identically distributed statistical sample. A finite-sample confidence band for its population cumulative distribution function isHere $\log$ is the natural logarithm. The DKW inequality impliesThis is simultaneous coverage over the whole real line, valid at each fixed sample size.^[ref]

To obtain the band, solve $2e^{-2n{\epsilon }^2}=\alpha$ for the positive $\epsilon$. Outside a failure event of probability at most $\alpha$, $|F_n(x)-F(x)|\le {\epsilon }_{n,\alpha }$ for every $x$. Since a cumulative distribution function already lies in $[0,1]$, clipping the endpoints to this interval preserves the coverage event. The guarantee is at least $1-\alpha$, and need not equal it.

Choose $\alpha =0.1$, giving at least $90\%$ simultaneous coverage under the stated model. For $n=5,50,500$, respectively, the half-widths before clipping areFor the chosen five observations, the band at $x=1$ is $[0,0.94733]$, while at $x=4$ it is approximately $[0.25267,1]$. At $x=3$ the unclipped band would be $[0.05267,1.14733]$, so clipping gives $[0.05267,1]$. The broad uncertainty is informative: five observations give a precise description of themselves, but a coarse bound for an unknown population cumulative distribution function.

A collection of $90\%$ pointwise confidence intervals only promises coverage separately at each preselected threshold; it does not promise $90\%$ coverage of an entire cumulative distribution function. Here the one simultaneous event permits inspecting any threshold after seeing the data. The probability refers to repeated sampling of the entire statistical sample, with $F$ fixed. It does not assign $90\%$ probability to a fixed $F$ after observing this particular band, and it is not a prediction interval for a future observation. The displayed $n=5,50,500$ comparison gives a guarantee for each fixed $n$ separately, not simultaneous validity over all sample sizes or an optional stopping rule. The chosen observations are an illustration; the coverage claim follows from the DKW inequality, not from their appearance or a simulation. Dependence or observations with different sampling laws require another argument.

At fixed $\alpha$, multiplying $n$ by $100$ divides the half-width by $10$. Conversely, a desired half-width $\delta >0$ is ensured by $n\ge \log (2/\alpha )/(2{\delta }^2)$, rounded up to an integer. For $\alpha =0.1$ and $\delta =0.1$, this sufficient sample size is $150$. This makes the confidence band useful for planning how many observations are needed before estimating threshold probabilities to a chosen simultaneous accuracy.

For independent and identically distributed real random variables with cumulative distribution function $F$, the Dvoretzky-Kiefer-Wolfowitz inequality gives, for every $n\ge 1$ and $\epsilon >0$,The supremum is the Kolmogorov distance: a single event controls all thresholds simultaneously. The bound holds for discrete, continuous, and mixed probability distributions.^[ref]

For one fixed threshold, Hoeffding's inequality gives the same expression as an upper bound on $\mathbb{P}(|F_n(x)-F(x)|>\epsilon )$. That fixed-threshold result alone does not justify moving the supremum inside the probability. The DKW inequality is the stronger statement needed for a simultaneous empirical distribution confidence band.