Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Why do stars flicker and planets don't? At least this is what I've read online and seen on the night sky. I've heard that it has to do something with the fact that stars emit light and planets reflect it. But I don't get it, isn't this light, just "light"? What happens to the reflected light that it doesn't flicker anymore?

I was thinking that it has to do something with Earth's atmosphere, different temperatures or something (if this has any role at all).

share|cite|improve this question
up vote 12 down vote accepted

Here is a nice answer, taken from http://www.enchantedlearning.com/subjects/astronomy/stars/twinkle.shtml

The scientific name for the twinkling of stars is stellar scintillation (or astronomical scintillation). Stars twinkle when we see them from the Earth's surface because we are viewing them through thick layers of turbulent (moving) air in the Earth's atmosphere. Stars (except for the Sun) appear as tiny dots in the sky; as their light travels through the many layers of the Earth's atmosphere, the light of the star is bent (refracted) many times and in random directions (light is bent when it hits a change in density - like a pocket of cold air or hot air). This random refraction results in the star winking out (it looks as though the star moves a bit, and our eye interprets this as twinkling). Stars closer to the horizon appear to twinkle more than stars that are overhead - this is because the light of stars near the horizon has to travel through more air than the light of stars overhead and so is subject to more refraction. Also, planets do not usually twinkle, because they are so close to us; they appear big enough that the twinkling is not noticeable (except when the air is extremely turbulent). Stars would not appear to twinkle if we viewed them from outer space (or from a planet/moon that didn't have an atmosphere).

share|cite|improve this answer
    
Thanks you! I think I got it now. – user25857 Jun 16 '13 at 2:42

(This question is a duplicate of What is the physical origin of scintillation?) The answer is in the link below. Short version: Refractive gradients in the mixed atmosphere, thousands of meters above the surface, are far too small to act as lenses and prisms, and scintillation is not a geometric optics phenomenon. It results from interference effects of plane-wave light (spatially coherent light, like starlight or laser light) as the wavefront is distorted by tiny refractive gradients associated with Kolmogorov turbulence. This accounts for both the intensity and color fluctuations seen by eye, and also explains why the larger planets, whose light is far from coherent, do not scintillate. Little's paper anticipates by a decade the works of Kolmogorov, Tatarski and Rytov. It also anticipates speckle, seen in larger apertures, where the distorted wavefront breaks up the Airy disc into tiny pieces that fluctuate on millisecond time scales.

http://adsabs.harvard.edu/full/1951MNRAS.111..289L

Enchanted Learning, like other simple refraction/dispersion explanations, is simply wrong. Little's paper requires some understanding of Fresnel diffraction, but it's well worth the read.

share|cite|improve this answer
    
Why is starlight coherent? There aren't any lasers in stars. – Peter Shor Jan 1 '14 at 23:43
1  
At our vantage point, the light from a star has spread out into a vast sheet, i.e. a plane wave, and the phase everywhere on the wavefront is the same over distances of many meters. Starlight doesn't have temporal coherence, which many lasers do, and obviously it isn't monochromatic. The sky is crisscrossed with plane waves which your eye transforms into a pattern of dots. – John Kuehne Jan 2 '14 at 1:09
    
For planets, the light from each photon is also essentially a plane wave. And planets don't twinkle. – Peter Shor Jan 2 '14 at 1:12
    
The spatial coherence length of light from planets visible to the eye is much smaller than the dissipation range of turbulence in the atmosphere that perturbs the light, i.e. see "Laser Beam Scintillation with Applications" by Andrews, Phillips, Hopen. The coherence length for the sun and moon are even smaller, and they don't scintillate either. – John Kuehne Jan 2 '14 at 1:43
    
See book sample that explains the coherence of starlight. (The author uses the term "coherence length" in the temporal direction, and I mean across the wavefront.) Pages 153-156 explain why starlight is coherent, gives an example for the Sun and Betelgeuse, and explains why diffraction patterns we see in telescopes look like patterns from lasers. books.google.com/… – John Kuehne Jan 2 '14 at 15:58

This is just a sidenote to Nijankowski's nice answer: This twinkling of stars caused by atmospheric turbulence was a major problem for the earlier reflecting telescopes when astronomers tried to look deep into the sky.

Placing the telescope over mountains solved only a part of the problem. A good solution was brought up in two ways in the 1990s. First, sending telescopes to space - Hubble Space Telescope was carried to orbit where it took astonishingly sharp images like the Hubble Deep Field due to the absence of atmosphere's interaction. Then arrived the serious flaw in its mirror. Repairing the telescope (by orbiting in space) was a very big problem as many equipments have to be replaced by service missions.

The second solution was adaptive optics. In telescopes like the Large Binocular Telescope, a secondary mirror was placed which is readily deformable and the shape is modified according to the incoming light source by a number of hydraulic pistons behind, thus correcting for some amount of atmospheric distortion.

share|cite|improve this answer

Much closer than stars are the distant lamps (polychromatic or monochromatic), say 2 to 3km away, also twinkle. this cannot be related to change of index of refraction due to temperature's variations , since the frequency of this twinkling does not vary strongly with air turbulence. the accepted explanation for this is: light is formed of photons (light particles) which spreads out from the source in all directions, thus as the distance increases, the spherical surface area increases with the square of the radius, hence photon's flux decreases, and the number of photons reaching the eye become less frequent. at the times of absence of photons entering the eye, the image of the source disappear. Moreover, it is known from astronomy, that forming an image of a very far celestial object needs long time, and large lenses or mirrors. this shows that: in order to form an image, an enough number of photons is needed, which requires long time to collect them.
By Sami Kheireddine

share|cite|improve this answer
    
Welcome on Physics SE. As you can see from the other upvoted and accepted answers, this is not a canonical, accepted answer. Could you please provide a scientific source for your statements? – Sanya yesterday
    
It's worth noting that large telescopes with multi-element main mirrors can (and do) do active compensation for atmospheric distortions which they measure in real time by tracking the back-scatter of pilot lasers pointed parallel to the telescope line of sight. The fact that this works demonstrates that turbulence lensing is the dominate effect (that is, the accepted answer is right). – dmckee yesterday
    
If star twinkling is due to change of direction of the path of light, then stars would have been observed OSCILLATING , not twinkling. – Sami Kheireddine yesterday
    
IT should be noted that forming an image of a far away celestial object does NOT require a long time. Many astronomers will take a long time to try to reduce noise in the image, but an image can be formed in real-time as well. – Jim yesterday
    
Deep field observations are long-lasting observations of a particular region of the sky intended to reveal faint objects by collecting the light from them for an appropriately long time. The 'deeper' the observation is (i.e. longer exposure time), the fainter are the objects that become visible on the images. Astronomical objects can either look faint because their natural brightness is low, or because of their distance. In the case of the Hubble Deep and Ultra Deep Fields, it is the extreme distances involved which make them faint from: spacetelescope.org/science/deep_fields – Sami Kheireddine yesterday

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.