Questions on the Doppler effect and red shift

Question 1

State what is meant by the red shift of a galaxy's spectrum, and what it tells you about the galaxy's motion.

Accepted Answer

Every line in the spectrum arrives at a longer wavelength than the same line measured in the laboratory, the whole pattern stretched by one factor (1). The galaxy is receding, with z = Δλ/λ giving v = zc for speeds well below c (1).

Question 2

State Hubble's law, and the meaning of the Hubble constant.

Accepted Answer

The recession speed of a galaxy is proportional to its distance: v = Hd (1). H is the gradient, about 65 km s⁻¹ Mpc⁻¹: every megaparsec of distance adds 65 km/s of recession speed (1).

Question 3

Name the two key pieces of observational evidence for the Big Bang, other than the expansion itself.

Accepted Answer

The cosmic microwave background, a 2.7 K black-body glow from every direction, the cooled afterglow of the hot early universe (1); and the roughly three-to-one hydrogen-helium mass ratio, matching fusion in the universe's first minutes (1).

Question 4

Write down the relationship between the fractional shift in the frequency of light and the speed of its source, state the condition under which it applies, and define the shift so that its sign is unambiguous.

Accepted Answer

Δf/f = v/c in magnitude (equivalently z = Δλ/λ) (1); it holds only for speeds much less than the speed of light (1). Taking Δf as the observed frequency minus the rest frequency, a receding source gives a negative Δf and a positive Δλ, so Δf/f = −v/c and Δλ/λ = +v/c for recession speed v.

Question 5

The absorption lines from a nearby star arrive at slightly shorter wavelengths than the same lines measured in the laboratory. State what this shows about the star's motion, and how its speed would be found.

Accepted Answer

The star is approaching: motion towards the observer squeezes the wavelengths (a blue shift) (1). The fractional shift gives the line-of-sight speed, v = cΔλ/λ (1).

Question 6

State why the wavelength of a spectral line must also be known from laboratory measurements before a galaxy's red shift can be found.

Accepted Answer

z = Δλ/λ compares the observed wavelength with the unshifted one, so the laboratory value is the baseline for the shift (1).

Question 7

Hydrogen's 486.1 nm line arrives from a galaxy at 490.0 nm. Calculate the galaxy's red shift and its recession speed.

Accepted Answer

z = (490.0 − 486.1)/486.1 (1) z = 8.0 × 10⁻³ (1) v = zc = 2.4 × 10⁶ m s⁻¹, receding (1)

Question 8

A spectral line from one star of a binary system swings periodically by up to 0.080 nm either side of 550 nm. Calculate the star's orbital speed, and state what the period of the swing represents.

Accepted Answer

v = c Δλ/λ = 3.0 × 10⁸ × 0.080/550 (1) v = 4.4 × 10⁴ m s⁻¹ (1) The period of the wavelength oscillation is the orbital period of the binary (1)

Question 9

A galaxy recedes at 8800 km s⁻¹. Taking H = 65 km s⁻¹ Mpc⁻¹, calculate its distance.

Accepted Answer

d = v/H = 8800/65 (1) d = 135 Mpc (1)

Question 10

Explain why the clean textbook treatment of a spectroscopic binary specifies that the orbit is observed in its own plane.

Accepted Answer

The Doppler shift responds only to the velocity component along the line of sight (1). Edge-on, each star's full orbital speed alternately points towards and away from Earth, giving the largest, cleanest wavelength swing (1); tilt the orbit and the measured shifts shrink (1).

Question 11

The 486.1 nm hydrogen line is received at 484.2 nm from galaxy G1 and at 492.0 nm from galaxy G2. Deduce which galaxy's distance can be estimated from Hubble's law, and calculate that distance for H = 65 km s⁻¹ Mpc⁻¹.

Accepted Answer

G1's light is blue-shifted: it is approaching, so Hubble's law does not apply to it. G2, red-shifted, is receding (1). z = (492.0 − 486.1)/486.1 = 0.0121 (1), so v = zc = 3.64 × 10⁶ m s⁻¹ = 3640 km s⁻¹ (1). d = v/H = 3640/65 ≈ 56 Mpc (1).

Question 12

Hydrogen in a distant galaxy emits radio waves at a frequency of 1.420 GHz; they are received at 1.402 GHz. Calculate the galaxy's speed and state its direction of motion.

Accepted Answer

The size of the shift is |Δf| = 1.420 − 1.402 = 0.018 GHz (1). v = c|Δf|/f = 3.0 × 10⁸ × 0.018/1.420 (1) = 3.8 × 10⁶ m s⁻¹, receding, since the received frequency is the lower (1). With Δf defined as observed minus rest, the shift itself is −0.018 GHz, and that negative sign is what recession means.

Question 13

A galaxy lies 220 Mpc away. Taking H = 65 km s⁻¹ Mpc⁻¹, calculate its recession speed, and predict the wavelength at which hydrogen's 656.3 nm line will be received from it.

Accepted Answer

v = Hd = 65 × 220 = 14300 km s⁻¹ = 1.43 × 10⁷ m s⁻¹ (1). z = v/c = 0.0477 (1). Δλ = 0.0477 × 656.3 = 31.3 nm (1), so the line arrives at 656.3 + 31.3 ≈ 688 nm (1).

Question 14

State the Cosmological principle in its three parts, and explain which of those parts is being relied on whenever the red shift of a distant galaxy is used to find its recession speed.

Accepted Answer

Averaged over a large enough scale the universe is homogeneous: its composition and density are the same everywhere, so no place is special (1). It is isotropic: it looks the same in whichever direction it is observed, so no direction is special (1). The laws of physics are universal, the same in a distant galaxy as in a laboratory here (1). A red shift is found by comparing the wavelength arriving from the galaxy with the laboratory wavelength of the same line, which takes for granted that hydrogen billions of light years away emits and absorbs at the wavelengths hydrogen emits and absorbs here: that is the universal-laws part doing the work (1).

Question 15

Assuming every galaxy has always receded at its present speed, estimate the age of the universe for H = 65 km s⁻¹ Mpc⁻¹, with 1 Mpc = 3.08 × 10²² m. Give the answer in years.

Accepted Answer

t = d/v = 1/H for every galaxy (1)
In SI units H = 65 000/(3.08 × 10²²) (1)
H = 2.11 × 10⁻¹⁸ s⁻¹ (1)
t = 1/H = 4.7 × 10¹⁷ s (1)
Dividing by 3.156 × 10⁷ s per year gives 1.5 × 10¹⁰ years, about 15 billion years (1)

Question 16

A student describes the Big Bang as an explosion flinging galaxies outward through space from a central point. Give the better description, and one observation the student's picture fails to explain.

Accepted Answer

Space itself expands, carrying the galaxies apart: there is no centre and no edge (1), and an observer in any galaxy sees the same Hubble law (1). The student's picture cannot explain why the recession looks the same in every direction from here, and would from anywhere, nor why the microwave background arrives uniformly from the whole sky (1).

Question 17

A quasar shows z = 0.30. Estimate its recession speed and distance (H = 65 km s⁻¹ Mpc⁻¹), and state why the speed is only approximate.

Accepted Answer

v = zc = 0.30 × 3.0 × 10⁸ (1)
v = 9.0 × 10⁷ m s⁻¹ (1)
d = 90 000/65 = 1.4 × 10³ Mpc (1)
At nearly a third of light speed the small-shift approximation z = v/c is strained, so both figures are estimates (1)

Question 18

Explain how the properties of the cosmic microwave background support the Big Bang model.

Accepted Answer

The early universe was hot, dense and filled with radiation (1). The expansion of space has stretched (red-shifted) that radiation, cooling it (1); it now has a black-body spectrum at 2.7 K, peaking near one millimetre, as the model predicts for the cooled afterglow (1). It arrives almost uniformly from every direction, as radiation filling all of space should, rather than coming from any single source (1).

Question 19

A binary star system is observed edge-on, in the plane of the stars' orbit. Both stars are bright enough to contribute a detectable absorption line, making this a double-lined spectroscopic binary. Describe and explain fully how one absorption line from the system appears and changes over one orbital period, and state what can be deduced from the observations.

Accepted Answer

Each star alternately approaches and recedes along the line of sight as it orbits (1). Its line is blue-shifted while it approaches and red-shifted while it recedes, so the wavelength oscillates about the laboratory value (1). The two stars always move oppositely, and since each contributes a detectable copy of the line, it splits into two components that separate and merge as the stars orbit; were only one star bright enough to register, a single oscillating line would be seen instead (1). The period of the oscillation equals the orbital period of the binary (1). The maximum shift gives the orbital speed through v = cΔλ/λ (1). Because the orbit is edge-on, the full orbital speed lies along the line of sight; a tilted orbit would give smaller shifts (1).

Question 20

A student measures two galaxies. Galaxy A recedes at 4550 km s⁻¹ at a distance of 70 Mpc; galaxy B recedes at 9750 km s⁻¹ at 150 Mpc. Determine whether the measurements are consistent with Hubble's law, and find the value of H they give.

Accepted Answer

H = v/d: for A, 4550/70 = 65 km s⁻¹ Mpc⁻¹ (1); for B, 9750/150 = 65 km s⁻¹ Mpc⁻¹ (1). Both give the same constant, so speed is proportional to distance, consistent with Hubble's law with H = 65 km s⁻¹ Mpc⁻¹ (1).

Question 21

For OCR: describe, stage by stage, the story of the universe from the Big Bang up to the release of the microwave background, giving the reason each stage had to wait until the universe had cooled to the temperature stated for it. Then show that the roughly 1100-fold stretch of wavelengths since that release turns the ≈3000 K temperature at which atoms formed into the microwave background temperature measured today.

Accepted Answer

Below about 10¹² K, roughly a microsecond in, collisions are no longer energetic enough to blast quarks apart the instant they meet, so quarks bind into protons and neutrons (1)
Primordial nucleosynthesis, fusing those nucleons into helium nuclei, has to wait until about 10⁹ K, one second to twenty minutes in, because above that temperature photons carry enough energy to break apart any nucleus as soon as it forms (1)
Atoms form only once the temperature falls to about 3000 K, some 380 000 years in: only then are photons too cool to immediately strip an electron back off a nucleus, so nuclei can finally hold on to electrons (1)
Before that point free electrons scatter photons constantly; once electrons are bound into atoms the scattering stops and light travels freely for the first time, and that freed light is the microwave background (1)
Expansion has stretched every wavelength in that light by a factor of about 1100 since it was released (1)
3000 K / 1100 ≈ 2.7 K, matching the microwave background measured today (1)