... I have repeatedly demonstrated that this person who calls himself "kayman80" has been blatantly dishonest about past conversations that have occurred here on Slashdot and elsewhere. And that he has a habit of deliberately distorting what other people say, for reasons of his own. I have ceased feeding the troll. I recommend that you do so as well. [Jane Q. Public, 2014-10-13]

Instead of endlessly accusing me of blatantly dishonest deliberate distortions, a real skeptic would write down an energy conservation equation before wrongly "cancelling" terms. Actually, a real skeptic would've done that months ago, but better late than never.

Jane, I believe in you. I believe you can learn how conservation of energy works, but first you have to take a baby step of your own by writing down an energy conservation equation before wrongly "cancelling" terms. Just try it. You might learn something. On the other hand, endlessly accusing me of dishonesty probably isn't very educational.

I haven't expended ANY energy to avoid writing anything down. I've written down the proper and necessary equations not just once but many times now. [Jane Q. Public, 2014-10-13]

Ironically, Jane's still trying hard to avoid writing down his energy conservation equation before wrongly "cancelling" terms. If he'd try to write down that equation just once, he might realize that the nonsensical equation he's written down many times isn't proper or necessary.

I don't need to write down a "conservation of energy equation" in regard to Spencer's experiment. I don't refuse to do it because I can't, as you have clearly implied. I refuse to do it because this is a dead issue. You were proved wrong weeks ago, and your demands for additional proof from me are just laughable. [Jane Q. Public, 2014-10-13]

If Jane actually could write down an energy conservation equation before wrongly "cancelling" terms, Jane would see that "radiative power from the walls" can't cancel out.

Once again, the only way Jane's final term could cancel with the radiative power in term "(e*s)*T4^4" to obtain Jane's final equation would be if "radiative power from chamber walls, re-emitted back out" equals "(e*s)*T4^4". But it's being emitted by the source, which is at temperature T1. If reflections confuse you, just remember that the gray body equation has to reduce to the black body equation where there aren't any reflections at all. In that case, all that power is being absorbed and re-emitted, not reflected.

If Jane would write down an energy conservation equation and think about it, he might realize that he's been endlessly crowing about "proving me wrong" using Sky Dragon Slayer nonsense that violates conservation of energy and/or the Stefan-Boltzmann law.

But since Jane's Slayer brainwashing is so thorough that he can't bring himself to write down that equation, Jane will probably keep endlessly crowing about "proving me wrong".

Ironically, if Jane's Slayer nonsense was right, Jane would also have "proven wrong" Prof. Brown, Dr. Joel Shore, the National Academies of Science, the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, the European Physical Society, etc.

... YOU are the one going against "established" physics here. ... If you could actually show how the physics textbook idea of heat transfer was wrong, you would be world famous by now. ... [Jane Q. Public, 2014-10-06]

No, I'd have to get in line behind all those other physicists who agree that adding CO2 warms Earth's surface, which is equivalent to saying that enclosing a heat source warms it. This is probably the most fascinating part of Jane's delusion. Not only does Jane completely misunderstand fundamental physics, Jane seems to earnestly believe that his crackpot Slayer conspiracy theory represents "established" physics. Fascinating.

Jane's power in = electrical heating power + radiative power in from chamber walls

NO, it doesn't, and fucking well STOP claiming that it is. If YOU want to assert that, go ahead, but stop putting my name on it. I did not say that, and I do not say that, so stop putting my name on it. DO YOU UNDERSTAND??? Holy fuck, you're a dimwit. [Jane Q. Public, 2014-10-13]

Oh, okay. So Jane completely denies that the chamber walls emit radiation in through a boundary around the source. ...

Jane objected:

NO, I VERY CLEARLY AND REPEATEDLY EXPLAINED THAT I DENY NO SUCH THING. I don't have any patience for your lying anymore. Goodbye. I will record any responses, at least for a while, but I won't reply. Jesus, you're an ass. I mean the most incredible ass I've ever had the misfortune to meet online. I mean that very, very sincerely. [Jane Q. Public, 2014-10-13]

Only if "VERY CLEARLY AND REPEATEDLY EXPLAINED" means that Jane/Lonny Eachus clearly and repeatedly explained that he would never write down an energy conservation equation, and that only dimwits would claim that Jane thinks that "radiative power in from chamber walls" should be included in "Jane's power in".

It's fascinating how much effort Jane/Lonny Eachus has expended just to avoid writing down the power flowing into and out of a boundary around the heat source. If Jane/Lonny Eachus is so sure that he's right, why not just write down that obviously correct energy conservation equation?

No. Once again, gray body equations have to reduce to black body equations where there are no reflections.

No, they don't, because you still have an emissivity (which is the same as absorptivity, in a gray body). You're trying to have it both ways again. There is also a "scattering" term which you're ignoring, which is not the same as reflection. [Jane Q. Public, 2014-10-13]

Gray bodies have emissivities between 0 and 1. So black bodies are one limit of gray bodies. Black bodies don't scatter or reflect radiation, they only absorb it.

Jane's power in = electrical heating power + radiative power in from chamber walls

NO, it doesn't, and fucking well STOP claiming that it is. If YOU want to assert that, go ahead, but stop putting my name on it. I did not say that, and I do not say that, so stop putting my name on it. DO YOU UNDERSTAND??? Holy fuck, you're a dimwit. [Jane Q. Public, 2014-10-13]

Oh, okay. So Jane completely denies that the chamber walls emit radiation in through a boundary around the source. That's what I suspected from the beginning, but Jane kept coyly saying things like this:

... net radiative power from the wall THROUGH A BOUNDARY between the heat source and the wall would be zero... the net POWER IN from the wall across that boundary is zero... radiative power from the walls through that boundary cancels itself out, leaving a net across that boundary from the wall = 0. ... [Jane Q. Public, 2014-10-11]

... net radiation across the boundary from the wall is zero, because it just goes right back out. ... [Jane Q. Public, 2014-10-11]

These statements made me think Jane was rational enough to see that "power in" through a boundary around the source would have to include radiative power from the chamber walls. Jane just seemed to wrongly think it cancelled, because Jane kept refusing to write down the energy conservation equation. But now Jane completely denies that radiation from the chamber walls passes in through a boundary around the heat source.

Fascinating.

Jane is saying what Jane already actually said, not this distorted nonsense of yours. [Jane Q. Public, 2014-10-13]

Jane already actually said:

... net radiative power from the wall THROUGH A BOUNDARY between the heat source and the wall would be zero... the net POWER IN from the wall across that boundary is zero... radiative power from the walls through that boundary cancels itself out, leaving a net across that boundary from the wall = 0. ... [Jane Q. Public, 2014-10-11]

... net radiation across the boundary from the wall is zero, because it just goes right back out. ... [Jane Q. Public, 2014-10-11]

Jane's already actually described the following energy conservation equation:

Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

If Jane didn't mean to describe that equation, Jane would've written down his actual equation. But Jane hasn't, because Jane can't.

... none (NET) is absorbed in the first place. Again, this is what I have been saying all along. It is reflected or scattered. As I have stated before, this is a requirement of the Second Law of Thermodynamics. ... [Jane Q. Public, 2014-10-13]

No. Once again, gray body equations have to reduce to black body equations where there are no reflections.

... Are you trying to claim that radiation from the chamber walls is absorbed, and NOT re-emitted? ... [Jane Q. Public, 2014-10-13]

No, I'm claiming that any real physicist would write down an energy conservation before wrongly "cancelling" terms. Jane refuses to do that, so:

Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

... radiation inward from the chamber walls does cancel, because it is reflected or scattered and goes right back out. (A small part of it misses the inner sphere completely.) So no matter how you look at it, it is still a zero sum. ... [Jane Q. Public, 2014-10-13]

No, Jane. Once again, gray body equations have to reduce to black body equations where there are no reflections.

Once again, if Jane would simply write down his energy conservation equation before wrongly "cancelling" terms, he would see that they can't cancel.

You've tried to claim that POWER IN to the heat source is somehow magically dependent on the chamber walls. And the justification you gave for this was a heat transfer equation, as I described above. [Jane Q. Public, 2014-10-13]

It's physics, not magic. Radiation from the chamber walls passes in through a boundary around the heat source:
power in = electrical heating power + radiative power in from the chamber walls
power out = radiative power out from the heat source

Jane, however, seems to say this:

Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

Jane's saying that "radiative power from the walls through that boundary cancels itself out" so Jane claims those terms cancel to produce Jane's final energy conservation equation:

Jane's power in = electrical heating power
Jane's power out = radiative power out from source

The only way Jane's final term could cancel with the radiative power in term "(e*s)*T4^4" to obtain Jane's final equation would be if "radiative power from chamber walls, re-emitted back out" equals "(e*s)*T4^4". But it's being emitted by the source, which is at temperature T1. If reflections confuse you, just remember that the gray body equation has to reduce to the black body equation where there aren't any reflections at all. In that case, all that power is being absorbed and re-emitted, not reflected.

The acknowledged formula for finding radiative power from temperature is just (sigma epsilon)T^4. There are no other factors involved... [Jane Q. Public, 2014-09-05]

That's why radiation re-emitted by the source at temperature T1 is (e*s)*T1^4. There are no other factors involved. The source can't re-emit radiation at (e*s)*T4^4, so those terms in Jane's equation can't cancel. And the last term double-counts radiation emitted by the source, so it's zero.

You insist that the radiant power output calculation of the heat source has to take into account the cooler temperature of the chamber walls. [Jane Q. Public, 2014-10-13]

No, I've repeatedly agreed that radiative power out only depends on emissivity and temperature.

Once again, I'm just saying that "radiative power out" is different than "electrical heating power".

That's why Jane will never write down his "energy conservation" equation before wrongly "cancelling" terms. If Jane ever did, he'd have to face the fact that Jane's terms don't cancel.

How ironic. Jane's terms would only cancel if radiant power output of the heat source took into account the cooler temperature of the chamber walls. But that's impossible because it would violate the Stefan-Boltzmann law.

That's ridiculous, Jane. Notice that "net radiative power out" equals negative "net radiative power in". Since Jane seems to agree that "net radiative power out" is positive, "net radiative power in" can't be zero. It has to be negative, which just means more radiative power is flowing out than flowing in.

Now you've just gone off the deep end. And by "deep end" I mean the deep end of the pit full of BS you've dug yourself. Just no. Any spherical boundary you draw within this system has additional input: your vaunted electrical power. I'm amazed that you finally got so caught up in your own bullshit that you made a mistake quite THAT fundamental. Get stuffed, troll. For that and actually quite a pile of other reasons that have built up over time, I still don't believe you're a real physicist. [Jane Q. Public, 2014-10-12]

Electrical power isn't radiative power, so it wouldn't be included in net radiative power.

... I have written down all I need to write down to answer Spencer's challenge. I solved for the correct temperature, and showed your own answer to be utterly wrong. ... [Jane Q. Public, 2014-10-11]

Once again, Jane's solution halved the electrical heating power. Jane didn't notice this because he calculated net transfer incorrectly, which led him to the absurd conclusion that Jane was only off by about 0.1% when Jane was actually off by ~100%.

So Jane hasn't written down all he needs to give the correct answer to Spencer's challenge. To give the correct answer, Jane has to draw a boundary around the heat source:
power in = electrical heating power + radiative power in from chamber walls
power out = radiative power out from source

This is the same answer that Prof. Brown and Dr. Joel Shore tried to explain to Jane. It's also the same answer that underlies the positions taken by the National Academies of Science, the American Institute of Physics, the American Physical Society, the Australian Institute of Physics, and the European Physical Society, etc.

... YOU are the one going against "established" physics here. ... If you could actually show how the physics textbook idea of heat transfer was wrong, you would be world famous by now. ... [Jane Q. Public, 2014-10-06]

No, I'd have to get in line behind all those other physicists who agree that adding CO2 warms Earth's surface, which is equivalent to saying that enclosing a heat source warms it. This is probably the most fascinating part of Jane's delusion. Not only does Jane completely misunderstand fundamental physics, Jane seems to earnestly believe that his crackpot Slayer conspiracy theory represents "established" physics. Fascinating.

Maybe that's why Jane won't even take a few seconds to write down an energy conservation equation before he wrongly "cancels" terms. Deep down, maybe Jane suspects that he's wrong and mainstream physicists are right.

... net radiative power from the wall THROUGH A BOUNDARY between the heat source and the wall would be zero... the net POWER IN from the wall across that boundary is zero... I did NOT claim the net radiative power through the boundary was zero. What I wrote was that the radiative power from the walls through that boundary cancels itself out, leaving a net across that boundary from the wall = 0. ... The actual radiant power through that boundary isn't zero, because the radiant power output of the heat source still goes through it of course. Leaving a NET positive transfer of energy OUTWARD through the boundary. ... [Jane Q. Public, 2014-10-11]

This is just more dishonest out-of-context nonsense again. I clearly told you that the context of my statement was radiation from the walls through a boundary. I did not claim the net radiation across that boundary was zero. I claimed the net radiation across the boundary from the wall is zero, because it just goes right back out. This is a wonderful example of how you distort context, in order to make it appear someone else is saying something they actually did not. That is a form of lying. [Jane Q. Public, 2014-10-11]

Since Jane keeps bolding "from the wall" and claiming that "radiative power from the walls through that boundary cancels itself out," Jane seems to be saying:

Jane's power in = electrical heating power + radiative power in from chamber walls
Jane's power out = radiative power out from source + radiative power from chamber walls, re-emitted back out

My energy conservation equation is this: electrical power in = (epsilon * sigma) * T^4 * area = radiant power out [Jane Q. Public, 2014-10-08]

It certainly seems like that's what Jane's saying. If "radiative power from the walls through that boundary cancels itself out" then those terms cancel to produce Jane's final energy conservation equation:

Jane's power in = electrical heating power
Jane's power out = radiative power out from source

Jane, is that what you're saying by "net radiation across the boundary from the wall is zero"?

... net radiative power from the wall THROUGH A BOUNDARY between the heat source and the wall would be zero... the net POWER IN from the wall across that boundary is zero... I did NOT claim the net radiative power through the boundary was zero. What I wrote was that the radiative power from the walls through that boundary cancels itself out, leaving a net across that boundary from the wall = 0. [Jane Q. Public, 2014-10-11]

Again, Jane must using some kind of Sky Dragon Slayer definition of the word "net". In physics, "net radiative power out" means "radiative power out" minus "radiative power in". This is only zero when the source and chamber walls are at the same temperature.

Similarly, "net radiative power in" means "radiative power in" minus "radiative power out". Again, this is only zero when the source and chamber walls are at the same temperature.

The actual radiant power through that boundary isn't zero, because the radiant power output of the heat source still goes through it of course. Leaving a NET positive transfer of energy OUTWARD through the boundary. [Jane Q. Public, 2014-10-11]

That's ridiculous, Jane. Notice that "net radiative power out" equals negative "net radiative power in". Since Jane seems to agree that "net radiative power out" is positive, "net radiative power in" can't be zero. It has to be negative, which just means more radiative power is flowing out than flowing in.

So Jane must not be using the physics definition of "net". What's the Sky Dragon Slayer definition of "net"? And how is it possible for the "net power in" to be zero when the source is hotter than the chamber walls?

... An object that is radiating at a certain black-body temperature WILL NOT absorb a less-energetic photon from an outside source. This is am extremely well-known corollary of the Second Law. ... [Jane Q. Public, 2013-05-30]

... I have NOT been claiming that no radiation from a cooler body is absorbed by a warmer body. ... Energy can be absorbed and re-emitted... [Jane Q. Public, 2014-09-28]

... I do not deny that some radiation is absorbed; but then it's just re-emitted. ... [Jane Q. Public, 2014-10-03]

... Here is a fundamental principle of thermodynamics, as related to radiant energy: net incoming radiation from cooler bodies is ALL either reflected, transmitted, or scattered. Any absorption and re-transmission is part of the "transmitted" term. ... [Jane Q. Public, 2014-10-10]

Jane can't quote a textbook stating this "fundamental principle" because it's nonsense. For instance, the "transmitted" term describes a body's transparency, not its absorption and re-emission. Here's an introduction:

"A body's behavior with regard to thermal radiation is characterized by its transmission t, absorption a, and reflection p. ...

An opaque body is one that transmits none of the radiation that reaches it, although some may be reflected. That is, t = 0 and a + p = 1

A transparent body is one that transmits all the radiation that reaches it. That is, t = 1 and a = p = 0."

Jane, absorption and re-emission isn't part of the "transmitted" term. They're totally different. The "transmitted" term is zero for opaque bodies like aluminum or blackbodies. If absorption and re-emission were part of the "transmitted" term, blackbodies would be transparent because they absorb all radiation that hits them. If absorption and re-emission were part of the "transmitted" term then both terms would equal 1. But once again, blackbodies can't be transparent.

Also, it's bizarre that Jane insists he's accounting for absorbed and re-emitted radiation in a "transmitted" term that isn't even in his energy conservation equation.

... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. ... that ZERO of the radiative power output ... THE NET IS ZERO. ... [Jane Q. Public, 2014-10-10]

... First, there is no NET radiative power absorbed by a body at one thermodynamic temperature from another body at a lower temperature. ... Those are the statements I made. Anything else is a logical extension of those two principles. ... [Jane Q. Public, 2014-10-10]

That's a serious problem, because Jane's first principle is wrong. Net radiative power would only be zero if the source and the chamber walls were at the same temperature.

Jane might consider replacing his incorrect first principle with "conservation of energy" which means power in = power out through a boundary where nothing inside is changing.

... No NET radiative input from chamber walls means anything crossing your precious boundary inward goes right back out. As I have explained to you many times now. ... [Jane Q. Public, 2014-10-10]

Jane, it's not my "precious" boundary. It's a general principle called "conservation of energy". Drawing a boundary is needed to apply conservation of energy. Here are some introductions: example (backup), example (backup), example (backup).

Note that drawing a boundary is needed to apply conservation of energy in all those introductions, just like I've repeatedly explained.

If that's what you think, could you take a few seconds to write down the energy conservation equation (before cancelling terms) that you think is correct?

I could, but I will not. ... I'm not going to take valuable time out of my day ... I'm not going to waste my time. ... [Jane Q. Public, 2014-10-09]

If Jane really could write down an energy conservation equation before wrongly "cancelling" terms, that would only take a few seconds. Endlessly screaming in ALL CAPS and accusing me of lying wastes more of Jane's valuable time than writing down an energy conservation equation for a boundary around the source:

Jane's power in = ?
Jane's power out = ?

... there is no NET radiative power absorbed by a body at one thermodynamic temperature from another body at a lower temperature. Doing so would violate the Second Law of Thermodynamics. I'm using the standard definition of "net", which is to say "all inputs minus all outputs". ... [Jane Q. Public, 2014-10-10]

If net radiative power is "all inputs minus all outputs" then net radiative power is only zero if all the inputs equal all the outputs. That only happens if the source is at the same temperature as the chamber walls.

But Jane claims that net radiative power is zero when the source is hotter than the chamber walls.

NOWHERE did I state "zero net radiative output". I don't believe I used that phrase at all, but if I did, you present it here out of context. [Jane Q. Public, 2014-10-10]

Hmm...

... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. ... that ZERO of the radiative power output ... THE NET IS ZERO. ... [Jane Q. Public, 2014-10-10]

... no NET radiative power ... [Jane Q. Public, 2014-10-10]

Again, net radiative power is only zero if the source is at the same temperature as the chamber walls.

NOWHERE did I state "zero net radiative output". I don't believe I used that phrase at all, but if I did, you present it here out of context. [Jane Q. Public, 2014-10-10]

Oh, so you're saying net radiative power isn't zero? For some odd reason I thought you were saying it was zero.

... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. ... that ZERO of the radiative power output ... THE NET IS ZERO. ... [Jane Q. Public, 2014-10-10]

By the way, just in case it wasn't obvious from the fact that I was responding to Jane's claims of zero net radiation absorbed and zero net radiative power output, I was talking about net radiative power because that's what Jane seemed to be talking about. That's why my equation only has radiative terms. Here's a less ambiguous version:

So you're not going to retract your claim that net radiative power is zero when the source is warmer than the chamber walls?

Is Jane using some kind of special Sky Dragon Slayer definition of the word "net"? In physics, net radiative power through a boundary around the source = "radiative power out" minus "radiative power in".

So you're not going to retract your claim that net power is zero when the source is warmer than the chamber walls?

... when this is the hottest body in the room, that figure is ZERO. Zero net radiation absorbed from other, cooler bodies. This is a requirement of the Second Law of Thermodynamics. Now, this is NOT the same as saying "no radiation absorbed at all". But when you put the two points above together, what it does mean is that ZERO of the radiative power output from the above equation is coming from other bodies. THE AMOUNT OF POWER OUTPUT IN THIS EQUATION DOES NOT NEED TO ACCOUNT FOR POWER FROM THE WALLS, BECAUSE THE NET IS ZERO. ... [Jane Q. Public, 2014-10-10]

Is Jane using some kind of special Sky Dragon Slayer definition of the word "net"? In physics, net power through a boundary around the source = "radiative power out" minus "radiative power in".