3
$\begingroup$

So I have often read that, at least in e.g. northern Europe, in the colder seasons, there is not enough UV (-B) light arriving from the sun, so many people have not enough vitamin-D from that.

At first I thought it was simply due to the sun "shining" for only a much shorter period of time in winter compared to summer and hence less possible exposure (not to mention that most of the skin area is covered then).

But I just had a thought coming to my mind, thinking about that in the mornings and evenings, we mostly see red light here, the higher end of the visible spectrum not getting through. I am not familiar with the physics behind that phenomenon, but thought that the higher-end of the spectrum like the invisible UV light may not be getting through here for even longer parts of the day towards and away from high noon, and that in winter, the part of the day where UV gets through is maybe very narrow and that's why it's said not to be enough.

Is that correct? And how exactly does this work physically?

|cite|improve this question
$\endgroup$
  • $\begingroup$ Thanks for the answers so far, sorry I didn't select one yet - I need to find some time to really understand what's written, it seems all helpful at a glance, I wouldn't know which to select as "the" reply right now... $\endgroup$ – user1847129 Jul 19 '16 at 22:30
4
$\begingroup$

The reddening of the sun has to do with Rayleigh scattering as the sun passes through more atmosphere. (see picture). This is in a sense, related to less energy but not the primary cause.

enter image description here

The reason we get less solar energy per square meter is that the angle of the sun in the sky affects how spread out the light is. (see updated picture). Ignoring atmospheric effects, it's the sin of the angle times peak energy. 90 degrees or directly overhead, figuring peak solar energy is 1,369 Watts per square meter (that also varies with distance), but the energy from the sun is mostly governed by the sin of the angle.

45 degrees: 1,369 * sin(45) W/m^2 or 71% of overhead or Zenith. 20 degrees above horizon, 1,369 * sin (20), just 34% of peak solar energy. Winter corresponds with the sun being lower in the sky, sunlight is more spread out. There is measurably less energy hitting the same area when the sun is low in the sky.

Passing through more atmosphere amplifies that somewhat, but the angle of the sun is the primary cause.

enter image description here

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ What are the $342W/m^2$? Is that the 24 hour average of the solar constant? $\endgroup$ – CuriousOne Jul 17 '16 at 9:33
  • 1
    $\begingroup$ @CuriousOne it's the 24 hour average, so the chart I posted has that wrong. (it answers the question simply, but you're correct, and it's not. It should say 1,369 per square meter, when the sun is overhead). home.iprimus.com.au/nielsens/solrad.html I could try to find a picture without that error, but some things are explained far easier with pictures than words. I'll edit my answer. $\endgroup$ – userLTK Jul 17 '16 at 9:42
  • 1
    $\begingroup$ I e-mailed NESTA and pointed out that error you caught. windows2universe.org/earth/climate/sun_radiation_at_earth.html FYI. $\endgroup$ – userLTK Jul 17 '16 at 10:05
  • 1
    $\begingroup$ I like your answer, but I'm not sure its the primary cause. It assumes all surfaces are flat. I can point my face directly towards the Sun for the hour before sunset, and for an hour at high noon. One of them will burn my face. What about the sun reflecting of the atmosphere more and more with increasing angle (relative to the earth surface normal), isn't that true for general linear materials? +1 $\endgroup$ – R. Rankin Jul 17 '16 at 11:59
  • $\begingroup$ @R.Rankin I don't have a good understanding of UV rays in the atmosphere, but I did mention that passing through more atmosphere adds to this. UV rays, I think, even more so. It's in a sense, the "you don't get sunburned at the Dead Sea" argument, and that's just 6% more atmosphere when the sun is overhead. With the Sun at an angle, it's measurably more than 6% more atmosphere the photons have to pass through, but not being an expert in UV rays, and Curious one's answer covering that, I think I'm better off not making an update. $\endgroup$ – userLTK Jul 17 '16 at 17:36
5
$\begingroup$

For a definition of the UV index and a rough discussion of influencing factors see e.g. https://en.wikipedia.org/wiki/Ultraviolet_index.

You can compare the model shown there against actually measured data for the US: http://www.cpc.ncep.noaa.gov/products/stratosphere/uv_index/uv_annual.shtml.

Both sources will support your statement that UV exposure in summer is significantly higher (by up to almost an order of magnitude) than in winter.

This graph

enter image description here

shows the difference in irradiance between the top and the bottom of the atmosphere. For the UV-B range of 280-315nm over 50% of the radiation is being absorbed while going trough the atmosphere at right angle, i.e. the extinction will be significantly larger than that when the sun is low over the horizon. I would agree that the geometric angle makes a significant difference in the available irradiance. Combine that with the time we spend outside (or, do not spend outside) and the fact that we are almost completely covered up in European latitudes, and our actual UV exposure can only be a tiny fraction of that in summer. Having said that, unfiltered summer exposure of more than half an hour is considered dangerous.

|cite|improve this answer
$\endgroup$
2
$\begingroup$

But I just had a thought coming to my mind, thinking about that in the mornings and evenings, we mostly see red light here, the higher end of the visible spectrum not getting through.

It does get through, at least to some extent. You see a very blue sky when the Sun is at its reddest. That blue sky can give you a sunburn if you go skiing in the mountains in the winter.

That said, even including the blue sky effect (which obviously has a significant ultraviolet component) the amount of ultraviolet light that does get through the atmosphere is indeed strongly reduced in winter compared to summer. The reason is the amount of atmosphere that the incoming solar radiation has to pass through.

For visible light, this is quantified by "air mass". The amount of visible light the reaches the surface of the Earth in winter is significantly reduced compared to summer because of the amount of atmosphere through which light from the Sun has to travel.

That said, the Earth's atmosphere, and in particular, the stratosphere, does an even better job of absorbing ultraviolet light than it does compared to visible light. The ultraviolet equivalent of "air mass" is even more strongly dependent on solar zenith angle than is visible light.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Full disclosure: I was the scientific programmer who was responsible for NASA's Nimbus 7 satellite not seeing the ozone hole almost 40 years ago. I received a large number of phone calls and personal contacts from high up in NASA 35 years ago or so. While I did do exactly what the PI demanded, I take full responsibility. I would not do that now. I have learned since then that being vociferous at the right time can be a virtue. For more, see physicsforums.com/insights/software-never-perfect . $\endgroup$ – David Hammen Jul 17 '16 at 14:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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