OCR物理 天体物理 光度 黑体辐射 红移
天体物理是 OCR A Level 物理 A（H556）模块 5 里的 5.5 节，和热学、圆周运动、引力场放在同一个模块，所以它不是一门孤立的天文常识，而是把引力、辐射与能量守恒搬到恒星尺度上去用。这一节的计算题出现在 Paper 1 与 Paper 3 上，题目通常把光度、黑体辐射、红移与哈勃定律串成一条链，前一问的答案往往就是后一问的已知量。很多学生以为这一节靠背结论，实际上拉开差距的是公式选择与单位换算。下面按考纲顺序把公式、例题与常见失分点过一遍。
Astrophysics and cosmology form section 5.5 of Module 5 in OCR A Level Physics A, sitting alongside thermal physics, circular motion and gravitational fields. Questions appear in Component 01 and Component 03, and they usually chain luminosity, black-body radiation, redshift and Hubble's law together so that one answer becomes the next input. What separates grades is rarely the vocabulary; it is choosing the right equation and converting units cleanly.
卷面位置：5.5 在 H556 上的分值 / Where 5.5 sits in H556
OCR 官方 spec（H556，2015 年 9 月起施教，2017 年首次考试）把内容分成六个模块，5.5 属于模块 5 Newtonian world and astrophysics。Component 01 Modelling physics 考模块 1、2、3、5，2 小时 15 分、100 分、占 37%；Component 02 考模块 1、2、4、6，同为 2 小时 15 分、100 分、37%；Component 03 Unified physics 综合全部模块，1 小时 30 分、70 分、占 26%。天体物理因此只出现在 Paper 1 与 Paper 3 上，Paper 2 不会考。Paper 1 里的 5.5 常是一道多小问的计算题，从黑体辐射问到红移再问到距离。很多学生以为只有最后答案算对才得分，阅卷时每一条公式代值都单独给分。
In the OCR specification for H556, first teaching September 2015 and first assessment 2017, the content divides into six teaching modules and 5.5 sits inside Module 5, Newtonian world and astrophysics. Component 01 covers modules 1, 2, 3 and 5 over 2 hours 15 minutes for 100 marks and 37 per cent, Component 02 covers modules 1, 2, 4 and 6 on the same terms, and Component 03 pulls all six modules together in 1 hour 30 minutes for 70 marks and 26 per cent. Cosmology therefore never appears on Component 02, and in practice a 5.5 question arrives as a multi-part calculation where the mark scheme credits each correct substitution.
光度与辐射通量强度 / Luminosity and radiant flux intensity
光度 L 是恒星每秒辐射的总能量，单位 W；辐射通量强度 F 是单位面积接收到的功率，单位 W m⁻²。两者差一个球面面积，恒星发出的功率摊到半径等于距离 d 的球面上，于是 F = L / (4πd²)。OCR 数据手册印的是 L = 4πr²σT⁴，σ 取 5.67 × 10⁻⁸ W m⁻² K⁻⁴。拿太阳验算：L 取 3.8 × 10²⁶ W、r 取 6.96 × 10⁸ m，反解 T 约 5760 K，与太阳表面约 5800 K 的说法一致。换方向，一颗 L = 3.8 × 10²⁶ W 的恒星放到 10 pc（3.1 × 10¹⁷ m）外，F 约 3.2 × 10⁻¹⁰ W m⁻²。学生最常把 r 与 d 用混，一个描述恒星多大、一个描述它多远，两个平方项不能互换；把 F 的单位写成 W 也会直接丢分。
Luminosity is the total energy a star radiates each second in watts, while radiant flux intensity is the power received per unit area in W m⁻², so the two differ by the sphere's surface area: F = L / (4πd²). The data booklet gives L = 4πr²σT⁴ with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, and the Sun's L = 3.8 × 10²⁶ W with r = 6.96 × 10⁸ m returns T close to 5760 K; at 10 pc, or 3.1 × 10¹⁷ m, the same star delivers about 3.2 × 10⁻¹⁰ W m⁻². Examiners see the same slip every year, because a stellar radius gets squared where a distance was needed, and the flux is then quoted in watts rather than W m⁻².
黑体辐射与 Wien 位移定律 / Black-body radiation and Wien's displacement law
恒星的连续谱接近黑体谱，峰值波长 λmax 与表面温度成反比，温度越高峰值越短、越偏蓝。数据手册只给比例式 λmax ∝ 1/T，没给带常数的等式，所以考题里最稳的是比值法 λmax1 T1 = λmax2 T2。太阳峰值在 500 nm，用 2.9 × 10⁻³ m K 反算得 5800 K，与 Stefan 定律的结果互相印证。双星题常问 A 星峰值 290 nm、B 星峰值 580 nm 时温度之比，波长差一倍，温度就差一倍，答案 2:1。真实恒星的吸收线会让光谱凹陷，峰值附近的位置仍可用来估计温度。这一步最容易出错的是单位，290 nm 要写成 2.9 × 10⁻⁷ m，直接用纳米代入会差 10⁹ 倍。
A star's continuous spectrum resembles a black-body curve whose peak wavelength falls as temperature rises, which is why hot stars look blue-white and cool ones red. The booklet gives the proportionality λmax ∝ 1/T rather than an equation with a constant, so the ratio λmax1 T1 = λmax2 T2 is the safe route: peaks at 290 nm and 580 nm imply a temperature ratio of 2:1. Absorption lines do distort a real spectrum, but the peak still locates the temperature, provided every wavelength is converted into metres before substitution.
用 Stefan 定律反推恒星半径 / Finding a stellar radius with Stefan's law
从光谱峰值拿到温度、从通量强度与距离拿到光度之后，半径由 r = sqrt(L / (4πσT⁴)) 反解。例题：一颗红巨星的光度是太阳的 4 倍、表面温度 3000 K，先写 L = 4 × 3.8 × 10²⁶ = 1.52 × 10²⁷ W，(3000)⁴ = 8.1 × 10¹³，于是 4πσT⁴ 约 5.8 × 10⁷ W m⁻²，相除得 2.6 × 10¹⁹ m²，开方 r 约 5.1 × 10⁹ m，约 7 倍太阳半径。这解释了 H-R 图上巨星为什么在同样温度下更亮：表面积大。顺序不能颠倒，先温度、再光度、最后半径，任何一步用错量，后面的答案全部作废。很多学生以为半径可以直接由视直径测出，A Level 阶段除了少数近距离恒星，只有这条间接路径。
With temperature from the spectrum and luminosity from flux and distance, the radius follows from r = sqrt(L / (4πσT⁴)). A red giant of four solar luminosities at 3000 K gives (3000)⁴ = 8.1 × 10¹³, so 4πσT⁴ is about 5.8 × 10⁷ W m⁻², and 1.52 × 10²⁷ W divided by that leaves 2.6 × 10¹⁹ m²; the square root is roughly 5.1 × 10⁹ m, near seven solar radii. Taking temperature, then luminosity, then radius protects you, because one wrong quantity at the start ruins every later part.
恒星演化与 H-R 图 / Stellar evolution and the H-R diagram
H-R 图横轴是表面温度、向右递减，纵轴是光度，主序带、巨星与超巨星、白矮星各占一片区域。质量与太阳接近的恒星在主序上聚变氢，核心氢耗尽后膨胀成红巨星，外层抛掉形成行星状星云，核心冷却成白矮星，白矮星的质量上限约 1.4 倍太阳质量。质量大得多的恒星走得更快，聚变到铁核后坍缩引发超新星爆发，核心可能留下中子星或黑洞。主序寿命大致按 M / L 缩放，而光度大致按 M^3.5 增长，所以质量越大寿命反而越短。白矮星落在图的左下角，温度与主序星相近，光度却低得多，因为它的半径只有地球量级。很多学生以为 H-R 图是同一颗恒星在时间上的轨迹，实际上图上的位置是同一时刻不同恒星的状态，演化路径只是叠加上去的一条线。
On a Hertzsprung-Russell diagram temperature runs along the horizontal axis decreasing to the right and luminosity up the vertical axis, with main sequence stars, giants, supergiants and white dwarfs in separate bands. A star like the Sun fuses hydrogen on the main sequence, expands into a red giant, sheds a planetary nebula and leaves a white dwarf capped near 1.4 solar masses, whereas a heavier star burns through to an iron core and ends in a supernova that may leave a neutron star or a black hole. Main sequence lifetime scales roughly as M / L, and because luminosity grows about as M^3.5 the heaviest stars die first; reading the diagram as one star's journey over time is a common misreading.
核聚变与质能方程 / Fusion and the mass-energy equation
恒星的能量来自四个质子聚变成氦-4 的质子-质子链，过程中放出两个正电子与两个中微子。质量并不守恒：四个质子共 4 × 1.00728 u = 4.02912 u，氦-4 核约 4.00151 u，亏损 0.02761 u，约占初始质量的 0.7%。亏损换成千克是 0.02761 × 1.661 × 10⁻²⁷ 约 4.6 × 10⁻²⁹ kg，代入 E = mc² 得约 4.1 × 10⁻¹² J，这就是一次聚变释放的能量。NASA 的太阳数据给出每秒消耗约 6 × 10¹¹ kg 氢，按上面的效率折算出 3.8 × 10²⁶ W 的输出量级。常见的错误是把质量亏损说成质量消失了，它其实以能量形式离开恒星。阅卷时 E = mc² 里三个量都要写清来源与单位，跳步只写数字拿不到过程分。
Stars run on the proton-proton chain, in which four protons fuse into a helium-4 nucleus and release two positrons and two neutrinos. Four protons total 4.02912 u against about 4.00151 u for the helium nucleus, a deficit of 0.02761 u, or 0.7 per cent of the original mass; converting that to 4.6 × 10⁻²⁹ kg and applying E = mc² gives roughly 4.1 × 10⁻¹² J per fusion event. NASA's solar figures put hydrogen consumption near 6 × 10¹¹ kg per second, which converts to the familiar 3.8 × 10²⁶ W of output, and mass lost in the process leaves the star as energy rather than disappearing.
视差与秒差距 / Parallax and the parsec
秒差距由观测定义：地球绕太阳的基线为 1 天文单位时，恒星的周年视差为 1 角秒，距离就是 1 秒差距。关系式是 d(pc) = 1 / p(角秒)，1 pc 约 3.1 × 10¹⁶ m（数据手册值）、约 3.26 光年、约 2.06 × 10⁵ 天文单位。视差 0.05 角秒对应 20 pc，即 6.2 × 10¹⁷ m；0.01 角秒对应 100 pc。地面望远镜受大气抖动限制，能测到的视差大约到 0.01 角秒这一档，视差法因此只覆盖太阳附近。ESA 的 Hipparcos 与 Gaia 把测量搬到太空，可测距离被推远，更远处就必须换标准烛光。学生最常把角秒与度混用，1 角秒是 1/3600 度，忘记换算会让距离小掉三个量级。
The parsec is defined observationally: a star with an annual parallax of one arcsecond, measured against a baseline of one astronomical unit, lies one parsec away, so d in parsecs equals 1 / p in arcseconds. One parsec is about 3.1 × 10¹⁶ m in the data booklet, roughly 3.26 light years and some 2.06 × 10⁵ astronomical units, which puts a parallax of 0.05 arcseconds at 20 pc, or 6.2 × 10¹⁷ m. Ground-based parallax stops near 0.01 arcsecond because of atmospheric turbulence, and the ESA Hipparcos and Gaia missions moved the measurement into space, so anything much further out needs a standard candle.
标准烛光与造父变星 / Standard candles and Cepheid variables
标准烛光是光度已知的天体，测到通量强度 F 就能用 d = sqrt(L / (4πF)) 反推距离，这也是视差够不着的距离区间里唯一的办法。造父变星最好用：1912 年 Leavitt 统计麦哲伦云中的造父变星时发现，光变周期越长光度越高，于是测出周期就能读出光度。周期-光度关系必须先由视差或星团校准，再用到更远的星系，这条距离阶梯每上一级都会带进前一级的误差。Ia 型超新星的峰值光度足够一致，被用来测到几十亿光年之外，也是 1998 年发现加速膨胀所依赖的观测。常见的错误是直接拿视星等当亮度比较，星际尘埃削弱并红化星光，距离会被高估。很多学生以为标准烛光必须有完全相同的亮度，实际上只要光度能由某个可观测量唯一确定就够了。
A standard candle is an object of known luminosity, so once its flux intensity is measured the distance follows from d = sqrt(L / (4πF)), and beyond a few hundred parsecs nothing else works. Cepheid variables are the classic case, because Leavitt found in 1912 that longer pulsation periods go with greater luminosity in the Magellanic Clouds, so the period alone yields the luminosity once the relation has been calibrated with parallax or a star cluster. Type Ia supernovae stretch the same method to billions of light years and supplied the data behind the discovery of accelerating expansion in 1998, while dust that reddens and dims a star is what pushes a distance estimate too high.
红移与哈勃定律 / Redshift and Hubble's law
远离我们的星系光谱整体向长波方向移动，红移量 z = Δλ / λ，在速度远小于光速时 Δλ / λ ≈ Δf / f ≈ v / c，这正是数据手册给的那串等式。1929 年 Hubble 把红移与距离作图得到 v = H₀d，退行速度与距离成正比，试卷给的 H₀ 是 2.2 × 10⁻¹⁸ s⁻¹，也可以写成 km s⁻¹ Mpc⁻¹。例题：氢线 656.3 nm 移到 660.0 nm，Δλ = 3.7 nm，z = 5.6 × 10⁻³，v = c z 约 1.7 × 10⁶ m s⁻¹。再除以 H₀ 得 7.7 × 10²³ m，约 25 Mpc。宇宙学红移来自空间本身的膨胀，与星系在静态空间里穿行的多普勒效应只是形式上相似。这一步最容易错在 z 写成百分数，5.6 × 10⁻³ 写成 5.6 会让速度放大一千倍。
Light from a receding galaxy shifts towards longer wavelengths, and for speeds well below c the booklet's chain Δλ / λ ≈ Δf / f ≈ v / c applies, leading to Hubble's 1929 result v = H₀d with the examination value H₀ = 2.2 × 10⁻¹⁸ s⁻¹. A hydrogen line observed at 660.0 nm instead of 656.3 nm gives a shift of 3.7 nm, so z = 5.6 × 10⁻³, v = c z ≈ 1.7 × 10⁶ m s⁻¹, and a distance near 7.7 × 10²³ m, or 25 Mpc. Cosmological redshift arises from the expansion of space rather than from a galaxy travelling through static space, and writing z as 5.6 instead of 5.6 × 10⁻³ inflates the velocity a thousandfold.
宇宙年龄与微波背景 / Age of the universe and the microwave background
把哈勃定律反过来读，宇宙年龄可以用 t = H₀⁻¹ 估算，数据手册直接给的正是这个式子。代入 H₀ = 2.2 × 10⁻¹⁸ s⁻¹ 得 4.5 × 10¹⁷ s，用 1 年约 3.16 × 10⁷ s 换算约 1.4 × 10¹⁰ 年。ESA 的 Planck 2018 结果给出宇宙年龄 13.797 ± 0.023 十亿年、哈勃常数 67.4 ± 0.5 km s⁻¹ Mpc⁻¹，两者在量级上对得上。差别来自加速膨胀，H₀⁻¹ 假设膨胀速度一直等于今天的值，实际早期更慢、晚期更快。大爆炸的主要证据是星系红移、微波背景辐射与轻元素丰度，微波背景是 1965 年 Penzias 与 Wilson 调试天线时发现的，温度约 2.7 K 且高度各向同性。Planck 后来测出十万分之一量级的温度涨落，这些起伏正是星系团后来得以形成的种子。
Read backwards, Hubble's law estimates the age of the universe as t = H₀⁻¹, which for H₀ = 2.2 × 10⁻¹⁸ s⁻¹ gives 4.5 × 10¹⁷ s, about 1.4 × 10¹⁰ years using the booklet's year of 3.16 × 10⁷ s. ESA's Planck 2018 results put the age at 13.797 ± 0.023 billion years with H₀ = 67.4 ± 0.5 km s⁻¹ Mpc⁻¹, and the gap against that simple estimate comes from accelerated expansion, since H₀⁻¹ assumes today's rate has always applied. Evidence for a hot big bang rests on galactic redshift, the cosmic microwave background and the light-element abundances, and the microwave background itself was found accidentally in 1965 by Penzias and Wilson at about 2.7 K, later resolved by Planck into ripples of roughly one part in a hundred thousand.
暗物质与暗能量 / Dark matter and dark energy
星系旋转曲线在外围几乎不随半径下降，可见质量算出的引力不足以维持这样的速度，这是暗物质最早的论据，1970 年代 Rubin 等人的观测把它推到了台面上。独立支持来自星系团质量、引力透镜与微波背景涨落，Planck 2018 测得暗物质密度参数 Ωc h² = 0.120 ± 0.001，物质总量 Ωm = 0.315 ± 0.007。1998 年两个独立团队用 Ia 型超新星测距离，发现膨胀在加速而不是减速，2011 年诺贝尔物理学奖授予 Perlmutter、Schmidt 与 Riess。加速膨胀需要一个压强为负的成分，通常称为暗能量，Planck 2018 给出的状态方程参数 w = -1.03 ± 0.03 与宇宙学常数一致。OCR 的 5.5 只要求定性理解这些证据与结论。错得最多的是把暗物质与暗能量当成一件事：前者让结构聚起来，后者让膨胀加速，观测手段也完全不同。
Galaxy rotation curves stay nearly flat at large radii, so luminous matter cannot supply the gravity required, which is the original argument for dark matter and one that Vera Rubin and colleagues pushed into the mainstream in the 1970s. Independent support comes from cluster masses, gravitational lensing and microwave background ripples, and Planck 2018 measures a dark matter density parameter Ωc h² = 0.120 ± 0.001 inside a total matter density Ωm = 0.315 ± 0.007. In 1998 two independent teams using Type Ia supernovae found that the expansion is accelerating, work recognised by the 2011 Nobel Prize in Physics, and the required negative-pressure component, dark energy, has an equation-of-state parameter w = -1.03 ± 0.03 in Planck 2018; OCR's 5.5 asks only for a qualitative grasp of this evidence.
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