galaxies expansion and Big Bang model

Observations show that almost every galaxy outside the Local Group has a spectrum whose lines are shifted toward longer wavelengths. For small redshifts the shift is converted into a recession velocity:

$$z = \frac{\lambda_{\text{obs}} - \lambda_{\text{emit}}}{\lambda_{\text{emit}}}, \qquad v \approx c z .$$

Hubble's law links that velocity to a distance obtained from standard candles such as Cepheids and Type Ia supernovae:

$$v = H_0 d, \qquad H_0 \approx 70\ \mathrm{km\,s^{-1}\,Mpc^{-1}} .$$

The linearity of this relation means the expansion is uniform: an observer in any galaxy sees every distant galaxy receding with a velocity proportional to its separation. Reading the slope as a rate of expansion gives the Hubble time $t_H = 1/H_0 \approx 14\ \text{billion years}$, a rough estimate of the age of the Universe. Extrapolating the expansion backwards leads to a hot, dense initial state — the Big Bang model — supported by the cosmic microwave background at $T \approx 2.7\ \mathrm{K}$ and by the measured abundances of hydrogen and helium. At large $z$ the simple law $v = cz$ fails, and one must use the relativistic redshift together with a cosmological model.

A galaxy's spectrum has redshift $z = 0.050$. Taking $c = 3.0\times 10^{5}\ \mathrm{km/s}$ and $H_0 = 70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$, find its recession velocity from $v \approx cz$ and its distance from Hubble's law.

Estimate the age of the Universe as the Hubble time $t_H = 1/H_0$ for $H_0 = 70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$. First convert $H_0$ to SI units ($\mathrm{s^{-1}}$), then find $t_H$ in seconds and in years.

A quasar has redshift $z = 0.50$. First apply $v = cz$ and Hubble's law naively to get a velocity and a distance. Then use the relativistic relation $v = c\frac{(1+z)^2 - 1}{(1+z)^2 + 1}$ and explain why the linear formulas give misleading results at this $z$.

Stretching-band expansion model (safe household measurement). Take an elastic band about 30 cm long. Mark it every 1 cm and label the marks $A, B, C, \dots$; let mark $A$ be "our observer". Stretch the band so its length increases, say, by a factor of 2, and measure again the separations between $A$ and every other mark.

1. Record the initial and final separations in a table.
2. Treating the stretch as happening over the same "time" $\Delta t$, compute the "recession speed" of each mark as $v = \Delta d / \Delta t$.
3. Plot $v$ against the initial distance $d$ and check that the points lie on a straight line through the origin; its slope is the "Hubble constant" of your model.
4. Repeat with mark $B$ as the observer and compare the slopes. State the conclusion: every observer sees the same pattern of recession.

Extension: find a cosmic microwave background map online (for example the Planck data) and describe the temperature fluctuations $\Delta T \sim 10^{-5}\ \mathrm{K}$ that appear on it. Do not use heated or luminous objects, and never point instruments at the Sun.

Show that Hubble's law is a straight line on the distance–velocity graph, that the slope of this line is $H_0$, and that stretching a grid of distances gives the same recession pattern for every observer.

Graph plus sketch. The main panel is a Hubble diagram: horizontal axis is distance $d$ in $\mathrm{Mpc}$, vertical axis is recession velocity $v$ in $\mathrm{km/s}$; galaxy data points are plotted with the best-fit line $v = H_0 d$ through them and the slope labelled $H_0 \approx 70\ \mathrm{km\,s^{-1}\,Mpc^{-1}}$. An inset shows three successive frames of a stretching grid: identical squares scaled by 1, 1.5 and 2, with labelled nodes $A$, $B$, $C$ and velocity arrows $v_B < v_C$ whose lengths are proportional to the distance from $A$. A small separate inset is a false-colour cosmic microwave background map with a colour bar of $\pm 200\ \mu\mathrm{K}$.

At a stadium a training radar operates at frequency $f_0=24.15$ GHz. A ball flies toward the goal at speed $v=30$ m/s. The radar measures the frequency shift $\Delta f = 2 v f_0/c$. With $c=3.00\cdot10^8$ m/s we get $\Delta f = 2\cdot30\cdot24.15\cdot10^9/(3.00\cdot10^8) \approx 4.83\cdot10^3$ Hz $=4.83$ kHz. The receiver mixes the reflected wave with the source wave and extracts this beat frequency.

The frequency shift is the Doppler effect: relative motion changes the observed frequency. In cosmology the same principle is applied to light: redshift $z=\Delta\lambda/\lambda$. Hubble found that lines in galaxy spectra are shifted to longer wavelengths and connected this with recession of galaxies and expansion of space.

An ambulance moves away at speed $v=30$ m/s. The siren frequency is $f_0=700$ Hz, the speed of sound is $c_s=343$ m/s. The observed frequency is $f' = f_0 c_s/(c_s+v) = 700\cdot343/373 \approx 644$ Hz, so the pitch drops by about $56$ Hz.

The observed frequency and wavelength depend on the relative speed of source and observer. In astrophysics, galaxy spectral lines shift to the red; from the value $z$, for small $z$, one estimates the recession speed $v \approx c z$, and then tests Hubble's law $v=H_0 d$.

A rectangular piece of dough is $10$ cm long. Two raisins are $3.0$ cm apart. After $30$ min the dough is $15$ cm long, and the separation of the raisins is $4.5$ cm. The average recession speed of the raisins is $(4.5-3.0)\text{ cm}/(1800\text{ s}) \approx 8.3\cdot10^{-4}$ cm/s. The ratio $v/d \approx (8.3\cdot10^{-4})/(3.0) \approx 2.8\cdot10^{-4}$ s$^{-1}$.

This is homogeneous expansion: all distances increase by the same factor, so distant raisins recede faster than nearby ones. In the Big Bang model space stretches between galaxies; galaxies themselves do not expand, and Hubble's law $v=H_0 d$ follows from the scale factor $a(t)$.