GRE · GRE Subject Test · GRE Physics

A planet accelerates even when its speed is constant: velocity includes direction.

Prerequisites: 47.

• Apply Newtonian, Lagrangian 拉格朗日量 and Hamiltonian descriptions
• Analyse rotation, central forces and oscillations
• Use conservation laws with their conditions

Newton’s second law is a vector statement. Choose coordinates and identify constraints before components.

Energy conservation applies when work from nonconservative forces is accounted for; momentum conservation needs zero net external impulse.

For small oscillations, linearise around a stable equilibrium. A restoring term proportional to displacement gives harmonic motion.

Lagrange’s equations use L=T−V and generalised coordinates. This undergraduate formalism is distinct from school-level force substitution.

Choose an inertial frame and a coordinate x measured from spring equilibrium. For a conservative spring, the Lagrangian avoids solving constraint forces.

With m = 2.0 kg and k = 18 N/m, Positive stiffness gives stable motion. The units of k/m are inverse seconds squared.

Conservation of kinetic energy does not hold in every collision, even when momentum is conserved.

Lagrangian: Kinetic energy minus potential energy for the standard mechanical system.

angular momentum 角动量: The moment of linear momentum about a reference point.

Zero electric potential 电势 at a point does not imply zero electric field there.

Prerequisites: 13, 14, 15, 29, 30.

• Use electrostatic potentials, fields and Gauss’s law
• Analyse circuits, magnetism and induction
• Apply Maxwell equations and electromagnetic wave relationships

Electric field is the negative gradient of potential. A value and a spatial change are different quantities.

Use Gauss’s law with symmetry. A known total flux 通量 alone does not give the field at every surface point.

Kirchhoff’s laws track charge and energy in circuits. Induced emf follows changing magnetic flux with the Lenz-law sign.

Maxwell equations link electric and magnetic fields. In vacuum a wave has E/B=c and carries energy.

A spherical shell has radius R and uniform charge Q. Choose zero potential at infinity. Spherical symmetry lets Gauss's law 高斯定律 determine the field outside.

Inside, enclosed charge is zero and symmetry gives E = 0. Continuity gives The potential is nonzero and constant inside. A zero gradient, rather than a zero value, gives a zero field. The radial field jumps at the charged surface.

Do not use Gauss’s law to claim a uniform field on an arbitrary nonsymmetric surface.

flux: The surface integral of a field’s normal component.

potential: Potential energy per unit charge.

Two individually bright waves can cancel at a point if their phases oppose.

Prerequisites: 16, 17, 18, 31, 32, 33.

• Analyse wave superposition and interference
• Use diffraction 衍射, polarisation and optical imaging
• Relate phase, frequency and dispersion

Add wave amplitudes, then calculate intensity. Intensity is proportional to squared amplitude for the same medium.

A path difference gives phase difference 相位差 2πΔ/λ. Coherent sources need stable relative phase.

For a single slit, minima satisfy a sin θ=mλ for nonzero integers m. A double slit has a different interference condition.

Geometric optics needs a stated sign convention. Wave effects matter when apertures are comparable to wavelength.

Coherent waves have a stable phase difference. Add amplitudes before squaring.

For equal individual intensities I0 and a path difference of one quarter wavelength, This is neither a maximum nor a minimum. Random relative phase removes the cross term in a long time average; stable coherence 相干性 is a physical condition, not a drawing convention.

The central single-slit maximum is not obtained by setting m=0 in the minima condition.

coherence: A stable phase relationship between waves.

diffraction: Wave spreading and interference caused by an aperture or obstacle.

Two engines can obey energy conservation, yet only one obeys the entropy 熵 limit on efficiency.

Prerequisites: 19, 20, 34, 35, 36.

• Apply the first and second laws with a sign convention
• Relate ensembles, Boltzmann factors and entropy
• Analyse ideal gases, heat engines and quantum statistics

State the first-law sign convention: ΔU=Q−W when W is work done by the system.

For a reversible engine between reservoirs, efficiency is bounded by 1−Tc/Th using absolute temperatures.

Boltzmann weights are exp(−E/kT). Probabilities require division by the partition function 配分函数.

Distinguish Maxwell–Boltzmann, Bose–Einstein and Fermi–Dirac assumptions. Fermions cannot share an identical single-particle state.

Use work W done by the gas as positive. The first law 热力学第一定律 is

For reversible isothermal expansion of an ideal gas, U depends only on T, so For n = 0.50 mol, T = 300 K and doubled volume, The gas expands and receives heat. The reservoir loses the same entropy in this reversible limit.

Celsius temperature ratios cannot be used in the Carnot efficiency formula.

entropy: A state quantity connected to multiplicity and reversible heat transfer.

partition function: The sum of statistical weights used to normalise probabilities.

A particle in a box cannot have zero ground-state kinetic energy. Its confinement changes what states are allowed.

Prerequisites: 24, 37, 38, 39.

• Use wavefunctions, operators and measurement probabilities
• Solve standard bound-state models and tunnelling reasoning
• Apply angular momentum, spin and atomic spectral principles

A state wavefunction 波函数 must satisfy the model’s boundary conditions and have total probability one: integrate |ψ(x)|² over the allowed domain. For orthonormal states φ_j, a superposition ψ=Σc_jφ_j has norm squared Σ|c_j|² because the cross terms integrate to zero. Thus an equal superposition of N orthonormal states has coefficient magnitude 1/sqrt(N). The probability of finding one constituent energy is |c_j|², not c_j. Relative phases may affect position probabilities even when energy probabilities are unchanged.

An observable is represented by a Hermitian operator 算符. Its eigenvalues are real, and a normalised state gives expectation value ⟨A⟩=∫ψ* Aψ dx. The expectation is an average over repeated preparations, not necessarily the result of one measurement. If A satisfies a polynomial identity, apply it to an eigenstate: for A⁴=I, a real eigenvalue must be +1 or −1. A complex fourth root is excluded by Hermiticity even though it solves the polynomial.

For an infinite well 0<x<L, ψ(0)=ψ(L)=0 allows sine modes sin(nπx/L), with n=1,2,… . Their energies are E_n=n²π²$\hbar$²/(2mL²), so the ground state is not zero and doubling L divides every energy by four. The equivalent h expression is n²h²/(8mL²); never substitute h for $\hbar$ without its 2π factor. A finite barrier instead permits exponential tails, and a travelling wave in a classically allowed region has oscillatory rather than exponentially decaying spatial dependence.

Operators commute when [A,B]=AB−BA=0; compatible observables can have a common eigenbasis. Since kinetic energy p²/(2m) is a function of p, it commutes with p. A position-dependent potential generally makes H=p²/(2m)+V(x) fail to commute with p. Angular momentum obeys [J_x,J_y]=i$\hbar$J_z and cyclic permutations, while J² commutes with each component. Atomic transitions exchange positive photon energy equal to a level difference; angular momentum and mechanism-specific selection rules restrict which transitions occur.

An infinite well extends from 0 to L. Vanishing wavefunction at both ends gives

Prepare $\psi=(\psi_1+i\psi_2)/\sqrt2$. Orthogonality makes its norm one. Energy measurements give E1 or 4E1 with equal probability: This average is not an allowed single energy result. The relative phase changes in time, so a superposition of different energies is not a stationary state.

A stationary state can have a nonzero energy even though its probability density is time-independent.

normalisation 归一化: Making total probability equal to one.

operator: A mathematical action representing an observable.

A precise instrument can give consistently wrong results when its calibration is shifted.

Prerequisites: 21, 22, 42, 43, 44.

• Apply special-relativistic energy and spacetime relations
• Propagate measurement uncertainty and analyse experimental data
• Recognise nuclear, particle, condensed-matter and astrophysical concepts

Special relativity uses γ=1/sqrt(1−v²/c²). Rest energy is mc² and total energy is γmc².

For independent small uncertainties, combine contributions in quadrature; correlated uncertainties require covariance 协方差 terms.

Random scatter affects precision 精密度; calibration bias affects accuracy. A log plot can reveal a power law.

Nuclear, particle, condensed-matter and astrophysics questions require undergraduate vocabulary and models. School physics alone is not a complete preparation course.

Proper time 固有时 is measured between events at one position in a frame. For a particle speed 0.60c,

If its mean proper lifetime is 4.0 microseconds, A lifetime is a mean of a distribution; it does not predict an individual decay time.

Adding percentage uncertainties linearly gives a worst-case bound, not the independent statistical standard uncertainty.

precision: Closeness of repeated measurements to each other.

covariance: A measure of joint variation used for correlated uncertainties.

すべてのエネルギー差が強い電気双極子線を生じるわけではない。

前提条件：23、40、41。

• 原子エネルギーとスペクトル線の関係
• 角運動量量子数の適用
• 許容遷移と禁止遷移の区別（電気双極子遷移）

水素様一電子イオンにおいて、非相対論的モデルで束縛エネルギーは −13.6 Z²/n² eV に比例する。放射には下向き遷移と正の光子エネルギーが必要である。

軌道角運動量は l=0,…,n−1 および m_l=−l,…,l である。電子スピンは s=1/2 を加え、全角運動量は軌道とスピンの寄与を合成する。

電気双極子の選択則には Δl=±1 および Δm=0,±1 が含まれる。これらの規則は特定の遷移メカニズムに関するものであり、「禁止」がすべてのメカニズムで不可能を意味するわけではない。

微細構造、スピン-軌道効果、ゼーマン分裂は単純な水素模型を精密化する。多電子原子では遮蔽効果や電子間相互作用により、単純な水素エネルギーの代入は成立しない。

非相対論的一電子イオンについて、微細構造を無視すると、

Z = 2 の場合、n = 3 から n = 2 への放射は正の光子エネルギーを与える： エネルギー差だけでは電気双極子遷移を保証しない。 関与する軌道状態は選択則を満たす必要がある。

放射される光子について、最終状態エネルギーから初期状態エネルギーを引いた値が負にならないようにする。

選択則：特定の相互作用メカニズムに対する遷移の制限。

ゼーマン分裂：磁場による原子エネルギー準位の分裂。

半導体の伝導度は、単に材料内の電子数だけでなく、利用可能な状態とキャリアにも依存する。

前提条件：25、45、46。

• バンドおよびキャリア記述の活用
• 放射性減衰および結合エネルギーの概念を適用する
• 粒子反応における保存則を確認する

充填されたバンドは、近傍に利用可能な状態がない限り通常の電流を運べない。金属は部分充填または重なり合うバンドを持ち、半導体は小さなバンドギャップを持つ。ドナーとアクセプターのドーピングはキャリア濃度を変化させる。

フェルミ・ディラック占有確率は 1/[exp((E−μ)/(kT))+1] で与えられる。零温ではフェルミエネルギーで段関数となり、熱拡散によってすべての状態が半分占有されるわけではない。

放射性核種数は N=N0e^(−λt) に従い、放射能は λN、半減期は ln2/λ である。結合エネルギーは質量欠損に c² を掛けたものから得られ、安定性を議論する際は核子あたりの結合エネルギーを比較する。

核反応および粒子反応では、電荷、エネルギー、運動量、および適用すべき量子数が保存されなければならない。ベータ崩壊にはニュートリノまたは反ニュートリノが含まれ、電子のみによる二体崩壊では連続的なエネルギースペクトルを説明できない。

結合エネルギーとは、原子核を自由な核子に分離するために必要な正のエネルギーのことである。 反応においては、総初末静止質量を用い、一貫した質量の定義に従うこと。

0.0040 amu の質量減少により 正のQ値はエネルギー放出を意味する。電荷、核子数、エネルギー、運動量は 依然として釣り合わねばならず、Q値だけでは反応速度は決定されない。

活度と生存核種数はどちらも減少するが、単位は異なる：活度は秒あたりの遷移数を示す。

バンドギャップ: バンド模型において、許容される bulk 電子状態が存在しないエネルギー間隔。

半減期: 放射性核種の数が半分になるまでの時間。

Energy methods connect a spring oscillator and an orbit, but their coordinates and physical assumptions differ.

Prerequisites: 26, 41.

• Construct a simple Hamiltonian
• Use canonical equations
• Apply gravitational and radiative scaling

The canonical momentum 正则动量 is p=∂L/∂qdot. The Legendre transform H=p qdot−L gives a Hamiltonian when velocities can be expressed using coordinates and momenta.

Hamilton’s equations are qdot=∂H/∂p and pdot=−∂H/∂q. For an ordinary oscillator H=p²/(2m)+kq²/2, these reproduce Newton’s equation.

For a circular gravitational orbit v²=GM/r and period²=4π²r³/(GM). The assumptions include a dominant central mass and a circular approximation; elliptical orbits use the semimajor axis in Kepler’s law. For a circular satellite of mass m, angular momentum magnitude is mrv=m sqrt(GMr), so identical satellites have L proportional to sqrt(r). A radius ratio 9 therefore gives angular-momentum ratio 3, while the period ratio is 27. This distinction follows from the same centripetal-force relation; do not use period scaling for angular momentum.

Luminosity 光度 and received flux obey F=L/(4πd²) for isotropic radiation. A blackbody has L=4πR²σT⁴; its spectral peak shifts inversely with temperature. Distinguish intrinsic luminosity from observed brightness.

For a regular Lagrangian $L=m\dot q^2/2-kq^2/2$, canonical momentum is

Substitute the velocity into the Legendre transform: Combining these gives $\ddot q=-(k/m)q$. The inversion of momentum to velocity is required; it cannot be assumed for every constrained or singular Lagrangian.

A Hamiltonian equals total mechanical energy only under the relevant system assumptions; do not infer this universally from its name.

canonical momentum: The derivative of the Lagrangian with respect to a generalised velocity.

luminosity: Total emitted power, distinct from flux at an observer.

Two carts approach from different directions. After they stick, their momentum can be conserved while their kinetic energy falls.

Prerequisites: 1.

• Compute vector momentum and energy loss in sticking collisions
• Compare fixed-force springs and use work–energy conditions
• Derive small-oscillation periods and oscillator energies

Choose an isolated system during the short collision. With negligible external impulse 外力冲量, conserve total momentum separately in every Cartesian direction. If masses m1 and m2 stick, their common velocity is (m1 v1+m2 v2)/(m1+m2). Find the magnitude only after adding vectors. Initial kinetic energy is the sum of each ½m|v|²; final kinetic energy uses the total mass and the common speed. Their difference becomes internal energy or deformation, not missing momentum. A perfectly inelastic collision means sticking, not final rest.

The work–energy theorem ΔK=∫F·dr uses net force along displacement. For a constant parallel force over distance s, ΔK=Fs. Speed doubling quadruples kinetic energy. Conservation of mechanical energy is a separate statement requiring no unaccounted nonconservative work. At a spring displacement x, U=½kx². Under the same applied force F, equilibrium extension is F/k and stored energy is F²/(2k): a stiffer spring stretches less and stores less energy. Under the same extension, it stores more. Identify which condition is held fixed.

Near a stable equilibrium x0, expand U≈U(x0)+½U″(x0)(x−x0)²; the effective stiffness is positive U″(x0). The motion obeys m xddot=−k_eff(x−x0), with ω=sqrt(k_eff/m) and period 2π/ω. A simple pendulum obeys θddot+(g/L)sinθ=0. For small angles sinθ≈θ, so T=2πsqrt(L/g). This approximation explains period–length scaling; a large amplitude requires a correction, and the mass cancels for an ideal pendulum.

For an undamped harmonic oscillator, total energy is ½m v²+½kx²=½kA². At equilibrium all its energy is kinetic; at a turning point 转折点 its speed is zero and all its energy is potential. Angular frequency is not ordinary frequency: ω=2πf. If the same oscillator has equilibrium speed vmax, A=vmax/ω. Distinguish the instant at which speed is measured from the release displacement, and keep SI units when a millijoule answer is requested.

Two carts stick with negligible external impulse. Add momentum vectors 动量矢量, then divide by total mass. Known: m1 = 2 kg, v1 = (3,0) m/s; m2 = 1 kg, v2 = (0,3) m/s.

The 6 J difference becomes internal energy. Kinetic energy is not conserved in sticking.

Do not add incoming speed magnitudes as momenta, or conserve kinetic energy in a sticking collision. Fixed-force and fixed-extension spring comparisons have opposite answers.

external impulse: Time integral of the net force from outside the chosen system.

turning point: An extreme oscillator position where its instantaneous speed is zero.

Turning a floating block upside down can change how much wood is submerged even though the total weight stays the same.

Prerequisites: 10.

• Use pivot torque and rotational inertia consistently
• Partition rolling kinetic energy with the no-slip constraint 无滑动约束
• Balance buoyancy and drag for composite bodies

Torque about a fixed pivot is r×F; its magnitude uses the perpendicular lever arm. For a uniform rod of length L pivoted at one end, gravity acts at its centre L/2 and the moment of inertia 转动惯量 is ML²/3. If its angle θ is measured from vertical, gravitational torque magnitude is Mg(L/2)sinθ, so angular-acceleration magnitude is 3g sinθ/(2L). Define the positive rotation direction before assigning a sign. Using the centre-of-mass inertia ML²/12 without the parallel-axis correction gives a wrong acceleration.

An object rolling without slipping satisfies v=Rω. Write total kinetic energy as ½Mv²+½Iω². With I=βMR², the rotational fraction is β/(1+β); for a solid disk β=1/2 and the fraction is 1/3. A hoop has β=1 and fraction 1/2. Static friction can supply the torque required for rolling without dissipating energy at an instantaneously stationary contact on a fixed surface. A sliding object does not satisfy the no-slip relation automatically.

For a floating composite, buoyancy equals the weight of displaced fluid, summed over every submerged part. Let a wood block have volume V and a dense stone volume Vs. If the stone is above the water, only submerged wood contributes: ρwater g fV equals total weight. If the stone is attached below and fully submerged, its displaced volume reduces the wood’s required submerged fraction to f−Vs/V. The stone does not cease to weigh anything; the water now supplies some of its support. Check whether the assumed orientation and flotation are physically possible.

Terminal speed 终端速度 means zero acceleration, not zero speed or zero force. With quadratic drag C v² and a downward weight Mg, neglecting buoyancy gives vt=sqrt(Mg/C). Two otherwise identical balls share the drag coefficient C; doubling mass then multiplies terminal speed by sqrt(2). If buoyancy matters, replace Mg by (M−ρfluid V)g. A linear-drag model instead gives a different mass scaling. State the specified drag law and compare force balances, rather than transferring a formula between models.

A body rolls without slipping on a fixed surface. The no-slip constraint is v = R omega.

For a solid disk, beta = 1/2, so the rotational share is For a uniform rod pivoted at one end, use $I=ML^2/3$, not its centre-of-mass inertia. About this pivot $\tau=-Mg(L/2)\sin\theta$, for theta measured from downward vertical.

Use the inertia about the chosen pivot. In flotation, include the stone’s displaced volume as well as its weight. Terminal speed requires a specified drag law.

moment of inertia: Mass-weighted squared perpendicular distance from a specified rotation axis.

terminal speed: Steady speed at which opposing forces balance.

A load released from an aircraft keeps the aircraft’s horizontal velocity. Relative to the aircraft, its initial motion can therefore be purely downward.

Prerequisites: 1, 11.

• Solve coupled-body equations using one acceleration constraint 约束
• Transform velocities between inertial frames
• Check force and energy limits for constrained motion

For a mass m1 on a frictionless horizontal table connected over an ideal massless pulley to a hanging mass m2, choose table motion and downward hanging motion as positive. A taut inextensible string gives one acceleration magnitude. Write T=m1a and m2g−T=m2a. Adding eliminates the internal tension, so a=m2g/(m1+m2) and T=m1m2g/(m1+m2). Thus T<m2g when both masses are positive. Setting tension equal to hanging weight silently assumes zero acceleration.

The common tension assumption requires an ideal massless string and pulley with negligible friction and inertia. A pulley with rotational inertia can have different tensions, with (T2−T1)R=Iα and a=Rα if the string does not slip. Draw each body separately: a force internal to the combined system remains an external force on one chosen body. Constraint forces can cancel from a system equation while still being needed for individual motion.

For two inertial frames with constant relative velocity 相对速度 V, Galilean velocity transformation is v_relative=v_ground−V. A payload released by a level aircraft initially has the aircraft’s horizontal speed. With negligible air resistance, horizontal ground velocity remains constant and vertical velocity becomes −gt if upward is positive. Relative to that aircraft, horizontal velocity is zero and downward speed is gt. Position is a different question: vertical displacement is −½gt². Do not confuse ground speed with relative speed.

Check limiting cases after solving. As m1 approaches zero, the ideal coupled acceleration approaches g and tension approaches zero; as m1 grows very large, acceleration approaches zero and tension approaches m2g from below. An energy derivation gives m2g s=½(m1+m2)v² for release from rest and an ideal pulley, agreeing with v²=2as. This agreement checks the equations; it does not permit energy conservation if friction or other unaccounted work is present.

A table mass m1 and hanging mass m2 share an acceleration through a taut ideal string. Choose rightward table motion and downward hanging motion positive.

T is less than the hanging weight 20 N. A nonzero net downward force is needed for acceleration.

A massless ideal pulley equates tension, not tension and weight. Subtract frame velocities component by component before taking a speed magnitude.

constraint: A condition linking allowed positions or motions of connected bodies.

relative velocity: Velocity of one object measured in another moving frame.

A charged conductor can have zero field inside its material while its potential remains nonzero.

Prerequisites: 2, 47.

• Add Coulomb forces as vectors using the actual geometry
• Compute flux from symmetry or solid angle 立体角 with a stated orientation
• Find conductor fields and potentials using induced surface charges

For a point charge Q at position r0, E(r)=Q(r−r0)/(4πε0|r−r0|³). Superpose vectors, not field magnitudes. For two positive charges attracting an electron, forces in the same direction add; opposite directions subtract; perpendicular components combine by Pythagoras. Define the observation point and the direction of each force first. Potential is the scalar sum V=ΣQ/(4πε0r) with zero at infinity. A zero potential does not generally mean a zero field, since E=−∇V.

Gauss’s law is the closed-surface integral ∮E·dA=Q_enclosed/ε0. It determines a local field simply only when symmetry makes the normal field uniform or makes other flux contributions zero. A charge near an infinite plane sends half its total flux through that plane in magnitude: the plane subtends solid angle 2π out of 4π. The result is independent of distance and lateral position, but its sign depends on the chosen plane normal and charge sign. An open plane does not enclose a charge; using closed-surface wording for it is incorrect.

In electrostatic equilibrium 静电平衡, the electric field inside conducting material is zero and the conductor has constant potential. Place a charge Q at the centre of a conducting spherical shell with inner radius a, outer radius b and shell net charge q. A Gaussian surface inside the material requires inner-surface charge −Q, so the outer surface has q+Q. Spherical symmetry makes the external field that of total Q+q at the centre. The material’s potential, zero at infinity, is (Q+q)/(4πε0b), independent of the material observation radius.

In the cavity of that centred-charge shell, the field is Q/(4πε0r²) radially for 0<r<a. Potential includes both the point-charge contribution and constant shell contributions; continuity holds across each surface even though the normal field jumps at a surface charge. Keep cavity, conductor material and exterior separate. An off-centre charge still induces total inner charge −Q, but its cavity field is no longer the simple centred radial field. Zero interior conductor field follows equilibrium, not an assumption that every surface charge distribution is uniform.

A centred point charge Q lies in a conducting spherical shell of net charge q. Electrostatic equilibrium makes the field zero inside the metal. A Gaussian surface in the metal encloses zero net charge, so $Q_{inner}=-Q$. Charge conservation gives $Q_{outer}=q+Q$. With outer radius b and zero potential at infinity,

Here Q = +2 nC and q = -1 nC. The cavity is not conducting material; its field need not vanish.

Do not confuse the open-plane half-flux result with enclosed charge. A conductor’s zero field fixes a constant potential, not necessarily zero potential.

solid angle: Angular area subtended by a surface, measured in steradians.

induced surface charge 感应表面电荷: Charge rearranged on a conductor to satisfy electrostatic equilibrium.

A resistor in a parallel branch sees the branch voltage, which can differ from the battery voltage because another resistor is in series.

Prerequisites: 2, 30.

• Solve resistor networks and distinguish rms from peak AC quantities
• Apply flux linkage 磁通链, Lenz direction and the current-loop field
• Use charge-to-mass ratios and balance crossed-field forces

For steady ideal resistor circuits, series resistances add and parallel conductances add. Reduce a network 电路网络 while preserving which nodes share the same voltage. With a 24 V source, a 2 Ω series resistor and parallel 3 Ω and 6 Ω branches, the parallel equivalent is 2 Ω. Total current is 6 A, branch voltage is 12 V and branch currents are 4 A and 2 A. The current through every component is not necessarily the total current; verify Kirchhoff current and voltage sums after reduction.

For sinusoidal current I=Imax sin(ωt−φ), Irms=Imax/sqrt(2). A series RLC circuit has impedance magnitude sqrt(R²+(ωL−1/(ωC))²) and average dissipated power Irms²R=Vrms Irms cosφ. Ideal inductors/capacitors exchange stored energy with zero average dissipation. If rms current is already given, do not insert another factor 1/2. An ideal rectifying diode conducts in its forward direction and blocks in reverse: a suitably oriented series diode clips the negative half of a sinusoidal resistor voltage; it does not create a full-wave rectifier by itself.

Magnetic flux is ∫B·dA, or BA cosθ for a uniform field with θ measured from the area normal. Faraday emf is −N dΦ/dt. For changing perpendicular field and fixed loop geometry, its average magnitude is NA|ΔB|/Δt. Determine the sign or current direction using opposition to the flux change, not opposition to the field itself. The magnetic field at the centre of one circular loop is μ0I/(2R), obtained from Biot–Savart: every element contributes in the same axial direction. An N-turn compact coil multiplies this result by N.

For a nonrelativistic charge with velocity perpendicular to uniform B, magnetic force |q|vB provides centripetal force mv²/r. Thus r=mv/(|q|B) and cyclotron angular frequency is |q|B/m, independent of speed in this approximation. Charge sign sets rotation direction. A deuteron has approximately twice proton mass and the same charge, while an alpha particle has four times its mass and twice its charge; their frequencies are each half the proton value. In crossed E and B fields, undeflected motion requires electric and magnetic forces opposite and v=E/B. Check vector directions before using this magnitude.

A 24 V ideal source drives 2 ohms in series with parallel 3 and 6 ohm branches. Identify nodes before reducing the network.

The other branch carries 4 A; the branch currents sum to the source current.

Do not apply battery voltage to every branch or halve an rms-current power again. Flux angle uses the area normal; a selector also needs opposing force directions.

root mean square 均方根: Square root of the mean squared value, used for effective AC current or voltage.

flux linkage: Sum of magnetic flux through a coil’s turns.

電荷を帯びたコンデンサは、ギャップを伝導電流が通過しない場合でも、ギャップ内の電場を変化させることができる。

前提知識: 13, 29, 47。

• 変位電流および媒体の波速関係式を用いる
• 波の場方向および放射されるエネルギーの流れを決定する
• 場の成分を混同せずにマクスウェルの境界条件を適用する

真空中のアンペール-マクスウェルmyは ∮B·dl=μ0(I_conduction+ε0 dΦE/dt) である。追加された変位電流項は、選択した表面を貫通する電気フラックスの時間変化率に依存する。これは、張る表面がコンデンサの導線を横切るか、極板の間を通るかによって一貫した結果を保証する。変位電流は磁気フラックスや過去の電気フラックスの積分ではない。均一な板状電場の場合、ギャップ寄与は ε0A dE/dt であり、符号は向付けられた表面に従う。

透磁率 μ と誘電率 ε を持つ均質で線形かつ無損失な媒質において、マクスウェル方程式より波速 v=1/sqrt(με)、屈折率 n=c/v=sqrt(μrεr) となる。非磁性材料では μr≈1 なので、v=c/sqrt(εr) である。分散が重要な場合は周波数依存性のある媒質定数を使用すること。静的な誘電定数はすべての光学周波数を記述するとは限らない。真空中の波では E/B=c；この単純な媒質ではその関係は E/B=v となる。

単位ベクトル n に沿って伝播する平面波について、B=(n×E)/v である。両方の場は伝播方向および互いに垂直である。位相 kz−ωt は +z 方向へ伝播し、kz+ωt は −z 方向へ伝播する。E が x+y 方向にある場合、伝播が +z なら B は −x+y 方向になる（なぜなら z×x=y かつ z×y=−x だから）。ポインティングベクトル S=E×H（この媒質では E×B/μ）はエネルギー輸送の方向を指す。加速電荷の放射領域では、エネルギーフラックスは源から外向きであり、必ずしも電荷運動の瞬時方向とは一致しない。

マクスウェルの発散方程式 ∇·B=0 は、界面をまたぐ法線成分 B の連続性を示す：薄いピルボックス内に磁気電荷はない。接線成分 H のジャンプは面電流に関係しており、ゼロである必要はない。理想的な超伝導マイスナー状態モデル（内部 B=0）において、境界における外部の法線成分もゼロでなければならないため、残りの nonzero な外部 B は表面に沿った接線成分となる。この結論は提示された zero-interior モデルに基づくものであり、外部の場すべてがゼロであると仮定して導かれるものではない。静電学においては、法線成分 D は自由面電荷によってジャンプすることがある。

平面波は、線形で無損失かつ非磁性の媒質内を正の z 方向に伝播する。 相対誘電率が 9 の場合、

E が正の x 方向を向いているとき、$\mathbf B=\hat{\mathbf z}\times\mathbf E/v$ は正の y 方向を向く。 E0 = 200 V/m のとき、 $\mathbf E\times\mathbf H$ は伝播方向を向く。媒質定数が波の周波数において適用されることを確認せよ。

変位電流は変化する電気束を用いる。内部Bがゼロであることは、外部の法線成分を固定するものであり、外部のすべての成分を固定するわけではない。波の伝播と粒子の運動を区別する。

変位電流: アンペール-マクスウェルmyにおける電気フラックスの時間微分寄与。

ポインティングベクトル: 単位面積・単位時間あたりの電磁気エネルギーフラックス。

A driver hears an echo from a stationary wall at a shifted pitch: the outbound and returning sound each acquire a Doppler factor 多普勒因子.

Prerequisites: 3.

• Apply source and observer Doppler factors in the medium frame
• Derive allowed pipe frequencies from displacement boundary conditions
• Distinguish frequency, wavelength and boundary changes

For sound in a stationary medium, separate source motion from observer motion. A source approaching at speed u_s compresses wavefront spacing and produces received frequency f c/(c−u_s) at a stationary observer. An observer approaching a stationary source at speed u_o meets more wavefronts per second and measures f(c+u_o)/c. Speeds are measured relative to the medium; the acoustic source and observer formulas are not symmetric under exchanging roles. Receding motion reverses the appropriate sign. These expressions assume subsonic motion along the propagation line.

For a siren moving toward a stationary reflecting wall with speed u, the wall first receives f_wall=f c/(c−u). Reflection from a stationary wall preserves frequency in the medium frame. The moving driver then approaches the returning wavefronts and receives f_echo=f_wall(c+u)/c=f(c+u)/(c−u). At small u/c the fractional shift is approximately 2u/c, but the exact expression has different numerator and denominator. A moving reflecting surface needs its own Doppler step; do not reuse the stationary-wall result blindly.

At an ideal open pipe end, air displacement is an antinode and pressure variation is a node. At a rigid closed end, displacement is a node and pressure is an antinode. For both ends open, length L contains n half-wavelengths, giving f_n=nc/(2L), n=1,2,… . For one end closed, it contains an odd number of quarter-wavelengths, giving f_n=(2n−1)c/(4L). Real pipes may need end corrections; the ideal GRE model uses the stated length without inventing an adjustment.

Closing one end of a previously both-open pipe halves its fundamental and leaves only odd multiples of that new fundamental. Its old frequencies were integer multiples of c/(2L), which are even multiples of c/(4L), so none is an allowed frequency of the new ideal one-closed spectrum. This differs from simply deleting even harmonics while retaining the old fundamental. When the medium is unchanged, wave speed is unchanged; frequency and wavelength change together to satisfy the new boundary geometry.

A siren approaches a stationary wall in still air. Treat outward and return journeys separately. The wall first receives $f_w=fc/(c-u)$; reflection keeps this frequency in the medium frame. The moving listener then receives $f_e=f_w(c+u)/c$.

This Doppler effect 多普勒效应 uses velocities relative to the medium. It is not the light-wave formula.

An echo needs both Doppler factors. Closing a pipe changes the fundamental; odd harmonics are counted from the new fundamental, not the old one.

displacement antinode 位移波腹: A standing-wave position where the air displacement amplitude is maximal.

Doppler factor: Frequency multiplier produced by a specified source or observer motion.

A gas cell in one arm of a Michelson interferometer 干涉仪 is crossed twice. Counting only one passage halves the inferred refractive-index change.

Prerequisites: 3, 33.

• Calculate telescope magnification and grating resolving power 分辨本领
• Relate Michelson fringe counts to double-pass optical path changes
• Trace surface normals before applying Snell and reflection laws

For an ideal astronomical telescope adjusted for relaxed viewing at infinity, the objective forms its focal-plane image and the eyepiece collimates the emerging light. Lens separation is f_objective+f_eyepiece; angular magnification magnitude is f_objective/f_eyepiece. The usual two-converging-lens telescope produces an inverted image, represented by a negative signed angular magnification under a consistent convention. A finite final-image distance changes lens separation; do not use the infinity adjustment without checking the condition.

Resolving power R=λ/Δλ describes the smallest distinguishable wavelength separation near λ. For an ideal diffraction grating in order m with N illuminated slits, R=mN. This is a resolution criterion, not simply the angular position formula d sinθ=mλ. Increasing illuminated slit count sharpens the principal peaks; increasing slit spacing mainly changes their angular locations. A measured λ=600 nm and Δλ=3 nm gives R=200, independent of converting both lengths to metres because their ratio uses matching units.

Michelson interference depends on the optical path difference between the arms. If one arm contains a gas cell of geometric length L and index changes from n to 1 upon evacuation, a double passage changes path by 2L(n−1). If N fringes pass, the magnitude of this change is Nλ, so n−1=Nλ/(2L). For a moving mirror instead, displacement Δx gives path change 2Δx. Fringes count phase cycles; one fringe is one wavelength of optical path, not one wavelength of mirror motion.

Snell’s law n1 sinθ1=n2 sinθ2 uses angles from the local surface normal. Reflection has equal incoming and outgoing angles from that same normal. In a rectangular block, normals of neighbouring faces are perpendicular: an incidence angle measured from one normal is complementary to the angle of that same ray from the other. Draw the ray and normals before substitution. For incidence from index n into air, total internal reflection occurs only when sinθ>1/n; equality is the critical grazing case. A stated partly refracted ray must satisfy the transmission condition.

A Michelson interferometer sends light twice through a gas cell of length L. Evacuation changes path by $2L(n-1)$. N fringes correspond to path change $N\lambda$.

Thus n = 1.00030. Moving a mirror is also double-pass; a fringe is not one wavelength of mirror motion.

Use optical path, not just geometric distance. Refraction angles are measured from the normal of the actual surface, and telescope focal-length addition assumes a final image at infinity.

resolving power: Wavelength divided by the smallest resolvable wavelength separation.

optical path length 光程: Geometric path weighted by refractive index along the ray.

A waveform can contain many harmonics while having no sine terms at all. Reflection symmetry tells you which coefficients vanish before integration.

Prerequisites: 3, 47.

• Use orthogonality 正交性 to identify Fourier coefficients
• Exploit even and odd parity without discarding the constant term
• Relate component amplitudes to interference and physical waveforms

For a sufficiently regular real 2π-periodic function, write f(x)=a0/2+Σ[a_n cos(nx)+b_n sin(nx)]. The coefficients are a_n=(1/π)∫from−πtoπ f(x)cos(nx)dx and b_n=(1/π)∫from−πtoπ f(x)sin(nx)dx, with a0 using n=0. Cosines and sines are orthogonal on a full period. The mean is a0/2, not a0; retain the constant term even when all nonzero-frequency coefficients of one family vanish.

If f is even about the chosen origin, f(x)sin(nx) is odd and its symmetric integral is zero, so all b_n vanish. If f is odd, its mean and all a_n vanish. Symmetry depends on the origin: shifting the same physical signal can mix sine and cosine coefficients while leaving its harmonic frequencies unchanged. A nonnegative triangular waveform symmetric about x=0 can therefore have cosine harmonics and a nonzero mean without any sine harmonics.

For the 2π-periodic extension of f(x)=|x| on [−π,π], the mean is π/2. Integrating x cos(nx) by parts on [0,π] gives a_n=2[(-1)^n−1]/(πn²), so even-n cosine coefficients are zero and odd-n coefficients are −4/(πn²). All sine coefficients are zero. The 1/n² decay reflects a continuous function with a slope discontinuity. A jump discontinuity often gives slower coefficient decay and partial-sum overshoot; do not infer the same convergence behaviour for every waveform.

A Fourier sum adds amplitudes with their phases. For equal coherent monochromatic amplitudes A meeting with phase difference φ, resultant squared amplitude is 2A²(1+cosφ), so intensity is 2I0(1+cosφ). It is 4I0 in phase and zero at φ=π. Incoherent averaging removes the cross term, yielding 2I0. Harmonics at different frequencies can construct a shape over time; their instantaneous sum is not the sum of their individual intensities. Identify whether the question concerns a waveform, time-average power or coherent interference.

For a real periodic function, use $f=a_0/2+\sum(a_n\cos nx+b_n\sin nx)$. Even symmetry 偶对称 makes all sine coefficients zero because the integrand is odd. For $f(x)=|x|$ on $[-\pi,\pi]$,

Integration by parts supplies the second expression. The mean is $a_0/2=\pi/2$; only odd cosine harmonics survive. A cusp does not justify dropping the constant term.

Even symmetry removes sine coefficients, not the constant term or every cosine harmonic. Check the symmetry origin and whether fields are coherent before adding intensities.

orthogonality: A zero integral of the product of distinct basis functions over the specified interval.

Fourier coefficient 傅里叶系数: Weight of a sine, cosine or constant basis component in a periodic expansion.

A refrigerator 制冷机 can move more heat than the work supplied to it. That ratio is a coefficient of performance 性能系数, not an efficiency that must be below one.

Prerequisites: 4.

• Apply first-law work and heat signs to gas processes
• Compare reversible entropy balances and P–V cycle areas
• Distinguish engine efficiency from refrigerator coefficient of performance

Use ΔU=Q−W, where Q is heat entering the system and W is work done by it. Quasistatic boundary work is ∫P dV, positive during expansion. For a fixed amount of ideal gas, internal energy depends only on temperature. At constant volume W=0, so ΔU=Q; for an isothermal ideal-gas process ΔU=0, so Q=W. Adiabatic means Q=0, not constant temperature. Over a cycle, the state returns and ΔU=0, so net Q equals net W even though individual legs have different heat transfers.

For reversible isothermal expansion of an ideal gas, PV is constant. With constant heat capacities, a reversible adiabatic path obeys PV^γ=constant, with γ=Cp/Cv>1; through the same initial state its pressure falls faster as volume increases. An isobaric path is horizontal on a P–V plot. A clockwise closed cycle has positive ∮P dV; reversing direction changes the sign. For a rectangular loop the magnitude is ΔPΔV, and a triangular loop has half the corresponding bounding-rectangle area. Use Pa and m³ for joules, not an unconverted litre value.

For reversible heat transfer, dS=δQ_rev/T; entropy is a state function while heat is path dependent. A reversible system-plus-environment process has zero total entropy change, but the system’s entropy alone can increase or decrease. Irreversible spontaneous processes produce nonnegative total entropy. A reversible isothermal expansion increases the gas entropy by nR ln(V2/V1); the reservoir loses the same amount. Reversibility does not require constant system temperature, zero work or zero internal-energy change.

A reversible engine between TH and TC has efficiency W/QH=1−TC/TH. A reversible refrigerator instead has COP=QC/W=TC/(TH−TC), and a heat pump has QH/W=TH/(TH−TC). Use absolute kelvin temperatures. On a reversible T–S diagram, heat magnitude along an isotherm is T times the entropy change; the Carnot rectangle’s area is the work magnitude. Refrigeration moves heat from cold to hot by consuming work, reversing the engine cycle. COP can exceed one without violating conservation because the moved heat is not supplied solely by the work.

A reversible refrigerator transfers entropy Delta S from Tc to Th.

COP above one is possible because heat is moved, rather than all produced from work.

Zero total entropy production 熵产生 does not mean zero system entropy change. A refrigerator uses TC/(TH−TC), not the engine efficiency formula.

entropy production: Nonnegative total entropy generated by irreversibility.

coefficient of performance: Useful heat transferred divided by work input for a refrigerator or heat pump.

A gas can have zero mean velocity while its particles have substantial average speed. Opposite directions cancel in a vector mean; speed has no sign.

Prerequisites: 4, 35.

• Normalise continuous speed densities and interpret their units
• Distinguish vector means, speed means and most probable speeds
• Use Boltzmann weights with degeneracy and a partition function 配分函数

For a continuous speed probability density 概率密度 f(v), f(v)dv approximates probability in a narrow interval and ∫from0to∞ f(v)dv=1. The density has inverse-speed units and is not itself a probability. If P(v)dv counts particles in that interval, its area is the particle count N and f=P/N. A trapezoid rising over width v0, remaining flat over width 2v0, then falling over width v0 has area 3av0 for height a. Normalise by area before computing a mean or interpreting a plotted height.

In an isotropic equilibrium gas without bulk flow, each velocity-component distribution is symmetric, so ⟨v_vector⟩=0. Speed v=|v_vector| is nonnegative and has positive mean. For a classical ideal gas, the Maxwell speed density is proportional to v²exp[−mv²/(2kBT)]. Its mode is sqrt(2kBT/m), its mean is sqrt(8kBT/(πm)) and its rms speed is sqrt(3kBT/m). These are three different quantities. The v² factor makes speed density zero at v=0, even though the velocity-vector density is largest at the zero vector.

For a continuous distribution, the probability of exactly one specified speed is zero; nonzero probabilities refer to intervals. This statement does not imply there are no particles with arbitrarily small speeds, or that a finite-resolution detector cannot record a zero bin. Temperature changes the scale of the Maxwell distribution: characteristic speeds are proportional to sqrt(T/m). Doubling temperature does not double the speed, and heavier particles are slower on average at the same temperature.

In a canonical ensemble 正则系综 at temperature T, a state of energy E_i has weight exp(−E_i/(kBT)). Divide by partition function Z=Σexp(−E_i/(kBT)) to obtain probabilities. If an energy level has degeneracy g_i, its level probability is g_i exp(−E_i/(kBT))/Z. Equal energy per state does not imply equal probability per level when degeneracies differ. For a ground level of degeneracy 1 and an excited level of degeneracy 2 at Δ=kBT ln2, the excited total weight is 2e^(−ln2)=1: the two levels each have probability 1/2. Classical Maxwell–Boltzmann assumptions differ from Bose–Einstein and Fermi–Dirac quantum occupations.

A canonical ensemble has ground degeneracy 1 and excited degeneracy g at gap Delta.

With g = 4 and $\Delta=k_BT\ln2$, This is probability for the entire excited level. Each of its four states has probability 1/6.

A density height is not a probability. Mean velocity, mean speed and mode differ; include degeneracy before normalising level probabilities.

probability density: Probability per unit of a continuous variable, whose integral gives interval probability.

partition function: Sum of statistical weights used to normalise equilibrium state probabilities.

A detector with ten percent efficiency does not register exactly ten photons out of every hundred. The count fluctuates even with stable efficiency.

Prerequisites: 47.

• Use binomial and Poisson count means and fluctuations
• Propagate small uncertainties with stated correlation assumptions
• Solve dimensional exponent constraints and check a model’s units

For N independent incident particles each detected with probability p, count X is binomial: mean Np and variance Np(1−p). The standard deviation is sqrt[Np(1−p)], not the variance itself. With large N and small p, a Poisson approximation 泊松近似 has mean λ=Np and variance λ, so deviation sqrt(λ). For N=200 and p=0.05, the exact mean is 10 and deviation sqrt(9.5)≈3.08, while Poisson gives sqrt(10)≈3.16. Neither model promises a fixed count. Correlated detections or dead time can invalidate independence.

For a smooth measured function y(x1,…), linearise changes using its partial derivatives. Independent small standard uncertainties combine in quadrature: σ_y²≈Σ(∂y/∂xi)²σ_i². Correlated inputs require covariance cross terms. For y=x^a, fractional standard uncertainty 标准不确定度 is approximately |a|σ_x/|x|. Thus kinetic energy K=½mv² with negligible mass uncertainty has fractional uncertainty twice that of speed. With independent mass uncertainty, combine (σ_m/m)²+(2σ_v/v)². These approximations need small errors and a suitable local linear model.

Random scatter measures precision; a common calibration bias affects accuracy and does not disappear by averaging repeats. For independent repeated readings, the standard uncertainty of the mean falls as 1/sqrt(n), but shared systematic error does not. The uncertainty of a physical spread and uncertainty of its estimated mean are different quantities. State whether a quoted percentage is a standard uncertainty, confidence interval or worst-case bound. Summing absolute contributions is a conservative bound, not the independent-standard-error quadrature rule.

Dimensional analysis equates powers of mass, length and time, rather than numerical sizes. For a Planck-length form G^a $\hbar$^b c^d, dimensions are [G]=L³/(MT²), [$\hbar$]=ML²/T and [c]=L/T. Requiring length gives −a+b=0, 3a+2b+d=1 and −2a−b−d=0. Hence a=b=1/2 and d=−3/2, so length is sqrt(G$\hbar$/c³). Dimensional analysis cannot determine an arbitrary dimensionless coefficient or prove that the chosen constants are physically sufficient. Reject a formula with wrong units before inserting numbers.

For independent small standard uncertainties in $K=mv^2/2$, differentiate first.

With relative uncertainties 3% in mass and 2% in speed, A worst-case bound would use a sum, not this quadrature. Correlations require covariance terms.

Do not confuse mean count with guaranteed count, variance with standard deviation, or a random standard error with a calibration bias. Dimensional consistency is necessary but not sufficient.

standard uncertainty: Uncertainty expressed as a standard deviation under a stated measurement model.

Poisson approximation: Rare independent-event count model with variance equal to its mean.

A fast unstable particle can travel farther in the laboratory than its rest-frame lifetime multiplied by its speed would suggest.

Prerequisites: 6, 42.

• Relate proper lifetime to laboratory time and distance
• Use invariant energy–momentum and relativistic kinetic energy
• Infer longitudinal recession speed from wavelength ratio

For relative speed v, define β=v/c and γ=1/sqrt(1−β²). Proper time 固有时 is measured along the particle’s worldline by a clock at rest with it. If its proper mean lifetime is τ0, its laboratory mean lifetime is γτ0 and mean travel distance is vγτ0 at constant speed. The decay is statistical: mean lifetime is not a guaranteed decay time for each particle. At β=0.8, γ=5/3. Compute spacetime events consistently in one frame; proper time and coordinate time are different quantities.

Total energy 总能量 is E=γmc², momentum p=γmv and invariant E²−p²c²=m²c⁴. When E and pc are expressed in the same energy units, find mc²=sqrt(E²−(pc)²), not E−pc. The nonnegative square root is required for positive rest mass. For E=13 GeV and pc=12 GeV, rest energy 静能 is 5 GeV and mass is 5 GeV/c². A massless particle has E=pc but need not have zero energy or momentum. These relations concern isolated-particle four-momentum, not classical mv at relativistic speed.

Work accelerating a particle from rest equals kinetic energy K=E−mc²=(γ−1)mc². At β=0.6, γ=1.25 and K=0.25mc², while the classical ½mv² gives 0.18mc². Classical kinetic energy is the low-speed expansion and becomes inaccurate near c. Finite acceleration work increases γ rather than allowing a massive particle to reach or exceed c. Keep total energy, rest energy and kinetic energy distinct when interpreting answer units.

For purely longitudinal relative recession in special relativity, wavelength ratio r=λ_observed/λ_emitted=sqrt((1+β)/(1−β)). Rearranging gives β=(r²−1)/(r²+1). A ratio r=2 gives β=3/5, not c times r−1 from a low-speed approximation. Blueshift uses r<1 and a negative recession parameter under this convention. The formula assumes the shift is entirely kinematic; cosmological expansion or gravitational redshift requires a different model. State the question’s stipulated model before interpreting a spectral ratio.

Total energy and momentum determine invariant rest energy.

For E = 17 GeV and pc = 15 GeV, Rest mass is 8 GeV/c squared. Do not subtract momentum directly from energy.

Do not use the proper lifetime as laboratory time, subtract pc from E to find rest mass, or apply the classical Doppler approximation to a large wavelength ratio.

proper time: Time recorded by a clock moving with the object along its worldline.

rest energy: Energy mc² associated with an object’s rest mass.

等間隔の電流低下は、異なる原子状態の遷移ではなく、同一の励起エネルギーの損失が繰り返されたことを示すことができる。

前提知識：7。

• 光電効果閾値と特性X線を区別する
• 換算質量によるスペクトルスケール化と多電子系におけるスピン充填を適用する
• フランク＝ヘルツのエネルギー間隔を新しい能準を仮定せずに解釈する

光電効果の最大運動エネルギーは Kmax=hν−ϕ で、光子周波数が値 ν0=ϕ/h を超える場合に成立する。停止電圧の大きさ eVs=Kmax を満たすため、Vs は閾値以上で ν に比例する。一定周波数での強度増加は利用可能な光子数を増やし、通常は光電流を増大させるが、最大光電子エネルギーは変化しない。閾値未満では、単一光子モデルにより、強度に関わらず光電子は放出されない。仕事関数は表面固有の性質であり、照明周波数に比例するものではない。

特性X線は、電子が内殻の欠陥を埋める際に放出される光子が壳層間のエネルギー差に等しいときに生じる。その鋭利な線は標的原子に依存する。ブレームストラhlungは、核電場中での高エネルギー電子の減速によって発生し、連続的な背景スペクトルを生む。このスペクトルの上限エネルギーは入射電子エネルギーによって決まる。連続背景と離散線が同時に現れることもあり、片方の存在が他方を排除することはない。これらのメカニズムは可視蛍光、フォノン散乱、粒子捕獲とは異なる。

水素様二体原子においては、電子質量を換算質量 μ=m_eM/(m_e+M) に置き換える。単純なクーロンモデルでは、能準エネルギーは −μZ²/n² に比例するため、スペクトル周波数は μZ² に比例し、波長はこれに反比例する。陽電子分子（ポシトリウム）では M=m_e、μ=m_e/2 となり、そのライドバーグ定数は無限の核質量の場合の半分となる。重核では μ は m_e に近づく。多電子原子においては、水素モデルを用いず、軌道充填とフント則に従う：縮重軌道は、対形成を行う前に平行スピンで単に占められる。炭素の 2p² は2つの不対電子を持ち、全スピン S=1 である。酸素の 2p⁴ も2つの不対電子を持ち、同様に S=1 である。閉じた対は正味のスピン寄与ゼロとなる。

フランク＝ヘルツ実験において、加速電子は原子の閾値に達した時点で非弾性励起を通じてエネルギーを失う。加速電圧における電流低下やピークの等間隔は、同一の励起エネルギーが1回、2回、あるいはそれ以上失われたことを識別するために用いられる。間隔 ΔV はエネルギー eΔV に相当する。4,8,12 V のピークが必ずしも3つの異なる励起能準を表しているわけではない。接触電位や逆方向電界が絶対的なピーク位置をずらすことがあるため、間隔と装置条件を参照して使用すべきである。励起原子がそのエネルギーを持つ1つの光子として脱励起する場合、波長は hc/(eΔV) となる。

単一光子光電効果においては、表面の仕事関数 phi を用いる。

光子エネルギーが 4.5 eV で phi = 2.0 eV の場合、 停止電圧の大きさは 2.5 V である。同じ周波数での強度の増加 は利用可能な電子の速度を変化させるものであり、この最大エネルギーは変わらない。

電流強度と停止電圧は異なる問いに対する答えである。換算質量と軌道の縮重度を含め、すべての極大を新しい準位として扱うのではなく、フランク＝ヘルツの間隔を用いるべきである。

仕事関数：指定された表面から電子を引き出すために必要な最小エネルギー。

換算質量：相対運動に対する有効質量 m1m2/(m1+m2)。

量子波動は、古典粒子にとって転回点がない下向きのポテンシャル段でも反射することがある。

前提条件: 5。

• 有限ポテンシャル変化における伝播波とフラックスを一致させる
• 等方調和振動子の縮重状態の数を数える
• パウリ積を評価し、反交換と同一性を区別する

一定ポテンシャルVおよびエネルギーE>Vにおいて、空間解はexp(±ikx)の伝播因子であり、k=sqrt[2m(E−V)]/$\hbar$である。同じ粒子質量を持つ有限ステップでは、波動関数およびその第一微分は連続である。右側からの入射ビームがない左入射散乱問題において、遠方の右側解には右方向伝播因子Ae^(ikx)のみが含まれる。有限ポテンシャル井戸が元の外部ポテンシャルに戻った場合、内部の波数は異なっても、透過された外部の波数は入射波数と等しい。指数関数的に減衰する解はE<V領域を記述するものであり、すべてのポテンシャル井戸ではない。

V1からV2へのステップで両領域が古典的に許容される場合、領域1に入射加反射振幅を、領域2に透過振幅を表す。連続性よりr=(k1−k2)/(k1+k2)、t=2k1/(k1+k2)となる。反射確率はR=|r|²、透過はT=(k2/k1)|t|²である。これは確率電流が波数に依存するためである。よって、損失のないステップに対してR+T=1となる。確率電流係数なしで単に二乗した透過振幅をRに加えてはいけない。k2=3k1の場合、下降ステップであってもR=1/4、T=3/4となる。

三次元等方調和振動子は非負整数n_x, n_y, n_zを持つx, y, zモードに分離する。エネルギーは(N+3/2)$\hbar$ωであり、N=n_x+n_y+n_zである。スピンゼロの識別可能な単一粒子について、空間的な縮重度はそのような三重項の数である: (N+1)(N+2)/2。これは単純な順序付けされない分割だけでなく、異なる割り当ての数を数えるものである。N=2の場合、6つの状態は(2,0,0)および(1,1,0)の順列である。N=3の場合、縮重度は10である。追加のスピンまたは同一粒子の制約は counting 問題を変化させる。

パウリ行列はσ_x=[[0,1],[1,0]]、σ_y=[[0,−i],[i,0]]、σ_z=[[1,0],[0,−1]]である。各々は単位行列に二乗して等しくなる。直接乗算によりσ_xσ_y=iσ_z、σ_yσ_x=−iσ_zとなり、異なるパウリ行列は反交換する。x→y→zの循環的積は正のiを持ち、順序を逆転させると符号が変わる。一般的な恒等式はσ_iσ_j=δ_ijI+iΣε_ijkσ_kである。行列の順序は重要であり、σ_xσ_z=−iσ_yであり、iσ_yや単にσ_yではない。これらの無次元行列はスピン演算子S_i=$\hbar$σ_i/2となり、物理的单位と係数を加える。

有限ポテンシャルステップにおいて、同じ粒子質量の波動関数および微分を一致させる。

透過確率は振幅 alone ではなく電流を用いる。 $k_2=3k_1$、$r=-1/2$、$t=1/2$のために。 下降ステップであっても反射は発生する。確率を取る際に位相の符号は消滅する。

散乱、縮重およびパウリ演算子: GRE教育用原図。

透過確率は波数の確率流比を必要とする。調和振動子の縮重度は順序付きモード triplets の数を数える。パウリ行列の順序を逆にする場合は、符号も変更すべきである。

反射係数: 入射電流に対する反射確率電流の割合。

縮重度: 指定されたエネルギーを共有する独立した状態の数。

Measuring conductivity alone does not tell you whether electrons or holes dominate transport. A magnetic Hall measurement provides a sign-sensitive test.

Prerequisites: 8, 46.

• Compare binding energy per nucleon 平均核子结合能 and reaction energy
• Distinguish leptons, mesons and baryons with conservation constraints
• Infer dominant carrier sign from the one-carrier Hall model

Binding energy B is the energy needed to separate a nucleus into its constituent free nucleons, equal to the corresponding mass deficit times c². Binding energy per nucleon B/A is the relevant comparison across different mass numbers; a heavier nucleus can have greater total B but lower B/A. The broad B/A maximum is near the iron/nickel region. Light-nucleus fusion and very-heavy-nucleus fission can release energy when products have greater total binding. For a reaction, Q=(initial rest mass−final rest mass)c²; positive Q means released energy. Always compare the actual specified isotopes and final products, not just their element names.

Leptons such as electrons, muons and neutrinos are not made of constituent quarks in the Standard Model. A negative muon has charge −e and spin 1/2, like an electron, but has a different mass and belongs to a different flavour family; it is not a meson despite its historical name. Hadrons contain quarks: ordinary mesons have one quark and one antiquark, and ordinary baryons have three quarks. Corresponding antibaryons have three antiquarks. These elementary classification models do not imply every composite particle is a simple baryon or meson.

Check charge, energy, momentum, baryon number and relevant lepton numbers in a proposed process. A proton’s uud quarks sum to charge +e; neutron udd sums to zero. Beta-minus decay n→p+e−+antineutrino conserves electric charge and includes the antineutrino to balance lepton number and kinematics. A particle’s mass alone does not classify it as a quark, lepton or hadron. Reaction thresholds may require kinetic energy beyond an exothermic rest-mass difference when other constraints apply.

In a simple one-carrier conductor, transverse magnetic Lorentz force builds a Hall electric field until transverse carrier force balances. With a stated tensor/sign convention, Hall coefficient 霍尔系数 is R_H=1/(nq); negative q gives electron-like negative coefficient, positive q gives hole-like positive coefficient. Resistivity magnitude alone does not reveal this sign. Set current, magnetic field and voltage directions consistently before reading raw polarity. Semiconductors with electrons and holes both contributing require a mobility-weighted two-carrier model; the simple coefficient need not equal the inverse total carrier density.

In a one-carrier Hall effect 霍尔效应 model, use $R_H=1/(nq)$ with a fixed lead convention. For electrons, q = -e, n = $5.0\times10^{22}$ per cubic metre,

Negative coefficient indicates electron-like carriers under this model, not a negative density. Raw voltage polarity is meaningful only with declared field and lead directions.

Compare binding per nucleon, not just total binding. A muon is a lepton, and Hall sign requires the declared measurement convention and carrier model.

binding energy per nucleon: Nuclear binding energy divided by nucleon count.

Hall coefficient: Signed proportionality relating transverse Hall field to current density and magnetic field under a stated convention.

A pendulum tension changes throughout a swing, but its angular equation can be derived without solving for tension.

Prerequisites: 1, 47.

• Derive Euler–Lagrange motion after imposing a holonomic constraint 完整约束
• Use a cyclic coordinate 循环坐标 to identify conserved canonical momentum
• Linearise a stable equilibrium and check the resulting frequency

A constraint removes an independent coordinate before you form the kinetic energy. For a fixed-length pendulum choose θ from the downward vertical: x=l sinθ and y=−l cosθ. Differentiating both coordinates gives v²=l²θdot², not l² sin²θ θdot². Thus T=ml²θdot²/2 and V=mgl(1−cosθ), with zero potential at the bottom. The fixed length is a holonomic constraint: it is a relation among coordinates and possibly time. A rolling velocity constraint needs its own analysis; do not automatically treat every constraint as a coordinate substitution.

For ideal constraints and the usual conservative system, the stationary-action equation is d/dt(∂L/∂qdot)−∂L/∂q=0 with L=T−V. The variations vanish at the two time endpoints. A pendulum gives ∂L/∂θdot=ml²θdot and ∂L/∂θ=−mgl sinθ, so ml²θddot+mgl sinθ=0. The momentum derivative acts on every time-dependent factor: for a varying radius, d(mr²θdot)/dt contains 2mr rdot θdot. The variational method does not mean mechanical energy is conserved in a time-dependent system.

For planar central motion, L=m(rdot²+r²θdot²)/2−V(r). The angle θ is cyclic because L has no explicit θ dependence, even though L depends on θdot. Its canonical momentum pθ=mr²θdot is constant. The radial equation is m rddot=mrθdot²−dV/dr. The first term is part of the coordinate acceleration, not a new outward real force in an inertial frame. Conservation of pθ makes angular speed increase as r decreases; constant angular speed is a different, externally driven situation.

Near a stable equilibrium q0, expand V to second order and use a constant local inertia M: V≈V(q0)+V″(q0)(q−q0)²/2. Then the small-displacement frequency is sqrt(V″/M). For a pendulum sinθ≈θ in radians, so ω=sqrt(g/l); the approximation requires small amplitude. A negative V″ gives instability, not an oscillation with an imaginary measurable frequency. Check the full nonlinear equation first, then state the approximation and compare units: V″/M must have units of inverse time squared.

For a fixed pendulum length l, use theta from downward vertical: x = l sin theta, y = -l cos theta. Both Cartesian velocities give $T=ml^2\dot\theta^2/2$ and $V=mgl(1-\cos\theta)$. The Euler-Lagrange equation 欧拉—拉格朗日方程 is

Divide by $ml^2$ before linearising: The approximation uses radians. Endpoint variations vanish in the stationary-action derivation.

A cyclic coordinate is absent from L itself; its velocity need not be absent. Do not discard the time derivative of a changing r² factor.

holonomic constraint: A constraint expressible as a relation among coordinates and possibly time.

cyclic coordinate: A coordinate absent explicitly from the Lagrangian, with conserved canonical momentum under the Euler–Lagrange equation.

A person standing still on a turning platform needs a real inward force, although their platform-relative acceleration is zero.

Prerequisites: 26.

• Transform acceleration with translating and rotating frame 旋转参考系 terms
• Determine centrifugal, Coriolis and Euler directions from cross products
• Separate apparent forces from real interactions and test inertial limits

In a translating frame with origin acceleration A, Newton’s equation becomes m a′=F_real−mA. An upward-accelerating elevator therefore has N−mg=ma in the ground frame, or N−mg−ma=0 for its stationary passenger in the elevator frame. These are the same prediction. The additional term is an apparent force caused by the chosen accelerating coordinates; it is not a new contact with another body. Uniform translation with A=0 changes velocity but needs no apparent force.

For a rotating basis, differentiating a vector adds Ω×that vector. Applying this twice gives a=a_origin+a′+2Ω×v′+Ω×(Ω×r)+Ωdot×r. Here r and v′ are measured relative to the moving origin in its rotating axes; all vectors in a calculation must be expressed in the same basis at the same instant. Move the last three rotational terms to the force side to obtain Coriolis −2mΩ×v′, centrifugal −mΩ×(Ω×r), and Euler −mΩdot×r. A constant rotation removes the Euler term, not the Coriolis term.

Take Ω along +z, an anticlockwise platform viewed from above, and a particle at r along +x. Centrifugal force 离心力 is along +x with magnitude mΩ²r. If the particle moves outward with v′ along +x, then Ω×v′ is +y, so Coriolis force 科里奥利力 points −y. Reversing the relative velocity reverses Coriolis; keeping v′=0 makes it vanish. Coriolis is perpendicular to v′ and does no instantaneous work on that relative motion. Centrifugal force is outward from the rotation axis, not necessarily outward from the chosen origin in an arbitrary three-dimensional position.

A body at rest on the platform has a′=v′=0. With constant rotation and a fixed origin, its real force must be mΩ×(Ω×r), inward, cancelling the outward apparent term in the rotating equation. If rotation changes, a tangential real force must also balance the Euler term. Always check Ω→0 and A→0: ordinary inertial Newtonian motion must return. Do not mix the real inward centripetal requirement with an added outward real reaction on the same body; interaction partners belong in separate free-body diagrams.

In a rotating frame, Coriolis force uses the relative velocity.

For m = 2 kg, Omega = 2 rad/s along z, r = 1 m along x and v' = 3 m/s along x, Euler force is zero only because Omega is constant. These are coordinate terms, not extra physical contacts.

The relative velocity v′ belongs in the Coriolis term. Using the full inertial velocity counts rotation twice.

Coriolis force: The apparent rotating-frame force −2mΩ×v′ caused by relative motion.

centrifugal force: The apparent rotating-frame force −mΩ×(Ω×r), directed away from the rotation axis.

A narrow pipe carries the same steady flow rate as a wide pipe, but its pressure behaviour depends on both speed and viscous losses.

Prerequisites: 10, 47.

• Combine incompressible continuity with pressure, height and kinetic energy
• Apply hydrostatic and Bernoulli relations only under their stated conditions
• Use viscosity, Poiseuille scaling and Reynolds number to distinguish flow regimes

For steady flow, the mass passing successive cross-sections per unit time is equal: ρAv is constant. An incompressible liquid has nearly constant ρ, so Q=Av is the volume flow rate 体积流量. Halving pipe radius quarters area and multiplies mean speed by four for the same Q. Pressure does not determine Q without a model of the rest of the system. In a stationary liquid, dp/dz=−ρg with z upward; therefore pressure increases by ρgh a distance h below a surface. Use absolute or gauge pressure consistently on both sides.

For steady, incompressible, inviscid motion along a streamline with no pump or dissipative loss, Bernoulli gives p+ρv²/2+ρgz=constant. These three terms are energy per volume and have pressure units. At equal height, larger speed requires smaller static pressure under these assumptions. A higher outlet also uses pressure/kinetic energy to gain gravitational energy. Bernoulli along one streamline need not imply the same constant on different streamlines in rotational flow. A stagnation point has v=0 and converts local speed energy to a pressure rise in the ideal model.

Viscosity transports momentum between neighbouring fluid layers. For steady fully developed laminar flow of a Newtonian incompressible liquid in a circular tube, Q=πR⁴Δp/(8ηL). Here Δp is the pressure drop along the tube, η dynamic viscosity 动力黏度 and L tube length. Doubling radius at fixed Δp, η and L multiplies Q by sixteen. Doubling Q at fixed geometry needs twice the pressure drop. This viscous pressure loss cannot be added to a loss-free Bernoulli equation as though nothing changes; include a dissipative pressure/head loss or use the viscous model.

The Reynolds number Re=ρvD/η compares inertial and viscous effects using characteristic speed v and length D. For a pipe use mean speed and internal diameter, not radius. Re is dimensionless: density times speed times length has the same units as dynamic viscosity. Small Re favours viscous dominance; transition to turbulence depends on geometry and disturbances, so a single threshold is not a universal law. Poiseuille scaling is not safe after assuming a turbulent flow. Check volume continuity, sign of pressure change and the regime before selecting a formula.

A steady incompressible horizontal ideal flow has areas 4 and 1 square centimetres. At the first section speed is 1 m/s. Continuity and Bernoulli give

These relations assume no pump, viscosity loss or height change. A real narrow tube may violate them.

Continuity keeps Q constant across one steady pipe; Poiseuille compares Q between different systems at a specified pressure drop. State what is held fixed before comparing radius powers.

volume flow rate: Volume crossing a section per unit time, Q=Av for mean speed v.

dynamic viscosity: The coefficient relating shear stress to velocity gradient in a Newtonian fluid.

コンデンサに誘電体を入れる際、電源が接続されたかどうかに応じて電場の変化が異なります。

前提知識: 13, 15。

• 自由電荷と束縛電荷をD、E、分極との関連で説明する
• 一定電荷条件と一定電圧条件における誘電体の違いを比較する
• 磁気本質的関係式と自由電流の境界条件を用いる

電気的分極Pは単位体積あたりの電気双極子モーメントであり、C/m²で測定される。D=ε₀E+Pと定義すると∇·D=ρ_freeとなり、束縛電荷を含む全電荷は依然として∇·E=ρ_total/ε₀に現れる。束縛体積電荷密度はρ_b=−∇·P、束縛表面電荷密度はσ_b=P·n（nは材料の外側を向く法線ベクトル）である。均一なPは内部には束縛電荷を生じないが、異なる符号の表面電荷を生じる可能性がある。誘電体は理想導体ではなく、分極によって内部の電場がゼロになる必要はない。

線形等方性誘電体の場合、P=ε₀χ_e EおよびD=εE（ただしε=ε₀(1+χ_e)=ε₀ε_r）となる。これらの単純なスカラー関係式は、応答が線形であることと異方性、強い分散、非線形効果の無視を前提としている。端効果が無視できる充分に大きい平行板コンデンサでは、極板に垂直なDは自由表面電荷密度に等しい。σ_freeが一定であれば、εを増加させるとE=σ_free/εは減少する。束縛電荷は印加電場に対抗する方向に働き、その大きさが必ずしも自由極板電荷に等しいわけではない。

極板面積A、間隔dに対してC=εA/dである。電源を切断して自由電荷Qを一定にすると、V=Q/Cは減少し、誘電体の挿入によりCが増加すると保存エネルギーU=Q²/(2C)も減少する。理想的な電圧源を接続したままにするとVが一定になり、Q=CVおよび保存エネルギーU=CV²/2はともに増加する。電源からエネルギーが供給され、誘電体の移動に伴って機械的仕事が行われる可能性があるため、提示された電気エネルギーの量を実験全体で比較し、コンデンサ単体が孤立系であると仮定してはいけない。物質界面においてn·(D₂−D₁)=σ_freeであり、静的な接線成分Eは連続である。透磁率の変化に伴い通常、垂直成分Eも変化する。

磁化Mは単位体積あたりの磁気双極子モーメントであり、A/mで測定される。SI単位系ではB=μ₀(H+M)であり、線形等方性応答の場合、M=χ_m HおよびB=μ₀(1+χ_m)Hとなる。BとHは異なる単位と役割を持つ。長い均一ソレノイドにおいて端効果や消磁効果が無視できる場合、H≈nIとなり、物質の応答によってBが変化する。自由表面電流はn×(H₂−H₁)=K_freeを満たし、垂直成分Bは連続である。強磁性体のヒステリシスや飽和は、すべての磁場範囲で一つの定数χ_mで表現することはできない。

コンデンサは相対透磁率4の線形誘電体を挿入する前に電源から切断されている。 自由電荷は一定に保たれ、静電容量は元の値の4倍になる。

もし理想的な電池が接続されたままなら、Vは一定である： 電源とのエネルギー交換が生じる。孤立した電荷条件と一定電圧条件を同一の実験として比較してはいけない。

D の式には自由電荷を、E の式には全電荷を使用せよ。固定 Q と固定 V を混同したり、B を数値的に H と同一視したりしないこと。

電気的分極: 単位体積あたりの電気双極子モーメントPであり、束縛電荷の寄与をする。

磁化: 単位体積あたりの磁気双極子モーメント M。B=μ₀(H+M) に用いられる。

A capacitor voltage and an inductor current cannot change instantly under finite ordinary circuit drives, but their steady-state behaviour is different.

Prerequisites: 14.

• Solve RC charging and discharging with stated initial conditions
• Solve RL current response and account for stored energy
• Use complex impedance, phase and resonance in sinusoidal circuits

For a resistor R in series with a capacitor C and a DC source V_s, Kirchhoff’s voltage law is R dq/dt+q/C=V_s. Define V_C=q/C and τ=RC. After a switch to a constant source, V_C(t)=V_s+(V_C(0)−V_s)exp(−t/τ). An initially uncharged capacitor therefore has V_C=V_s(1−exp(−t/τ)) and charging current I=(V_s/R)exp(−t/τ). With the source removed and the resistor connected across it, V_C=V_C(0)exp(−t/τ). Capacitor voltage is continuous across the switch if there is no impulsive current; the resistor current may change instantly.

In a series RL circuit, Kirchhoff’s law gives L dI/dt+RI=V_s and τ=L/R. The current after a constant drive is I(t)=V_s/R+(I(0)−V_s/R)exp(−t/τ). Inductor current is continuous for a finite voltage; its voltage may change abruptly as switching changes dI/dt. At late times an ideal inductor in a DC circuit acts as a zero-voltage connection, while an ideal capacitor has zero DC current. Stored energies are LI²/2 and CV_C²/2. During decay into a resistor, the initially stored energy becomes resistor heat rather than disappearing when the source is disconnected.

Use the declared phasor convention exp(iωt). A resistor has impedance R, an inductor iωL and a capacitor 1/(iωC)=−i/(ωC). A series RLC circuit therefore has Z=R+iX with X=ωL−1/(ωC). Divide the source voltage phasor by Z to obtain current. Its magnitude is V_rms/sqrt(R²+X²) when rms quantities are used. The impedance phase φ satisfies tanφ=X/R; current lags voltage for X>0 and leads it for X<0. Adding the scalar magnitudes R, ωL and 1/(ωC) loses the vector phase information.

Series resonance occurs at ω₀=1/sqrt(LC) when X=0. Current is maximal for a fixed voltage in this ideal series model, and the voltage/current phase difference is zero. Mean real power is V_rms I_rms cosφ=I_rms²R; ideal L and C exchange energy but dissipate no average power. This differs from transient natural frequency: a damped series circuit can oscillate at sqrt(1/(LC)−(R/(2L))²) when underdamped. Large resistance removes such free oscillation but does not change the condition X=0 of the ideal driven series impedance. State whether a question asks for a step response, free decay or sinusoidal steady state.

An RC circuit switches to fixed source voltage Vs. Charge conservation and Kirchhoff's law give

For R = 2 kilohms, C = 1 millifarad, Vs = 12 V and initial voltage zero, The time constant 时间常数 controls the remaining difference from the final state.

RC has τ=RC, while RL has τ=L/R. Do not confuse current amplitude with rms current, or resonance of a driven impedance with damped free-oscillation frequency.

time constant: The exponential response scale, RC for an RC circuit and L/R for an RL circuit.

impedance: The complex voltage-to-current phasor ratio in sinusoidal steady state.

Inserting a third polariser 偏振片 between crossed filters can increase the transmitted intensity, even though every filter removes energy.

Prerequisites: 3.

• Calculate successive ideal analyser transmissions from the immediate input state
• Distinguish linear, circular and elliptical field motion using relative phase
• Use Brewster incidence with refractive-index and plane-of-incidence conventions

For propagation along z, a transverse electric field can have x and y components. Linear polarisation means the field oscillates along one fixed line; an ideal analyser transmits the projection along its axis. Field amplitude becomes E cosθ, so intensity becomes I cos²θ. Here θ is between the incoming polarisation and that analyser, and I is the intensity immediately before it. An unpolarised beam is a statistical mixture of transverse orientations; an ideal first polariser passes half its mean intensity. The outgoing beam is then linearly polarised along the filter axis. Do not apply a new factor of one half at every later filter, or use one angle to the original source for the whole chain.

After each analyser, update both the intensity and the polarisation direction. For an initially unpolarised beam and ideal axes α₁, α₂, α₃, the final intensity is (I₀/2)cos²(α₂−α₁)cos²(α₃−α₂). Two crossed filters transmit zero in this model; a middle oblique axis changes the direction before the final projection. This increase relative to the crossed pair does not create energy: every step still has transmission between zero and one. Real filters have absorption, imperfect extinction and wavelength dependence. Use the ideal law only when those losses are excluded or separately specified.

Write E_x=A cosωt and E_y=B cos(ωt+δ) at a fixed point. Equal or opposite phases give a line, including a line at an oblique angle; equal nonzero amplitudes and a quarter-cycle phase difference give a circle. Unequal amplitudes at quarter-cycle phase give an ellipse. A quarter-wave plate adds a relative phase of π/2 between its principal axes under its design conditions. A linear input at 45° supplies equal components, so the outgoing field can be circular. A linear input along a principal axis has only one component and remains linear. For circular input, every ideal linear analyser passes half the intensity; handedness requires a declared viewing direction and phase convention.

For incidence from transparent nonmagnetic medium n₁ into n₂, Brewster’s angle measured from the normal satisfies tanθ_B=n₂/n₁. At that angle, the reflected p component, whose electric field lies in the plane of incidence, vanishes in the ideal dielectric model. Reflected light from unpolarised input is then s polarised, perpendicular to that plane. Snell’s law gives a refracted angle complementary to θ_B. This is distinct from the critical-angle condition sinθ_c=n₂/n₁, which needs n₁>n₂ and concerns total internal reflection. Brewster reflection does not imply that the entire incident intensity is reflected or that the transmitted beam is completely polarised.

An ideal first polariser transmits half of unpolarised incident intensity I0. Each later analyser uses the angle from the immediately preceding axis. For axes 0, 30 and 90 degrees,

The field after each filter points along that filter's axis. Using only the first and last axes would predict zero incorrectly.

Malus intensity uses cos² of the angle to the immediately preceding polarisation; a phase plate changes relative phase, while a polariser removes a field component.

Malus’s law 马吕斯定律: Ideal linear-analyser intensity law I_out=I_in cos²θ for linearly polarised input.

Brewster angle 布儒斯特角: Incidence angle at which the reflected p component vanishes for the ideal dielectric interface.

Two equal slits can have the right path difference for constructive interference and still deliver no intensity at that angle.

Prerequisites: 3, 17.

• Derive and evaluate the far-field single-slit intensity using amplitude superposition
• Separate double-slit interference spacing from finite-aperture envelope 包络线 width
• Identify missing orders and state the limits of far-field and small-angle formulas

For a uniformly illuminated slit of width a, far-field contributions across its opening arrive with a phase gradient k sinθ. Add their complex amplitudes before squaring: the normalised integral over x from −a/2 to a/2 is sinβ/β, where β=πa sinθ/λ. The intensity ratio is (sinβ/β)². At θ=0 take the limit sinβ/β→1, rather than calling the centre undefined or dark. Zeros occur at a sinθ=mλ with nonzero integer m. The central peak lies between the first zeros and is twice as wide as one adjacent zero-to-zero interval in sinθ. Side-peak maxima are not exactly halfway between their zeros.

At small angles on a distant screen L away, y≈Lθ and sinθ≈θ give first zeros y≈±Lλ/a. Thus the central width is 2Lλ/a. Use metres consistently: a millimetre slit and a nanometre wavelength differ by six powers of ten. For a wider angle, use θ=asin(mλ/a) and y=L tanθ; the small-angle y expression then becomes inaccurate. Far-field conditions need nearly parallel rays from different parts of the aperture, for example L much greater than a²/λ, or the equivalent focal-plane arrangement. Near-field Fresnel patterns cannot be assigned this intensity law blindly.

For two coherent identical uniformly illuminated slits of width a and centre separation d, the normalised pattern is (sinβ/β)² cos²α, with α=πd sinθ/λ and central intensity as the normalisation. The cos² factor produces interference orders d sinθ=nλ; the single-slit factor gives the broader envelope zeros. Small-angle neighbouring interference spacing is λL/d. Increasing d narrows fringe spacing; increasing a narrows the envelope. These are separate changes. The equal-height narrow-slit interference formula alone does not predict the diminishing brightness or missing orders of finite apertures. Incoherent sources do not maintain the same phase-dependent cross term.

A missing order 缺级 occurs when nλ/d=mλ/a simultaneously, giving n=m d/a. If d/a is an integer r, orders ±r, ±2r and so on are cancelled by aperture zeros. Do not count an order at an envelope boundary as a visible bright fringe. In the central envelope, the nominal interference-order centres satisfy |n|<d/a; when r is an integer there are 2r−1 such centres. For finite slit width, exact local maxima are shifted slightly by the changing envelope, so the interference-order locations are an approximation to observed peak centres. Identify what quantity is being requested before treating every cos² maximum as an exact maximum of the product.

For two finite slits, separate interference spacing from the single-slit envelope.

If d = 5a, the first envelope zero $a\sin\theta=\lambda$ coincides with order n = 5. Nominal order centres in the central envelope are n = -4 through +4: nine in total. These are interference centres; the envelope can shift the exact product maxima slightly.

Add field amplitudes before squaring, distinguish a from d, and exclude dark boundary orders. An interference maximum alone does not guarantee a maximum of the full pattern.

diffraction envelope 衍射包络: The aperture-dependent intensity factor that modulates the interference pattern.

missing order: An interference order cancelled because it coincides with an aperture intensity zero.

A negative image distance is a geometric result that locates a virtual image 虚像, not a failed calculation to replace by its absolute value.

Prerequisites: 3.

• Locate paraxial thin-lens and spherical-mirror images with a declared sign convention
• Use signed magnification to classify orientation and real or virtual character
• Track successive image locations as the next element’s object without resetting geometry

Use the real-is-positive convention for this unit: a real object sends diverging incident rays into the element, so s>0; an already converging incident beam can represent a virtual object, s<0. A real image has s′>0, while a virtual image has s′<0. The paraxial equation is 1/f=1/s+1/s′ and transverse magnification 横向放大率 M=−s′/s. A converging lens 会聚透镜 has f>0 and a diverging lens f<0. For left-to-right light and a real object at x<0 relative to a lens at x=0, a real image lies at x=s′>0; a negative s′ puts its virtual image on the input side. State the convention before substitution, as other signed-coordinate conventions give different-looking equations.

For a converging thin lens, a ray parallel to the axis exits through the far focal point, and a ray through the ideal optical centre continues undeviated. Their intersection identifies a real image; backwards extensions can locate a virtual image. The paraxial approximation uses rays close to the axis and neglects thickness and aberrations. With s>f>0, s′ is positive and M is negative: the image is real and inverted. With 0<s<f, s′ is negative and M is positive: the image is virtual, upright and enlarged. At s=f, emerging rays are parallel and no finite image plane exists. A diverging lens with a real object produces an upright reduced virtual image.

For a spherical mirror in the same real-is-positive convention, concave f>0 and convex f<0, with f=R/2 in the paraxial limit. A real reflected image is in front of the mirror on the incoming-light side; a virtual image is behind it. Thus a positive image distance has a different physical side for a mirror than for a transmitting lens. A parallel ray reflects through the concave focus; a ray aimed through the centre of curvature retraces its path. Convex reflected rays diverge as if from the focus behind the mirror. Use the same equation and M=−s′/s with signed values, rather than silently combining mirror geometry with a lens coordinate diagram.

For a sequence of separated lenses, first calculate the actual image coordinate from the first lens. Relative to the next lens, decide whether the incoming rays diverge from a point before it or are still converging toward a point beyond it; these are real and virtual objects respectively. Only then assign its signed object distance and solve again. Total transverse magnification is the product of the individual signed magnifications for aligned paraxial elements. Element separation is not automatically the second object distance. Geometric real/virtual character depends on convergence, not on whether the final image is magnified. These ideal results do not model wave-optical resolution or spherical/chromatic aberration.

Use real-is-positive distances and positive focal length for a converging lens. With f = 12 cm and real object distance s = 8 cm,

Thus s' = -24 cm and $M=-s'/s=+3$. The virtual image is upright, on the object's side. Signed distances identify the physical side; a negative distance is not an arithmetic failure.

Keep signs until the geometry is interpreted; a positive mirror image lies on the incident side, while a positive transmitting-lens image lies on the outgoing side.

virtual image: An apparent image located by backwards ray extensions rather than actual outgoing-ray convergence.

transverse magnification: Signed image-height to object-height ratio, −s′/s in the declared convention.

Two wall layers in series carry the same steady heat rate 热流率, but their temperature drops need not be equal.

Prerequisites: 4.

• Combine conductive thermal resistances and distinguish heat rate from heat flux
• Balance sensible and latent heat 潜热 with stated isolation and heat-capacity conditions
• Calculate free thermal expansion and constrained thermal stress with their limits

For one-dimensional steady conduction through a uniform slab, Fourier’s law gives heat flux q_x=−κ dT/dx. With constant conductivity κ, area A, length L and hot-to-cold temperature difference ΔT, heat rate is Qdot=κAΔT/L. Define thermal resistance 热阻 R_th=L/(κA), measured in K/W; temperature drop is Qdot R_th. Series layers without internal sources carry the same Qdot and their resistances add. Parallel paths at the same endpoint temperatures have heat rates that add, so inverse resistances add. Heat flux is rate per area, W/m², and need not match between layers of different area. Interface contact resistance and heat leakage must be included if specified.

Heat capacity C=dQ/dT depends on the thermodynamic path and is not the same as specific heat c per mass. For a material interval without a phase transition, Q=∫mc(T)dT, or mcΔT when c is constant. In an isolated calorimeter, sum the energy changes of all objects, including the container when relevant, and set the sum to zero. At a phase transition at its equilibrium temperature, added energy can change phase fraction rather than temperature: Q=mℓ uses latent heat ℓ in J/kg. First supply the sensible heat to reach the transition, then budget latent heat. Do not let a simple weighted-temperature average predict an impossible temperature when melting or freezing is part of the process.

For a freely expanding rod and small temperature change, ΔL≈αL₀ΔT with linear expansion coefficient α. For an isotropic solid with small strain, each dimension acquires factor 1+αΔT, so ΔA/A≈2αΔT and ΔV/V≈3αΔT. A hole expands with the surrounding material as if its missing region had expanded too. These relations assume a nearly constant coefficient and no mechanical constraint; anisotropic crystals need directional coefficients. If an elastic rod is prevented from changing length, its mechanical strain cancels thermal strain. With tensile stress positive, σ≈−YαΔT during heating, where Y is Young’s modulus. This small-strain estimate requires elastic response without yielding or buckling.

Conduction transfers energy through a temperature gradient, convection transports it with moving matter, and thermal radiation can cross a vacuum. For a surface at T facing large surroundings at T_env, a simple grey-body model gives net radiative rate εσ_SB A(T⁴−T_env⁴). Use absolute kelvin in these fourth powers, rather than Celsius values. A lumped body whose internal temperature remains nearly uniform can obey C dT/dt=−hA(T−T_env) under Newton cooling with constant h. Then the temperature excess decays with τ=C/(hA), rather than the entire Celsius or kelvin temperature decaying toward zero. This approximation requires sufficiently small internal gradients; a conduction-limited body needs a spatial temperature model.

Two slabs in series carry the same steady heat rate. Let resistances be 0.08 and 0.24 K/W. Boundary temperatures are 60 and 20 degrees Celsius.

The interface is 50 degrees Celsius when slab 1 touches the hot side. Heat rate has watt units; flux divides it by area.

Keep W separate from W/m², include the calorimeter or latent energy when needed, and do not use free-expansion length together with constrained-stress assumptions.

thermal resistance: Temperature difference per steady heat-transfer rate, measured in K/W.

latent heat: Energy transferred during a phase change at its transition temperature; specific latent heat is per mass.

A thermal energy gap can keep heat capacity small at both low and high temperatures, even while the excited-state probability keeps increasing.

Prerequisites: 20.

• Obtain canonical mean energy and free energy from a fixed energy spectrum
• Relate energy variance to constant-volume heat capacity with fixed-spectrum conditions
• Evaluate entropy and low/high-temperature limits for an original finite-level model

For a system exchanging energy with a bath at T while N and V remain fixed, let β=1/(kBT). Sum Z=Σ_i g_i exp(−βE_i) over energy levels with their degeneracies. Level probability is g_i exp(−βE_i)/Z; mean energy U is the probability-weighted energy sum. Differentiating a temperature-independent spectrum gives U=−∂lnZ/∂β. Keep the energy gap and degeneracies fixed during the derivative. Writing Δ=kBT ln3 to describe one evaluation temperature must not be read as making Δ change with T. For N independent distinguishable identical two-level subsystems, Z_total=z^N and U_total=N u. Indistinguishable particles and interactions need their own state counting rather than this product assumption.

The Helmholtz free energy 亥姆霍兹自由能 is F=−kBT lnZ. Canonical entropy follows from S=kB(lnZ+βU), equivalently −kBΣ p_j ln p_j over individual microstates. Constant-volume heat capacity is C_V=(∂U/∂T)_V,N. These quantities carry different units and describe different derivatives. Adding a constant energy offset ε to every state multiplies Z by exp(−βε), increases U and F by ε, and leaves probabilities, entropy and heat capacity unchanged when ε is independent of T. A negative chosen mean energy is therefore not evidence of a negative heat capacity. Use the same energy reference in the probabilities and thermodynamic expressions.

The second β derivative of lnZ gives variance Var(E)=⟨E²⟩−U². For the same fixed-spectrum canonical model, C_V=Var(E)/(kBT²), so C_V/kB=β²Var(E). Nonnegative variance implies nonnegative C_V within these conditions. The denominator includes kB, not kB², when heat capacity retains its ordinary J/K units. This fluctuation formula concerns the canonical energy distribution; it does not say each particle has exactly the mean energy. For independent subsystems variances add, while means add too. Relative energy fluctuations typically shrink like 1/√N when the mean and per-subsystem variance remain finite and nonzero.

For one ground state at E=0 and one excited state at fixed E=Δ>0, put x=Δ/(kBT). Then z=1+e^(−x), p_exc=1/(1+e^x), u=Δp_exc and C/kB=x²e^x/(1+e^x)². At low T, excitation and heat capacity vanish exponentially. At high T, p_exc tends to one half and energy saturates, so heat capacity tends to zero again. Entropy rises from zero for the unique ground state toward kB ln2 as the two states become equiprobable. The resulting finite-temperature heat-capacity peak is specific to the finite-level model; it is not the constant classical oscillator value. Ground-state degeneracy or additional levels would change the limiting entropy and temperature response.

Hold the two energy levels 0 and Delta fixed when differentiating the partition function 配分函数.

At $\beta\Delta=\ln3$, excited probability is 1/4, mean energy Delta/4, variance $3\Delta^2/16$ and $C_V/k_B=3(\ln3)^2/16=0.2263$. Degeneracy, if present, belongs in Z before differentiation.

Hold Δ fixed when differentiating, include degeneracies in Z, and use energy variance rather than the square of mean energy in the heat-capacity formula.

energy fluctuation 能量涨落: Canonical spread of energy about its ensemble mean, quantified by the energy variance.

Helmholtz free energy: Thermodynamic potential F=U−TS, equal to −kBT lnZ for the canonical ensemble.

A fermionic energy level can contain several particles when it contains several distinct states; the exclusion rule applies per complete state.

Prerequisites: 20, 35, 38.

• Compute mean occupation 平均占据数 per complete quantum state for fermions and bosons
• Count allowed identical-particle occupations and identify the classical dilute limit
• Compare classical quadratic-mode heat capacity with a quantum oscillator response

For noninteracting particles in thermal and particle exchange equilibrium, use chemical potential μ and x=(ε−μ)/(kBT). Mean occupation of one complete state is n_F=1/(e^x+1) for fermions and n_B=1/(e^x−1) for bosons. Fermionic occupation of that state is zero or one; its mean lies between them. Bosonic occupation can exceed one. A level of degeneracy g has total mean g times the single-state occupation when its states share the same energy. Count spin as part of a complete state. The Bose denominator requires ε>μ for the ordinary finite expression, with the ground-state limit treated separately. For equilibrium photons μ=0 because photon number is not conserved; do not set μ=0 for every material particle gas.

Count occupation patterns rather than labelling identical particles. Two identical fermions distributed among four distinct complete states have choose(4,2)=6 allowed patterns. Two identical bosons among those states have choose(4+2−1,2)=10 patterns because both may share one state. Two labelled distinguishable particles would have 4²=16 assignments. These are different counting models, not three interchangeable answers to the same specification. If a question supplies spin degeneracy, first decide whether its stated number counts complete states or just orbital levels. A Pauli prohibition on two identical complete states does not prohibit opposite-spin fermions in one spatial orbital.

When x is large and positive, occupation is small and both denominators are dominated by e^x: n_F≈n_B≈e^(−x), the Maxwell–Boltzmann dilute limit. At x=ln4 the means are 1/5 for fermions, 1/3 for bosons and 1/4 in the classical approximation; the difference is still significant. A large total particle number alone does not justify classical statistics: density, temperature and accessible states control occupation. For fermions at low T, states below μ become nearly occupied and those above nearly empty. At ε=μ a fermionic state has mean one half; inserting that value into the ordinary Bose formula would instead produce a divergent denominator and requires different limiting treatment.

Classical equipartition assigns kBT/2 of mean energy to each independent quadratic term in an equilibrated Hamiltonian. A monatomic ideal gas has three translational terms, giving U=3NkBT/2 and C_V=3NkB/2. One classical one-dimensional harmonic oscillator has kinetic and potential terms, giving mean kBT and C=kB. For a quantum oscillator with fixed spacing ε=$\hbar$ω, the thermal energy above its temperature-independent zero point is ε/(e^x−1), now x=ε/(kBT), and C/kB=x²e^x/(e^x−1)². At high T it approaches the classical value; at low T excitation freezes out and C→0. A zero-point energy ε/2 shifts U but not C. Molecular rotational/vibrational contributions likewise need their energy scales checked before assigning classical quadratic terms.

Occupation number 占据数 refers to one complete single-particle state. For $x=(\epsilon-\mu)/(k_BT)=\ln4$,

The classical dilute approximation gives $e^{-x}=1/4$; it is not exact here. A g-fold level has mean total occupation $g\bar n$, so a fermionic level can contain more than one particle while each complete state still obeys exclusion.

Exclude two fermions from one complete state, not from an entire degenerate energy level. State which x is used; chemical-potential occupations and fixed oscillator excitation formulas are different models.

mean occupation: Ensemble average number of particles in one complete quantum state or a specified group of states.

equipartition: Classical equilibrium rule assigning kBT/2 to each independent quadratic Hamiltonian term.

A zero first-order energy shift 一阶能量修正 can coexist with a changed state and a nonzero higher-order shift.

Prerequisites: 5, 24.

• Calculate nondegenerate first-order energy shifts as normalised expectation values
• Use symmetry and matrix elements to assess weak-coupling limits
• Diagonalise the perturbation within an exactly degenerate subspace 简并子空间

Let H=H₀+λW, where λ is a small dimensionless parameter and the eigenstates of H₀ are known and normalised. For a nondegenerate level n, the first-order energy correction is λ⟨n|W|n⟩. In position space this is λ∫ψ_n*(x)W(x)ψ_n(x)dx for a multiplicative potential, with integration over the allowed domain. A potential value at one point is not an expectation value; the state’s probability density weights the whole domain. For an infinite well 0<x<L, ψ_n=√(2/L)sin(nπx/L). Reflection about its centre makes ⟨x⟩=L/2, so a weak added potential γx has shift γL/2 for every nondegenerate well level at first order. γ has units of energy per length.

For a perturbation odd about the centre, such as γ(x−L/2), the probability density of an unperturbed well eigenstate is even, so its diagonal expectation vanishes. A centred quadratic perturbation η(x−L/2)² instead gives shift ηL²[1/12−1/(2π²n²)], which is positive for η>0 and depends on n. Parity makes the first-order integral zero only for the specified state and operator symmetry. Off-diagonal matrix elements can still change the wavefunction. Do not turn a symmetry cancellation into a claim that every energy correction vanishes or that the original state remains exact.

For a nondegenerate state, the leading admixture of another unperturbed state m is proportional to λW_mn/(E_n⁰−E_m⁰). The useful smallness condition therefore compares coupling matrix elements with the relevant level separations, not just with an arbitrary absolute energy zero. To second order, the energy correction contains Σ_(m≠n)|λW_mn|²/(E_n⁰−E_m⁰). For the lowest nondegenerate state, all these denominators are negative, so the second-order correction is nonpositive in this model. Near a degeneracy, a small denominator defeats the nondegenerate expansion; use a coupled subspace instead of dividing by zero.

If a level of H₀ is exactly degenerate, choose an orthonormal basis within that subspace and form the Hermitian matrix of the perturbation there. Its eigenvalues are the first-order energy shifts, and its eigenvectors specify the combinations that diagonalise the leading splitting. For a two-state subspace with perturbation ε[[2,1],[1,2]], the normalised symmetric and antisymmetric combinations have shifts 3ε and ε. Reading only the two diagonal entries would incorrectly predict two shifts of 2ε. A common scalar multiple of the identity shifts both states equally and does not split their degeneracy; off-subspace couplings can matter at higher order.

A degenerate subspace requires matrix diagonalisation before first-order shifts. Suppose the perturbation restricted to two states is $\epsilon\begin{pmatrix}2&1\\1&2\end{pmatrix}$.

Shifts are epsilon and 3 epsilon. Normalised states are $(1,-1)/\sqrt2$ and $(1,1)/\sqrt2$. Reading only the diagonal entries would miss the splitting. Couplings to other levels must remain weak relative to their energy gaps.

Use normalised state weights, keep perturbation units, and diagonalise a degenerate block. First-order cancellation and exact invariance are different claims.

first-order energy shift: Leading weak-perturbation correction given by the unperturbed state’s expectation value.

degenerate perturbation theory 简并微扰理论: Method that first diagonalises the perturbation within an unperturbed degenerate subspace.

Two electrons can share one spatial orbital if their complete spatial-plus-spin state has the required antisymmetry.

Prerequisites: 5, 36.

• Construct normalised symmetric and antisymmetric 反对称 two-orbital spatial states
• Combine spatial and spin symmetry to produce allowed two-electron states
• Count complete-state occupations and interpret exchange-induced correlations

For identical particles, exchanging every coordinate and spin label changes a fermionic total wavefunction by a minus sign and leaves a bosonic one unchanged. Mathematical slots 1 and 2 label arguments, not permanently distinguishable particles. For two distinct orthonormal orbitals a and b, spatial combinations are Ψ_±=[a(1)b(2)±b(1)a(2)]/√2. Orthogonality makes these combinations normalised and gives exchange eigenvalues ±1. If the two orbitals are identical, the antisymmetric combination is identically zero and the displayed symmetric formula is not correctly normalised; the double-occupation spatial state is simply a(1)a(2). Recheck normalization whenever orbitals or their overlaps change.

Two spin-1/2 particles have one spin singlet 自旋单态 (↑↓−↓↑)/√2 with total spin S=0; it is antisymmetric under exchange. The three triplets ↑↑, (↑↓+↓↑)/√2 and ↓↓ have S=1 and are symmetric. Electrons require an antisymmetric total state, so a symmetric spatial state pairs with the singlet, while an antisymmetric spatial state pairs with a triplet. Two electrons in the same spatial orbital can form the singlet, but cannot form a triplet there in this simple two-electron state. Pauli exclusion prevents occupation of the same complete single-particle state, including spin. Opposite-spin electrons in one orbital are two different complete states; describing exclusion as one electron per orbital discards this distinction.

For two spatial orbitals a,b and two spin states each, there are four complete single-particle states. Two identical electrons have choose(4,2)=6 occupation patterns: two double-occupation singlets, one different-orbital singlet, and three different-orbital triplets. For three spatial orbitals, there are six complete states and choose(6,2)=15 patterns. Counting ordered assignments overcounts identical particles. Spinless bosons in two orbitals instead have three occupations: both in a, one in each, both in b. These counts assume no additional energy restriction and the stated accessible single-particle states; restricting total energy or spin projection changes the allowed subset.

An antisymmetric spatial state satisfies Ψ_−(x,x)=0, so its joint position density vanishes on the coincidence line. A symmetric spatial state need not vanish there. The difference arises from interference between exchanged amplitudes, even for noninteracting particles; it is not a separately imposed classical repulsive force. For a well of length L, choose a(x)=√(2/L)sin(πx/L), b(x)=√(2/L)sin(2πx/L). At x₁=L/4 and x₂=3L/4, the antisymmetric spatial amplitude is −2/L and its density is 4/L²; the symmetric amplitude is zero. These are joint probability densities per two lengths, not dimensionless probabilities at exact points. Physical spin compatibility still decides which total electronic state uses each spatial combination.

For orthonormal different orbitals a and b, symmetric and antisymmetric spatial states are

A two-electron total state must be antisymmetric under simultaneous exchange of position and spin. Thus symmetric space pairs with the spin singlet; antisymmetric space pairs with triplet spin. If both electrons occupy a, the spatial product is symmetric and only singlet spin is allowed. The different-orbital normalisation formula cannot simply be reused for a = b.

Exchange all spatial and spin arguments. Do not confuse antisymmetric space with antisymmetric total state, or use the distinct-orbital √2 factor for two identical orbitals.

spin singlet: Antisymmetric two-spin-1/2 state with total spin S=0.

Slater determinant 斯莱特行列式: Antisymmetric fermionic construction from complete single-particle states; duplicate states make it vanish.

A bound wavefunction can extend outside a finite well 有限深势阱 even though the particle’s energy is below the external potential.

Prerequisites: 5, 24, 37.

• Match decaying finite-well states and derive parity-dependent quantisation conditions
• Interpret dimensionless bound-state roots and distinguish confinement from scattering
• Use hydrogenic quantum numbers, angular momentum and radial probability measures

Declare a symmetric well V=−V₀ for |x|<a and V=0 outside, with V₀>0 and the same mass throughout. A bound energy lies between −V₀ and zero. Inside, k=√[2m(E+V₀)]/$\hbar$ gives oscillatory solutions; outside, κ=√(−2mE)/$\hbar$ gives decaying tails. Reject growing exponentials to obtain a normalisable state. Reflection symmetry permits even interior cos(kx) or odd sin(kx) states. At finite boundaries without delta interactions, both ψ and ψ′ are continuous. Even matching gives k tan(ka)=κ; odd matching gives −k cot(ka)=κ. A finite well does not require ψ to vanish at its edges as an infinite wall does.

Introduce z=ka and ρ=a√(2mV₀)/$\hbar$. Then κa=√(ρ²−z²) and 0<z<ρ. Solve z tan z=√(ρ²−z²) for an even state, or −z cot z=√(ρ²−z²) for an odd one, on intervals with the appropriate sign and away from tangent poles. Energy follows as E/V₀=z²/ρ²−1. For ρ=1, the ground root lies in 0<z<1 and the even equation is equivalent there to z=cos z. Bisection or a converged root finder gives z≈0.739085 and E/V₀≈−0.453753. Since ρ<π/2, no odd bound-state branch fits. In one dimension an attractive square well has an even bound ground state however shallow it is; an E=0 threshold tail is not square integrable.

For an even bound state 束缚态 write ψ=A cos(kx) inside and ψ=A cos(ka)exp[−κ(|x|−a)] outside. Continuity sets the relative tail amplitude; integrate |ψ|² over both the interior and tails to determine A. Probability outside the well is nonzero and depends on its depth, width and the selected state. Bound states have discrete energies and normalisable decaying asymptotes. A scattering state with E>0 has propagating external waves and is normalised or interpreted using a continuum/flux convention. A decaying tail is not evidence that the state violates energy conservation, and raw tail amplitude is not the outside probability until it has been integrated and normalised.

For a one-electron Coulomb ion with nuclear charge Ze, use the nonrelativistic central-potential model and a heavy nucleus approximation unless reduced mass is specified. E_n≈−13.6 Z²/n² eV; allowed orbital numbers are n≥1, l=0,…,n−1 and m_l=−l,…,l. L² has eigenvalue l(l+1)$\hbar$² and L_z=m_l$\hbar$; the ground 1s orbital has l=0, despite the historical Bohr circular-orbit rule. Spatial degeneracy at fixed n is Σ_l(2l+1)=n² before spin and fine-structure corrections. If ψ=R_nl(r)Y_lm with angular part normalised, radial probability in dr is r²|R|²dr, not just |R|²dr. For hydrogen 1s, the radial density is 4r²e^(−2r/a₀)/a₀³: it peaks at r=a₀ and has mean 3a₀/2, while the three-dimensional point density is largest at the origin. Distinguish these measures before locating a most probable radius.

A finite well has V = -V0 for $|x|<a$ and zero outside. A bound energy lies strictly between -V0 and zero. Define

Even parity gives an interior cosine and decaying tails. Matching value and derivative at a gives Odd parity instead gives $-k\cot(ka)=\kappa$. Solve only within valid parity intervals; a root at a tangent pole is not a bound state.

Use the declared potential zero, match ψ′ as well as ψ, and include the radial shell measure. A bound tail and a continuum travelling wave have different energy regimes.

bound state: Normalisable stationary state with confined probability and discrete energy in the specified model.

radial probability density 径向概率密度: Probability per radial distance, including the spherical-shell measure after angular integration.

The same magnetic field needs different quantum labels when its energy scale crosses the spin-orbit splitting.

Prerequisites: 7, 37.

• Calculate orbital and spin magnetic shifts in the appropriate coupling regime
• Find photon-energy shifts from allowed upper-minus-lower level changes
• Contrast parity cancellation with degenerate electric-field splitting

An electron has a negative magnetic moment: μ_L=−μ_B L/$\hbar$ and μ_S≈−2μ_B S/$\hbar$. With B along positive z, the interaction −μ·B therefore gives H_Z=μ_B B(L_z+2S_z)/$\hbar$. μ_B≈5.788×10⁻⁵ eV/T. In an uncoupled basis with orbital projection m_l and spin projection m_s=±1/2, the shift is μ_B B(m_l+2m_s). Use this uncoupled rule when the magnetic interaction is large relative to the relevant spin-orbit splitting but still small relative to orbital energy gaps, so quadratic diamagnetic and orbital restructuring effects can be neglected. Large and small are comparisons of energies, not universal Tesla thresholds. In the opposite weak-field regime, m_l and m_s are not independently conserved labels for an LS-coupled eigenstate.

When LS coupling dominates a weak magnetic perturbation, J=L+S and m_J label the multiplet. The first-order shift is μ_B B g_J m_J, with g_J=1+[J(J+1)+S(S+1)−L(L+1)]/[2J(J+1)] in the approximation g_s=2. L,S,J here are dimensionless quantum numbers. For L=1,S=1/2, J=3/2 gives g_J=4/3 and shifts −2,−2/3,+2/3,+2 times μ_B B. J=1/2 gives g_J=2/3 and shifts ±μ_B B/3. Do not substitute J=0 into the singular formula; a J=0 state has no first-order vector projection shift. Intermediate fields require a coupled Hamiltonian rather than mixing weak-field and uncoupled formulas.

A photon’s energy is the upper atomic energy minus the lower energy. Its field-induced change is therefore ΔE_γ=ΔE_upper−ΔE_lower. Count allowed transitions using the stated selection rules as well as the two level patterns. Several transitions can share one frequency; the number of split states is not automatically the number of spectral lines. In a spin-neglected orbital model, an upper L=1 triplet has shifts m_l μ_B B with m_l=−1,0,+1, and a lower L=0 state has zero shift. Electric-dipole Δm_l=0,±1 permits three photon shifts −μ_B B,0,+μ_B B. This normal-triplet model does not describe every real atom or fine-structure multiplet. Polarisation and observation direction can change which components are detected.

In a uniform electric field F along z, an electron’s perturbation is +eFz for the stated electrostatic potential convention. A nondegenerate parity eigenstate has ⟨z⟩=0, so its first-order diagonal Stark shift vanishes, although a quadratic shift can remain. Exactly degenerate opposite-parity states instead allow an off-diagonal dipole matrix element. In an original two-state model, suppose the dipole matrix is [[0,d],[d,0]], with real d in charge×length units. The perturbation F times this matrix has eigenvalues ±dF and opposite superpositions of the two states. Changing a basis phase changes the off-diagonal sign but not the splitting 2|dF|. Actual near-degenerate atoms require comparison with fine structure and other small splittings before an exactly degenerate approximation is adopted; this supplied matrix is not a universal hydrogen coefficient.

In weak LS-coupled magnetic fields, the Zeeman effect 塞曼效应 uses J, not uncoupled ml and ms.

For L = 1, S = 1/2 and J = 3/2, $g_J=4/3$. Select the field regime before inserting quantum numbers. Photon shifts subtract the lower-level shift.

Declare the coupling regime, keep the negative electron moment sign, and subtract lower-level shifts. Zero diagonal dipole expectation does not exclude degenerate mixing or quadratic shifts.

Landé factor 朗德因子: Projection factor relating a weak-field LS-coupled magnetic shift to g_J m_J.

Stark splitting 斯塔克分裂: Electric-field-induced separation of atomic energy levels, with symmetry and degeneracy controlling leading order.

A spectral peak, a total radiation flux and an atomic transition energy are three different observables.

Prerequisites: 7, 9, 34.

• Distinguish wavelength spectral peak, integrated radiation and photon energy
• Derive Bohr radius and energy scaling under one-electron assumptions
• Calculate Coulomb-ion transitions, series limits and ionisation thresholds

For ideal thermal equilibrium radiation, Planck’s wavelength spectral radiance 光谱辐亮度 is B_λ=2hc²/{λ⁵[exp(hc/(λk_B T))−1]}. B_λ is per wavelength interval and per solid angle, not a total power. A blackbody’s hemispheric surface flux spectrum is M_λ=πB_λ; integrating over all wavelengths gives σ_SB T⁴. All temperatures are absolute Kelvin. The wavelength peak obeys λ_max T≈2.898×10⁻³ m·K. A spectrum per frequency has a different peak because B_ν dν and B_λ dλ include a Jacobian; c/λ_max is not the peak frequency of B_ν. Increasing T moves the wavelength peak shorter and increases the integrated flux by T⁴, not by the peak-position ratio alone.

For an original uniform grey surface of area A and wavelength-independent emissivity ε, net radiative power to a large uniform environment is εσ_SB A(T⁴−T_env⁴) under the stated view-factor assumptions. Use σ_SB≈5.670×10⁻⁸ W·m⁻²·K⁻⁴. Spectrally varying emissivity or incomplete surroundings requires a more detailed model. A photon at a specified wavelength has E=hc/λ, conveniently about 1240 eV·nm/λ_nm. A thermal spectrum contains many photon energies; a photon at the wavelength peak is not the mean energy of every photon. For T=3000 K, λ_max≈966 nm; its photon energy is about 1.284 eV. A surface with A=2×10⁻⁴ m² and ε=0.5 emits 459.27 W to a negligibly cold environment.

In the historical Bohr one-electron model with a heavy nucleus of charge Ze, Coulomb force m_e v²/r=Ze²/(4πε₀r²) and angular momentum m_e vr=n$\hbar$ lead to r_n=a₀n²/Z, v_n=Zαc/n and E_n≈−13.6Z²/n² eV. The radius scales with n²/Z while binding energy scales with Z²/n²; do not use the same charge power in both. These circular-orbit assumptions are a historical scaling model. Wave-mechanical angular momentum is √[l(l+1)]$\hbar$ with l=0,…,n−1, so a 1s orbital has zero orbital angular momentum rather than the Bohr n$\hbar$ value. Reduced-mass corrections replace m_e with μ: Coulomb radius scales as 1/μ and energy as μ. Multielectron screening, large-Z relativistic corrections and fine structure are outside the simple model.

For an emission ni→nf with ni>nf, E_γ≈13.6Z²(1/nf²−1/ni²) eV. Convert this positive difference to λ≈1240/E_γ nm. Ionisation from n requires energy 13.6Z²/n² eV to reach the continuum zero. Absorption reverses the level ordering and needs the corresponding positive incoming photon energy. For a series ending at fixed nf, the largest bound-bound photon energy occurs as ni→∞, giving E_limit=13.6Z²/nf² and the shortest series wavelength. For one-electron helium Z=2, n=3→2 gives 7.5556 eV and λ≈164.12 nm. Its Bohr n=3 radius is 4.5a₀; this radius describes the historical orbit, not a universal radial mode for every l at n=3.

Wien's law 维恩定律 refers to the wavelength-density peak of a blackbody spectrum.

A photon at that wavelength has energy The source emits a distribution of photon energies. Its integrated surface flux is $\sigma T^4$, and the frequency-density peak is not obtained by simply using c divided by this peak wavelength.

Declare a wavelength or frequency density, use Kelvin for fourth powers, and distinguish transitions from ionisation. A Bohr circular radius and angular momentum are not every wave orbital’s radius and L.

spectral radiance: Radiation intensity per projected area, solid angle and stated spectral interval.

series limit 谱线系限: Limiting bound-bound photon energy or wavelength as the initial level approaches the continuum for a fixed final level.

Two observers can disagree about which of two events happened first, and both can be right.

Prerequisites: 22.

• Transform event separations and identify proper length 固有长度 versus simultaneous endpoint readings
• Apply relativistic velocity addition and check a four-vector invariant
• Classify the interval sign to order events and rule out causal contact

Use the Lorentz transformation with declared frame motion: if frame S′ moves at +v along x relative to S, then x′=γ(x−vt) and t′=γ(t−vx/c²) with γ=1/sqrt(1−v²/c²). Apply it to event separations Δx and Δt, not only to single coordinates. Two events simultaneous in S (Δt=0) at different places have Δt′=−γvΔx/c²: the event further in the +x direction happens earlier in S′. Simultaneity is frame-dependent, not a bookkeeping error.

A moving rod length must be read from endpoint positions at the same lab time (Δt=0). Then Δx′=γΔx gives L=L0/γ with L0 the rest-frame (proper) length. Reading the endpoints simultaneously in the rod frame instead does not produce the lab length; the two reading conditions answer different questions. State the frame and the reading condition before quoting any length.

The interval s²=c²Δt²−Δx² (mostly-minus convention, declared once) is invariant. s²>0 is timelike: a frame exists with the events co-located, and Δτ=s/c is the proper time between them. s²<0 is spacelike: a frame exists with the events simultaneous, and sqrt(−s²) is the proper distance. s²=0 is lightlike. Timelike order is the same in every frame; spacelike order can swap, and no signal can join the events because |Δx/Δt|>c.

Collinear velocities add as u=(u′+v)/(1+u′v/c²); the denominator keeps u<c for any u′,v<c, and u′=c gives u=c exactly. Energy and momentum form a four-vector with invariant E²−p²c²=m²c⁴: compute it in any convenient frame. Check each result in the nonrelativistic limit, where the addition rule reduces to u=u′+v and the invariant reduces to the rest energy.

Two lab events have Delta x = 900 m and Delta t = 2 microseconds. Their spacetime interval 时空间隔 is spacelike because c Delta t = 600 m is less than Delta x. A frame making them simultaneous satisfies

Their separation in that frame is $\sqrt{\Delta x^2-c^2\Delta t^2}=670.8$ m. No frame can put these spacelike events at the same position.

Quoting L=L0/γ without the simultaneous-endpoint condition, or adding velocities as u=u′+v at relativistic speeds. A simultaneity difference is physical, not a mistake.

spacetime interval: The invariant combination c²Δt²−Δx² under the declared sign convention, whose sign classifies an event separation.

proper length: The endpoint separation measured in the object's rest frame; any other frame must read both endpoints simultaneously in its own time.

A voltmeter 电压表 changes the circuit it measures; the reading is a divider output, not the open-circuit voltage.

Prerequisites: 14, 21, 30.

• Calculate measurement loading from finite meter and source impedances
• Determine simple amplifier gain and first-order filter response
• Interpret detector efficiency, resolution, dead time 死时间 and calibration bias

Model the measured circuit as a Thevenin source V_th with resistance R_th and the instrument as input resistance R_m. The reading is V_th·R_m/(R_m+R_th), with the same sign as V_th and magnitude no greater than |V_th| for positive resistances. A 10 V source with R_th=100 kΩ read by a 1 MΩ meter gives 10×(1/1.1)=9.09 V, a 9% low bias. A 10× oscilloscope probe raises the effective input resistance and divides the signal by ten; quote both effects.

An ideal non-inverting op-amp stage has gain 1+R2/R1; the inverting stage has −R2/R1, with the sign carried explicitly. A first-order RC low-pass has cutoff f_c=1/(2πRC); its amplitude ratio is 1/sqrt(1+(f/f_c)²) and its phase is −arctan(f/f_c). At f=f_c the amplitude is $1/\sqrt{2}$ (the −3 dB point), not one half. For R=1 kΩ and C=100 nF, f_c≈1.59 kHz.

Detector efficiency ε is detected events divided by incident events; energy resolution is quoted as FWHM/E, such as 13.2 keV on a 662 keV line, about 2%. A non-paralyzable detector with dead time τ records m=n/(1+nτ) from true rate n; inverting gives n=m/(1−mτ), which fails as mτ→1. State which dead-time model you assume; paralyzable behaviour differs.

A linear calibration against known standards can correct offset and scale error; it does not remove noise or guarantee correction of nonlinearity. A calibration curve maps indicated to true values; interpolation between points assumes local smoothness. Report resolution and efficiency separately: sharp peaks with poor efficiency still miss events, and high efficiency with broad resolution still mixes nearby lines.

A voltmeter of finite input resistance forms a divider with the source resistance.

For a nonparalysable detector with observed rate m and dead time tau, Efficiency, dead time and spectral resolution are different corrections; one cannot replace the others.

Treating the $1/\sqrt{2}$ cutoff as one half, ignoring that a finite meter reads low through the divider, or applying the non-paralyzable correction beyond its range.

dead time: The minimum interval after one recorded event during which the detector cannot record another.

input impedance 输入阻抗: The effective load a measuring instrument presents to the circuit under test.

A photon beam has no definite range, yet a heavy charged particle does; shielding calculations fail if the two are swapped.

Prerequisites: 7, 17, 21.

• Distinguish charged-particle stopping and range from exponential photon attenuation
• Compute attenuated intensity and detected counts with stated coefficients
• Explain stimulated emission, population inversion 粒子数反转 and optical cavity resonance conditions

Heavy charged particles lose energy continuously through many small collisions, so they have an approximate range: under a constant-loss model a particle with initial energy E0 and loss rate dE/dx stops after E0/(dE/dx). Electrons straggle more and radiate. Photons instead interact in single events, so a narrow beam attenuates exponentially as I=I0 e^(−μx) with no definite maximum depth; the half-value layer 半值层 is ln2/μ.

Apply the exponential only to the stated narrow-beam geometry: μ=0.2 /cm gives a half-value layer of 3.47 cm and I/I0=e^(−2)≈0.135 after 10 cm. Multiply by detector efficiency for recorded counts: with μx=ln4 and ε=0.25 the recorded fraction is 0.25×1/4=1/16. Build-up from scattered photons makes broad-beam shielding transmit more than the narrow-beam exponential; state the geometry.

Stimulated emission produces a photon matching the stimulating photon in frequency, direction and phase, giving coherent amplification. It competes with absorption; net gain needs population inversion N2/g2>N1/g1 (N2>N1 for equal degeneracies). Positive-temperature equilibrium has N2/g2<N1/g1 by the Boltzmann factor, so a two-level system in equilibrium cannot lase continuously; practical lasers pump a third level or a metastable state.

A linear cavity of length L supports standing modes with L=mλ/2, i.e. frequencies ν_m=mc/(2L) with spacing c/(2L): for L=30 cm the spacing is 500 MHz. Gain must exceed the round-trip losses for oscillation. Interferometers such as Michelson use the same coherence: fringe counts track optical path changes, as in the gas-cell measurement.

A narrow beam has attenuation $I=I_0e^{-\mu x}$ and half-value layer $x_{1/2}=\ln2/\mu$.

Three half-value layers transmit 1/8. This model excludes scattered build-up into the detector. For a linear vacuum laser cavity, adjacent longitudinal modes are separated by $c/(2L)$. Net stimulated gain requires $N_2/g_2>N_1/g_1$; the simpler $N_2>N_1$ assumes equal degeneracies.

Assigning photons a definite range or charged particles a single exponential law, and forgetting that equilibrium two-level populations cannot invert. Check which interaction model the beam species requires.

half-value layer: The material thickness that halves an exponentially attenuated beam, equal to ln2/μ.

population inversion: A non-equilibrium state with more population in the upper laser level than the lower, enabling net stimulated emission.

Rotating a crystal in an X-ray beam maps its atomic planes, because constructive reflection happens only at discrete angles.

Prerequisites: 8, 36.

• Apply the Bragg condition 布拉格条件 to find plane spacings or diffraction angles
• Estimate free-electron Fermi energy 费米能 and Fermi temperature
• Distinguish electronic and lattice heat-capacity contributions

Bragg reflection from parallel planes spaced d interferes constructively when 2d sinθ=nλ, with θ measured from the plane surface (the glancing angle), not from the normal. First order n=1 gives the smallest angle. For λ=0.154 nm and a peak at θ=20°, d=0.154/(2 sin20°)≈0.225 nm. If 2d<λ no order exists; the sine must stay at or below 1.

In a cubic crystal the (hkl) plane spacing is d=a/sqrt(h²+k²+l²) with lattice constant a; the (110) planes give a/√2. A measured Bragg angle therefore measures the lattice constant. Systematic absences from the basis (for example body-centred lattices missing h+k+l odd) carry structure information beyond the spacing formula.

The free-electron model packs N conduction electrons into states up to the Fermi wavevector k_F=(3π²n)^(1/3) for number density n, giving E_F=$\hbar$²k_F²/(2m). For n=8.5×10²⁸ /m³, k_F≈1.36×10¹⁰ /m, E_F≈7.05 eV and T_F=E_F/k_B≈8.2×10⁴ K. Since T_F far exceeds room temperature, only a fraction of order T/T_F of the electrons are thermally excited.

Metals therefore have two low-temperature heat-capacity terms: an electronic term γT linear in T and a lattice (Debye) term βT³. Plotting C/T against T² gives intercept γ and slope β. At room temperature the lattice part approaches the classical Dulong–Petit value and dominates; the electronic term matters at low T, where the T³ lattice term collapses faster.

Bragg diffraction uses angle theta from the crystal plane, not from its normal.

Low-temperature metal heat capacity has separate electronic and lattice terms: A plot of C/T against T squared has intercept gamma and slope beta. A straight line identifies this model's regime.

Measuring θ from the plane normal instead of the plane, or assigning the whole low-T heat capacity to electrons. The sine condition and the two-term decomposition are where these calculations fail.

Bragg condition: Constructive reflection from lattice planes when 2d sinθ=nλ, with θ measured from the plane.

Fermi energy: The highest occupied electron energy at zero temperature in the free-electron model.

One atomic mass unit is 931.5 MeV, so a reaction energy is a small difference between large masses.

Prerequisites: 8, 25, 21.

• Compute reaction energy from complete atomic or nuclear mass bookkeeping
• Calculate activity from half-life and a simple decay-chain balance
• Compare fission and fusion binding changes with charge and nucleon conservation

The reaction energy is Q=(m_initial−m_final)c²; Q>0 releases kinetic energy. Track every species, and keep the mass convention consistent: tabulated atomic masses include electrons. For β− decay and for α decay with a neutral helium product, atomic electron counts balance directly; β+ or electron capture needs explicit electron-mass terms. Use 1 u=931.5 MeV/c².

Alpha decay of ²³⁸U with atomic masses 238.0508 u → 234.0436 u + 4.0026 u gives Δm=0.0046 u and Q≈4.28 MeV. Fusion D+T with 2.0141+3.0161−4.0026−1.0087=0.0189 u gives Q≈17.6 MeV. A reaction also requires charge and nucleon-number balance: ¹⁴N+α→¹⁷O+p balances 7+2=9 and 14+4=18.

Activity is A=λN with decay constant λ=ln2/T½; one becquerel is one decay per second. For T½=2 h and N=6×10¹², A=6×10¹²×ln2/7200≈5.8×10⁸ Bq. After three half-lives both N and A fall by 1/8. Equal activities do not imply equal atom counts: shorter half-life at fixed N gives larger A.

In a chain A→B→C with a long-lived parent (λ_A≪λ_B), B accumulates until its activity approaches the parent activity (secular equilibrium); write the balance dN_B/dt=λ_A N_A−λ_B N_B. The binding-energy-per-nucleon curve peaks near iron, so heavy fission (about 7.6 to 8.5 MeV per nucleon over roughly 240 nucleons, of order 200 MeV) and light fusion both release energy; compare total binding, not nucleon counts.

Activity is $A=\lambda N$, with $\lambda=\ln2/T_{1/2}$. For N = $6.0\times10^{12}$ and half-life 2 hours, convert time before calculating.

A decay chain requires both production and loss terms: $\dot N_B=\lambda_AN_A-\lambda_BN_B$. Equal activities do not imply equal populations. Atomic-mass beta-plus Q values require subtracting $2m_ec^2$.

Mixing atomic and nuclear masses without tracking electrons, or reporting a positive Q for a reaction that needs energy input. Balance charge and nucleon number first, then check the sign of Q.

Q value Q值: The rest-mass energy converted to kinetic energy in a reaction, (m_initial−m_final)c².

activity: The decay rate λN of a radioactive sample, measured in becquerel.

Writing the Laplacian of a radial function as f″ drops a term; the operator belongs to the coordinates, not the symbol.

Prerequisites: Algebra, differentiation and vector components.

• Use the divergence 散度 and Laplacian in the coordinate system selected by the symmetry
• Solve a simple separated boundary-value mode problem
• Check convergence, units and stability of an original numerical update

Choose coordinates by symmetry before writing operators. For a radial function in spherical coordinates, ∇²f=f″+(2/r)f′, and the divergence of a radial field A=A_r r̂ is (1/r²)d(r²A_r)/dr. Check: A=k r̂/r² has divergence (1/r²)dk/dr=0 away from the origin, and f=1/r satisfies ∇²f=f″+2f′/r=2/r³−2/r³=0 away from the origin. Cartesian forms apply only to Cartesian dependences.

Separation of variables 分离变量法 turns a boundary-value problem into ordinary modes. For a string fixed at both ends, write u=X(x)T(t); the wave equation gives X″/X=(1/c²)T″/T=−k², the fixed ends force X=sin(nπx/L), and the allowed frequencies are ω_n=nπc/L, f_n=nc/(2L). Each mode must satisfy the boundary conditions before any sum over modes.

An explicit Euler update for y′=−λy is y_{n+1}=(1−λh)y_n, bounded when |1−λh|≤1, i.e. 0<h≤2/λ for λ>0; asymptotic decay needs 0<h<2/λ. With λ=10 /s, h must not exceed 0.2 s; h=0.25 s multiplies by −1.5 each step and diverges with alternating sign. Stability is a property of the recurrence, not of the true solution.

Check a numerical answer three ways before trusting it: units of every term (λh must be dimensionless), convergence under refinement (halve h and compare; explicit Euler error is first order), and a limiting case with a known exact answer. For y′=−4y with y0=8 and h=0.2, five steps give 8×0.2⁵=0.00256 versus exact 8e^(−4)≈0.1465: stable but inaccurate, so refine.

A spherical radial field has divergence

For $A_r=Cr$, it gives 3C, not C. Fixed string ends require $X(0)=X(L)=0$, hence $X_n=\sin(n\pi x/L)$ with positive integer n. Euler for $\dot y=-\lambda y$ gives $y_{n+1}=(1-\lambda h)y_n$. For positive h, bounded absolute stability permits $h\leq2/\lambda$; decay to zero requires the strict inequality $0<h<2/\lambda$. At the upper endpoint the numerical solution alternates without decay. Stability does not establish accuracy.

Applying the Cartesian Laplacian to a radial function, summing modes that violate the boundary conditions, or calling a bounded-looking run converged without refining the step. Match operator, boundaries and stability condition 稳定性条件 to the actual problem.

separation of variables: Solving a partial differential equation by writing the unknown as a product of single-variable factors.

stability condition: The step-size restriction that keeps a numerical recurrence from amplifying errors.