Global Mean Temperature

[Lagboltz]

Some of the first paragraphs of your post are physics 101. You demonstrate that you know the names of various elementary forms of EM radiation for no particular reason, and you even mention W and Z particles probably to impress people.

But then you put other physics 101 terms into sentences that make no physical sense. Yes, you can successfully copy equations into your script, but have no idea what they mean.

if we rearrange the equation, we get:

This is clearly a subtraction of two EM fields. ...

This is another amazing display of total misunderstanding. Pale, the T in the Stefan_Boltzman law is temperature. Then you say that temperature raised to the fourth power is an EM field. You went through the trouble of touting your knowledge of the names of various EM fields, then you demonstrate that you don't even know what they are.

Then you continue on with grade school science and finally get back to the topic and say

I repeat, no energy transfer exists between the cooler atmosphere and the warmer surface of the earth as predicted by the second law of thermodynamics, the law of conservation of energy, and the Stefan-Boltzman law.

Yes, I agree and said the same thing several times, concerning energy transfer which is different than photon transfer.

At every point where our photons are in opposition, we must subtract those vectors.

Next you prefer to deal with Boltzmann radiation as fields. That's perfectly valid. But then you screw up again and say that EM fields from two different radiators form vectors that subtract.

...we will subtract the magnitude of my EM vectors from the magnitude of your EM vectors

Pale, Boltzmann radiation is incoherent. It does not subtract. Look up the properties of incoherent radiation.
[/quote]
You continue to confuse the concepts of heat flow and radiation. The rest of your post is of course a demonstration of your lack of understanding of radiation equilibrium.

 

[palerider]

Some of the first paragraphs of your post are physics 101. You demonstrate that you know the names of various elementary forms of EM radiation for no particular reason, and you even mention W and Z particles probably to impress people.

You don't really read for comprehension, do you? Did I not clearly state that quantum was DEFINED as?

But then you put other physics 101 terms into sentences that make no physical sense. Yes, you can successfully copy equations into your script, but have no idea what they mean.

Sorry guy, but is you who has the fundamental misunderstanding. It is you who absolutely agreed with me when I pointed out that the second law states clearly that energy won't flow from a cooler body to a warmer body, that energy specifically being IR radiated away from the surface of the earth. Then it was you who said you agree insofar as IR energy goes, but not in the case of electromagnetic energy as if infrared energy were not electromagnetic energy. It is you who had no idea that we were talking about photons the whole time and believed that you were introducing something new. Yes, there is a terribly flawed understanding of physics here, but it belongs to you and climate science, not me.

This is another amazing display of total misunderstanding. Pale, the T in the Stefan_Boltzman law is temperature. Then you say that temperature raised to the fourth power is an EM field. You went through the trouble of touting your knowledge of the names of various EM fields, then you demonstrate that you don't even know what they are.

Actually I say no such thing and the fact that you think that I have is another indication that you really don't grasp the topic. Temperature is nothing more than a number that represents one system's ability to transfer energy to another system. In the case of the Stefan-Boltzman equations

is just a statement on the ability of the radiator to transfer energy to the background. Energy radiated by the emitter is proportional to the fourth power of the absolute temperature. The exponent has nothing to do with whether or not the energy being emitted is an EM field or not, the exponent is a requirement of the use of the Stefan-Boltzman law. I didn't inject an exponent into the equation to turn the energy into an EM field. It represents an EM field anyway.

Yes, I agree and said the same thing several times, concerning energy transfer which is different than photon transfer.

Sorry guy, but energy transfer and photon transfer are one in the same. Your confusion arises from your fundamental misunderstanding of what a photon is. Clearly you believe that a photon is an independent particle that goes zipping around without regard to the laws that govern electromagnetic radiation and energy transfer. Simply stated, you are wrong. Let me repeat the defintion of what a photon is for you . Do take the time to actually read the definition this time and apply it to what you believe a photon is.

Photon - The subatomic particle that carries the electromagnetic force and is the quantum of electromagnetic radiation..

The key phrase here lagboltz, and the description of what a photon actually is, is "THE QUANTUM OF ELECTROMAGNETIC RADIATION". A photon isn't some "thing" that goes zooming around independent of the physical laws that govern the behavior of EM radiation. A photon is nothing more, I repeat, NOTHING MORE than the smallest possible bit of an EM field that retains the characteristics of the EM field. When you reach the point in an EM field at which you can no longer divide the energy into a smaller part, you have reached the level of the photon. Divide further and you lose the characteristic of the EM field it is part of. Not a free agent, but the smallest possible unit of the EM field. I gave you the definition of quantum for a reason. When you went off on your claim that you agreed that the atmosphere couldn't transfer energy to the ground but it could transfer photons, it was clear that you had no idea what a photon actually was so I gave you a definition which you clearly ignored in favor of what you believed a photon to be.

So, at this point we have photon defined as the quantum of electromagnetic radiation. Clearly the word quantum doesn't mean anything to you so again, let me give you the definition of quantum: (CAUTION: the terms W and Z particles appear in this defintion and are there because they are part of the definition and not an attempt to impress anyone) as if I need to impress you.

Quantum - A discrete, indivisible manifestation of a physical property, such as a force or angular momentum. Some quanta take the form of elementary particles; for example, the quantum of electromagnetic radiation is the photon, while the quanta of the weak force are the W and Z particles.

Now apply that defintion to what you believe a photon to be. The smallest possible part of an EM field is called a photon. The photon is the smallest bit of electromagnetic energy possible. The laws that apply to the EM field apply to the photon because the EM field is photons. Apply this to the vector subtraction of electromagnetic fields. You have two EM fields in opposition to each other. You subtract one from the other and the difference describes the magnitude of the resulting field and the direction in which it is propagated. When you subtract the fields, what precisely do you believe you are subtracting? I can only guess that you aren't aware that you are subtracting photons. You are whittling away the EM field. Reducing its magnitude. If you remove the smallest part of the EM field possible, you remove a photon because a photon is nothing more than the smallest possible part of the EM field that retains the characteristics of the EM field.

It is precisely the sort of fundamental misunderstanding that you demonstrate that is the cause of the error cascade that climate science is suffering from except that the error isn't simply not understanding what a photon is but a complete disregard for the physical laws. A photon is a bit of energy, nothing more. A bit of energy along with a great many other bits of energy that make up an electromagnetic field.

Taking that into consideration, now look at the second law of thermodynamics again.

"It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object."

Contrary to what you believe, the second law isn't just talking about heat. It explicitly states that energy will not spontaneously flow from a low temperature object to a higher temperature object and photons are, by definition the smallest possible bit of energy in an electromagnetic field.

(continued)

 

[palerider]

(continuation)

Next you prefer to deal with Boltzmann radiation as fields. That's perfectly valid. But then you screw up again and say that EM fields from two different radiators form vectors that subtract.

I can only guess that you believe that due to your complete misunderstanding of what a photon is. The Stefan-Boltzman equation is clearly the subtraction of two EM fields.

Where you are subtracting the EM field represented by the background from the EM field radiated by the blackbody (emitter). The resulting number describes the magnitude of the remaining EM field and its direction of propagation.

Pale, Boltzmann radiation is incoherent. It does not subtract. Look up the properties of incoherent radiation.

You are really grasping now lagboltz. White light is considered to be incoherent radiation. Are you telling me now that visible light waves are incapable of interfering with or cancelling each other out? Here is an example of interference in incoherent radiation, from wiki since you apparently trust the source:

Figure 11: Spectrally incoherent light interferes to form
continuous light with a randomly varying phase and amplitude

Just so we don't have to go through the whole discussion revolving around what you believe interference to be, here is the definition from the science dictionary. Do read it and try to apply it to what you believe..

Interference - "The superposition of two or more waves propagating through a given region. Depending on how the peaks and troughs of the interacting waves coincide with each other, the resulting wave amplitude can be higher or smaller than the amplitudes of the individual waves. ◇ When two waves interact so that they rise and fall together more than half the time, the amplitude of the resulting wave is greater than that of the larger wave. This is called constructive interference. ◇ When two waves interact such that they rise and fall together less than half the time, the resulting amplitude is smaller than the amplitude of the stronger wave. This interference is called destructive interference. It is possible for two waves of the same magnitude to completely cancel out in destructive interference where their sum is always zero, that is, where their peaks and troughs are perfectly opposed.

Interference describes the addition, subtraction, or cancellation of EM fields and it can happen with coherent and incoherent radiation. Again, you exemplify the sort of misunderstanding of basic physics that has made climate science the victim of an error cascade of such magnitude. Have you ever looked at an atmospheric physics book used by a cliamte science cirriculum? It will stray from the classical physics being taught to students of the hard sciences (engineering, chemistry, physics, astrophysics, etc. on a number of topics.

You continue to confuse the concepts of heat flow and radiation. The rest of your post is of course a demonstration of your lack of understanding of radiation equilibrium.

Sorry guy, but the confusion is all yours. I am sure that your view of yourself as the smartest guy in the room won't allow you to actually accept that reality, but there it is. The definitions and statements of physical laws support, and predict my position, not yours. In order to accept your position, one must disregard the second law of thermodynamics, the law of conservation of energy, the Stefan-Boltzman law, the accepted definition of what constitutes a photon, the defintion of quantum, and the very defnition of electromagnetic radiation. Who do you really believe is right here?

And not to worry, I am surprised that you attempted to address as much of my post as you did. Don't worry about not addressing the rest of it. Doing so would have just made this rebuttal that much longer.

 

[Lagboltz]

Sorry guy, but is you who has the fundamental misunderstanding. It is you who absolutely agreed with me when I pointed out that the second law states clearly that energy won't flow from a cooler body to a warmer body, that energy specifically being IR radiated away from the surface of the earth.

Sorry guy, I didn't simply agree with you. Yes to the Stefan-Boltzmann equation. No to the photons or fields canceling out.

In the case of the Stefan-Boltzman equations is just a statement on the ability of the radiator to transfer energy to the background. Energy radiated by the emitter is proportional to the fourth power of the absolute temperature. The exponent has nothing to do with whether or not the energy being emitted is an EM field or not, the exponent is a requirement of the use of the Stefan-Boltzman law. I didn't inject an exponent into the equation to turn the energy into an EM field. It represents an EM field anyway.

I know what the Stefan-Boltzman equation is but you don't. You can't wiggle out of that. You correctly showed the Stefan-Boltzman law as the subtraction of two temperatures raised to the 4th power and IMMEDIATLY made this incorrect statement:
"This is clearly a subtraction of two EM fields."
That is wrong wrong wrong, no matter what power they are raised to (including 1st power).

The key phrase here lagboltz, and the description of what a photon actually is, is "THE QUANTUM OF ELECTROMAGNETIC RADIATION". A photon isn't some "thing" that goes zooming around independent of the physical laws that govern the behavior of EM radiation. A photon is nothing more, I repeat, NOTHING MORE than the smallest possible bit of an EM field that retains the characteristics of the EM field. When you reach the point in an EM field at which you can no longer divide the energy into a smaller part, you have reached the level of the photon. Divide further and you lose the characteristic of the EM field it is part of.

The first part is the usual physics 101, but divide a photon further??? A physicist would not let that thought go through their head.

So, at this point we have photon defined as the quantum of electromagnetic radiation ..... [More elementary fill-in until we get this:] ...... I can only guess that you aren't aware that you are subtracting photons.

That is too funny. First you talk about dividing them. Now subtract them? You can't subtract photons. Are you making this up as you go along?

Here is an optics experiment for you to see if you can prove your point about fields or photons subtracting and canceling out. Set two 60 watt bulbs one foot apart. Tungsten in the two bulbs will have the same spectral emissivity, so the radiation cancellation should be fairly complete at all frequencies if it works. If they are both at the same temperature heat energy will not flow between them. Will the radiation cancel out?

To verify the Stefan-Boltzmann-Palerider version of the theory the following will happen: Turn the bulbs on and see if you can find a black column where the EM fields cancel out between the two bulbs. To prove blackness you might have to blow smoke between the bulbs. But don't inhale.

Please try it and send me a picture.

If that doesn't work, you might try two laser pens which have coherent radiation.

It is precisely the sort of fundamental misunderstanding blah blah blah, ... now look at the second law of thermodynamics again. ........

We have already gone through this many times.

I can only guess that you believe that due to your complete misunderstanding of what a photon is. The Stefan-Boltzman equation is clearly the subtraction of two EM fields.

There you go again. Please find a source that says that S-B is about a subtraction between two fields. See my 60 watt light bulb experiment above.

Where you are subtracting the EM field represented by the background from the EM field radiated by the blackbody (emitter). The resulting number describes the magnitude of the remaining EM field and its direction of propagation.

Not so. You are really grasping now Pale. Please do the light bulb experiment.

You are really grasping now lagboltz. White light is considered to be incoherent radiation. Are you telling me now that visible light waves are incapable of interfering with or cancelling each other out?

Nope. Not saying that.

Spectrally incoherent light interferes to form continuous light with a randomly varying phase and amplitude

Just so we don't have to go through the whole discussion revolving around what you believe interference to be, here is the definition from the science dictionary.

Interference - " [...typical text book definition...] It is possible for two waves of the same magnitude to completely cancel out in destructive interference where their sum is always zero, that is, where their peaks and troughs are perfectly opposed.

"Perfectly opposed". What percentage of black body radiation do you think is perfectly opposed? The radiation is incoherent. Did you ever look up how the amplitudes of incoherent waves at the same frequency add up?

Here is a homework problem for you:
Given 100 waves from a black body that happen to be emitted in exactly the same direction with exactly the same wavelength, and exactly the same amplitude, and "A", roughly what will the final amplitude be? You should be able to do it in your head. No fair looking it up on the web.

And yes you have nice pictures of interferences, but remember that the waves must be traveling the same direction for the peaks and troughs to be "perfectly opposed." That doesn't happen to waves heading toward each other between two black bodies even if they are at the same temperature.

Interference describes the addition, subtraction, or cancellation of EM fields ... [more basic physics 101, and a scoff at climate scientists again.]

Yes, I understand wave interference, and you do too. What you don't understand is that black body radiation between two sources can't interfere to the extent that they reduce the EM field, let alone cancel out anything. Let me say that again. In black body radiation the fields will not cancel out.

Sorry guy, but the confusion is all yours. I am sure that your view of yourself as the smartest guy in the room won't allow you to actually accept that reality, but there it is. The definitions and statements of physical laws support, and predict my position, not yours. In order to accept your position, one must disregard the second law of thermodynamics, the law of conservation of energy, the Stefan-Boltzman law, the accepted definition of what constitutes a photon, the defintion of quantum, and the very defnition of electromagnetic radiation. Who do you really believe is right here?

I think you might be one of those guys that thinks reality is a nice place to visit, but you wouldn't want to live there. Please try the light bulb electric field cancellation experiment and let me know who is really right here.

If it doesn't work, don't say I misquoted or didn't understand you. You argued cancellation of fields too many times in too many ways.

 

[palerider]

Sorry guy, I didn't simply agree with you. Yes to the Stefan-Boltzmann equation. No to the photons or fields canceling out.

One in the same guy...one in the same.

I know what the Stefan-Boltzman equation is but you don't. You can't wiggle out of that. You correctly showed the Stefan-Boltzman law as the subtraction of two temperatures raised to the 4th power and IMMEDIATLY made this incorrect statement:
"This is clearly a subtraction of two EM fields."
That is wrong wrong wrong, no matter what power they are raised to (including 1st power).

Temperature of what? Again, temperature is just a word that describes the ability of one system to transfer energy to another. Temperature represents the magnitude of the EM field being radiated by the emitter and the magnitude of the field being radiated by the background. What do you suppose the temperatures being subtracted are, just numbers that don't relate to anything physical?

The first part is the usual physics 101, but divide a photon further??? A physicist would not let that thought go through their head.

I said that when you reach the point where the EM field is no longer divisible, you have reached the level of a photon. I never suggested that you could divide a photon. You seem to have a fundamental need to take my statements out of context or otherwise alter them and then use your alteration in an attempt to somehow show that I am wrong. That is a poor and somewhat dishonest method of debate and it doesn't help your argument at all.

That is too funny. First you talk about dividing them. Now subtract them? You can't subtract photons. Are you making this up as you go along?

First, I never said that you could divide a photon, and when you diminish an EM field, considering that it is made up of photons, what excactly do you suppose is being taken away?

Here is an optics experiment for you to see if you can prove your point about fields or photons subtracting and canceling out. Set two 60 watt bulbs one foot apart. Tungsten in the two bulbs will have the same spectral emissivity, so the radiation cancellation should be fairly complete at all frequencies if it works. If they are both at the same temperature heat energy will not flow between them. Will the radiation cancel out?

Yes, the radiation cancels out, or will eventually. If you measure the temperature of the two filaments and they are the same, then the radiation between the two are cancelling out. If it were not, then the filaments would be absorbing radiation from the other and by defintion would become warmer and in turn radiate more which would then be absorbed which would cause the filament to become warmer and in turn radiate more which would then be absorbed which would cause the filament to become warmer and in turn radiate more which would cause the .....again, see where we are going. You keep describing an infinite positive feedback loop which is the engine required for a perpetual motion machine.

There you go again. Please find a source that says that S-B is about a subtraction between two fields. See my 60 watt light bulb experiment above.

Again, temperature is just a word that represents one system's ability to transfer energy to another. What do you believe the temperatures being subtracted represent?

Not so. You are really grasping now Pale. Please do the light bulb experiment.

Why don't you try it and see if they get progressively hotter without increasing the amount of electricity going into them. If they do, then you have the engine for a perpetual motion machine. Along the vectors where radiation from one bulb is in opposition to radiation from the other bulb, if the EM fields are of the same magnitude, then yes, they are cancelling out.

"Perfectly opposed". What percentage of black body radiation do you think is perfectly opposed? The radiation is incoherent. Did you ever look up how the amplitudes of incoherent waves at the same frequency add up?

Tell me lagboltz, what frequency do you believe the atmosphere might be radiating towards the earth that isn't present in the radiation outbound from the surface of the earth?

And yes you have nice pictures of interferences, but remember that the waves must be traveling the same direction for the peaks and troughs to be "perfectly opposed." That doesn't happen to waves heading toward each other between two black bodies even if they are at the same temperature.

And yet, the Stefan -Bopltzman equation is describing the subtraction of two EM fields. Of course, in the Stefan-Boltzman equation there is only one black body and in the real world neither the surface of the earth, nor the atmosphere are blackbodies. If you can't relate what you believe you know to the real world, of what value, exactly, is it?

Yes, I understand wave interference, and you do too. What you don't understand is that black body radiation between two sources can't interfere to the extent that they reduce the EM field, let alone cancel out anything. Let me say that again. In black body radiation the fields will not cancel out.

Of course they can. Ask an electrical engineer who sets up microwave towers, or optical transmitters, or radio, or television towers or any communications equipment. They most certainly can and do cancel out.

I think you might be one of those guys that thinks reality is a nice place to visit, but you wouldn't want to live there. Please try the light bulb electric field cancellation experiment and let me know who is really right here.

Again, I suggest that you try it. If they keep getting progressively hotter and you aren't increasing the amount of electricity going into them, let me know. I have a mechanical engineer friend who will show you how to turn that increase in energy being emitted without an increase in power going in into a bazillion dollars overnight.

The second law and the law of conservation of energy predict that the filaments won't get even the smallest fraction of a degree warmer and if either absorbed energy from the other then they would necessarily become warmer and in turn radiate more.

You are stuck on the idea that perpetual motion exists and I am afraid you are wrong.

 

[Lagboltz]

Temperature of what? Again, temperature is just a word that describes the ability of one system to transfer energy to another. Temperature represents the magnitude of the EM field being radiated by the emitter and the magnitude of the field being radiated by the background. What do you suppose the temperatures being subtracted are, just numbers that don't relate to anything physical?

Ah, you got it now. Temperature is a magnitude, and of course that is not a vector. To others that may be reading this tedious interminable dialog, the following is a way to look at the concepts.

Speed is what you see on the meter on your dashboard. It is a scalar - just a number. In physics, the word "velocity" is a vector and is reserved to include the direction you are going. You need a compass to see direction.

The temperature gauge on your car gives a number that tells how hot your engine is, but it doesn't say which way heat is flowing.

Pale, I'm sure you know all this about velocity, but you did not previously apply that concept to temperature correctly.

I said that when you reach the point where the EM field is no longer divisible, you have reached the level of a photon. I never suggested that you could divide a photon. You seem to have a fundamental need to take my statements out of context or otherwise alter them and then use your alteration in an attempt to somehow show that I am wrong. That is a poor and somewhat dishonest method of debate and it doesn't help your argument at all.

First, I never said that you could divide a photon, and when you diminish an EM field, considering that it is made up of photons, what excactly do you suppose is being taken away?

I didn't say that you personally thought you could divide a photon, I just said, "A physicist would not let that thought go through their head." It is a sign of poor thinking in the physical world.

Yes, the radiation cancels out, or will eventually. If you measure the temperature of the two filaments and they are the same, then the radiation between the two are cancelling out. If it were not, then the filaments would be absorbing radiation from the other and by defintion would become warmer and in turn radiate more which would then be absorbed which would cause the filament to become warmer and in turn radiate more which would then be absorbed which would cause the filament to become warmer and in turn radiate more which would cause the .....again, see where we are going. You keep describing an infinite positive feedback loop which is the engine required for a perpetual motion machine. .....

Why don't you try it and see if they get progressively hotter without increasing the amount of electricity going into them. If they do, then you have the engine for a perpetual motion machine. Along the vectors where radiation from one bulb is in opposition to radiation from the other bulb, if the EM fields are of the same magnitude, then yes, they are cancelling out.

Both bulbs would get very slightly hotter. You must consider the solid angle subtended by each bulb. (Look up solid angle on wiki.) Each bulb absorbs a very small percentage of radiation of the other bulb proportional to the small solid angle of the adjacent bulb, but it reradiates the absorbed energy in every direction and not just to the adjacent bulb. This happens until the temperatures reach a new equilibrium. The final temperature rise would probably be in micro degrees, and hard to measure.

I have a better experiment that has higher accuracy for the temperature measurement. Replace the light bulbs with hotplates. Set a 1000 watt hot plate on it's side and plug it in. Let it come to thermal equilibrium and measure the temperature in the middle of the hotplate. Put an identical hot plate on its side with its hot surface one inch from the first. Plug it in and let the ensemble of the two hotplates come into thermal equilibrium. The S-B equation says they will not further exchange heat energy. Now measure the temperature at the same spot. I guarantee it will be hotter. To see this intuitively, think of one hotplate at 1000 watts. It is hot. Think of a second hotplate with the surface in close proximity to the first pouring out 1000 watts -- a total of 2000 watts in close proximity. The subtended solid angle is much larger than the light bulb experiment. The whole thing will be measurably hotter.

Finally, lay the hotplates back to where they were intended, and make lunch.

Of course they can. Ask an electrical engineer who sets up microwave towers, or optical transmitters, or radio, or television towers or any communications equipment. They most certainly can and do cancel out.

Radio frequencies are coherent and have lots of interference. Black body radiation is not coherent. Big difference. You need a better example. The interference of 100 waves of BB radiation in the example I gave would lead to a factor of the square root of 100, which is a 10 fold increase in amplitude of the EM field because the phases are random. A laser would create those waves in phase, and the amplitude would increase by 100. There is a big difference in the behavior of coherent light and incoherent radiation in the real world. In fact one way of creating radar waves is by using a MASER -- same principle as the laser.

Again, I suggest that you try it. If they keep getting progressively hotter and you aren't increasing the amount of electricity going into them, let me know. I have a mechanical engineer friend who will show you how to turn that increase in energy being emitted without an increase in power going in into a bazillion dollars overnight.

The second law and the law of conservation of energy predict that the filaments won't get even the smallest fraction of a degree warmer and if either absorbed energy from the other then they would necessarily become warmer and in turn radiate more.

As I said earlier, the bulbs will get a very small fraction hotter, and the hotplates, a more tightly coupled experiment (due to the increase in solid angle), will have a measurable increase in temperature.

You are stuck on the idea that perpetual motion exists and I am afraid you are wrong.

If you want to think of the infinite series in temperature rise until thermal equilibrium is reached, don't think it is linear. For example 1/2 + 1/4 + 1/8 + 1/16 ... is equal to 1, not infinity.

If the ensemble of 2000 watts of hotplates does not get hotter, where does that extra 1000 watts of heat go? If you don't believe in the hotplate "thought" experiment, then you have no recourse but to try it.

 

[palerider]

Ah, you got it now. Temperature is a magnitude, and of course that is not a vector. To others that may be reading this tedious interminable dialog, the following is a way to look at the concepts.

Temperature is a magnitude, and vector is a direction. No radiation from a cooler object is absorbed by a warmer object. Do keep in mind that photons are radiation.

[/quote]Both bulbs would get very slightly hotter.[/quote]

Neither filament would get hotter. The filament is the emitter and if the two bulbs are emitting an EM field of precisely the same magnitude, neither will absorb radiation from the other. If one is radiating an EM field that is slightly greater in magnitude then it will not absorb a single photon from the cooler bulb.

I have a better experiment that has higher accuracy for the temperature measurement. Replace the light bulbs with hotplates. Set a 1000 watt hot plate on it's side and plug it in. Let it come to thermal equilibrium and measure the temperature in the middle of the hotplate. Put an identical hot plate on its side with its hot surface one inch from the first. Plug it in and let the ensemble of the two hotplates come into thermal equilibrium. The S-B equation says they will not further exchange heat energy. Now measure the temperature at the same spot. I guarantee it will be hotter. To see this intuitively, think of one hotplate at 1000 watts. It is hot. Think of a second hotplate with the surface in close proximity to the first pouring out 1000 watts -- a total of 2000 watts in close proximity. The subtended solid angle is much larger than the light bulb experiment. The whole thing will be measurably hotter.

If they are radiating EM fields of the same magnitude, neither will get warmer. If they did, it is off to the infinite positive feedback loop again. If they are radiating at a slightly different magnitude, then no radiation from the cooler will be absorbed by the warmer. Subtract the two fields and the difference will tell you the magnitude of the resulting field and its direction of propagation.

As I said earlier, the bulbs will get a very small fraction hotter, and the hotplates, a more tightly coupled experiment (due to the increase in solid angle), will have a measurable increase in temperature.

You can say it till the cows come home but the fact is that neither filament will get warmer if they are both radiating an EM field of the same magnitude. Both the second law of thermodynamics and the law of conservation of energy predict this result. You seem not to have much respect for the laws of physics because to accept your ideas is to reject the very laws of physics.

If the ensemble of 2000 watts of hotplates does not get hotter, where does that extra 1000 watts of heat go? If you don't believe in the hotplate "thought" experiment, then you have no recourse but to try it.

It goes off in vectors that are not subtracted. If you have 2 hotplates radiating 1000 watts each, 2000 is all you have. For one of the plates to get warmer would require that one of the plates be radiating more than 1000 watts.

 

[palerider]

Lagboltz, I recommend that you read the Unified Theory of Climate by Ned Nikolov and Karl Zeller. These two physicists have put together a theory that is both supported and predicted by the laws of physics, explains the temperature on our planet without any need for a greenhouse effect as described by climate science, and unlike the models upon which modern climate science operates, when this theory is applied to the other planets within the solar system, you actually get accurate results. When you apply the physics upon which modern models operate to other planets within the solar system, their predictions of temperature aren't even close telling a thinking person that their resemblence to earth is purely coincidental because they are operating on a terribly flawed understanding of physics.

A link to the paper is below. It describes a new paridgm in thinking about the physics of energy within our atmosphere and will eventually be accepted because it is correct regardless of the impact it has on the CO2 reduction business.

http://www.scribd.com/doc/76694199/...enhouse-Effect-Using-Thermodynamic-Principles

 

[Lagboltz]

Lagboltz said: ↑

If the ensemble of 2000 watts of hotplates does not get hotter, where does that extra 1000 watts of heat go?

It goes off in vectors that are not subtracted.

So you think. Let's suppose the 1000 watt hotplates are ideally insulated fully at the back and fully at the sides. This is not an unreasonable goal in hotplate design. Suppose the surface area of each hotplates is 100 sq inches, and after the hotplates are up to their full temperature, (say 150 deg C), and at thermal equilibrium, the IR radiation flux is simple: 1000 watts per 100 sq inches is 10 watts per square inch of IR radiation.

Now suppose the hotplates, laid on their sides are moved together by .004 inches (the thickness of a human hair.) The perimeter gap area is .004 multiplied by the perimeter which is about 36 inches. That amounts to .144 sq inches. That .144 sq inch hotplate gap considerably reduces the possible paths of the IR to the outside world.

The radiation can only escape through that gap because the sides and backs of the hotplates are ideally insulated. That means that the combined 2000 watts are radiating out through a gap of .144 sq inches. (You refer to these as "vectors that are not subtracted.") That is equivalent to 14,000 watts per sq inch escaping through the gap. However, according to the Stefan-Boltzman-Palerider theory each hot plate will remain at 150 deg C because the radiation between them cancels out. So the Stefan-Boltzman-Palerider theory says that 2000 watts of energy escapes at a radiation density of 14,000 watts per sq inch from surfaces at 150 deg C, which was already shown to only radiate 10 watts per sq inch. There is a big flaw here.

As you move the plates even closer and closer the watts per sq inch radiated through the gap diverges to infinity. The above analysis avoids what you think is an infinite positive feedback loop, but you introduce another infinity.

I say the temperatures of the two hotplates will both rise until they are hot enough that the BB radiation at the higher temperature becomes 14,000 watts per centimeter. That will again achieve thermal equilibrium, and the total radiation flux will be 2000 watts, and the system will no longer rise in temperature. In reality the whole thing will melt before equilibrium happens.

One further note: Black body radiation cannot use a classical model of EM vector fields. In fact if you try to use classical field theory in computing BB radiation, you get what was called the Rayleigh–Jeans catastrophe. Max Planck solved the problem in the early days only by assuming light was quantized, i.e. photons. So you, Palerider have to explain your ideas in terms of photons. Photons cannot annihilate (subtract.) They don't even interact unless you go to energies available only at large enough accelerators, such as the ones at CERN. (Look up two photon interaction.)

You gave a reference by Ned Nikolov, et. al. Seriously Palerider, if this is where you get your concepts of radiation physics I don't see why I would want to read it.

 

[palerider]

Imagine that. You pay the electric company for enough electricity to radiate 10 watts per square inch and you end up radiating 14,000 watts per square inch. The energy concerns of the world are over. We can power the world by placing hotplates on thier sides and run perpetual motion machines off the excess wattage created.

And I can understand why you might not want to read Nikolov. No fiction there. No perpetual motion, no creation of energy, no cold objects further warming warm objects. And now I understand how you guys believe the surface of the earth can absorb more energy from the atmosphere than it does from its actual energy source.

As to photons; can you show me any hard evidence for their existence? Can you prove they exist or is it a matter of faith for you like backradiation from cool objects further warming warm objects?

 

[Lagboltz]

Imagine that. You pay the electric company for enough electricity to radiate 10 watts per square inch and you end up radiating 14,000 watts per square inch. The energy concerns of the world are over. We can power the world by placing hotplates on thier sides and run perpetual motion machines off the excess wattage created.

Pale, you have to watch your units when you do physics calculations. The world can't be powered by watts per square inch. It has to be powered by watts. Watts is a unit of power. Watts per square inch is a measure of power density.

So the answer is:
14,000 watts per sq inch x .144 sq inches is roughly 2000 watts. (roughly, not exactly because of previous round-off error)
No perpetual motion there. You can't generate perpetual motion since you are not correctly using of the meaning of the physics units.

Here is another example similar to what you are doing. If someone says, "the density of water is 8.34 lb/gallon", would you reply, "What? you are saying all the oceans weigh only 8.34 lb?" We both know you wouldn't, but I'm only stressing how physics units must be correct and can't be taken out of context.

And I can understand why you might not want to read Nikolov. No fiction there. No perpetual motion, no creation of energy, no cold objects further warming warm objects. And now I understand how you guys believe the surface of the earth can absorb more energy from the atmosphere than it does from its actual energy source.

No fiction? Are you also referring the fact that they don't use your fictional version of radiation physics? If so, it might be an interesting read. Remember, you are the one who is creating the fiction of perpetual motion by misunderstanding the physics units involved.

As to photons; can you show me any hard evidence for their existence? Can you prove they exist or is it a matter of faith for you like backradiation from cool objects further warming warm objects?

Your argument is now reduced to the existence of photons? Do you doubt quantum mechanics? In prior posts you showed disdain for mathematical modeling of the observational world. All of the hard sciences, (including photons,) involve mathematical modeling. Are you a scientific Luddite?

 

[palerider]

Pale, you have to watch your units when you do physics calculations. The world can't be powered by watts per square inch. It has to be powered by watts. Watts is a unit of power. Watts per square inch is a measure of power density.

Oddly, the energy budget upon which you believe says the earth is powered in watts per square meter.

Your argument is now reduced to the existence of photons? Do you doubt quantum mechanics? In prior posts you showed disdain for mathematical modeling of the observational world. All of the hard sciences, (including photons,) involve mathematical modeling. Are you a scientific Luddite?

Not at all. My argument has been that you don't know what a photon is. You have some fictional notion of what they are but can't show even the smallest bit of proof to support that belief. I have said that when you reduce the magnitude of an EM field, you are reducing photons since photons is what the field is made up of. Photon is just a word that describes the smallest possible bit of the field that still retains the characteristics of that field.

 

[Lagboltz]

Oddly, the energy budget upon which you believe says the earth is powered in watts per square meter.

Sure, you can use metric units for the area. It seems you are talking about the energy density of the sun on the earth, and that can be around a kilowatt per square meter on a good day.

Previously you were talking about electric companies powering the world. In that case the units are watts.

Not at all. My argument has been that you don't know what a photon is. You have some fictional notion of what they are but can't show even the smallest bit of proof to support that belief. I have said that when you reduce the magnitude of an EM field, you are reducing photons since photons is what the field is made up of. Photon is just a word that describes the smallest possible bit of the field that still retains the characteristics of that field.

Sure, you can look at photons that way if you want. In the theory of photons in physics you need a lot more information than that, such as momentum, wave length, spin, etc.

 

[palerider]

Previously you were talking about electric companies powering the world. In that case the units are watts.

Semantics much?

Sure, you can look at photons that way if you want. In the theory of photons in physics you need a lot more information than that, such as momentum, wave length, spin, etc.

I don't look at photons the way I "want". I look at them the way the laws of physics dictate. You, however, are off the reservation entirely. Paying the electric company for enough electricity to turn up your hot plates to the point where they are radiating 10 watts per square inch and believing that somehow you can get them to radiate 14,000 watts per square inch without paying for any more power. Like I said, you represent an explanation of how warmers are able to believe that the surface of the earth receives more energy from the atmosphere than it does from that big ball of fire in the sky. How you got to that point is, a terrible shame and merits a very close look at our educational system.

 

[Lagboltz]

I don't look at photons the way I "want". I look at them the way the laws of physics dictate. You, however, are off the reservation entirely. Paying the electric company for enough electricity to turn up your hot plates to the point where they are radiating 10 watts per square inch and believing that somehow you can get them to radiate 14,000 watts per square inch without paying for any more power. Like I said, you represent an explanation of how warmers are able to believe that the surface of the earth receives more energy from the atmosphere than it does from that big ball of fire in the sky. How you got to that point is, a terrible shame and merits a very close look at our educational system.

I am so surprised you still don't understand the radiation physics involved. I tried to make it as simple as possible. You are still confusing watts per square inch with watts. After thermodynamic equilibrium the hotplates still only absorb 2000 watts from the electric company, and still radiate 2000 watts to the world, otherwise there would be no thermodynamic equilibrium and it would violate conservation of energy.

The important factor in this discussion is the total radiation area exposed to the outside world. If the area is large, the energy density can be small. If the area is small, the energy density must be large to maintain the 2000 watts radiated.

Perhaps you can explain what you are thinking physically in terms of the energy balance. Please be quantitative and compute the energy densities of the hot plates separated and then the densities in very close proximity. You don't need to use any subtle principles. All that is needed is to look at the configuration of the system and think in terms of total energy input, total energy output, and energy density for the output.

Earlier you said photons can "subtract" each other, please explain what you think is the physical mechanism for photons to do that.