Rayleigh Scattering Conceptual Question about the Color of the Sky

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SUMMARY

The discussion centers on Rayleigh scattering and its effect on the color of the sky, specifically addressing why the Sun appears red/orange when viewed directly despite blue light being scattered more than red light. Participants clarify that while blue light is scattered more, the direct line of sight to the Sun allows more red light to reach the observer, resulting in a red appearance. The concept of total cross-section is introduced, emphasizing that the interaction probability between light and air molecules is greater for blue light, yet the atmosphere diminishes blue light more effectively, leading to the observed colors.

PREREQUISITES
  • Understanding of Rayleigh scattering principles
  • Familiarity with light wavelength concepts
  • Knowledge of light interaction with atmospheric molecules
  • Basic grasp of cross-section calculations in physics
NEXT STEPS
  • Study Rayleigh scattering cross-section calculations
  • Explore the perception of color in relation to light wavelengths
  • Investigate atmospheric effects on light propagation
  • Examine why the sky does not appear purple despite violet light scattering
USEFUL FOR

Physics students, educators, and anyone interested in atmospheric science and optical phenomena will benefit from this discussion on Rayleigh scattering and its implications for sky color perception.

PhDeezNutz
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Homework Statement
Why when we look at the sun directly is it red/orange? and why does the sky appear blue?
Relevant Equations
For Rayleigh Scattering we have the scattering cross section

##\frac{\ d \sigma}{d \Omega} = \left( \frac{k_0^2 \alpha}{4 \pi}\right)^2 \left( 1 - \left|\hat{r} \cdot \hat{e}_0 \right|^2 \right)##

Where ##\hat{r} = \hat{k}## (the scattered wave propagates radially). ##\hat{e}_0## is the unit vector of the incident electric field. For the sake of simplicity I'm going to assume normal incidence (see picture below)

The ##k_0^4## dependence in the numerator implies a ##\frac{1}{\lambda^4}## dependence of ##\frac{d \sigma}{d \Omega}##.
Image 5-27-20 at 7.58 AM.jpg


We know ##\lambda_{blue} < \lambda_{red}## so citing the formula above (in the relevant equations) it's apparent that blue light is scattered more than red light. But presumably this would hold for ##\theta = 0## as well but when we look at the sun head on we don't see blue, we see red/orange.

Am i interpreting this formula wrong?
 
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I think it's because of the following. Even when ##\theta = 0## blue light is scattered more and red light is scattered less and that's why the "inner part" is red and the "outer part" is blue.

I think it's just that simple.
 
What do you mean by inner part and outer part?
 
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vela said:
What do you mean by inner part and outer part?

inner part - apparent angular size of the sun

outer part - the rest of the blue sky
 
Your answer feels incomplete to me. In the first post, you said blue light is scattered more than red light in the forward direction and asked why this didn't mean the Sun appeared blue. In the second post, you said blue light is scattered more than red light in the forward direction and concluded this is why the Sun appears red. I don't really see an explanation in there.
 
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vela said:
Your answer feels incomplete to me. In the first post, you said blue light is scattered more than red light in the forward direction and asked why this didn't mean the Sun appeared blue. In the second post, you said blue light is scattered more than red light in the forward direction and concluded this is why the Sun appears red. I don't really see an explanation in there.

yeah I think I was trying to reconcile it anyway I could...and it fell short. Still open to suggestions.
 
It might be easier to think about the total cross section, ##\sigma = \int (d\sigma/d\Omega)\,d\Omega##. You can interpret ##\sigma_{\rm blue} > \sigma_{\rm red}## to mean that the air molecules are a "bigger" target than for red light. In other words, the probability of an interaction between the light and an air molecule is greater for blue light than for red light.

Now think of light along your line of sight coming from the Sun to your eyes. For simplicity, let's assume equal amounts of blue and red light are emitted by the Sun. If you were on the Moon, equal amounts of blue and red light would reach your eyes because there's nothing that gets in the way between you and the Sun. If you're on the Earth, however, the atmosphere is there. What's going to happen?
 
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Not all the "forward" light is scattered. The stuff that is not scattered goes on through but diminished preferentially of its blue light. If you work it out, the total scattering of sunlight is about 10-5 per meter of air at sea level density. Depending on your interest you could calculate the total Rayleigh cross section (per atom) and figure what percentage of the sunlight is Rayleigh scattered. Of course the path gets longer as the sun sets but even at dusk it is only a fraction.
 
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A good follow up question is: why doesn't the sky appear purple.
 
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nrqed said:
A good follow up question is: why doesn't the sky appear purple.

That's an interesting question.

Based on the Rayleigh scattering cross-section formula

$$\sigma_s \sim \frac{1}{\lambda^4}$$

We indeed have ##\sigma_{violet} > \sigma_{blue}## so one would a priori expect to see the sky purple. The answer has to do with how we perceive violet
 
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