Monster92 Posted April 7, 2011 Share Posted April 7, 2011 Hello I'm studying for a course in Environmental Studies and I wondered if anyone could provide any help. So I've been given the handbook of what is supposed to be known as well as the textbook. However, the text book does not provide sufficient information regrading the following. Quoted from the guide 1. The wave nature of electromagnetic radiation and the electromagnetic spectrum 1A. Wavelength characteristics, and the environmental importance of ultraviolet light, visable light and infra red light 2. The characteristics of insolation and factors which cause it to to change... at the outer limits of the atmosphere as it passes through the atmosphere when it reaches the Earth's surface. I've tried google of course but I either come up with websites that explain the latter in great complexity or information that is too simplistic. I suppose no advanced physics is needed as it is environmental studies after all. Plus, what does it mean when the guide refers to 'insolation? Yeah, the textbook sucks lol thanks Link to comment Share on other sites More sharing options...
mooeypoo Posted April 7, 2011 Share Posted April 7, 2011 You need to be a bit more specific as to how deep you want an explanation for it... The electromagnetic spectrum is a division of electromagnetic waves to sections according to frequencies and function. http://imagine.gsfc.nasa.gov/docs/science/know_l1/emspectrum.html http://en.wikipedia.org/wiki/Electromagnetic_spectrum Visible light is only a little part of it, and it explains what the main differences are between high-frequency and low-frequency EM waves. Here's a helpful picture: As to the effects of EM waves on the atmosphere, it's a bit more complicated than that. The *general* idea is that EM waves carry energy with them, and they bounce off (reflect) and interact with (refract) the atmosphere. These supply heat and affect the colors we see in the skies among other things. It is also involved in the greenhouse effect. I found this list of general facts from what APPEARS to be a similar course to yours. It might help: http://eesc.columbia.edu/courses/ees/climate/lectures/radiation/index.html This seems helpful too: http://en.wikipedia.org/wiki/Sunlight Hope it helps... ~mooey Link to comment Share on other sites More sharing options...
swansont Posted April 8, 2011 Share Posted April 8, 2011 There's also [math]E=h\nu = hc/\lambda[/math], telling you that the energy of light depends on its frequency (or wavelength), and that a blackbody will radiate EM waves in a spectrum that depends on T^4. So light from the hot sun has a spectrum that is peaked in the visible, while the cooler earth radiates a spectrum that has longer wavelengths. Link to comment Share on other sites More sharing options...
lemur Posted April 8, 2011 Share Posted April 8, 2011 There's also [math]E=h\nu = hc/\lambda[/math], telling you that the energy of light depends on its frequency (or wavelength), and that a blackbody will radiate EM waves in a spectrum that depends on T^4. So light from the hot sun has a spectrum that is peaked in the visible, while the cooler earth radiates a spectrum that has longer wavelengths. Shouldn't the Earth actually be reflecting more radiation than it is absorbing and radiating? Link to comment Share on other sites More sharing options...
swansont Posted April 8, 2011 Share Posted April 8, 2011 Shouldn't the Earth actually be reflecting more radiation than it is absorbing and radiating? Why should it? The average albedo is 0.3 - 0.4; it's higher in snowy/icy/desert areas, and cloudy regions and lower in heavily forested areas. http://earthobservatory.nasa.gov/IOTD/view.php?id=5484 Link to comment Share on other sites More sharing options...
lemur Posted April 8, 2011 Share Posted April 8, 2011 Why should it? The average albedo is 0.3 - 0.4; it's higher in snowy/icy/desert areas, and cloudy regions and lower in heavily forested areas. http://earthobservat...iew.php?id=5484 Based on the pic posted by Mooeypoo, it looks like most of the infrared spectrum is blocked along with microwaves, x-rays, and most of UV. I guess, though, that the blocking could refer to atmospheric absorption as well as reflection. I guess I was just assuming that anything blocked must be reflected. Link to comment Share on other sites More sharing options...
swansont Posted April 8, 2011 Share Posted April 8, 2011 No, that includes absorption, and the spectrum isn't flat. More than half of the energy of a blackbody at 6000K (i.e. like the sun) is in the visible and Near-IR part of the spectrum (0.4 - 1.0 micron) where the transmission is high. Link to comment Share on other sites More sharing options...
mooeypoo Posted April 8, 2011 Share Posted April 8, 2011 Light both refracts, reflects *and* is absorbed in the atmosphere. This is a good resource image, actually: Source: http://amazing-space...on/basics/g17b/ You can see not only how much of the EM radiation is absorbed into the atmosphere, but approximately how high up it takes to to get absorbed. Also, the energy from the absorbed particles is transfered to the particles in the atmosphere, which helps in various phenomena we witness on the surface (and in the air). Here's a good collection of answers to what happens to the X-ray and gamma-ray EM waves when they hit the earth's atmosphere: http://www.newton.de...05/phy05042.htm Excerpt: Question - What happens to the energy of a gamma ray or X-raywhen it is absorbed by the atmosphere? Robert,The energy of any absorbed electromagnetic wave (gamma, X-ray, ultraviolet, even visible light) is transferred to the particles that the waves hit. The waves are made of little pieces called photons. Each photon can either hit something or pass through. The particles can be whole molecules, individual atoms, or sometimes just electrons. This energy can be held in these particles, causing the atmosphere to warm>up. The energy can be released in another direction, never reaching th Earth. The energy can be released as several lower energy photons, making>it less dangerous. Different particles are better at absorbing different energies. A great deal of visible light gets through because the atmosphere is not really good at absorbing visible light. The sky looks blue because the sky can absorb blue light, but cannot hold it. The result is scattering. Blue light from the sun can go off to one side, scatter in the atmosphere, and bounce back to your eyes. Red light travels in a straight line, so you only see suc colors at or near the sun. In certain situations, the atmosphere scatters>red light. At these times the sky looks red. Dr. Ken Mellendorf Physics Instructor Illinois Central College Link to comment Share on other sites More sharing options...
lemur Posted April 8, 2011 Share Posted April 8, 2011 Nice references, Mooeypoo. So gamma and xray radiation actually get absorbed by the atmosphere by gradually stepping down to longer wavelengths, according to the linked discussion. That means that some of the visible light and infrared reaching the ground were first superluminous radiation (is there a better word for that, btw?). That also leads me to wonder if the visible light from the sun also steps down in frequency as it penetrates the atmosphere. Is there a difference between this stepping down process for different wavelengths of radiation? It almost seems like you could call this absorption-redshift if it was a universal phenomenon for all radiation to decrease in wavelength as is passes through a medium gradually giving away energy to electrons. Link to comment Share on other sites More sharing options...
swansont Posted April 8, 2011 Share Posted April 8, 2011 Nice references, Mooeypoo. So gamma and xray radiation actually get absorbed by the atmosphere by gradually stepping down to longer wavelengths, according to the linked discussion. That means that some of the visible light and infrared reaching the ground were first superluminous radiation (is there a better word for that, btw?). That also leads me to wonder if the visible light from the sun also steps down in frequency as it penetrates the atmosphere. Is there a difference between this stepping down process for different wavelengths of radiation? It almost seems like you could call this absorption-redshift if it was a universal phenomenon for all radiation to decrease in wavelength as is passes through a medium gradually giving away energy to electrons. It's not universal. The light has to interact, and the probability of that is not constant with energy. There are a number of scattering processes involved. Link to comment Share on other sites More sharing options...
mooeypoo Posted April 8, 2011 Share Posted April 8, 2011 Nice references, Mooeypoo. So gamma and xray radiation actually get absorbed by the atmosphere by gradually stepping down to longer wavelengths, according to the linked discussion. That means that some of the visible light and infrared reaching the ground were first superluminous radiation (is there a better word for that, btw?). That also leads me to wonder if the visible light from the sun also steps down in frequency as it penetrates the atmosphere. Is there a difference between this stepping down process for different wavelengths of radiation? It almost seems like you could call this absorption-redshift if it was a universal phenomenon for all radiation to decrease in wavelength as is passes through a medium gradually giving away energy to electrons. The sun emits EM radiation in all spectrum, that later interacts with the atmosphere. Some is refracted, some reflected and absorbed. The resulting visible light we see on Earth is the result of all three of those. I would not call it "redshift" at all, since that phenomenon is completely different, has a different cause and does not result in the same effect. I repost my earlier link to this site: http://eesc.columbia.edu/courses/ees/climate/lectures/radiation/index.html It has a lot of explanations on how the sun's radiation interacts and affects the Earth's atmosphere and climate. Also, this page: http://eesc.columbia.edu/courses/ees/climate/lectures/radiation/heat_xfer.html Which explains a bit of the energy concepts regarding electromagnetic radiation from the sun that interact with the atmosphere on the Earth. As swansont said, this isn't universal. The atmosphere isn't equally distributed and the effect is not 1:1 everywhere. Interaction with areas with less or more of particular chemicals in the atmosphere will produce different absorption reactions. That's part of the reason we have much more adverse effects on the ground above the "holes" in the Ozone layer. Also, light that reaches the surface can (and is) reflected BACK towards the atmosphere, where it AGAIN either reflects back to the ground ("greenhouse effect", somewhat) or is absorbed. There are areas where light is reflected more out from the ground like swansont's link showed above, mainly the poles (where the ground is mostly snow/ice). That creates complicated reactions that affect what we eventually get on the surface as "visible light" too. ~mooey Link to comment Share on other sites More sharing options...
lemur Posted April 8, 2011 Share Posted April 8, 2011 I would not call it "redshift" at all, since that phenomenon is completely different, has a different cause and does not result in the same effect. I suppose the fact that the wave loses energy as it "shifts" is different from redshifting in that redshifting conserves the wave energy as it expands. I just thought it sounded like a universal phenomenon of light going through a medium; and if it was that it would make sense that it would have a name (if it doesn't already). As swansont said, this isn't universal. The atmosphere isn't equally distributed and the effect is not 1:1 everywhere. Interaction with areas with less or more of particular chemicals in the atmosphere will produce different absorption reactions. That's part of the reason we have much more adverse effects on the ground above the "holes" in the Ozone layer. That makes sense, but the part that is really interesting to me is the question of how much sunlight at sea-level differs from sunlight prior to atmospheric contact. I.e. how much does the sunlight change and what happens to its energy? I used to look at high frequency radiation as wasted energy as far as solar collectors are concerned, but I thought it was blocked (i.e. reflected) by the atmosphere completely and I never considered that it might make it to the ground at a lower frequency. I also didn't know that any infrared was blocked so I assumed that all the heat felt from sunlight was the total amount of heat the sun is emitting. Now I wonder how much hotter sunlight is in orbit. Also, light that reaches the surface can (and is) reflected BACK towards the atmosphere, where it AGAIN either reflects back to the ground ("greenhouse effect", somewhat) or is absorbed. There are areas where light is reflected more out from the ground like swansont's link showed above, mainly the poles (where the ground is mostly snow/ice). That creates complicated reactions that affect what we eventually get on the surface as "visible light" too. This question may be too far an extension of this topic, but it's one I've been trying to figure out. How much does visible light get absorbed as heat, as opposed to getting reflected off visible objects? I guess the color and hue of the object basically tells you how much light it's absorbing, but is a certain amount of all visible light absorbed instead of reflected and we just don't notice it because its constant for all visible objects? Link to comment Share on other sites More sharing options...
mooeypoo Posted April 8, 2011 Share Posted April 8, 2011 I suppose the fact that the wave loses energy as it "shifts" is different from redshifting in that redshifting conserves the wave energy as it expands. I just thought it sounded like a universal phenomenon of light going through a medium; and if it was that it would make sense that it would have a name (if it doesn't already). Light from stars does not go through any medium when the "redshift" is observed. It's not the same phenomenon at all. That makes sense, but the part that is really interesting to me is the question of how much sunlight at sea-level differs from sunlight prior to atmospheric contact. I.e. how much does the sunlight change and what happens to its energy? I used to look at high frequency radiation as wasted energy as far as solar collectors are concerned, but I thought it was blocked (i.e. reflected) by the atmosphere completely and I never considered that it might make it to the ground at a lower frequency. I also didn't know that any infrared was blocked so I assumed that all the heat felt from sunlight was the total amount of heat the sun is emitting. Now I wonder how much hotter sunlight is in orbit. I posted some links that give you those answers. Swansont posted a link with the actual answer to these questions in them visually. This question may be too far an extension of this topic, but it's one I've been trying to figure out. How much does visible light get absorbed as heat, as opposed to getting reflected off visible objects? I guess the color and hue of the object basically tells you how much light it's absorbing, but is a certain amount of all visible light absorbed instead of reflected and we just don't notice it because its constant for all visible objects? Swansont answered that question already and posted a link that answered it further. ~mooey Link to comment Share on other sites More sharing options...
lemur Posted April 8, 2011 Share Posted April 8, 2011 Light from stars does not go through any medium when the "redshift" is observed. It's not the same phenomenon at all. I already conceded you were right based on the fact redshift doesn't involve any loss of energy in changing to a shorter wavelength. However, if you're going to beat the dead horse to see if there's any kick left in it, there is still a little:) Namely, I still think there's some analogical validity in the fact that gravitational redshifting is caused by field-force interaction with the light, which is sort of similar to light getting (partially) absorbed and re-emitted by the electrons of an atom. Do you think that's a baseless analogy? I posted some links that give you those answers. Swansont posted a link with the actual answer to these questions in them visually. I read all those diagrams both of you posted as treating each wavelength as an isolated timeline from sun to absorption. I didn't see anything about partial absorption and shifting of remaining energy to lower frequencies until the links you posted. Link to comment Share on other sites More sharing options...
mooeypoo Posted April 9, 2011 Share Posted April 9, 2011 the images we posted were simplified. If you truly want to know the more advanced principles, you need to do some reading. The links we both supplied should give you a good start. Then, you can get into the physics of specific phenomena you're interested in within those subjects. Here's another interesting resource to go over:http://atoc.colorado.edu/~dcn/ATOC1060/Members/Lectures/06_Greenhouse.pdf Link to comment Share on other sites More sharing options...
lemur Posted April 9, 2011 Share Posted April 9, 2011 the images we posted were simplified. If you truly want to know the more advanced principles, you need to do some reading. The links we both supplied should give you a good start. Then, you can get into the physics of specific phenomena you're interested in within those subjects. Here's another interesting resource to go over:http://atoc.colorado..._Greenhouse.pdf Thanks for the slideshow. I still wasn't able to extrapolate the answer to my question about how different wavelengths translate into different breakdowns to shorter wavelengths. I guess that's the reason I should raise it as an issue for discussion in a forum. Maybe it would work better as a series of specific questions; i.e. "what happens to gamma radiation in the atmosphere?," "what happens to UV radiation in the atmosphere?" and "what happens to blocked-infrared and blocked microwaves in the atmosphere?" Link to comment Share on other sites More sharing options...
swansont Posted April 9, 2011 Share Posted April 9, 2011 If the light is non-ionizing, it's absorbed and then can be re-emitted. The emission direction is basically random, so it can send a photon back out into space. Otherwise, the energy is lost through nonradiative means, and this results in an increase in temperature of the atmosphere. It's the same basic process as the loss of microwaves in your microwave oven. If the light ionizes an atom or molecule, the electron loses energy by scattering and emitting bremsstrahlung. Recombination of an ion will give off a cascade of photons as it de-excites. Link to comment Share on other sites More sharing options...
lemur Posted April 10, 2011 Share Posted April 10, 2011 If the light is non-ionizing, it's absorbed and then can be re-emitted. The emission direction is basically random, so it can send a photon back out into space. Otherwise, the energy is lost through nonradiative means, and this results in an increase in temperature of the atmosphere. It's the same basic process as the loss of microwaves in your microwave oven. If the light ionizes an atom or molecule, the electron loses energy by scattering and emitting bremsstrahlung. Recombination of an ion will give off a cascade of photons as it de-excites. So the process of wavelength-decrease is really caused by absorption of the photon, whose energy gets divided into a combination of kinetic energy and re-emission. So if the particle gets ionized, some of the energy ends up as free-electron momentum while the rest may either end up changing the momentum of the particle, re-radiating at a frequency appropriate to the energy-level change, or some combination. So just pretending I know the energy-quantities to check my logic, take an example where an atom gets hit with 100 units of EM radiation. 20 units could be carried off as momentum of a liberated electron, 30 units could be expressed as additional kinetic energy to the atom's motion, and the remaining 50 units would be re-emitted as a wavelength with half the energy of the light originally absorbed? Does this sum up the logic and are their any other outlets for the energy that conserves the total amount absorbed? Link to comment Share on other sites More sharing options...
farmboy Posted April 10, 2011 Share Posted April 10, 2011 Apologies if this is totally wrong lol (I've mentioned before that this is not my normal subject area plus im a little drunk lol) but it just seems to me that the way you are thinking about ''wavlength decrease'' as if it is somehow analagous to redshift just isn't really correct. Perhaps it is because you are looking at light like a wave in this context, when it would be better to look at it like a particle? So for example the light coming from the source (in this instance the sun) may include all the wavelegths that constitute the visible spectrum, but when we consider how this interacts with the atmosphere I'm not sure we should look at it like a continuous spectrum with a mean wavelength which changes after it enters the atmosphere, but rather a series of discrete particles of different energies the mean/median of which will change after it has passed through the atmosphere Apologies if this is totally wrong lol (I've mentioned before that this is not my normal subject area plus im a little drunk lol) but it just seems to me that the way you are thinking about ''wavlength decrease'' as if it is somehow analagous to redshift just isn't really correct. Perhaps it is because you are looking at light like a wave in this context, when it would be better to look at it like a particle? So for example the light coming from the source (in this instance the sun) may include all the wavelegths that constitute the visible spectrum, but when we consider how this interacts with the atmosphere I'm not sure we should look at it like a continuous spectrum with a mean wavelength which changes after it enters the atmosphere, but rather a series of discrete particles of different energies the mean/median of which will change after it has passed through the atmosphere Link to comment Share on other sites More sharing options...
swansont Posted April 10, 2011 Share Posted April 10, 2011 So the process of wavelength-decrease is really caused by absorption of the photon, whose energy gets divided into a combination of kinetic energy and re-emission. So if the particle gets ionized, some of the energy ends up as free-electron momentum while the rest may either end up changing the momentum of the particle, re-radiating at a frequency appropriate to the energy-level change, or some combination. So just pretending I know the energy-quantities to check my logic, take an example where an atom gets hit with 100 units of EM radiation. 20 units could be carried off as momentum of a liberated electron, 30 units could be expressed as additional kinetic energy to the atom's motion, and the remaining 50 units would be re-emitted as a wavelength with half the energy of the light originally absorbed? Does this sum up the logic and are their any other outlets for the energy that conserves the total amount absorbed? Something like that, yes. Link to comment Share on other sites More sharing options...
lemur Posted April 10, 2011 Share Posted April 10, 2011 Something like that, yes. So atoms, electrons, and photons are like billiard balls breaking apart from the triangle only with more complicated forces constraining their motion. I'm sure that would be a misleading analogy if applied in other ways, but it seems like an energy-conserved system that would basically depict the breaking apart of the electron(s) from the atom along with the photon(s) radiating away from both. Link to comment Share on other sites More sharing options...
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