Its weird that someone would say that earth is 33° “warmer than it should be”, especially when they already reduced the amount of heat by 30% before the calculation started. My approach is that everything is exactly what it should be, and if someone says that there is heat coming from an ice-cold atmosphere, my first thought is that someone miscalculated. .

We start by taking a look at the differences between the 2 dimensional spherical surface of the absorbing and emitting blackbody, and the 3 dimensional volume of Earth absorbing solar radiation.

To find the emitted intensity of the surface of the inner spherical shell, we need to account for absorption of the received intensity from solar irradiation, TSI, over the surface area of the disc that is the shadow of earth, but irradiation is only on the hemisphere 2(πr²).

Then we need to account for absorption in depth of the volume within the system. Actually, two shells of concentric volumes needs to be accounted for. Atmosphere and solid earth, which gives:

Which is equal to a surface temperature of 286.6 Kelvin.

When the surface emission has been found, we can use the Stefan-Boltzmann equation for radiative heat transfer from the solar constant to the surface:

But the the transferred heat from the solar constant must follow the same laws as the emitted radiation, so with the inverse square law we find:

Which gives us the effective temperature of 256.3 Kelvin.

If we use radiative heat transfer for the emitted intensity of the surface and the effective temperature:

We can see that it fits the temperature at the tropopause with satisfying accuracy.

The first transformation of solar radiation through the volume of the outer shell, using the full value of TSI, gives a result close to direct irradiation at zenith on the surface:

And using the solar constant and the effective temperature for radiative heat transfer to the system gives a good match to the total solar irradiation, direct and diffuse:

Both are logical results, as the transformation through the outer shell gives a value that should represent what arrives at the surface from the bottom of the volume of the atmosphere, and the heat transfer to the troposphere should include both what is diffused/scattered in that volume, and what directly arrives at the surface.

We now have a main structure of the system determined by only small modifications of the blackbody model, heat transfer and the inverse square law.

Now we get to the surprising part. I did not expect to find this.

I want to address the difference between the emitted effective temperature, and the true blackbody temperature of a perfect absorber and emitter of the solar constant.

If we use the solar constant to find the effective temperature that should be emitted from a perfect blackbody, we get:

The difference is:

If the difference is assumed to be the amount of work performed by the system to keep the atmosphere in place, according to the first law of thermodynamics where deltaU=Q-W, we can use this for gravity. Consider gravity as the force that acts in a point at the center of mass in a parcel of air, that has a surface area of 1m², and the mass of 1kg, laying statically above the surface of Earth. The force acting in the point at the center of mass in that parcel of air is:

If we use units of Nm^2, that are used for thermal resistance, stress and pressure, the force acting on the surface of the parcel is:

The source strength of the body emitting the power needed to raise the force radiated into the surroundings of the source according to the inverse square law is:

The source strength matches the surface emission of heat in the model. The difference between the true blackbody intensity of earth and the effectively radiated intensity, the missing heat which according to the first law must be work, is found to be equal to the force of gravity. Also, when comparing source strength, it equals exactly the emissive power of the earth surface..

The earth core is said to have a temperature of about 6000 kelvin, I assume that it has the same temperature as the surface of the sun.

If we use the same approach as for the surface and atmosphere, the internal layers decrease by 3/4 for each surface and another 3/4 for the volume above that surface, as heat travel from the core. If the core is 5780 Kelvin and there are 4 pairs of surfaces and volumes it seems to be a simple problem to solve. I´m not fully comfortable with this calculation, but it could maybe be improved with closer study of data on subsurface stratification.

Greenhouse effect? A cold fluid that heats it´s own glowing heat source by being wet and cold? How is that different to delusion? Since when does a 15C surface get hotter from -18C fluid packed with water vapor and a small fraction dry ice?

Climate scientists promoting co2-driven alarmist models are the Islamic State of science. The deceptive behaviour, manipulation of data, hiding of data, lying and oppressing sceptic voices with personal attacks and unscientific references to consensus-bullshit, is repulsive and unmatched in history. It´s about time they are held responsible for their abuse of public trust.

I think that we should stop making shit up, and use what we know is true.

There is one more way of doing it.

If we use the relationship:

There is another solution:

/LIT

Why are you saying this:

”Then we need to account for absorption in depth of the volume within the system. Actually, two shells of concentric volumes needs to be accounted for. Atmosphere and solid earth, which gives:”

Are you saying that the sunlight is absorbed by the atmosphere and passes through it to be absorbed by the surface?

GillaGilla

I write what I mean. I calculate the optimized flow in a massless cavity with no other restrictions than geometry. And as you can see, there is a reduction of heat which is close to the ill-defined greenhouse albedo. Observations agree with my results, 1000W/m^2 is a common value on almost the whole surface. (And no, I didn´t write all the time, everywhere)

My intention was to build a clean frame to the model, so I used a non-interacting spherical cavity. I was as surprised as anyone else when I saw that it matches observation perfectly. But thinking about for some time, it is logical.

GillaGilla

How does one calculate heat without a mass and dimensions? After all, heat is a measure of energy from one mass to another. Seems nonsensical to talk about heat and energy loss when there is no mass to interact with the radiation.

GillaGilla

Heat is energy in transfer. If the big bang is a real event in history, there was lots of heat and no mass. Mass is equal to energy, so there is no point in separating them when considering heat.

Heat flow is actually independent of mass. The draper point is proof of that. At 798K (practically) all solids glow. It means that the intensity of heat flow(the glow) doesn´t depend on the mass it flows through. The state of mass depends on the power of heat flow. Just think about what would happen at 0K. The properties of mass depends on the internal state, the temperature, which is equal to the amount of heat that flows through it. Check out Adrian Bejan and his constructal theory, where flows dissipate and do work on mass to optimize the flow. That is the same principle. Mass obeys heat, not the other way around. When thinking in those terms E/m=c^2 makes a whole lot more sense.

GillaGilla

I’ll also add real quick, that both temperatures in the above calculation are lower than the observed surface temperatures.

GillaGilla

So let’s start here. Easier for me to see each step one number at a time. We seem to be arriving at different numbers but without going a step at a time I’m not sure where/why that is happening.

Observed temperature of the sun is 5780K.

Luminosity (L) is equal to emissive power * surface area. While it is accurate that the sun is a sphere, for the calculation we only consider the cross section of the source that is perpendicular from our point of view. Radius of the sun is approx 696 * 10^6 m. So in this case

L(sun) = 4pi R(sun)^2 * SBc * 5780 ^ 4 = 3.85 * 10^26

The irradiance after it has traveled the distance from sun to our orbit path is calculated by dividing L(sun) by the area of a sphere with a radius of our distance from the sun (149.598 * 10^9)

I(at earth) = L(sun) / (4pi * (149.598*10^9)^2) = 1370

Next we get to radiant flux absorbed by earth. Again, while it is a sphere we are only concerned with the cross section of the earth as the absorption surface. So we multiply the irradiance * the surface area. Also, assuming that the earth is in equilibrium means that the outbound flux of the earth as a sphere = the inbound flux which will look like this

(outbound) = (inbound)

4pi R(earth)^2 * SBc * T(earth)^4 = I * pi * R(earth)^2

Now solving for T(earth) we get

T(earth) = 279K

This is the calculated temperature of earth, using SB, if the earth were a blackbody. We of course know that it’s not. This is where I’m a little confused and trying to learn and understand better myself. In physics they will talk about effective temperature. Effective temperature makes the assumption that the object emits as a blackbody. This is often used when the absorptivity/emissivity value can’t be determined. In the case of earth, the measured albedo is .3, equating to an absorptivity value of .7. So if we add in a .7 (A) for absorptivity and leave emissivity at 1, then the equation looks like

(outbound) = (inbound)

4pi R(earth)^2 * SBc * T(earth)^4 = I * pi * R(earth)^2 * A

This leads to a T(earth, low A) = 255K

So there it is. Calculating the temp of earth based on irradiance from the sun, you get an effective temp of 255K – 279K. Again, I’m not entirely understanding with the concept of assigning an emissivity of 1 to the earth while simultaneously giving it an absorptivity of .7. Kirschoff’s law states that absorptivity=emissivity. I think the catch is that a=e for all wavelengths only for gray bodies. In reality, it’s not unheard of for an object to be reflective one particular wavelength but highly absorbant to another. In the case of earth this would mean it is 30% reflective to visible but high abosorptivity, and thus high emissivity, to IR.

These numbers seem a bit different than yours. So maybe we should start here in discussing the difference in the numbers we are arriving at before moving forward on anything else.

GillaGilla

First, thank you for taking the time to have a look. Now let´s have a look at this:

I(at earth) = L(sun) / (4pi * (149.598*10^9)^2) = 1370

I use observed value of 1360.8W/m^2. But 1370 works as well if you want. For inclusion of gravity I found that 1360.8 produce an exact match to the value of surface acceleration given by NASA. But I started my investigation using 1368W/m^2. Here is a reference:

http://onlinelibrary.wiley.com/doi/10.1029/2010GL045777/full

”Next we get to radiant flux absorbed by earth. Again, while it is a sphere we are only concerned with the cross section of the earth as the absorption surface. So we multiply the irradiance * the surface area. Also, assuming that the earth is in equilibrium means that the outbound flux of the earth as a sphere = the inbound flux which will look like this…”

This is one major point of my calculations. You are not including the full geometry, that is why your calculations don´t get the right answer, and you think that the surface temperature is ”higher than it should be”. I use the shell theorem for irradiation and emission, and it is a valid method to use for these things. The accuracy of the results confirm it.

https://en.wikipedia.org/wiki/Shell_theorem

Your calculation:

4pi R(earth)^2 * SBc * T(earth)^4 = I * pi * R(earth)^2 * A

is for a perfect blackbody, emitting from a twodimensional surface, infinitely thin and perfectly black. As I point out in this post, I account for depth and volume of the sphere as well, and it produce a very accurate solution.

”In the case of earth, the measured albedo is .3, equating to an absorptivity value of .7. So if we add in a .7 (A) for absorptivity and leave emissivity at 1, then the equation looks like”

I think you should read more about the uncertainty in albedo measurements. It is not a well understood property of earth. What my calculations show is that you get 25% albedo from just geometry. Geometric albedo is a real thing. But, more importantly, variations in albedo does not regulate temperature, temperature regulates variations of albedo. Albedo is logically an effect of temperature, not a cause. The only cause is the sun, we observe the effects.

”Again, I’m not entirely understanding with the concept of assigning an emissivity of 1 to the earth while simultaneously giving it an absorptivity of .7. Kirschoff’s law states that absorptivity=emissivity. ”

That´s some good critical thinking there, keep that spirit. But it will make you lose your faith in the greenhouse;)

”I’ll also add real quick, that both temperatures in the above calculation are lower than the observed surface temperatures.”

Yours? In that case I am aware of that, I´ve done those calculations myself.

Have you read my other posts? There are not that many, and there is some additional information and explanations. For example:

https://lifeisthermal.wordpress.com/2017/06/21/electric-field-equations-gaussian-surface-gaussian-gravity/

My point of view is: the accuracy and large number of correlations I found with this approach, is way beyond what the greenhouse theory can dream of producing. The gh-theory can´t even give us correct surface temperature without claiming that dry ice in a cold fluid increase the temperature of it´s own heat source. This theory works for real, to describe the earth system. Not only heat, but it includes the force and work of gravity as well. I think it has alredy won, in terms of being functional.

GillaGilla

”This is one major point of my calculations. You are not including the full geometry, that is why your calculations don´t get the right answer, and you think that the surface temperature is ”higher than it should be”. I use the shell theorem for irradiation and emission, and it is a valid method to use for these things. The accuracy of the results confirm it.

https://en.wikipedia.org/wiki/Shell_theorem”

In physics, you always have to start with deciding whether or not your system is static (in balance) or dynamic (accelerating or changing in someway). While it would be accurate to say that the temperature of the earth varies over large periods of time, in regards to incoming and outgoing radiation, it is said to be in balance. If it weren’t in balance then we would either be a) heating up until outgoing radiation balances with incoming or b) cooling down until outgoing radiation balances with incoming. So you will have to decide whether or not you want to accept the earth to be in balance, warming due to incoming radiation exceeding outgoing, or cooling due to outgoing radiation exceeding incoming. There are only those 3 options. Since you don’t agree that outgoing surface radiation matches incoming radiation absorbed at the surface, is the surface warming or cooling? Second, systems with unchanging inputs will always seek to stabilize over time. So since the input (radiation from sun) is relatively unchanged over a long period of time we should see observable data indicating that we are trending toward a stabilized temperature, either warmer or cooler. Again which is it? I could hypothesize that at the formation of the planet temperatures would have been much much higher than currently. That means we would still be over our target equilibrium temperature, if you want to take the stance that we aren’t in equilibrium. So again, can you provide data or backing that suggests that we are on an eon long cooling streak. There have been plenty of periods of cooling and warming to suggest that we are in equilibrium.

In regards to your above quote you say that I think the surface temperature is higher than it should be. I’ve only used SB to calculate solar irradiance and turned that into an effective temperature. Please explain how this has given a higher than it should be value.

”Your calculation:

4pi R(earth)^2 * SBc * T(earth)^4 = I * pi * R(earth)^2 * A

is for a perfect blackbody, emitting from a twodimensional surface, infinitely thin and perfectly black. As I point out in this post, I account for depth and volume of the sphere as well, and it produce a very accurate solution.”

On the left side the equation is the emittance for a perfect blackbody sphere surface_area (4pi*R) * SBc * T. That is my calculation for emittance, that is the surface area for a sphere. So I can only assume that when you speak of emitting for a 2D surface you are referring to the right side, the irrdiance of the sun. When evaluating total outgoing radiation it is the product of the surface area and temp as already discussed with the calculation of earths outgoing. But if we are to measure how much of the is reaching an observer, we have to integrate over the surface area to account for the difference in the angle from perpendicular to the surface and the observer. The observer only gets the full effect from a surface when that surface is perpendicular to him. Surfaces not directly facing the observer will still emit radiation towards the observer, but at a lesser amplitude. When we integrate over the hemisphere of the sun toward the earth (observer) it equates to transmitting from a circle with the radius of the sun that is perpendicular to us. That is why the irradiance on the right side is multiplied by the surface area of the disc of the sun. No different than when we calculate the incoming radiation as being absorbed by the disc of the earth, rather than the hemisphere of the earth.

Have some comments in regards to the albedo but have to get back to work for now.

GillaGilla

Rest of my response here:

”I think you should read more about the uncertainty in albedo measurements. It is not a well understood property of earth. What my calculations show is that you get 25% albedo from just geometry. Geometric albedo is a real thing. But, more importantly, variations in albedo does not regulate temperature, temperature regulates variations of albedo. Albedo is logically an effect of temperature, not a cause. The only cause is the sun, we observe the effects.”

It’s not an exact measure. There are a huge number of variables that go into the conditions that can lead to reflection of radiation back to space. Cloud cover, snow/ice are the most common. All that said though, it is very accepted that the albedo of the earth is approx .3. The albedo is really a collection of variables and scenarios, all changing constantly, that cause reflection. But the approx value of this on average is .3.

To say that albedo isn’t well understood is a very inaccurate statement. There is plenty of available information and measurements supporting an albedo of .3 There’s an entire wiki page if you need more info:

https://en.wikipedia.org/wiki/Albedo

Geometry is one thing that could impact albedo, especially given that albedo can change with angle of incidence. Outside of that though I’m not sure what you mean by geometry. Angle of incidence will have a big impact of the albedo, to say, oceans but very little impact to fresh snow covered land.

You also say ”albedo is logically an effect of temperature.” Can you link someone that defines it as such? Because all I see is albedo is defined as a scale of how much radiation is reflected by a surface. It ranges from 0 (also known as black surface) to 1 (also known as white surface). There is no mention of temperature anywhere in it definition.

” The gh-theory can´t even give us correct surface temperature without claiming that dry ice in a cold fluid increase the temperature of it´s own heat source.”

If you place a pot of water on the stove, will it heat quicker and come to a boil faster or slower with a lid on it? The top of the lid will be the coldest part of the system, expecially after the water has began to warm. Yes this lid helps to reduce outgoing LW radiation. According to you, how is it then that the lid helps to keep in heat if it is the coolest part of the system? How does your jacket that you wear outside during the winter help keep your body warm given that the outside of the jacket is only marginally warmer than the outdoor temps? These devices help to inhibit the outbound LW radiation. This technically happens because:

1) they absorb outbound radiation

2) because they are colder their rate of emission is lower

3) for a layer of atmosphere, half of the emission heads toward space, the other half toward the ground.

Radiated energy is radiated energy. 1 W/m^2 of 15micrometer flux will warm the surface all the same, whether it came from an ice cold atmosphere or a 5800K sun.

GillaGilla