the Air Vent

Because the world needs another opinion

Circular Reasoning on Rectilinear Propagation of Electromagnetic Radiation

Posted by Jeff Id on January 26, 2012

I’ve been spending time working on improved sea ice code.  Anthony Watts asked for something and Steve McIntyre helped find the right R function to get it going. I want to also overlay SST data as well as flow direction in the videos. Hopefully it will lead us to some statistical analysis.

In the meantime JWL has been discussing backradiation on the previous thread. It is an old argument which doesn’t have a lot of entertainment value for some of us. He has put together a pdf which alleges to prove that backradiation (radiation from cold to hot) is impossible in the climate system and has made the claim that nobody has critiqued his math.  I’m going to help him out.

What he does here though is completely circular reasoning.  We have two surfaces and a flow between them.  f is a variable representing absorption of the medium in between the two surfaces.  Surface one emits energy at a power Θ1 which is equal to σT1^4.  T is the plate temp and σ is a constant.  Now classic physics agrees that emission power is proportional to T^4 so there is no problem with that.   Lets look at the assumptions he makes though:

The heat transfer q from i->j is equal to the emission at i minus the emission of j in watts per meter squared.   The negative term in equation 1 is the exact magnitude we would expect of back radiation.  However, the author is already writing text that the back-radiation is non-physical.  How do we know?

We then proceed to the equations at the top of this extensive PDF. Figure 1a includes the emission now from a surface.  Joseph has apparently re-written the laws of physics for this as now the emission of the first surface to the second is equal to f(θ1 – θ2) where theta is the flux.  I asked how this equation came about and was pointed to equation 1.  Equation 1 though doesn’t say anything about the source of the flux, it only says that the sum of the flux is σ(Ti^4-Tj^4).  Of course that is not the assumption stated but the math doesn’t know anything about what assumption you state.  My question was and still is, how does the surface know to emit less energy when the second surface exists in a different timeframe? If the second surface vanishes instantly, does the first return to emitting a proper amount of radiation?

Very confusing.

So then we proceed to the next section of the treatise.

So for system 1 we have in steady state – input = output or f(θ1 – θ2) = f θ2 which with some easy algebra is  fθ1 = 2fθ2.  In system 2 we have the same thing fθ1 = 2fθ2.  They are exactly the same.  Now for system 2 Joseph states that plate 2 of model B, having two hand drawn arrows, is emitting twice as much as the first.  But wasn’t the original assumption that plate 1 changed its emission based on some spooky knowledge of plate 2?  This assumption was based on nothing I can see but it included the plate 1 emission correction which happened to be exactly equal to the emission of plate 2 back to plate 1. He then concludes that B is non-physical based on absolutely NO evidence other than his original (and obviously false) assumption.

The logic is totally circular and the math for both cases is identical.

Joseph hasn’t made the claim that the second plate won’t absorb the photons from the first, he is making the claim that the photons don’t even exit the first surface.  As this example of heat is identical to two incandescent lamps, this is physically identical to stating that if you shine a bright light on a dim light the second will dim further.

Many real world examples prove Joseph wrong.  Lets assume though that most of us already know this and look at it from a spacetime standpoint – because it is also fun.

Each surface, from its own point of reference, sees the other surface as behind itself in time.  The amount each sees the other as behind (X) is equal to the time light takes to travel across the distance between the objects.  Imagine that one plate suddenly vanished.  The other plate wouldn’t realize it for X seconds and in Josephs world, would stubbornly continue to emit at a reduced level until it became aware of the vanished counterpart.  These reduced output photons would pass through the space where the first plate had been for a full 2X the number of seconds.  This is because the first plate would become aware that the second had vanished after (X seconds) and then began emitting a greater amount across the same distance (another X seconds)!  A sensor behind the second plate would see a reduced output for 2x seconds after the plate disappeared, followed by a sudden jump in brightness of the first plate.

Now imagine the plates were a light-day apart and you had enough time to stick a sensor between the surfaces.

That is unphysical in my book.

So many examples prove radiation does flow from cold to warm.  My favorite, presented here before, is one of an uncooled infrared camera looking at an image of ice. Now we know the camera’s sensor is room temp so if radiation does not emit from cold to warm, everything cooler than the camera sensor, should appear midnight black. As this video progresses the color scale shifts so the table – room temp – is significantly warmer than many other objects in the image.  Especially the edges of the ice.  The manufacturer seems to also believe the camera can measure to minus 20 C.   Much to the chagrin of AGW advocates (and Joseph), that would be well below room temp in my room.

Remember, light and IR radiation are the same thing.  Your flashlight does not change intensity when you shine it at the sun.  Nor does the sun, however the energy from the flashlight will interact with the plasma of the solar surface 8 minutes after you shine it.

After this, I am done with this topic.  I won’t mind trying to explain where people go wrong in their thermodynamics, but really this is very much elementary and life is short.  I have spent a good deal of time trying to straighten out this form of “skepticism” which unfortunately is flatly not reasonable.  However, some are honestly confused and when they read a bunch of math letters piled together with words, they can’t always understand what the author is saying.


61 Responses to “Circular Reasoning on Rectilinear Propagation of Electromagnetic Radiation”

  1. Jeff Condon said

    My high school physics teacher used to enjoy saying rectilinear propagation of electromagnetic radiation. He used to demonstrate his 3d images from his projector using polarizing glasses as he had independently discovered the technique. Also fun.

  2. steve Fitzpatrick said

    Jeff,

    Sorry, I can’t make myself read the nonsense… in might induce vomiting.

  3. Brian H said

    Lesson: Don’t derive vast conclusions from half-vast assumptions.

  4. Genghis said

    It is extremely hard to change what someone intuitively knows is correct. Forcing someone to defend their position, generally just hardens their belief. Someone once said that, ” It is not possible to reason anyone out of a position that they did not reason themselves into in the first place.”

    That is the purpose of the Zen Koans, to disconnect the normal emotional responses (what most people think is thinking), and make them actually reason for a while. But koans can be used in reverse, Hansen used a koan (a greenhouse) to promote his ‘truth’, and allowed people to follow their natural instinct and trust intuition.

    So how do you change what someone knows to be true? Anyone can go outside and feel the heat of the sun, and at night feel that the moon doesn’t provide any heat at all, but it does illuminate. To make matters worse we are trying to make people believe that under a freezing cold, clear sky at night, that invisible gases are actually radiating down 250 W/m2. I can make at ice at night in the summer where I live : )

    In the end, this is what makes the AGW debate so interesting. This debate won’t be ended until there is a general shift in thinking, we need a good koan.

  5. Carrick said

    The causality argument is a good one, Jeff.

    Even if you assume “sentient photons” that somehow can get emitted from the colder object and know to not get absorbed by a hotter object (or even better his nonsensical interpretation, being ethical photons, they know better than to ever get emitted, if they’re going to eventually get absorbed), for any finite separation, they would have to know acausally that the other surface is warmer.

  6. JWR said

    @4 Genghis

    Your satements are very true.
    It works in the two directions of the people debating here back-radiation!

    @Jeff

    We now agree that you studied my paper. You missed however the definition of “f”. There is vacuum between the plates, and “1-f” represents holes in the plate.
    But you were reading it with the intention to say that one-way heat flow is wrong. (see post 4 Genghis).
    Not every body will agree that I am doing circular reasoning, in fact you mix up conclusions from implementation B to argue it in implementation A!
    I separated them on purpose. I found that the temperature distribution for the two different implementations of the one-slab model were equal.
    But I found a discreapancy in the absorption!

    In fact I do not have to say more than I wrote already in the paper. Every body can judge for his own.

    To you I can say, suppose that implementation A was not shown to you.
    You than had only implementation B.
    What does plate B absorb, and, in steady state conditions, emits immediately?
    Is it 2fq/(2-f)? Or for a completely opaque plate with f=1, is it 2q?
    And for a stack of N completely opaque plates , would the lower plate absorb 2Nq?
    It is shown in the paper in the section following the one-slab section.

    If you agree with me that physically it can be argued that in the one-slab model plate 2 absorbs q and not 2q,
    then something is wrong with implementation B. At least you should agree with that.
    All the 60 one- slab models you can find in text books ( see compilation of Hans Schreuder) teach students something wrong:
    a model of a one-slab atmosphere where the atmosphere absorbs and re-emits twice the heat.

  7. steveta_uk said

    My question was and still is, how does the surface know to emit less energy when the second surface exists in a different timeframe? If the second surface vanishes instantly, does the first return to emitting a proper amount of radiation?

    Each surface, from its own point of reference, sees the other surface as behind itself in time. The amount each sees the other as behind (X) is equal to the time light takes to travel across the distance between the objects.

    Ah, but you’re ignoring relatavistic effects.

    The emitted photon from the hotter surface is of course moving at the speed of light (they like to do that). Due to infinite time dilation, it exists for no time whatsoever in it’s own space-time frame.

    Therefore, it can prevent itself from leaving the hotter surface if the destination is unable to receive it. Since the message back occured in zero time, in the photon’s time frame, then there is no causality issue.

    And if you think this is rubbish, try explaining entanglement.

  8. steveta_uk said

    Just had a thought. Anyone with a child knows what a fever is like, and knows that when you hold your hand close to the feverish child’s forhead, you can easily feel the higher temperature.

    You can even do this to youself, with no fever involved, though it’s harder to detect.

    Since your hand is normally several degrees cooler than your forehead, and certainly cooler than a feverish forehead, what exactly are you feeling?

  9. Carrick said

    Steveta_uk:

    Therefore, it can prevent itself from leaving the hotter surface if the destination is unable to receive it. Since the message back occured in zero time, in the photon’s time frame, then there is no causality issue.

    Therefore, it can prevent itself from leaving the hotter surface if the destination is unable to receive it. Since the message back occured in zero time, in the photon’s time frame, then there is no causality issue.

    I think it is rubbish, sorry. Acausal events happening at two physically separated points completely generally don’t have a well-defined temporal order. So in one frame of reference, the photon would be querying before hand whether the temperature of the individual molecule or atom that it desires to tangle with has a higher mean kinetic energy than it will have at the point when it encounters it, and in another frame of reference it won’t be querying until would it already have to have left in order for it to travel at the speed of light to meet its potential future molecular or atomic partner.

    Such a theory is manifestly in violation of causality, even if you could come up with a “smart photons” that know which molecules and atoms they are allowed to jeehaw with and which they are not.

    In quantum entanglement, causality is preserved, so that is not a counter example where this sort of thing could happen.

  10. Carrick said

    JWR: The actual theory of infrared radiation and absorption is at the atomic/molecular/photonic level. If you are getting a theory that violates causality, that should be a tripwire for you to let you know you’ve made a mistake:

    What you have done is wrong, and we know it for many reasons, a) we know cooler photons can absorb into a warmer body because we can measure it, b) there simply is no physical mechanism by which the individual photons could be prohibited to prevent them from being absorbed and c) your theory violates causality.

    It is sufficient to show your theory is impossible and vastly in conflict with measurement to eliminate it, pointing out the actual errors and getting you to admit to them is not necessary.

    I don’t need to find your math mistakes for you, nor does Jeff or anybody else. We needn’t work for free. You should realize they are there and fix them, or concede you aren’t competent to do the math.

  11. Jeff Condon said

    Steveta,

    Entanglement is a different effect which is not immune to the speed of light either. There is a reason that quantum entanglement does not communicate information instantaneously as your comment incorrectly assumes. I have my own ideas but am not an expert. Maybe someday I’ll write something on that and we can all rip it apart.

    JWR,

    I am making the same statements here that I made in the comment thread before. It is not because I have read further. Your conclusion is present in your primary assumption, you can simply skip the math part as it is identical in both cases. My definition of f sounds exactly the same as yours. Holes in plate or whatever physical definition you want to give it.

    this is where you are flat wrong:

    “If you agree with me that physically it can be argued that in the one-slab model plate 2 absorbs q and not 2q, then something is wrong with implementation B.”

    It is a logic error. The difference in absorption – which was your assumption – is simply a difference in absorption. It does not negate B any more than it negates A. Why you assume it negates B rather than A is a mystery. We have to look elsewhere from your math to find answers.

    Besides the thousands of devices which wouldn’t work at all if you were correct, or the endless measurements which wouldn’t give the same results (see the camera above), your argument really does violate causality. Using this logic, you could mathematically argue that the light output causes the lamp filament to glow. The math doesn’t care but we are creatures who progress forward in time.

  12. steveta_uk said

    Carrick – I knew that – I was just hoping someone could explain entanglement ;(

  13. steve fitzpatrick said

    Carrick #10,
    Yes, the key is to look at the physical events associated with the absorption or emission of a photon. I mean, a transition from one energy state to another (in the case of infrared, usually molecular vibrational states) is always associated with absorption or emission. For emission from a gas, the direction of emission is randomized by the completely random nature of molecular orientations. When someone’s says the emission process is controlled and directed by the temperature of a specific target molecule located in a distant object (perhaps light years distant!), I can only roll my eyes in disbelief. And what about radiant loss to empty space? Where is the controlling molecule located, and how is it communicating with the emitting molecule?

  14. Genghis said

    Maybe I can reconfigure the problem : )

    Lets take two perfectly conducting plates that are touching so that kinetic energy (heat) can be transferred between the plates. We place a heater in plate one and heat it up to 100˚, plate two because it is in contact with plate one will be at 100˚ also by the time plate one heats up to 100˚ via kinetic energy transfer. Both plates will be radiating exactly equal radiation in all directions. Now lets create a vacuum between the 100˚ plates, both plates will be radiating 1099 W/m2 between them and both plates will stay at 100˚ as long as plate one is continuously heated.

    There is no conceptual difference between two plates touching and exchanging kinetic energy or spaced apart and exchanging radiation energy.

  15. Bruce said

    Steveta_UK: Just had a thought. Anyone with a child knows what a fever is like, and knows that when you hold your hand close to the feverish child’s forhead, you can easily feel the higher temperature.”

    Good analogy.

    And if you stand outside at night and put your palm towards the 400W/sqm backradiation supposedly flowing down to earth … what do you feel?

    Nothing.

    If you put a 400W heater in a room and put your hand towards it what do you feel?

    Heat.

  16. Anonymous said

    Bruce you feel about 35°C warmer than you would have were there no back radiation.

    The total weight of the air above you is about 15-pounds per square inch (100,000 Newtons per meter squared if you prefer).

    If I put a 15-lb bowling ball in your hand, can you feel it?

    Why can’t you feel the weight of that column of air that really is sitting on top of you, shoving you down?

    (Bonus question: why is it that it is actually pushing you up slightly, not shoving you down?)

  17. Bruce said

    Anonymous: “Bruce you feel about 35°C warmer than you would have were there no back radiation.”

    So if you put say 1 foot of the best insulator known to man between you and the night sky, your hand would freeze?

    Cool. But not believable.

    But is there 400W per square meter of back radiation at night?

    Why do deserts drop as much as 50C at night if all that “back radiation” is flowing down warming up the desert?

  18. Jeff Condon said

    Actually, I like this:

    “And if you stand outside at night and put your palm towards the 400W/sqm backradiation supposedly flowing down to earth … what do you feel?”

    You feel the same thing on both sides of your hands. Is it reasonable to say that there is no wattage coming from the earth surface? The question should be to those who don’t beleive backradiation exists – Why don’t you feel a difference?

  19. A. C. Osborn said

    Genghis said
    January 27, 2012 at 11:28 am

    Maybe I can reconfigure the problem : )
    Not according to Back radiation, the non electrically heated plate will heat up the electrically heated plate to more than 100C.
    Which will heat up the non electrically heated plate even more, which will heat up the electrically heated plate even more, hence runaway global warming.

  20. A. C. Osborn said

    Mr Condon, as the 400W can heat the earth, can you explain why it can’t be used to do any work.
    In fact whenever anyone tries to use it it actually causes “Cooling”?

  21. Carrick said

    I see the comparison with a bowling bowl whizzed over Bruce’s head. Friggin’ bowling ball and he still can’t follow it. I swear a bowling ball is less dense….

    AC Osborn, you can only do work when you have a net exchange of heat energy. Secondly, the 400 W isn’t an energy source, it represents a deficit in heat lost. So for the hot plate example, runaway is not possible, because the feedback has to be less than one.

    Let’s see if you can work out the remaining answers for yourself. The first step is to accept your own understanding is grossly flawed.

  22. A. C. Osborn said

    Mr Condon, using the backradiation calculations, on a cloudless night sky which will warmer, a Car exposed to the radiation or a Car protected by a carport roof?

  23. Jeff Condon said

    #22,

    The carport would be warmer.

  24. Bruce said

    Jeff Condon: “You feel the same thing on both sides of your hands. ”

    Not necessarily. Imagine it is 30 minutes after sunset and you hold you hand over a sidewalk. The palm faces the night sky. Do you feel warmth on the back of your hand? Of course you do. Because the heat radiating from that warmed up the sidewalk is real.

    Again, back to the insulation example carrick (and others) pointedly ignored (by throwing out childish insults)..

    You put your hand outside palm up. It is cold. One foot above your hand there is 1 foot of the best insulation you can buy.

    Does your hand cool less because of the imaginary “deficit in heat lost” isn’t available?

  25. Jeff Condon said

    Bruce,

    I agree with your comment about heat from the sidewalk being greater. So what is the big deal?

    This whole issue doesn’t seem very emotionally charged to me. It is a simple concept which is misunderstood by climate skeptics far too often. We can see the camera working, nobody can deny that it is detecting radiation from a surface cooler than itself. The cooler surface just sends less radiation than a warm one.

    I wonder why those who don’t believe in backradiation from cold to hot – not sure of your position – don’t find that camera amazingly perplexing.

  26. Eric Anderson said

    One point, perhaps relevant here to Bruce’s question. Your hand (body generally) doesn’t detect temperature per se. It detects rate of transfer to or away from your body. When you wake up in the morning and step on the floor barefoot, a tile floor feels significantly colder than carpet, even though they are both at exactly the same room temperature. In the one case there is a quick and significant flow from the feet to the tile; with carpet there is a much slower and less significant flow.

  27. Jeff Condon said

    Eric,

    I have also used the example of opening the freezer door while standing outside of the range of convection. You can feel the cool on your face immediately.

  28. kuhnkat said

    Jeff Condon,

    electron entanglement is instantaneous. Explain why this wouldn’t be??

  29. Carrick said

    Kuhnkat:

    electron entanglement is instantaneous. Explain why this wouldn’t be??

    It’s not instantaneous because that would violate causality, and further there is no way to build a self-consistent theory in which entanglement is truly instantaneous.

    If you have event A happen at time T1 and position P1, and event B happen at time T2 and position P2, such that |P2 – P1| ≥ c |T1 – T2|, then depending on the frame of reference T1 > T2 or T1 T1 in our frame of reference. There exists another frame of reference where the order of the events is reversed, and B will appear to happen in that frame of reference before A. Cause and effect got scrambled: The effect comes first in this other relativistic frame of reference.

    There is some confusion because people use Shrödinger’s formulation of quantum mechanics to do these sorts of calculations, but this formulation is specifically non-relativistic, and of course entanglement looks instantaneous in that frame of reference.

  30. Carrick said

    A slightly easier way of seeing this… let’s assuming the entanglement is instantaneous in our (assumed to be inertial, that is non-accelerating) frame of reference.

    That means in our frame of reference, T1 = T2. This is nothing more than the definition of instantaneous. For this we need only require P1 ≠ P2 for the system to be acausal.

    There exist relativistic inertial frames of reference for which T1 T2…. that is, not instantaneous in other frames of reference.

  31. joshv said

    A cooler surface cannot warm a warmer surface via black body radiation, it can however cause the warmers surface to cool more slowly.

  32. Jeff Condon said

    Joshv,

    I think your statement represents the problem that frustrates Carrick about thermodynamic terminology. While what you say is in practice usually correct, it is often misinterpreted because it is also true that a cooler surface does add radiant energy to a warmer surface. The warmer surface in your concept radiates more than the cooler one adds to it, so the result is to cool more slowly.

    However —

    Consider a case where the warm sphere body in an infinite 0K vacuum, it has a fixed thermal energy input and has reached an equilibrium temp of 300K. We .add a nearby sphere in the same condition having a stable temp of 100K. The second cooler sphere will then increase the temp of the previously temp-stable warmer sphere because its radiant energy impinges on, and is absorbed by, the warmer sphere’s surface. More net input energy = higher temp.

  33. Genghis said

    Joshv,

    The cooler surface is emitting radiation (everything is emitting radiation), when that radiation is absorbed by the warmer surface and converted into kinetic energy it does indeed warm the warmer surface. That is precisely how it causes the warmer surface to cool more slowly.

  34. Anonymous said

    Let me start by hitting the nuances — Clausius’s Principle is not being violated when you have radiation being emitted by a cooler object and absorbed by the warmer one, because the warmer object transfers more heat energy (internal kinetic energy) to the cold object than the cold object does to the warm one.

    Clausius’s Principle states that “No process is possible whose sole result is the transfer of heat from a body of lower temperature to a body of higher temperature.”

    Since the sole result is a net transfer of heat energy from the body of higher temperature to the cold one.

    Because of that transfer of energy, the warmer object is losing net heat energy and is therefore still cooling, but at a slower rate than it would have were the cold object not present.

    Now to the nuance, what does it mean to “warm”?

    I use the definition “warm” to mean to increase the temperature of an object. So if I am “warming” it, I am causing its temperature to increase.

    If I am “cooling” it, I am causing its temperature decrease.

    If an object is “warmer” than a nearby object, that means its temperature is higher than a surrounding object.

    If an object causes another object to cool less quickly that means the object is simply acting as an insulator for the other object.
    _____

    With that pedantry out of the way, in Jeff’s example, the warmer object warms up not because heat energy transfer from the colder object is warming the warmer object, but because the warmer object has its own heat energy source, and the colder object is acting merely as an insulator that helps reduce the rate of heat energy loss by the hotter object.
    _____

    Back to the sink analogy where you have the faucet turned on, and you’ve constricted the drain causing the water level to rise, the extra water, regardless of mechanism of constriction, comes from the faucet, not the source of constriction.

  35. Carrick said

    Sorry that “anonymous” post was me. Somehow my fields got reset on this computer.

    But you can see why this statement is not quiet correct “when that radiation is absorbed by the warmer surface and converted into kinetic energy it does indeed warm the warmer surface”… it doesn’t actual warmer the warmer surface because the warmer surface is emitting more infrared radiation than it is receiving , hence it still is cooling (not warming), just not as rapidly.

  36. Bruce said

    You put your hand outside palm up. It is cold. One foot above your hand there is 1 foot of the best insulation you can buy.

    Does your hand cool less because of the imaginary “deficit in heat lost” isn’t available?

    Where does the “backradiation” appear in the above scenario?

    Could you detect a difference if the insulation was not 1 foot above your hand?

  37. Carrick said

    Bruce:

    Where does the “backradiation” appear in the above scenario?

    As warmer air around you. Again, it’s not a heat source, it represents an additional source of insulation against heat loss.

  38. Carrick said

    The ground radiating off stored thermal energy at night is your heat source.

  39. A. C. Osborn said

    So how does a Car under a Flat Surface like a Car Port remain Warmer than a Car right next to it exposed to DWLIR?
    Please do not say that the surface is reflecting the IR from the Ground because with the slightest of breezes any heat under there will be mixed with the surrounding air, probably even before it reaches the surface.
    Also the Ground is supposed to be emmitting less than 400w and would also be warming the Car not under the Car port.
    Also why does air cut off from the Ground get colder than air that is not and is only exposed to DWLIR, as has been shown by many experiments, including the one by Dr Spencer Snr?

  40. A. C. Osborn said

    Can UWIR pass through a car to make the Roof warmer?

  41. Carrick said

    AC Osborn:

    Please do not say that the surface is reflecting the IR from the Ground because with the slightest of breezes any heat under there will be mixed with the surrounding air, probably even before it reaches the surface.

    Back radiation from the top of the car port, which is still there, even when there is wind.

    Generally when there is wind, the surface air actually warms up at night, by the way. That’s because there’s a temperature inversion and the air say 10-m up is hotter than the air near the ground, and the turbulence associated with the wind mixes these together, warming up the air near the ground, but cooling it further up.

  42. Carrick said

    Can UWIR pass through a car to make the Roof warmer?

    The car emits long-wave infrared radiation too, more or less isotropically, and part of it gets absorbed by the roof.

    So in an indirect way, “yes”.

  43. Jeff Condon said

    This concept is a lot easier than people make it.

    Nearly everything with a temperature lights up (emits photons).
    The light isn’t always visible to our limited sensors – eyes – but it is still there.
    The light has a probability of being absorbed by whatever it hits. Quantum probability.
    After absorption of the light, the absorbing object is warmer than it would have been without the light.

    A dim (low output) light transfers less energy than a bright one. It still transfers energy. The net is from bright to dim.

    the end.

    All this talk of cooling slower, hot to cold etc, are all mutations (sometimes wrong) of the basic concept.

  44. Carrick said

    Jeff, the not so simple part of the problem is working out where the infrared photons are coming from that impinge upon our skin, as we sit in the surface boundary layer, blanketed by a layer of (relatively) high-humidity air.

    I can guarantee you that the majority of the long-wavelength down-welling photons impinging on your hand did not get emitted at the top of the atmosphere.

  45. Yes, Jeff, everything with a temperature lights up (emits photons).

    The light has a probability of being absorbed by whatever it hits. After absorption, the absorbing object is warmer than it would have been without the light.

    Thus the cooler, dimmer object transfers less energy than a brighter, hotter one. But it still transfers energy.

    The net result (in the absence of lens to focus or disperse light) is heat energy flows from the brighter to the dimmer object.

  46. harrywr2 said

    Jeff Condon said
    January 27, 2012 at 6:13 pm

    It is a simple concept which is misunderstood by climate skeptics far too often

    It’s a concept that is misunderstood by many of every perspective imaginable.

    Net Heat always flows from Warm to Cold. I myself was taught wrongly that ‘heat always flows’ from warm to cold.

    Hence, ones hand is always going to feel cold when raised to the night sky because the hand is emitting more heat to the the night sky then the night sky is emitting to the hand.

    Of course if we went to the moon, where there is no atmosphere I hand would feel really,really cold at night and probably freeze and fall off.
    .

  47. Genghis said

    Carrick said

    “But you can see why this statement is not quiet correct “when that radiation is absorbed by the warmer surface and converted into kinetic energy it does indeed warm the warmer surface”… it doesn’t actual warmer the warmer surface because the warmer surface is emitting more infrared radiation than it is receiving , hence it still is cooling (not warming), just not as rapidly.”

    You are “kind of” wrong : ) (emphasis on the “kind of”) Radiation and heat are not directly related.

    Let me illustrate. Let’s take two hollow perfect black body spheres, one with a radius of 1 and the other with a radius of 2, and put them at the same distance from the sun such that they are both receiving 400 W/m2 from the sun. They will both be 290K and radiating 400 W/m2 in all directions.

    Now let’s put the smaller sphere inside the larger sphere. The larger sphere has 4 times area or the smaller sphere and the smaller sphere will be receiving/transmitting 1600 W/m2. Now the outer spheres temperature is still 290K, but the inner spheres temperature is 410K. This is indeed a case of a cooler body heating up a warmer body. The net energy difference for the systems is zero. This is also the green house effect.

    This example also illustrates why the current AGW theory is wrong, the earth is not a super conducting sphere.

  48. Bruce said

    Carrick: “As warmer air around you”

    The air may be warm around my hand, but the heat loss has ZERO to do with the imaginary backradiation.

  49. Genghis said

    Bruce said – “The air may be warm around my hand, but the heat loss has ZERO to do with the imaginary backradiation.”

    You do understand that whether or not the air feels warm to your hand has absolutely nothing (almost anyway) to do with radiation of any kind?

  50. curious said

    47 Genghis –

    “Let me illustrate. Let’s take two hollow perfect black body spheres, one with a radius of 1 and the other with a radius of 2, and put them at the same distance from the sun such that they are both receiving 400 W/m2 from the sun. They will both be 290K and radiating 400 W/m2 in all directions.”

    Sorry Genghis I must be missing something here. If I have a 1m2 conducting bb plate aligned normally to the sun at a distance where the SC equals 400W/m2, won’t it stabilise at a temp. equivalent to outgoing radiation of 200Wm/2 as it will radiate from both sides?

  51. Genghis said

    curious said – “Sorry Genghis I must be missing something here. If I have a 1m2 conducting bb plate aligned normally to the sun at a distance where the SC equals 400W/m2, won’t it stabilise at a temp. equivalent to outgoing radiation of 200Wm/2 as it will radiate from both sides?”

    You would think so wouldn’t you? The total energy absorbed must equal the total energy radiated. So how can a plate that is receiving 400 W/m2 on one side radiate 400 W/m2 on both sides? That is 800 W/m2, double what it is receiving and obviously it can’t do that. An interesting conundrum : )

    Let’s go back to the two plates that are first heated together and then separated just far enough that only radiation can transfer energy between them. Now instead of a single plate transferring 400 W/m2 left and right we now have two plates transferring 400 W/m2 both left and right. The secret though is that the net energy flow between the two plates is zero. The plate on the left side is simultaneously absorbing and radiating 400 W/m2 from/to the plate on the right which is doing the reverse. This is that mysterious back radiation : )

    The net result is that the plate on the left is transmitting 400 W/m2 to the left and the plate on the right is transmitting 400 W/m2 to the right, with 800 W/m2 or zero between them, take your pick. It also doesn’t matter which plate has the electric heater in it.

  52. kuhnkat said

    Carrick,

    I notice you only speak of photons as many modern types do. Do you have a reference you could give me that explains the 2 slot experiments with photon only?? My limited reading only covers duality.

  53. Carrick said

    Kunhkat, I’m really not sure what you’re looking for here because you didn’t provide enough information.

    Individual potons have a “wavelike/particle like duality” even for single photons unless you do something to “collapse the wavefunction”, just as with electrons or any other particle.

    The two-slit interference pattern was first discovered in the context of continuous light (enough photons superimposed so as to look continuous). It wasn’t until the photoelectric effect was explained by Einstein that the particle/wave nature was understood. Later it was demonstrated you could get exactly the same effect of a two-slit interference by reducing the rate of photon emission to say one every 15-seconds. (That rate btw is roughly the threshold of vision in for a dark-adapted human retina).

  54. Carrick said

    By the way, the photon is a an example of an elementary particle that you can’t treat with Shrödinger’s equation nor does Heisenberg’s uncertainty principle (which is non-relativistic) apply.

    The reason is simple, the photon is massless, and therefore intrinsically relativistic. There is no frame of reference where it is at rest.

  55. steve fitzpatrick said

    Carrick,
    “the photon is massless”.

    Humm… E = MC^2. The energy of a photon is clearly defined by its wavelength, so it seems you can calculate an energy equivalent mass (M = E/C^2). Light is bent by gravitation fields, and can’t escape from a black hole (indicating a gravitational attraction). Indeed, light leaving any massive object is red shifted due to the energy required to “escape” the object’s gravitational field….. some of the energy content was ‘used’ to do the work required for remaining energy equivalent mass of the photon to escape the object’s gravitational field. Photons have physical momentum (as shown by Einstein) of E/C, which is of course exactly equal tot he momentum of a particle with a mass of (E/C^2) multiplied by its velocity C. So while photons are intrinsically relativistic, I am not sure you can say they have no mass; they behave in a lot of ways as if they have mass. I guess it depends a little on how you want to define mass :-)

  56. Carrick said

    Let me rephrase that Steve.

    The photon has a zero REST MASS.

  57. Carrick said

    And whilst you were being tongue in cheek, and verily I did chuckle, the term massless particle has a particular technical meaning.

    The only observed “massless particle” is the photon. The other one is the gluon which is “known” to exist, but never observed directly (other than as gluon jets).

    Interestingly enough, the unitary generalization of Schrödinger’s equation… the Dirac Equation, also does not properly treat the photon, since it deals with spin 1/2 particles. The other equation, the Klein-Gordon equation, with spin-less particles (photons are spin one so cross it off already) has the problem of being non-unitary, so no direct correspondence between its solution and the solutions of Schrödinger equation are possible.

    “Negative probability is a bitch” I think Feynman was known to say.

  58. steve fitzpatrick said

    Carrick,

    I was being tong in cheek of course, but I think it important to see the smooth connection between relativistic and non-relativistic physics… there is only one. The wavelength of a photon can be calculated from its energy equivalent mass just like the wavelength of a particle with a non-zero rest mass, like an electron: you only have to convert the particle’s rest mass to energy content and the wavelength calculation becomes identical for both photons and non-zero mass particles.

    For photons
    lamda = (h * C)/E
    where E is the photon energy, C the speed of light, and h is Plank’s constant

    For other particles (with rest mass)
    lamda = h / (M * V)
    where M is the particle mass and V the particle velocity

    but M = E/C^2, so lamda = {(h * C)/E} * (C/V)

    which is identical in form to the photon: lamda = {(h * C)/E} * (C/C)

    So the two equations are the same when the rest mass is converted to energy and the ratio of the actual particle velocity to the speed of light is explicitly stated… for photons, the ratio is 1. Of course, the mass of any particle other than a photon grows at relativistic speeds! :-)

  59. Carrick said

    Steve, of course I knew you were being tongue in cheek. None the less there are always fun things to mulligrub about here.

    There is a classical physics equivalency one can discuss between the photon as particle as wave and a quantum mechanical equivalency between non-zero-rest-mass particle as particle or wave.

    However, when one goes to truly quantum mechanical systems, it turns out you need to use full glory of quantum field theory to adequately describe photons:

    You can’t use Schrödinger’s equation nor any of its first quantized cousins, e.g., the DIrac or Klein-Gordan equations.

    All of this was a bit of a surprise at the time. Parts of this continue as a mysterious not yet fully sorted out (mathematically or interpretationally.)

  60. steve fitzpatrick said

    Carrick,

    All of this was a bit of a surprise at the time.

    And I think, intellectually, it remains so. I understand the apparent disconnect between the Heisenberg uncertainty principle and particle wavelength. Yet it seems intuitively correct that a particle’s ‘wavelength’ must convey information about the uncertainty pairs that Heisenberg described. My purely intuitive guess: we are thinking about it the wrong way, and there is probably a connection that is not yet apparent.

  61. Brian H said

    steve fitzpatrick said
    January 30, 2012 at 8:40 pm | Reply w/ Link

    there is probably a connection that is not yet apparent.

    Wrong. I’m throwing googolplex dice googolplex times per second.
    Wah-ha-ha-ha-ha!

    – God

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