Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Blackbody radiation confusion

  1. Dec 20, 2009 #1
    Ive heard a black body described as 'a body which absorbs all the radiation that falls upon it'. This seems to me to contradict the idea that a black body emits exactly as well as it absorbs light.

    I can understand that if you shine light matching a blackbody curve (at a given temperature) on a black body that it should be capable of absorbing all that light. Furthermore it would absorb the light as heat and immediately re-mit the exact same profile of light.

    But imagine instead a profile of light that is greater in intensity than the blackbody curve for every wavelength. Indeed, consider a profile that does not even match the 'shape' of the blackbody curve i.e a flat distribution of light. The first definition would suggest that the black body would be able still to absorb all the light. Is this the case? If so, the temperature of the black body would rise and re-emit the light as a new blackbody curve for the higher temperature. If not, then are we saying a black body can 'reflect' some of the excess light that it is not capable of absorbing?

    So, what do we mean when we say a black body absorbs as well as it emits?

    Is it, that the black body can only absorb the exact same profile of light as it emits at a given temperature, or is it the weaker statement that the black body aborbs all profiles of light, but that it then re-emits a blackbody profile that has the same 'total power' (over all wavelengths) as the incident light?
     
    Last edited: Dec 20, 2009
  2. jcsd
  3. Dec 21, 2009 #2
    Anyone got any ideas on this? Is my question clear?
     
  4. Dec 21, 2009 #3

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor
    2016 Award

    Not really- a blackbody is a material object which, at thermal equilibrium, emits a spectrum of light identical to blackbody radiation.

    Blackbody radiation is a very specific spectrum of light that can be assigned a thermodynamic temperature.

    Emission = absorption due to the requirement of equilibrium.
     
  5. Dec 21, 2009 #4

    clem

    User Avatar
    Science Advisor

    At a fixed temperature, any object emits energy at the same rate it absorbs energy.
    Therefor, a body that absorbs the best will emit the best.
     
  6. Dec 21, 2009 #5
    This is true I believe: "..... the black body aborbs all profiles of light, but that it then re-emits a blackbody profile that has the same 'total power' (over all wavelengths) as the incident light? "
     
  7. Dec 21, 2009 #6
    A blackbody is one of those physics idealizations, like a frictionless surface or a massless string, that help us to better understand a phenomenon. I'm sure you know that a hot metal rod will glow red at high temperatures and will glow white when heated to even higher temperatures. It turns out that all bodies with temperature emit electromagnetic waves. If its not at absolute zero, a body will give off some sort of light because of its temperature. Most glow at frequencies which are not visible to humans...infrared, radio, microwave, ultraviolet, etc. Physicists wanted to understand the mathematical relationship between temperature and the frequency of the light given off. Most objects also reflect some of the light that hits them...this is what allows us to see objects. A blackbody is an idealization where none of the observed light from the object is due to reflection...all of it is generated from within by its temperature. Conflict between theoretical prediction and experimental observation of the frequencies given off and the temperature of blackbodies early in the 20th century helped lead to the development of quantum physics.
     
  8. Dec 21, 2009 #7
    Yes, but we know that already. The quite interesting question is ;

    "Is it, that the black body can only absorb the exact same profile of light as it emits at a given temperature, or is it the weaker statement that the black body aborbs all profiles of light, but that it then re-emits a blackbody profile that has the same 'total power' (over all wavelengths) as the incident light? "

    If we put a continuous red laser light generator inside a "black body", it would seem to me obvious that the body would heat up and remit at the black body profile.

    If we fire the laser light at "a body which absorbs all the radiation that falls upon it", I suppose we are in the same situation, but i'm not sure
     
  9. Dec 22, 2009 #8

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor
    2016 Award

    Since laser radiation is nonthermal light, that question is not well posed.
     
  10. Dec 22, 2009 #9

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    No - that is the mistake in the original question

    Yes

    Imagine a steel furnace, you can heat the steel with microwaves or RF radiation at long wavelengths but the steel will glow in the visible emitting the blackbody curve of a material at it's melting point.
     
  11. Dec 22, 2009 #10

    fluidistic

    User Avatar
    Gold Member

    I'm curious about the meaning of your sentence. Does that mean that I can't heat a black body (or gray body) using a laser?
     
  12. Dec 22, 2009 #11

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Yes you can, but you migth confuse yourself mixing statements about a blackbody and a non-blackbody source.

    For example, you can only heat a blackbody target using a blackbody source to the same temperature as the source. So however much you concentrate the light with lenses/mirrors you can't use the sun to heat something to a higher temperature than the surface of the sun. Because at that point the target would emit back at the sun.

    But you can use a non-blackbody source, such as a laser or an RF transmitter to radiatnly heat something to a higher temperature and then that something would emit as a blackbody.
    So a piece of steel at room temperature is only emitting very little infrared, you can use an infrared 10.6um CO2 laser to cut through the steel by heating it to it's melting point, where it will emit in the visible as any blackbody at 1100C would.
     
  13. Dec 22, 2009 #12
    A blackbody will absorb all of the incident light regardless of the distribution, be it a blackbody distribution or a discrete laser output. The blackbody emission will only depend on the temperature of the object, so if it absorbs energy from incident light such that its steady state temperature is 320K the peak emitted wavelength will be at 9 micron (as per Wien's displacement law) and the distribution will be according to Planck's law.
     
  14. Dec 22, 2009 #13

    fluidistic

    User Avatar
    Gold Member

    Ah thank you both, I get it. By the way I'm not the OP.
     
  15. Dec 22, 2009 #14
    Thanks for all the replies, very illuminating (please excuse the pun!)

    I have followed the thread and I think it has cleared up my understanding. I was taught this stuff years ago before the internet really got going. At the time I must have thought I had understood it but since I started this thread and after reading many articles I see that I was confused on a lot of subtle issues.

    I have another post in this broad area that I would welcome any comments on :

    https://www.physicsforums.com/showthread.php?t=365018
     
  16. Dec 22, 2009 #15
    "you can only heat a blackbody target using a blackbody source to the same temperature as the source. So however much you concentrate the light with lenses/mirrors you can't use the sun to heat something to a higher temperature than the surface of the sun. Because at that point the target would emit back at the sun"

    Are you sure of this? My first rection would be to say if you focused the light from a large 1000K black body onto a small target, then the target would rapidly rise to a temperature well above the source.

    But you may well be right if we can say that whatever the power of the incident radiation, the target will re-emit more and more radiation (not nececessarily back at the source) and attain thermal equilibrium at the same temperature profile as the incident radiation. If this were true it would mean that if we doubled, and doubled again, the power falling on the target, its equilibrium temperature would remain unchanged.

    I would have thought the target would have to rise to a higher temperature each time to be able to re-emit the additional incoming power, and so could finish at a temperature well above that of the source..
     
  17. Dec 22, 2009 #16
    As I understand it from following the previous discussion, the profile of the incident light isnt the thing that determines the equilibrium temperature of the black body, it is the total power (the integrated spectrum) of the incident light. The black body will come to the temperature at which it is emitting black body radiation which has the same total integrated power.

    Considering focussing light from one black body onto another, consider this experiment :

    I can imagine an idealised small spherical black body of radius r and temperature t, sat inside a larger hollow sphere that has its inner surface being a black body. If we also assume that no radiation comes out of the system as a whole (i.e. out of the external surface of the enclosing hollow sphere), then by equating the total energy transferred between the two surfaces we then can have radiative equilibrium at different temperatures right? implying

    r * t^2 = R * T^2

    This suggests that you cant heat a large black body to a temperature higher than that of a smaller black body in a closed system.

    This seems strange as I am used to thermodynamic equilibrium meaning things at the same temperature? Its late maybe I am confused again?
     
    Last edited: Dec 22, 2009
  18. Dec 22, 2009 #17

    mgb_phys

    User Avatar
    Science Advisor
    Homework Helper

    Correct
    Yes

    No there is no reason for an equal power transfer, they will come to the same temperature but the power will depend on the relative areas.
     
  19. Dec 22, 2009 #18

    Andy Resnick

    User Avatar
    Science Advisor
    Education Advisor
    2016 Award

    This conceptual error is the root cause for the persistent belief in Archimedes' "death ray". It's fundamentally not possible due to the source being of finite size- sunlight (or your 'large 1000k blackbody') cannot be focused beyond a certain limit.
     
  20. Dec 23, 2009 #19
    It's fundamentally not possible due to the source being of finite size- .....the light from a source....cannot be focused beyond a certain limit.

    It can. See "Integrating Sphere" used to concentrate near 100% of the power of a light source onto a sensor placed just ouside a hole in the sphere.
     
  21. Dec 23, 2009 #20
    Excuse me, my example of an integrated sphere is not relevant. All the photons produced by the source placed inside a perfect Integrating Spere have to come out of a small hole at the surface of a sphere but they come out at all angles to the tangent. So they would not, and if I'm not mistaken, could not be focussed onto a target placed some distance from the sphere.
    'All angles', I don't know what the angular distibution to the tangent would be, it might be cosine, I've never asked myself this question, maybe somebody knows already, that too is an interesting question but straying from the subject.

    Please ignore my previous message.
     
    Last edited: Dec 23, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Blackbody radiation confusion
  1. Blackbody Radiation (Replies: 2)

Loading...