Absorption Spectrography vs New Proof of Planck’s Law

· blackbody radiation, physics
Authors

Absorption spectroscopy can be used as experimental support of the new derivation of Planck’s Law presented in Mathematical Physics of Blackbody Radiation and Computational Blackbody Radiation and discussed in the recent post Two Proofs of Planck’s Law vs Backradiation.

The experimental setup is illustrated in the above picture showing the recording of two spectra: An absorption line spectrum originating from a continuous spectrum (e.g. a blackbody spectrum) after passage through a cooler gas and an emission line spectrum from the cool gas after being lighted by the source:

The absorption line spectrum is the continuous spectrum with certain frequencies being absorbed by the cool gas which reappear in the emission line spectrum from the cool gas:

  • absorption spectrum + emission spectrum  = continuous source spectrum.

The absorption line spectrum gives information about the constituents of the gas.

Notice that the gas is cooler than the light source, which means the continuous spectrum reaches beyond the cut-off frequency of the gas. Some of these frequencies resonate with the gas molecules and cause heating of the gas until they are re-emitted with original source amplitude.

If the gas is warmer than the source, then the absorption lines will be missing and the spectrography will not give any information about the gas.

In the case of IR radiation from the Earth looking down from above the atmosphere, the absorption/emission line spectrum is represented by the dip in the spectrum displayed in The Empty Postulate of CO2 alarmism.

Absorption spectrography shows the uni-directional aspect of the interaction between two blackbodies displayed in the new derivation of Planck’s Law with transfer of heat energy from warm to colder, without possibility of backradiation with heat transfer the other way.

This adds to the advantages of the new derivation of Planck’s law as compared to that by Planck: In Planck’s derivation absorption is followed by emission and the case of absorption with emission replaced by intermal heating is not really an option. But this is the case in absorption spectrography which thus is covered by the new proof but not really by the old.

This post was stimulated by the comments given by Doug Cotton on Climate Etc. concerning the Sky Dragon, blackbody radiation and backradiation.

12 Comments

Comments RSS
  1. iceskaterfinland

    How about we do this kind of test with the following light source:

    🙂

    Thats a brightly glowing fungus. How hot can it be?

  2. Tor

    Can there be excluded that the absorption lines are there, just not that it is impossible to resolve them if the intensity from the gas goes up after heating?

    Do you have some references to real measurements?

  3. iceskaterfinland

    Tor, The absorption lines in b, are the relatively low intensity emissions of the gas superimposed upon the total absorption by the gas. There appearance is one of blackness.

    These lines are Fraunhofer lines when they are seen in Sunlight. These lines are only visible when the hot source of the sun passes thru a colder area of itself. If the sun was not colder away from itself you would not see these lines.

    The experiment says nothing about the cold light from a fungus being able to be absorbed by a 200C plate of green glass. 🙂

  4. iceskaterfinland

    Interestingly an infrared camera uncooled detector works by:

    1. The detector having a reasonable temperature dependant resistance so that the temperature of the detector is known.
    2. Building an array of such detectors that ‘observe’ the IR image coming from the target where each detector creates a pixel of information
    3. Maximising absorption per pixel. Absorption is about .55 per pixel once .8 total absorption is reduced by thermally separating detectors.
    4. Arranging the detector so that the dominant cooling mechanism of the detector is by radiation to the surroundings where the detector is built upon a chip that acts as a heat sink.
    5. Rapidly heating and cooling the detectors by 2C, about 50 times a second. The heating period is 65 microseconds and the detectors cool down to ambient temperature between a completed heating cycle called a ‘frame’. This surprisingly consistant heating and cooling enables a difference temperature of zero to be obtained for each successive frame of observation when the cameras sees no target.
    6. IR absorptions from the image are then observed as positive temperature differences after subtraction between successive frames.

    If people are interested i have references for this and surprisingly detailed calculations and theory using all of the well known radiation laws.

    It seems very convincing that those warm detectors really are absorbing a colder targets radiation. Then there is that fungus.

  5. Tor

    Interesting.

    Another interesting question about the spectroscopy.

    Can one see an emission spectra when the gas get hotter?

    • iceskaterfinland

      Maybe it is possible to play with this to prove that a colder blackbody light from a ‘hot’ source can be absorbed by a hotter source of light? When i thought about this yesterday using the same method, i could not think of a way to do it.

      If you observe a bright line in C as your reference, and observe the same area of B then you have a measure of relative intensity or a difference signal.

      In fact they will have similar intensities when the gas is colder. The black line that you see is a contrast effect caused by the human eye/brain. So the difference signal will be near zero.

      I am pretty sure that if the gas is hotter, the emission from B will appear as areas of brightness arranged between the lines of the spectrum of the colder heated black body. At the moment that totally makes sense anyway. 🙂 For example a hot sodium light has a couple of intense yellow lines and these would appear brighter against a feinter source of yellow that had yellow in a wider range of yellow.

      If the hot gas cannot absorb the emissions from the colder gas then the area of brightness will have some relationship to the sum of the two brightnesses, and the plot of the difference signal will have a different slope once the gas is hotter.

      If the hot gas can absorb the emissions from the colder gas then the difference signal slope will not alter once the gas is warmer.

      So all you need to do i think is get a colder flame with some copper wire in it glowing less dim that it glows in a hotter flame and then observe the green colour produced by the wire. That sounds very simple to do.

      Somebody must have done this before.

      Assuming the thought experiment so far is not total rubbish! 🙂

      • iceskaterfinland

        Correction: You need to have a source of warm heat that produces emissions continuously across the reference wavelength produced by the hotter subset of emissions. So having the same material in both flames is not going to achieve that.

        A bluish cold flame will have emissions absorbed by the bluish green of copper though.

        Visible colour though is not a relevant factor but rather wavelength. Different flames will no doubt have the different absorptions and emissions needed to test this if it can be done.

  6. iceskaterfinland

    James Clerk Maxwell in 1871 describes a proof that a hotter gas will absorb the radiation coming from a colder gas.

    http://books.google.fi/books?id=DqAAAAAAMAAJ&pg=PA220&dq=prevost+theory+of+exchanges&hl=en&sa=X&ei=mYRcT7zWL7KK4gSAh6CfDw&sqi=2&ved=0CCwQ6AEwAA#v=onepage&q=prevost%20theory%20of%20exchanges&f=false

    This displays page 220

    Towards the bottom of page 220 Maxwell begins a discussion of Prevosts theory of Exchanges

    By the top of page 225 you have the experiment shown on the web page of this blog showing a continuous spectrum coming from very hot Lime (Limelight) passing thru a colder gas of sodium vapour heated by a low temperature flame where sodium mainly produces only two intense lines

    As i suggested above if the ‘cold gas’ flame is hotter than the continuous spectrum flame, the sodium lines will appear as two bright lines for the same arrangement as shown on this blog.

    Maxwells experiment is simpler than what i proposed.

    Without a spectrograph the light of a hotter sodium flame will appear yellow on a screen just like the light coming from a yellow sodium street light.

    Therefore if the hotter flame producing the yellow light of sodium is placed after the prism that produces the colder continuous spectrum of Lime then the screen will show the spectrum of Lime *plus* the light from the hotter sodium flame as two amounts of brightness everywhere *apart* from the area where the spectrum of the colder lime is absorbed by the hotter sodium, where only the light of sodium will be shown and “the dark lines will be seen distinctly”.

  7. nerd

    Hi everyone, it’s my first pay a quick visit at this site, and post is really fruitful designed for me, keep up posting these types of content.

  8. Dirk

    I have been browsing online greater thn three
    hours as of late, but I never found any interesting article like yours.
    It is lovely worth enough foor me. Personally, iif all website owners and bloggers
    made excellent content material as you probably did, the nett will likely be a lot more useful than ever
    before.

  9. defensive driving

    Excellent way of explaining, and nice paragraph to obtain facts rewgarding my
    presentation focus, which i am going to deliver in academy.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: