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Wednesday, 28 June 2017 00:14

Francis Berthomieu

"EAAE Summerschools" Working Group

C.L.E.A. Comité de Liaison Enseignants Astronomes (France)


The Doppler-Fizeau effect is a valuable tool in astrophysics.

In this workshop, we will explain what it consists in and describe the mathematic consequences as formulas.

We will then use it through the interpretation of some strange photographs, such as Saturn and its rings spectra.


1 - Doppler-Fizeau effect

Everybody could hear one day the acoustic Doppler Effect: When a car is moving towards you with its crying klaxon, you can hear a more acute sound than when the same car is going away.

A sound is a periodic phenomenon, with its frequency N and its period T. It propagates itself with a celerity C. These parameters allow us to define a wavelength λ = C.T or λ = C/N.

How can we explain the apparent modification of these parameters when the sound source is moving in relation with the observer?

Let's imagine three points placed on a same straight line Ox. A and B do not move, while C is moving with constant velocity V from A toward B. Let's assume that C emits some sound characterised by its frequency N0 and period T0.

Figure 1.Figure 1.

In fig.1 you can see a graphic representation of this situation in correlation with the time t. We can assume that C emits one "beep" every T seconds, as shown by points in the drawing. As sound propagates with a celerity C, it lasts some time to go from C to A (fig.2a) or B (fig.2b), depending on the distance it must travel through.

In these drawings, we can represent that propagation with straight lines, and we can see at what time the sound of a "beep" reaches A or B.

Figure 2a.Figure 2a.

Figure 2b.Figure 2b.

It appears clearly that the period T of the phenomenon, as received in A, is longer than T0, while as received in B it is shorter.

We can so deduce that:

If the sound source is moving away from the observer, the period of the sound seems longer than the original one. Its wavelength is also longer than the original one. Its frequency seems thus shorter and the sound seems louder.

If the sound source is moving toward the observer, the period of the sound seems shorter than the original one. Its wavelength is also shorter than the original one. Its frequency seems thus longer and the sound seems more acute.

Let's have a look at fig.3a and 3b, corresponding respectively to each of these situations.

Figure 3a.Figure 3a.

Figure 3b.Figure 3b.

We can there do a comparison between T and T0, and have an easy way to calculate V and C. The calculation shows that ΔT/T = V/C. As a logic consequence, and using the relation λ = C.T we can also deduce Δλ/λ = V/C

As sound, light is also a periodic phenomenon. Its celerity is much bigger than that of sound and its visible wave lengths quite shorter. Mathematically the equations are similar. If we consider a light source giving a monochromatic light, it has only one wavelength.

So, we can deduce that:

If the light source is moving away from the observer, the associated period seems longer than the original one and the associated wavelength is also longer: that is to say that it moves towards the red part of the spectrum. This is told "red-shift"

If the light source is moving toward the observer, the associated period seems shorter than the original one and the associated wavelength is shorter: that is to say that it moves towards the blue part of the spectrum and this is told "blue-shift"

What happens in case of a light source of "compound" light, such as white light? Its decomposition by a prism or a grid gives a coloured spectrum, including each one of the visible wavelengths, and if the light source moves in relation with the observer, each of them moves towards blue or red side of the spectrum.

Let's have a look to the light sent by a star, such as the Sun. Its spectrum is not a continuous one: we can observe there some black lines, whose wavelengths correspond to the elements the light had to pass through while going from the star to the observer eye. We can see the same spectrum if we use the light coming from a planet: this light is the Sun light, reflected by the planet surface, acting as a mirror, and we can observe a lot of black lines, having each of them a characteristic wavelength.

Spectrum of Sun.Spectrum of Sun.

Now, what happens if the light source is moving in relation with the observer? Each of its black lines wavelengths seems to be modified and shifted towards the red or blue side the spectrum whether the source is moving towards the observer or moving away. If we want to do some measurement of this displacement, it will be necessary to use a reference well known spectrum.

2 - Saturn spectrum

Have a look to this spectrum. It was taken at "Observatoire de Haute Provence" on July 24, 1962. At this time, Saturn was quite in opposition with the Sun, and its rings plane was doing a very small angle with the Observer-Saturn direction. Its distance from Earth was 1,3 x 109 km. Fig.4 shows how the spectroscope slit was placed:

Figure 4.Figure 4.

The slit passes through a diameter of the planet (that can be considered as the equator of the planet) and both extremities of its rings. So it allows obtaining on the same image 3 bands. In the middle, the larger band is the spectrum of the light coming from Saturn globe. On each side appear first the spectra of both extremities of the rings and a "comparison spectrum", with the well known emission lines of iron, obtained with the same spectroscope, doing it easy to determine any wavelength. Some of these characteristic wavelengths are given.

As you can see, the lines of the three spectra are inclined, while those of the reference iron spectrum are not. This can be explained by the Doppler-Fizeau effect and the radial movement of each point, considered as a mirror reflecting the light coming from the Sun. The Saturn globe is rotating around its own axis so that part of its equator is moving toward us while the other part is moving away, each point of it moving with a different radial velocity relatively with us, looking at them from the Earth.

The rings are also moving around the planet, each of the blocks that constitute them having a different velocity. It is easy to understand that we will have to use the radial velocity of each light source.

Caution! These objects act as a mirror for the Sun light. If such a point is moving in relation with us at a radial velocity Vr, the image it gives of the Sun seems to move with a 2Vr velocity! So the value to be used in the Doppler-effect formulae will be V=2Vr!

Radial velocity and real velocity

In this drawing, you can understand easily the mathematical relation between this two values.

Some complementary data

Saturn radius is D = 64.104 km

Radius of Saturn Orbit: R = 9, 54 A.U.

Now, look at this photo and find the answers to some questions ...


  1. Why are the lines of Saturn globe spectrum inclined?
  2. Why are these lines straight lines?
  3. What is the relation between this slope and Saturn rotation velocity?
  4. Can you deduce the period of rotation of planet Saturn?
  5. Do the rings rotate as a solid body?
  6. Why are the lines of the rings spectra inclined in the opposite way?
  7. Deduce the velocity of an external point of the rings.
  8. Which is the radius of the external side of the rings?
  9. Can you deduce the period of rotation of the external side of the rings?

More questions

  1. From the previous results, and using Newton laws, deduce the mass M of planet Saturn.
  2. Compare Saturn and Earth density. Is it true that Saturn would float over water?
  3. Compare Saturn and Earth equatorial velocity.
  4. Show that Saturn and Earth gravity are quite equal.