📚 A-Level Physics: Key Formula Derivations for IAL Unit 5 (PH05) | A-Level 物理：国际AL Unit 5 (PH05) 核心公式推导
In Edexcel International A-Level Physics Unit 5 (PH05), examiners often ask candidates to derive fundamental equations from thermodynamics, radioactivity, oscillations, and cosmology. Mastering these derivations helps you write high-scoring example responses and deepens your physical understanding. Below we cover the essential derivations step by step, highlighting the reasoning markers look for.
在 Edexcel 国际 A-Level 物理 Unit 5（PH05）中，考官经常要求考生推导热力学、放射性、振动和宇宙学中的基本方程。掌握这些推导不仅能帮助你写出高分范例答案，还能加深物理理解。以下我们逐步讲解核心推导，突出阅卷老师关注的推理要点。
1. Kinetic Theory and the Ideal Gas Equation | 气体分子运动论与理想气体方程
We start by modelling a single gas particle of mass m moving with speed c in a cubic box of side L. It collides elastically with a wall perpendicular to the x‑axis. The momentum change is 2mc_x, and the time between collisions with the same wall is 2L/c_x. Force on the wall from one particle is (2mc_x) / (2L/c_x) = m c_x² / L.
我们从一个质量为 m、速率为 c 的气体分子在边长为 L 的正方体盒子中运动开始。它与垂直于 x 轴的器壁发生弹性碰撞。动量变化为 2mc_x，与同一器壁两次碰撞的时间间隔为 2L/c_x。因此单个分子对器壁的作用力为 (2mc_x) / (2L/c_x) = m c_x² / L。
Summing over all N particles, the total force on the wall is F = (m/L) ∑ c_x². Since the motion is random, the mean square speed components are equal: ⟨c_x²⟩ = ⟨c_y²⟩ = ⟨c_z²⟩ = ⅓ ⟨c²⟩, where ⟨c²⟩ is the mean square speed. Hence total force F = (m/L) N × ⅓ ⟨c²⟩.
对所有 N 个分子求和，总作用力为 F = (m/L) ∑ c_x²。因为运动是随机的，各方向均方速率分量相等：⟨c_x²⟩ = ⟨c_y²⟩ = ⟨c_z²⟩ = ⅓ ⟨c²⟩，其中 ⟨c²⟩ 为均方速率。因此总力 F = (m/L) N × ⅓ ⟨c²⟩。
Pressure p = F / A = F / L². Substituting F gives p = (Nm ⟨c²⟩) / (3 L³). Since volume V = L³, we obtain the ideal gas equation from kinetic theory:
压强 p = F / A = F / L²。代入 F 得到 p = (Nm ⟨c²⟩) / (3 L³)。由于体积 V = L³，我们就从分子运动论得到了理想气体方程：
pV = ⅓ N m ⟨c²⟩
This links a macroscopic quantity (pV) to the microscopic mean kinetic energy of the particles.
这个方程将宏观量 (pV) 与微观的分子平均动能联系起来。
2. Linking Temperature and Kinetic Energy | 温度与分子动能的联系
Compare the kinetic‑theory result with the experimental ideal gas law pV = nRT, where n = N/Nₐ and Nₐ is Avogadro’s number. Write pV = N k T, with k = R/Nₐ the Boltzmann constant.
将分子运动论的结果与实验得到的理想气体定律 pV = nRT 相比较，其中 n = N/Nₐ，Nₐ 为阿伏伽德罗常数。写为 pV = N k T，k = R/Nₐ 是玻尔兹曼常数。
Equating the two expressions for pV:
令 pV 的两个表达式相等：
⅓ N m ⟨c²⟩ = N k T
Cancel N and multiply by 3/2 to find the mean translational kinetic energy per particle:
消去 N 并乘以 3/2，得到每个分子的平均平动动能：
½ m ⟨c²⟩ = 3⁄2 k T
This result shows that temperature is a direct measure of the average random kinetic energy of gas particles. The total internal energy U of a monatomic ideal gas is the sum of the translational kinetic energies: U = 3⁄2 N k T = 3⁄2 nRT.
这个结果表明温度是气体分子无规则运动平均动能的直接量度。单原子理想气体的总内能 U 就是所有平动动能之和：U = 3⁄2 N k T = 3⁄2 nRT。
3. Work Done and the First Law of Thermodynamics | 做功与热力学第一定律
When a gas expands by a small volume dV against an external pressure p, the work done BY the gas is δW = p dV. Therefore the work done ON the gas is –p dV.
当气体反抗外压强 p 发生微小体积膨胀 dV 时，气体对外做的元功为 δW = p dV。因此外界对气体做的功是 –p dV。
The first law of thermodynamics states that the increase in internal energy ΔU equals the sum of the heat supplied to the system Q and the work done ON the system W:
热力学第一定律指出，内能增量 ΔU 等于系统吸收的热量 Q 与外界对系统做的功 W 之和：
ΔU = Q + W (our sign convention, W = work done ON gas)
With this sign convention, for an adiabatic process (Q = 0), we have dU = –p dV. This relationship is the starting point for deriving the adiabatic condition pVγ = constant.
采用这种符号规定，对于绝热过程 (Q = 0)，有 dU = –p dV。这一关系是推导绝热条件 pVγ = 常数的出发点。
4. Derivation of the Adiabatic Equation | 绝热方程的推导
For an ideal gas, internal energy depends only on temperature: dU = n C_V dT, where C_V is the molar heat capacity at constant volume. Combine with the first law for an adiabatic change: n C_V dT = –p dV.
对于理想气体，内能只依赖于温度：dU = n C_V dT，其中 C_V 为定体摩尔热容。结合绝热变化的第一定律：n C_V dT = –p dV。
Write the ideal gas equation pV = nRT, differentiate it: p dV + V dp = nR dT. Eliminate dT using dT = (–p dV)/(n C_V) and the relation R = C_P – C_V. After algebra, we obtain:
写出理想气体方程 pV = nRT，微分得：p dV + V dp = nR dT。利用 dT = (–p dV)/(n C_V) 和 R = C_P – C_V 消去 dT。代数整理后得到：
p dV + V dp = –(C_P – C_V)/C_V p dV
Introducing the adiabatic index γ = C_P/C_V, this simplifies to γ p dV + V dp = 0, or dp/p + γ dV/V = 0. Integrating yields ln p + γ ln V = constant, hence:
引入绝热指数 γ = C_P/C_V，上式简化为 γ p dV + V dp = 0，即 dp/p + γ dV/V = 0。积分得到 ln p + γ ln V = 常数，从而：
p Vγ = constant
Equivalent forms, using pV = nRT, are TVγ‑1 = constant and T p(1‑γ)/γ = constant. Mark schemes award credit for clear justification of the integration step and the use of R = C_P – C_V.
利用 pV = nRT 可以得到等效形式 TVγ‑1 = 常数 和 T p(1‑γ)/γ = 常数。阅卷标准会对清晰的积分步骤以及 R = C_P – C_V 的使用给予加分。
5. Radioactive Decay Law and Activity | 放射性衰变定律与活度
The rate of decay is proportional to the number of undecayed nuclei N present. Introducing a proportionality constant λ (decay constant), we write:
衰变速率与当前尚未衰变的原子核数 N 成正比。引入比例常数 λ（衰变常数），写作：
dN/dt = –λ N
Separate variables: dN/N = –λ dt. Integrate with limits N₀ at t = 0 to N at time t:
分离变量：dN/N = –λ dt。积分，t = 0 时 N = N₀，t = t 时 N = N：
∫N₀N dN/N = –λ ∫0t dt
This gives ln(N/N₀) = –λt, and exponentiating both sides yields the exponential decay law:
得到 ln(N/N₀) = –λt，两边取指数即得指数衰减定律：
N = N₀ e‑λt
Activity A is defined as the number of decays per second, A = |dN/dt| = λ N. Therefore the activity also follows A = A₀ e‑λt, where A₀ = λ N₀. In PH05 example responses, you must show the integration steps and define all symbols clearly.
活度 A 定义为每秒衰变数，A = |dN/dt| = λ N。因此活度同样遵循 A = A₀ e‑λt，其中 A₀ = λ N₀。在 PH05 范例答案中，你必须展示积分步骤并清楚地定义所有符号。
6. Half‑Life and the Decay Constant | 半衰期与衰变常数
The half‑life T₁/₂ is the time taken for the number of nuclei (or activity) to fall to half its initial value. Set N = N₀/2 in the decay law:
半衰期 T₁/₂ 是原子核数（或活度）减少到初始值一半所需的时间。在衰变定律中令 N = N₀/2：
N₀/2 = N₀ e‑λ T₁/₂ → 1/2 = e‑λ T₁/₂
Taking natural logarithms: ln(1/2) = –λ T₁/₂, and since ln(1/2) = –ln 2, we obtain:
取自然对数：ln(1/2) = –λ T₁/₂，又因 ln(1/2) = –ln 2，于是得到：
T₁/₂ = ln 2 / λ
This relation is frequently used in numerical questions and required derivations. Remember that λ has units of s⁻¹, so T₁/₂ is in seconds.
这个关系式经常在数值计算和推导题中使用。请记住 λ 的单位是 s⁻¹，因此 T₁/₂ 的单位是秒。
7. Simple Harmonic Motion – Displacement Equation | 简谐运动 – 位移方程
For a body moving in a circle of radius A with constant angular speed ω, the projection onto a diameter executes SHM. If at t = 0 the radius makes an angle φ with the x‑axis, the displacement x is:
一个物体在半径为 A 的圆周上以恒定角速度 ω 运动，其直径上的投影做简谐运动。如果 t = 0 时半径与 x 轴的夹角为 φ，则位移 x 为：
x = A cos(ωt + φ)
Usually we set φ = 0 for simplicity, giving x = A cos ωt. The defining characteristic of SHM is that acceleration is proportional to displacement and directed towards equilibrium:
为简单通常取 φ = 0，得 x = A cos ωt。简谐运动的定义特征是加速度与位移成正比且指向平衡位置：
a = –ω² x
Differentiating x twice confirms this: v = dx/dt = –Aω sin ωt, and a = dv/dt = –Aω² cos ωt = –ω² x. The relationship a = –ω² x leads to ω = √(k/m) for a mass‑spring system, where k is the spring constant.
对 x 求二阶导即可验证：v = dx/dt = –Aω sin ωt，a = dv/dt = –Aω² cos ωt = –ω² x。由 a = –ω² x 可导得弹簧振子的 ω = √(k/m)，其中 k 是劲度系数。
8. Velocity and Energy in SHM | 简谐运动的速度与能量
From the displacement x = A cos ωt, the velocity is v = –Aω sin ωt. Using sin²θ = 1 – cos²θ, we can express speed in terms of displacement:
由位移 x = A cos ωt，速度为 v = –Aω sin ωt。利用 sin²θ = 1 – cos²θ，可将速率用位移表示：
v = ± ω √(A² – x²)
The kinetic energy Eₖ = ½ m v² becomes ½ m ω² (A² – x²). The potential energy stored in an oscillator (e.g., elastic potential energy in a spring) is ½ k x², and since ω² = k/m, this equals ½ m ω² x².
动能 Eₖ = ½ m v² 变为 ½ m ω² (A² – x²)。谐振子中储存的势能（例如弹簧的弹性势能）为 ½ k x²，由 ω² = k/m 可知这等于 ½ m ω² x²。
The total mechanical energy is the sum:
总机械能为二者之和：
Etotal = Eₖ + Ep = ½ m ω² (A² – x²) + ½ m ω² x² = ½ m ω² A²
Thus total energy is constant and proportional to the square of the amplitude, a central result for PH05 oscillations questions.
因此总能量恒定，且与振幅的平方成正比，这是 PH05 振动类问题的一个核心结论。
9. Deriving the Period of a Simple Pendulum | 单摆周期的推导
For a pendulum of length l, the restoring force along the arc is –mg sin θ. For small angles (θ < 10°), sin θ ≈ θ in radians. The arc displacement s = lθ, so the tangential acceleration is a = l d²θ/dt².
对于摆长为 l 的单摆，沿圆弧切线方向的回复力为 –mg sin θ。在小角度（θ < 10°）条件下，sin θ ≈ θ（弧度）。弧长位移 s = lθ，切向加速度 a = l d²θ/dt²。
Using Newton’s second law:
应用牛顿第二定律：
m l d²θ/dt² = –mg θ → d²θ/dt² = –(g/l) θ
This matches the SHM form a = –ω² x with ω² = g/l. Since T = 2π/ω, we obtain:
这与简谐运动形式 a = –ω² x 一致，其中 ω² = g/l。由 T = 2π/ω 得到：
T = 2π √(l/g)
Marks are awarded for stating the small‑angle approximation and correctly identifying the restoring force.
评分标准会对小角度近似的陈述以及正确找出回复力给予分数。
10. Period of a Mass‑Spring Oscillator | 弹簧振子的周期
A mass m attached to a spring of stiffness k experiences a restoring force F = –k x when displaced by x. Newton’s second law gives m a = –k x, hence a = –(k/m) x.
质量为 m 的物体连接在劲度系数为 k 的弹簧上，当位移为 x 时受到回复力 F = –k x。由牛顿第二定律 m a = –k x，得 a = –(k/m) x。
Again comparing with a = –ω² x gives ω² = k/m. Therefore the period is:
再次与 a = –ω² x 对比，得到 ω² = k/m。因此周期为：
T = 2π √(m/k)
Note that this period does not depend on the amplitude, meaning the oscillations are isochronous. In Unit 5, you may be asked to combine this with energy conservation or with the spring constant of a series/parallel combination.
注意该周期与振幅无关，即振动是等时的。在 Unit 5 中，你可能需要将此与能量守恒或弹簧的串／并联劲度系数结合起来。
11. Hubble’s Law and Cosmological Redshift | 哈勃定律与宇宙学红移
Observations show that distant galaxies recede with a speed v proportional to their distance d: v = H₀ d, where H₀ is the Hubble parameter. For non‑relativistic speeds, the redshift z is related to velocity by the Doppler shift:
观测表明，遥远星系的退行速度 v 与其距离 d 成正比：v = H₀ d，其中 H₀ 为哈勃参数。在非相对论速度下，红移 z 与速度的关系可由多普勒频移给出：
z = Δλ / λ₀ ≈ v / c (v << c)
Combining gives d = (c / H₀) z. Inverting, the age of the Universe can be estimated as 1/H₀ if the expansion rate has been constant. This simple derivation is often required in PH05 cosmology example responses; remember to state the assumption that v ≪ c.
联立两式可得 d = (c / H₀) z。反过来，如果膨胀速率恒定，宇宙年龄可估算为 1/H₀。这个简单推导经常出现在 PH05 宇宙学的范例答案中；切记要说明假定 v ≪ c。
12. Connecting Adiabatic Work and Temperature Change | 绝热功与温度变化的联系
Many PH05 questions ask you to link pVγ = constant back to temperature changes. Starting from pV = nRT and pVγ = constant, eliminate p to obtain TVγ‑1 = constant, or eliminate V to obtain T p(1‑γ)/γ = constant. This allows you to calculate, for example, the temperature after an adiabatic compression.
许多 PH05 题目要求将 pVγ = 常数与温度变化联系起来。从 pV = nRT 和 pVγ = 常数出发，消去 p 得到 TVγ‑1 = 常数，或消去 V 得到 T p(1‑γ)/γ = 常数。这样就能计算例如绝热压缩后的温度。
An alternative derivation for the work done in an adiabatic process uses ΔU = W (since Q = 0). Because ΔU = n C_V (T₂ – T₁), the work done by the gas is W = n C_V (T₁ – T₂). This result often appears alongside the pV relation.
绝热过程做功量的另一种推导使用 ΔU = W（因 Q = 0）。由于 ΔU = n C_V (T₂ – T₁)，气体对外做功 W = n C_V (T₁ – T₂)。这一结果常与 pV 关系一同出现。
Practise these derivations until you can reproduce them fluently; marks are awarded for clear algebraic steps, correct sign conventions, and explicit linking of the macroscopic and microscopic pictures.
反复练习这些推导，直至你能流畅地写出它们；评分会给思路清晰的代数步骤、正确的符号规定，以及宏观与微观图像的明确联系。
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