That’s the way my high school physics teacher used to title our weekly physics bonus quizzes – Phun Phriday Physics Challenge! They were a lot of fun, actually.
Phun Physics does not support Greenhouse Funny Physics
Let’s set up a classic, high school/undergraduate physics scenario that has power, flux, temperature, heat, all combined in one question:
Q: An electrical circuit containing a resistor with resistance R, has a voltage V applied to it. The resistor is cylindrical with radius r and length l with l being much larger than r. Calculate the temperature of the resistor if it has unit emissivity.
A: The power utilized by the circuit is P = V2/R. Assuming this power is dissipated by the resistor and the only means of dissipation is radiant, the surface flux F from the resistor will be F = P/A where A is the surface area of the resistor, A = 2πlr. (Since l >> r, the ends of the resistor can be ignored for simplicity; otherwise add 2π(r2 – rw2) to A where rw is the radius of the wire going into the resistor). Therefore F = V2/(2πlrR). Since the surface of the resistor has unit emissivity, then via the Stefan-Boltzmann equation F = σTR4, we have TR4 = V2/(2σπlrR) for the temperature of the resistor.
Take the fourth root of the right side to get the temperature.
The Climate Pseudophysics Greenhouse Solution
Note that the above solution doesn’t depend on the greenhouse effect of climate science. That might be a surprising statement, since, why would it? However, in climate science greenhouse argumentation the supposition is made that if you have a powered source holding an object’s temperature, and this temperature is higher than the surroundings, then if you raise the temperature of the cooler surroundings the object also has to raise in temperature in order to maintain the heat flow differential so that the object can still emit the same amount of energy supplied to it from the circuit.
Well then how would that work? If the resistor is in an ambient radiative environment like the Cosmic Microwave Background, say, except at a temperature we can specify and its related flux, then that ambient flux would have to be added in to the previous solution so that the total output from the resistor still has the differential of the original solution relative to its environment.
Therefore, the climate science solution to the above problem would be σTR4 = V2/(2πlrR) + FAmb where FAmb is the ambient radiant flux, FAmb. = σTAmb4.
Simplifying slightly: TR4 = V2/(2σπlrR) + TAmb4 is the climate science solution to the original problem, because you have to factor in its version of greenhouse effect mechanics and heat flow, which is universal.
Electrical Circuits Function at Room Temperature
Any number of electrical appliances you have on your person all day are designed to operate at room temperature. I have a Samsung Galaxy S3 running continuously, for example. How can an electrical appliance be designed to operate at room temperature? You can go look up on an electrical components supply website for any number of resistors with a specified size, power usage, operating temperature, etc.
So, according to climate science greenhouse physics, how do you get a resistor that operates at room temperature? If the electrical circuit is “running” and the resistor is at room temperature, and the ambient environment is at room temperature surrounded by walls etc. with unit emissivity, then what is the raw power input to the circuit according to climate science? Power is P = V2/R so solve for that expression out of the climate science solution: TR4 = P/(2σπlr) + TAmb4, and so P = 2σπlr(TR4 – TAmb4).
So, since the resistor is operating at ambient room temperature, or in other words, if you want to design a resistor to operate at room temperature in room temperature, you design it to utilize zero electrical power, since TR = TAmb and so the power required is P = 0.
Clear as mud?
Let’s try the converse example. We have a resistor that is pre-designed to operate at room temperature (meaning it will rise to +22C or 295K when the appropriate electrical circuit is engaged) using the original solution. However, we now have to take into account that it might be used in a room at, you guessed it, room temperature. So, in this case, the actual temperature the resistor will get to is TR4 = V2/(2σπlrR) + TAmb4 = 2TAmb4, so that TR = 351K = 78C.
So climate science says that it is impossible to get a resistor to operate at room temperature, unless you don’t apply any power to the resistor at all, in which case the electrical circuit will still somehow function even though it has zero voltage applied to it, or, unless you’re never in room temperature and always below it. Why? Because to make a resistor that operates at room temperature, in an environment at room temperature, you can’t apply any voltage to the circuit at all. Or, you can only have a resistor operate at room temperature, if the ambient environment is below room temperature. Climate science says that there is no such thing as a circuit that operates at room temperature, in an ambient environment at room temperature.
Let’s consider this again in a different way. Say we’ve selected a resistor that will rise to 30C when applied with a given voltage, and so this puts out 478 W/m2. These are the ratings if the resistor is in a 0K vacuum since climate science says that needs to be specified. But now we run the circuit inside an igloo so that there’s an ambient radiative environment at, say, -10C. Well, -10C is 263K and so this makes a radiative environment of 271 W/m2. So now, using climate science “physics”, the resistor has to reach a temperature equivalent to 478 W/m2 + 271 W/m2 which is 339K or 66C. So, by placing a warm resistor at 30C inside a cold igloo at -10C, the resistor gets to 66C.
One more scenario would be taking the resistor circuit from the last example, but placing it in an oven with a warmer ambient radiative temperature at, say, 60C, which is 697 W/m2. Again, using climate science physics, the resistor has to emit 478 W/m2 + 697 W/m2 which is 379K or 106C. So, by placing a resistor that normally operates at 30C, in an oven at 60C, the resistor will actually get to 106C.
Actually, Heat only flows from Hot to Cold
So, the correct answer to the problem being discussed is of course only the first answer, that doesn’t use climate science pseudophysics. Just think about it practically. If climate pseudophysics were true, such physics would be something we’ve always lived with, that we’ve always had to contend with, to factor in to our equations always, to mathematically account for, to utilize when creating electrical circuits and thermal power systems, etc. Climate pseudophysics, being physics, would be universal, everywhere, and part and parcel of the basis of physics education. So, the fact that we never use climate pseudophysics and are never educated on it, and have never heard about it in all these other areas of real physics which climate pseudophysics should also exist and be influencing, by default indicates it is a fraud.
I mean if you’re a physicist you can just look at the math and the solution climate pseudophysics gives for this simple problem and the paradox it creates…you cannot design a resistor or any electrical circuit that dissipates heat to operate at room temperature, in room temperature. You can only get a circuit to operate at room temperature if it is in a room cooler than room temperature. Etc.
What really happens when you put a warm resistor in a cooler environment? Physics gives the answer, not climate pseudophysics: the warmer resistor will transfer heat to the cooler environment and thereby warm it up, and that’s it. At best, the environment would warm up to the temperature of the resistor. Thermodynamics is really that simple.
Let us, again, use this example to explore the climate pseudophysics solution to a warmer resistor being in a cooler environment. First, we saw that climate pseudophysics claims that the resistor has to reach a temperature equal to its own flux output from the power supply of its circuit, plus the flux of the ambient radiative environment. So just by putting a warmer resistor in a cooler environment, it immediately gets warmer still.
But now what happens? Now the resistor will heat its environment, since the environment is cooler. This will warm up, i.e. raise the temperature of, the environment, which means the environment will have a higher ambient flux. However, now that the environment has a higher ambient flux, then the resistor has to again warm up by the equivalent amount in order to offset the environment so that it can still emit its intrinsic power relative to the new warmer environment. More briefly, as the resistor heats up something cooler, the resistor heats up by proportion.
That is, something hot becomes hotter still in the process of, and because of, heating something cooler. This process would never end, by its logic, its physics and its math. Climate pseudophysics doesn’t just give silly static answers, its answers as such actually indicate that there should be no asymptotic equalizing heat transfer anywhere in the universe, that all heat transfer should cause temperatures to run-away exponentially before “some other process” causes them to stop.
Really, Heat only flows from Hot to Cold
Going back to a previous example, what happens when you bring a cooler resistor circuit into a warmer ambient environment? Climate pseudophysics says that the cooler resistor circuit has to jump to a temperature equivalent to the circuit’s own flux output, plus the flux from the warmer ambient environment, in order to maintain the differential to its environment so that it can still emit its power. So, a cool object brought into a warmer environment becomes much warmer than the warmer environment.
What really happens? Obviously, the warmer environment will simply warm up the cooler resistor, and that’s it. At best, the resistor would warm up to the ambient environment temperature. Thermodynamics really is that simple.
Let us follow the “logic” of climate pseudophysics a little further in this example. A cool resistor circuit is brought into a warmer environment, and then it jumps up in temperature to be hotter than the warm environment. So now, the resistor is warm, and the environment is cool. And this always has to happen. With the resistor now warmer and the environment cooler, the resistor will then heat up the environment and raise its temperature. And this leads back once again to the mutual-self-heating of the previous example: as warm heats cool, warm becomes warmer by proportion because it heated cool. This is Climate Pseudophysics 101, but you won’t find it in any physics textbook in the world.
Heat and Energy not the Same
Climate pseudophysics lays its foundation in not distinguishing between heat and energy. Climate pseudophysics claims that cold must heat hot just as equally as hot heats cold, because “you can’t prevent the energy going from cool to hot”. Well actually, thermodynamics does prevent just this, because thermodynamics is about distinguishing heat and energy. Heat and energy have the same units, Joules, but they’re not quite the same thing.
Heat flow is what causes temperature to change. Heat flow can only occur if there is a temperature differential. Heat flow occurs only in the direction from higher temperature to lower temperature.
Let’s just look at the (simplified) equation for radiant heat flow:
Q = σ(Th4 – Tc4) = σTh4 – σTc4
where Q is the heat flow, and Th & Tc are the hot and cool temperatures. Look at the individual terms for that equation: σTh4 is the full energy emitted by the warmer body; σTc4 is the full energy emitted by the cooler body. These full energy outputs oppose each other with only the stronger one having some energy “left over after the battle” to continue on as heat flow to the cooler object. If the energy from the hot body is “2″ and the energy from the cool body is “1″, then there’s only “1″ left over to flow as heat (Q) from hot to cool. There is no heat that flows from cool to hot, at all. There is energy flowing from cool to hot, because the cool side balances the equation, but this energy is not heat because heat, which can cause temperature to change, is only the balance of the energy that’s left over, flowing from hot to cool.
Climate pseudophysics and the pseudoscientists who try to promote it refuse to distinguish, let alone learn, the thermodynamic definitions and concepts of heat, energy, and heat flow and how these differentiate from each other. It is a gold-mine for semantic sophistry, and a grave-yard for common-sense and basic physics.
It is very easy to use thermodynamics and physics to demonstrate that the postulate of radiative self-forcing (climate pseudophysics’ alternative version of the greenhouse effect) is incorrect, violating the definition of heat flow. It is trivial, and always has been for anyone paying attention, right from the beginning.
Let us simply use the heat flow equation!
Q = σTh4 – σTc4
So, the hot body is the body with a powered source, the resistor circuit say, that we’ve been discussing. There is no ambient cool environment, except, some process that sends the thermal radiant emission from the hot body, back to the hot body, back to itself. A mirror say. What happens? Climate pseudophysics says that the hot body heats itself up because of this radiation. What does the actual equation for heat flow say?
The second “cool” σTc4 term in the heat flow equation then becomes representative of the backradiation, in units of flux (W/m2). Well, what is the flux of the backradiation? The backradiation, by definition, from the source, at best has a flux equal to its source, which is the first-term of the equation, σTh4. What is the flux of the backradiation? At best, it is as strong as…itself. At best, the backradiation is the radiation from the source.
What does that do for heat flow? It makes Q = 0. And with no heat flow then the source, powered or not, can not increase in temperature. Thermodynamics always says that heat only flows from hot to cool, but it never actually states that heat doesn’t flow into itself – it hasn’t needed to say that explicitly, because it follows on from the previous statement by default. The climate pseudoscientists have taken a semantic cheap advantage of this.
Actually, thermodynamics does state it generally, the climate pseudophysicists simply ignore it: Heat flow can only occur if there is a temperature differential.
By definition, there is no temperature differential between an object and itself, therefore an object can not heat itself. It doesn’t matter whether the object is “powered” or not. There is no differential between a surface and its own radiation, hence no heat flow, hence no heating, hence no “backradiation greenhouse”. Of course, the “backradiation greenhouse” is not how a real greenhouse operates, even though it should. See the last post for more detail. This alternative version of the “greenhouse effect” invented by climate pseudophysics isn’t even found where it should be found, and there is no other empirical demonstration that has ever shown it.
Photons don’t Conserve like Matter
If you had a high-pressure gas line and sealed the output end into a cavity, the cavity will expand and blow up. If you shine a light into a cavity, it does nothing.
Matter, such as gas molecules, push against each other, push each other around, force each other to “find a way out”, etc. All that stuff. Fermions. Photons don’t do this. Photons pass through each other without noticing, other than some possible constructive and destructive interference. Bosons. Interference doesn’t change the temperature spectrum and energy quantity of the photons.
So what happens to the photons that might constitute “backradiation” (which is a useless pseudoscientific phrase of climate pseudophysics) bouncing around inside a cavity that would set up such a situation, that we easily(!) explained and solved with the heat flow equation?
Well they do what bosons…do. Pass through each other without caring. Bounce off the heated surface without transferring themselves back into it as heat…because they can’t, because they don’t have the energy, because their energy is the energy of their source, and so they have a zero flux differential to their source.
In practise there is no such thing as a perfect 100% reflective mirror. Some of the energy does get absorbed in and by the mirror. So 100% flux can never come back to the source anyway, and so the heat flow will always be outward (and at best could be zero). What happens in the purely unrealistic situation where there was 100% backradiation? What happens to the energy? Easy, it does exactly what is specified by the heat flow equation and what it does when backradiation is not 100%: the output energy is used to balance the radiation field inside the cavity. That is what the energy is doing. Just like the portion of energy from the hotter object which is equal to the cooler object is used to balance the cooler object, which leaves a balance to flow as heat. And remember, only the balance is heat flow. Except in this case there would be no balance left over, and there is certainly no balance left over to flow back inside as heat. The low absorptivity of the mirror also comes with a low emissivity as per Kirchhoff’s Law, and finally the mirror will conduct energy to whatever is on the other side of it, or it will radiate the energy if there is nothing on the other side of it.
I hope to see the word pseudophysics added to the lexicon. Its origin will be climate “science” and its alternative, fraudulent versions of the greenhouse effect and thermodynamics and heat flow.