Let each side of a symmetric abrupt p-n junction have thickness $d$, area $\mathcal{A}$ and doping density $N$. Partial depletion gives $w^2={\epsilon }_s(V_{\mathrm{bi}}-V)/(eN)$. Both sides become fully depleted at $V_{\mathrm{fd}}=V_{\mathrm{bi}}-eN{d}^2/{\epsilon }_s$. Beyond that point, the ionized dopant charges are fixed, but an incremental terminal voltage changes the electric field uniformly across the full thickness $2d$. Gauss's law then gives $d{Q}_{\mathrm{terminal}}={\epsilon }_s\mathcal{A}dV/(2d)$, up to terminal orientation. The measured positive capacitance approaches $C_{\mathrm{fd}}$ rather than vanishing. This geometrical result assumes ideal contacts, negligible mobile carriers, no breakdown and no other series capacitances.

A Josephson junction can make a collective quantum phase behave as a measurable dynamical coordinate. In current-biased Josephson phase dynamics, the Gauge-invariant Josephson phase $\varphi$ moves in a tilted Josephson washboard potential; the junction's capacitance supplies its effective inertia. A trapped phase gives zero mean electric potential difference, while sustained phase advance produces a nonzero mean electric potential difference. Quantized phase levels in a Josephson junction and quantum escape of a current-biased Josephson junction test quantum mechanics for this collective coordinate.

Begin with a stipulated parallel electrical circuit: an ideal sinusoidal Josephson junction, capacitance $C>0$, and effective electrical resistance $R>0$. The Kirchhoff node law and Josephson effect giveHere $I_b$ is the constant bias electric current, $I_{\mu }$ is a deliberately applied microwave electric current, and $I_N$ represents unwanted fluctuating electric current. Multiplication by $\hslash /(2e)$ puts the undriven equation into the mechanical formThe effective inertia is a coefficient for a dimensionless quantum phase, not a literal material mass moving through the insulating layer. The capacitor's energy, $C{V}^2/2=M_{\varphi }{\dot{\varphi }}^2/2$, provides a direct check of that identification. The effective $R$ includes loading by the external electrical circuit; a constant parallel $R$ is an approximation over the relevant frequency range.

The useful design freedom is that $I_b$ changes the confinement while $C$ changes its inertia. The same manufactured device can therefore connect a classical oscillation, discrete energy levels, and escape to an electrical readout. The Martinis-Devoret-Clarke circuit experiments separate these questions experimentally. John Clarke, Michel H. Devoret and John M. Martinis shared the 2025 Nobel Prize in Physics for circuit experiments establishing macroscopic quantum tunnelling and discrete energy levels.^[ref]

In the ideal sinusoidal Josephson junction, the pair electric current and quantum phase obeyHere $e>0$ is the magnitude of the electron electric charge, $\hslash =h/(2\pi )$ uses the Planck constant, and $I_c$ is the junction's superconducting critical current under the specified conditions. Define the signs as follows: positive conventional electric current flows from terminal 2 to terminal 1, and $V$ is the voltage of terminal 2 relative to terminal 1. The Gauge-invariant Josephson phase below fixes the matching orientation of $\varphi$. The DC Josephson effect and AC Josephson effect follow from these relations. ^[ref]

A superconducting weak link is a connection between regions with superconductivity whose coupling is weak enough that a substantial difference of quantum phase can develop across the connection. An insulating tunnel barrier is one realization; a narrow conducting constriction can also couple the regions. The current-phase relation of a superconducting weak link depends on the connection. A sinusoid is the leading tunnel-junction approximation, not a universal law for every superconducting weak link. ^[ref]

For fixed circular eigenvectors ${\mathbf{u}}_{\pm }=(1,\mp i{)}^{\mathsf{T}}/\sqrt{2}$, let $P_{\pm }={\mathbf{u}}_{\pm }{\mathbf{u}}_{\pm }^{†}$. A uniform optical medium with amplitude propagation coefficients ${\gamma }^{\pm }$ has the displayed Jones matrix. Put $\gamma =({\gamma }^{+}+{\gamma }^{-})/2$ and $\delta =({\gamma }^{+}-{\gamma }^{-})/2$. ThenThe sum of the two modal exponentials supplies a hyperbolic cosine, not its reciprocal. The determinant is $e^{-2\gamma z}$, and $\mathcal{J}(z_1+z_2)=\mathcal{J}(z_1)\mathcal{J}(z_2)$. Real modal coefficients give circular dichroism; purely imaginary coefficients give lossless circular birefringence. Complex modal labels do not have an intrinsic numerical ordering.

For a radial weight $F$ with the required absolutely integrable moments, rotational symmetry makes the second and fourth moments isotropic tensors. The second moment is $A{\delta }_{ij}$; its trace gives $3A=4\pi {\int }_0^{\infty }{r}^{4}F(r)dr$. Complete symmetry of the fourth moment forces equal coefficients of the three Kronecker delta pairings. Its double tensor contraction gives $15B=4\pi {\int }_0^{\infty }{r}^{6}F(r)dr$, proving the displayed expression. Odd moments vanish under $x\mapsto -x$. Choosing $F$ to be the indicator function of a spherical shell gives all the corresponding shell moments without separate angular integrations.

For $y(0)=y(1)=0$, write $A={\int }_0^{1}ydx$ and $w=x(1-x)$. Integration by parts gives $2A={\int }_0^1{y}^{\prime }{w}^{\prime }dx$, while ${\int }_0^1{w}^{\prime 2}dx=1/3$. The Cauchy-Schwarz inequality therefore gives the displayed sharp bound, with equality exactly when $y=Kw$. A prescribed positive budget $B={\int }_0^1(y+y^{\prime 2})dx$ consequently givesThis proves the unique global optimum, rather than only stationarity from a Lagrange multiplier.

For a single-valued curve below the chord joining $(-a,0)$ to $(a,0)$, with length $2a<L\le \pi a$, the maximum enclosed area is attained by the lower minor circular arc. Equivalently it minimizes ${\int }_{-a}^{a}ydx$ at fixed length. Its equation and length areTo prove global minimality for $R>a$, augment the objective by $R\int \sqrt{1+y^{\prime 2}}dx$. The augmented integrand is linear in $y$ and strictly convex in $y^{\prime }$. The displayed arc satisfies its Euler-Lagrange equation, so the supporting-line inequality for the convex slope dependence, followed by integration by parts, proves it minimizes the augmented functional among curves with the prescribed endpoints. On the fixed-length class the multiplier term is constant. The semicircle endpoint follows by a limit; its endpoint tangents are vertical but its length is finite.

For uniform linear mass density $\mu$, a chain minimizing $\mu \int \Phi (z)ds$ at fixed length has Beltrami identity $(\mu \Phi -\lambda )/\sqrt{1+z^{\prime 2}}=C$. In the domain $z<h$, the displayed potential has downward gravitational acceleration $-{\Phi }^{\prime }(z)=-\kappa /(h-z{)}^2$. A circular arc $z=h-\sqrt{a^2-x^2}$ satisfies the identity because $\sqrt{1+z^{\prime 2}}=a/(h-z)$.

This arc is a global minimum among admissible competitors of the same length with the same endpoints in $z<h$. Set $w=h-z>0$. The variable part of the energy is $\mu \kappa \int ds/w$, a hyperbolic length in the Poincaré half-plane model. The calibration for semicircular hyperbolic geodesics $S=\mathrm{arsinh}(x/w)$ has $|\nabla S|=1/w$. The Cauchy-Schwarz inequality gives $\int ds/w\ge S(B)-S(A)$, with equality along the circle traversed with increasing $x$. The constant-potential part is unchanged by the fixed-length constraint.

The following original model chooses piecewise constant oxygen consumption rates in a closed, well-mixed $2\mathrm{mL}$ suspension. It assumes no gas exchange, negligible addition volume, a corrected sensor and sufficient oxygen throughout. The curves are invented teaching examples, not measured traces, fitted data or a reconstruction of Figure 6.

Start at $c(0)=220\mu \mathrm{mol}{\mathrm{L}}^{-1}$. On successive one-minute intervals, choose positive slope magnitudes $5,25,5,25,6,32,1\mu \mathrm{mol}{\mathrm{L}}^{-1}\mathrm{mi}{\mathrm{n}}^{-1}$. Integration gives the endpoint concentrations $220,215,190,185,160,154,122,121\mu \mathrm{mol}{\mathrm{L}}^{-1}$. This is just the recurrence $c_{i+1}=c_i-r_i\Delta t$, with $\Delta t=1\min$; the step plot shows $r_i=-dc/dt$. The trace remains above zero by construction.

During the second ADP interval, the total consumption rate isWith a stipulated mitochondrial protein mass of $0.20\mathrm{mg}$, that is $250\mathrm{nmol}\mathrm{mi}{\mathrm{n}}^{-1}\mathrm{m}{\mathrm{g}}^{-1}$. Using a denominator twice as large would halve this reported rate without changing the raw oxygen trace. Comparisons therefore need both loading and the chosen denominator.

For a second bookkeeping check, the modeled post-antimycin A slope of $1\mu \mathrm{mol}{\mathrm{L}}^{-1}\mathrm{mi}{\mathrm{n}}^{-1}$ corresponds to $2\mathrm{nmol}\mathrm{mi}{\mathrm{n}}^{-1}$ background. If that background is additive and unchanged in earlier windows, subtraction gives $48$, $10$ and $62\mathrm{nmol}\mathrm{mi}{\mathrm{n}}^{-1}$ for excess ADP, oligomycin and FCCP, respectively. The oligomycin-sensitive difference is $48-10=38\mathrm{nmol}\mathrm{mi}{\mathrm{n}}^{-1}$. It is a difference in oxygen consumption under the chosen interventions, not an ATP production rate. Converting it into one requires additional coupling assumptions or an independent ATP measurement; the FCCP interval shows why slope alone cannot supply that conversion.

Across the whole seven-minute example, consumption is $V(c(0)-c(7))=198\mathrm{nmol}$, also the sum of the seven interval consumptions. This equality checks the rate-to-amount accounting independently of how the intervals are named. Actual records require attention to calibration, mixing transients, background stability and decreasing oxygen availability; a line that steepens or flattens alone does not identify the cause.

Take a cycle graph with four labeled graph vertices and four distinct edges $\{1,2\},\{2,3\},\{3,4\},\{4,1\}$. Set $h=0$, $J>0$, and impose periodic boundary conditions by writing ${\sigma }_5={\sigma }_1$. The square drawing represents a one-dimensional ring: it has no diagonal interactions and is not a two-dimensional lattice.

Call an edge between opposite Ising spins a domain wall. If $d$ is their number, the four bond products sum to $4-2d$, so the energy isEach domain wall costs $2J$ relative to an aligned bond. Returning around the ring must return to the starting Ising spin, so $d$ is even: equivalently, ${\prod }_{i=1}^4({\sigma }_i{\sigma }_{i+1})=1$. Choosing an even set of $d$ wall edges and then choosing ${\sigma }_1$ determines the entire pattern. The state degeneracy is therefore $g_d=2\left(\genfrac{}{}{0ex}{}{4}{d}\right)$, using a binomial coefficient.

The twelve middle microstates consist of eight patterns with one exceptional Ising spin and four patterns with two adjacent $+1$ Ising spins and two adjacent $-1$ Ising spins. The two alternating patterns have four domain walls. All $2+12+2=16$ microstates are shown below.

Writing $K=\beta J$, the partition function and the probability of each energy class areFor an original exact example, choose $K=(\ln 2)/4$. Then $e^{4K}=2$, $Z=17$, and $P(d=0)=4/17$, $P(d=2)=12/17$, $P(d=4)=1/17$. Each aligned microstate has probability $2/17$, each middle microstate $1/17$, and each alternating microstate $1/34$: multiply by state degeneracy before comparing whole classes. As $K\to 0$, all sixteen microstates become equally likely, but the class probabilities approach $1/8,3/4,1/8$. As $K\to \infty$, the two all-aligned microstates each have limiting probability $1/2$.

Take a cold, unmagnetized, collisionless plasma with uniform equilibrium electron density $n_e$, fixed neutralizing ions, and a small electromagnetic wave that is a transverse wave. Cold means that thermal-motion corrections are neglected. For electron mass $m_e$, elementary charge magnitude $e$ and electric field $\mathbf{E}=\mathrm{Re}(\hat{\mathbf{E}}{e}^{-i\omega t})$, the linear Newton's second law equation is $m_e\dot{\mathbf{v}}=-e\mathbf{E}$. Thus $\hat{\mathbf{v}}=-ie\hat{\mathbf{E}}/(m_e\omega )$ and the current density is $\hat{\mathbf{J}}=i{n}_e{e}^2\hat{\mathbf{E}}/(m_e\omega )$. Writing the induced electric polarization as $\hat{\mathbf{J}}=-i\omega \hat{\mathbf{P}}$ gives a negative electric susceptibility. With vacuum permittivity ${ϵ}_0$,The plasma frequency ${\omega }_p$ controls the response. For $\omega >{\omega }_p$, the propagating branch has real refractive index $0<n<1$. For $\omega <{\omega }_p$, this homogeneous model has an evanescent wave, not an ordinary propagating wave with a negative real refractive index. In a slowly varying vertical profile, a normally incident ray approaches a turning region where ${\omega }_p(z)=\omega$; the geometrical-ray approximation fails right at that region. At oblique incidence the conserved horizontal wave vector changes the turning condition. A smooth density gradient can therefore return a skywave without a mirror or a layer of metal.

For a homogeneous propagating section, direct differentiation of the dispersion relation givesA phase velocity greater than $c$ does not send information faster than $c$. Here a narrow wave packet has group velocity $v_g$, while the full causal response must still respect special relativity. Even this stripped-down plasma is strongly dispersive near its plasma frequency. In the actual ionosphere, collisions produce absorption, and the magnetic field of Earth changes the response with direction and wave polarization; a single real scalar $n$ cannot capture all of that behavior.

On 11 December 1924, Edward Victor Appleton and Miles Aylmer Fulton Barnett perform an experiment: the Bournemouth broadcasting radio transmitter slowly changes wavelength; an Oxford radio receiver records alternating maxima and minima from ground wave-skywave wave interference. The inferred equivalent height is about $90\mathrm{km}$. A separate comparison of simultaneous loop and vertical antenna signals tests the downward arrival direction. The E layer is established first; the higher F layer is found in winter 1926–27. These dates describe different findings.^[ref]

Why change the wavelength deliberately? Ordinary fading alone could arise from changing emission, receiver drift or a moving propagation path. A controlled sweep predicts a specific progression of phase difference between two arrivals. The directional antenna comparison supplies an independent kind of evidence for an overhead contribution. Agreement supports an atmospheric return; fitting a distance alone would not establish a unique electron density profile. In the model below, comparable received amplitudes make the interference fringes easier to see. A practical repetition would also calibrate the frequency sweep, track radio transmitter output and receiver gain, and compare repeated sweeps with fixed-frequency intervals to detect time drift. These are controls inferred from the measurement model, rather than additional procedures attributed to the historical account.

The radio receiver does not need to resolve two pulses to detect two paths. If one path takes longer, the two contributions reaching the receiver during a slow frequency sweep carry slightly different emission frequencies and can produce a slow beat in the detected amplitude. At strictly fixed frequency with stationary paths, the arrivals instead have a fixed phase difference: there is no continuing beat merely because one path is longer. This distinction connects the historical observation to ionospheric interference fringe spacing.

A skywave is a radio wave returned toward Earth through interaction with the ionosphere. In a smoothly varying plasma, the ray can bend continuously through refraction before returning: there need be no sharp material surface at the apparent point of reflection. Different frequencies, launch directions and wave polarizations can have different paths and losses.