📚 Essential Formula Derivations from Pearson’s IB Physics HL Textbook | IB 物理 HL Pearson 教材核心公式推导
Understanding the mathematical origins of key equations is fundamental to mastering IB Physics HL. The Pearson Baccalaureate textbook presents derivations that link physical concepts with rigorous algebra and calculus. This article walks through ten essential derivations, explaining each step in both English and Chinese to strengthen your conceptual and analytical skills.
理解关键方程的数学源头是掌握 IB 物理 HL 的基础。Pearson Baccalaureate 教材提供了将物理概念与严谨的代数和微积分联系起来的推导过程。本文逐一讲解十个核心推导，用中英双语解释每一步，以强化你的概念与分析能力。
1. Deriving the SUVAT Equations | 匀加速运动方程推导
Starting with the definition of constant acceleration a, we have a = (v – u)/t, where u is initial velocity, v is final velocity, and t is time. Rearranging gives the first SUVAT equation.
从恒定加速度 a 的定义出发，有 a = (v – u)/t，其中 u 为初速度，v 为末速度，t 为时间。移项得到第一个运动学方程。
v = u + at
The average velocity for uniform acceleration is v_avg = (u + v)/2. Displacement s equals average velocity multiplied by time, so s = ((u + v)/2)t. Substituting v = u + at yields s = ut + ½at².
匀加速运动的平均速度 v_avg = (u + v)/2。位移 s 等于平均速度乘以时间，因此 s = ((u + v)/2)t。代入 v = u + at 得到 s = ut + ½at²。
s = ut + ½at²
Combining the first two equations by eliminating t gives v² = u² + 2as, which is particularly useful when time is not involved.
联立前两个方程并消去 t 得到 v² = u² + 2as，这在无需时间的场景中尤其有用。
v² = u² + 2as
2. Deriving Centripetal Acceleration a = v²/r | 向心加速度公式推导
Consider an object moving at constant speed v in a circle of radius r. In a short time Δt, the object moves through a small angle Δθ. The velocity vector changes from v₁ to v₂, both of magnitude v. The change in velocity Δv points toward the centre, and its magnitude is approximately vΔθ for small angles.
考虑物体以恒定速率 v 在半径为 r 的圆周上运动。在短时间 Δt 内，物体转过小角度 Δθ。速度矢量从 v₁ 变为 v₂，两者大小均为 v。速度变化量 Δv 指向圆心，小角度下其大小约为 vΔθ。
The arc length traveled is Δs = rΔθ, and since Δs = vΔt, we have Δθ = vΔt / r. Substituting into Δv ≈ vΔθ gives Δv ≈ v²Δt / r. Acceleration a = Δv/Δt, hence a = v²/r directed radially inward.
走过的弧长 Δs = rΔθ，又因为 Δs = vΔt，可得 Δθ = vΔt / r。代入 Δv ≈ vΔθ 得到 Δv ≈ v²Δt / r。加速度 a = Δv/Δt，因此 a = v²/r，方向沿半径指向圆心。
a = v² / r
3. Deriving Orbital Speed from Universal Gravitation | 万有引力推导轨道速度
For a satellite in a stable circular orbit, the gravitational force provides the necessary centripetal force. Newton’s law of gravitation states F = GMm/r², where M is the central mass, m is the satellite mass, and r is the orbital radius.
对于稳定圆轨道上的卫星，万有引力提供所需的向心力。牛顿万有引力定律给出 F = GMm/r²，其中 M 为中心天体质量，m 为卫星质量，r 为轨道半径。
Equating gravitational force to centripetal force mv²/r gives GMm/r² = mv²/r. Canceling m and multiplying both sides by r yields v² = GM/r, so the orbital speed is v = √(GM/r).
令万有引力等于向心力 mv²/r，得到 GMm/r² = mv²/r。消去 m 并两边乘以 r 得 v² = GM/r，因此轨道速度 v = √(GM/r)。
v = √(GM / r)
4. Deriving the Ideal Gas Pressure Equation | 理想气体压强公式推导
Imagine a cubical container of side L containing N gas molecules, each of mass m, moving with root-mean-square speed cᵣₘₛ. Consider one molecule striking a wall perpendicular to the x-axis. Its velocity component changes from +uₓ to -uₓ, so the momentum change per collision is 2muₓ.
想象一个边长为 L 的立方体容器，内装 N 个质量均为 m 的气体分子，均方根速率为 cᵣₘₛ。考虑一个分子撞击垂直于 x 轴的器壁：其速度分量从 +uₓ 变为 -uₓ，每次碰撞动量改变为 2muₓ。
The time between collisions with the same wall is 2L/uₓ. Average force exerted by this molecule is F = Δp/Δt = (2muₓ)/(2L/uₓ) = muₓ²/L. Summing over all N molecules and noting the mean square speeds are equal in all three directions (uₓ² = uᵧ² = u_z² = c²/3), total force on the wall is F_total = Nm(c²/3)/L.
与同一器壁的两次碰撞间隔时间为 2L/uₓ。该分子施加的平均力为 F = Δp/Δt = (2muₓ)/(2L/uₓ) = muₓ²/L。对所有 N 个分子求和，并注意三个方向均方速度相等（uₓ² = uᵧ² = u_z² = c²/3），器壁总受力 F_total = Nm(c²/3)/L。
Pressure p is force per unit area A = L², so p = (Nm(c²/3)/L)/L² = Nmc²/(3L³) = (1/3)ρc², where ρ = Nm/V is density.
压强 p 是单位面积受力 A = L²，故 p = (Nm(c²/3)/L)/L² = Nmc²/(3L³) = (1/3)ρc²，其中 ρ = Nm/V 为密度。
p = ⅓ ρ ⟨c²⟩
5. Deriving the Charging Equation for an RC Circuit | RC 电路充电方程推导
In a series RC circuit connected to a battery of emf ε, Kirchhoff’s loop rule gives ε – IR – Q/C = 0, where Q is the charge on the capacitor and I = dQ/dt is the current. Substituting I yields ε – R dQ/dt – Q/C = 0.
在串联 RC 电路连接至电动势 ε 的电池时，基尔霍夫回路定律给出 ε – IR – Q/C = 0，其中 Q 为电容器电荷，I = dQ/dt 为电流。代入 I 得 ε – R dQ/dt – Q/C = 0。
Rearrange to separate variables: dQ/dt = (εC – Q)/(RC). Integration from Q = 0 at t = 0 gives ∫₀ᵠ dQ/(εC – Q) = ∫₀ᵗ dt/(RC), resulting in -ln(εC – Q) + ln(εC) = t/(RC). Simplify to Q = εC(1 – e^{-t/RC}).
整理分离变量：dQ/dt = (εC – Q)/(RC)。从 t=0 时 Q=0 积分得到 ∫₀ᵠ dQ/(εC – Q) = ∫₀ᵗ dt/(RC)，得出 -ln(εC – Q) + ln(εC) = t/(RC)。化简得 Q = εC(1 – e^{-t/RC})。
Q(t) = Q₀(1 – e^{-t/RC})
The product RC is the time constant τ, representing the time taken for charge to reach about 63% of its final value Q₀ = εC.
乘积 RC 为时间常数 τ，表示电荷达到最终值 Q₀ = εC 的约 63% 所需的时间。
6. Deriving the SHM Differential Equation and Solution | 简谐运动微分方程及解推导
Simple harmonic motion occurs when the restoring force is proportional to displacement and opposite in direction: F = -kx. By Newton’s second law, ma = -kx. Acceleration a is the second derivative of displacement x, so m d²x/dt² = -kx.
当回复力与位移成正比且方向相反时，物体做简谐运动：F = -kx。由牛顿第二定律，ma = -kx。加速度 a 是位移 x 的二阶导数，因此 m d²x/dt² = -kx。
Defining angular frequency ω = √(k/m), the differential equation becomes d²x/dt² = -ω²x. The general solution is x = A cos(ωt) + B sin(ωt) or equivalently x = x₀ sin(ωt + φ).
定义角频率 ω = √(k/m)，微分方程化为 d²x/dt² = -ω²x。其通解为 x = A cos(ωt) + B sin(ωt)，或等效写为 x = x₀ sin(ωt + φ)。
x = A sin(ωt + φ)
Velocity and acceleration are then v = dx/dt = Aω cos(ωt + φ) and a = dv/dt = -Aω² sin(ωt + φ) = -ω²x, confirming the defining relationship.
速度和加速度则为 v = dx/dt = Aω cos(ωt + φ) 以及 a = dv/dt = -Aω² sin(ωt + φ) = -ω²x，验证了定义关系。
7. Deriving the Doppler Effect for Sound | 多普勒效应公式（声波）推导
When a source emitting sound of frequency f moves towards a stationary observer at speed v_s, the wavelength in front of the source is compressed. In source’s frame, the wave travels at speed c, but in one period T = 1/f, the source moves v_s T. The observed wavelength λ’ = cT – v_s T = (c – v_s)/f.
当声源以速度 v_s 向静止观察者移动并发出频率为 f 的声波时，声源前方的波长被压缩。在声源参考系中，波速为 c，但在一个周期 T = 1/f 内声源移动 v_s T。观察到的波长 λ’ = cT – v_s T = (c – v_s)/f。
Using the wave equation f’ = c/λ’, the observed frequency is f’ = c f/(c – v_s). This implies a higher pitch as the source approaches. For a stationary source and moving observer, the formula becomes f’ = f (c + v_o)/c for an observer moving toward the source.
应用波速公式 f’ = c/λ’，观察频率为 f’ = c f/(c – v_s)。声源靠近时音调变高。对于静止声源和运动观察者，观察者朝向声源运动时公式为 f’ = f (c + v_o)/c。
f’ = f (c ± v_o)/(c ∓ v_s)
8. Deriving the Radius of a Charged Particle in a Magnetic Field | 磁场中带电粒子半径推导
A charged particle q moving with speed v perpendicular to a uniform magnetic field B experiences a magnetic force F = qvB. This force acts as the centripetal force, F = mv²/r. Equating gives qvB = mv²/r, hence r = mv/(qB).
带电粒子 q 以速度 v 垂直于匀强磁场 B 运动时，受到洛伦兹力 F = qvB。该力充当向心力 F = mv²/r。令两式相等得 qvB = mv²/r，因此 r = mv/(qB)。
r = mv / (qB)
The radius is proportional to momentum and inversely proportional to both charge and magnetic field strength. The period T of the circular motion is T = 2πr/v = 2πm/(qB), which is independent of speed.
半径与动量成正比，与电荷和磁感应强度成反比。圆周运动的周期 T = 2πr/v = 2πm/(qB)，与速度无关。
9. Deriving Electric Potential Energy and Potential | 电势能和电势的推导
For two point charges Q and q separated by distance r, the Coulomb force is F = kQq/r². The work done to bring q from infinity to a distance r against this force is W = ∫∞ʳ F dr = ∫∞ʳ kQq/r² dr = -[kQq/r]∞ʳ = kQq/r. This work is stored as electric potential energy U.
两点电荷 Q 与 q 相距 r 时，库仑力为 F = kQq/r²。将 q 从无穷远处移到距 Q 为 r 的位置时克服库仑力做功 W = ∫∞ʳ F dr = ∫∞ʳ kQq/r² dr = -[kQq/r]∞ʳ = kQq/r。该功储存为电势能 U。
U = k Qq / r
The electric potential V at a point is defined as the potential energy per unit charge: V = U/q = kQ/r. For a uniform electric field E between parallel plates separated by distance d, potential difference ΔV = Ed arises from integrating E along the displacement.
电场中某点的电势 V 定义为单位电荷的电势能：V = U/q = kQ/r。对于间距为 d 的平行板间的匀强电场 E，电势差 ΔV = Ed 来自沿位移积分 E 的结果。
10. Deriving the Single-Slit Diffraction Minimum Condition | 单缝衍射极小条件推导
When light of wavelength λ passes through a single slit of width a, each point across the slit acts as a secondary wavelet source. Consider two rays from the top and the centre of the slit that meet at angle θ on a distant screen. The path difference between these two rays is (a/2)sinθ.
波长为 λ 的光通过宽度为 a 的单缝时，缝上每一点都可看作次级子波源。考虑来自缝顶端和缝中心的两条光线，它们以角度 θ 在远处屏幕相遇。这两条光线的光程差为 (a/2)sinθ。
If this path difference equals λ/2, the two waves arrive out of phase and cancel. Similarly, the entire slit can be paired up to produce complete destructive interference. This leads to the condition for the first minimum: a sinθ = λ. In general, minima occur at a sinθ = nλ, where n = 1, 2, 3,…
若该光程差等于 λ/2，两列波到达时反相并相消。同样，整个狭缝可以配对产生完全相消干涉。于是得到第一级极小的条件：a sinθ = λ。一般情况下，极小出现在 a sinθ = nλ，n = 1, 2, 3,…
a sinθ = nλ
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