Free coursewareRewritten from MIT OpenCourseWareCC BY-NC-SA 4.0
Open CosmologyPhysics & cosmology: four years, plus a graduate year

The Doppler effect and special relativity

Why a moving source shifts the wavelength of sound and light, what redshift z means, and the three effects of special relativity that cosmology relies on.

You need this first

Full reading path (2 pages)
  1. Vectors (8.01)
  2. Position and velocity in one dimension (8.01)
On this page · 7 sections
  1. Doppler shift for sound
  2. Light has no medium
  3. Time dilation
  4. The relativistic Doppler shift
  5. Accelerating clocks
  6. The three kinematic effects
  7. Summary

Almost everything we know about the expansion of the universe comes from one measurement: how much the light of distant galaxies has been shifted toward the red. Before we can read that shift, we need to understand the Doppler effect, first for sound and then for light, where special relativity changes the rules.

Doppler shift for sound#

Sound is a wave in air. Its crests travel at a fixed speed uu relative to the air, whoever made the sound. Let ΔtS\Delta t_S be the time between crests at the source (one period) and ΔtO\Delta t_O the time between crests arriving at the observer. The question is how they are related when something is moving.

Case 1: the source moves away

The observer and the air are at rest; the source moves away at speed vv. Between emitting one crest and the next, the source moves a distance Δℓ=v ΔtS\Delta\ell = v\,\Delta t_S farther away. The second crest therefore has an extra distance Δℓ\Delta\ell to travel, which takes an extra time Δℓ/u\Delta\ell/u:

Sound, source moving
ΔtO=ΔtS+v ΔtSu⟹ΔtOΔtS=1+vu\Delta t_O = \Delta t_S + \frac{v\,\Delta t_S}{u} \quad\Longrightarrow\quad \frac{\Delta t_O}{\Delta t_S} = 1 + \frac{v}{u}
In words

capital delta t sub O equals capital delta t sub S plus v capital delta t sub S over u which gives capital delta t sub O over capital delta t sub S equals 1 plus v over u

Source: 8.286 Lecture 3 transcript, derived on the board, pp. 1–2; diagrams on slides 1–5

Astronomers describe the stretching by the redshift zz, defined so that z=0z = 0 means no shift at all:

Definition of redshift
1+z≡λOλS=ΔtOΔtS1 + z \equiv \frac{\lambda_O}{\lambda_S} = \frac{\Delta t_O}{\Delta t_S}
In words

1 plus z is defined as lambda sub O over lambda sub S equals capital delta t sub O over capital delta t sub S

Source: 8.286 Lecture 3 transcript, pp. 2–3

So for a moving source of sound, z=v/uz = v/u.

Case 2: the observer moves away

Now the source and the air are at rest and the observer moves away at vv. Between receiving one crest and the next, the observer moves Δℓ=v ΔtO\Delta\ell = v\,\Delta t_O, and the second crest must chase it. That gives an equation with ΔtO\Delta t_O on both sides:

Sound, observer moving
ΔtO=ΔtS+v ΔtOu⟹1+z=11−v/u,z=v/u1−v/u\Delta t_O = \Delta t_S + \frac{v\,\Delta t_O}{u} \quad\Longrightarrow\quad 1 + z = \frac{1}{1 - v/u}, \qquad z = \frac{v/u}{1 - v/u}
In words

capital delta t sub O equals capital delta t sub S plus v capital delta t sub O over u which gives 1 plus z equals 1 over 1 minus v/u , z equals v/u over 1 minus v/u

Source: 8.286 Lecture 3 transcript, derived on the board, pp. 3–4; diagrams on slides 6–10

For slow motion both answers are close to z≈v/uz \approx v/u. They differ by (v/u)2/(1−v/u)(v/u)^2/(1 - v/u): for v/u=1/1000v/u = 1/1000 the difference is about one part in a million. They differ at all because the air is a real, physical frame. Moving through the air and having the air move past you are genuinely different situations.

Worked example

A passing siren

An ambulance siren sounds at 440 Hz. The speed of sound is 343 m/s. What frequency do you hear (a) if the ambulance drives away from you at 30 m/s, (b) if you drive away from a parked ambulance at 30 m/s?

Source moving

1+z=1+v/u=1+30/343=1.08751 + z = 1 + v/u = 1 + 30/343 = 1.0875. Frequency is one over the period, so fO=440/1.0875=405f_O = 440/1.0875 = 405 Hz.

Observer moving

1+z=1/(1−30/343)=1.09581 + z = 1/(1 - 30/343) = 1.0958, so fO=440×(1−30/343)=402f_O = 440 \times (1 - 30/343) = 402 Hz.

Compare

Both drop the pitch by about a semitone and a half, but by slightly different amounts. For sound, it matters who is moving.

Light has no medium#

Light is different. In the 1860s Maxwell completed the equations of electromagnetism, and they predict a definite speed of light cc — but relative to what? For decades physicists assumed an invisible medium, the ether, with light moving at cc relative to it. Einstein's insight in 1905 was to drop the ether and instead change how measurements in different frames are related, so that light moves at cc for every inertial observer. The rules for that change are the Lorentz transformations of special relativity.

Without a medium, "the source moves" and "the observer moves" are the same physical situation seen from two frames. So the two calculations must agree. Checking that they do is a good test of special relativity.

Time dilation#

The key ingredient is that a moving clock runs slow. If the moving clock itself records a time Δtclock\Delta t_\text{clock} between two of its ticks, the frame it moves through measures a longer time between those ticks:

Time dilation
Δtframe=γ Δtclock,γ≡11−β2,β≡vc\Delta t_\text{frame} = \gamma\,\Delta t_\text{clock}, \qquad \gamma \equiv \frac{1}{\sqrt{1 - \beta^2}}, \qquad \beta \equiv \frac{v}{c}
In words

capital delta t sub frame equals gamma capital delta t sub clock , gamma is defined as 1 over the square root of 1 minus beta squared , beta is defined as v over c

Source: 8.286 Lecture 3 slides, slide 11

Guth stresses what "appears" means here. It is not what your eye sees at one instant, which is affected by light travel times. It is what is measured by a whole frame of rulers with synchronised clocks at every point. With that care, a clock moving at speed vv ticks once every γ\gamma of the frame's seconds.

The relativistic Doppler shift#

Source moving. Repeat case 1 with u=cu = c, but the source's clock now runs slow: in the observer's frame, the time between emissions is γ ΔtS\gamma\,\Delta t_S, and the source moves Δℓ=γv ΔtS\Delta\ell = \gamma v\,\Delta t_S in that time. Then ΔtO=γ ΔtS (1+β)\Delta t_O = \gamma\,\Delta t_S\,(1 + \beta).

Observer moving. Repeat case 2 in the source's frame, where the time between receptions is Δt′=ΔtS/(1−β)\Delta t' = \Delta t_S/(1 - \beta). The observer's clock runs slow, so it records only ΔtO=Δt′/γ\Delta t_O = \Delta t'/\gamma.

Using 1−β2=(1+β)(1−β)1 - \beta^2 = (1+\beta)(1-\beta), both give exactly the same answer:

Relativistic Doppler shift
1+z=ΔtOΔtS=1+β1−β1 + z = \frac{\Delta t_O}{\Delta t_S} = \sqrt{\frac{1 + \beta}{1 - \beta}}
In words

1 plus z equals capital delta t sub O over capital delta t sub S equals the square root of 1 plus beta over 1 minus beta

Source: 8.286 Lecture 3 transcript, pp. 15–19; Δℓ = γvΔt_S on slide 14, Δt_O = Δt′/γ on slide 18

For small β\beta this is z≈β+β2/2z \approx \beta + \beta^2/2, between the two sound results. As β→1\beta \to 1 the redshift grows without limit.

The three Doppler formulas, computed. They agree for slow motion and separate at high speed.
Worked example

From a measured line to a speed

The hydrogen line at 656.3 nm is observed in a galaxy at 662.9 nm. Find zz and the recession speed. Then: what speed would give z=1z = 1?

Redshift

1+z=662.9/656.3=1.01011 + z = 662.9/656.3 = 1.0101, so z=0.0101z = 0.0101.

Speed (slow motion)

zz is small, so β≈z\beta \approx z and v≈0.0101 c≈3000v \approx 0.0101\,c \approx 3000 km/s.

z = 1 needs the full formula

(1+z)2=4=(1+β)/(1−β)(1+z)^2 = 4 = (1+\beta)/(1-\beta) gives β=3/5=0.6\beta = 3/5 = 0.6. A naive v=zcv = zc would have said the speed of light.

A warning for cosmology

For distant galaxies, the next lectures show that redshift is better understood as the stretching of space, 1+z=a(tO)/a(tS)1 + z = a(t_O)/a(t_S), not as a velocity in one inertial frame. The Doppler formula is the special case of a universe with no gravity.

Accelerating clocks#

Special relativity is sometimes said to fail for accelerated motion. Guth corrects this: acceleration is handled by special relativity plus the relevant forces; general relativity is a theory of gravity. What special relativity cannot say is how a real clock responds to being shaken — a wristwatch survives, a clock smashed into a wall stops. Physicists therefore define an ideal clock: one whose rate at each moment equals that of an unaccelerated clock moving alongside it, so only its speed matters.

The three kinematic effects#

Everything special relativity says about measuring times and distances comes down to three effects:

  1. Time dilation: a moving clock runs slow by a factor γ\gamma.
  2. Lorentz–FitzGerald contraction: a moving rod is shorter by a factor γ\gamma along its motion; lengths across the motion are unchanged.
  3. Relativity of simultaneity: two clocks at the ends of a moving rod of rest length ℓ0\ell_0, synchronised in the rod's own frame, are not synchronised in yours — the trailing clock reads later.
Relativity of simultaneity
trailing clock ahead byβ ℓ0c\text{trailing clock ahead by}\quad \frac{\beta\,\ell_0}{c}
In words

trailing clock ahead by beta ell sub 0 over c

Source: 8.286 Lecture 3 slides, slide 21

The third effect resolves an apparent paradox: two observers moving past each other each find the other's clocks running slow. There is no contradiction, because they disagree about which distant events are simultaneous.

Summary#

  • Redshift is defined by 1+z=λO/λS=ΔtO/ΔtS1 + z = \lambda_O/\lambda_S = \Delta t_O/\Delta t_S.
  • For sound, source motion gives z=v/uz = v/u and observer motion z=(v/u)/(1−v/u)z = (v/u)/(1 - v/u): the medium makes them different.
  • For light there is no medium; with time dilation both cases give 1+z=(1+β)/(1−β)1 + z = \sqrt{(1+\beta)/(1-\beta)}.
  • Special relativity's kinematics is three effects: time dilation, length contraction and the relativity of simultaneity.

Symbols on this page

Show the 13 symbols used here
β\beta beta
Speed as a fraction of the speed of light, β = v/c. Say it: “beta” (sounds like “bay-ta or bee-ta”).
γ\gamma gamma
The Lorentz factor, γ = 1/√(1 − β²). Say it: “gamma”.
Δ\Delta capital delta
A change or difference in a quantity, e.g. Δx = x₂ − x₁. Say it: “capital delta”.
λ\lambda lambda
Wavelength. Say it: “lambda”.
ℓ\ell script l
A length or distance; ℓ_p is a physical distance and ℓ_c a comoving one. Say it: “ell”.
≈\approx approximately equal
Is approximately equal to. Say it: “is approximately”.
≡\equiv identical to
Is defined as (or is identically equal to). Say it: “is defined as”.
→\to arrow
Goes to (tends to), or becomes. Say it: “goes to”.
⇒\Rightarrow implies
Implies: what follows is a consequence of what came before. Say it: “which gives”.
×\times times
Times (multiplication). Between two vectors: the cross product. Say it: “times”.
aa a
The scale factor a(t): how much the universe has expanded. Physical distance = a × comoving distance (Guth's units: metres per notch). Say it: “a”.
zz z
Redshift: 1 + z = λ_O/λ_S, how much a wave's wavelength is stretched between source and observer (on this page by the source's motion). Say it: “z”.
cc c
The speed of light, c ≈ 2.998 × 10⁸ m/s. Say it: “c”.

Tap or click a symbol in a formula to see the same explanation. All symbols: symbols and notation.

Where this leads

Pages that build on this

Check your understanding

  1. A spectral line normally at 500 nm is observed at 550 nm. What is the redshift zz?
  2. For sound, does it matter whether the source or the observer is moving (at the same relative speed)?
  3. For light, why must the source-moving and observer-moving calculations give the same answer?
  4. A light source recedes at β=0.6\beta = 0.6. What is its Doppler redshift?
  5. Which of these is not one of the three kinematic effects of special relativity?
0 of 5 answered

Flashcards

1 / 5