📚 Derivation of Key Formulae in PH03 International Physics A | PH03国际物理A卷关键公式推导
The PH03 International Physics A paper tests your ability to apply fundamental principles to unfamiliar situations. A deep understanding of where key formulae come from, not just how to use them, is essential for top marks. In this article, we derive the most important equations step by step, linking physical reasoning with mathematical rigour.
PH03国际物理A卷考察的是你将基本原理应用到陌生情境的能力。要想拿到高分，不仅要会用公式，更要深刻理解这些关键公式从何而来。本文将一步步推导最重要的方程，把物理推理与数学严谨性结合起来。
1. Introduction to the PH03 Paper and Formula Derivation | PH03试卷与公式推导导言
The PH03 paper rewards logical structure. When you derive a formula, you show examiners that you can connect definitions, laws, and boundary conditions. Every derivation in this article follows a clear pattern: state the starting principle, justify each algebraic step, and interpret the final result physically.
PH03试卷看重逻辑结构。当你推导一个公式时，你是在向考官展示你有能力将定义、定律和边界条件串联起来。本文中每一个推导都遵循清晰的模式：陈述起始原理，论证每一步代数操作，并从物理角度诠释最终结果。
We will cover kinematics, dynamics, energy, fields, oscillations, and quantum phenomena – the core topics of the syllabus. Read each section as a worked example and then attempt similar derivations on your own.
我们将涉及运动学、动力学、能量、场、振动和量子现象——这些是教学大纲的核心主题。请把每一节当做一个完整的例题来读，然后自己尝试类似的推导。
2. Kinematics Equations for Uniform Acceleration | 匀加速运动学方程
Start with the definition of acceleration as the rate of change of velocity: a = (v – u) / t, where u is initial velocity, v is final velocity, and t is time taken. Rearranging gives the first equation:
从加速度是速度变化率的定义出发：a = (v – u) / t，其中u为初速度，v为末速度，t为所用时间。重新整理就得到第一个方程：
v = u + at
To find displacement s, use average velocity: with constant acceleration, average velocity = (u + v) / 2. Displacement is average velocity times time:
为求位移s，使用平均速度：匀加速时平均速度 = (u + v) / 2。位移等于平均速度乘以时间：
s = (u + v) t / 2
Substitute v from the first equation into the displacement expression:
将第一个方程中的v代入位移表达式：
s = (u + u + at) t / 2 = (2u + at) t / 2 = u t + ½ a t²
Hence the second kinematics equation: s = u t + ½ a t². 因此第二个运动学方程：s = u t + ½ a t²。
To eliminate t, rearrange v = u + at to t = (v – u)/a and substitute into s = (u+v)t/2:
消去t：将v = u + at改写为t = (v – u)/a，代入s = (u+v)t/2：
s = (u + v)(v – u) / (2a) = (v² – u²) / (2a)
Thus, v² = u² + 2 a s. This set of three equations is fundamental for solving uniformly accelerated linear motion.
于是得到 v² = u² + 2 a s。这三个方程是解决匀加速直线运动问题的基础。
3. Newton’s Second Law and Impulse-Momentum Theorem | 牛顿第二定律与冲量-动量定理
Newton’s second law in its most useful form is F = d p / d t, where p = m v is the linear momentum. For constant mass, this becomes F = m (d v / d t) = m a. But the F = d p / d t formulation is more general, applying to systems where mass changes, such as a rocket ejecting fuel.
牛顿第二定律最有用的形式是F = d p / d t，其中p = m v是线动量。当质量不变时，它变成F = m (d v / d t) = m a。但F = d p / d t的形式更普遍，适用于质量变化的系统，比如火箭喷射燃料。
Rearrange and integrate over a time interval Δt: ∫ F d t = Δ p. The left side is defined as the impulse J. Therefore, J = Δ p = m v – m u. This is the impulse-momentum theorem: the impulse delivered by a net force equals the change in momentum of the object.
重新整理并对时间间隔Δt积分：∫ F d t = Δ p。左边定义为冲量J。因此，J = Δ p = m v – m u。这就是冲量-动量定理：合力提供的冲量等于物体动量的变化量。
4. Work-Energy Theorem and Kinetic Energy | 功-能定理与动能
Work done by a net force F over a displacement Δs in one dimension is W = F Δ s. Using F = m a and the kinematic result v² = u² + 2 a s, we can eliminate acceleration a.
合力F在一维位移Δs上做的功为W = F Δ s。利用F = m a以及运动学结果v² = u² + 2 a s，可以消去加速度a。
From v² – u² = 2 a s we get a s = (v² – u²) / 2. Substituting gives:
由v² – u² = 2 a s可得a s = (v² – u²) / 2。代入得：
W = m × (v² – u²) / 2 = ½ m v² – ½ m u²
Define kinetic energy K = ½ m v². Then the work-energy theorem states: W = Δ K. This theorem holds for constant forces in one dimension and is readily generalised in integral form W = ∫ F · d s = Δ K for variable forces.
定义动能K = ½ m v²，那么功-能定理即为：W = Δ K。此定理对一维恒力成立，对于变力，可以很方便地推广成积分形式W = ∫ F · d s = Δ K。
5. Gravitational Potential Energy and Escape Velocity | 引力势能与逃逸速度
Gravitational force between two point masses M and m separated by distance r is F = – G M m / r², where the negative sign indicates attraction. The change in potential energy when moving from separation r₁ to r₂ is the negative of the work done by the gravitational field:
两个点质量M与m相距r时的引力为F = – G M m / r²，负号表示吸引力。当两者间距从r₁变到r₂时，势能的变化量等于引力场做功的负值：
Δ U = – ∫_{r₁}^{r₂} F d r = – ∫_{r₁}^{r₂} (– G M m / r²) d r = G M m ∫_{r₁}^{r₂} r⁻² d r
Integrate and choose the zero of potential at infinity, U(∞) = 0. Then for any r,
积分，并规定无穷远处势能为零，U(∞) = 0，则对任意r，
U(r) = – G M m / r
Escape velocity vₑ is the minimum speed an object needs to escape a planet’s gravity from its surface (radius R). Use conservation of energy: total energy at surface equals total energy at infinity (where K = 0, U = 0).
逃逸速度vₑ是物体从行星表面（半径R）出发、刚好能摆脱引力的最小速率。利用能量守恒：物体在表面的总能量等于无穷远处的总能量（此时K = 0, U = 0）。
½ m vₑ² + (– G M m / R) = 0 ⇒ vₑ = √(2 G M / R)
This result is independent of the escaping object’s mass.
这一结果与逃逸物体的质量无关。
6. Simple Harmonic Motion Displacement Equation | 简谐运动的位移方程
Simple harmonic motion (SHM) arises when the restoring force is proportional to displacement and directed towards equilibrium: F = – k x. Using Newton’s second law, m a = – k x, giving the defining differential equation:
当回复力与位移成正比且指向平衡位置时，就发生简谐运动（SHM）：F = – k x。利用牛顿第二定律m a = – k x，得到定义性的微分方程：
d²x/dt² = – (k/m) x
Define the angular frequency ω = √(k/m), so the equation becomes d²x/dt² = – ω² x. A general solution is x = A sin(ω t) + B cos(ω t), or equivalently, x = A₀ sin(ω t + φ) where amplitude A₀ and phase constant φ depend on initial conditions.
定义角频率ω = √(k/m)，方程变为d²x/dt² = – ω² x。其一般解为x = A sin(ω t) + B cos(ω t)，或者等价地写成x = A₀ sin(ω t + φ)，其中振幅A₀和初相φ由初始条件决定。
If at t = 0, the particle is at maximum displacement A with zero velocity, we obtain x = A cos(ω t). The period T satisfies ω T = 2π, hence T = 2π/ω = 2π √(m/k).
如果t = 0时物体位于最大位移A处且速度为零，就得到x = A cos(ω t)。周期T满足ω T = 2π，因此T = 2π/ω = 2π √(m/k)。
7. Capacitor Discharge Equation | 电容器放电方程
Consider a capacitor C discharged through a resistor R. The charge Q on the capacitor decreases with time. The current I is the rate of decrease of charge: I = – d Q / d t. Ohm’s law across the resistor gives V = I R, and the capacitor voltage is V = Q / C.
考虑电容器C通过电阻R放电。电容器上的电荷Q随时间减少。电流I是电荷减少的速率：I = – d Q / d t。通过电阻的欧姆定律给出V = I R，而电容器电压为V = Q / C。
Equating voltages around the closed loop: Q / C = I R = – R d Q / d t. Separate variables:
对于闭合回路，电压相等：Q / C = I R = – R d Q / d t。分离变量：
d Q / Q = – (1 / R C) d t
Integrate from initial charge Q₀ at t = 0 to charge Q at time t:
从t = 0时初始电荷Q₀积分到t时刻电荷Q：
ln(Q / Q₀) = – t / (R C)
Exponentiate both sides to obtain the exponential decay law:
两边取指数得到指数衰减规律：
Q = Q₀ e^(− t / R C)
The product R C is called the time constant, after which the charge drops to about 37% of its initial value.
乘积R C称为时间常数，经过一个时间常数后，电荷下降到初始值的约37%。
8. Magnetic Force on a Moving Charge | 运动电荷在磁场中的受力
Experimental evidence shows a moving charge q in a magnetic field experiences a force perpendicular to both its velocity v and the magnetic field B. The Lorentz force law states:
实验证据显示，运动电荷q在磁场中会受到一个既垂直于速度v、又垂直于磁场B的力。洛伦兹力定律表述为：
F = q (v × B)
The magnitude of this force is F = q v B sin θ, where θ is the angle between v and B. The direction is given by the right-hand rule for positive charges (and opposite for negative charges).
此力的大小为F = q v B sin θ，其中θ是v与B之间的夹角。对于正电荷，方向由右手定则判定（负电荷则相反）。
If the velocity is entirely perpendicular to a uniform magnetic field, θ = 90° and sin θ = 1, so F = q v B. This force provides the centripetal force for circular motion: q v B = m v² / r, giving the radius r = m v / (q B) and period T = 2π m / (q B), independent of speed.
若速度完全垂直于匀强磁场，θ = 90°且sin θ = 1，则F = q v B。这个力充当圆周运动的向心力：q v B = m v² / r，由此得半径r = m v / (q B)，周期T = 2π m / (q B)，与速率无关。
9. Faraday’s Law of Electromagnetic Induction | 法拉第电磁感应定律推导
Faraday’s law states that the induced emf in a loop is proportional to the negative rate of change of magnetic flux: ε = – d Φ / d t. Consider a conducting rod of length L moving at speed v perpendicular to a uniform magnetic field B. The magnetic force on a free electron is F = e v B, causing charge separation. The electric field builds up until e E = e v B, so the induced emf across the rod is ε = E L = B L v.
法拉第定律声明，回路中的感应电动势与磁通量的负变化率成正比：ε = – d Φ / d t。考虑一根长度为L的导体棒以速率v垂直于匀强磁场B运动。自由电子受到磁力F = e v B，引起电荷分离。电场累积直到e E = e v B，因此棒两端的感应电动势为ε = E L = B L v。
Now, the flux swept out by the moving rod in time Δ t is Δ Φ = B × (L v Δ t), so Δ Φ / Δ t = B L v. In the limit Δ t → 0, d Φ / d t = B L v. The minus sign in Faraday’s law relates to Lenz’s law: the induced current opposes the change in flux that produced it.
现在，在时间Δ t内，运动棒扫过的磁通量为Δ Φ = B × (L v Δ t)，因此Δ Φ / Δ t = B L v。当Δ t → 0时，d Φ / d t = B L v。法拉第定律中的负号与楞次定律相关：感应电流的方向总是对抗产生它的磁通量变化。
10. Photoelectric Effect Equation | 光电效应方程
Einstein explained the photoelectric effect by proposing that light consists of photons, each with energy E = h f (where h is Planck’s constant and f is frequency). When a photon strikes a metal surface, its energy can be absorbed by an electron. If the photon energy exceeds the work function φ of the metal, the electron is ejected with maximum kinetic energy:
爱因斯坦解释光电效应时提出，光由光子组成，每个光子的能量为E = h f（其中h是普朗克常数，f是频率）。当光子撞击金属表面时，其能量可能被电子吸收。如果光子能量大于金属的逸出功φ，电子就会被击出，并具有最大动能：
Kₘₐₓ = h f – φ
The work function φ is the minimum energy needed to remove an electron from the metal. The stopping potential Vₛ is related by e Vₛ = Kₘₐₓ, so e Vₛ = h f – φ. A graph of Vₛ against f gives a straight line with slope h / e and intercept – φ / e on the Vₛ axis, confirming the photon model.
逸出功φ是从金属中移出一个电子所需的最小能量。遏止电压Vₛ满足e Vₛ = Kₘₐₓ，故e Vₛ = h f – φ。画出Vₛ对f的图像是一条直线，斜率为h / e，在Vₛ轴上的截距为– φ / e，这证实了光子模型。
11. Summary – Derivation as a Learning Tool | 小结 – 让推导成为学习利器
The derivations shown here are not merely mathematical exercises; they reinforce the conceptual links that PH03 exam questions frequently target. By working through each step, you sharpen your ability to manipulate symbols, set up integral boundaries, and recognise the physical meaning behind every term.
以上推导并非单纯的数学练习；它们强化了PH03试题常考的概念联系。通过逐步推演，你能提高符号运算能力、学会设定积分边界，并理解每一项背后的物理含义。
We recommend that you reproduce each derivation from memory, then adapt the arguments for slightly different scenarios (e.g., a charged particle in crossed electric and magnetic fields, or a pendulum with large amplitude). This active practice is the surest route to exam success.
我们建议你先凭记忆重现每个推导，然后再将推导思路稍作调整，适应略微不同的情境（例如，带电粒子在正交电磁场中，或者大振幅摆动）。这种主动练习是通往考试成功的最可靠途径。
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