Lecture 7: Doppler effect for light -1 Doppler effect: phenomena where the measured frequency f (or wavelength λ) of a

wave is changed by relative motion of source and observer.

Acoustic Doppler effect: • The medium supporting the wave is air

• Sound has well defined velocity relative to medium

• See University Physics 16.8

• Source moving, detector stationary wrt medium

• Source stationary, detector moving wrt medium

• Source and detector moving wrt medium

Relativistic Doppler effect for light

Consider longitudinal case: relative motion of source and observer is along line joining them.

Source at rest at 𝑥 = 0 on S.

Emits 1st crest at 𝑡 = 0, 2nd crest at 𝑡 = 𝑇.

𝑇 is the period as measured in S, 𝑇 =1

𝑓

Doppler Effect - 2

Now consider observer, O’, who moves at speed v’ in x-direction relative to S (O’ receding from 0).

What frequency, 𝑓′, does he measure?

Suppose O’ is at 𝑥 = 𝑥0 at 𝑡 = 0, when first crest is emitted at O.

2nd crest has further to travel. ∴ 𝑇 ↑, 𝑓 ↓.

t

xO𝑥0

O’

𝑇

𝑠𝑙𝑜𝑝𝑒 = 1/𝑐

𝑡1

𝑡2

Doppler Effect - 3

How much do 𝑇 and 𝑓 change? How do we proceed?

First find times and positions of 1st and 2nd crests at O’ as measured by O.

Then use LT:

𝑡′ = 𝛾 𝑡 −𝑣𝑥

𝑐2

Arrival at O’ given by 𝑥1, 𝑡1 , (𝑥2, 𝑡2) as measured in S.

Doppler Effect - 4

𝑥1 = 𝑐𝑡1 = 𝑥0 + 𝑣𝑡1 ①

𝑥2 = 𝑐 𝑡2 − 𝑇 = 𝑥0 + 𝑣𝑡2 ②

①, ② ⇒ 𝑥0 = 𝑐𝑡1 − 𝑣𝑡1 = 𝑐 𝑡2 − 𝑇 − 𝑣𝑡2

⇒ 𝑡2 − 𝑡1 =𝑐𝑇

𝑐−𝑣③

Also using ①, ②: 𝑥2 − 𝑥1 = 𝑣(𝑡2 − 𝑡1) =𝑣𝑐𝑇

𝑐−𝑣④

Doppler Effect - 5

But we need to find time difference as measured in S’, i.e. 𝑡2′ − 𝑡1

′ .

Use LT:

𝑡′ = 𝛾 𝑡 −𝑣𝑥

𝑐2

𝑡2′ − 𝑡1

′ = 𝛾 𝑡2 − 𝑡1 −𝑣 𝑥2 − 𝑥1

𝑐2

Sub using ③, ④: 𝑡2′ − 𝑡1

′ = 𝛾𝑐𝑇

𝑐−𝑣−

𝑣

𝑐2∙𝑣𝑐𝑇

𝑐−𝑣

=𝛾𝑐𝑇

𝑐 − 𝑣1 −

𝑣2

𝑐2

⇒ 𝑡2′ − 𝑡1

′ =1

𝛾𝑡2 − 𝑡1

Doppler Effect - 6

𝑡2′ − 𝑡1

′ =1

𝛾𝑡2 − 𝑡1

Note that 𝑡2′ − 𝑡1

′ is proper time interval measured in S’ but 𝑡2 − 𝑡1 measured in S is not.

Put 𝛽 =𝑣

𝑐. Then the period as measured in S’ is:

𝑇′ = 𝑡2′ − 𝑡1

′

=𝛾 1 − 𝛽2

1 − 𝛽𝑇

= 𝛾 1 + 𝛽 𝑇

But 𝛾 = 1 − 𝛽2 −1

2

∴ 𝑇′ =1 + 𝛽

1 − 𝛽

1/2

𝑇

Doppler Effect - 8

𝑇′ =1 + 𝛽

1 − 𝛽

1/2

𝑇

Time difference between reception of successive pulses.

Time at source of emission of successive pulses.

Note that 𝑇′ ≠ 𝛾𝑇 since we don’t measure the time interval between the same pair of events.

Also:

𝑓 =1

𝑇⇒ 𝑓′ =

1 − 𝛽

1 + 𝛽

1/2

𝑓

Doppler Effect – approximation

If 𝛽 ≪ 1, we can use Binomial Expansion:

1 + 𝑥 𝑛 = 1 + 𝑛𝑥 +𝑛 𝑛 − 1

1 ∙ 2𝑥2 +⋯

1 + 𝛽

1 − 𝛽

1/2

≅ 1 + 𝛽 ∴ 𝑇′ ≅ 1 + 𝛽 𝑇

1 − 𝛽

1 + 𝛽

1/2

≅ 1 − 𝛽 ∴ 𝑓′ ≅ 1 − 𝛽 𝑓

Doppler Effect – approximation

But 𝜆 =𝑐

𝑓= 𝑐𝑇, so 𝜆 changes also, proportionally to 𝑇:

𝜆′ =1 + 𝛽

1 − 𝛽

1/2

𝜆 ∴ 𝜆′ ≅ 1 + 𝛽 𝜆

If S’ is receding from S:

𝑣 is positive, 𝛽 is positive, you get a red shift:

𝑣′ < 𝑣, 𝜆′ > 𝜆

If S’ is approaching S:

𝑣 is negative, 𝛽 is negative, you get a blue shift:𝑣′ > 𝑣, 𝜆′ < 𝜆

Another viewpoint

Two observers: O stationary in S

O’ and source moving at 𝑣 w.r.t. S

Emission of pulse 1 Emission of pulse 2

Measured by O’ (𝑡1′ , 𝑥′) (𝑡2

′ , 𝑥′)

Measured by O (𝑡1, 𝑥1) (𝑡2, 𝑥2)(N.B. – These are not the same events as earlier!)

𝑡2 − 𝑡1 = 𝛾 𝑡2′ − 𝑡1

′ = 𝛾𝜏0

𝑡2 − 𝑡1 > 𝑡2′ − 𝑡1

′

(The moving clock runs slow)

Reception of pulse 1 Reception of pulse 2

Measured by O (𝑡3, 𝑥) (𝑡4, 𝑥)

Emission of pulse 1 Emission of pulse 2

Measured by O’ (𝑡1′ , 𝑥′) (𝑡2

′ , 𝑥′)

Measured by O (𝑡1, 𝑥1) (𝑡2, 𝑥2)

𝑡4 − 𝑡3 = 𝜏(Period seen by O)

𝜏 =1 + 𝛽

1 − 𝛽

½

𝜏0

Also, (𝑡4 − 𝑡3) > (𝑡2−𝑡1)

since the second pulse has further to travel than the first.

Doppler effect is an example of looking at a moving clock.

Recessional Red ShiftA famous example of the Doppler effect is the red shift of light from distant galaxies.

Atoms at rest emit narrow spectral lines,

e.g. H 1s-2p : 𝜆0 = 121.6𝑛𝑚 (Lyman α)

But spectral lines from distant galaxies are shifted to longer λ, “red shifted”.

This implies those galaxies are receding from us!

In fact, the spectrum of light is almost continuous but with some weak absorption lines – produced when escaping radiation passes through a cooler region.

e.g. H and K lines in ionised Ca

𝛽 =

𝜆′𝜆

2

− 1

𝜆′𝜆

2

+ 1

, 𝛽 =𝑣

𝑐

Edwin Hubble (1889-1953)

From observations of the redshift of 46 galaxies, Hubble was able to conclude that the speed at which a galaxy recedes is proportional to its distance from us

𝑣 = 𝐻0𝑑

Where 𝐻0 = 72 ± 8 𝑘𝑚 𝑠−1/𝑀𝑝𝑐 (1999)(1𝑝𝑐 = 3.26 𝑙𝑖𝑔ℎ𝑡 𝑦𝑒𝑎𝑟𝑠)

𝑣 = the velocity of the receding galaxy

𝑑 = the distance from us, in megaparsecs

Redshift

Astronomers define redshift, ‘𝑧’, as follows:

𝑧 =𝜆𝑜𝑏𝑠 − 𝜆0

𝜆0(𝑧 + 1)2=

1 + 𝛽

1 − 𝛽

𝜆0 = 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑒𝑚𝑖𝑡𝑡𝑒𝑑 𝑙𝑖𝑔ℎ𝑡𝜆𝑜𝑏𝑠 = 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ 𝑤𝑒 𝑟𝑒𝑐𝑜𝑟𝑑 𝑜𝑛 𝐸𝑎𝑟𝑡ℎ

Current record: z=10

𝐻 𝐿𝑦𝑚𝑎𝑛 𝛼: 𝜆0 = 121.6𝑛𝑚 → 1337.6𝑛𝑚

β = 0.9836

Looking at moving clocks -1

t’

x’

S’

t

x

S

A set of synchronised clocks in S

As the moving clock travels its reading is compared with a stationary clock at the same point.

Observer O, in S, measures the moving clock to be running slow by a factor 𝛾 = (1 − 𝛽2)½

Looking at moving clocks -2What happens if O looks at the moving clock?

What does O see?

Suppose the moving clock sends out light pulses at equal intervals 𝜏0 of its proper time.

What we see at time 𝑡 (on our clock) was the reading on the moving clock at earlier time 𝑡 −

𝑟

𝑐, where 𝑟 is the distance of the moving clock at that earlier time.

O sees signals at later time when they reach him – This is just the Doppler effect!

Looking at moving clocks -3

If at some instant we see the moving clock reading 𝑡.

At time τ’ later (as measured by us!) we see the moving clock reading 𝑡 + 𝜏0

If clock is moving on straight line through our own position, then:

𝜏′ =1 + 𝛽

1 − 𝛽

½

𝜏0

Moving clock appears to be running slow.

If clock is moving towards us β is –ve and clock will appear to be running fast, not slow!

If the clock is a collection of moving atoms – blueshift!

Moral: Be specific about event or process.

Do not confuse observing with seeing

(seeing involves finite time for transit of light)