Stellar radii · 恒星半径
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| temperature/ˈtemprɪtʃə/ | 温度 | wēndù |
| Wien's displacement law/viːnz dɪˈspleɪsmənt lɔː/ | 维恩位移定律 | wéi ēn wèi yí dìng lǜ |
| Stefan-Boltzmann law/ˈstefən ˈbɒltsmən lɔː/ | 斯特藩-玻尔兹曼定律 | sī tè fān - bō ěr zī màn dìng lǜ |
| blackbody/ˈblækbɒdi/ | 黑体 | hēi tǐ |
| wavelength/ˈweɪvleŋθ/ | 波长 | bō cháng |
| Stefan-Boltzmann constant/ˈstefən ˈbɒltsmən ˈkɒnstənt/ | 斯特藩-玻尔兹曼常量 | sī tè fān - bō ěr zī màn cháng liàng |
No telescope can see a star as a disc
- Every star except the Sun is a point in even the largest telescope. Its disc is far too small to resolve.
- Yet the radius of a star thousands of light years away is a routine calculation, and it is done from two things you can measure: its colour and its brightness.
- Colour gives the surface temperature. Brightness and distance give the total power. Those two together give the size.
- This lesson is Wien's displacement law 维恩位移定律 and the Stefan-Boltzmann law 斯特藩-玻尔兹曼定律, and the three-step chain that joins them.
没有哪台望远镜能把恒星看成一个圆面
- 除太阳之外,每一颗恒星在再大的望远镜里都只是一个点。它的圆面小得根本分辨不出来。
- 可是算出几千光年之外一颗恒星的半径却是家常便饭,而所用的只是两样可测的东西:它的颜色和它的亮度。
- 颜色给出表面温度。亮度加距离给出总功率。两者合起来就给出大小。
- 这一课讲维恩位移定律(Wien's displacement law)与斯特藩-玻尔兹曼定律(Stefan-Boltzmann law),以及把它们串起来的那三步。
Wien's displacement law
- A hot body radiates a continuous blackbody 黑体 spectrum whose peak sits at a wavelength 波长 set by its temperature 温度:
- The wavelength at which the intensity is a maximum is inversely proportional to the thermodynamic temperature. Say "wavelength of maximum intensity", not just "the wavelength".
- The temperature must be in kelvin, and the law describes the peak of the continuous curve, not the spectral lines.
- Hotter means bluer: a red star near $3000\ \text{K}$ peaks in the infrared, the Sun near $5800\ \text{K}$ peaks around $500\ \text{nm}$, and a blue-white star near $20000\ \text{K}$ peaks in the ultraviolet.
Hotter, taller, and further left
维恩位移定律
- 热的物体辐射出连续的黑体(blackbody)谱,其峰所在的波长(wavelength)由它的温度(temperature)决定:
- *__强度取最大值__处的波长与热力学温度成反比。*要说"强度最大处的波长",不能只说"波长"。
- 温度必须用开尔文,而这条定律描述的是连续曲线的峰,不是谱线。
- 越热越蓝:约 $3000\ \text{K}$ 的红星峰在红外,约 $5800\ \text{K}$ 的太阳峰在 $500\ \text{nm}$ 附近,约 $20000\ \text{K}$ 的蓝白星峰在紫外。
更热、更高,而且更靠左
Hotter stars have their spectral peak (Wien's law) at a ____ wavelength. · 更热的恒星(维恩定律)的光谱峰值在 ____ 的波长处。
$\lambda_{\max}T = b$, so a higher $T$ means a smaller $\lambda_{\max}$ — toward the blue. · $\lambda_{\max}T = b$,所以更高的 $T$ 意味着更小的 $\lambda_{\max}$——移向蓝端。
Which belong in a statement of Wien's displacement law? Select all · 所有 that apply. · 哪些属于维恩位移定律的表述?选出所有适用的。
It is about the peak of the continuous black-body curve, not the lines. Saying only the wavelength loses the mark; say the wavelength of maximum intensity. · 它讲的是连续黑体曲线的峰,不是谱线。只说波长会丢分;要说强度最大处的波长。
Worked example: taking a star's temperature
- A star's spectrum peaks at $500\ \text{nm}$. Find its surface temperature.
- $T = \dfrac{b}{\lambda_{\text{max}}} = \dfrac{2.90\times10^{-3}}{500\times10^{-9}} = 5800\ \text{K}$.
- Convert nanometres to metres first. The constant is in metre kelvin, so a wavelength in nm gives a temperature out by $10^9$.
- This one measurement is the whole reason a star's colour is useful: colour is temperature.
例题:量一颗恒星的温度
- 某恒星的谱峰在 $500\ \text{nm}$。求它的表面温度。
- $T = \dfrac{b}{\lambda_{\text{max}}} = \dfrac{2.90\times10^{-3}}{500\times10^{-9}} = 5800\ \text{K}$。
- 先把纳米换成米。常量的单位是米开,波长用 nm 会让温度差 $10^9$ 倍。
- 这一次测量正是恒星的颜色之所以有用的全部理由:颜色就是温度。
A star's spectrum peaks at $\lambda_{\max} = 500\ \text{nm}$. What is its surface temperature? ($b = 2.9 \times 10^{-3}\ \text{m}\cdot\text{K}$) · 一颗恒星的光谱在 $\lambda_{\max} = 500\ \text{nm}$ 处达到峰值。它的表面温度是多少?($b = 2.9 \times 10^{-3}\ \text{m}\cdot\text{K}$)
$T = \dfrac{b}{\lambda_{\max}} = \dfrac{2.9 \times 10^{-3}}{500 \times 10^{-9}} \approx 5800\ \text{K}$ — about the Sun. · $T = \dfrac{b}{\lambda_{\max}} = \dfrac{2.9 \times 10^{-3}}{500 \times 10^{-9}} \approx 5800\ \text{K}$——大约是太阳。
Match each star to where its spectrum peaks. · 把每类恒星与它的谱峰所在配对。
Hotter peaks shorter. This one measurement is why a star's colour is useful: colour is temperature. · 越热峰越短。这一次测量正是恒星颜色之所以有用的原因:颜色就是温度。
The Stefan-Boltzmann law
- Treat the star as a blackbody sphere of radius $r$ at surface temperature $T$:
- $\sigma$ is the Stefan-Boltzmann constant 斯特藩-玻尔兹曼常量.
- Two dependences, and they are very different in strength. $L \propto r^2$: twice the radius, four times the luminosity. $L \propto T^4$: twice the temperature, sixteen times the luminosity.
- So a small error in temperature is a large error in luminosity, which is why the peak wavelength must be read carefully.
斯特藩-玻尔兹曼定律
- 把恒星当作半径 $r$、表面温度 $T$ 的黑体球:
- $\sigma$ 是斯特藩-玻尔兹曼常量(Stefan-Boltzmann constant)。
- 两种依赖关系,强弱大不相同。$L \propto r^2$:半径加倍,光度变四倍。$L \propto T^4$:温度加倍,光度变十六倍。
- 所以温度上一点小误差就是光度上的大误差,这就是峰值波长必须仔细读的原因。
A star's luminosity and radius · 恒星的亮度与半径
L ∝ r²
For a given surface temperature, a star's luminosity grows with the SQUARE of its radius (Stefan's law). · 对于给定的表面温度，恒星的亮度随其半径的平方增长（斯特藩定律）。
The Stefan–Boltzmann law gives a star's luminosity as: · 斯特藩–玻尔兹曼定律给出一颗恒星的光度为:
$L = 4\pi\sigma r^{2}T^{4}$ — strongly dependent on both radius and temperature. · $L = 4\pi\sigma r^{2}T^{4}$——强烈依赖于半径和温度。
Using $L = 4\pi\sigma r^2T^4$, which are true? Select all · 所有 that apply. · 用 $L = 4\pi\sigma r^2T^4$,哪些是对的?选出所有适用的。
L goes as r^2 and T^4. That fourth power is why the peak wavelength must be read carefully and corrected for redshift. · L 按 r^2 和 T^4 变化。正是那个四次方,使峰值波长必须仔细读并对红移作修正。
Worked example: the luminosity of a Sun-like star
- A star has radius $7.0\times10^{8}\ \text{m}$ and surface temperature $5800\ \text{K}$. Find its luminosity.
- $L = 4\pi\sigma r^2 T^4 = 4\pi(5.67\times10^{-8})(7.0\times10^{8})^2(5800)^4 = 3.9\times10^{26}\ \text{W}$.
- Square the radius and raise the temperature to the fourth power. On a calculator, do $T^4$ as its own step and check its magnitude: $(5800)^4 \approx 1.1\times10^{15}$.
例题:一颗类太阳恒星的光度
- 某恒星半径 $7.0\times10^{8}\ \text{m}$,表面温度 $5800\ \text{K}$。求它的光度。
- $L = 4\pi\sigma r^2 T^4 = 4\pi(5.67\times10^{-8})(7.0\times10^{8})^2(5800)^4 = 3.9\times10^{26}\ \text{W}$。
- 半径要平方,温度要取四次方。用计算器时把 $T^4$ 单独算一步并核对量级:$(5800)^4 \approx 1.1\times10^{15}$。
If a star's temperature doubles (same radius), its luminosity multiplies by: · 如果一颗恒星的温度加倍(半径不变),它的光度变为原来的多少倍:
$L \propto T^{4}$, so doubling $T$ gives $2^{4} = 16$ times the luminosity. · $L \propto T^{4}$,所以 $T$ 加倍给出 $2^{4} = 16$ 倍的光度。
Same luminosity, opposite stars
- Because $L$ depends on both $r$ and $T$, two completely different stars can have the same luminosity.
- A red giant is cool, so its $T^4$ is small, but it is enormous, and the $r^2$ makes up for it.
- A white dwarf is very hot but tiny: a huge $T^4$ against a minute $r^2$.
- Whenever an answer looks surprising, check it this way. A star $140$ times as luminous as the Sun but only $1.6$ times as hot must be considerably bigger, since $L \propto r^2T^4$.
光度相同,恒星却相反
- 由于 $L$ 同时取决于 $r$ 和 $T$,两颗完全不同的恒星可以有相同的光度。
- 红巨星很冷,所以它的 $T^4$ 很小,但它极其庞大,$r^2$ 把差额补了回来。
- 白矮星非常热却很小:巨大的 $T^4$ 对上微小的 $r^2$。
- 每当答案看起来出人意料,就用这个办法核对。一颗光度是太阳 $140$ 倍而温度只有 $1.6$ 倍的恒星必定大得多,因为 $L \propto r^2T^4$。
Put the three steps for estimating a star's radius in order. · 把估算恒星半径的三步按顺序排列。
Each step is one line. Keep full precision of L and T until the end, because T is raised to the fourth power. · 每一步都是一行。L 和 T 要保持全部精度到最后,因为 T 要取四次方。
Estimating a radius, in three steps
- One. Measure $\lambda_{\text{max}}$ and get $T$ from Wien's law.
- Two. Get $L$, usually from the flux and distance: $L = 4\pi d^2 F$.
- Three. Solve the Stefan-Boltzmann law for the radius:
- Each step is one line. Keep the full precision of $L$ and $T$ until the end, because $T$ is raised to the fourth power and rounding it early is expensive.
三步估算半径
- **一。**测出 $\lambda_{\text{max}}$,由维恩定律得到 $T$。
- **二。**得到 $L$,通常由通量和距离:$L = 4\pi d^2 F$。
- **三。**把斯特藩-玻尔兹曼定律解出半径:
- 每一步都是一行公式。$L$ 和 $T$ 要保持全部精度到最后,因为 $T$ 要取四次方,过早取近似代价很大。
Combining Wien's law and the Stefan–Boltzmann law lets us estimate a star's radius. · 把维恩定律和斯特藩–玻尔兹曼定律结合起来,能让我们估计一颗恒星的半径。
Wien gives $T$, the flux and distance give $L$, then $r = \sqrt{\dfrac{L}{4\pi\sigma T^{4}}}$. · 维恩给出 $T$,通量和距离给出 $L$,然后 $r = \sqrt{\dfrac{L}{4\pi\sigma T^{4}}}$。
Sunlight has flux 1370 W/m^2 at 1.50e11 m and the Sun's spectrum peaks at 500 nm. What is the Sun's radius, in units of 1e8 m? · 阳光在 1.50e11 m 处的通量为 1370 W/m^2,太阳谱峰在 500 nm。太阳的半径是多少(以 1e8 m 为单位)?
L = 4 pi d^2 F = 3.87e26 W, T = 5800 K, then r = sqrt(L / (4 pi sigma T^4)) = 6.9e8 m against an accepted 6.96e8 m. Three one-line formulae, no telescope resolution needed. · L = 4 pi d^2 F = 3.87e26 W,T = 5800 K,再 r = sqrt(L / (4 pi sigma T^4)) = 6.9e8 m,而公认值是 6.96e8 m。三行公式,不需要望远镜的分辨本领。
Worked example: the radius of the Sun, from scratch
- Sunlight at the Earth has flux $1370\ \text{W/m}^2$ at $1.50\times10^{11}\ \text{m}$, and the Sun's spectrum peaks at $500\ \text{nm}$. Estimate the Sun's radius.
- Luminosity: $L = 4\pi d^2 F = 4\pi(1.50\times10^{11})^2(1370) = 3.87\times10^{26}\ \text{W}$.
- Temperature: $T = 2.90\times10^{-3}/(500\times10^{-9}) = 5800\ \text{K}$.
- Radius: $r = \sqrt{\dfrac{3.87\times10^{26}}{4\pi(5.67\times10^{-8})(5800)^4}} = 6.9\times10^{8}\ \text{m}$.
- The accepted value is $6.96\times10^{8}\ \text{m}$. Three one-line formulae, no telescope resolution required.
例题:从零开始算太阳的半径
- 地球处的阳光通量为 $1370\ \text{W/m}^2$,距离 $1.50\times10^{11}\ \text{m}$,太阳的谱峰在 $500\ \text{nm}$。估算太阳的半径。
- 光度:$L = 4\pi d^2 F = 4\pi(1.50\times10^{11})^2(1370) = 3.87\times10^{26}\ \text{W}$。
- 温度:$T = 2.90\times10^{-3}/(500\times10^{-9}) = 5800\ \text{K}$。
- 半径:$r = \sqrt{\dfrac{3.87\times10^{26}}{4\pi(5.67\times10^{-8})(5800)^4}} = 6.9\times10^{8}\ \text{m}$。
- 公认值是 $6.96\times10^{8}\ \text{m}$。三行公式,完全不需要望远镜的分辨本领。
In $L = 4\pi\sigma r^2T^4$ and $F = L/(4\pi d^2)$, the lengths $r$ and $d$ mean the same thing. · 在 $L = 4\pi\sigma r^2T^4$ 和 $F = L/(4\pi d^2)$ 中,长度 $r$ 与 $d$ 指的是同一个东西。
r is the star's own radius; d is its distance from us. Swapping them is the commonest error in this topic, and the two differ by many orders of magnitude. · r 是恒星自身的半径;d 是它到我们的距离。弄混是这个主题最常见的错误,而两者相差许多个数量级。
Marks that slip away
- Wien's law is about the wavelength of maximum intensity, and $T$ must be in kelvin.
- Convert $\lambda_{\text{max}}$ to metres; the constant $b$ is in metre kelvin.
- In $L = 4\pi\sigma r^2T^4$, $r$ is the star's radius; in $F = L/(4\pi d^2)$, $d$ is the distance to us. Two different lengths, and swapping them is the commonest error in this topic.
- Keep full precision until the last line, because $T$ is to the fourth power.
- For a receding galaxy, correct the observed peak before using Wien's law, or the temperature comes out too low.
容易丢掉的分
- 维恩定律说的是强度最大处的波长,而 $T$ 必须用开尔文。
- 把 $\lambda_{\text{max}}$ 换成米;常量 $b$ 的单位是米开。
- $L = 4\pi\sigma r^2T^4$ 里的 $r$ 是恒星的半径;$F = L/(4\pi d^2)$ 里的 $d$ 是到我们的距离。两个不同的长度,弄混是这个主题最常见的错误。
- 保持全部精度到最后一行,因为 $T$ 要取四次方。
- 对退行的星系,用维恩定律之前要先修正观测到的峰,否则温度会算得偏低。
You've got it
- Wien's displacement law: the wavelength of maximum intensity is inversely proportional to the thermodynamic temperature, $\lambda_{\text{max}}T = 2.90\times10^{-3}\ \text{m K}$
- Stefan-Boltzmann: $L = 4\pi\sigma r^2T^4$, so luminosity goes as the square of the radius and the fourth power of the temperature
- a radius comes in three steps: $\lambda_{\text{max}} \to T$, then $F$ and $d \to L$, then $r = \sqrt{L/(4\pi\sigma T^4)}$
- keep $r$ (the star) and $d$ (to us) apart, and hold full precision until the end because of the $T^4$
你掌握了
- 维恩位移定律:强度最大处的波长与热力学温度成反比,$\lambda_{\text{max}}T = 2.90\times10^{-3}\ \text{m K}$
- 斯特藩-玻尔兹曼:$L = 4\pi\sigma r^2T^4$,所以光度按半径的平方、温度的四次方变化
- 求半径分三步:$\lambda_{\text{max}} \to T$,再由 $F$ 与 $d \to L$,最后 $r = \sqrt{L/(4\pi\sigma T^4)}$
- 把 $r$(恒星)与 $d$(到我们)分清楚,并因为那个 $T^4$ 而把全部精度保持到最后