Hubble's law and the Big Bang · 哈勃定律与大爆炸
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| redshift/ˈredʃɪft/ | 红移 | hóng yí |
| Hubble's law/ˈhʌblz lɔː/ | 哈勃定律 | hā bó dìng lǜ |
| Big Bang/bɪɡ bæŋ/ | 大爆炸 | dà bào zhà |
| Doppler effect/ˈdɒplə ɪˈfekt/ | 多普勒效应 | duō pǔ lè xiào yìng |
| recession/rɪˈseʃn/ | 退行 | tuì xíng |
| blueshifted/ˈbluːʃɪftɪd/ | 蓝移 | lán yí |
| Hubble constant/ˈhʌbl ˈkɒnstənt/ | 哈勃常数 | hā bó cháng shù |
| cosmic microwave background/ˈkɒzmɪk ˈmaɪkrəʊweɪv ˈbækɡraʊnd/ | 宇宙微波背景 | yǔ zhòu wēi bō bèi jǐng |
The fingerprint is right, but stretched
- Hydrogen in a laboratory always emits a line at $656.3\ \text{nm}$. In the light of a distant galaxy the same line arrives at $660.9\ \text{nm}$.
- The pattern of lines is unmistakably hydrogen. Every line has simply moved to a longer wavelength, by the same fraction.
- Something has stretched the light on its way here, and doing that arithmetic on galaxy after galaxy produced the largest conclusion in physics.
- This lesson is redshift 红移, Hubble's law 哈勃定律, and how they lead to the Big Bang 大爆炸.
指纹没错,只是被拉长了
- 实验室里的氢总在 $656.3\ \text{nm}$ 处发出一条谱线。而在一个遥远星系的光中,同一条线到达时却在 $660.9\ \text{nm}$。
- 谱线的图样无疑就是氢。每一条线只是按同样的比例挪到了更长的波长上。
- 有什么东西在光来的路上把它拉长了,而对一个又一个星系做这道算术,得出了物理学中最宏大的结论。
- 这一课讲红移(redshift)、哈勃定律(Hubble's law),以及它们如何引向大爆炸(Big Bang)。
Redshift
- The observed wavelength of the radiation from a source is longer than the wavelength emitted, because the source is moving away from the observer.
- It is the Doppler effect 多普勒效应 for light. For $v \ll c$:
- $\Delta\lambda = \lambda_{\text{observed}} - \lambda_{\text{emitted}}$, and $v$ is the speed of recession 退行.
- Divide by the emitted, laboratory wavelength, not the observed one. At small $v/c$ the difference is slight, but the mark scheme is specific.
The same pattern, slid to the right
红移
- 观测到的辐射波长比发出时的波长更长,因为源正在远离观察者。
- 这就是光的多普勒效应(Doppler effect)。对 $v \ll c$:
- $\Delta\lambda = \lambda_{\text{observed}} - \lambda_{\text{emitted}}$,而 $v$ 是退行(recession)速率。
- 要除以发出时的实验室波长,不是观测到的那个。在 $v/c$ 很小时两者差别甚微,但评分标准写得很明确。
同样的图样,整体右移
Redshift means a galaxy's spectral lines are shifted to: · 红移意味着一个星系的谱线移向:
A receding source stretches the wavelengths — a redshift, with $\dfrac{\Delta\lambda}{\lambda} \approx \dfrac{v}{c}$. · 一个退行的源把波长拉伸——红移,其中 $\dfrac{\Delta\lambda}{\lambda} \approx \dfrac{v}{c}$。
In $\Delta\lambda/\lambda \approx v/c$, the $\lambda$ on the bottom is the observed wavelength. · 在 $\Delta\lambda/\lambda \approx v/c$ 中,分母上的 $\lambda$ 是观测到的波长。
It is the EMITTED, laboratory wavelength. At small v/c the difference is slight, but the mark scheme is specific about it. · 是发出时的实验室波长。在 v/c 很小时差别甚微,但评分标准对此写得很明确。
Worked example: how fast, and how far
- A hydrogen line at $656.3\ \text{nm}$ in the laboratory is observed at $660.9\ \text{nm}$. Find the recession speed and, with $H_0 = 2.3\times10^{-18}\ \text{s}^{-1}$, the distance.
- $\Delta\lambda = 660.9 - 656.3 = 4.6\ \text{nm}$.
- $v = c\dfrac{\Delta\lambda}{\lambda} = (3.00\times10^{8})\dfrac{4.6}{656.3} = 2.1\times10^{6}\ \text{m/s}$.
- $d = \dfrac{v}{H_0} = \dfrac{2.1\times10^{6}}{2.3\times10^{-18}} = 9.1\times10^{23}\ \text{m}$, about $100$ million light years.
- The nanometres cancel inside the ratio, so there is no need to convert them, as long as both wavelengths are in the same unit.
例题:多快,多远
- 实验室中 $656.3\ \text{nm}$ 的氢线被观测到在 $660.9\ \text{nm}$。求退行速率,并用 $H_0 = 2.3\times10^{-18}\ \text{s}^{-1}$ 求距离。
- $\Delta\lambda = 660.9 - 656.3 = 4.6\ \text{nm}$。
- $v = c\dfrac{\Delta\lambda}{\lambda} = (3.00\times10^{8})\dfrac{4.6}{656.3} = 2.1\times10^{6}\ \text{m/s}$。
- $d = \dfrac{v}{H_0} = \dfrac{2.1\times10^{6}}{2.3\times10^{-18}} = 9.1\times10^{23}\ \text{m}$,约一亿光年。
- 纳米在比值里约掉了,所以不必换算,只要两个波长用的是同一单位。
A galaxy's lines show $\dfrac{\Delta\lambda}{\lambda} = 0.020$. What is its recession speed, as a fraction of $c$? · 一个星系的谱线显示 $\dfrac{\Delta\lambda}{\lambda} = 0.020$。它的退行速度是 $c$ 的几分之几?
For · 支持 $v \ll c$, $\dfrac{v}{c} = \dfrac{\Delta\lambda}{\lambda} = 0.020$. · 当 $v \ll c$ 时,$\dfrac{v}{c} = \dfrac{\Delta\lambda}{\lambda} = 0.020$。
Worked example: the other way round
- A galaxy recedes at $21400\ \text{km/s}$. At what wavelength is its $656.3\ \text{nm}$ hydrogen line seen?
- $\Delta\lambda = \lambda\dfrac{v}{c} = 656.3 \times \dfrac{2.14\times10^{7}}{3.00\times10^{8}} = 46.8\ \text{nm}$.
- So the line appears at $656.3 + 46.8 = 703\ \text{nm}$, moved out of the red and almost into the infrared.
- Every line and the continuous spectrum are stretched by the same factor $(1 + v/c) = 1.071$. Redshift moves the whole spectrum, not one line.
- Convert $\text{km/s}$ to $\text{m/s}$ first. A factor of $1000$ here is a very visible wrong answer.
例题:反过来算
- 某星系以 $21400\ \text{km/s}$ 退行。它的 $656.3\ \text{nm}$ 氢线会在什么波长被观测到?
- $\Delta\lambda = \lambda\dfrac{v}{c} = 656.3 \times \dfrac{2.14\times10^{7}}{3.00\times10^{8}} = 46.8\ \text{nm}$。
- 所以这条线出现在 $656.3 + 46.8 = 703\ \text{nm}$,已经移出红光、几乎进入红外。
- 每一条线以及连续谱都被同一个因子 $(1 + v/c) = 1.071$ 拉长。红移移动的是整个光谱,不是某一条线。
- 先把 $\text{km/s}$ 换成 $\text{m/s}$。这里差 $1000$ 倍会是个非常扎眼的错误答案。
A galaxy recedes at 21400 km/s. At what wavelength is its 656.3 nm hydrogen line observed, in nm? · 某星系以 21400 km/s 退行。它的 656.3 nm 氢线在什么波长被观测到(nm)?
delta lambda = lambda v/c = 46.8 nm, so the line appears at 703 nm. Convert km/s to m/s first; a factor of 1000 here is a very visible wrong answer. · delta lambda = lambda v/c = 46.8 nm,所以谱线出现在 703 nm。先把 km/s 换成 m/s;这里差 1000 倍会是个非常扎眼的错误答案。
Why this means the Universe is expanding
- The three-mark argument runs: (1) the spectral lines from almost all distant galaxies are shifted to longer wavelengths; (2) so by the Doppler effect those galaxies are moving away from us; (3) and the further away a galaxy is, the greater its redshift and so its speed.
- The third point is what makes it cosmological rather than local. That pattern is what an observer in any galaxy would see if the space between all galaxies were stretching.
- So the Universe as a whole is expanding, rather than everything fleeing from the Earth in particular.
- A few nearby galaxies are blueshifted 蓝移 by their own local motion, which is consistent: the expansion shows in the average over large distances.
这为什么意味着宇宙在膨胀
- 三分的论证是:(1)几乎所有遥远星系的谱线都移向更长的波长;(2)于是由多普勒效应,那些星系正在远离我们;(3)而且星系越远,红移越大、速率也越大。
- 第三点才使它成为宇宙学的而非局域的结论。若所有星系之间的空间都在拉伸,那么任何一个星系里的观察者都会看到这个图样。
- 所以是整个宇宙在膨胀,而不是万物特别地从地球逃开。
- 少数近邻星系因自身的局域运动而蓝移(blueshifted),这并不矛盾:膨胀体现在大尺度上的平均。
Almost all distant galaxies are redshifted, which shows the Universe is expanding. · 几乎所有遥远的星系都发生红移,这表明宇宙在膨胀。
They are moving apart on average — the space between them is stretching. · 它们平均而言在彼此远离——它们之间的空间在拉伸。
Put the three-mark argument that the Universe is expanding in order. · 把宇宙在膨胀的三分论证按顺序排列。
The third point is what makes it cosmological. Without it you have only shown that things are moving away from the Earth in particular. · 第三点才使它成为宇宙学的结论。没有它,你只证明了东西在特别地离开地球。
The trap with Wien's law
- If you feed the observed peak wavelength of a receding galaxy straight into Wien's law, the temperature comes out too low, because the observed peak is longer than the emitted one.
- Correct it first: $\lambda_{\text{emitted}} = \lambda_{\text{observed}}/(1 + v/c)$, then use Wien's law on the emitted peak.
- In the standard exam version, emitted $4.62\times10^{-7}\ \text{m}$ and observed $4.91\times10^{-7}\ \text{m}$ give $T = 6300\ \text{K}$ from the emitted peak. Using the observed one gives $5900\ \text{K}$: too low by $400\ \text{K}$.
- Asked to sketch the observed spectrum against the emitted one, draw the same shape shifted to longer wavelengths, peak included.
The whole curve slides, so the peak lies
维恩定律的陷阱
- 若把退行星系观测到的峰值波长直接代入维恩定律,温度会算得偏低,因为观测到的峰比发出时的更长。
- 要先修正:$\lambda_{\text{emitted}} = \lambda_{\text{observed}}/(1 + v/c)$,再对发出时的峰用维恩定律。
- 在标准的考试版本里,发出 $4.62\times10^{-7}\ \text{m}$、观测 $4.91\times10^{-7}\ \text{m}$,由发出峰给出 $T = 6300\ \text{K}$。用观测峰则给出 $5900\ \text{K}$:低了 $400\ \text{K}$。
- 若要求在同一坐标上画出观测谱与发出谱,就画同样的形状向长波方向平移,峰也一起移。
整条曲线都在滑,所以那个峰会骗人
Hubble's law · 哈勃定律
v = H₀·d
Recession speed is proportional to distance — the gradient is Hubble's constant. · 退行速度与距离 成正比——斜率是哈勃常数。
Using the observed peak wavelength of a receding galaxy in Wien's law gives a temperature that is: · 把退行星系观测到的峰值波长代入维恩定律,得到的温度会:
Correct the peak first with lambda_emitted = lambda_observed/(1 + v/c). In the standard version this is the difference between 6300 K and 5900 K. · 先用 lambda_emitted = lambda_observed/(1 + v/c) 修正峰值。在标准版本里,这就是 6300 K 与 5900 K 之差。
Hubble's law
- The speed of recession of a galaxy is directly proportional to its distance from the observer:
- $v$ is the recession speed, $d$ the distance, and $H_0$ the Hubble constant 哈勃常数, about $2.3\times10^{-18}\ \text{s}^{-1}$.
- Identify every symbol when asked. The exam quotes $H_0$ in SI units, so $v$ must be in $\text{m/s}$ and $d$ in metres, never kilometres per second per megaparsec.
- On a graph of $v$ against $d$ the data lie on a straight line through the origin of gradient $H_0$.
Further means faster, in proportion
哈勃定律
- 星系的退行速率与它到观察者的距离成正比:
- $v$ 是退行速率,$d$ 是距离,$H_0$ 是哈勃常数(Hubble constant),约 $2.3\times10^{-18}\ \text{s}^{-1}$。
- 被问到时每个符号都要说明。考卷用国际单位给出 $H_0$,所以 $v$ 要用 $\text{m/s}$、$d$ 要用米,绝不用千米每秒每兆秒差距。
- 在 $v$ 对 $d$ 的图上,数据落在一条斜率为 $H_0$ 的过原点直线上。
越远越快,而且成正比
Hubble's law: recession speed $= H_0 \times$ ____. · 哈勃定律:退行速度 $= H_0 \times$ ____。
$v = H_0 d$ — speed grows in proportion to distance. · $v = H_0 d$——速度与距离成正比地增大。
Galaxy B is 3 times as far away as galaxy A. By Hubble's law, B recedes how many times as fast? · 星系 B 比星系 A 远 3 倍。根据哈勃定律,B 退行的速度是 A 的几倍?
$v \propto d$, so 3 times the distance means 3 times the recession speed. · $v \propto d$,所以 3 倍的距离意味着 3 倍的退行速度。
Which are required when using $v = H_0 d$ with $H_0$ in per second? Select all · 所有 that apply. · 当 $H_0$ 以每秒为单位使用 $v = H_0 d$ 时,需要哪些?选出所有适用的。
The exam quotes H0 in SI units, so everything must be SI. Kilometres per second per megaparsec belongs to the astronomy literature, not to this paper. · 考卷用国际单位给出 H0,所以一切都必须是国际单位。千米每秒每兆秒差距属于天文文献,不属于这份卷子。
From Hubble's law to the Big Bang
- If speed is proportional to distance, then running the expansion backwards brings every galaxy together at the same moment.
- All distances shrink to zero at $t = -1/H_0$: the Universe was once a tiny, enormously dense, hot point. That is the Big Bang.
- For steady expansion the age of the Universe is
- The main evidence, worth listing: the expansion itself, the redshift of galaxies, the cosmic microwave background 宇宙微波背景, and the observed hydrogen and helium abundances.
从哈勃定律到大爆炸
- 如果速率正比于距离,那么把膨胀倒着放,每个星系都会在同一时刻汇到一起。
- 所有距离在 $t = -1/H_0$ 处缩为零:宇宙曾经是一个极小、极致密、极热的点。这就是大爆炸。
- 若膨胀速率恒定,宇宙的年龄约为(即约 140 亿年):
- 主要证据值得列出来:膨胀本身、星系的红移、宇宙微波背景(cosmic microwave background),以及观测到的氢和氦的丰度。
The approximate age of the Universe is: · 宇宙的大致年龄是:
Running the expansion back at a steady rate, all distances reach zero after a time $\dfrac{1}{H_0} \approx 14$ billion years. · 以稳定的速率把膨胀倒放,所有距离在 $\dfrac{1}{H_0} \approx 140$ 亿年后都变为零。
Which are evidence for the Big Bang? Select all · 所有 that apply. · 哪些是大爆炸的证据?选出所有适用的。
Standard candles are a measuring tool, not evidence. 1/H0 is an ESTIMATE of the age assuming a steady rate of expansion, and the assumption should be stated. · 标准烛光是测量工具,不是证据。1/H0 是在膨胀速率恒定假设下对年龄的估计,那个假设应当说出来。
Marks that slip away
- Divide $\Delta\lambda$ by the emitted wavelength, and put $v$ in $\text{m/s}$ and $d$ in metres for $H_0$ in $\text{s}^{-1}$.
- The expansion argument needs the third point: further galaxies recede faster. Without it you have only shown things are moving away from us.
- Correct a receding galaxy's peak wavelength before using Wien's law.
- Space itself is stretching. The galaxies are not flying through space away from a centre, and the Earth is not at one.
- $1/H_0$ is an estimate of the age assuming a steady rate of expansion. Say the assumption.
容易丢掉的分
- $\Delta\lambda$ 要除以发出时的波长,而 $H_0$ 用 $\text{s}^{-1}$ 时 $v$ 要用 $\text{m/s}$、$d$ 要用米。
- 膨胀的论证必须有第三点:越远的星系退行越快。没有它,你只证明了东西在离开我们。
- 用维恩定律之前先修正退行星系的峰值波长。
- 是空间本身在拉伸。星系并不是在空间中从某个中心飞散开,地球也不在什么中心上。
- $1/H_0$ 是在膨胀速率恒定这一假设下对年龄的估计。要把这个假设说出来。
You've got it
- redshift is an observed wavelength longer than the emitted one, with $\Delta\lambda/\lambda \approx \Delta f/f \approx v/c$ for a receding source
- almost every distant galaxy is redshifted and the further ones more, which is what an observer in any galaxy would see if space were expanding
- Hubble's law is $v \approx H_0 d$ in SI units, a straight line through the origin of gradient $H_0$
- running the expansion backwards gives the Big Bang and an age of about $1/H_0 \approx 4.3\times10^{17}\ \text{s}$, supported also by the cosmic microwave background
你掌握了
- 红移是观测波长比发出波长更长,对退行的源有 $\Delta\lambda/\lambda \approx \Delta f/f \approx v/c$
- 几乎每个遥远星系都红移,而且越远红移越大,这正是若空间在膨胀、任何星系中的观察者都会看到的景象
- 哈勃定律是国际单位下的 $v \approx H_0 d$,是一条斜率为 $H_0$ 的过原点直线
- 把膨胀倒放就得到大爆炸,年龄约为 $1/H_0 \approx 4.3\times10^{17}\ \text{s}$,宇宙微波背景也支持这一点