📚 IB Physics: Formula Derivation | IB 物理：公式推导
In IB Physics, being able to derive key equations from fundamental principles is a core skill that deepens understanding and prepares you for Paper 1 multiple-choice questions, Paper 2 data analysis, and the internal assessment. This article walks you through the most important derivations, step by step, so you can see how theoretical models connect to the real world.
在 IB 物理中，从基本原理推导关键方程是一项核心能力，它既能加深理解，也有助于应对试卷一的选择题、试卷二的数据分析以及内部评估。本文逐步解析最重要的推导过程，让你看清理论模型与现实世界之间的关联。
1. Deriving the SUVAT Equations | 推导匀变速直线运动方程
The four SUVAT equations emerge directly from the definition of acceleration and the area under a velocity–time graph. Start with constant acceleration a = (v − u)/t, where u is initial velocity, v is final velocity, and t is time.
四个匀变速直线运动方程来源于加速度的定义以及速度−时间图下的面积。从恒定加速度 a = (v − u)/t 入手，其中 u 是初速度，v 是末速度，t 是时间。
Rearranging gives v = u + at. Next, note that displacement s equals the average velocity multiplied by time: s = ½ (u + v) t. Substituting v = u + at into this average-velocity expression yields s = ut + ½ at².
整理可得 v = u + at。接下来，位移 s 等于平均速度乘以时间：s = ½ (u + v) t。将 v = u + at 代入这个平均速度表达式，得到 s = ut + ½ at²。
Eliminating t from the first two equations produces v² = u² + 2as. And eliminating a gives s = ½ (u + v) t again. These four equations form the backbone of linear kinematics.
从前两个方程中消去 t 可推出 v² = u² + 2as；消去 a 则再次得到 s = ½ (u + v) t。这四个方程构成了直线运动学的核心。
- v = u + at
- s = ½ (u + v) t
- s = ut + ½ at²
- v² = u² + 2as
Typical derivation flow: a = (v − u)/t → v = u + at → s = ut + ½ at².
典型推导流程：a = (v − u)/t → v = u + at → s = ut + ½ at²。
2. Deriving Centripetal Acceleration a = v²/r | 推导向心加速度 a = v²/r
Consider an object moving in a circle of radius r with constant speed v. In polar coordinates, the position vector is r(cosθ i + sinθ j) with θ = ωt. Differentiate twice to find acceleration, or use geometry: in a short time Δt, the velocity vector rotates through angle Δθ.
考虑一个物体以恒定速率 v 在半径为 r 的圆周上运动。在极坐标中，位置矢量为 r(cosθ i + sinθ j)，其中 θ = ωt。对其求两次导数可得到加速度；也可以利用几何方法：在很短的时间 Δt 内，速度矢量转过角度 Δθ。
The magnitude of the velocity change is Δv ≈ vΔθ, and the arc length is rΔθ = vΔt. Thus Δθ = vΔt / r, giving Δv = v (vΔt / r) = v²Δt / r. The acceleration a = Δv / Δt → a = v² / r, directed toward the centre.
速度变化量的大小约为 Δv ≈ vΔθ，弧长 rΔθ = vΔt，因此 Δθ = vΔt / r，进而 Δv = v (vΔt / r) = v²Δt / r。加速度 a = Δv / Δt → a = v² / r，方向指向圆心。
Using angular velocity ω = v/r, we also write a = ω²r. For the IB, be ready to show this derivation with clear vector diagrams.
利用角速度 ω = v/r，还可以写成 a = ω²r。IB 考试中需要能配合清晰的矢量图展示该推导。
a = v² / r = ω²r
3. Newton’s Law of Gravitation and g = GM/r² | 万有引力定律与重力加速度 g = GM/r²
Newton’s law states that any two point masses M and m attract each other with force F = GMm / r², where r is their separation. For a small mass m near Earth’s surface, the weight mg must equal the gravitational force on m.
牛顿定律指出，两质点 M 和 m 之间的引力为 F = GMm / r²，r 为它们的间距。在地表附近的小质量 m，其重力 mg 必须等于 m 所受的引力。
Thus GMm / R² = mg, where R is Earth’s radius and M its mass. Cancelling m gives g = GM / R². This shows that gravitational field strength equals the gravitational force per unit mass and is independent of the test mass.
因此 GMm / R² = mg，其中 R 为地球半径，M 为地球质量。约去 m 得到 g = GM / R²。这表明重力场强度等于单位质量的引力，且与检验质量无关。
For a point at height h, g = GM / (R + h)². This derivation is fundamental to understanding satellite motion and gravitational potential.
在高度 h 处，g = GM / (R + h)²。这一推导对于理解卫星运动和引力势至关重要。
4. Deriving Escape Velocity v = √(2GM/r) | 推导逃逸速度 v = √(2GM/r)
Escape velocity is the minimum speed needed for an object to leave a planet’s gravitational influence without further propulsion. At the planet’s surface, the total mechanical energy is ½ mv² − GMm/r.
逃逸速度是物体无需后续推进就能脱离行星引力束缚的最小速度。在行星表面，总机械能为 ½ mv² − GMm/r。
At infinity, both kinetic and potential energy approach zero for the borderline case. By conservation of energy, ½ mv² − GMm/r = 0. Solve for v: v = √(2GM/r).
在逃逸的临界情况下，无穷远处的动能和势能都趋于零。根据能量守恒，½ mv² − GMm/r = 0。解出 v：v = √(2GM/r)。
Notice that m cancels out; escape velocity depends only on the mass and radius of the planet. For Earth, plugging in M ≈ 5.97×10²⁴ kg and R ≈ 6.37×10⁶ m gives about 11.2 km s⁻¹.
注意 m 被约去了；逃逸速度只依赖于行星的质量和半径。对于地球，代入 M ≈ 5.97×10²⁴ kg，R ≈ 6.37×10⁶ m，得出约 11.2 km s⁻¹。
v_esc = √(2GM / r)
5. Kinetic Theory Derivation of pV = ⅓ N m | 气体动理论推导 pV = ⅓ N m
Model a gas as N identical molecules of mass m moving randomly in a cubic box of side L. Consider one molecule hitting a wall perpendicularly; its momentum change is 2mv_x.
将气体建模为 N 个质量为 m 的相同分子，在一个边长为 L 的立方盒内随机运动。考虑一个分子垂直撞击墙壁，其动量变化为 2mv_x。
The time between collisions with the same wall is 2L/v_x, so the average force on that wall is Δp/Δt = (2mv_x) / (2L/v_x) = mv_x²/L. Summing over all N molecules and dividing by wall area L² gives pressure p = (m / L³) Σ v_x².
与同一面墙壁碰撞的间隔时间为 2L/v_x，因此对该壁的平均作用力为 Δp/Δt = (2mv_x) / (2L/v_x) = mv_x²/L。对所有 N 个分子求和，再除以墙壁面积 L²，得到压强 p = (m / L³) Σ v_x²。
Because the motion is random, the mean square speeds are equal:
由于运动是随机的，方向方均值相等：
Connecting
结合理想气体定律将
pV = ⅓ N m
6. Deriving a = −ω²x for Simple Harmonic Motion | 推导简谐运动的 a = −ω²x
SHM is defined by a restoring force proportional to displacement: F = −kx. Applying Newton’s second law, ma = −kx, so a = −(k/m)x. Defining ω² = k/m gives a = −ω²x.
简谐运动的定义是回复力与位移成正比：F = −kx。应用牛顿第二定律，ma = −kx，因此 a = −(k/m)x。定义 ω² = k/m，得到 a = −ω²x。
The solution to this differential equation is x = x₀ sin(ωt) or x = x₀ cos(ωt). Taking derivatives verifies that v = ωx₀ cos(ωt) and a = −ω²x₀ sin(ωt) = −ω²x.
这个微分方程的解是 x = x₀ sin(ωt) 或 x = x₀ cos(ωt)。通过求导可以验证 v = ωx₀ cos(ωt) 以及 a = −ω²x₀ sin(ωt) = −ω²x。
This derivation highlights why the period T = 2π/ω is independent of amplitude, a hallmark of isochronous oscillations.
该推导揭示了为什么周期 T = 2π/ω 与振幅无关——这是等时振荡的标志。
a = −ω²x, with ω = 2πf
7. Time Period of a Simple Pendulum, T = 2π√(l/g) | 单摆周期 T = 2π√(l/g)
For a simple pendulum of length l, the tangential restoring force is mg sinθ. For small angles, sinθ ≈ θ = x/l, so F ≈ −mg x/l. This fits the SHM form F = −kx with k = mg/l.
对于摆长为 l 的单摆，切向回复力为 mg sinθ。小角度下 sinθ ≈ θ = x/l，因此 F ≈ −mg x/l。这符合 SHM 的形式 F = −kx，且 k = mg/l。
Using the SHM period formula T = 2π√(m/k) and substituting k gives T = 2π√(m / (mg/l)) = 2π√(l/g). The mass cancels, showing period depends only on length and gravitational field strength.
利用 SHM 周期公式 T = 2π√(m/k) 并代入 k，得 T = 2π√(m / (mg/l)) = 2π√(l/g)。质量被约去，表明周期只取决于摆长和重力场强度。
For IB, you should be able to derive this from first principles, referencing the small-angle approximation and the SHM condition a ∝ −x.
IB 要求能从基本原理推导此式，说明小角度近似和 a ∝ −x 的简谐条件。
T = 2π√(l/g)
8. Doppler Effect Formula for Sound: f’ = f (v ± uₒ)/(v ± uₛ) | 声波多普勒效应公式：f’ = f (v ± uₒ)/(v ± uₛ)
When a source or observer moves relative to the medium, the observed frequency shifts. Derive separately for moving observer and moving source, then combine.
当波源或观察者相对介质运动时，观测频率会改变。可分别推导观察者运动和波源运动的情况，再合并。
For a stationary source and observer moving toward it at speed uₒ, the relative speed of sound is v + uₒ, so observed frequency f’ = (v + uₒ)/λ = f (v + uₒ)/v. Away gives f’ = f (v − uₒ)/v.
波源静止、观察者以速度 uₒ 朝向波源运动时，声音的相对速度为 v + uₒ，因此观测频率 f’ = (v + uₒ)/λ = f (v + uₒ)/v。远离时 f’ = f (v − uₒ)/v。
For a moving source, the wavelength ahead becomes λ’ = (v − uₛ)/f, so f’ = v/λ’ = f v/(v − uₛ) when approaching. Combining both motions yields the general formula: f’ = f (v ± uₒ)/(v ± uₛ), with signs chosen appropriately.
对于运动波源，前方的波长变为 λ’ = (v − uₛ)/f，从而当接近时 f’ = v/λ’ = f v/(v − uₛ)。将两种运动结合，得到一般公式：f’ = f (v ± uₒ)/(v ± uₛ)，符号需恰当选取。
This equation is fundamental for solving problems involving sirens, radar speed traps, and the redshift of light.
该公式是解决警笛、雷达测速以及光波红移等问题的基础。
f’ = f (v ± uₒ) / (v ± uₛ)
9. Single-slit Diffraction Minima: nλ = a sinθ | 单缝衍射暗纹条件：nλ = a sinθ
When monochromatic light passes through a slit of width a, secondary wavelets interfere. Minima occur when wavelets from the top and middle of the slit are out of phase by λ/2, causing cancellation.
单色光通过宽度为 a 的狭缝时，次级子波发生干涉。当从狭缝顶部和中点发出的子波相位差为 λ/2 时，相消干涉形成暗纹。
The path difference between rays from the top edge and the centre is a/2 sinθ. Setting this equal to λ/2 gives a sinθ = λ for the first minimum. More generally, dividing the slit into n pairs results in nλ = a sinθ, where n = 1, 2, 3, …
狭缝上边缘与中心的光线之间的程差为 a/2 sinθ。令其等于 λ/2 得到一级暗纹条件 a sinθ = λ。更一般地，将狭缝分为 n 对子波，可得 nλ = a sinθ，n = 1, 2, 3, …
This derivation explains the central bright fringe and the rapid decrease in intensity of higher-order maxima.
该推导解释了中央亮纹的形成以及高级次明纹强度迅速下降的原因。
nλ = a sinθ, n ∈ ℕ
10. Deriving the Radioactive Decay Law N = N₀ e^(−λt) | 推导放射性衰变定律 N = N₀ e^(−λt)
Radioactive decay is a random process where the activity A = −dN/dt is proportional to the number of undecayed nuclei: dN/dt = −λN, where λ is the decay constant.
放射性衰变是一个随机过程，其活度 A = −dN/dt 与尚未衰变的原子核数成正比：dN/dt = −λN，λ 为衰变常数。
Separating variables: dN/N = −λ dt. Integrating both sides from N₀ at t = 0 to N at t yields ln(N/N₀) = −λt. Exponentiating gives the exponential decay law N = N₀ e^(−λt).
分离变量：dN/N = −λ dt。从 t = 0 时 N = N₀ 积分到 t 时刻的 N，得到 ln(N/N₀) = −λt。取指数后得到指数衰变定律 N = N₀ e^(−λt)。
The half-life T₁/₂ is defined when N = N₀/2, leading to T₁/₂ = ln 2 / λ. This law also appears in capacitor discharge and damped oscillations, highlighting the unifying power of exponential mathematics.
半衰期 T₁/₂ 定义为 N = N₀/2 的时刻，代入得 T₁/₂ = ln 2 / λ。该定律也出现在电容器放电和阻尼振荡中，凸显了指数数学的统一力量。
N = N₀ e^(−λt), T₁/₂ = ln 2 / λ
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