Standard candles · 标准烛光
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| luminosity/ˌluːmɪˈnɒsɪti/ | 光度 | guāng dù |
| inverse-square law/ɪnˈvɜːs skweə lɔː/ | 平方反比定律 | píng fāng fǎn bǐ dìng lǜ |
| standard candle/ˈstændəd ˈkændl/ | 标准烛光 | biāo zhǔn zhú guāng |
| total power/ˈtəʊtl ˈpaʊə/ | 功率 | gōng lǜ |
| energy/ˈenədʒi/ | 能量 | néngliàng |
| radiant flux intensity/ˈreɪdɪənt flʌks ɪnˈtensɪti/ | 辐射通量密度 | fú shè tōng liàng mì dù |
| Cepheid variables/ˈsefiɪd ˈveərɪəblz/ | 造父变星 | zào fù biàn xīng |
| Type Ia supernovae/taɪp wʌn ˈeɪ ˌsuːpəˈnəʊviː/ | 超新星 | chāo xīn xīng |
Two stars, equally bright, nothing alike
- Pick two stars out of the sky that look exactly as bright as each other. One may be a modest star nearby; the other a monster a hundred times further away.
- Brightness on its own tells you nothing. It is the product of two unknowns: how much light the star makes, and how far it has had to spread.
- Break that deadlock and you have measured the distance to another galaxy. The trick is to find an object whose output you already know.
- This lesson is luminosity 光度, the inverse-square law for flux, and the standard candle 标准烛光.
两颗一样亮的星,毫无相似之处
- 在天上挑两颗看起来一样亮的星。其中一颗可能是近处一颗平平常常的星;另一颗则是远上一百倍的庞然大物。
- 单看亮度什么也说明不了。它是两个未知量的乘积:恒星发出多少光,以及那些光被摊开了多远。
- 打破这个僵局,你就量出了到另一个星系的距离。诀窍是找到一个你已经知道其输出的天体。
- 这一课讲光度(luminosity)、通量的平方反比定律,以及标准烛光(standard candle)。
Luminosity
- The luminosity of a star is the total power 功率 of radiation emitted by the star, in watts. Equivalently, the total energy 能量 emitted per unit time.
- It is a property of the star alone. How bright it looks from Earth is not.
- Asked for two reasons one star appears brighter than another, give exactly these two: a greater luminosity, and a smaller distance.
- Anything else, including colour or size, only matters through those two.
光度
- 恒星的光度是该恒星发出的辐射的总功率(total power),单位是瓦。等价地说,是单位时间发出的总能量(energy)。
- 它是恒星本身的属性。而它从地球上看起来有多亮则不是。
- 若问一颗星看起来比另一颗亮的两个原因,就给这两个:光度更大,以及距离更近。
- 别的一切,包括颜色和大小,都只能通过这两条起作用。
The luminosity of a star is: · 恒星的光度是：
Luminosity $L$ is the energy radiated per second in all directions (watts). · 光度 $L$ 是指向四面八方每秒辐射的能量（瓦特）。
Why might one star appear brighter than another? Select all · 所有 that apply. · 为什么一颗恒星看起来比另一颗更亮？选择所有适用项。
Colour and size only matter through the luminosity they produce. The two marked reasons are greater luminosity and smaller distance. · 颜色和大小仅通过它们产生的光度起作用。标记的两个原因是更高光度和更短距离。
Flux and the inverse-square law
- At a distance $d$ that power has spread over a sphere of area $4\pi d^2$, so the radiant flux intensity 辐射通量密度 is
- Units: $\text{W/m}^2$. This is the inverse-square law 平方反比定律. Double the distance and the flux falls to a quarter.
- A telescope measures $F$. If $L$ is known too, the distance follows:
Same light, four times the area
通量与平方反比定律
- 在距离 $d$ 处,那份功率已经摊在面积为 $4\pi d^2$ 的球面上,所以辐射通量密度(radiant flux intensity)是
- 单位:$\text{W/m}^2$。这就是平方反比定律(inverse-square law)。距离加倍,通量降到四分之一。
- 望远镜测的是 $F$。若 $L$ 也已知,距离就出来了:
同样的光,四倍的面积
Standard candle distance lab · 标准烛光距离实验室
brightness proportional to 1 / distance^2 · 亮度与 1 / distance^2 成正比
Move distance and see why brightness falls quickly. · 移动距离并观察亮度为何迅速下降。
The flux received from a star falls as 1 over the distance ____. · 来自恒星的通量随距离1而下降。
$F = \dfrac{L}{4\pi d^{2}}$ — the inverse-square law. · $F = \dfrac{L}{4\pi d^{2}}$ —— 平方反比定律。
Doubling the distance to a star reduces the radiant flux intensity to one half. · 将恒星距离加倍会使辐射通量强度减半。
To one QUARTER. The power spreads over a sphere of area 4 pi d^2, so flux goes as 1/d^2. Using pi d^2 instead is out by a factor of four. · 变为四分之一。功率分布在面积为 4 pi d^2 的球面上，因此通量与 1/d^2 成正比。使用 pi d^2 会导致结果相差四倍。
Worked example: sunlight at the Earth
- The Sun's luminosity is $3.8\times10^{26}\ \text{W}$ and the Earth is $1.5\times10^{11}\ \text{m}$ away. Find the radiant flux intensity here.
- $F = \dfrac{L}{4\pi d^2} = \dfrac{3.8\times10^{26}}{4\pi(1.5\times10^{11})^2} = 1.4\times10^{3}\ \text{W/m}^2$.
- That is the number behind every solar-panel calculation you will ever do.
- The $4\pi d^2$ is the area of a sphere, not a circle. Using $\pi d^2$ or $\pi r^2$ is the standard slip and it is out by a factor of four.
例题:地球处的阳光
- 太阳的光度是 $3.8\times10^{26}\ \text{W}$,地球在 $1.5\times10^{11}\ \text{m}$ 之外。求这里的辐射通量密度。
- $F = \dfrac{L}{4\pi d^2} = \dfrac{3.8\times10^{26}}{4\pi(1.5\times10^{11})^2} = 1.4\times10^{3}\ \text{W/m}^2$。
- 这就是你今后每一道太阳能电池板计算背后的那个数。
- $4\pi d^2$ 是球的面积,不是圆的。写成 $\pi d^2$ 或 $\pi r^2$ 是标准失误,会差四倍。
A star gives a flux of $100\ \dfrac{\text{W}}{\text{m}^2}$ at distance $d$. What is the flux at $3d$? · 一颗恒星在距离 $100\ \dfrac{\text{W}}{\text{m}^2}$ 处产生的流量为 $d$。该处的流量是多少 $3d$?
Inverse-square: $\dfrac{100}{3^{2}} = \dfrac{100}{9} \approx 11\ \dfrac{\text{W}}{\text{m}^2}$. · 平方反比：$\dfrac{100}{3^{2}} = \dfrac{100}{9} \approx 11\ \dfrac{\text{W}}{\text{m}^2}$。
The graph that gives you $L$
- Plot $F$ against $1/d^2$ for one star and the inverse-square law becomes a straight line through the origin.
- Its gradient is $L/4\pi$, so $L = 4\pi \times \text{gradient}$.
- A more luminous star gives a steeper line. The origin matters: a line that misses it is not obeying an inverse-square law.
Straighten the curve and read off the luminosity
那张能给出 $L$ 的图
- 对一颗恒星,把 $F$ 对 $1/d^2$ 作图,平方反比定律就变成一条过原点的直线。
- 它的斜率是 $L/4\pi$,所以 $L = 4\pi \times$ 斜率。
- 光度更大的星给出更陡的线。过原点这一点很要紧:不过原点的直线并不服从平方反比。
把曲线拉直,就能读出光度
A graph of flux against $1/d^2$ is a straight line through the origin whose gradient equals the luminosity divided by . · 通量对 $1/d^2$ 的图是一条过原点的直线，其斜率等于光度除以。
F = (L/4pi)(1/d^2), so L = 4pi x gradient. A more luminous star gives a steeper line, and a line missing the origin is not obeying an inverse-square law. · F = (L/4pi)(1/d^2)，所以 L = 4pi x 斜率。光度越高的恒星产生越陡的直线，不过原点的直线则不遵循平方反比定律。
Worked example: a solar panel on a probe
- The Sun has radius $6.96\times10^{8}\ \text{m}$ and surface temperature $5780\ \text{K}$. A probe $4.5\times10^{10}\ \text{m}$ from the Sun's centre carries a $2.0\ \text{m}^2$ panel facing the Sun. Find the power falling on it.
- Luminosity: $L = 4\pi\sigma r^2 T^4 = 4\pi(5.67\times10^{-8})(6.96\times10^{8})^2(5780)^4 = 3.85\times10^{26}\ \text{W}$.
- Flux there: $F = \dfrac{L}{4\pi d^2} = \dfrac{3.85\times10^{26}}{4\pi(4.5\times10^{10})^2} = 1.5\times10^{4}\ \text{W/m}^2$, eleven times the flux at the Earth.
- Power: $P = FA = 1.5\times10^{4}\times2.0 = 3.0\times10^{4}\ \text{W}$.
- Flux is power per unit area perpendicular to the radiation. A panel tilted at $\theta$ receives $FA\cos\theta$, which is why the question says "facing the Sun".
例题:探测器上的太阳能板
- 太阳半径 $6.96\times10^{8}\ \text{m}$,表面温度 $5780\ \text{K}$。一台距太阳中心 $4.5\times10^{10}\ \text{m}$ 的探测器带着一块 $2.0\ \text{m}^2$、正对太阳的板。求落在板上的功率。
- 光度:$L = 4\pi\sigma r^2 T^4 = 4\pi(5.67\times10^{-8})(6.96\times10^{8})^2(5780)^4 = 3.85\times10^{26}\ \text{W}$。
- 那里的通量:$F = \dfrac{L}{4\pi d^2} = \dfrac{3.85\times10^{26}}{4\pi(4.5\times10^{10})^2} = 1.5\times10^{4}\ \text{W/m}^2$,是地球处的十一倍。
- 功率:$P = FA = 1.5\times10^{4}\times2.0 = 3.0\times10^{4}\ \text{W}$。
- 通量是垂直于辐射方向的单位面积上的功率。倾斜 $\theta$ 角的板接收到的是 $FA\cos\theta$,这就是题目要说"正对太阳"的原因。
A probe 4.5e10 m from the Sun (L = 3.85e26 W) carries a 2.0 m^2 panel facing the Sun. What power falls on it, in kW? · 一个距太阳 4.5e10 m（L = 3.85e26 W）的探测器携带一块面向太阳的 2.0 m^2 面板。落在上面的功率是多少（单位 kW）？
F = L/(4 pi d^2) = 1.5e4 W/m^2, so P = FA = 3.0e4 W = 30 kW. A tilted panel receives FA cos(theta), which is why the question says the panel faces the Sun. · F = L/(4 pi d^2) = 1.5e4 W/m^2，所以 P = FA = 3.0e4 W = 30 kW。倾斜的面板接收 FA cos(theta)，这就是题目说面板面向太阳的原因。
Standard candles
- A standard candle is an object whose luminosity is known from what type of object it is.
- Find one in a distant galaxy, measure the flux $F$ we receive from it, and the distance falls out of $d = \sqrt{L/(4\pi F)}$.
- This is the only way to reach galaxies far too distant for parallax. The whole distance ladder rests on it.
- The logic is worth stating cleanly in an answer: known $L$, measured $F$, therefore $d$.
标准烛光
- 标准烛光是这样一类天体:凭它属于哪一类,就已知它的光度。
- 在遥远星系中找到一个,测出我们收到的通量 $F$,距离就由 $d = \sqrt{L/(4\pi F)}$ 算出来。
- 这是抵达那些远到无法用视差测量的星系的唯一办法。整个距离阶梯都建立在它上面。
- 这条逻辑值得在答题时干净地写出来:已知 $L$,测得 $F$,于是得到 $d$。
A standard candle is an object whose ____ is known from its type. · 标准烛光是其____可通过类型已知的天体。
Knowing $L$ (without first knowing distance) lets you get the distance from the measured flux. · 已知 $L$（无需先知道距离）可让你根据测得的通量计算出距离。
Put the use of a standard candle in order. · 按顺序排列标准烛光的使用步骤。
Known L, measured F, therefore d. This is the only way to reach galaxies far too distant for parallax. · 已知 L，测得 F，因此求 d。这是唯一能触及视差无法测量的遥远星系的方法。
The two standard candles to name
- Cepheid variables 造父变星 are pulsating stars whose pulsation period is tightly linked to their luminosity. Time the pulsation and you know $L$.
- Type Ia supernovae 超新星 are white dwarfs that explode on reaching a critical mass, so they always have very nearly the same peak luminosity.
- Both are extremely bright, which is what lets them be seen in other galaxies at all.
- Name the mechanism, not just the object: "the period gives the luminosity", "they all explode at the same mass".
要说得出名字的两种标准烛光
- 造父变星(Cepheid variables)是脉动的恒星,其脉动周期与光度紧密相关。给脉动计时,就知道了 $L$。
- Ia 型超新星(Type Ia supernovae)是白矮星达到临界质量时爆炸,所以它们的峰值光度几乎总是相同。
- 两者都极其明亮,这正是它们能在别的星系中被看到的原因。
- 要说出机制,而不只是名字:"周期给出光度","它们都在同样的质量上爆炸"。
A Cepheid variable's pulsation period tells you its luminosity. · 造父变星的脉动周期告诉了你它的光度。
The period–luminosity relation makes Cepheids excellent standard candles. · 周光关系使造父变星成为极佳的标准烛光。
Match each standard candle to why its luminosity is known. · 将每种标准烛光与其光度已知的原因匹配。
Name the mechanism, not just the object. Both are extremely bright, which is what lets them be seen in other galaxies at all. · 命名机制，而不仅仅是物体。两者都极其明亮，这正是它们能被观测到其他星系中的原因。
Marks that slip away
- The area in the inverse-square law is $4\pi d^2$, a sphere. Not $\pi d^2$.
- Luminosity is a property of the star; flux is what we receive. Questions swap the words deliberately.
- The two reasons one star looks brighter are greater luminosity and smaller distance. Nothing else.
- A standard candle needs a known luminosity. Saying it is "a very bright star" is not the point.
- Keep the distance in metres throughout, and remember flux is measured perpendicular to the radiation.
容易丢掉的分
- 平方反比定律里的面积是 $4\pi d^2$,一个球面。不是 $\pi d^2$。
- 光度是恒星的属性;通量是我们收到的东西。题目会故意把这两个词换来换去。
- 一颗星看起来更亮的两个原因是光度更大和距离更近。没有别的。
- 标准烛光需要的是已知的光度。说它"是一颗很亮的星"没有说到点子上。
- 距离全程用米,并记住通量是在垂直于辐射的面上量的。
You've got it
- luminosity is the total power of radiation emitted by a star, a property of the star alone
- radiant flux intensity is $F = L/(4\pi d^2)$, so doubling the distance quarters the flux, and $F$ against $1/d^2$ is a straight line of gradient $L/4\pi$
- a standard candle has a known luminosity, so a measured flux gives the distance $d = \sqrt{L/(4\pi F)}$
- Cepheid variables give $L$ from their pulsation period, and Type Ia supernovae all peak at nearly the same luminosity
你掌握了
- 光度是恒星发出的辐射的总功率,是恒星本身的属性
- 辐射通量密度是 $F = L/(4\pi d^2)$,距离加倍通量降到四分之一,而 $F$ 对 $1/d^2$ 是斜率为 $L/4\pi$ 的直线
- 标准烛光的光度已知,所以测得通量就给出距离 $d = \sqrt{L/(4\pi F)}$
- 造父变星由脉动周期给出 $L$,而 Ia 型超新星的峰值光度几乎都相同