Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

We are able to look directly at the sun near sunset and sunrise, which clearly demonstrates the fact that our atmosphere attenuates visible light. Let's imagine it follows the typical attenuation profile.

$$ I = I_0 \, e^{-(\mu/\rho)\rho \ell} $$

Where $(\mu/\rho)$ is the mass-attenuation coefficient in units of $(m^2/kg)$, and $\rho l$ is the mass-thickness (or the area-density, I think it has a few names) in units of $(kg/m^2)$.

The effect of the "soft" sunlight is then predicted as a result of the fact that the mass-thickness of the atmosphere between our eyes and the sun diverges fairly fast as the angle to the sun falls to zero degrees above the horizon.

Let's say that a person is standing on (perfectly spherical) Earth, with their eyes at a known elevation, looking at the sun which lies at a known angle above the horizon. What is the expression for the mass-thickness of air in that line of sight?

The reason I find this non-trivial is that I can't figure out if the density profile of the atmosphere should matter or not. You could reduce it to simple geometry and get an answer, but is there a coherent argument for that being correct? With a clear expression, I'm actually kind of curious if you could measure the mass-attenuation coefficient with just a digital picture. The sun's intensity starts out constant over the circle, and you know the angle between the top and bottom of the sun exactly. So if you could extract intensity data over the vertical diameter maybe you could then do a least-squares function fit to extract out that attenuation coefficient, and even do it for each of the 3 colors. I don't plan on doing that, but it would be a cool science project.

share|improve this question
For a decent estimate of the density profile in the atmosphere, you could use the Standard Atmosphere –  OSE Apr 18 '13 at 17:08
There is a complication in that the attenuation coefficient is not constant: witness how very different sunsets can be from day to day depending on the nature of the tropospheric air mass that's just west of you. –  dmckee Apr 18 '13 at 17:08
@dmckee Well yeah, but how different would it be between a line-of-sight to the bottom vs the top of the sun? I totally believe that the coefficient will change due to climate conditions. Page 281 in this link simulatedvision.co.uk/V&A_Chap15.pdf sums up the differences pretty well. It's the aerosols that complicate the issue. I'm concerned about the difference between upper vs lower atmosphere, since aerosol attenuation is not proportional to density. I'm not sure if the different lines will spend different fractions of mass-thickness in upper vs lower atmosphere... –  Alan Rominger Apr 18 '13 at 18:09

2 Answers 2

First of all, atmospheric attenuation in the visible region is primarily due to scattering, not molecular absorption as in the infrared and microwave regions. This is perhaps not that important to your question, but a good thing to keep in mind. The ligth doesn't disappear, it only changes direction.

If you have the density profile of the atmosphere, let's denote it $\rho(h)$, where $h$ is the height above sea level, you can calulate what you call mass-thickness, (I will call it slant mass column and denote it $C$), using this integral:

$$ C=\int^{\infty}_{0}\frac{\rho(h)dh}{\sqrt{1-\left(\frac{\cos{\theta_0}}{1+h/R}\right)^2}} $$

or this one:

$$ C=\int^{\pi/2}_{\theta_0}\rho\left(R\left(\frac{\cos{\theta_0}}{\cos{\theta}}-1\right)\right)\frac{R\cos{\theta_0}}{\cos^2{\theta}}d\theta $$

where $\theta_0$ is the elevation angle of the sun and $R$ is the earth radius. These formulas are for an obeserver located at sea level.

This assumes that the light travels straight through the atmosphere. This assumption works reasonably well for high elevation angles, but for lower angles (think sunsets) you have to take atmospheric refraction into account and perhaps also some additional scattering effects.

Your digital camera experiment might work in theory but it will perhaps not be as easy as you would like. First of all you need to account for refraction when calculating the slant mass column as I explained above. Refraction is wavelength-dependent so you will have to do it for each color. Secondly, only Rayleigh scattering will be proportinal to $C$. You will also have Mie scattering due to aerosols. This will depend on the aerosol loadings in the atmosphere which will be variable. Mie scattering will also be more important for low elevation angles since a larger part of the path through the atmosphere will be in the lower parts, which have higher aerosol loadings.

share|improve this answer
If there was significant wavelength-dependent refraction going on, wouldn't that make the apparent image of the sun non-circular? I mean, at any given wavelength it would be circular, but there would be a red sun displaced above a blue sun, or something like that. –  Alan Rominger Apr 18 '13 at 18:47
@AlanSE The phenomen you describe does exist (but red is at the bottom) and is often referred to as green rim which can cause a green flash at sunset. The rim is green because most of the blue ligth is lost in scattering (plus the background is blue so it is harder to see a blue rim). This displacement of the different colors is small compared to the apperent size of the sun, so it is hard to see unless you experience a green flash. The difference in pathlength through the atmosphere for the different colors might still be significant though. –  jkej Apr 18 '13 at 19:04
$AlanSe An elaboration on that last sentence: The displacement of the colors is smaller than the apparent size of the sun, but it is still on similar scales, and hence the difference in pathlength through the atmosphere will be on a simlar scale as the difference between upper and lower edge of the sun. You are correct to assume that this is probably not the largest effect but that doesn't mean that it is insignificant. –  jkej Apr 18 '13 at 19:17

I'm going to assume the atmosphere is constant density and solve the problem. We'll pick up the pieces from this assumption later. Next, I'll trash the specification that the observer is at a nonzero height. Not a big deal, just put the camera on the ground.

Since the atmosphere is constant density at all heights, it has a defined boundary. The height of the top of the atmosphere is then easy to calculate. Just divide the mass of the atmosphere by the density and surface area of the Earth.

$$ h = \frac{5.1480 \times10^{18} \text{ kg}}{\left( 1.3 \frac{\text{ kg}}{m^3} \right) 4 \pi \left( 6,378.1 \text{ km} \right)^2} = 7.746 \text{ km} $$

Additionally, we assume we know the angle of sight. That gives enough information to formulate the triangle with the law of cosines.

sunset ray trace

The equation is written from using the law of cosines combined with inspection. The side opposite of the large angle is the radius of the Earth plus the height of the atmosphere. One other side of the triangle is the radius of the Earth. The final side is the unknown we want.

$$ \left( R+ h \right)^2 = R^2 + x^2 - 2 R x \cos{ \left( \theta + \frac{\pi}{2} \right) } $$

$$ x = -R \sin{\theta} + \sqrt{ R^2 (\sin{\theta})^2 + 2 R h + h^2 } $$

Now we have enough information to plot this.


As predicted, this gets the height of the atmosphere as the angle goes to pi/2. As the angle goes to zero the value goes to $314.4 km$. To get the mass-thickness multiply this by the density.

Why this might be a decent approximation

Now you will say, "but the atmosphere isn't constant density". I think I've come up with a good counter argument.

Let's say the density is half of the sea-level value. That will increase the height of the atmosphere by a factor of two. So I resolve the above equation with that new number. It's not exactly linear with $h$, but we will find that the value of $x$ is about twice its former value. But then we multiply by $\rho$ which is half the sea level value. That means that we arrive at our original number! In fact, I plotted this for all densities. This is for example purposes, so I supposed an arbitrary angle from the horizon of 0.1 radians.


Probably less than 1/20th of the atmosphere is below a density of $0.1 kg/m^3$. Maybe the above graph is sufficiently flat in that range? ...but maybe not. The method isn't as good as I had hoped. The very high altitude distribution of our upper atmosphere really does call for a more exact accounting, but since the pressure falloff is exponential I'm sure that would be difficult.

share|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.