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Do black objects get hotter than white objects in the sun?

  1. Apr 29, 2015 #1
    I understand that white objects 'reflect' all frequencies of visible light, and black objects 'absorb' all frequencies of visible light.

    But I thought that 'reflecting' light means that the electrons in the atoms are caused to oscillate, and 'absorbing' means the electrons in the atoms are not caused to oscillate.

    If this is the case, wouldn't white objects get hotter in the sun, because their electrons are caused to oscillate?
     
  2. jcsd
  3. Apr 29, 2015 #2

    HallsofIvy

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    Your "understanding" is wrong- electrons don't "oscillate" in either case. When an atom absorbs light (energy) an electron jumps to a higher energy level. If the light is then "reflected" the electron goes back to the lower energy level. But the simplest way to think about it is not to look at atoms- if the light is absorbed, its energy becomes part of the energy of the object, raising its temperature. If the light is reflected, that light energy does not add to the energy of the object.
     
  4. Apr 29, 2015 #3
    But fundamentally any EM radiation is caused by accelerating electric charge. Maybe 'oscillate' is the wrong word.

    I don't understand this. Light is only reflected off an object if the electrons in that object are caused to accelerate. Wouldn't this acceleration of electrons heat up the object?

    If an object absorbs light, that means its electrons are not being caused to accelerate, and thus should not heat up.
     
  5. Apr 29, 2015 #4

    DaveC426913

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    Step back a bit and look at it in bulk.
    Photons are energy. Add them to a material and the material's energy is increased - which manifests as heat.
    Bounce them off the material and the energy is not retained by the material. It does not get warmer.
     
  6. Apr 29, 2015 #5

    davenn

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    What acceleration

    you even quoted hallsofivy's earlier comment about it seems you didn't really read it ....
    here it is again....


    Dave
     
  7. Apr 29, 2015 #6
    Yeah but photons don't bounce off anything. They cause force on electric charges.
     
  8. Apr 29, 2015 #7

    Vanadium 50

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    Look in a mirror and say that.
     
  9. Apr 29, 2015 #8

    Nugatory

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    The answer to the question, "Do black objects get hotter in the sun?" is most certainly "yes". It's easy enough to test this for yourself, by leaving a black stone and a white stone side by side in bright sunlight for a few hours... and solar water heaters are painted black because that way they absorb more heat... and if you'll won't find many dark-colored roofs being built in sunny climates... and many more examples.

    So the question isn't whether black objects get hotter in the sun, the question is why they do.

    Light-colored object: Incoming photon interacts with an electron, increasing its energy by one photon's worth of energy and exciting it to a higher energy level. The structure of the material is such that the when the electron dumps this energy by dropping back to a lower energy level, it releases a photon just like the incoming one. This photon flies away from the material, taking one photon's worth of energy with it so there is no net increase in the energy of the object - one photon's worth of energy came in, one photon's worth of energy left.
    This process happens so quickly that we say that there's just one photon that came in and bounced off; and as Vanadium50 points out above, something like a mirror is really good at making this happen.

    Dark-colored object: Incoming photon interacts with the electron, increasing its energy by one photon's worth of energy and exciting it to a higher level. The structure of the material is such that when the electron dumps this energy by dropping back to a lower level, it doesn't release a photon that flies away. Instead, the electron directly or indirectly interacts with its nucleus and its neighboring atoms to set them vibrating - and that's heat. One photon's worth of energy came in, one photon's worth of energy didn't go out; and the total amount of energy in the object has to go up.

    In practice, these interactions are complicated and very difficult to model correctly. For example, a smooth surface will usually be more reflective than a rough surface even if the the material is the same; it is possible to calculate this effect exactly, and it's been done, but it's a LOT of work. Thus, we generally use the approach that DaveC426913 described above: Add up the energy of the incoming photons, subtract the energy of the outgoing photons, and the difference has to be amount of energy that's left behind to heat up the object.
     
  10. Apr 30, 2015 #9

    DaveC426913

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    You're missing the point. That's why I asked you to step back and look at the bigger picture, rather than the mechanisms.

    As a unit of energy, if the photon stays in the material, the material has more energy. If the photon leaves the material the material does not have that energy.
    That is an inarguable fact, so the flaw must be elsewhere in your logic.
     
  11. Apr 30, 2015 #10
    The real answer has to do with energy levels in the atom and QM. But it seems reasonable that we should be able to somewhat reconcile the energy transfer using classical mechanics. EM wave comes in and we pretend that the atom is an electron acting as a harmonic oscillator or like a planet in orbit. Caveat: this is not true, but qualitatively reproduces some of the same interactions.

    If the EM wave/photon hits the oscillator, the electron begins to oscillate with more energy. In classical mechanics the oscillating electron will always radiate, but in QM it may not. It could absorb the energy and move to a higher level. In the way that you are thinking this might be like a positive acceleration where the system gains energy. If the surface reflects the light then the electron drops back down to a less energetic "orbit". You could think about this as a negative acceleration where energy is being lost to the environment in the form of EM radiation.

    Of course this isn't quite right because an electron should emit radiation for both positive and negative acceleration. But that shouldn't really come as a surprise since you would arrive at this conclusion by studying classical electrodynamics, not quantum. I think everyone else gave correct answers, but I am just hoping that this will help you understand why your thinking isn't correct. Really the main point is that acceleration isn't quite as well defined in QM.
     
  12. Apr 30, 2015 #11
    Yeah I understand the big picture explanation, I want to know why the big picture explanation works.
    I didn't mean to imply that I was correct and everyone else was incorrect, I was just trying to convey my point of view so someone can tell me exactly which part I am misunderstanding.

    My current understanding of reflection is that photons don't actually 'bounce off' objects, rather they interact with the electrons in the atoms on the surface of the object, increasing their energy, which causes the electron to release new photons.

    So since reflection is electrons being given more energy, I don't understand why white objects, which are reflecting all frequencies of visible light, stay cooler than black objects, which do not reflect any frequencies of visible light.
     
    Last edited: Apr 30, 2015
  13. Apr 30, 2015 #12

    Nugatory

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    Reflection is not electrons being given more energy, it is electrons momentarily increasing their energy and then giving the entire increase back in the form of reflected light, for no net increase in energy. By comparison, the electrons in the darker object also momentarily increase their energy, but they don't give the increase back in the form of reflected light.
     
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