📚 International AS and A Level Physics Formula Derivations | 国际AS与A Level物理公式推导
Understanding how key formulas are derived is fundamental to mastering A Level Physics. Rather than memorising equations in isolation, seeing the logical steps behind them strengthens your ability to apply concepts to unfamiliar problems. This article walks through essential derivations from mechanics, waves, electricity, and quantum physics, all of which appear regularly in International AS and A Level specifications. Each derivation is broken down into clear steps with paired English and Chinese explanations to support bilingual learning.
理解核心公式的推导过程是掌握 A Level 物理的基础。孤立地记忆方程而不知其所以然，难以灵活应对陌生题目。本文将带你走过力学、波动、电学和量子物理中常见的重要推导，这些内容频繁出现在国际 AS 与 A Level 考纲中。每个推导都分解为清晰的步骤，并配有中英双语解释，助你扎实掌握。
1. Deriving the SUVAT Equations | 匀变速运动公式推导
For motion with constant acceleration a, we start with the definition: a = (v − u) / t, where u is initial velocity, v is final velocity, and t is time. Rearranging gives the first SUVAT equation.
对于加速度 a 恒定的运动，我们从定义出发：a = (v − u) / t，其中 u 为初速度，v 为末速度，t 为时间。整理后得到第一个 SUVAT 方程。
v = u + a t
Average velocity when acceleration is constant is s / t = (u + v) / 2. Substituting v from above gives s = ((u + u + a t)/2) t, which simplifies to s = u t + ½ a t².
匀加速运动中的平均速度为 s / t = (u + v) / 2。代入上述 v 的表达式得到 s = ((u + u + a t)/2) t，化简后即为 s = u t + ½ a t²。
s = u t + ½ a t²
Alternatively, eliminating t by expressing t = (v − u)/a from the first equation and substituting into s = (u + v)/2 × t yields v² = u² + 2 a s.
另一种方法，从第一个方程得 t = (v − u)/a，代入 s = (u + v)/2 × t 消去 t，可得 v² = u² + 2 a s。
v² = u² + 2 a s
These three core equations assume uniform acceleration in a straight line.
这三个核心方程均假设物体沿直线做匀加速运动。
2. Derivation of Kinetic Energy Formula | 动能公式推导
Consider a constant net force F acting on an object of mass m over a displacement s. The work done is W = F s. Using Newton’s second law F = m a and the SUVAT relation v² = u² + 2 a s, we can eliminate a and s.
考虑一个恒定的净力 F 作用在质量为 m 的物体上，位移为 s。力所做的功为 W = F s。利用牛顿第二定律 F = m a 和运动学关系 v² = u² + 2 a s，可消去 a 和 s。
From v² = u² + 2 a s, we have a s = (v² − u²) / 2. Therefore W = m a s = ½ m (v² − u²).
由 v² = u² + 2 a s 可得 a s = (v² − u²) / 2。因此 W = m a s = ½ m (v² − u²)。
If the object starts from rest (u = 0), work done equals ½ m v². This quantity is defined as kinetic energy Eₖ.
若物体从静止开始 (u = 0)，则功等于 ½ m v²。这个量被定义为动能 Eₖ。
Eₖ = ½ m v²
3. Centripetal Acceleration Derivation | 向心加速度公式推导
An object moving at constant speed v in a circle of radius r undergoes centripetal acceleration directed towards the centre. Consider a short time interval Δt. The velocity vector changes direction but not magnitude.
物体以恒定速率 v 在半径为 r 的圆周上运动时，会产生指向圆心的向心加速度。考虑极短时间 Δt，速度矢量方向改变但大小不变。
The two velocity vectors separated by angle Δθ form an isosceles triangle. The change in velocity Δv has magnitude v Δθ for small angles. The distance travelled is v Δt = r Δθ, so Δθ = v Δt / r.
两个速度矢量间隔角度 Δθ 构成等腰三角形。对于极小角度，速度变化量 Δv 的大小为 v Δθ。经过的路程为 v Δt = r Δθ，故 Δθ = v Δt / r。
Acceleration magnitude a = Δv / Δt = (v Δθ) / Δt = v × (v / r) = v² / r. Using v = ω r gives a = ω² r.
加速度大小 a = Δv / Δt = (v Δθ) / Δt = v × (v / r) = v² / r。代入 v = ω r 得 a = ω² r。
a = v² / r = ω² r
4. Derivation of Gravitational Potential Energy | 引力势能公式推导
The gravitational potential energy U of two point masses M and m separated by distance r is defined as the work done to bring them from infinity to that separation. The gravitational force is F = G M m / r².
两个点质量 M 与 m 相距 r 时的引力势能 U，定义为将它们从无穷远处移至该距离时外力克服引力所做的功。引力大小 F = G M m / r²。
Work done against gravity moving a small distance dr is dW = F dr = (G M m / r²) dr. Integrating from r = ∞ to r = R gives U = −G M m / R.
克服引力移动微小距离 dr 所做的功为 dW = F dr = (G M m / r²) dr。从 r = ∞ 积分至 r = R 得到 U = −G M m / R。
U = − G M m / r
The negative sign indicates that work is done by the gravitational field as masses come together; potential energy decreases.
负号表示当质量相互靠近时，引力场对外做正功，势能减小。
5. Simple Harmonic Motion Equations | 简谐运动方程推导
SHM occurs when the restoring force is proportional to displacement from equilibrium and opposite in direction: F = −k x. Using F = m a, we obtain a = − (k/m) x = −ω² x.
当恢复力与离开平衡位置的位移成正比且方向相反时，物体做简谐运动：F = −k x。由 F = m a 得 a = − (k/m) x = −ω² x。
The solution to this differential equation is x = A cos(ω t + φ) or x = A sin(ω t + φ₀). Differentiating twice confirms it satisfies a = −ω² x.
这个微分方程的解为 x = A cos(ω t + φ) 或 x = A sin(ω t + φ₀)。求导两次验证其满足 a = −ω² x。
x = A cos(ω t + φ)
Velocity v = dx/dt = −A ω sin(ω t + φ), and maximum speed is vₘₐₓ = ω A. Acceleration a = −A ω² cos(ω t + φ) = −ω² x.
速度 v = dx/dt = −A ω sin(ω t + φ)，最大速率为 vₘₐₓ = ω A。加速度 a = −A ω² cos(ω t + φ) = −ω² x。
6. Capacitor Discharge Formula | 电容器放电公式推导
For a capacitor of capacitance C discharging through a resistor R, the potential difference V and charge Q are related by Q = C V. From Kirchhoff’s voltage law, V = I R with I = −dQ/dt (negative because charge decreases).
对于电容 C 通过电阻 R 放电，电压 V 与电荷 Q 满足 Q = C V。由基尔霍夫电压定律，V = I R，且 I = −dQ/dt（负号表示电荷减少）。
Substituting gives Q / C = −R (dQ/dt). Rearranging: dQ / Q = − (1 / RC) dt. Integrating both sides yields ln Q = − t / (RC) + constant.
代入得 Q / C = −R (dQ/dt)。整理得 dQ / Q = − (1/RC) dt。两边积分：ln Q = − t/(RC) + 常数。
At t = 0, Q = Q₀, so the constant is ln Q₀. Hence Q = Q₀ e⁻ᵗ/ᴿᶜ. The same exponential decay applies to current and voltage.
在 t = 0 时 Q = Q₀，故常数为 ln Q₀。因此 Q = Q₀ e⁻ᵗ/ᴿᶜ。同样的指数衰减规律适用于电流和电压。
Q = Q₀ e⁻ᵗ/ᴿᶜ, V = V₀ e⁻ᵗ/ᴿᶜ, I = I₀ e⁻ᵗ/ᴿᶜ
7. Deriving the Diffraction Grating Equation | 衍射光栅方程推导
A diffraction grating with slit spacing d causes constructive interference when the path difference between adjacent slits equals an integer multiple of wavelength λ. Consider two parallel rays incident normally.
缝间距为 d 的衍射光栅，当相邻狭缝的光程差等于波长 λ 的整数倍时发生相长干涉。考虑正入射的两束平行光线。
The path difference for light diffracted at angle θ to the normal is d sin θ. For a maximum, d sin θ = n λ, where n = 0, 1, 2, …
衍射角为 θ（与法线的夹角）时，光程差为 d sin θ。极大值条件为 d sin θ = n λ，其中 n = 0, 1, 2, …
d sin θ = n λ
This equation allows calculation of wavelength or grating spacing from measured angles of bright fringes.
利用此方程，可从测得的亮纹角度计算波长或光栅常数。
8. Derivation of Magnetic Force on a Current-Carrying Wire | 载流导线所受磁力公式推导
A straight wire of length L carrying current I in a uniform magnetic field B experiences a force. Current is flow of charge: I = Q / t. If charges drift with velocity v, then Q = n e A L and t = L / v, so I = n e A v.
长 L 的直导线载有电流 I，置于匀强磁场 B 中会受到安培力。电流即电荷流动：I = Q / t。若电荷漂移速度为 v，则 Q = n e A L，t = L / v，因此 I = n e A v。
The total number of charge carriers in the wire is N = n A L. Each carrier experiences a Lorentz force F₀ = e v B for perpendicular v and B. Total force F = N e v B = (n A L) e v B.
导线中的总载流子数为 N = n A L。每个载流子受到洛伦兹力 F₀ = e v B（当 v 与 B 垂直时）。总力 F = N e v B = (n A L) e v B。
Substituting I = n e A v simplifies to F = I L B. If the wire is at an angle θ to the field, the perpendicular component gives F = B I L sin θ.
代入 I = n e A v 化简得 F = I L B。若导线与磁场成 θ 角，则垂直分量给出 F = B I L sin θ。
F = B I L sin θ
9. Deriving Transformer EMF Equation | 变压器电动势公式推导
A transformer works on the principle of electromagnetic induction. An alternating current in the primary coil creates a changing magnetic flux Φ. According to Faraday’s law, the induced emf per turn is ε = − dΦ/dt.
变压器基于电磁感应原理工作。初级线圈的交变电流产生变化的磁通量 Φ。根据法拉第定律，每匝线圈的感应电动势为 ε = − dΦ/dt。
If the same flux links both coils (ideal transformer), the primary emf Vₚ has Nₚ turns and secondary Vₛ has Nₛ turns. Thus Vₚ = − Nₚ dΦ/dt and Vₛ = − Nₛ dΦ/dt.
若相同磁通量穿过两线圈（理想变压器），初级电压 Vₚ 涉及 Nₚ 匝，次级电压 Vₛ 涉及 Nₛ 匝。因此 Vₚ = − Nₚ dΦ/dt，Vₛ = − Nₛ dΦ/dt。
Dividing the two equations eliminates dΦ/dt, yielding the turns ratio relationship.
两式相除消去 dΦ/dt，得到匝数比关系。
Vₚ / Vₛ = Nₚ / Nₛ
For an ideal transformer with no power loss, Vₚ Iₚ = Vₛ Iₛ, so Iₛ / Iₚ = Nₚ / Nₛ.
对于无功率损耗的理想变压器，有 Vₚ Iₚ = Vₛ Iₛ，因此 Iₛ / Iₚ = Nₚ / Nₛ。
10. Photoelectric Effect Equation Derivation | 光电效应方程推导
Einstein explained the photoelectric effect by proposing that light consists of photons, each with energy E = h f. When a photon strikes a metal surface, its energy is used to overcome the work function φ and give kinetic energy to the emitted electron.
爱因斯坦用光子假说解释了光电效应，提出光由光子组成，每个光子能量 E = h f。光子撞击金属表面时，其能量一部分用于克服功函数 φ，剩余部分转化为出射电子的动能。
By energy conservation: h f = φ + Kₘₐₓ, where Kₘₐₓ = ½ m v²ₘₐₓ is the maximum kinetic energy of photoelectrons.
根据能量守恒：h f = φ + Kₘₐₓ，其中 Kₘₐₓ = ½ m v²ₘₐₓ 是光电子的最大动能。
The stopping potential V₀ satisfies e V₀ = Kₘₐₓ. Hence h f = φ + e V₀, enabling experimental determination of h and φ.
遏止电压 V₀ 满足 e V₀ = Kₘₐₓ。因此 h f = φ + e V₀，可用于实验测定普朗克常数 h 和功函数 φ。
h f = φ + ½ m v²ₘₐₓ
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