Bob Ashworth had this to say about the reality of what the atmosphere actually does:
“Just remember that the earth is the temperature it is because of the sun. Even if CO2 had a warming effect on earth, which it doesn’t, CO2 is so small (400 ppmv) mans activities (only 12 ppmv) you couldn’t even measure the effect.
First Law of Thermodynamics: Energy can be changed from one form to another, but energy cannot be created or destroyed. The total amount of energy and mass in the Universe is always constant; it merely changes from one form to another.
Second Law of Thermodynamics: Heat may only be transferred from a hotter to cooler body, never vice versa. A cooler atmosphere cannot heat the earth’s warmer surface. The greenhouse effect is a myth because its premise clearly violates the second law of thermodynamics! Although all bodies above absolute zero radiate and absorb radiant energy, the warmer body always provides more energy to the cooler body than it receives back from the cooler body. A cooler body therefore can never heat up a warmer body; the cooler body warms and the warmer body cools, never vice-versa.
The Scientific Truth
The truth of the matter is that any mass between you and a radiant energy source will provide cooling. Stand near a fireplace that is burning and feel the warmth of the radiant energy, then have two people drape a blanket between you and the fireplace – you will feel cooler! This is like standing outside on a sun shiny day and when a cloud goes over and shields you from the direct rays of the sun, you feel cooler. A child knows this. Regarding the earth, our atmosphere provides cooling in the same manner: Nitrogen, oxygen, water vapor, carbon dioxide and any dust that is in the atmosphere all provide cooling.
Why is this? It is very simple. If there were no atmosphere, all of the radiant energy from the sun would hit the earth. With an atmosphere, a portion of the incoming sun’s rays are reflected back toward the sun by striking the gaseous molecules and dust particles, so less radiant energy hits the earth and the earth is cooler because of its atmosphere, see the figure below.
Everyone also knows that cloud cover at night (more insulation) prevents the earth from cooling off as fast as it does when there are no clouds. However, on a relatively clear night if a cloud goes overhead you cannot feel any warming effect of the cloud, so this insulating effect is shown to be minimal compared to the daytime effect.
Anthropogenic Global Warming – more energy out than in – good trick, but violation of First Law of Thermodynamics. The US Patent Office would never patent such a concept.“
This ties in to a couple of things. The first is to ask why the bottom of the atmosphere is warmer than the top of the atmosphere. One answer is that warming occurs at the atmosphere’s bottom in the first place with the very high-temperatures generated by Sunlight on the ground surface. As we saw in my paper and real-world data, the surface is heated to almost 90oC(!) by sunlight on the day-side for a very large fraction of the surface of the Earth. But this isn’t the entire picture.
On Venus, the clouds are so thick that hardly any direct sunlight actually even makes it to the surface. But we also know that the bottom of the Venusian atmosphere is incredibly hot – much warmer than its equivalent radiative temperature as seen from space. The Venusian atmosphere also has a very high reflectivity (albedo), and so its equivalent temperature as seen from space is actually cooler than the Earth’s. So, if atmospheres reduce the amount of potential heat generation by filtering out and reflecting sunlight, then why do atmospheres also get to be warmer at the bottom than at the top? All planets with atmospheres, even gas giants like Jupiter, follow this pattern of the gas getting warmer and warmer with depth.
The answer for this is of course the “lapse rate”. One of the most useful ideas in the history of thermodynamics is the concept of “local thermodynamic equilibrium”, or LTE. When a parcel of gas is in LTE, its energy content is constant, so that its temperature is stable. We know that any gas in a gravitational field changes in temperature with depth, and this is knowledge which simply comes from observation, such as that in Earth’s atmosphere but also from that of stellar (star) atmospheres. The problem to figure out, is, if the gas is thermally stable, but the temperature is not uniform, then how do we explain that the temperature of the gas is a function of depth (or altitude) in the atmosphere?
It turns out to be very simple: Since thermodynamics is all about energy content (all of physics is really about energy content!), and the gas is in a gravity field, then what is the energy of a section of gas given its temperature and its gravitational potential energy? The gravitational energy is simply a function of altitude, and so we can calculate the gravitational energy (EG) for a “slice” of the gas column at a particular height ‘h’:
EG = m*g*h
where ‘m’ is the mass of the slice of atmosphere, ‘g’ is the strength of gravity, and ‘h’ is the altitude above the surface.
The thermal energy (ET) from the temperature ‘T’ of the same mass ‘m’ of gas is:
ET = m*CP*T
where ‘CP‘ is the thermal energy capacity of the gas. The total energy (ETotal) of the slice of gas, adding together its thermal energy and its gravitational energy, is thus
ETotal = m*g*h + m*CP*T
We’re now going to do something really amazing using calculus – we’re going to relate changes in the total energy to changes in all of the other parameters. For this slice of gas, however, the mass does not change and neither does its thermal capacity; the only thing that can change is its height ‘h’ and temperature ‘T’. In calculus, the prefix ‘d‘ (which stands for “differential”) is used to simply indicate a change in the value of a parameter, and since we know which parameters are changing and which are not, the equation becomes
dETotal = m*g*dh + m*CP*dT
Now we can return to the idea of local thermodynamic equilibrium (LTE), which states that a slice of gas at a particular height ‘h’, and temperature ‘T’, will have a constant (i.e. unchanging) total energy when the whole column is thermally and hydrostatically stable. But if the total energy of the slice of gas is constant, then the total energy does not change, and so the change of the total energy (dETotal) is equal to zero. Thus, the above equation is equal to zero:
0 = m*g*dh + m*CP*dT
Remember, what we wanted to explain was why the temperature of a gas in a gravity field changes with altitude. What we have in the above equation is ‘dh’, a change in altitude, and ‘dT’, a change in temperature, so now we can just rearrange the equation:
m*CP*dT = - m*g*dh
dT/dh = -g/CP
What this says is that the rate of change of temperature with respect to change in altitude (dT/dh), is equal to the strength of gravity divided by the thermal energy capacity of the gas. It is really so simple, and, it matches exactly what is observed for dry air! Isn’t that amazing how real mathematics and real quantification of physics explains reality? No doubt this is why the greenhouse effect never quantifies itself in terms of an equation, math, physics, or thermodynamics!
Because the right-hand side of the equation is negative, this says that the rate of change of temperature with altitude is negative; or in other words, the temperature decreases with altitude, exactly as we expected. The dry-air value of -g/CP is about -9.74 Kelvin per kilometer (i.e. for every kilometer you go up in altitude, the temperature drops by 9.74 K).
The presence of water vapor modifies the “dry” air calculation by adding heat to the atmosphere via condensation; this was described and calculated in my paper on pages 7 through 9, resulting in a global average atmospheric lapse rate of about -6.5 K/km. What happens is that, on average, water is constantly being evaporating at the ground surface, and because water vapor (evaporated water) is lighter than air (and also warmer than air if it is evaporated by the high-temperature action of sunlight), then it naturally rises to high altitude, condensing back out as liquid droplets as it cools with altitude. There is an equal amount of condensation of water vapor occurring which balances out the rate of evaporation, so that the total amount of water in the atmosphere is roughly constant. Thus, a cycle is set up which preferentially adds heat to the atmosphere, and adds heat to a rising parcel of gas which thereby slows down its rate of cooling from -9.7 K/km to -6.5 K/km. As we saw in a previous post, the latent heat of water also serves to transport heat to the poles, away from the equator, without requiring additional radiative input from the Sun. Thus, if there is anything causing a warmer planet than expected, it is the latent heat in water, and the sophistry of the greenhouse effect is in ascribing to “backradiation” what latent heat is actually doing.
Let’s denote the average lapse rate with “Lave“: Lave = -6.5 K/km. Now, it is great to know that we have successfully calculated a major feature of the atmosphere, but the lapse rate itself isn’t actually a function of temperature with height, but is simply the rate at which temperature changes with height. If we want to know the actual function of temperature vs. height, we have to take the differential equation we derived for the lapse rate,
dT/dh = -g/CP = Lave (when factoring in water vapor)
and anti-differentiate it with respect to ‘h’. This results in the equation
T(h) = Lave*(h – h0) + T0
where ‘h0‘ and ‘T0‘ are constants of differentiation. This equation is exactly like the equation for a straight line, y = m*(x-x0) + b. Usually, the term ‘m*x0‘ is combined with ‘b’ to make a “new” constant ‘b’, so that the equation can be written as y = m*x + b. Writing it as we did here, denoting the ‘x0‘ term (i.e. h0) specifically, is more physically correct because ‘h0‘ and ‘T0‘ can then represent constant characteristics of the system – a characteristic height ‘h0‘ and a characteristic temperature ’T0‘.
The characteristic height and temperature are not as easy to define as you might think, but, it would be nice to peg these constant values to something we are familiar with. Well, we are familiar with the equivalent radiative temperature of -18oC (from 240 W/m2) for the Earth discussed previously, so, let’s use that value for ’T0‘. Recall from that post (previous link), that there is ambiguity in how an “equivalent radiative temperature” actually corresponds to a kinetic temperature measured with a thermometer, and that it is this fundamental ambiguity which creates the sophistry of the backradiation greenhouse effect. We can now actually resolve that ambiguity!
If we say that the characteristic temperature ’T0‘ is equal to -18oC, for the atmosphere, then we can calculate the corresponding characteristic height ’h0‘ if we simply know the average kinetic temperature at a single other height. And we do know this, from the same ambiguity which creates the sophistry of the backradiation greenhouse effect: at an altitude of h = 0, i.e. the sea-level surface, the average kinetic temperature is measured at T = +15oC. Plugging it all into the previous equation from above:
T(h) = Lave*(h – h0) + T0
+15C = -6.5K/km*(0 - h0) – 18oC
h0 = 5 km.
Thus, the equation for the average temperature of the atmosphere with height above the ground is:
T(h) = Lave*(h – 5km) - 18oC
The ambiguity of the difference between the equivalent radiative output temperature and the kinetic surface temperature is therefore found in the fact that the “equivalent radiative temperature” of a radiative output of 240 W/m2 is not kinetically found at the ground surface, but in the atmosphere at altitude. This does not mean that 240 W/m2 comes from an altitude of 5km, but just that the equivalent kinetic temperature is found somewhere around that altitude. This is similar to the radiating atmosphere of stars, which I am familiar with from my work in astrophysics, where the kinetic temperature of the star’s atmosphere is close to its equivalent radiative output temperature at a characteristic average optical depth.
What the greenhouse effect and supporters of it try to argue, is that if the atmosphere didn’t have a greenhouse effect from backradiation, then the near-surface air would have an average kinetic temperature of -18oC. Let’s think about that for a moment: if the average temperature was found at the surface, and we know from the lapse rate that the temperature will naturally decrease in altitude above the surface, then the average temperature will in fact be found at a boundary of the whole atmosphere (the surface boundary), and lower temperatures will thus be found above this boundary. In other words, the warmest area of the atmosphere will be one of its boundaries, and this boundary will also be the average.
This is a plain logical contradiction. If the boundary is the warmest, then it can not also be the average, because the average of a whole “ensemble” is necessarily smaller in number than the largest number of the ensemble. What greenhouse effect advocates imply is that the average of the sequence
-18, -24.5, -31, -37.5, -44,
must be equal to -18! They say that without a greenhouse effect, -18oC would be found at the ground surface, with the temperature decreasing by -6.5 K/km above it, and that the average would still be -18oC. It makes no sense at all.
The fact that the atmosphere naturally has a lapse rate necessitates that the average or characteristic temperature of the atmosphere can not be be found at the atmosphere’s lower (or upper for that matter) boundary, but has to be found somewhere “in the middle” of the atmosphere by the simple definition and requirement of pure logical mathematics. The average is found “somewhere in the middle” by the definition of what an average is – an average is not the boundary and highest value of a sequence! Warmer temperatures will have to be found below the altitude of the characteristic average temperature, guaranteeing that the surface boundary will be the warmest part of the ensemble, and this has nothing to do with backradiation or a greenhouse effect, but everything to do with the natural lapse rate. For the Earth, it is also a given that the bottom of the atmosphere will be warmer than the average of -18oC, since the heating occurs at the bottom of the atmosphere at temperatures approaching 90oC on a large fraction of the daytime hemisphere.
Of course, you will find greenhouse effect advocates thus trying to argue that the greenhouse effect either is, or creates, the lapse rate, but you saw the derivation for the lapse rate here above, and it has nothing to do with discussing backradiation or a greenhouse effect. Their arguments are just another attempt at sophistry and obfuscation.
What effect does carbon dioxide have on the lapse rate, g/CP? It doesn’t change gravity, obviously. And its effect on CP is almost negligible, given that 1) the CP of carbon dioxide is nearly equal to that of standard air, 2) carbon dioxide is a trace gas that barely even factors in to CP in the first place. The effect of changes in CO2 on the atmospheric average thermal capacity is almost zero.
Also note what effect water vapor has on the lapse rate – water vapor is said to be the “strongest greenhouse gas” due to its radiative properties and because it is several dozen times more prevalent than carbon dioxide But its effect on the lapse rate, which is a well-known and well-measured effect, makes no reference whatsoever to its specific radiative abilities, and rather only to its thermal capacity and latent heat. And calculating its effect on the lapse rate due to its latent heat and thermal capacity, without any reference or inclusion of any radiative effects, is entirely successful at predicting the results, thus proving that there are no additional radiative heating effects. If the supposed radiative effects from water vapor have no effect on the temperature distribution in the atmosphere, then the same supposed radiative effect from carbon dioxide will likewise have no effect.
The reason for that is because the radiative effects from water vapor and carbon dioxide do not represent additional energetic input to the system! Note that the only effect water vapor has is through its release of latent heat, which specifically is additional energetic input to the system. This is yet another indication that the vaunted effects of backradiation are entirely fabricated and at least badly imagined.
I said at the start that Ashworth’s quote tied in to a couple of things; the second thing relates to this statement “The US Patent Office would never patent such a concept”, and I will discuss it in the next post with reference to what “else” the GHE is said to be.