📚 IB Physics HL Study Guide: Formula Derivation | IB物理HL学习指南：公式推导
Formula derivations sit at the heart of IB Physics HL – they connect fundamental principles to the equations you apply in problems. This guide walks you through the essential derivations step by step, reinforcing both your conceptual understanding and your ability to reproduce them under exam conditions.
公式推导是 IB 物理高水平课程的核心——它们将基本原理与解题中使用的方程联系起来。本指南将逐步带你完成关键推导，加深概念理解，同时提升你在考试中再现推导过程的能力。
1. Derivation of Kinematic Equations for Constant Acceleration | 匀加速直线运动方程的推导
The definitions of velocity and acceleration as rates of change allow us to derive the kinematic equations when acceleration is uniform. We start from the definition of instantaneous acceleration.
将速度和加速度定义为变化率，使我们可以推导出加速度恒定时的运动学方程。我们从瞬时加速度的定义开始。
Step 1: Acceleration is the rate of change of velocity: a = dv/dt. For constant acceleration, integrate from initial time 0 to t: ∫₀ᵗ dv = a ∫₀ᵗ dt, giving v − u = at.
第一步：加速度是速度的变化率：a = dv/dt。对于恒加速度，从初始时刻 0 到 t 积分：∫₀ᵗ dv = a ∫₀ᵗ dt，得到 v − u = at。
v = u + at
Step 2: Velocity is the rate of change of displacement: v = ds/dt. Substitute v = u + at and integrate s from s₀ to s: ∫ ds = ∫₀ᵗ (u + at) dt, yielding s − s₀ = ut + ½at².
第二步：速度是位移的变化率：v = ds/dt。代入 v = u + at，并对 s 从 s₀ 积分到 s：∫ ds = ∫₀ᵗ (u + at) dt，得到 s − s₀ = ut + ½at²。
s = ut + ½at²
Step 3: Eliminate t between v = u + at and s = ut + ½at². From the first, t = (v − u)/a. Substitute into the second to obtain v² = u² + 2a(s − s₀).
第三步：在 v = u + at 和 s = ut + ½at² 之间消去 t。由第一式得 t = (v − u)/a，代入第二式得到 v² = u² + 2a(s − s₀)。
v² = u² + 2as
2. Deriving Centripetal Acceleration | 向心加速度的推导
When an object moves in a circle at constant speed, its velocity vector changes direction continuously. The acceleration directed toward the centre is found from the geometry of the velocity change.
当物体以恒定速率做圆周运动时，其速度矢量方向不断变化。指向圆心的加速度可通过速度变化量的几何关系求出。
Consider a particle moving from point A to B through a small angle Δθ. The velocity vectors at A and B have equal magnitude v but are separated by Δθ. The change in velocity Δv is the base of an isosceles triangle, giving |Δv| ≈ v Δθ for small Δθ.
考虑质点从 A 运动到 B，转过小角度 Δθ。A 点和 B 点的速度矢量大小均为 v，但方向相差 Δθ。速度变化量 Δv 是等腰三角形的底边，对于小 Δθ 近似为 |Δv| ≈ v Δθ。
The time taken is Δt = (arc length AB) / v = (r Δθ) / v. Therefore the magnitude of acceleration a = |Δv|/Δt = v Δθ / (r Δθ / v) = v²/r.
所用时间为 Δt = (弧长 AB) / v = (r Δθ) / v。因此加速度大小 a = |Δv|/Δt = v Δθ / (r Δθ / v) = v²/r。
ac = v²/r = ω²r
The direction of Δv points toward the centre of the circle, so the centripetal acceleration is radially inward.
Δv 的方向指向圆心，因此向心加速度沿径向向内。
3. Work-Energy Theorem Derivation | 功-动能定理推导
The work done by the net force on a particle equals its change in kinetic energy. This follows directly from Newton’s second law and the definition of work.
合力对质点做的功等于其动能的变化量。这直接来自牛顿第二定律和功的定义。
Start with the net force F = ma = m(dv/dt). The work done for a small displacement dx is dW = F dx = m (dv/dt) dx. Using the chain rule, dv/dt = (dv/dx)(dx/dt) = v dv/dx, so dW = m v dv.
从合力 F = ma = m(dv/dt) 出发。对微小位移 dx 做的功为 dW = F dx = m (dv/dt) dx。利用链式法则，dv/dt = (dv/dx)(dx/dt) = v dv/dx，因此 dW = m v dv。
Integrate from initial speed u to final speed v: W = ∫ dW = ∫uᵛ m v dv = ½m v² − ½m u² = ΔKE.
从初速度 u 到末速度 v 积分：W = ∫ dW = ∫uᵛ m v dv = ½m v² − ½m u² = ΔKE。
Wnet = ½mv² − ½mu²
This result holds for any net force, whether constant or variable, as long as the integration is performed correctly.
该结果对任何合力都成立，无论恒力还是变力，只要正确积分即可。
4. Gravitational Potential Energy Derivation | 引力势能的推导
Near the Earth’s surface we use ΔU = mgΔh, but for large distances we need the general expression U = −GMm/r. This is obtained by integrating the gravitational force from infinity to a point r.
在地球表面附近我们使用 ΔU = mgΔh，但对于大距离需要通用表达式 U = −GMm/r。这是通过从无穷远积分引力直到距离 r 得到的。
The gravitational force is F = −GMm/r² (negative sign shows attraction). To bring a mass m from infinity to a point a distance r from M, the work done by an external agent against gravity is W = ∫∞ʳ (−F) dr = ∫∞ʳ (GMm/r²) dr.
引力为 F = −GMm/r²（负号表示吸引）。要将质量 m 从无穷远移到距离 M 为 r 的点，外力克服引力做的功为 W = ∫∞ʳ (−F) dr = ∫∞ʳ (GMm/r²) dr。
Evaluating the integral: W = [−GMm/r]∞ʳ = −GMm/r − (−GMm/∞) = −GMm/r. Since the change in potential energy is the work done, and we define U(∞) = 0, we have U(r) = −GMm/r.
计算积分：W = [−GMm/r]∞ʳ = −GMm/r − (−GMm/∞) = −GMm/r。由于势能变化等于外力做功，且定义 U(∞) = 0，得到 U(r) = −GMm/r。
U = −GMm/r
5. Pressure of an Ideal Gas (Kinetic Model) | 理想气体的压强推导（分子动理论）
The pressure exerted by an ideal gas can be derived from the momentum change of molecules colliding elastically with the walls of a container.
理想气体施加的压强可通过分子与容器壁弹性碰撞的动量变化推导出来。
Consider a cubical container of side L. A molecule of mass m₀ moving with velocity vx toward a wall perpendicular to x rebounds elastically, changing momentum by Δp = 2m₀vx. The time between collisions with the same wall is Δt = 2L / vx, so the average force on the wall is F = Δp/Δt = m₀vx² / L.
考虑边长为 L 的立方体容器。一个质量为 m₀ 的分子以速度 vx 向垂直于 x 的器壁运动并弹性反弹，动量变化为 Δp = 2m₀vx。与同一器壁两次碰撞的时间间隔为 Δt = 2L / vx，因此对器壁的平均力为 F = Δp/Δt = m₀vx² / L。
For N molecules, the total force on this wall is F = (m₀/L) Σ vxi². Using the average square speed ⟨v²⟩, and noting that by isotropy ⟨vx²⟩ = ⅓⟨v²⟩, the force becomes F = (N m₀ / L) (⅓⟨v²⟩).
对于 N 个分子，该器壁上的总力为 F = (m₀/L) Σ vxi²。利用方均速率 ⟨v²⟩，并由各向同性得 ⟨vx²⟩ = ⅓⟨v²⟩，力变为 F = (N m₀ / L) (⅓⟨v²⟩)。
Pressure is p = F / L², so p = ⅓ (N m₀ ⟨v²⟩) / L³ = ⅓ ρ ⟨v²⟩, where ρ = N m₀ / V. Thus pV = ⅓ N m₀ ⟨v²⟩.
压强 p = F / L²，因此 p = ⅓ (N m₀ ⟨v²⟩) / L³ = ⅓ ρ ⟨v²⟩，其中 ρ = N m₀ / V。因此 pV = ⅓ N m₀ ⟨v²⟩。
pV = ⅓ N m₀ ⟨v²⟩
Comparing with the ideal gas equation pV = nRT reveals that the average translational kinetic energy is ⟨KE⟩ = ³/₂ kBT.
与理想气体状态方程 pV = nRT 比较，可得平均平动动能 ⟨KE⟩ = ³/₂ kBT。
6. Energy Stored in a Capacitor | 电容器储能公式推导
Charging a capacitor requires work to move charge against the increasing potential difference. The total energy stored equals the area under the V–q graph.
给电容器充电需要反抗不断升高的电势差移动电荷做功。储存的总能量等于 V–q 图下的面积。
At an instant the potential difference is V = q / C. The work done to add a small charge dq is dW = V dq = (q / C) dq.
在某一时刻，电势差为 V = q / C。增加微小电荷 dq 所做的功为 dW = V dq = (q / C) dq。
Integrate from q = 0 to the final charge Q: W = ∫₀ᵟ (q / C) dq = ½ Q² / C. Since the final voltage is V = Q / C, we can write the energy as:
从 q = 0 积分到最终电荷 Q：W = ∫₀ᵟ (q / C) dq = ½ Q² / C。因为最终电压 V = Q / C，所以能量可写为：
E = ½ QV = ½ CV² = ½ Q²/C
This energy is stored in the electric field between the plates.
该能量储存在两极板之间的电场中。
7. Magnetic Force: From Moving Charge to Current-Carrying Wire | 磁力：从运动电荷到载流导线
The force on a single charge moving in a magnetic field is given by F = qvB sinθ. For a straight conductor carrying many charges, this becomes the macroscopic Ampère force F = BIL sinθ.
单个运动电荷在磁场中所受的力为 F = qvB sinθ。对于含有大量运动电荷的直导体，这变为宏观的安培力 F = BIL sinθ。
A segment of wire of length L and cross-sectional area A has n charge carriers per unit volume, each with charge q. The total number of carriers in the segment is N = n A L. If they drift with speed v, each experiences a force qvB (assuming B perpendicular to v).
一段长度为 L、横截面积为 A 的导线，单位体积内有 n 个电荷载体，每个电荷为 q。该线段中载流子总数 N = n A L。如果它们以漂移速度 v 运动，每个受到力 qvB（设 B 垂直于 v）。
The total force on the wire is F = N q v B = (n A L) q v B. The current is I = n A q v, so F = (n A q v) L B = I L B.
导线所受总力 F = N q v B = (n A L) q v B。电流 I = n A q v，因此 F = (n A q v) L B = I L B。
F = BIL sinθ
where θ is the angle between B and the direction of current.
其中 θ 是 B 与电流方向之间的夹角。
8. Motional EMF and Faraday’s Law | 动生电动势与法拉第定律
When a conductor moves through a magnetic field, an emf is induced. This can be derived from the magnetic force on free electrons, leading to the same result as Faraday’s law.
当导体在磁场中运动时，会感应出电动势。这可以从自由电子所受的磁力推导出来，并得到与法拉第定律相同的结果。
Consider a straight rod of length L moving with speed v perpendicular to a uniform magnetic field B. Free electrons experience a magnetic force FB = −e v B, driving them to one end, creating a charge separation. An electric field E builds up until the electric force eE equals the magnetic force.
考虑一根长度为 L 的直棒，以速度 v 垂直于均匀磁场 B 运动。自由电子受到磁力 FB = −e v B，被推向一端，产生电荷分离。建立起电场 E，直到电场力 eE 与磁力平衡。
At equilibrium, eE = e v B, so E = v B. The potential difference along the rod is ε = E L = B L v. This is the induced emf.
平衡时 eE = e v B，因此 E = v B。沿棒的电位差为 ε = E L = B L v。这就是感应电动势。
In terms of flux: the rod sweeps out area ΔA = L v Δt, so the flux change is ΔΦ = B L v Δt. Then ε = ΔΦ/Δt = B L v, consistent with Faraday’s law ε = −dΦ/dt.
用磁通量表示：棒扫过的面积 ΔA = L v Δt，因此磁通量变化 ΔΦ = B L v Δt。于是 ε = ΔΦ/Δt = B L v，与法拉第定律 ε = −dΦ/dt 一致。
ε = B L v
9. Single-Slit Diffraction Minima Condition | 单缝衍射极小条件推导
The positions of dark fringes in a single-slit diffraction pattern are found by pairing wavelets from different parts of the slit that interfere destructively.
单缝衍射图样中暗纹的位置可通过将狭缝不同部分的子波配对发生相消干涉来求得。
Divide the slit of width a into two halves. For the first minimum, consider a wavelet from the top of the slit and one from the midpoint. Their path difference to a distant screen is (a/2) sinθ. If this equals λ/2, the waves cancel.
将宽度为 a 的狭缝分成两半。对于第一极小，考虑从狭缝顶端和中点发出的子波。它们到远处屏幕的光程差为 (a/2) sinθ。若等于 λ/2，则两波相消。
Thus (a/2) sinθ = λ/2 ⇒ a sinθ = λ. For the n-th minimum, we pair points separated by a/2n, leading to path difference a sinθ = nλ (n = ±1, ±2, …).
因此 (a/2) sinθ = λ/2 ⇒ a sinθ = λ。对于第 n 级极小，将相隔 a/2n 的点配对，得光程差 a sinθ = nλ（n = ±1, ±2, …）。
a sinθ = nλ, n = ±1, ±2, ±3, …
These angles give the directions of destructive interference and hence the minima of the intensity pattern.
这些角度对应相消干涉方向，因此是强度图样的极小值位置。
10. Photoelectric Effect Equation | 光电效应方程推导
Einstein proposed that light consists of photons, each with energy hf. The photoelectric effect follows from energy conservation when a photon is absorbed by a metal surface.
爱因斯坦提出光由光子组成，每个光子能量为 hf。当光子被金属表面吸收时，由能量守恒可得出光电效应方程。
A photon of energy hf strikes an electron in the metal. The electron must use a minimum energy φ (the work function) to escape the surface. The remainder becomes the electron’s maximum kinetic energy, KEmax.
能量为 hf 的光子撞击金属中的电子。电子必须至少消耗能量 φ（逸出功）才能脱离表面，剩余能量成为电子的最大动能 KEmax。
hf = φ + KEmax
Since KEmax = eVs, where Vs is the stopping potential, the equation becomes eVs = hf − φ. This linear relation between f and Vs provided strong evidence for the photon model.
由于 KEmax = eVs，其中 Vs 为遏止电压，方程变为 eVs = hf − φ。f 与 Vs 之间的线性关系为光子模型提供了有力证据。
11. Period of a Mass-Spring System in SHM | 弹簧振子简谐运动周期推导
A mass attached to a spring obeys Hooke’s law, leading to simple harmonic motion. The period can be derived by relating the acceleration to the displacement or by using the auxiliary circle method.
弹簧上的重物遵循胡克定律，产生简谐运动。通过将加速度与位移关联，或使用参考圆法，可推导出周期。
From Newton’s second law and Hooke’s law (F = −kx), we have ma = −kx, so the acceleration a = −(k/m)x. This matches the SHM condition a = −ω²x, giving ω² = k/m.
由牛顿第二定律和胡克定律 F = −kx，有 ma = −kx，因此加速度 a = −(k/m)x。这与简谐运动条件 a = −ω²x 相符，得到 ω² = k/m。
Alternatively, consider uniform circular motion projected onto a diameter. The acceleration of the projected point is a = −ω²x, and the restoring force is F = −mω²x. Comparing with F = −kx yields ω² = k/m.
另一种方式，考虑匀速圆周运动在直径上的投影。投影点的加速度为 a = −ω²x，回复力为 F = −mω²x。与 F = −kx 比较，得 ω² = k/m。
Since the period T = 2π/ω, we obtain:
由于周期 T = 2
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