The Science of Doom

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35 Responses

1. on March 28, 2011 at 4:27 pm | Reply UncleArtie

   I recently discovered science of doom and really like it.

   Could you please educate on atmospheric oxygen; or maybe what I am asking is outside the scope of this site.

   With all the hydrocarbons we burn it is my understanding that atmospheric oxygen concentration has been steadily decreasing.

   Years ago I read an article about how seawater sprays are broken down into H2 and O2 by the interaction of salt and UV. Do you know anything about this? If this process exists I presume it would consume a portion of sunlight energy.

   
   
2. on April 7, 2011 at 1:29 am | Reply macha

   Hi, I am new to this site.
   I guess being a chemist, I like the focus on the science rather than the political.

   In relation to the Greenhouse Effect (Theory), whats your take on this paper?

   http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf

   best regards,

   
   
3. on April 7, 2011 at 1:33 am | Reply scienceofdoom

   macha:

   On the Miseducation of the Uninformed by Gerlich and Tscheuschner (2009) – a review of the very bad but very popular On Falsification Of The Atmospheric CO2 Greenhouse Effects by two people who probably know a lot about highly theoretical thermodynamics, but nothing about atmospheric physics.

   This followed – On Having a Laugh – by Gerlich and Tscheuschner (2009) – in the first foray with interest sadly dying out before the real bit due to so many early flaws.

   
   
4. on May 3, 2011 at 9:08 pm | Reply scienceofdoom

   This page of symbols shows how to easily add Greek letters, maths symbols and so on. I tried a few, e.g.:

   ⇒ ∂p/∂τ → 0

   and all you have to do is type & followed by things like tau; – don’t forget the ; at the end..

   
   
5. on May 6, 2011 at 10:35 am | Reply Neal J. King

   Hi, SoD:

   I understand and fully support your effort to pin down Miskolczi and his argument. This takes confrontation.

   However, I recommend a slightly gentler tone. “You must do this” and “You must do that” come across as a bit ham-fisted.

   Splashing the water doesn’t catch the fish; the crucial point is whether the hook is in the mouth.

   Best regards,
   Neal

   
   
6. on May 14, 2011 at 9:49 pm | Reply Darren

   SOD:

   This maybe a very simplistic question, but if CO2 molecules absorbed and re-radiated out as heat, wouldn’t that mean that there would need to be a greater amount of energy being expelled than was absorbed initially?

   According to thermodynamic theory, each absorption and emission event would result in a very small amount of entrophy, the loss of useable energy. The greater the number on times radiation is absorbed and reemitted, the greater the entropy. Is my reading on this correct? And if so how can this cause heat?

   
   
   • on January 5, 2012 at 12:44 am | Reply Matt J.

     @Darren I see no one has answered this, but it is a good question, so I will give it a shot.

     No, no entropy is released by a single atomic emission event. Entropy is a thermodynamic quantity (a category that also includes pressure and temperature), which qua “thermodynamic quantity” assumes a large number of particles. So single-particle events have no relevance to thermodynamic quantities.

     See, for example, Kubo’s “Thermodynamics”, which explains in detail this vital distinction between “thermodynamic quantities” and other physical quantities in the very first chapter (sec. 4).

     
     
7. on May 14, 2011 at 11:10 pm | Reply scienceofdoom

   Darren:

   The key is that the energy source is the sun. Without the sun, regardless of “greenhouse” gases, the earth would cool down to the background temperature of the universe, close to absolute zero.

   The short explanation (over-simplifying some concepts):

   The earth’s surface absorbs solar radiation, and with no atmosphere it would radiate this out directly to space. The surface of the earth would be a lot colder than it is now.

   With an atmosphere that can absorb terrestrial radiation the surface radiation is absorbed in the atmosphere and some of it is re-emitted back to the surface. Some of it is re-emitted out to space. (No energy is created or destroyed).

   Think of this as reducing the cooling rate.

   If CO2 could create heat from nowhere then it would violate thermodynamic laws. But instead water vapor, CO2 and other “greenhouse” gases just affect the “passage of radiation” as it travels through the atmosphere.

   As a poor example, but possibly a useful mental picture, think of a lagged (insulated) hot water pipe. The lagging makes the water hotter than it would be without the lagging, but no one claims or believes that the lagging creates heat. It just stops heat escaping as effectively.

   In this poor example, the heat transfer mechanism is conduction. In the climate system, the heat transfer mechanism responsible for the higher surface temperatures is radiation.

   The reason the example is useful is that the energy source is from within the pipe. In the climate because the atmosphere is (mostly) transparent to solar radiation the climate is effectively heated at the surface, with the atmosphere acting to reduce the cooling to space.

   I hope these simple explanations provide a starting point.

   To understand the longer version, a couple of articles might help. Have a read of them and feel free to ask more questions -

   Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part One;

   and for entropy – The Three Body Problem.

   
   
8. on May 15, 2011 at 7:18 pm | Reply Darren

   SOD,

   I was reading one of your articles and I noticed the post stating:

   “What does this mean? [It means that we can confident] of the amount of energy:

   Arriving from the sun at the top of atmosphere and at the surface
   From the sun that is reflected back into space by the atmosphere or the earth’s surface
   Emitted by the earth”

   The sentence in brackets seems to be incomplete. Just thought you might want to look to correct.

   
   
9. on May 16, 2011 at 1:43 pm | Reply Darren

   SOD,

   You have stated that the radiative force equation is 240W/m2, but Wikipedia is showing.

   “The current increase to 386 ppm from 280 ppm causes a radiative forcing of 1.66 W/m^2, and 1.34 W/m^2 from increases in other gases, totaling 3.00 W/m^2″
   http://en.wikipedia.org/wiki/Carbon_dioxide_in_Earth%27s_atmosphere

   How do you account for the discrepancy?

   
   
   • on May 16, 2011 at 7:31 pm | Reply scienceofdoom

     Darren:

     Please can you make the comment in the relevant article – it will help other readers as the context will be clear.

     Radiative forcing is totally different from the absorbed solar radiation.

     You have to understand what radiative forcing is – which you can find in CO2 – An Insignificant Trace Gas? Part Seven – The Boring Numbers.

     And have a read of The Earth’s Energy Budget – Part One.

     
     
10. on July 22, 2011 at 10:00 pm | Reply David

    I recently discovered this and have found both your explanations and the comments fascinating.

    I’m a chemist, not a physicist or atmospheric scientist, so my interest was mainly in the thermodynamics of atmospheric science.

    What I found surprising, and a little disturbing, was how profound some of the misconceptions about thermodynamics were, even among people who seem to have quite strong science backgrounds

    
    
    • on February 28, 2012 at 11:23 am | Reply Abdo

      Michael D Smith (00:57:13) :Clouds and coioensatndn are the balancing outgoing delivery mechanism of heat on this planet, and overwhelm the radiative effect with convection, and as a bonus also block incoming radiation, especially in the tropics, leading to a natural, self regulating thermostat effect. This results in the lower climate sensitivity we are seeing in many recent studies like Lindzen?s. The notion of an H2O positive feedback (which probably is present on a clear day) is squashed by this process.While warmer air can hold exponentially more water vapor, presumably increasing greenhouse effects (an process the IPCC hangs its collective hat on), it is also this exact same property that vastly improves the chances of convective and phase change heat transport by thunderstorms. Once triggered, the radiative effects of H2O are completely overwhelmed by the storms, resulting in a very strong localized negative feedback. Cumulus clouds will have the same effect, but more in balance with the positive effects, resulting in less negative net feedback, but with the same result, much lower climate sensitivity than the IPCC would have you believe.I realize that climate sensitivity is not usually discussed as a local phenomenon, but it should be, since it is the integral of all local phenomena. I could not agree more. I discuss this at length on another thread here at . The greenhouse effect exists, but it does not control the temperature of the planet. That honor goes to clouds and thunderstorms.

      
      
11. on July 23, 2011 at 2:38 am | Reply DeWitt Payne

    David,

    If you have the time, please make detailed comments about what you consider misconceptions and why you think so in a relevant thread. A statement as general as yours provides no real help.

    
    
12. on August 24, 2011 at 12:32 pm | Reply Doug Cotton

    See Dr Thomas” posts at

    http://theconversation.edu.au/the-jones-weightless-invisible-gas-trial-2955

    
    
13. on October 28, 2011 at 5:38 pm | Reply JWR

    I was puzzled by back radiation.
    Started to make models of 1 slab without back radiation, next 2, next3….and now as many as Iwant, programmed in Matlab.
    The results are coherent.
    What you expect from the atmosphere.
    Noback radiation is showing up.
    Read it yourself
    http://www.ilovemycarbondioxide.com/pdf/IRabsW27102011.pdf

    
    
14. on November 24, 2011 at 9:53 pm | Reply Svend Ferdinandsen

    Dear SOD.

    I read about the greenhouse effect from time to time, and have a question:

    How would the temperature be on ground if all CO2 was removed, or if the CO2 was halfed instead of doubled.

    I think you would still have most of the geenhouse effect because of water vapor, clouds and lapse rate, the only missing part would be the CO2.

    
    
    • on November 30, 2011 at 12:21 am | Reply RW

      Svend,

      About a little under 1/3rd of atmospheric absorption is from CO2 and other GHGs and over 2/3rds is from water vapor. If you consider that the Earth is also about 2/3rds cloud covered and any CO2 or other GHG absorption would not matter because the clouds would be absorbing the energy anyway, over 90% of the GHE is from water vapor and/or clouds and less than 10% is from CO2 and other GHGs.

      0.3*(1-0.667) = 0.10 or 10%

      There is also H2O overlap in the absorption bands of CO2, which could diminish CO2′s net absorption somewhat too I suppose.

      I think it’s safe to say that about 90-95% of the GHE is from water in some form.

      
      
      • on March 1, 2012 at 3:39 am Subodh

        Gary Hladik (20:01:30) :Monckton of Brenchley (15:30:32) : ??giving a final calimte sensitivity parameter close to 0.9 K/W/m2, around five times too big. Discuss.?This one really has me confused. Up to now I?ve only seen ?calimte sensitivity? defined as the temperature increase expected from a doubling of atmospheric CO2. I assumed all the temperature rise was attributed to ?radiative forcing? directly and indirectly increased by CO2. I thought the magnitude of the radiative increase was in dispute, but the effect of a known change in radiation wasn?t. Now it is? WUWT?BTW, according to the Stefan-Bolzmann Equation, a sensitivity of 0.9 K/W/m2 implies the average temperature of the earth is about 170 Kelvins, or -103 degrees Celsius.I?m putting another blanket on the bed tonight. The UN IPCC gives climat sensitivity of ~ 3K per doubling of CO2. Doubling of CO2, in turn, gives a UN IPCC value of 3.7 W/m2 of forcing.Dividing the first by the second gives a calimte sensitivity of 0.8K/W/m2. Same number, different incarnation. Monckton is just using a different way of expressing the same thing.Finally, the sensitivity is not constant, it is only valid at a certain temperature range. This is because the relationship between temperature and radiation is far from linear.

        
        
15. on November 30, 2011 at 12:27 am | Reply RW

    As to how much cooler it would be with no CO2 – that’s hard to say.

    Maybe SOD can offer and estimate?

    
    
16. on November 30, 2011 at 3:28 am | Reply Rodney Horton

    The concept of a “greenhouse gas” is in error, false.

    A greenhouse is cooled by exhaust fans blowing hot moist air out of the greenhouse. Local, ambient air is allowed in, bringing CO2 with it

    The plants sequester carbon. As a result, the exhaust air has a lower concentration of CO2 than the incoming air.

    
    
17. on November 30, 2011 at 6:43 am | Reply scienceofdoom

    Umm, “everyone” knows that the inappropriately-named “greenhouse” effect is not working like a greenhouse.

    On this blog I always put it in quotes.

    Take a look at the series Atmospheric Radiation and the “Greenhouse” Effect.

    The inappropriately-named “greenhouse” effect is actually the effect of absorption/emission of IR radiation due to radiatively-active gases combined with the effect of declining temperature with height of the atmosphere (lapse rate).

    Much more pithy.

    
    
18. on November 30, 2011 at 12:52 pm | Reply JWR

    I was puzzled by back radiation.
    Started to make models of 1 slab without back radiation, next 2, next3….and now as many as Iwant, programmed in Matlab.
    The results are coherent.
    What you expect from the atmosphere.
    Noback radiation is showing up.
    Read it yourself
    A new version is now available.
    http://www.tech-know.eu/uploads/IRabsW23112011.pdf

    
    
19. on December 26, 2011 at 4:27 pm | Reply Ross Firestone

    I just discovered your excellent blog so I missed your explanation of how quantum mechanics is involved with atmospheric physics. Would you please refer me to an article in which you discussed this. Thank you.

    
    
20. on December 27, 2011 at 8:12 am | Reply scienceofdoom

    Ross Firestone:

    The principles of absorption and emission of radiation by various atmospheric trace gases like water vapor and CO2 rely on the theory of quantum mechanics.

    These principles have been heavily used in many articles – for example:

    - Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations
    - Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Nine

    But I haven’t developed any radiation theory from quantum mechanics if that is what you are looking for.

    
    
    • on January 5, 2012 at 1:01 am | Reply Matt J.

      Well, yes, it is a little unclear what Ross is asking for. My own guess is that rather than the completely general “how is quantum mechanics involved with absoprtion/emission”, that he wants to know how it is relevant to the Earth’s energy balance.

      If that is correct I would say the answer is: in addition to explaining the basic blackbody radiation correctly (avoiding ultraviolet catastrophe among other problems of the classical explanation), it explains why atmospheric gases have absorption peaks; this in turn explains why the atmosphere is transparent to incoming solar radiation but opaque to IR (longwave) radiation from the earth.

      Of course the latter is a key point in understanding the admittedly poorly named “greenhouse effect”. It is also key to understanding how it is certain gases that cause the trouble, such as CO2 and methane.

      Now in fact, since the “ultraviolet catastrophe” as the name suggests, concerns only higher frequencies, one could do pretty with with Rayleigh-Jeans instead of Planck’s Law, but why bother?

      
      
21. on January 13, 2012 at 1:12 pm | Reply Kevin Parker

    Dear SoD,
    I agree with all the other commentators that this is a great informative site. I’m a chemist, and I’m trying to check my understanding of some of the concepts before developing some simple teaching material. I’ve read much of the arguments around downward radiation here and also the discussion on whether the CO2 band is saturated.

    Q1 Surely if CO2 is saturated, we would not expect to see an increase in downward radiation (at 15 microns) as CO2 levels increase? However if CO2 is not saturated, we would expect the downward radiation to increase? Wouldn’t this be a simple empirical test of whether CO2 is saturated? Or am I missing something?

    
    
22. on January 13, 2012 at 1:15 pm | Reply Kevin Parker

    PS if there is a better thread to ask the question, send me to it and I’ll post it there. It’s just that the discussion on the ‘CO2 trace gas’ and other apposite threads seems to date from 2011 or earlier.

    
    
23. on January 13, 2012 at 8:43 pm | Reply scienceofdoom

    Kevin Parker,

    Good questions.

    Some resources under Atmospheric Radiation and the “Greenhouse” Effect, and you should check out especially Parts 4, 8, 9 & 10.

    Also CO2 – An Insignificant Trace Gas? – Part Eight – Saturation – explaining “saturation” in more detail.

    New comments do get seen by everyone in the right hand comments pane, so even old threads can get quickly revived.

    There are two important points to note in your question.

    First, the CO2 band is very wide. Some parts are saturated, some parts are not. It is the parts that are not that have the major effect.

    i) In the center of the band 95% of radiation (at that specific wavelength) is absorbed in just 1m of the atmosphere ii) in the “wings” of the band, the absorption through the entire lower atmosphere might range from 10% – 50% , depending on wavelength (you can see the transmissivity of the whole band in Atmospheric Radiation and the “Greenhouse” Effect – Part Nine

    Second, downward radiation is not really the important issue (I only write so much about it because so many people think it doesn’t exist or can’t affect the surface temperature). The important issues is the emission of radiation to space – and at what height this takes place from – that causes the surface temperature to change. (The real answer is more complicated, has more caveats and explanations, but this is a good start).

    You can see this shown in a fictious and simplified example in Atmospheric Radiation and the “Greenhouse” Effect – Part Four – but good to work through earlier parts in this series before reading that one.

    Feel free to ask questions. It is not a simple subject – at least, it is easy to trip up if you don’t understand some important points.

    
    
24. on January 22, 2012 at 2:41 pm | Reply Joe Born

    Suggestion for a new post: follow-up on the tall-rooms discussion of the summer before last.

    Wait, before you dismiss that horse as having already been flogged unmercifully, I’d like to bring to your attention an alternative theory, one between the isothermal and the adiabatic-lapse rate theories. It was advanced in a paper seventeen years ago, and it says that, yes, the distribution of an isolated air column in a gravitational field approaches isothermal, but only in the infinite-number-of-molecules limit. Short of that, there’s a small but non-zero lapse rate.

    The papers on which this is based are mentioned here at WUWT: http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-870078 and here at Tallbloke’s Talkshop: http://tallbloke.wordpress.com/2012/01/04/the-loschmidt-gravito-thermal-effect-old-controversy-new-relevance/#comment-13608.

    Unfortunately, neither of those threads was directed at the Velasco et al. and Roman et al. papers in which this theory was advanced, so, in my view, more heat than light resulted. And that would largely be the case here unless you request that your readers concentrate on which of the equations they base their theory on is wrong.

    
    
25. on February 11, 2012 at 4:39 pm | Reply Andrejs Vanags

    You site is so comprehensive (very good thing), that unfortunately I cant find my own comments. I try to comment on the threads and posts that are relevant, but when I try to find a response, or think about it some more and want to continue the comment or my line of thought I cant find them again. Is there a way to search for posts by a particular poster?

    
    
26. on February 12, 2012 at 6:26 am | Reply scienceofdoom

    Andrejs Vanags,

    There’s no way for a commenter to find them I think.
    But I can search via the wordpress backend.

    Here are the posts you have commented in, once inside each just use the search function in your browser (e.g. CTRL-F) to find your name:

    Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Four

    Venusian Mysteries

    The Mystery of Tau – Miskolczi – Part Five – Equation Soufflé

    Radiative Forcing and the Surface Energy Balance

    What’s the Palaver? – Kiehl and Trenberth 1997

    Paradigm Shifts in Convection and Water Vapor?

    Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Nine

    Kramm & Dlugi On Dodging the “Greenhouse” Bullet

    Measuring Climate Sensitivity – Part One

    Plus the latest article, Potential Temperature.

    
    
27. on February 29, 2012 at 8:45 am | Reply scienceofdoom

    For those checking out recent comments via the sidebar..

    Many comments in the last 1-2 days have been trapped by the spam filter. It is a law unto itself and can occasionally be overzealous.

    Very frustrating I’m sure if that happens, so please email me directly if your comment disappears. scienceofdoom -you know what goes here – at gmail.com

    
    
28. on June 29, 2012 at 2:38 pm | Reply Mike

    Hi. Reading your blog. Enjoying it. A point on determination of turbulent conditions in air. You state that turbulence is determined by Renold’s Number greater than 5×10^5, however conventionally, transition from laminar to turbulent flow regime is determined at Reynold’s numbers between 2 and 4×10^3. So while the general idea of turbulence in air is correct, your calculation underestimates the tendency for turbulent flow.

    Using the Reynolds number above in your calculation, you can see that turbulent flow in air would happen (at a height of 1m above ground) at a velocity of 0.0315 m/s. Due to the low density and (dynamic) viscosity of air, it is almost impossible for moving air to be in a laminar flow regime.

    
    
29. on May 3, 2013 at 2:57 pm | Reply phychai

    Do you plan to discuss something about the scattering process by aerosol sometime?
    After reading your blog, I’m quite certain that CO2 has a big influence in shaping our climate. But what about aerosol and clouds? IPCC is quite uncertain on these topics. How do we estimate their influences?