Unit E.5: Fusion and Stars单元 E.5：聚变与恒星
The closing super-topic of Theme E "Nuclear and quantum physics". Nuclear fusion and why fusing light nuclei releases energy (the binding-energy-per-nucleon curve), the Coulomb barrier and the temperature and density needed to overcome it, the proton-proton chain that powers the Sun, the hydrostatic balance that keeps a main-sequence star stable, luminosity as power output with the Stefan-Boltzmann law, apparent brightness and the inverse-square law used to find stellar distances, and a brief overview of the main sequence and stellar end states. This unit ties the nuclear physics of E.1-E.4 to the astrophysics of how stars shine.主题 E"核物理与量子物理"的收尾超级专题。核聚变以及为何轻核聚变会释放能量（每核子结合能曲线）、库仑势垒以及克服它所需的温度与密度、为太阳供能的质子-质子链、维持主序星稳定的流体静力平衡、作为功率输出的光度与斯特藩—玻尔兹曼定律、用于求恒星距离的视亮度与平方反比律，以及主序与恒星归宿的简要概览。本单元把 E.1-E.4 的核物理与"恒星为何发光"的天体物理联系起来。
How to use this guide本指南使用说明
E.5 is a "two halves" unit. The first half is nuclear (fusion, the binding-energy curve, the Coulomb barrier, the proton-proton chain) and reuses the mass-defect ideas from E.1-E.4. The second half is astrophysical and almost entirely formula-driven: $L = \sigma A T^4$ and $b = L / (4\pi d^2)$ do most of the work. Marks come from clean unit handling (squaring temperatures, watching $\pi$ and the factor of 4) and from one-line qualitative explanations of why stars are stable and why heavier stars die faster.E.5 是"两半"单元。前半是核物理（聚变、结合能曲线、库仑势垒、质子-质子链），沿用 E.1-E.4 的质量亏损思想。后半是天体物理，几乎完全由公式驱动：$L = \sigma A T^4$ 与 $b = L / (4\pi d^2)$ 承担大部分工作。分数来自干净的单位处理（温度的四次方、注意 $\pi$ 与因子 4）以及"恒星为何稳定""为何越重的恒星寿命越短"等一句话定性说明。
Light nuclei fuse and release energy because binding energy per nucleon rises toward iron. Memorise $L = \sigma A T^4$ (use $A = 4\pi R^2$ for a sphere) and $b = L / (4\pi d^2)$. A star is stable because inward gravity balances outward pressure. The Sun runs on the proton-proton chain: four protons become one helium-4, releasing energy.
轻核聚变会释放能量，因为每核子结合能向铁方向上升。背熟 $L = \sigma A T^4$（球体取 $A = 4\pi R^2$）与 $b = L / (4\pi d^2)$。恒星稳定是因为向内的引力与向外的压力平衡。太阳依靠质子-质子链：四个质子变成一个氦-4，释放能量。
Explain the Coulomb barrier and why fusion needs both high temperature (kinetic energy to approach) and high density (collision rate). State the equilibrium as hydrostatic balance: at every radius, the pressure gradient supports the overlying weight. Be fluent rearranging $b = L/(4\pi d^2)$ for $d$, and explain qualitatively why a more massive star is far more luminous and far shorter-lived.
解释Coulomb barrier（库仑势垒），以及为何聚变同时需要高温（足够动能接近）和高密度（碰撞率）。把平衡表述为流体静力平衡：在每一半径处，压强梯度支撑上方物质的重量。熟练地把 $b = L/(4\pi d^2)$ 解出 $d$，并定性说明为何质量更大的恒星光度高得多、寿命短得多。
Nuclear Fusion and the Binding-Energy Curve核聚变与结合能曲线 E.5 SL+HL
Why energy is released. Plot binding energy per nucleon against nucleon number. The curve rises steeply for light nuclei and peaks near iron (${}^{56}\mathrm{Fe}$, $\approx 8.8\ \mathrm{MeV/nucleon}$). Fusing light nuclei moves up the curve, so the product is more tightly bound; the surplus binding energy is released as kinetic energy and radiation.
Mass-energy. The energy released equals the mass defect times $c^2$: from the data booklet,
E = mc², with $\Delta m$ the difference between reactant and product masses.
Conditions for fusion. Nuclei are positively charged and repel (the Coulomb barrier). Fusion needs high temperature (large kinetic energy to approach) and high density (frequent collisions). Both hold in stellar cores.
为何释放能量。把每核子结合能对核子数作图。曲线在轻核区急升，在铁附近达峰（${}^{56}\mathrm{Fe}$，$\approx 8.8\ \mathrm{MeV/核子}$）。轻核聚变沿曲线上行，产物结合得更紧；多余的结合能以动能与辐射形式释放。
质能关系。释放的能量等于质量亏损乘 $c^2$：数据手册中
E = mc²，$\Delta m$ 为反应物与产物质量之差。
聚变条件。核带正电会互斥（库仑势垒）。聚变需要高温（足够大的动能以接近）与高密度（频繁碰撞）。恒星核心两者皆满足。
Deuterium and tritium fuse: ${}^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \to {}^{4}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}$. Masses: ${}^{2}\mathrm{H} = 2.01410\ \mathrm{u}$, ${}^{3}\mathrm{H} = 3.01605\ \mathrm{u}$, ${}^{4}\mathrm{He} = 4.00260\ \mathrm{u}$, $\mathrm{n} = 1.00867\ \mathrm{u}$. Find the energy released (take $1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$).氘与氚聚变：${}^{2}_{1}\mathrm{H} + {}^{3}_{1}\mathrm{H} \to {}^{4}_{2}\mathrm{He} + {}^{1}_{0}\mathrm{n}$。质量：${}^{2}\mathrm{H} = 2.01410\ \mathrm{u}$、${}^{3}\mathrm{H} = 3.01605\ \mathrm{u}$、${}^{4}\mathrm{He} = 4.00260\ \mathrm{u}$、$\mathrm{n} = 1.00867\ \mathrm{u}$。求释放的能量（取 $1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^2$）。
Identify. Energy released $= \Delta m\, c^2$, where $\Delta m$ is the mass lost (reactants minus products), via E = mc².
识别。释放的能量 $= \Delta m\, c^2$，$\Delta m$ 为损失的质量（反应物减产物），用 E = mc²。
Set up. Total reactant mass: $2.01410 + 3.01605 = 5.03015\ \mathrm{u}$. Total product mass: $4.00260 + 1.00867 = 5.01127\ \mathrm{u}$.
列式。反应物总质量：$2.01410 + 3.01605 = 5.03015\ \mathrm{u}$。产物总质量：$4.00260 + 1.00867 = 5.01127\ \mathrm{u}$。
$$ \Delta m = 5.03015 - 5.01127 = 0.01888\ \mathrm{u}. $$Execute. $E = (0.01888)(931.5) \approx 17.6\ \mathrm{MeV}$.
执行。$E = (0.01888)(931.5) \approx 17.6\ \mathrm{MeV}$。
Evaluate. About $17.6\ \mathrm{MeV}$ per fusion — far more per unit mass than chemical burning or even fission per reaction relative to fuel mass. D-T is the easiest fusion to ignite, which is why it is the target for fusion reactors.
评估。每次聚变约 $17.6\ \mathrm{MeV}$——按单位质量计远超化学燃烧，相对燃料质量也胜过裂变。氘-氚最易点燃，故是聚变反应堆的目标反应。
Going deeper: why high temperature AND high density are both required深入：为何高温与高密度缺一不可
To fuse, two nuclei must approach within the range of the strong nuclear force ($\sim 10^{-15}\ \mathrm{m}$). At that separation the electrostatic potential energy is enormous:
要聚变，两核须接近到强核力作用范围内（$\sim 10^{-15}\ \mathrm{m}$）。在该间距处静电势能极大：
$$ U = \frac{k\, q_1 q_2}{r} = \frac{(8.99\times 10^{9})(1.6\times 10^{-19})^2}{10^{-15}} \approx 2.3\times 10^{-13}\ \mathrm{J} \approx 1.4\ \mathrm{MeV}. $$Temperature sets the average kinetic energy: $\bar{E}_k = \tfrac{3}{2} k_B T$. To give nuclei enough energy to climb most of the Coulomb barrier requires $T \sim 10^{7}\ \mathrm{K}$ (quantum tunnelling lets fusion proceed below the classical barrier height). Density sets the collision rate: even fast nuclei rarely fuse, so a star needs an enormous number of nuclei packed together so that, integrated over the core, enough fusions occur per second to balance the star's radiated power. High $T$ without high density gives too few collisions; high density without high $T$ gives collisions too gentle to overcome the barrier.
温度决定平均动能：$\bar{E}_k = \tfrac{3}{2} k_B T$。要使核获得足够能量攀越大部分库仑势垒，需 $T \sim 10^{7}\ \mathrm{K}$（量子隧穿使聚变可在经典势垒高度以下发生）。密度决定碰撞率：即使是高速核也很少聚变，故恒星需要海量核紧密堆积，使整个核心每秒发生足够多次聚变以平衡恒星辐射功率。高温而无高密度，碰撞太少；高密度而无高温，碰撞太弱难越势垒。
Fusion as the Stellar Energy Source作为恒星能源的聚变 E.5 SL+HL
The proton-proton (p-p) chain (qualitative, three steps).
- ${}^{1}\mathrm{H} + {}^{1}\mathrm{H} \to {}^{2}\mathrm{H} + e^{+} + \nu_e$ (two protons make a deuteron; one proton becomes a neutron).
- ${}^{2}\mathrm{H} + {}^{1}\mathrm{H} \to {}^{3}\mathrm{He} + \gamma$.
- ${}^{3}\mathrm{He} + {}^{3}\mathrm{He} \to {}^{4}\mathrm{He} + 2\,{}^{1}\mathrm{H}$.
质子-质子（p-p）链（定性，三步）。
- ${}^{1}\mathrm{H} + {}^{1}\mathrm{H} \to {}^{2}\mathrm{H} + e^{+} + \nu_e$（两质子生成氘核；一个质子变为中子）。
- ${}^{2}\mathrm{H} + {}^{1}\mathrm{H} \to {}^{3}\mathrm{He} + \gamma$。
- ${}^{3}\mathrm{He} + {}^{3}\mathrm{He} \to {}^{4}\mathrm{He} + 2\,{}^{1}\mathrm{H}$。
The Sun's luminosity is $L_\odot = 3.85\times 10^{26}\ \mathrm{W}$. At what rate is mass converted to energy in its core? (Use E = mc² with $c = 3.00\times 10^{8}\ \mathrm{m\,s^{-1}}$.)太阳光度 $L_\odot = 3.85\times 10^{26}\ \mathrm{W}$。其核心以多大速率把质量转为能量？（用 E = mc²，$c = 3.00\times 10^{8}\ \mathrm{m\,s^{-1}}$。）
Identify. Power radiated equals mass-energy converted per second: $L = \dfrac{dm}{dt}\,c^2$.
识别。辐射功率等于每秒转换的质能：$L = \dfrac{dm}{dt}\,c^2$。
Set up. Rearrange: $\dfrac{dm}{dt} = \dfrac{L}{c^2}$.
列式。变形：$\dfrac{dm}{dt} = \dfrac{L}{c^2}$。
$$ \frac{dm}{dt} = \frac{3.85\times 10^{26}}{(3.00\times 10^{8})^2} = \frac{3.85\times 10^{26}}{9.00\times 10^{16}} \approx 4.3\times 10^{9}\ \mathrm{kg\,s^{-1}}. $$Evaluate. The Sun loses about $4.3$ million tonnes of rest mass every second purely to energy — yet this is utterly negligible against its total mass of $2\times 10^{30}\ \mathrm{kg}$, which is why its luminosity is so steady.
评估。太阳每秒约损失 $430$ 万吨静止质量纯粹转为能量——但相对其 $2\times 10^{30}\ \mathrm{kg}$ 的总质量可忽略不计，故其光度极其稳定。
Going deeper: where the $26.7\ \mathrm{MeV}$ comes from深入：$26.7\ \mathrm{MeV}$ 从何而来
Take four hydrogen atoms ($4\times 1.00783\ \mathrm{u} = 4.03132\ \mathrm{u}$) and one helium-4 atom ($4.00260\ \mathrm{u}$). The mass defect is:
取四个氢原子（$4\times 1.00783\ \mathrm{u} = 4.03132\ \mathrm{u}$）与一个氦-4 原子（$4.00260\ \mathrm{u}$）。质量亏损为：
$$ \Delta m = 4.03132 - 4.00260 = 0.02872\ \mathrm{u}, \qquad E = (0.02872)(931.5) \approx 26.7\ \mathrm{MeV}. $$A small amount is carried away by the neutrinos (which escape the Sun directly), so the energy that heats the Sun and emerges as light is a little less. Note this is the net chain result; the individual p-p steps each release part of the total. The positrons quickly annihilate with electrons, adding their rest energy to the radiation.
其中一小部分被中微子带走（中微子直接逃离太阳），故加热太阳并最终以光形式逸出的能量略少。注意这是整条链的净结果；各 p-p 步骤各释放总量的一部分。正电子很快与电子湮灭，把其静能加入辐射。
Hydrostatic Equilibrium on the Main Sequence主序上的流体静力平衡 E.5 SL+HL
- If the core contracts slightly, it heats up, fusion speeds up, pressure rises and pushes back out.
- If the core expands slightly, it cools, fusion slows, pressure drops and gravity pulls back in.
- 若核心略微收缩，温度升高，聚变加快，压力上升把物质推回外。
- 若核心略微膨胀，温度下降，聚变减慢，压力下降，引力把物质拉回内。
Explain, in terms of the forces involved, why a main-sequence star such as the Sun neither collapses under its own gravity nor flies apart, and why a small contraction is self-correcting.用所涉及的力，解释为何像太阳这样的主序星既不会在自身引力下坍缩，也不会四散飞开，以及为何微小的收缩能自我纠正。
Identify the two agents. Inward: gravitation, pulling all the mass toward the centre. Outward: pressure (thermal gas pressure plus radiation pressure) generated by the hot, fusing core.
指出两个主体。向内：引力，把全部质量拉向中心。向外：压力（热气体压强加辐射压），由高温聚变的核心产生。
Equilibrium. On the main sequence these are equal at every radius, so the net force on each layer is zero and the star holds a steady size — this is hydrostatic equilibrium.
平衡。在主序上两者在每一半径处相等，故每一层所受合力为零，恒星保持稳定尺寸——即流体静力平衡。
Self-correction. Suppose gravity briefly wins and the core contracts. Compression raises the temperature, which raises the fusion rate (very temperature-sensitive), which raises the pressure. The increased outward pressure halts and reverses the contraction. The reverse chain damps any over-expansion. The star is therefore stable.
自我纠正。设引力短暂占优，核心收缩。压缩使温度升高，从而聚变率升高（对温度极敏感），从而压力升高。增大的向外压力阻止并逆转收缩。相反的链条则抑制过度膨胀。故恒星稳定。
Evaluate. This negative-feedback thermostat is what keeps the Sun's luminosity essentially constant over billions of years, until the core hydrogen runs low and the balance shifts (leading off the main sequence).
评估。正是这一负反馈"恒温器"使太阳光度在数十亿年间基本恒定，直到核心氢趋于耗尽、平衡改变（离开主序）。
Going deeper: the hydrostatic equilibrium equation深入：流体静力平衡方程
Consider a thin shell of gas at radius $r$, thickness $dr$, density $\rho$. The weight per unit area pulling it inward is balanced by the pressure difference across it pushing outward. This gives the hydrostatic equilibrium condition:
考虑半径 $r$、厚度 $dr$、密度 $\rho$ 的薄气壳。把它向内拉的单位面积重量与跨壳向外推的压强差相平衡，得到流体静力平衡条件：
$$ \frac{dP}{dr} = -\frac{G\, m(r)\, \rho}{r^2}, $$where $m(r)$ is the mass enclosed within radius $r$. The negative sign says pressure must fall outward (highest at the centre). The IB syllabus does not require you to derive or solve this, but the conceptual content — a pressure gradient supporting the overlying weight at every radius — is exactly the equilibrium you are expected to describe in words.
其中 $m(r)$ 为半径 $r$ 内所含质量。负号表明压强向外必须下降（中心最高）。IB 大纲不要求推导或求解此式，但其概念内容——在每一半径处由压强梯度支撑上方物质重量——正是你需要用文字描述的平衡。
Luminosity and Stefan-Boltzmann光度与斯特藩—玻尔兹曼定律 E.5 SL+HL
Stefan-Boltzmann law. Treating a star as a black body, from the data booklet
L = σAT⁴:
$$ L = \sigma A T^4, \qquad A = 4\pi R^2 \ \text{for a sphere}, $$
so $L = 4\pi R^2 \sigma T^4$. Here $T$ is the surface temperature in $\mathrm{K}$, $R$ the radius, and $\sigma = 5.67\times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$ (Stefan-Boltzmann constant, in the data booklet).
Watch. $T$ is to the fourth power — double the temperature, $16\times$ the luminosity. Use absolute temperature ($\mathrm{K}$), never $^\circ\mathrm{C}$.
斯特藩—玻尔兹曼定律。把恒星视为黑体，数据手册
L = σAT⁴：
$$ L = \sigma A T^4, \qquad A = 4\pi R^2 \ \text{（球体）}, $$
故 $L = 4\pi R^2 \sigma T^4$。其中 $T$ 为表面温度（$\mathrm{K}$），$R$ 为半径，$\sigma = 5.67\times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$（斯特藩—玻尔兹曼常量，见数据手册）。
注意。$T$ 是四次方——温度翻倍，光度变 $16$ 倍。用绝对温度（$\mathrm{K}$），绝不用 $^\circ\mathrm{C}$。
A star has radius $R = 7.0\times 10^{8}\ \mathrm{m}$ and surface temperature $T = 5800\ \mathrm{K}$. Find its luminosity. ($\sigma = 5.67\times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$.)某恒星半径 $R = 7.0\times 10^{8}\ \mathrm{m}$，表面温度 $T = 5800\ \mathrm{K}$。求其光度。（$\sigma = 5.67\times 10^{-8}\ \mathrm{W\,m^{-2}\,K^{-4}}$。）
Identify. Star as a black-body sphere: $L = \sigma A T^4$ with $A = 4\pi R^2$, via L = σAT⁴.
识别。恒星为黑体球：$L = \sigma A T^4$，$A = 4\pi R^2$，用 L = σAT⁴。
Set up. Surface area: $A = 4\pi (7.0\times 10^{8})^2 = 4\pi (4.9\times 10^{17}) \approx 6.16\times 10^{18}\ \mathrm{m^2}$.
列式。表面积：$A = 4\pi (7.0\times 10^{8})^2 = 4\pi (4.9\times 10^{17}) \approx 6.16\times 10^{18}\ \mathrm{m^2}$。
Execute. Fourth power of $T$: $(5800)^4 \approx 1.13\times 10^{15}\ \mathrm{K^4}$.
执行。$T$ 的四次方：$(5800)^4 \approx 1.13\times 10^{15}\ \mathrm{K^4}$。
$$ L = (5.67\times 10^{-8})(6.16\times 10^{18})(1.13\times 10^{15}) \approx 3.9\times 10^{26}\ \mathrm{W}. $$Evaluate. These are deliberately Sun-like figures, and the answer $\approx 3.9\times 10^{26}\ \mathrm{W}$ matches the Sun's luminosity — a good sanity check on the unit handling and the fourth power.
评估。这些数值有意取得接近太阳，结果 $\approx 3.9\times 10^{26}\ \mathrm{W}$ 与太阳光度吻合——对单位处理与四次方是个很好的自检。
Going deeper: comparing two stars with ratios深入：用比值比较两颗恒星
In exam problems you rarely need absolute values; ratios cancel $\sigma$ and $4\pi$. For two stars:
考题中很少需要绝对值；用比值可消去 $\sigma$ 与 $4\pi$。对两颗恒星：
$$ \frac{L_1}{L_2} = \left(\frac{R_1}{R_2}\right)^2 \left(\frac{T_1}{T_2}\right)^4. $$So a star with twice the radius and twice the surface temperature of another is $2^2 \times 2^4 = 4 \times 16 = 64$ times more luminous. This ratio form is the fastest route through most Stefan-Boltzmann comparison questions and avoids handling powers of ten.
故半径与表面温度均为另一颗两倍的恒星，光度为 $2^2 \times 2^4 = 4 \times 16 = 64$ 倍。这种比值形式是大多数斯特藩—玻尔兹曼比较题最快的路径，且避免处理 $10$ 的幂。
Apparent Brightness and the Inverse-Square Law视亮度与平方反比律 E.5 SL+HL
b = L/(4πd²):
$$ b = \frac{L}{4\pi d^2}. $$
Why inverse-square. The star's total power $L$ spreads over a sphere of radius $d$ (the distance to the observer), whose area is $4\pi d^2$. Dividing power by that area gives the power per unit area at distance $d$.
Finding distance. If $L$ is known (e.g. from the star's type) and $b$ is measured, rearrange: $$ d = \sqrt{\frac{L}{4\pi b}}. $$ Key distinction. $L$ is intrinsic (the star's own power); $b$ depends on distance. Two identical stars at different distances have the same $L$ but different $b$.
b = L/(4πd²)：
$$ b = \frac{L}{4\pi d^2}. $$
为何平方反比。恒星总功率 $L$ 散布在半径为 $d$（到观察者的距离）的球面上，其面积为 $4\pi d^2$。用功率除以该面积即得距离 $d$ 处的单位面积功率。
求距离。若已知 $L$（例如由恒星类型）且测得 $b$，变形： $$ d = \sqrt{\frac{L}{4\pi b}}. $$ 关键区别。$L$ 是内禀的（恒星自身功率）；$b$ 依赖距离。两颗相同的恒星处于不同距离，$L$ 相同而 $b$ 不同。
A star has luminosity $L = 4.0\times 10^{28}\ \mathrm{W}$ and is observed from Earth with apparent brightness $b = 8.0\times 10^{-9}\ \mathrm{W\,m^{-2}}$. Find its distance.某恒星光度 $L = 4.0\times 10^{28}\ \mathrm{W}$，从地球观测的视亮度 $b = 8.0\times 10^{-9}\ \mathrm{W\,m^{-2}}$。求其距离。
Identify. Inverse-square law $b = L/(4\pi d^2)$, via b = L/(4πd²); solve for $d$.
识别。平方反比律 $b = L/(4\pi d^2)$，用 b = L/(4πd²)；解出 $d$。
Set up. Rearrange: $d = \sqrt{\dfrac{L}{4\pi b}}$.
列式。变形：$d = \sqrt{\dfrac{L}{4\pi b}}$。
$$ d = \sqrt{\frac{4.0\times 10^{28}}{4\pi (8.0\times 10^{-9})}} = \sqrt{\frac{4.0\times 10^{28}}{1.005\times 10^{-7}}} = \sqrt{3.98\times 10^{35}} \approx 6.3\times 10^{17}\ \mathrm{m}. $$Evaluate. $6.3\times 10^{17}\ \mathrm{m} \approx 67$ light-years (divide by $9.46\times 10^{15}\ \mathrm{m/ly}$) — a plausible nearby-star distance. Always take the positive root; distance is positive.
评估。$6.3\times 10^{17}\ \mathrm{m} \approx 67$ 光年（除以 $9.46\times 10^{15}\ \mathrm{m/光年}$）——一个合理的邻近恒星距离。始终取正根；距离为正。
Going deeper: combining all three relations深入：三式联用
A common Paper 2 question chains the unit's relations. Given a star's surface temperature $T$, radius $R$, and measured apparent brightness $b$, find its distance:
Paper 2 常把本单元各式串联。已知恒星表面温度 $T$、半径 $R$ 与测得的视亮度 $b$，求距离：
First get the luminosity from Stefan-Boltzmann, then feed it into the inverse-square law:
先由斯特藩—玻尔兹曼求光度，再代入平方反比律：
$$ L = 4\pi R^2 \sigma T^4, \qquad d = \sqrt{\frac{L}{4\pi b}} = R\, \sqrt{\frac{\sigma T^4}{b}}. $$The compact form $d = R\sqrt{\sigma T^4 / b}$ shows the two $4\pi$ factors cancel cleanly. Recognising that cancellation saves arithmetic and reduces calculator error in the exam.
紧凑形式 $d = R\sqrt{\sigma T^4 / b}$ 表明两个 $4\pi$ 因子干净地相消。识别该相消可省去计算、减少考场计算器误差。
The Main Sequence, Mass, Lifetime and End States主序、质量、寿命与归宿 E.5 SL+HL
Mass sets everything.
- More massive stars have stronger gravity, hotter cores, much higher fusion rates and so much higher luminosity (roughly $L \propto M^{3.5}$). HL
- They are hotter and bluer at the surface.
- Lifetime falls with mass. Although massive stars have more fuel, they burn it disproportionately fast, so they live shorter lives ($t \propto M / L \propto M^{-2.5}$). HL
质量决定一切。
- 质量更大的恒星引力更强、核心更热、聚变率高得多，故光度高得多（大致 $L \propto M^{3.5}$）。HL
- 它们表面更热、更蓝。
- 寿命随质量减小。大质量恒星虽燃料更多，却消耗得不成比例地快，故寿命更短（$t \propto M / L \propto M^{-2.5}$）。HL
HL A main-sequence star has mass $M = 10\,M_\odot$. Using $L \propto M^{3.5}$ and lifetime $t \propto M / L$, estimate its main-sequence lifetime relative to the Sun's ($\approx 1.0\times 10^{10}$ years).HL 某主序星质量 $M = 10\,M_\odot$。用 $L \propto M^{3.5}$ 与寿命 $t \propto M / L$，估算其主序寿命相对太阳（$\approx 1.0\times 10^{10}$ 年）之比。
Identify. Lifetime scales as available fuel ($\propto M$) divided by burn rate ($\propto L \propto M^{3.5}$): $t \propto M / M^{3.5} = M^{-2.5}$.
识别。寿命正比于可用燃料（$\propto M$）除以消耗率（$\propto L \propto M^{3.5}$）：$t \propto M / M^{3.5} = M^{-2.5}$。
Set up. Ratio to the Sun ($M_\odot$, $t_\odot$): $\dfrac{t}{t_\odot} = \left(\dfrac{M}{M_\odot}\right)^{-2.5}$.
列式。对太阳之比（$M_\odot$、$t_\odot$）：$\dfrac{t}{t_\odot} = \left(\dfrac{M}{M_\odot}\right)^{-2.5}$。
$$ \frac{t}{t_\odot} = (10)^{-2.5} = \frac{1}{10^{2.5}} \approx \frac{1}{316} \approx 3.2\times 10^{-3}. $$Execute. $t \approx (3.2\times 10^{-3})(1.0\times 10^{10}) \approx 3\times 10^{7}$ years.
执行。$t \approx (3.2\times 10^{-3})(1.0\times 10^{10}) \approx 3\times 10^{7}$ 年。
Evaluate. About $30$ million years — roughly a third of a thousandth of the Sun's lifetime, despite having ten times the fuel. Massive stars live fast and die young.
评估。约 $3000$ 万年——尽管燃料是太阳的十倍，寿命却约为太阳的三十万分之一。大质量恒星"快活短命"。
Going deeper: reading a Hertzsprung-Russell diagram深入：读懂赫罗图
An H-R diagram plots luminosity (vertical, increasing upward) against surface temperature (horizontal, but increasing to the left by convention). Most stars lie on a diagonal band — the main sequence — running from hot, bright, blue stars (top-left) to cool, faint, red stars (bottom-right). Position along the main sequence is set by mass: massive stars at the top-left, low-mass stars at the bottom-right.
赫罗图以光度（纵轴，向上增大）对表面温度（横轴，但按惯例向左增大）作图。大多数恒星位于一条对角带上——主序——从又热又亮的蓝星（左上）延伸到又冷又暗的红星（右下）。沿主序的位置由质量决定：大质量星在左上，低质量星在右下。
Off the main sequence sit red giants and supergiants (upper-right: cool but very luminous because they are huge) and white dwarfs (lower-left: hot but faint because they are tiny). Stefan-Boltzmann ($L = 4\pi R^2 \sigma T^4$) explains these positions: at fixed $T$, a larger $R$ raises $L$, placing giants high and dwarfs low. These end states are the destinations stars reach after leaving the main sequence.
主序之外有红巨星与超巨星（右上：温度低但因体积巨大而极亮）以及白矮星（左下：温度高但因体积极小而暗淡）。斯特藩—玻尔兹曼（$L = 4\pi R^2 \sigma T^4$）解释这些位置：$T$ 不变时，$R$ 更大则 $L$ 更高，故巨星居高、矮星居低。这些归宿是恒星离开主序后到达的终点。
Exam Strategy and Common Pitfalls考试策略与常见陷阱
- Use $A = 4\pi R^2$ for the surface area of a spherical star, then $L = \sigma A T^4$. Forgetting the $4\pi$ (or using $\pi R^2$) is the most common error.
- 球形恒星的表面积用 $A = 4\pi R^2$，再代 $L = \sigma A T^4$。漏掉 $4\pi$（或误用 $\pi R^2$）是最常见错误。
- Temperature is to the fourth power and must be in kelvin. Convert $^\circ\mathrm{C}$ to $\mathrm{K}$ first; raise to the 4th, not the 2nd.
- 温度是四次方且必须用开尔文。先把 $^\circ\mathrm{C}$ 换成 $\mathrm{K}$；取四次方，不是二次方。
- Luminosity $L$ is intrinsic (total power radiated, in $\mathrm{W}$). Apparent brightness $b$ is what we measure at Earth (power per area, in $\mathrm{W\,m^{-2}}$), and depends on distance.
- 光度 $L$ 是内禀的（辐射总功率，$\mathrm{W}$）。视亮度 $b$ 是我们在地球测得的（单位面积功率，$\mathrm{W\,m^{-2}}$），且依赖距离。
- Link them with $b = L/(4\pi d^2)$. The $4\pi d^2$ is the area of the sphere over which $L$ has spread by the time it reaches you.
- 用 $b = L/(4\pi d^2)$ 连接二者。$4\pi d^2$ 是 $L$ 到达你时所散布球面的面积。
- "Why does a star release energy by fusion?" Light nuclei sit lower on the binding-energy-per-nucleon curve; fusing them gives a more tightly bound product, and the mass defect is released as energy.
- "恒星为何通过聚变释放能量？"轻核在每核子结合能曲线上更低；聚变得到结合更紧的产物，质量亏损以能量释放。
- "Why is the star stable?" Name both agents and their directions: inward gravitation balanced by outward radiation and gas pressure (hydrostatic equilibrium).
- "恒星为何稳定？"点名两个主体及其方向：向内的引力与向外的辐射压和气体压平衡（流体静力平衡）。
- For Stefan-Boltzmann comparisons, use $L_1/L_2 = (R_1/R_2)^2 (T_1/T_2)^4$. The constants cancel and you avoid powers of ten.
- 斯特藩—玻尔兹曼比较用 $L_1/L_2 = (R_1/R_2)^2 (T_1/T_2)^4$。常数相消，避免处理 $10$ 的幂。
- For brightness comparisons, use $b \propto L/d^2$. Equal-brightness questions reduce to $d \propto \sqrt{L}$.
- 亮度比较用 $b \propto L/d^2$。"等亮度"题化为 $d \propto \sqrt{L}$。
Flashcards闪卡
Unit E.5 Practice Quiz单元 E.5 练习测验
Readiness Checklist备考清单
Tick each item when you can do it cold, without notes, on your first attempt.
每一条都要"裸做"做对（不看笔记、一次过）才打勾。
- Explain, using the binding-energy-per-nucleon curve, why fusing light nuclei releases energy用每核子结合能曲线解释为何轻核聚变释放能量
- State the two conditions for fusion (high temperature and high density) and link them to the Coulomb barrier陈述聚变两条件（高温与高密度）并与库仑势垒联系
- Compute the energy released in a fusion reaction from masses using $E = \Delta m c^2$用 $E = \Delta m c^2$ 由质量算出聚变反应释放的能量
- Summarise the net proton-proton chain as $4\,{}^{1}\mathrm{H} \to {}^{4}\mathrm{He}$ and state it powers the Sun把净 p-p 链概括为 $4\,{}^{1}\mathrm{H} \to {}^{4}\mathrm{He}$ 并说明它为太阳供能
- Relate luminosity to mass-energy conversion rate via $L = (dm/dt)c^2$用 $L = (dm/dt)c^2$ 把光度与质量转能率联系
- State stellar equilibrium as inward gravitation balancing outward radiation and gas pressure把恒星平衡表述为向内引力平衡向外辐射压与气体压
- Explain the negative-feedback stability of a main-sequence star解释主序星的负反馈稳定性
- Define luminosity and compute it with $L = 4\pi R^2 \sigma T^4$ (Stefan-Boltzmann)定义光度并用 $L = 4\pi R^2 \sigma T^4$（斯特藩—玻尔兹曼）计算
- Compare two stars with $L_1/L_2 = (R_1/R_2)^2 (T_1/T_2)^4$用 $L_1/L_2 = (R_1/R_2)^2 (T_1/T_2)^4$ 比较两颗恒星
- Distinguish apparent brightness $b$ from luminosity $L$ and apply $b = L/(4\pi d^2)$区分视亮度 $b$ 与光度 $L$ 并应用 $b = L/(4\pi d^2)$
- Find a stellar distance from $L$ and $b$ using $d = \sqrt{L/(4\pi b)}$用 $d = \sqrt{L/(4\pi b)}$ 由 $L$ 与 $b$ 求恒星距离
- HL Explain qualitatively why a more massive main-sequence star is more luminous and shorter-lived定性解释为何质量更大的主序星光度更高、寿命更短
IB Paper-Style PracticeIB 试卷风格练习
E.5 Practice and Solutions are on the roadmap, to ship under Practice Questions/Unit_E5_*.html with the bilingual built-in pattern.
E.5 配套 Practice 与 Solutions 在排期，上线后位于 Practice Questions/Unit_E5_*.html。