the Air Vent

Because the world needs another opinion

Changing the Subject

Posted by Jeff Id on March 26, 2017

I’m not sure how to reply to PSI.  They’ve written a post which I don’t seem to be allowed to comment on or there is some technical problem.  Read at your own risk, if you find yourself agreeing with it, I recommend you wash your mind with some kind of acidic cleaner to get the nonsense out but people vote oddly so I don’t have any illusion that some won’t fall for it.  The post claims they have somehow been wronged by my argument, as I brought up the quantum nature of thermodynamics during a ridiculous discussion from literally 4 years ago!!   The discussion in question absolutely required quantum thermodynamics, but PSI is not a group widely cited for their knowledge of thermodynamics.  I will again, yes again, explain below.

They present an equation for radiative transfer of energy between two bodies:

Q ~ σ(Th4 – Tc4)

Us engineers have seen it. Radiative heat = “constant” times the fourth power of the temperature difference between two bodies – hot and cold.  Materials emit radiation at the fourth power of their temperature.  While the PSI group can sound reasonable to an untrained person, this equation tells the story of their own failure of understanding.

Most any high-school student can tell you that heat flows from hot to cold.  In large groups of atoms, such as a microscopic organism or larger pieces of flea dander, Thermodynamics becomes a bulk property science which represents billions of atoms interacting.   Good students can figure out that the heat flow between larger globs of atoms, is two directional between hot and cold, (it goes both ways Cali style) but the flow nets toward the cold in the exact amount predicted by the basic thermodynamic equations as shown in the radiative transfer equation above.  See each atom in the glob has its own unique energy state and physically can exist at a different energy level than its neighbor.  PSI members can’t figure that easy bit out, their members repeatedly claim that somehow energy emitted from a cooler body detects the “net” state of the warmer rather than the local state of the atom actually being affected, and somehow is prevented from transfer.  Completely non-physical nonsense.

Conduction or radiation, the effect is the same.  In any body there are atoms at different vibrational energy levels.  Vibration=heat.

Below is an example of heat movement in the wrong direction:

Imagine a tiny sphere of a billion atoms of pure solid hydrogen at absolute zero temperature, born in a universe where no other radiation was impinging upon it.  No vibration in the atoms whatsoever.   One photon of the correct wavelength strikes the sphere and is absorbed!  Does one photon absorbed into a hydrogen sphere cause all atoms in that sphere to rise in temperature, or just one atom?  The answer quantum mechanics gives is one atom.  That energy will then be transmitted in quantum nature to its neighbors dropping the original atom back to absolute zero and heating its neighbor to 1 arbitrary temp unit.  If the sphere of a billion atoms, and the energy didn’t radiate out to the surrounding universe, that same unit of energy would transfer from atom to atom, all around the sphere, with each different hydrogen atom rising in temperature and falling back to zero as it transfers the energy to the next absolute zero temperature neighbor.   Other unaffected atoms wouldn’t recognize any difference in their own temperatures as they are still at absolute zero energy state.

Now imagine having two of those one billion atom spheres, each one having absorbed two photons.  Two hot atoms inside of two billion atom spheres.  When measured with a properly sensitive instrument, both would have a non-zero temperature because the hot atoms would occasionally affect the amazingly good temperature sensor.  Assume sphere 1 emits a single photon by radiation and it impinges on sphere 2.   Now since these are billion atom spheres and only two of the atoms are energized, I’m going to assume that 999,999,998 in a billion which is really close to 1/1, chance of striking a zero temperature atom on sphere 2 came true, and that the photon struck an absolute zero temp atom on sphere 2 and was therefore absorbed into sphere 2.

Now here is the trick…

Before the photon left sphere 1, the equivalent spheres had 2 photonic units of energy and were equal temperature. The instant the photon left sphere 1, sphere 1 only had a single photon of heat instead of two which sphere two has.  It traveled across the between these spheres at that point going from a colder body to a hotter one, struck the hotter one and was absorbed by an atom at absolute zero temperature.  Heat energy moved from hot to cold.   If you say NO Jeff, it went from equal to equal, that’s fine too because the ENERGY moved from sphere 1 to sphere 2 and created heat.

The story gets worse!!!

Although sphere 1 only has a single photon left, and despite the 3:1 odds, not so amazing if you’ve ever rolled dice, it emits its last photon of heat before sphere two does!!!  Evil thing.

Anyway, that photon travels the short distance between the two spheres and also strikes the surface of sphere 2 which holds the quantum heat of 3 of these same photonic interactions.  3/1,000,000,000 atoms are energized, so again using basic probability, there is basically a 1/1 chance that the photon will strike an absolute zero atom.  In this example, the photon did not win the super-lotto with powerball and did in fact hit the surface of an unenergized atom and was absorbed.  4 to nuthin!! [edited for clarity]

Energy has again transferred from cold to hot [edited].  Jeff Id, god of basic high-school thermodynamics, has done the impossible!!  Who let the cats out!

PSI cannot accept this ridiculously basic thermodynamics, although every member I’ve run across has their own morphed version of thermo, so I won’t be surprised when one of the folk agrees with this and comes up with a different and equally mental argument.

How is a single photon relevant?  PSI can account for all of the photons from the cold source by using traditional physics, and the equations show that the cold source does not heat up a hotter source.

My bold.  My example follows the equation above but contradicts the bold of their own post0.  My example is also standard thermodynamics.  My example is accurate, standard, simple, explains how quantum physics is relevant and completely destroys most of the PSI member arguments, leaving room only for those who are more mentally muddled than others.   But I’m not done yet!

So PSI members of different ilk may be the problem.  We have had more than one of them even claim that somehow the photon knows it came from a colder body and will automatically reflect without absorption from the warmer body.  Truly silly stuff but it’s right there in the equation.

Q = σ(Th4)     — hot body “emission”

Q = σ(Tc4)     — cold body “emission”

Both are emitting photons!!!

Both emitting energy outward..

both receiving energy inward from both hot and cold bodies per the example above.

the difference in emission (assuming all energy strikes the other body) is the amount of heat energy transferred:

Q ~ σ(Th4) – σ(Tc4)

Q ~ σ(Th4 – Tc4)

And the probability lesson shown in the extreme above, is why photons which ARE emitted from cold objects, don’t “bounce off” of hot ones. At Earth temperatures (meaning not plasma hot) nearly 100% of the energy states potentially available for absorption of a photon are actually available at any one time.  Near 100%.  What that means is that cold body emissions, conductions etc… are absorbed into hotter bodies, making them warmer than they otherwise would be.  Cold to hot…

This is not a violation of thermodynamics.

The reality is though that despite nearly 100% of the absorptive sites being available, any non zero percentage of un-energized states in the atoms would allow for absorption from cold to hot bodies, and since the equation above is based on de-energization of an available site, they prove my case for me.

Energy can move from cold to hot!  Just don’t try to do any work with it — that is a different story and rather cute imho.

So in global warming science, the government claims consensus on global warming danger, but the government is lying by omission.  There is no significant controversy on the absorption wavelengths or emission wavelengths of different materials at different temperatures, so there is no real argument in science about the basic CO2 warming effect.   However, there is no consensus (good name for a blog) on dangerous warming, and most “scientific” people believe CO2 warming is NOT imminently dangerous.  Therefore there is absolute consensus with respect to the warming effect of CO2, it is basic optics/thermodynamics –  outside of the flat earth PSI types.

In the real world, there is no consensus that the effect is an actual problem and significant evidence that it is absolutely NOT a problem.


 

 

 


40 Responses to “Changing the Subject”

  1. Frank said

    Jeff: Before one can discuss whether heat is always transferred from hot to cold, one needs a definition of hot and cold. Thermodynamic temperature is proportional to the MEAN kinetic energy of a GROUP of rapidly colliding molecules. Therefore, single molecules can’t be hot or cold – though they can be “faster-moving” or “slower-moving”.

    A collision between two molecules can transfer kinetic energy from the slower-moving molecule to the faster moving molecule, but this doesn’t violate the 2LoT. In fact, this process produces a Boltzmann distribution of molecular speeds.

    Likewise, a photon can be emitted by a slower-moving molecule and absorbed by a faster-moving molecule without violating the 2LoT.

    Whenever one is dealing with two groups of colliding molecules large enough to be “hotter” or “colder”, the net flux of photons between them will always be from hot to cold. Heat is the NET radiative flux.

    • Jeff Id said

      Frank — Single molecules most definitely can be hot or cold.

      The rest I agree with completely.

      • 4kx3 said

        Jeff:

        Not to defend PSI, but Frank is correct. Temperature is a property attributed an ensemble in which is invested some amount of energy distributed through a variety of modes in equilibrium with each other. The temperature of a body is only defined rigorously when it is the same as that of some other body.

        Individual molecules can variously excited, and have a variety of motions, but the temperature of an isolated molecule cannot be defined, and even if it were it would change at the end of the radiative lifetime appropriate to its highest energy state.

        In fact whatever definition you chose might disprove your thesis as individual quanta come and go under their own rules, changing the molecules to higher and lower energy states which presumably you would use to represent temperature.

        In the atmosphere the condition of being decoupled from an ensemble produces NLTE radiation when the mean time between collisions exceeds the radiative lifetime. One might use this observation to assert that such molecules are cooler than their environment,
        yet they still radiate. The problem is that the temperature was not defined at the outset. Similarly, there is a phenomenon of anti-stokes flourescence where molecules radiate at a temperature higher then their surround. Again, these phenomena arise from systems far from equilibrium.)

        My suggestion is to avoid using “temperature” and “black bodies” as much as possible
        and concentrate on the Einstein radiation laws to the extent possible.

        All the best,

        • Jeff Id said

          First, I don’t believe Frank has the dubious affliction of PSI membership. Don’t really know that – but I don’t think so. He seems to smart for that.

          So Frank used the term ‘molecule’ and since the disagreement is pedantic enough, I think its fair to point out that molecules are quite large sometimes (salt crystal?, polypropylene?) and can definitely demonstrate bulk temperatures that can be easily measured, so Frank is incorrect in that aspect. I wondered if others would notice.

          However, with respect to atoms. Here is the way I think of temperature:

          I would also say that a single hydrogen atom at near C relative, is very hot in my frame of reference and not so hot in its own. You and Frank might state that it is only velocity, but it will feel an awful lot like temperature when hit with enough of them. Sometimes I wonder how good our thermodynamic units actually are.

          • jinghis said

            Jeff “I would also say that a single hydrogen atom at near C relative, is very hot in my frame of reference and not so hot in its own.”

            We are talking about Massless EM radiation that is already traveling at the speed of light (kind of a tautology) aren’t we? Isn’t our definition of temperature the rate and force at which atoms bounce off of something? I don’t think it is straightforward at all how to convert EM Radiation to Temperature.

          • Jeff Id said

            Sorry about that, I meant hydrogen atoms at near C rather than EM radiation in my reply. Temperature is an odd concept which is calculable both for a single atom at velocity and for a mass of atoms. It seems from the equations of temp, that all of these energies can contribute to measurable temperature. The leading surface of a high velocity object in atmosphere is a good example.

            It is possible to define the temperature of single particles and the concept is used at times in physics. I don’t know how useful the definition for the rest of us as our bulk equations don’t work so well on a quantum scale but it is what it is. Seems rather irrelevant to the post above.

            I wonder if there is some aspect of the post which you think I should edit?

            My comment calling a salt crystal a molecule is something I find interesting. I’ve heard the analogy before, but on looking it up after dinner tonight, I find the definition of molecule vastly more muddled than I realized. Perhaps someone could enlighten me on that.

          • 4kx3 said

            Jeff:

            I’m not sure that further discussion is helpful but you can judge. I do not agree with your statement “The answer quantum mechanics gives is one atom”. From your arguments I think you are missing some quantum mechanical ideas.

            Now, I think the following is all stated correctly. It is hard to construct a pithy course in molecular spectroscopy. You might find the paperback books by Planck, Heisenberg, and Tomanaga interesting. They are all inexpensive and readable. As I remember Pauling discusses the quantum oscillator.

            Reviewing, temperature is a statistic of a system, not an explicit scientific observable. Each mode of action has a temperature proportional to E/k. In reality we don’t usually know the energy in the modes, but we can know if net energy is being exchanged between a system and a reference system. The reference system is often called a “thermometer” and it can clearly be warmer or colder than some other system.

            Radiation exchange is only incidentally caused by temperature. It is caused by “structure” where “structure” means the disposition of all masses and charges. Structure is what determines radiative properties.

            Please research “coupling of quantum oscillators” which happens when two quantum systems interact. A consequence of coupling is that the excitation will be owned by the ensemble to an extent depending on the coupling. If the systems are identical, the coupling can be complete
            and all energy is delocalized.

            Since transitions involve change in charge density it is unusual for there to be no coupling. With weak coupling, the energy can stay localized.

            In the case of metals, presumably metallic hydrogen, the discrete electronic states contributed by individual atoms but coupled to each other, become property of the entire array. This include phonons (mechanical modes) as well as photons which interact with electronic modes. Metals reflect light because they interact strongly with electric fields.

            A system that is easy to understand and quantitatively analyze is that of the one dimensional crystal, the dye aggregate.

            https://www.pks.mpg.de/~eisfeld/wwww/pdf/Eisfeld_Briggs_2006_CP_376.pdf

            A dye aggregate consisting of molecules stacked like a deck of cards each of which in isolation absorbs at 450 nm can have a J band at 600 nm and which acts like a single molecule including the now longer radiative lifetime.

            An aggregate can also have a higher energy transition, or H band, because the interaction of the quantum system produces both a higher energy and a lower energy.

            So much for quantum coupling. Now back to temperature.

            The uncertainty principle limits the accuracy with which you can simultaneously measure all the variables necessary to define temperature. For a system with low molecular weight this uncertainty can be large compared to the energy of the system and very large compared to any realistic kT.

            Wiki talks about the temperature of a system with a small number of atoms each of which has two states. This would seem to be as close as you can get to a single molecule. https://en.wikipedia.org/wiki/Temperature

            The black body spectrum is not unique since other energy state distributions and/or multiple sources can give rise to similar spectral distributions. Estimating temperature from radiation requires knowledge of exactly what is radiating or absorbing for the reason noted above, that radiation is only incidently related to temperature.

            A small block of salt or a quantum dot can have the properties of a molecule. As the number of atoms involved gets large, the properties change, so the molecular weight of the molecule, dot, or block, is an issue as are the couplings between the atoms and the type of nucleii involved. He3(fermion) is different from He4(boson).

            You can let me know if this is not clear.

            All the best.

          • Jeff Id said

            4kx3,

            I stated that temperature can be applied to a particle. It has been done many times over the years. I did it in bachelor physics and have found a link here as to how it is done. Is that single particle temperature useful? That’s a different question. But if you have millions of the particles and calculate their interactions with a bulk material that none of us question the temperature of, the energy transfer is a known and done right, guess what value it shows up as?! The bottom line is that you can calculate a temp for a single atom, and I did not invent the technique – not my fault.

            Since you asked, there are many things which are not clear in your post which unfortunately meanders. I can’t find a specific intent, except you seem to think you need to teach me something so you wrote several things, most of which are not news and others are incomplete enough that I cannot understand. I don’t mind learning though so that is fine but I’m not at all sure what you are leading to.

            One small critique:

            You wrote — “Metals reflect light because they interact strongly with electric fields.” You could have equally said ‘glass transmits light because it interacts strongly with electric fields’ or ‘Black absorbs light because it interacts strongly with electromagnetic fields’, but all materials actually act strongly with electromagnetic fields. Their electrons have the same mass with different spring constants and react as any coupled second order differential equation would be expected to. The index of refraction is a really interesting chunk of information. Here’s a link on Lorentz that does a passable job but a short bit of work on your part will find better:

            https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-007-electromagnetic-energy-from-motors-to-lasers-spring-2011/lecture-notes/MIT6_007S11_lec22.pdf

            The complex index of refraction is one of the least appreciated bits of science.

      • Frank said

        Jeff and 4kx3: I think the best way to avoid controversy about the 2LoT is to say that individual gas molecules have kinetic energy, but are not hot or cold. That allows energy to be transferred from a slower-moving molecule to a faster-moving molecule via a collision or a photon. In statistical mechanics, the laws of thermodynamics are derived from the behavior of large numbers of molecules following the laws of QM. So I find it easy to draw a line between the macroscopic and molecular worlds and say that the laws and CONCEPTS (hot, cold, temperature, heat flow) of thermodynamics apply to the NET result produced by large numbers of molecules and photons. I find this dividing line can be understood by those who know that temperature is proportional to mean kinetic energy, that a Boltzmann distribution of molecular speeds exists, and who believe that “temperature” doesn’t change with every collision. LTE implies that a Boltzmann distribution of excited states exists. However, I’m not an authority in this field. If Jeff can deal with the 2LoT fanatics using other terminology, that is his business. This works for me.

        In the thermosphere, scientists refer to a thermodynamic temperature (kinetic energy) and Planck temperature (when the local radiation field has excited more molecules than the Boltzmann distribution predicts given the thermodynamic temperature). I don’t know what terminology is used with lasers or other situations where LTE doesn’t exist. When one is dealing with phenomena where more than one definition for temperature can be used, I simply say that temperature is not “well-defined” AND assert that these phenomena aren’t important in the lower atmosphere or to climate.

        The thermogravity folks also get lost crossing the boundary between the behavior of individual molecules and large collections of colliding molecules. They are applying conservation of energy (KE + PE) to a macroscopic world where collisions redistribute enormous amounts of kinetic energy much faster than tiny changes in PE occur between collisions. In the macroscopic world, additional forms of work/energy emerge and are critical: PdV and TdS. PE change is negligible.

        People also get lost transitioning between molecular behavior and macroscopic behavior when using Planck’s Law and the S-B eqn. These were derived FOR EQUILIBRIUM situations (bulk materials) without any knowledge of molecular behavior except quantized energy states. No Einstein coefficients and no absorption cross-sections. In the atmosphere, some wavelengths of radiation don’t interact strongly enough with GHGs to reach equilibrium. That produces endless confusion about optically-thick and -thin layers of gas and the emissivity of an atmosphere with an 100 K difference in temperature and 3X difference in T^4.

        If we were taught correctly about radiation, we would start with Einstein coefficients, the relationship between Einstein coefficients for systems in equilibrium produces the Planck function and absorption coefficients, derive the Schwarzschild equation and then learn that the Schwarzschild equation produces a blackbody spectrum (Planck’s Law) whenever radiation isn’t changing with distance traveled and absorption has come into equilibrium with emission. So, you don’t apply Planck’s Law and the S-B equation without equilibrium. (However, NO one agrees with me about this subject.)

        Jeff: Thanks for noticing that I’m “too smart” to follow PSI, but that doesn’t mean that I’m too smart to avoid making mistakes. Frank.

        • Jeff Id said

          I know that if we sat across a table and discussed temperature, kinetic energy of molecules, radiation or whatever concept came to mind, we would end up agreeing on most all of it. We are saying the same things. I could ask what is the difference between Kinetic energy and vibrational or rotational energy, and we would have little concern about disaggregation. It is very much the same with temperature– a concept invented for bulk properties, and extended by physics math to others. We both understand that the second law is probabilistic, and that the true intent of temperature is the same. Where we both also seem to agree is that the concept is imperfect.

          Relative to us, is a moving baseball (having in its timeframe a 23c temperature) warmer than a stationary one having a 23c temp in it’s own? Or is that energy somehow of a different sort?

          Did anyone else notice that 100% of the comments in this thread exceed the abilities of PSI? I sure did.

          • Frank said

            Jeff: My guess is that bulk motion doesn’t contribute to temperature. Temperature is “internal energy”, but T*S and P*V have units of energy also. I believe one can add chemical potential energy to these terms. So why not add in 1/2mv^2 for kinetic energy of bulk motion? Molecular motion creates pressure, but the motion of the baseball itself doesn’t create pressure.

            FWIW, I have been trying to develop a kinetic theory of gases in a gravitational field. In the traditional kinetic theory of gases, P = ρv^2.

            Taking the derivative with respect to z: dP/dz = 2ρv*(dv/dz) + v2*(dρ/dz)

            Hydrostatic equilibrium says: dP/dz = -ρg

            Eventually” -2ρv*(dv/dz) = v2*(dρ/dz) + ρg

            Alternatively P = ρRT and dP/dz = RT(dρ/dz) + ρR(dT/dz). Substituting T proportional to v^2 should get one into the same mathematics.

            If dv/dz is zero, the gas is isothermal. So far, I’ve done nothing to ensure that the gas is in equilibrium. Equilibrium requires the flux of upward-moving and downward-moving molecules must be equal at all z. I think that slightly fewer molecules are moving down slightly faster because of the acceleration from gravity between collisions. From that, I hope v2*(dρ/dz) can be proven to be -ρg. I can’t convince myself I’ve got this part right.

          • jinghis said

            Jeff – “Relative to us, is a moving baseball (having in its timeframe a 23c temperature) warmer than a stationary one having a 23c temp in it’s own? Or is that energy somehow of a different sort?”

            Yes, potential energy isn’t measured by a thermometer it has to be calculated. . . When that baseball collides with something its temperature will go up.

            I think this is one of the issues with the lapse rate, the atoms at the top of the column have a lot more potential energy that again isn’t measured by a thermometer.

          • Jeff Id said

            Jinghis,

            What is measured by a mercury thermometer and what is defined in physics as temperature are different things. When that baseball collides with something of the same temperature, the object it collides with will also experience an average temperature increase.

            Lapse rate is interesting isn’t it.

          • Jeff Id said

            Frank,

            Your comment – “My guess is that bulk motion doesn’t contribute to temperature.” and “Thermodynamic temperature is proportional to the MEAN kinetic energy of a GROUP of rapidly colliding molecules.”

            What is the difference? That is my point and the point of physicists calculating temperatures of single molecules — or how about seriously rarefied plasmas in the sun’s corona? The corona is millions of degrees and the gas travels mostly in the same direction. A lead object in the path of that wind will come to a stable temperature higher than measured by a thermometer traveling with the wind. The bulk velocity is known and the temp of the lead object will rise to a temperature which includes the bulk velocity. T1=T2 thermo style.

            That is why I said temperature in one reference frame vs the gas fields reference frame. Regarding chemical reactions, if they happened at impact, the thermodynamic equations would also fail to represent the stabilization temp. Not that some added energy couldn’t be put in to help, but temperature is an entirely human construct and the concept was created in a room without significant bulk motion. Not sure what quantitiy is perfect though either, mass?, time?, distance?. Certainly those seem better but that’s probably an issue with my own understanding.

          • Frank said

            One of Feynman’s books complains about teaching the names for things as a substitute for teaching an understanding of them. One topic he discusses is energy. His discussion of conservation of energy can be found here:

            https://smartsite.ucdavis.edu/access/content/user/00002774/Sears-Coleman%20Text/Text/nonS/Feynmanenergy.html

            “Second, the energy has a large number of different forms, and there is a formula for each one. These are: gravitational energy, kinetic energy, heat energy, elastic energy, electrical energy, chemical energy, radiant energy, nuclear energy, mass energy.”

            Notice that Feynman distinguishes between heat and kinetic energy even though he is aware of the kinetic theory of gases. This may let us distinguish between bulk motion and molecular motion within bulk (center of mass) that may or may not be moving. Looking from the center of mass frame of reference simplifies many problems. You may be aware of more sophisticated problems that can occur when using this approach (that I haven’t considered).

            The worst physics mistakes that plague climate science blogs always seem to come from people applying the laws of bulk materials (COE, Planck’s Law) to the behavior of individual molecules and photons. So, I find it easier to say individual molecules have kinetic energy, but not a temperature, and therefore we can’t apply the 2LoT to their behavior. This statement is perfectly correct given the definition of thermodynamic temperature and the fact that the 2LoT applies to the behavior of bulk materials and NET energy flux. Yes, the concept of temperature can be extended to situations where LTE doesn’t exist and where multiple definitions for temperature are possible, but that discussion isn’t very practical at a blog.

          • Jeff Id said

            Frank,

            I’m sorry for disappearing, life is busy. I’m not surprised to agree with your comment here. I often think about how these ‘different’ energies must be the same thing.

            I’m not smart enough to figure it out yet though, but that doesn’t stop me from trying.

  2. jinghis said

    Jeff “so there is no real argument in science about the basic CO2 warming effect.”

    I think everyone agrees that CO2 (and the atmosphere) is largely transparent to short wavelengths, it does not absorb visible light or 99% of the radiation emitted by the sun.

    The Ocean absorbs 84% of the suns radiation and the land absorbs 16%, roughly. The land re emits long wave radiation which CO2 can absorb which creates warming during the night. in other words CO2 raises the daily minimum temperature over land.

    The ocean on the other hand emits energy via evaporation, the net radiative loss is close to zero. That is where the green house effect that warms the world occurs, in the ocean. The Ocean temperature determines the atmospheric temperature via convection through H2O.

  3. Peter Fournier said

    Wow! I followed your link to the PSI website blog entry and read the whole thing. I also read Roy Spencer’s and Anthony Watts’ blog entries linked in the PSI article.

    May I offer my most heartfelt sympathy for your feeling you need to pay any attention to these ideas! Your blog entry and Watts’ and Spencer’s are all perfectly obvious — temperature is a statistical mean of all the energies of all the atoms in a mass — and individual atoms will have energies above or below the mean. Of course some energy is conveyed from cold to hot or you invalidate the entire mechanism for the transfer of energy! Sheesh! I remember the car ride when I figured this out at age 8 or 9 — very very exciting at the time.

    Perhaps the real problem is that the PSI folks imagine heat to be like water moving through a bed of sand. If you have blue-dyed water on one side at high pressure, and red-dyed water at low pressure on the other side, you can’t get red dye moving into the blue dye source.

    OOPS! Sorry that example is a mistake on my part. Both pressure and heat are the mean measurement of the state of many many atoms. On the other hand, if you look at the boundary between red and blue you will find it is fuzzy so even with the pressure example your point holds — there is no saran wrap boundary moving through the sand bed preventing molecules from crossing some imagined boundary, in both directions, from high-to-low pressure and low-to-high pressure or high-to-low temperature or low-to-high temperature.

    Still, I think the confusion is that that a net movement implies the equivalent of saran wrap in the water example.

    Sigh!

  4. Peter Fournier said

    “Still, I think the confusion is that that a net movement implies the equivalent of saran wrap in the water example.”

    should read

    “Still, I think the confusion FOR SOME PEOPLE is that that a net movement implies the equivalent of saran wrap in the water example.”

  5. M Simon said

    Lovely.

    And just to be obnoxious. As long as we are in the quantum realm.

    Zero point energy.

  6. Michael 2 said

    This is easily shown with an infrared thermometer. The freezer is already colder than the thermocouple or micro-bolometer in the remote reading thermometer. Despite being colder than the sensor, the sensor can “see” the infrared coming from inside the freezer.

    The freezer, of course, sees the sensor. Each radiates to the other until a balance is reached. The micro-bolometer is tiny and the freezer is large, so the energy input from bolometer to freezer is inconsequential. The freezer is large and thus sends many cold photons to the bolometer, but the energy of each photon can be absorbed only in the case that it hits an atom or molecule not already in an equal or higher energy state. So, the number of photons matters, but so does their energy state.

    What that means is they will equalize on temperature, not heat. Within a fraction of a second the micro-bolometer has reached the same temperature as the inside of the freezer, and without physical contact. Each radiates to the other.

    • jinghis said

      “What that means is they will equalize on temperature, not heat. Within a fraction of a second the micro-bolometer has reached the same temperature as the inside of the freezer, and without physical contact. Each radiates to the other.”

      That isn’t how a bolometer works. “The material used in the detector must demonstrate large changes in resistance as a result of minute changes in temperature.”

      The bolometer is essentially at ambient temperature and very slight temperature (radiation) differences are represented relative to the ambient temperature. The bolometer does not equalize in temperature with the source.

      EM radiation does not equal temperature, we just find wavelengths where it is close for selected materials.

  7. Michael 2 said

    I’ve never registered to comment there but as it appears they read this page here’s what I would say: Re John O’Sullivan March 23, 2017 at 2:03 pm |

    “here we see the Berkeley lab experiment on the issue where scientists took ‘greenhouse gases’ (SF6, CO2, NH3, and N2O) and some other gasses, put them between panes of glass and tested and measured their abilities to trap heat and/or inhibit heat loss.”

    So far so good but keep an eye on “between panes of glass”

    “Berkeley’s brightest and best reported that ‘the effect of the infrared radiation properties of CO2 is unnoticeable’.”

    Quite right. The glass blocks infrared so whatever effect exists in the gas between the glass panes is irrelevant and untested.

    “the ERBE satellite data shows no evidence of any back radiation signal. If it isn’t being detected then it doesn’t exist. QED”

    I’m not sure if that was meant to be sarcastic.

  8. jinghis said

    Jeff, with an induction cooktop I can heat up a pot of water to boiling while the radiation source stays the same temperature. That seems like a pretty straightforward example of a cold object heating up a hot object, and something of a ‘greenhouse’ effect.

    However I completely understand PSI’s claim that cold objects don’t warm hot objects. As long as they don’t claim that ‘cold’ objects can’t slow the rate of cooling of the warm object I don’t have a problem with that either.

    This is directly in line with my statement above that CO2 helps raise the minimum daily temperature (slowing the rate of cooling).

    While on the other hand I have observed that increasing H20 humidity levels, decreases the high daily temperatures and I think CO2 would have a similar effect (decreasing the high temperature) and for similar reasons. Dry deserts have both higher high daily temps and lower minimum daily temperatures than humid regions at similar latitudes.

    I see H20 and CO2 decreasing the temperature differential.

  9. Michael 2 said

    jinghis said “That isn’t how a bolometer works.”

    Thanks, I sit corrected!

    When pointing it at cold, the sensor radiates more than it gets back, but it does *get back* since otherwise all targets would read the same cold temperature. It is reading *something*, not nothing, and that’s the important point. That something impinges upon the bolometer making it warmer than it would otherwise be without the cold source radiating into it (as compared to a sink at absolute zero that radiates nothing).

  10. Michael 2 said

    I suppose I should add something that is implicit but not always obvious. The bolometer has a relatively constant input of heat energy from its environment, largely by conduction of the wires going to it but also radiation from its case elements.

    Consequently it cannot actually achieve the same temperature as a remote object, but its temperature will trend in that direction. How low it can read is at least partly a function of effective insulation of the bolometer from ambient temperature to permit a wider swing of sensor temperature. Mine can go down to -60 F and it’s an eye-opener to read the temperature of a cloud, even in winter, as compared to the night sky (where it heads to the lowest it can go; it is tuned to the optical window and does not see CO2 in the atmosphere but water droplets in clouds emit infrared in its sensing wavelengths).

    It is a good and simple tool to confirm or discover several principles; one of which is that IR does not go through glass, so filling a glass jar with CO2 then shining infrared on it thinking to prove something thereby is rather pointless.

    • jinghis said

      Michael 2, “It is a good and simple tool to confirm or discover several principles; one of which is that IR does not go through glass, so filling a glass jar with CO2 then shining infrared on it thinking to prove something thereby is rather pointless.”

      I have both an IR gun and FLIR on my boat and I too have a lot of fun pointing the IR Gun at things and comparing it to the FLIR display.

      About the Glass jar filled with CO2, the IR will heat up the glass because it is opaque to IR. According to AGW theory the warm glass should then radiate into the gas inside the jar and Greenhouse gases should warm faster than non greenhouse gases that don’t absorb in that frequency range. My problem with these experiments is that they are poorly designed and don’t account for the same number and ‘pressure’ of the molecules and need different size containers because one size does not fit all, and then you have big surface area issues.

      The most interesting thing I discovered with the IR gun was that the Surface of the ocean stays the same temperature regardless of day or night as long as the wind and humidity stay the same (seasonally adjusted of course).

      And at the same time the temperature of the water just below the surface can vary by quite a few degrees.

      • Jeff Id said

        “The most interesting thing I discovered with the IR gun was that the Surface of the ocean stays the same temperature regardless of day or night as long as the wind and humidity stay the same (seasonally adjusted of course).”

        What is happening to the boundary layer?

        • jinghis said

          Jeff,

          The boundary layer continuously evaporates leaving behind a cooler boundary layer. Consequently the surface interface is generally cooler than the water below it and the air above it.

          I think it has to do with the fact that humid air is less dense than dry air. When water evaporates it expands in volume and cools according to the gas law. Radiation doesn’t play much of a role in the water vapor cycle.

      • stevefitzpatrick said

        Odd, i’ve done the same thing and I routinely see changes in bolometer readings which follow changes in thermometer readings ~2 feet below the surface…. plainly visible from morning to nighttime on a sunny day. The readings are not exactly the same, but they do follow each other.

  11. Ed Bo said

    The PSI folk love to say that “heat can only flow from hot to cold” and claim that climate science has misinterpreted the 2nd law of thermodynamics as laid down by Clausius.

    So I love to quote Clausius on the subject:

    **********************************

    This principle, upon which the whole of the following development rests, is as follows: – Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time*.

    * [The principle may be more briefly expressed thus: Heat cannot by itself pass from a colder to a warmer body; the words “by itself,” (von selbst) however, here require explanation. Their meaning will, it is true, be rendered sufficiently clear by the expositions contained in the present memoir, nevertheless it appears desirable to add a few words here in order to leave no doubt as to the signification and comprehensiveness of the principle.

    In the first place, the principle implies that in the immediate interchange of heat between two bodies by conduction and radiation, the warmer body never receives more heat from the colder one than it imparts to it. The principle holds, however, not only for processes of this kind, but for all other by which a transmission of heat can be brought about between two bodies of different temperatures, amongst which processes must be particularly noticed those wherein the interchange of heat is produced by means of one or more bodies which, on changing their condition, either receive heat from a body, or impart heat to other bodies.

    On considering the results of such processes more closely, we find that in one and the same process heat may be carried from a colder to a warmer body and another quantity of heat transferred from a warmer to a colder body without any other permanent change occurring. In this case we have not a simple transmission of heat from a colder to a warmer body, or an ascending transmission of heat, as it may be called, but two connected transmissions of opposite characters, one ascending and the other descending, which compensate each other. It may, moreover, happen that instead of a descending transmission of heat accompanying, in the one and the same process, the ascending transmission, another permanent change may occur which has the peculiarity of not being reversible without either becoming replaced by a new permanent change of a similar kind, or producing a descending transmission of heat. In this case the ascending transmission of heat may be said to be accompanied, not immediately, but mediately, by a descending one, and the permanent change which replaces the latter may be regarded as a compensation for the ascending transmission.

    Now it is to these compensations that our principle refers; and with the aid of this conception the principle may be also expressed thus: an uncompensated transmission of heat from a colder to a warmer body can never occur. The term “uncompensated” here expresses the same idea as that which was intended to be conveyed by the word “by itself” in the previous enunciation of the principle, and by the expression “without some other change, connected therewith, occurring at the same time” in the original text. – 1864.]

    Clausius, R., The Mechanical Theory of Heat with its Applications to the Steam-Engine and to Physical Properties of Bodies, 1867, p 117.
    (Originally from Clausius’ Fourth Memoir, published in Philosophical Magazine, 1854)

    *********************************

    Note that Clausius has a specific term for the thermal transfer of energy from cold to hot: “ascending transmission of heat”!

  12. Anonymous said

    Question for y’all: If one placed a cool body, in a vacuum, surrounded on all directions by a very cold container, except that it is placed within a parabolic IR reflector, such that much of its thermal radiation is focused onto another body, which in turn is warmer, but free to radiate in any direction, what would happen?

    • jinghis said

      Ahh a parabolic IR reflector, that makes it essentially the same as a magnifying glass which will increase the temperature at the focal point.

    • Michael 2 said

      Re: Anonymous. Restating: Two objects exist. One is warm, one is cold. A reflector surrounds them in such a manner that most of the radiation of each is reflected to the other. No other source or sink of radiation exists. What will happen?

      Answer: Eventually these objects will reach the same temperature; the warm body will cool and the cool body will warm.

      During the process of equalization, the cold body will radiate a smaller number of lower energy photons to the warm body, and the warm body will radiate a larger number of higher energy photons to the cold body. When they have the same temperature, they will each radiate to the other the same number and energy of photons.

      Now if the warm body had no cool neighbor, but radiated into space (a perfect sink), it will cool faster than by having a cool neighbor. So it is that camping in a cold tent at night is still warmer than camping under a clear sky at high altitude in the desert of Wyoming where the sink temperature is -60 or colder. Zero is warmer than -60. Does the tent “warm” you? No, your body warms you; the tent merely helps maintain your temperature by reducing radiative (and convective) heat loss. But since nothing can stop you from radiating, at least you can get some of it back!

  13. page488 said

    Jeff – The chemical formula for one molecule of table salt is one Na (superscript +) and one Cl (superscript -). The formula is so written with the superscripts because the electron donated by the Na atom is loosely held between the two atoms, so each retains part of it’s original charge. In fact, once in solution, the atoms disassociate into ions (charged particles) altogether. It is the ionic nature of the molecule that allows it to form into face centered, cubic crystal lattices, which can combine with each other to form a number of shapes that can be different sizes (think of the difference between a grain of table salt versus a grain of the larger pretzel salt). [Other molecular info on NaCl: The atomic weight for one molecule of NaCl is given by adding the respective atomic weights of the two atoms: 22.9898 amu (Na) + 35.45 amu (Cl) = 58.44 amu for NaCl. One mole of the substance is also numerically 58.44, but the unit is grams.] Hope this helps!

    • stevefitzpatrick said

      Good grief. The electon is transfered just about 100%, forming two ions of opposite charge. There is no ‘loosely held’ electron between the atoms.

  14. gallopingcamel said

    As Ed points out the fundamental error of the PSI folks is their belief that “heat can only flow from hot to cold”.

    This is arrant nonsense so no scientist or engineer should take their ideas seriously.

    When I calculated the surface temperature of the Moon I used Stefan-Boltzman which quantifies the rate of heat transfer from cold bodies to hot ones as well as in the other direction, consistent with Planck quantum theory.

    It should be no surprise that the temperatures calculated based on the above theories agreed to better than 0.3 Kelvin with the measurements made by the Diviner Lunar Radiation Experiment:
    https://tallbloke.wordpress.com/2014/04/18/a-new-lunar-thermal-model-based-on-finite-element-analysis-of-regolith-physical-properties/
    https://tallbloke.wordpress.com/2014/08/27/extending-a-new-lunar-thermal-model-part-ii-modelling-an-airless-earth/

    Have the PSI folks ever tested their theory against measurements?

  15. Frank said

    Jeff: Here is an experiment that might detect a two-way flux of photons. Imagine a beam of ions (like those in mass spec) passing through a beam of light. Some of the ions will absorb a photon, be deflected to one side by the momentum of the absorbed photon and create two beams of ions that enter separate detectors. Hopefully the uncertainty principle won’t interfere with our ability to detect two beams.

    Now imagine light sources on both sides of the beam. Will we observe three beams? What happens if the two light sources come from splitting one laser source into two halves. Will interference prevent the beam from being split?

    This is called the “optical Stern-Gerlach effect”. The original Stern Gerlach effect involves atoms passing through an inhomogeneous magnetic field to separate quantized states with different angular momentum. Unfortunately, I don’t understand the optical version of this experiment because it is discussed from the point of view of an inhomogeneous electric field (with a standing wave) rather than streams of photons from each side of the beam.

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: