## Math vs. Physics

Some people say that “It’s all in the math!”, but the truth is that scientific mathematics as a form of received knowledge, applied without reason, is faith.  You can go wildly wrong and irrational with this type of mathematics.

If you started from a purely ontological mathematical starting-point, if you somehow had this correct foundation innate instead of having to go through empiricism first, then perhaps it would be possible to derive the grand unified mathematical theory of existence based purely on a rational logical approach without ever having to look outside.  But we’re not starting there as humans and so we have all sorts of contingent empiricist mathematics and mathematical methods being passed around as received knowledge, and then not being applied with the reason of physics which is required to give that type of mathematics meaning.  We shall proceed.

For back-story on this steel greenhouse debunking, see here, which has references to posts further back on this subject.

A commentator defended the steel greenhouse by saying that the flux from the shell must be added back to the sphere as the radiation from the sphere warms up the shell.  They presented a differential equation for this, which was partially correct, so I’ll show what that equation is here and the correct way to use it.

The differential equation for the temperature of the sphere is:

1]                                    mCp*dT/dt = ∑Q’i

where ‘m’ is the mass of the sphere, Cp is its thermal capacity, T is its temperature (and is actually T(t), a function of time, but I’ll just write T instead of T(t) for simplicity), and ∑Q’i is the sum of the heat flow into the sphere.

## Case 0

Case A is the “nought” scenario with the sphere being heated by an internal source of power P0, and then with surface emission given by the Stefan-Boltzmann Law, and nothing else is around. Therefore for this case:

2]                           ∑Q’i = Q’0 = P0 – 4πRsp2*σT04

Subbing that into equation 1 we get:

3]                             mCp*dT/dt = P0 – 4πRsp2*σT04

Now, thermal equilibrium is when the rate of change of temperature goes to zero, i.e., is when the temperature stops changing, and so the equilibrium temperature is:

4]                         mCp*dT/dt = 0 = P0 – 4πRsp2*σT04

Solving for T0:

5]                              T0 = [P0/(σ4πRsp2)]1/4

So obviously, the temperature of the surface of the sphere is given by the internal power input to the sphere, spread over its surface area.

## Case 1

Now we add a shell around the sphere, and for the subsequent cases this is the scenario too.

What the steel greenhouse people want to do is to say that when the shell gets heated by the sphere, then the shell radiates flux back inward according to its temperature and this adds back directly to the Q’ going in to the sphere.  So let’s try that.

Due to the inverse square law, the flux from the sphere gets reduced along its way to the shell, and so the flux from the sphere at the shell is a heating term of σT4*(Rsp/Rsh)2.  Because the ratio of radii is less than or equal to 1, then the temperature of the shell will generally be less than that of the sphere, or equal in the case where Rsp = Rsh.  The steel greenhouse people then say that the shell provides an ambient thermal environment surrounding the sphere that adds this ambient flux back to the sphere as an additional heat energy term Q’1 = 4πRsp2*σT4*(Rsp/Rsh)2.  Therefore the summation “heat flow” at the sphere is now:

6]                 ∑Q’i = Q’0 + Q’1 = P0 – 4πRsp2*{σT4 + σT4*(Rsp/Rsh)2}

and so

7]                     mCp*dT/dt = P0 – 4πRsp2*σT4*[1 – (Rsp/Rsh)2]

And then at thermal equilibrium when the temperature T1 stops changing:

8]                 mCp*dT/dt = 0 = P0 – 4πRsp2*σT14*[1 – (Rsp/Rsh)2]

Solving for T1 for this example,

9]             T1 = [P0/(σ4πRsp2{1 – (Rsp/Rsh)2})]1/4 = T0/[1 – (Rsp/Rsh)2]1/4

Given that the denominator on the right-hand-side is less than 1, then T1 > T0 and so the presence of the shell has increased the temperature of the sphere as compared to the case without the shell…

Or so they say.  Is the physics of the math correct?  What happens when Rsh = Rsp?  What happens in the denominator on the right-hand-side of equation 9, inside the square brackets?  When Rsh approaches Rsp then their ratio goes to one, and so the whole denominator goes to zero.  This is the runaway heating thing again because as the denominator goes to zero as the shell approaches the sphere, then the sphere temperature T1 goes to infinity.

Let’s back up to equation 7 and see what’s happening.  If we get rid of the entire right-most term since the square brackets go to zero when Rsh = Rsp, then solving for dT/dt we have:

10]                                dT/dt = P0/(mCp)

which has the solution

11]                             T(t) = {P0/(mCp)}*t

which means that the temperature keeps on increasing with ‘t’ until ‘t’ goes to infinity. That’s your runaway heating, once again.

This is the steel greenhouse debunking itself, and the climate science greenhouse effect, once again.  Something went wrong in thinking that that shell returns its heat energy to the sphere as the sphere warms up the shell.  What went wrong?

## Case 2

This case isn’t essential but I want to show it to walk you through getting to the final conclusion.  We’ll do something that seems to make sense here, but again with physics we will come to understand what we did wrong with the math.

∑Q’i is about summing all the heat flows.  In the first example the sum of the heat flow is given by the power input converted to flux minus the flux output from the surface of the sphere.  Those two terms, together, is what determines the heat flow into the sphere for that situation.

Now what about when we add the shell?  Is the flux from the shell a direct heat to be added to the sphere, or do you have to account for the balance of energy flow at the sphere relative to the shell to get the heat flow from the shell into the sphere?  Heat flow is all about the balance, about the subtraction of one flux from the other, as we always see in the heat flow equation between two sources Q’ = A*(σTB4 – σTA4).

If the shell provides a flux at the sphere of σT4*(Rsp/Rsh)2, then we should consider that the sphere is already emitting the σT4 itself, and thus you should subtract these to get the actual heat flow between the sphere’s radiation and the shell’s radiation.  And so Q’1 , the balance of heat flow from the shell to the sphere, is Q’1 = 4πRsp2*{σT4*(Rsp/Rsh)2 – σT4}.  Therefore for the total:

12]                 ∑Q’i = Q’0 + Q’1 = P0 – 4πRsp2*σT4* + 4πRsp2*{σT4*(Rsp/Rsh)2 – σT4}

so that:

13]                        mCp*dT/dt = P0 – 4πRsp2*σT4* + 4πRsp2*{σT4*(Rsp/Rsh)2 – σT4}

At thermal equilibrium when the temperature T2 stops changing:

14]                        mCp*dT/dt = 0 = P0 – 4πRsp2*σT24*[2 – (Rsp/Rsh)2]

Solving for T2 for this example,

15]                  T2 = [P0/(σ4πRsp2{2 – (Rsp/Rsh)2})]1/4 = T0/[2 – (Rsp/Rsh)2]1/4

Does the physics of the math we created end up making sense?  When Rsh = Rsp, then the denominator term of the square brackets goes to 1, and so the sphere temperature is just the original temperature without a shell, T0.  Nothing stands out as a problem there.  We didn’t get infinity like we did with the standard steel greenhouse solution!  What about if Rsh is much larger than Rsp (in physics we write this as Rsh >> Rsp)?  Then the denominator just goes to 2, which would mean that when the shell interior surface is infinitely distant from the sphere, the temperature of the sphere decreases to a value less than what it was in any case without a shell, T0.  In fact the sphere temperature would decrease from T0 as a function of the shell’s distance.  So now this makes no sense because the sphere temperature should be at least T0 at all times, indicating again that we must have done something wrong, even though we thought we were accounting for the heat fluxes correctly this time.

## The Solution

We haven’t accounted for the heat fluxes correctly yet.  Really, you can think of sensible heat for this exercise.  Empiricism is allowed.  If you face your hand to a block of ice, does the flux from the ice add to the energy from your hand and make your hand warmer?  All of this is about heat flow, and heat flow is about heat!  Heat feels hot!  That’s when it’s heating you!  Cold feels cold.  And that’s when you’re heating the cold.  Does the cold ice add heat to your hand?  Does having the ice there make your hand have to make itself hotter?  No.  For goodness’ sake, no.

So does the cooler ambient temperature of the shell add heat to the warmer sphere?  Is there a heat term coming at all from the shell going into the sphere?  No, there is no heat at all coming from the cooler shell going into the warmer sphere.  Because heat doesn’t flow from cool to hot.  A warmer thing does not feel or receive any heat energy at all from a cooler thing, thus a warmer thing doesn’t become warmer from a cooler thing nor does it have to make itself hotter because of the presence of a cooler thing.

The temperature that the shell develops due to the flux upon it from the sphere is simply

16]                                             Tsh = Tsp*(Rsp/Rsh)1/2

which is a cooler temperature than the sphere, due to the diminution of the sphere’s flux with distance.  If the shell is zero distance from the sphere, then Rsp = Rsh and the shell and the sphere have the same temperature.

The energy that the shell receives on its inside from the sphere goes into heating up the shell.  Whether the shell has a negligible thermal mass or not, the energy from the sphere that goes into heating the shell and sustains its temperature subsequently appears on the outside of the shell and then leaves at the exact same rate that it comes in when its thermal equilibrium is reached.

If you’re a good physicist when you solve this problem, you go through this thought-process:

First, what is the source of heat for the sphere?  Oh right, it’s the internal source with power Po.  That’s what heats the sphere.  Great.  Okay, there’s a shell too, but it doesn’t have an internal power source and is passive, and is absolute zero until the sphere heats it up.  So the shell isn’t a source of heat…ok great.  The shell will get heated up by the sphere then, from the sphere’s power output.  Actually, the shell is really just the ultimate exterior surface for the power source Po, so the shell will emit the full power from the source, and have a surface temperature according to that.  What about the sphere inside?  Well it too just emits the power from the source, but it has a smaller surface area and so will have a higher temperature according to that, and receives no heat from the shell.  The sphere gets a temperature from the power source, and then the shell absorbs the radiation from the thermal emission from the sphere.  The shell then gets heated up to the corresponding temperature of the flux that it receives, and then transfers that energy out to its exterior, where it fully emits all of the power from the power source it receives from the sphere.

That’s what a physicist would do.  That is what is intuitive, rational, logical, and doesn’t violate basic thermodynamic laws.

What these mathematical physicists are doing is accounting for all of the numbers that they can think of, but they’re not applying or considering any physical principles to their nature.  And that’s the cardinal sin if you’re a physicist.

Does cool air conductively heat the ground?  Does cool air make the warmer ground warmer still because of contact conduction?  Of course not.  All modes of heat transfer obey the conditions of thermodynamics in the same way.  Neither does radiation from a cooler object add any heat at all to a warmer object.

Their math leads to its own physically ridiculous results.  The steel greenhouse, and by extension the climate science greenhouse effect, is debunked.

## The True Problem

As Lyndon LaRouche always says, the pure, simple-minded mathematical approach isn’t sufficient, and is actually wrong!  No, this is not because mathematics is necessarily wrong, but it is because the human mind is limited in its grasp of ontological mathematics, i.e., of the true ontological nature of the behaviour of reality.  This is why we have the physical principles known as the laws of science, to guide our thinking, and is why the physicist reasons their way through a problem at a level beyond a simple-minded accounting of numbers.  Even with the ultimate ontological mathematical foundation, ideological “scientists” call still muck things up by arbitrarily inserting terms or improperly accounting for them, and then making up specious arguments to justify it.

Something very important has been exposed with climate alarm and its greenhouse effect regarding the nature of the scientific establishment and even its very method.  Specialization allows for, it creates a rich opportunity for, ideological groups to create simulacra, i.e. the simulated appearance of, an important branch of science.  And it receives all possible support in that peer-review can only occur within the speciality.

Because so few people of ‘the masses’ are able to penetrate and understand or critically assess the unique technical jargon from a supposed field of science – your average politician least of all! – this provides a lever to hijack governmental policy on pretty much any scale one might wish it.

I was asked how specialization can be used to hijack government.  Well, climate alarm is the example, but for an example of how to do it:

dU/dt + v*dU/dx = u*(d^2)U/dx^2 + Q(x,t)

and this equation means that we have a really bad problem with the water supply. It models the growth of a new strain of bacteria that we’ve recently found in the water supply, that doesn’t respond to conventional antibiotic treatments, probably because the water supply has had so many antibiotics flushed through it over the years, and so this strain has become immune. We’ve all heard of this problem of ineffective antibiotic treatments due to the evolution of bacteria. This poses a threat not only to our own health, but the health of our children and future generations. The science is very solid on this, and the math is known. The community has found the scientific consensus, and if we don’t begin to act now, our water supply, the stuff we need to live, may become totally undrinkable. We need a national-level policy to attack this problem, not a municipal/county-level approach. Our water-supply system is infected with this bacteria across the country…possibly even other countries. Federal funds are needed so that all communities can benefit from the treatment solutions.

That took me about two minutes to write.  Now imagine if there was a huge team of people and this is all they had to do, for decades, because they’re a bunch of disgusting ****ing parasites.  Or maybe they’re providing the opportunity to overcome…whatever.

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### 18 Responses to Gift of the Steel Greenhouse Keeps Giving

1. CW says:

JP: Excellent post! As an example of an organization that has within it a specialized scientific group of so-called climate specialists, which receives tremendous financial support, is the University of East Anglia. Ben Santer hails from this climate science haven of specialists. Unfortunately, MIT also produces similar specialized scientists, with support from government. There has been a wave over numerous years of professors and researchers (that have a vested interest in the GHE/AGW political agenda) from England and other Euro Institutions flowing into universities and other governmental agencies in the west. I find them on numerous blogs, all supporting the GHE. There are over 26,000 environmental groups, most of which push this con onto citizens everyday. Yahoo, Google, AOL, newspapers, NPR, and mass media in general constantly bombard the public with climate doom. Mankind has entered into another “Dark Age” of human knowledge, where scientific reasoning has been hijacked by individuals/organizations/governments to instill fear and helplessness onto a public that has limited understanding of the epistemology of events. Sad commentary on our human quest for truth based on rational, mathematical, physical understanding of the universe.

2. Yep. It can only be repeated, as Tim noted:

“When we allow science to become political then we are lost. We will enter the internet version of the Dark Ages, an era of stifling fears and wild prejudices, transmitted to people who don’t know any better.”

We may indeed be entering a new Dark Age.

3. Rosco says:

The really funny thing is that centuries ago Pictet conducted an experiment in radiation using polished concave metal mirrors. Each mirror was placed far enough from each other such that conduction/convection played little part in the experiment.

He initially placed a hot object at one focus and a thermoscope at the other focus.

The thermoscope indicated a temperature increase due to IR. He allowed for equilibrium with room temperature to re-establish.

He then placed a cold object – a container of snow/ice at the focus and – viola –

the thermoscope indicated a temperature decrease.

He introduced a “new” energy source – the flask of snow/ice – emitting say 235 W/sqm at say minus 18 C so remote from the thermoscope that the low conductivity of air played no part in either the hot or cold energy transfer BUT he PROVED that cold does not heat hot – quite the reverse occurred.

Modern physics explains this phenomenon by claiming the mirror screens the thermoscope from say up to 50% of the radiation from other objects in the room and the thermoscope is then radiating more to the flask than it gets back from the flask and hence decreases in temperature.

This explanation sounds reasonable but fails to account for the fact that the air around the thermoscope was unable to maintain equilibrium with the thermoscope ?

BUT any way you care to spin this the radiation from a cold object always results in a decrease in temperature of a hotter object.

This has been known for centuries until climate science invented their fantasies.

4. Ed Bo says:

[JP: Guys check this comment out. Check out how much these people will simply lie bald-faced!]

Joe:

You make the mistake that you accuse others of making: conserving flux, not total power.

[JP: From the OP: “Due to the inverse square law, the flux from the sphere gets reduced along its way to the shell, and so the flux from the sphere at the shell is a heating term of σT^4*(Rsp/Rsh)^2.” Thus, flux was used with the inverse square law, and I was not trying to conserve it like those others. Are you lying on purpose or just confused?]

If you are dealing with average flux and a constant area, it is a valid simplification. But you are analyzing the case of the shell having a larger area than the sphere, so it doesn’t work.

[JP: From the OP: “Because the ratio of radii [Rsp/Rsh] is less than or equal to 1, then the temperature of the shell will generally be less than that of the sphere.” The varying radii are right there! The area of neither is a necessary constant.]

Also, you believe that the shell will have a lower temperature than the sphere due to its larger surface area radiating the same total power. Yet in your analysis, you use the same temperature for the shell as the sphere.

[JP: From the OP: “The temperature that the shell develops due to the flux upon it from the sphere is simply Tsh = Tsp*(Rsp/Rsh)^(1/2), which is a cooler temperature than the sphere, due to the diminution of the sphere’s flux with distance“. Tsh = Tsp*(Rsp/Rsh)^(1/2) IS a cooler temperature than the sphere! I don’t use the same temperature for the shell as the sphere.]

These basic mistakes negate your whole analysis.

[JP: You are either lying on purpose, or not at the level of mathematical understanding I have been presuming. My apologies if the latter. The OP speaks for itself as do your comments here.]

5. Ed Bo says:

You can’t even see your own mistakes when they are explicitly pointed out to you. Look at your Equation 5. If Tshell is different from Tsphere, the whole expression for Tshell should be INSIDE the quantity taken to the 4th power. You have it outside. You fail basic algebra.

[JP: Equation 5 is T0 = [P0/(σ4πRsp2)]1/4. From the OP: “the temperature of the surface of the sphere is given by the internal power input to the sphere, spread over its surface area.” It is about the temperature of the sphere. Maybe you missed what was going on there.]

6. John Francis says:

Joe, I agree with your analysis, but please stop the name-calling and profanity.

7. Derek Alker says:

I went fishing. It helped. Didn’t catch owt though… LOL.

Ed Bo wrote –
“Also, you believe that the shell will have a lower temperature than the sphere due to its larger surface area radiating the same total power. Yet in your analysis, you use the same temperature for the shell as the sphere.”

LOL. Wonderful, but a dishonest, and I can only assume a deliberately misleading use of words by “Ed Bo”…

“Ed Bo” – “YOU BELIEVE the shell will have a lower temperature than the sphere due to it’s larger surface area.”
No, that is THE CASE unless one wants to create energy… This is WHY a two parallel plane model CAN NOT be used. It can not describe the actual situation. That, I assume is why Joe used the diagrams he produced in this thread,
https://climateofsophistry.com/2014/11/18/the-pseudoscientific-steel-greenhouse-debunks-the-climate-greenhouse-effect/
to clearly illustrate the two lines ARE NOT, and can not be of equal length.

“Ed Bo” – “Yet in your analysis, YOU USE the same temperature for the shell as the sphere.”
No, Joe REPORTS what the “theory” states is supposed to be the case, but it can not be, as the maths and physics used actually shows, as Joe shows how it should be CORRECTLY calculated. If done appropriately the reported answer IS NOT the answer the physics and maths SHOULD HAVE GIVEN…

Blinded faith, sophistry, deliberately lying, whichever, “Ed Bo” is certainly NOT CORRECT in his “interpretation” of how the physics and maths should be calculated, and Joe is CORRECT. Therefore the “theory” fails, by it’s own logic, physics, and maths. Not that people like “Ed Bo”, whoever he really is, want to see or admit that, ie, the obvious.

Why do people as obviously as “intelligent” as “Ed Bo” not want to see the obvious, why do they not want to admit the obvious??? There is THE question, and the answer probably has something to do with the fact that these people rarely, if ever, use their actual names. Dishonest? It is difficult to reach any other conclusion…

8. johnmarshall says:

Very good Joe, thanks.
There is a slight problem in that you have stated that heat radiates. this is not true in that heat is an intrinsic property of matter which has mass. Radiation is through photons which are massless. They cannot hold heat but can hold energy through vibration in that they adsorbe IR from a body above absolute zero. This energy flux can only add to other energy fluxes if it is at a higher level to the receiver and subtract from those at a higher level. ie., the GHE cannot work.

I think I’ve explained my thoughts correctly but comment will tell I expect.

9. OK John I’ll edit. I don’t mind language like that myself…I read it as righteous indignation.

10. Ed Bo says:

I repeat: You cannot see your own errors when they are repeatedly pointed out to you.

You say that Tsh = Tsp * (Rsp^2/Rsh^2)

[JP: That’s not what I have actually. At Equation 16 it lists: Tsh = Tsp*(Rsp/Rsh)1/2.]

But when you express the flux from the shell in the last term of Eqn 5, it should be:

sigma * Tsh^4 = sigma * [Tsp * (Rsp^2/Rsh^2)]^4

That is not what you have.

[JP: You forgot the fourth root on the bracketed term in your first equation. Thus: σTsh4 = σ[Tsp*(Rsp/Rsh)1/2]4 = σTsp4*(Rsp/Rsh)2 which is indeed what is seen at Equation 6 (your 5 meant 6…my mistake, I listed Equation 5 two times in a row).

But that is small potatoes. As I have already pointed out to you, if there is only a small difference in temperature between sphere and shell when separated by a vacuum, the net outward radiative flux of sigma * (Tsp^4 – Tsh^4) is orders of magnitude too small to keep the system in steady state, rejecting as much power as is input to it.

[JP: The difference in temperature between the sphere and the shell is given by Tsp – Tsh = Tsp – Tsp*(Rsp/Rsh)1/2 = Tsp*[1 – (Rsp/Rsh)1/2]. The shell’s temperature is maintained by the sphere once steady state is reached, without the sphere having to warm up as no heat energy is of course returned to it from the cooler shell, and the shell loses all the energy it receives from the sphere on its own exterior.]

11. Ed Bo says:

Joe:

Your analysis still doesn’t come anywhere near working. For a constant-power input in the sphere, you say that the surface temperature of the shell would be slightly lower than that of an uncovered sphere due to its slightly larger radius. So far, so good.

If a blackbody sphere radiating to ~0K a flux of 235 W/m^2, it would have a temperature of 253.7K. A shell with 0.1% larger radius (1.001x), as with one 6.4km above a 6400km radius sphere, with the same internal power (not flux) would have a slightly lower flux due to the larger surface area, and therefore a slightly lower temperature – in these particulars, 253.6K.

You maintain that the sphere would not increase in temperature due to the presence of the shell between it and space. With a vacuum between sphere and shell, radiative heat transfer is the only mode possible. The equation for the flux (which you have used often) is:

Q’ = sigma * (Tsphere^4 – Tshell^4) = 5.67E-8 * (253.7^4 – 253.6^4) = 0.37 W/m^2

This is the only method that the sphere has for rejecting the power it receives. We know that it must output 235 W/m^2 over its entire surface area for it to reject as much power as it receives, so this is not even close (and not explainable by any R1/R2 ratio).

If we start with the sphere and the shell at these temperatures, since the sphere is outputting far less power than it is taking in, its internal energy and therefore its temperature will increase until the Q’ output in the above equation hits the 235 W/m^2 input it receives from the source.

[JP: Equation 9 gives the answer for your idea here, according to your own math, and results in: T1 = T0/[1 – (1/1.0001)2]1/4 = T0*8.41. If T0 = 253.7, then T1 = 2133K…almost 2000 degree Celsius. If physics actually worked that way, then we would exploit that already and use it in steam-power generation stations and other applications. Imagine using a heat source at -18C and it generating 2000C. Reality doesn’t work that way.

Q’ does not represent a conservation of energy term. It only represent the heat flow. Conservation of energy is found by finding the output of the whole system, which is the emission from the exterior surface of the shell, and of course that surface emits all of the energy that it receives on its interior. The power from the core is conserved because the outside of the shell emits fully. Again, Q’ is not the term to use for conservation of energy, because the energy of the system is not conserved between the sphere and the shell, but to the outside entirely.

And in fact, Q’ is zero at the position of the interior of the shell, not the non-zero value you calculated, because the flux that it receives there from the core is equal to the temperature that has at equilibrium been induced on the interior of the shell. You didn’t factor in the difference of surface areas in your calculation, and you specified a difference in surface area between the sphere and shell (the 1.0001x value). At the point of thermal equilibrium when there is no more heat flow, it is simply energy transfer – the power that is lost by the shell on the outside is continually replaced by the power it receives from the sphere on the inside, and that is all of the sphere’s power. The vacuum between is simply the medium for energy transport. Q’ is zero from the sphere to the shell at the position of the interior of the shell when equilibrium is reached.

Flux from the sphere at the shell is: σT04*(Rsp/Rsh)2. Flux from the interior of the shell at the shell is σT14 = σ[T0*(Rsp/Rsh)1/2]4 = σT04*(Rsp/Rsh)2. And the difference between those two results is zero. No heat flows, which is the meaning of thermal equilibrium. The power however is totally conserved by it passing through the system and appearing fully on the outside of the shell.]

12. David Appell says:

Ed Bo says:
“You can’t even see your own mistakes when they are explicitly pointed out to you. Look at your Equation 5.”

The error is apparent before equation 5 — the units in equation 2 don’t balance. The units of the left-hand side are Joules/second (= Watts), while the units of the right-hand side are Watts/square-meter.

The RHS is energy flux (per unit area), but the LHS isn’t. This error carries through the rest of the equations.

[JP: The area is right there in Equation 2, it is under P0, and the area then follows through the rest of the equations like that. It doesn’t change any of the results.]

13. Ed Bo says:

Joe:

In equation [6], you simply ASSUME that Tsphere is the same with the shell as without. You don’t have the information to do that. Given the constraints of the problem, you can calculate Tshell – which we agree is slightly lower than what the temperature of the uncovered sphere would be, due to its slightly larger surface area.

[JP: Ed, in equation 6, the calculation is for the proposed increase in temperature of the sphere due to the shell’s presence using your math. The temperature is therefore obviously not being assumed constant. Your proposed solution for this debunks itself with its infinite result when the shell approaches the sphere.

Do you guys see how these frauds like Ed Bo work? They just make things up that make no sense whatsoever, and don’t even follow what was actually ever said or done. They invent a scenario of a complete reinterpretation or alternative reading of what was actually written, hope no one will notice, and then create Straw Men and Red Herrings with that sophistry. Isn’t that strange? In fact, isn’t that what all of climate science is about?]

But starting from this value of Tshell, you can only calculate Tsphere using energy balance calculations. This is where you make your next big mistake. You get completely confused between static equilibrium, where there is no heat flow and everything is at the same temperature, and dynamic steady state conditions, where there are constant heat flows and temperature differentials.

[JP: We start with the power supplied to the sphere, its effect on Tshell, and then calculate this effect back on Tsphere, using your proposed energy balance equations. That’s what was done, just as you asked, and it debunked the steel greenhouse and climate greenhouse effect with its infinite result when the shell approaches the sphere.

When thermal equilibrium is reached, then temperature stops changing, and Q’ = 0 between the sphere and the shell. The power flows through the system internally and is emitted as heat on the outside of the shell.]

For reasons I can’t fathom, you believe that a sphere without a shell must output as much power as it receives to stay in steady-state conditions, but once it is surrounded by a shell, it does not need to output any power at all in response to its input power to stay in thermal equilibrium. It makes no sense at all.

[JP: Another Straw Man, because nowhere is it said or implied that the sphere doesn’t output any power at all. It is stated explicitly that the internal power from the sphere gets emitted on the outside of the shell. The power from the sphere is what maintains the temperature of the shell when thermal equilibrium is reached; there is a steady flow of power. The flow of power goes from inside sphere, to surface of sphere, across the gap, into the shell, out of the exterior of the shell. The outside of the shell emits all of the power from the sphere, and so obviously the sphere is continually providing its power to the system. No heat flows back to the sphere from the shell because the shell is cooler, and hence, the sphere can’t get a higher temperature. This is really basic thermal analysis here Ed.]

You have created a scenario where the shell is receiving zero power from the sphere and outputting almost 235 W/m^2 to space and you maintain that its temperature is constant. Really?

[JP: Uhh, no, the shell is receiving full power from the sphere because the shell emits on its exterior; for the shell to not drop in temperature due to its external emission requires it to be constantly powered from the inside on its interior, which it is by the sphere.]

One of the first things you learn in an introductory thermodynamics class – nothing to do with climate – is that you can perform energy balance calculations on any system, or any subsystem within your system. This is simply a way of stating that the 1st LoT is universal – you cannot carve out exceptions like you have.

[JP: That’s quite funny. You’re trying to carve out an exception for the law of heat transfer, by having the cooler shell heat up the warmer sphere, and you’re still trying at it even after your own math for that proposition debunks itself with spurious infinite results when the shell approaches the sphere.

“First law of thermodynamics: When energy passes, as work, as heat, or with matter, into or out from a system, its internal energy changes in accord with the law of conservation of energy.”

Does the shell pass energy as either work, heat, or matter, to the sphere?

No.

Therefore I’m with the 1st LoT, and you are not. You suggesting that the shell causes the sphere to increase in temperature is a clear & direct violation of the 1st LoT. You, and climate science with its greenhouse effect, are the ones attempting to carve out an exception to the 1st Law of Thermodynamics.]