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A semiconductor has an electronic band gap and a conductivity strongly affected by temperature, impurities and carrier injection. Optical absorption can create mobile electrons and holes, enabling a photodiode.

The useful feature of a semiconductor is that its charge carrier population can be controlled. Semiconductor doping changes the supply of mobile electrons or electron holes; carrier injection changes their population away from thermal equilibrium. A transistor turns that control into an electronic amplifier: a small input signal controls a larger flow of supplied energy.

For a uniform semiconductor in a weak electric field, a simple drift model writes its electrical conductivity asHere $q>0$ is the elementary charge, $n,p$ are the number densities of mobile electrons and electron holes, and ${\mu }_n,{\mu }_p$ their positive charge carrier mobilities. The two contributions add because opposite electric charges drift in opposite directions and therefore carry electric current in the same direction. This model assumes fixed temperature and mobilities and omits diffusion from density gradients.

An original dimensionless example isolates the effect of carrier injection. Suppose initially $p\ll n=n_0$ and ${\mu }_n=2{\mu }_p$. Adding equal excess populations $\Delta n=\Delta p=0.1n_0$ gives $\Delta \sigma /{\sigma }_0=0.15$, even though the added electron holes remain a minority carrier population. A semiconductor can respond strongly to a small injected population without changing its underlying band gap.

At sufficiently low temperature, thermal ionization of shallow dopants becomes inefficient, reducing the mobile carrier density. An acceptor-doped sample may then have many neutral acceptors and few mobile electron holes. Extracting an activation energy requires a model of dopant ionization and compensation; the mobile density need not have the same simple activation exponent in every compensation regime.

A search in which isotope-enriched detector material is also the decay source. germanium detectors combine a contained energy deposit with energy resolution, pulse information and material-background controls. Background peaks and resolution tails must be tested before assigning a signal.

Equivalent chemical bonds in an intact molecule need not have equal successive bond-dissociation enthalpies. Once one chemical bond is removed, the next reactant has a different electronic quantum state and may relax to a different molecular geometry. The successive steps compare different species. This distinction also explains why a tabulated average chemical bond strength is not necessarily the cost of the first cleavage.^[ref]

For an invented species $A{B}_2$, stipulate the following gas-phase steps, with all enthalpies at the same temperature and the same standard states:The symbols stand for hypothetical species with consistently chosen electronic states. They are not values for water or another measured molecule. By Hess's law, the total enthalpy of atomization for $A{B}_2(g)\to A(g)+2B(g)$ is $700\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; the arithmetic mean per broken chemical bond is $350\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Neither individual step equals that mean. Nor must the second step be cheaper: here it is deliberately more expensive. An exact sequence sums exactly; substituting averages taken from other molecules generally gives only an estimate.

In a series RLC circuit, the same electric current $i=q^{\prime }$ flows through the resistor with electrical resistance $R$, the inductor, and the capacitor. Choose $q$ as the signed electric charge on the capacitor plate into which positive $i$ flows. With terminal drops measured in that direction, Kirchhoff's voltage law givesHere $V(t)$ is the applied electric potential difference; assume constant $L>0$, $C>0$, and $R>0$. With the applied source set to zero and the loop kept closed, the free response satisfiesThe angular frequency ${\omega }_0$ is the value for lossless free oscillation; $\alpha$ is the decay parameter. The three possibilities are underdamped RLC response for $\alpha <{\omega }_0$, critically damped RLC response for $\alpha ={\omega }_0$, and overdamped RLC response for $\alpha >{\omega }_0$. ^[ref].

For a fair comparison, start each circuit with the same electric charge $q(0)=q_0$ and zero electric current $q^{\prime }(0)=0$. Set $\tau ={\omega }_{0}t$ and $a=\alpha /{\omega }_0$; then $x=q/q_0$ obeys $d^{2}x/d{\tau }^2+2adx/d\tau +x=0$. The plot compares $a=0.5,1,2$. These dimensionless curves apply to any $L,C$ after rescaling time and electric charge.

The stored energy is the sum of magnetic energy and capacitor energy:Differentiating and using the circuit equation gives the complete power balance,In the free response, $E^{\prime }=-R{i}^2\le 0$: Joule heating accounts exactly for the lost energy. For the plotted initial conditions, $E(0)=q_0^2/(2C)$ and $E/E(0)=x^2+(dx/d\tau {)}^2$. Since every free solution tends to zero for $R>0$,In the lossless limit $R=0$, energy stays constant and shuttles between the capacitor and inductor. Thus an oscillation of electric charge and a decrease of total energy are compatible statements.

The analogy with a damped harmonic oscillator is exact within these ideal models: electric charge corresponds to displacement, electric current to velocity, inductance to mass, $1/C$ to spring constant, and $R$ to the coefficient of a resisting force proportional to velocity. The two stored energy terms play the roles of kinetic energy and elastic potential energy. ^[ref].

Differentiating the driven equation also gives $L{i}^{\prime \prime }+R{i}^{\prime }+i/C=V^{\prime }$. When using this equation to solve for electric current, retain the original constraint $L{i}^{\prime }(0)+Ri(0)+q(0)/C=V(0)$: the initial electric charge fixes the initial rate of change of electric current.

For a well of depth $V_0$ and full width $2a$, measured from zero exterior potential energy, even-state matching gives $q=k\tan ka$. If $a\sqrt{m{V}_0}/\hslash \ll 1$, then $ka\ll 1$ and $q\simeq 2m{V}_{0}a/{\hslash }^2$. Using $E_0=-{\hslash }^2{q}^2/(2m)$ gives the displayed leading binding energy. It agrees with the attractive delta-potential result for integrated strength $2a{V}_0$.

The Sherrington-Kirkpatrick model couples every distinct pair of Ising spins using independent standard normal random variables $z_{ij}$ for $i<j$, with $J_{ji}=J_{ij}$. Here $J_0$ and $\mathcal{J}$ have units of energy. David Sherrington and Scott Kirkpatrick introduced this scaling in 1975; their original replica-symmetric calculation gave an unphysical negative low-temperature entropy. The model and that initial proposed solution must be distinguished.^[ref]

The distance a downstream fluid element travels while losing a substantial fraction of its thermal energy. Multiplying the immediate downstream velocity by a radiative gas cooling time gives an order-of-magnitude length. A resolved layer instead uses $\ell =\int u(t)dt$, because mass density, temperature and speed change during cooling. It is distinct from the microscopic dissipative thickness of the shock wave.

The Shockley-Queisser limit is the radiative conversion-efficiency bound for an ideal single-junction solar cell, with one collected electron-electron hole pair per absorbed above-gap photon, carrier thermalization, and radiative recombination required by the reverse optical process. Photons below the band gap are not absorbed, while excess photon energy above it is thermalized. Maximizing the electrical power over the operating electric potential difference and then the band gap gives a ceiling of roughly thirty percent for ordinary unconcentrated sunlight; its precise value depends on the incident spectrum, cell temperature and emission geometry. The original calculation by William Shockley and Hans-Joachim Queisser obtained approximately $30\%$ at $1.1\mathrm{eV}$ for a $6000\mathrm{K}$ blackbody source and a cell at $300\mathrm{K}$.^[ref]

Precision X-ray emission spectroscopy turns closely spaced spectral lines into quantitative constraints on atomic energy levels. Bragg's law makes a crystal an angular ruler for wavelength: resolving nearby directions can reveal components hidden inside an apparently single line. Accurate line positions and the ability to separate two lines are related, but distinct, experimental achievements.

Karl Manne Georg Siegbahn received the 1924 Nobel Prize in Physics; his lecture was delivered on 11 December 1925.^[ref]

His son Kai Siegbahn received the 1981 Nobel Prize in Physics for electron spectroscopy.^[ref]

The Barkla characteristic X-ray experiment distinguished element-dependent K and L groups through absorption and excitation. Moseley's law related corresponding line frequencies across chemical elements to atomic number. Measuring the components within those groups asks a further question: which distinct atomic energy differences produce the neighbouring spectral lines?

Siegbahn's lecture organizes measured spectral lines into energy levels and resolved double levels.^[ref] The payoff is a map of atomic energy differences as well as an element's position in the periodic table.

Signal amplification is an increase in a specified downstream response relative to an initiating signal. In a molecular-count description, one activated component can generate or activate many downstream molecules. Repeated catalysis by an enzyme supplies a mechanism: the enzyme remains available for another chemical reaction. Successive stages can multiply the response. For a biological example, Cooper describes the epinephrine-to-glycogen phosphorylase pathway, where production of many cAMP molecules and repeated downstream protein phosphorylation provide opportunities for signal amplification. This is an example, not a claim that every signal transduction pathway has the same stages or gain. ^[ref]

Here is a counting toy extending the example under signal transduction. Suppose $N_0$ activated receptors each activate $a=5$ distinct relay proteins during a first interval. During a subsequent interval, each relay activates $b=20$ distinct target proteins. Assume enough inactive relays and targets, no shared targets, and no loss of activity during the count. ThenFor $N_0=1$, this gives $N_1=5$ and $N_2=100$: the target-to-receptor count ratio is $100$. These are assumed integer counts, not measured rates or a universal physiological amplification factor. The intervals are sequential; the equations do not specify a time course.

Availability limits the multiplication. As a second toy assumption, let there be only $T=250$ initially inactive target proteins shared by all branches, and suppose the branches can activate distinct targets until this pool is exhausted. ThenTwo activated receptors give $200$ active targets, whereas three give $250$, not $300$. At three activated receptors, the total count ratio is $250/3\simeq 83.3$; adding a fourth activates no additional targets. Thus total output per input and the increment caused by an extra input are different notions of gain. This finite-pool rule is an illustrative bound, not an enzyme kinetics law.

Signal amplification makes a small initiating signal capable of a larger response, but it does not supply the material or energy for downstream chemical reactions. In the second messenger transfer experiment, both stages still required added ATP and magnesium ions. Counts, changes in concentration, and changes in enzyme activity are different readouts; an amplification claim should name its input, output and observation interval. ^[ref]

Simple harmonic motion is motion in which the acceleration is proportional to the displacement from equilibrium and points back towards it:Here $x$ is measured from equilibrium, and $\omega$ is the constant angular frequency. For a point mass attached to an ideal spring, Hooke's law gives the restoring force $F=-kx$, with stiffness $k>0$, and Newton's second law yields $\omega =\sqrt{k/m}$. The minus sign matters: for $x>0$ the force is negative, and for $x<0$ it is positive. A restoring force need not oppose velocity; on the return towards equilibrium it increases speed.

The general solution isThe amplitude $A\ge 0$ and oscillation phase $\phi$ encode the initial conditions. We use $P$ for the oscillation period to distinguish it from kinetic energy $T$. For $A>0$, the oscillation period is independent of amplitude, the property of isochronous harmonic motion. At $A=0$ the body remains at equilibrium. ^[ref]

Choose $m=0.5\mathrm{kg}$, $k=2N/m$, and release the mass from rest at displacement $A=0.3\mathrm{m}$. Then $\omega =2{\mathrm{s}}^{-1}$, $P=\pi \mathrm{s}$, and $\phi =0$. Writing time $t$ in seconds,At $t=0$ the velocity vanishes, but the acceleration is $-1.2m/{\mathrm{s}}^2$: a turning point is not a state of rest that persists. At $t=\pi /4\mathrm{s}$, the body crosses equilibrium with velocity $-0.6m/s$ and zero acceleration. At $t=\pi /2\mathrm{s}$ it reaches the opposite turning point, $x=-0.3\mathrm{m}$.

Taking potential energy zero at equilibrium, the same solution givesAt each turning point all mechanical energy is potential energy; at each equilibrium crossing it is kinetic energy. Both repeat twice during one complete position cycle because squaring removes the sign. At $t=\pi /8\mathrm{s}$, for example, each contributes $0.045\mathrm{J}$.

The physical position trace covers a line segment, $-0.3\le x\le 0.3$ in metres. In the phase plane, the paired position and velocity instead trace the ellipseThis is precisely $m{v}^2/2+k{x}^2/2=E$, rewritten: the single conserved mechanical energy fixes the amplitude and the whole ellipse. It does not fix the starting oscillation phase. At $x=0$ both $v=-0.6m/s$ and $v=+0.6m/s$ lie on it; position alone cannot distinguish them. Replacing $v$ by momentum $p=mv$ gives the equivalent phase-space trajectory. The arrows show increasing time, clockwise in $(x,v)$ coordinates.

This closed ellipse assumes a time-independent spring and no damping or applied drive. Adding a resisting force $-bv$, $b>0$, gives $m\ddot{x}+b\dot{x}+kx=0$ and $dE/dt=-b{v}^2\le 0$. An underdamped damped harmonic oscillator spirals inward in the phase plane, crossing successively smaller energy contours. The ideal harmonic oscillator is also useful beyond springs: near a smooth potential energy minimum with $V^{\prime \prime }(0)>0$, its Taylor expansion begins with $V(0)+V^{\prime \prime }(0)x^2/2$. For sufficiently small displacements, this gives the same restoring force to leading order, explaining why small oscillations share one solvable pattern.

A mechanism that trades the magnitude or direction of an applied force against the displacement through which it acts. An ideal simple machine conserves work; it does not multiply energy.

Simulated annealing accepts worse feasible solutions to escape a local minimum. Minimize an objective function $E(x)$, analogous to energy. Effective temperature $T>0$ has the same units as $E$, absorbing the Boltzmann constant; it controls exploration, rather than a material's thermodynamic temperature.^[ref]

Given current feasible point $x$, draw $y$ from a proposal distribution $q(x,y)$. For a symmetric proposal distribution, $q(x,y)=q(y,x)$, use the Metropolis–Hastings acceptance probabilityAccept with this conditional probability; on rejection retain $x$. Every downhill or equal-cost proposal is accepted. Draw a fresh uniform random variable $U$ on $(0,1)$ and accept if $U<a_T(x,y)$. The off-diagonal transition probability $P_T(x,y)=q(x,y)a_T(x,y)$ includes proposing $y$.^[ref]

For a maximization problem with objective function $f$, set $E=-f$: every improving proposed neighbor is accepted, and a worsening move of loss $\Delta =f(x)-f(y)>0$ has conditional probability $\exp (-\Delta /T)$ of acceptance, under the same symmetry assumption. Likelihood-power annealing uses precisely this sign convention for a log-likelihood. Retain the best feasible solution seen, since the current state can worsen even after a global minimum has been visited.

If the proposal distribution is asymmetric, use the full Metropolis–Hastings acceptance probability, for $q(x,y)>0$,A missing reverse proposal gives zero acceptance. Even a downhill proposal need not be accepted with probability one in this case. Choosing uniformly among a variable number of neighbors is generally asymmetric: $q(x,y)=1/d(x)$ and $q(y,x)=1/d(y)$ give the correction $d(x)/d(y)$, where $d$ is vertex degree.^[ref]

The useful idea is to separate exploring from insisting on improvement: an uphill move can open a route to a better feasible solution. Its usefulness depends on the proposal distribution, the scale of the objective function, and the cooling schedule for simulated annealing. Finite practical schedules carry no general guarantee of finding a global minimum.

Take two stipulated oppositely charged daughters with equal rest mass $m=0.14\mathrm{GeV}/c^2$ and exactly opposite momenta of magnitude $p=0.30\mathrm{GeV}/c$. Their total momentum is zero, soThis is a constructed event, not an identification of a historical particle. Boosting both four-momenta by $\beta =0.60$ along the first daughter's direction gives total energy $0.82765\mathrm{GeV}$ and total momentum $0.49659\mathrm{GeV}/c$. Substitution recovers the same invariant mass, although the laboratory energy has increased.

The following independent simulation stipulates 700 signal combinations from a Cauchy distribution centered at $0.80\mathrm{GeV}/c^2$ with scale $0.020\mathrm{GeV}/c^2$, truncated to $0.50<M<1.10\mathrm{GeV}/c^2$, plus 2400 background combinations whose density decreases linearly across that interval. Each combination is converted to a back-to-back two-daughter event with $m=0.14\mathrm{GeV}/c^2$. Independent isotropic directions are chosen, each Cartesian momentum component receives normal distribution noise with standard deviation $0.012\mathrm{GeV}/c$, and invariant mass is reconstructed from the perturbed momenta. All inputs and counts are invented. No historical experimental points or fitted historical parameters appear.

Single-electron superconducting tunneling transfers individual electrons across a barrier into or out of quasiparticle states of a material with superconductivity. In a normal-metal–insulator–superconductor junction, abbreviated NIS, only one electrode has superconductivity. Such a junction can reveal a superconducting energy gap without supporting a DC Josephson effect between two superconducting reservoirs.

For an elementary interpretation, assume an opaque barrier, approximately constant normal-electrode density of states, and an energy-independent tunneling matrix element over the range probed. Write the superconducting density of states divided by its normal-state value as $n_s(E)$, with $E$ measured from its chemical potential. A signed voltage convention chosen for positive normal-electrode injection giveswhere $f$ is the Fermi-Dirac distribution at a common temperature and $R_N$ is the normal-state junction electrical resistance. This current convention is local to the NIS measurement. The expression counts occupied initial and empty final states; it is a quasiparticle transport model, not a current-phase relation of a superconducting weak link.

An original analytic illustration uses the ideal fully gapped BCS theory spectrum at zero temperature, with positive gap edge $\Delta$:For positive $V$, the occupation difference restricts the integral to $0<E<eV$. Since $d\sqrt{E^2-{\Delta }^2}/dE=E/\sqrt{E^2-{\Delta }^2}$, integration and odd extension giveThe threshold is $\Delta /e$ for this NIS model. In the corresponding zero-temperature single-electron process between two fully gapped superconductors, creating excitations on both sides instead requires $|eV|\ge {\Delta }_1+{\Delta }_2$. That quasiparticle threshold does not forbid coherent pair electric current below it. At finite temperature, thermally populated quasiparticles soften the simple threshold argument.

At zero temperature in the NIS approximation, $R_{N}d{I}_{\mathrm{qp}}/dV=n_s(eV)$ away from the singular gap edge. Thus a voltage sweep probes the excitation density of states. The ideal divergence is rounded in real measurements, and at nonzero temperature the derivative samples a thermally broadened range of energies. For $|eV|>\Delta$, the model has $V{I}_{\mathrm{qp}}>0$: electrical work creates excitations and is dissipated on relaxation. This distinguishes its steady transport from the conservative Josephson coupling energy.

Consider an original ideal one-dimensional monatomic harmonic oscillator chain with neighbor separation $a$, mass $M$ and nearest-neighbor restoring coefficient $K>0$. Its displacement equation isSubstitute the normal mode $u_r=u_0{e}^{i(qra-\omega t)}$. Since $e^{iqa}+e^{-iqa}-2=-4{\sin }^2(qa/2)$,Choose $a=4$ ångströms and ${\epsilon }_{\ast }=10\mathrm{meV}$. This is one acoustic phonon branch, distinct from the two-atom elastic example. It has $\epsilon (\pi /(2a))=5\sqrt{2}\mathrm{meV}$ and $\epsilon (\pi /a)=10\mathrm{meV}$.

For coherent creation of one phonon in a crystal, the excitation's wave vector $q$ and the measured transfer obey $Q=G+q$, with $G$ a reciprocal lattice vector. To connect the dispersion relation to an illustrative intensity, take zero temperature, longitudinal displacements and the first interval $Q=q\in (0,\pi /a]$, choosing $G=0$. Expand the scattering phase to first order, $e^{iQ(ra+u_r)}\simeq e^{iQra}(1+iQ{u}_r)$. The constant term supplies elastic scattering; the linear term creates one phonon. Its displacement matrix element is proportional to $1/\sqrt{\omega }$, so the squared amplitude has weight $Q^2/\epsilon (Q)$, apart from common factors. Normalize this weight to one at $Q=\pi /a$:The limit $w(0)=0$ follows from $\sin (\pi x/2)\sim \pi x/2$. Set $\sigma =0.4\mathrm{meV}$ and replace the ideal sharp line by a normalized Gaussian function:This chosen broadening illustrates readout width; it is not a measured resolution or a full spectrometer resolution function. Integrating over all $\Delta E$ gives $w(x)$. The small negative-energy tail is a broadening artifact, not a zero-temperature phonon-annihilation signal.

These panels separate three kinds of information: allowed reciprocal lattice positions, cancellation within a crystal basis structure factor, and an excitation's energy as a function of wave vector. The chain has no magnetic degrees of freedom, optical branches, damping or multi-phonon processes. It illustrates how one excitation could be read; it proves nothing about the completeness of a real material's excitation spectrum.

Starting in 1967, a collaboration at the Stanford Linear Accelerator Center and the Massachusetts Institute of Technology directed an electron beam at liquid hydrogen, providing proton targets. Jerome Isaac Friedman, Henry Way Kendall and Richard Edward Taylor participated in this experimental program.^[ref]

The 1969 paper Observed Behavior of Highly Inelastic Electron-Proton Scattering reports inclusive electron spectra at $6^{\circ }$ and $10^{\circ }$ with incident energies from $7$ to $17\mathrm{GeV}$. Its scaling comparison estimated ${\nu }_{\mathrm{lab}}{W}_2$ under a transverse-virtual photon assumption, finding approximate agreement as a function of $\omega =1/x$ for $Q^2>1{\mathrm{GeV}}^2$ and $\omega >4$. These are the paper's conditions, not a claim of exact Bjorken scaling at every momentum transfer.^[ref]

A movable magnetic spectrometer selected scattered electrons by momentum and scattering angle. Magnetic fields dispersed momenta at the focal plane. With relativistic electrons, the inferred momentum determines outgoing energy $E^{\prime }$. Commissioning included steering the beam through the spectrometer at zero angle to check its optics and accepted solid angle. Counters sorted events into momentum and angle bins. The acceptance entered the conversion from counts to differential scattering cross-sections.^[ref]

Radiative corrections matter because photon emission can move events from an elastic scattering peak into lower outgoing-energy bins. The analyses subtracted the elastic radiative tail before unfolding inelastic radiation. Independent MIT and SLAC procedures provided a check. An uncorrected low-$E^{\prime }$ tail alone therefore cannot establish deep inelastic scattering.^[ref]

First test rest. With no other tangential forces, the required uphill static friction is $f_s=mg\sin \theta$. Its maximum is ${\mu }_{s}N$, where ${\mu }_s$ is the coefficient of static friction. ThusThe tangent function is increasing on this angle range, and its inverse tangent determines the last permitted resting angle ${\theta }_c$. At equality, the ideal model still permits limiting rest; a larger angle excludes it. Below the threshold, setting $f_s={\mu }_{s}N$ would incorrectly impose the maximum static friction even when less is required. Static friction on an inclined plane and inclined-plane measurement of static friction develop this onset criterion further.

For a block already sliding downhill, uphill kinetic friction has magnitude $f_k={\mu }_{k}N$, where ${\mu }_k$ is the coefficient of kinetic friction. Taking downhill as positive, Newton's second law becomesEvery term in the force balance scales with mass; division by $m>0$ removes it. Doubling mass doubles both gravitational weight and the contact forces, while leaving this acceleration unchanged if the coefficients remain unchanged. A mass-independent result is a prediction of these assumptions, rather than a universal property of every ramp.

For an original numerical model, stipulate $m=3\mathrm{kg}$, $g=9.8m/{\mathrm{s}}^2$, ${\mu }_s=0.45$ and ${\mu }_k=0.30$. The critical angle is ${\theta }_c\simeq 24.23^{\circ }$. At $20^{\circ }$, the normal force is $27.627\mathrm{N}$; rest requires $10.055\mathrm{N}$ of uphill static friction, below its $12.432\mathrm{N}$ bound. A resting block can stay there. If it is already sliding downhill, however, the kinetic friction is only $8.288\mathrm{N}$, giving downhill acceleration $a\simeq 0.5891m/{\mathrm{s}}^2$. Specifying the initial state matters.

At $30^{\circ }$, the downhill component of gravitational weight is $14.700\mathrm{N}$, while $N\simeq 25.461\mathrm{N}$ and the static friction bound is only $11.458\mathrm{N}$. Rest is impossible. During downhill sliding,Released from rest on a sufficiently long ramp, this ideal block covers its first $1.00\mathrm{m}$ in $t=\sqrt{2L/a}\simeq 0.9218\mathrm{s}$, reaching speed $v=\sqrt{2aL}\simeq 2.1697m/s$. As an independent energy calculation, the lost gravitational potential energy is $mgL\sin \theta =14.700\mathrm{J}$ and kinetic friction removes $f_{k}L\simeq 7.638\mathrm{J}$ of mechanical energy. The remaining $7.062\mathrm{J}$ equals $m{v}^2/2$ by the work-energy theorem.

At $\theta =\arctan {\mu }_k$, downhill sliding has zero acceleration and therefore constant velocity. Below that angle, a downhill-moving block slows; when it stops, test the static friction bound again. For uphill sliding, kinetic friction points downhill, so the signed downhill acceleration instead is $g(\sin \theta +{\mu }_k\cos \theta )$. The sign follows contact slip, and the downhill formula cannot be used unchanged after a reversal.