LWIR, why so mysterious and difficult to measure
I M @ good guy
It is what cools the atmosphere, so it must be capable
of a substantial flux through a given cross section,
over time it must be capable of at least 80 percent
of solar insolation!
Here is a reference to a book, at first glance
it looks like upwelling LWIR or broadband IR is
measured and downward LWIR calculated,
This is starting to be difficult.
columbiaaccidentinvestigation
starting to be difficult."
laughing...
Rob Dekker
"I M @ good guy" <I...@good.guy> wrote in message news:6k34g59uq36j1k220...@4ax.com...
Chris Colose has a pretty good "introduction to scimatology" paper right here, dealing with long-wave (IR) radiation and a lot more
:
http://chriscolose.wordpress.com/2009/10/08/re-visiting-cff/#more-597
It's not so difficult to understand, if you just know a little algebra
Rob
Bill Ward
> It is what cools the atmosphere, so it must be capable of a substantial
> flux through a given cross section, over time it must be capable of at
> least 80 percent of solar insolation!
Energy is carried through the troposphere by convection, not radiation.
LWIR is absorbed by WV, resulting in a warmer gas in local thermodynamic
equilibrium (LTE).
Maybe this will help:
<http://en.wikipedia.org/wiki/Thermodynamic_equilibrium>
I M @ good guy
wrote:
Aren't you able to follow the discussion? The issue is
downward LWIR as causing the present atmosphere to be
warmer than if there was NO GHGs and no water.
The claim of downward LWIR from the stratosphere
where there is not much water has been claimed to be
substantial. In order to separate this radiation from
that of water vapor, either frequency filters are needed,
or measurements of the LWIR flux at various altitudes
to show where it is coming from.
I M @ good guy
<bw...@ix.REMOVETHISnetcom.com> wrote:
>On Mon, 16 Nov 2009 21:51:59 -0500, I M @ good guy wrote:
>
>> It is what cools the atmosphere, so it must be capable of a substantial
>> flux through a given cross section, over time it must be capable of at
>> least 80 percent of solar insolation!
>
>Energy is carried through the troposphere by convection, not radiation.
For AGW to have any merit at all, the claim of downward
LWIR from CO2 needs to be measured.
The only place this might be measurable is at altitude
below the stratosphere.
>LWIR is absorbed by WV, resulting in a warmer gas in local thermodynamic
>equilibrium (LTE).
And can be confined to a few dozen meters range.
>Maybe this will help:
>
><http://en.wikipedia.org/wiki/Thermodynamic_equilibrium>
LWIR from CO2 is the issue, if it can't be measured as
a cause for warming the mid or lower atmosphere, AGW is
pure hogwash.
columbiaaccidentinvestigation
> On Mon, 16 Nov 2009 21:51:59 -0500, I M @ good guy wrote:
>
> > It is what cools the atmosphere, so it must be capable of a substantial
> > flux through a given cross section, over time it must be capable of at
> > least 80 percent of solar insolation!
>
> Energy is carried through the troposphere by convection, not radiation. Â
> LWIR is absorbed by WV, resulting in a warmer gas in local thermodynamic
> equilibrium (LTE). Â
>
> Maybe this will help:
>
> <http://en.wikipedia.org/wiki/Thermodynamic_equilibrium>
>
>
>
> > Â Â Â Â Here is a reference to a book, at first glance
> > it looks like upwelling LWIR or broadband IR is measured and downward
> > LWIR calculated,
>
> >http://books.google.com/books?id=9M6hRQKvzasC&pg=PA265&lpg=PA265&dq=LWIR
>
> +and
> %20and%20lte&f=false
>
>
>
>
>
> > Â Â Â Â This is starting to be difficult.- Hide quoted text -
>
> - Show quoted text -
maybe you should explain how a pyrgeometer is calibrated...
BDR529
> On Mon, 16 Nov 2009 21:51:59 -0500, I M @ good guy wrote:
>
>> It is what cools the atmosphere, so it must be capable of a substantial
>> flux through a given cross section, over time it must be capable of at
>> least 80 percent of solar insolation!
>
> Energy is carried through the troposphere by convection, not radiation.
You are trying to fool everyone with silly statements, any first year
physics student knows that radiation, convection and conduction are all
possible in the atmosphere.
The Earth's outgoing longwave radiation will be absorbed everywhere
where there are molecules that can be exited between 2 and 100
micrometer. Water vapor does it, CO2 does it, Methane, etc etc.
There is also excellent software to do this calculation yourself, it is
called modtran for instance.
> LWIR is absorbed by WV, resulting in a warmer gas in local thermodynamic
> equilibrium (LTE).
That is all fine and well, but water vapor is not the only gas absorbing
the outgoing longwave IR mostly in the lower troposphere. There was
always water vapor, there is water vapor, and there always will be water
vapor, AND, it depends on the local conditions of the piece of the
planet you are looking at. With CO2 this is different, it is being added
slowly, on a global scale.
CO2 methane etc are however gases that mix pretty well throughout the
atmosphere. And therefor the energy absorption and re-radiation will
occur everywhere throughout the atmosphere. With water vapor this is not
the case, anyone with a little bit of knowledge knows that the
atmosphere is very arid at high altitudes.
Again, hinting to water vapor and LTE in the lower troposphere is not
the way to explain global warming which has a significant component in
the upper troposphere and stratosphere where there IS NO water vapor.
If you want to do the calculation yourself, then open your eyes once,
and look at the modtran code developed by the air force.
Modtran code follows from physical properties of gases measured in the
lab, it is able to explain observed radiometer spectra both on ground,
at high altitude and in space.
References can be provided on request.
Greetings,
Q
Rob Dekker
"I M @ good guy" <I...@good.guy> wrote in message
> On Mon, 16 Nov 2009 19:22:17 -0800, "Rob Dekker" <r...@verific.com>
> wrote:
>
>>
>>"I M @ good guy" <I...@good.guy> wrote in message
>>news:6k34g59uq36j1k220...@4ax.com...
>>>
>>>
>>> It is what cools the atmosphere, so it must be capable
>>> of a substantial flux through a given cross section,
>>> over time it must be capable of at least 80 percent
>>> of solar insolation!
>>>
>>> Here is a reference to a book, at first glance
>>> it looks like upwelling LWIR or broadband IR is
>>> measured and downward LWIR calculated,
>>>
>>> http://books.google.com/books?id=9M6hRQKvzasC&pg=PA265&lpg=PA265&dq=LWIR+and+lte&source=bl&ots=C40HIqCqaJ&sig=tLoRAUnUfN3QpyDKyO3JYAHueKs&hl=en&ei=5gwCS_XlLIjGlAekjLiICw&sa=X&oi=book_result&ct=result&resnum=6&ved=0CB4Q6AEwBQ#v=onepage&q=LWIR%20and%20lte&f=false
>>>
>>>
>>> This is starting to be difficult.
>>>
>>>
>>
>>Chris Colose has a pretty good "introduction to scimatology" paper right
>>here, dealing with long-wave (IR) radiation and a lot more
>>:
>>http://chriscolose.wordpress.com/2009/10/08/re-visiting-cff/#more-597
>>
>>It's not so difficult to understand, if you just know a little algebra
>>
>>Rob
>
> Aren't you able to follow the discussion? The issue is
> downward LWIR as causing the present atmosphere to be
> warmer than if there was NO GHGs and no water.
Mmm. And exactly how does downward LWIR change the temperature of the planet
?
Last time I checked, the planet as a whole cools by radiating IR to space.
That is : upward.
>
> The claim of downward LWIR from the stratosphere
> where there is not much water has been claimed to be
> substantial. In order to separate this radiation from
> that of water vapor, either frequency filters are needed,
> or measurements of the LWIR flux at various altitudes
> to show where it is coming from.
>
You are not making sense.
Of course, at any point inside the atmosphere, the radiation going up and
down is in close equilibrium.
Otherwise that particular layer of air would either warm up or cool very
quickly, which will then re-establish the quilibrium.
Bill Ward would call this an atmosphere in LTE (local thermal equilibrium).
No problem with that.
What matters for global warming or cooling, is how much radiation makes it
to space...
Did you read Colose's paper ?
>
>
>
>
>
>
Bill Ward
> Bill Ward wrote:
>> On Mon, 16 Nov 2009 21:51:59 -0500, I M @ good guy wrote:
>>
>>> It is what cools the atmosphere, so it must be capable of a
>>> substantial flux through a given cross section, over time it must be
>>> capable of at least 80 percent of solar insolation!
>>
>> Energy is carried through the troposphere by convection, not radiation.
>
> You are trying to fool everyone with silly statements, any first year
> physics student knows that radiation, convection and conduction are all
> possible in the atmosphere.
>
> The Earth's outgoing longwave radiation will be absorbed everywhere
> where there are molecules that can be exited between 2 and 100
> micrometer. Water vapor does it, CO2 does it, Methane, etc etc.
>
> There is also excellent software to do this calculation yourself, it is
> called modtran for instance.
>
>
>> LWIR is absorbed by WV, resulting in a warmer gas in local
>> thermodynamic equilibrium (LTE).
>
> That is all fine and well, but water vapor is not the only gas absorbing
> the outgoing longwave IR mostly in the lower troposphere. There was
> always water vapor, there is water vapor, and there always will be water
> vapor, AND, it depends on the local conditions of the piece of the
> planet you are looking at. With CO2 this is different, it is being added
> slowly, on a global scale.
There's enough WV to establish LTE wrt LWIR. Adding more CO2 makes no
significant change. The surface can radiate to space, but not in the
bands affected by WV or CO2.
> CO2 methane etc are however gases that mix pretty well throughout the
> atmosphere. And therefor the energy absorption and re-radiation will
> occur everywhere throughout the atmosphere. With water vapor this is not
> the case, anyone with a little bit of knowledge knows that the
> atmosphere is very arid at high altitudes.
>
> Again, hinting to water vapor and LTE in the lower troposphere is not
> the way to explain global warming which has a significant component in
> the upper troposphere and stratosphere where there IS NO water vapor.
So you've never seen anvil clouds? Towering cumulus can billow to above
40000 ft, well into the stratosphere.
I M @ good guy
wrote:
I am sorry to hear that after you claimed to find flaws
in a paper by prominent climate scientists, and must have
seen the graphs showing thermal transfer within the
atmosphere according to energy budget models, you still
don't know about the Earth's Energy Budget.
Have they changed it? Maybe?
http://en.wikipedia.org/wiki/File:LWRadiationBudget.gif
It sure looks like it;
http://en.wikipedia.org/wiki/File:57911main_Earth_Energy_Budget.jpg
Oh, no, here it is on page 10;
http://www.atmo.arizona.edu/students/courselinks/spring04/atmo451b/pdf/RadiationBudget.pdf
And the "back radiation is shown at;
http://www.windows.ucar.edu/tour/link=/earth/Atmosphere/earth_atmosph_radiation_budget.html
as 324 watts per square meter while the incoming solar
absorbed by the surface is only 168 watts per square meter.
The back radiation;
http://earthobservatory.nasa.gov/Features/EnergyBalance/page6.php
Its getting hotter, 333 watts per square meter back radiation
page 3;
http://polarmet.mps.ohio-state.edu/PolarMet/geog820_2009/Nicolas_2.pdf
>Last time I checked, the planet as a whole cools by radiating IR to space.
>That is : upward.
Right, mostly from the atmosphere.
But AGW seems to require "back radiation".
>> The claim of downward LWIR from the stratosphere
>> where there is not much water has been claimed to be
>> substantial. In order to separate this radiation from
>> that of water vapor, either frequency filters are needed,
>> or measurements of the LWIR flux at various altitudes
>> to show where it is coming from.
>>
>
>You are not making sense.
Can you read the graphs that show back radiation?
If there is supposed to be 324 watts per square meter
back radiation, then why can't it be measured at different
altitudes?
>Of course, at any point inside the atmosphere, the radiation going up and
>down is in close equilibrium.
That is nonsense according to the energy budget graphs.
>Otherwise that particular layer of air would either warm up or cool very
>quickly, which will then re-establish the quilibrium.
Oh really?
>Bill Ward would call this an atmosphere in LTE (local thermal equilibrium).
>No problem with that.
So you are ignoring back radiation now?
>What matters for global warming or cooling, is how much radiation makes it
>to space...
Right, GHGs cool the atmosphere, will more GHGs
cool the atmosphere more?
Will increasing atmospheric CO2 concentrations
cause an ice age?
>Did you read Colose's paper ?
Yes.
columbiaaccidentinvestigation
wrote:""
laughing, no reply from a person with nothing to offer.....
Crow
Nobody cares...
--
Crow.
columbiaaccidentinvestigation
cares..."
gosh, so should i care about your lame post, na. Now what was i
saying oh yeah, i was laughing at Imagoodguy because of his idiocy.
Now i can laugh at you as well, thanks...
BDR529
Sure, if you add all water on the planet in the equation than that is
probably true, but, if you refer to atmospheric water vapor then the
answer is simply no.
Water vapor is not a competing factor in the stratosphere.
>
>> CO2 methane etc are however gases that mix pretty well throughout the
>> atmosphere. And therefor the energy absorption and re-radiation will
>> occur everywhere throughout the atmosphere. With water vapor this is not
>> the case, anyone with a little bit of knowledge knows that the
>> atmosphere is very arid at high altitudes.
>>
>> Again, hinting to water vapor and LTE in the lower troposphere is not
>> the way to explain global warming which has a significant component in
>> the upper troposphere and stratosphere where there IS NO water vapor.
>
> So you've never seen anvil clouds? Towering cumulus can billow to above
> 40000 ft, well into the stratosphere.
Fortunately there isn't one over my head at the moment, so I'm not too
concerned.
But I can see from your reaction that you've absolutely nothing to
offer, water vapor does simply not explain global warming,
Good-day,
Q
Bill Ward
WV has already radiated its LWIR below the stratosphere. It doesn't need
to compete.
>
>
>>> CO2 methane etc are however gases that mix pretty well throughout the
>>> atmosphere. And therefor the energy absorption and re-radiation will
>>> occur everywhere throughout the atmosphere. With water vapor this is
>>> not the case, anyone with a little bit of knowledge knows that the
>>> atmosphere is very arid at high altitudes.
>>>
>>> Again, hinting to water vapor and LTE in the lower troposphere is not
>>> the way to explain global warming which has a significant component in
>>> the upper troposphere and stratosphere where there IS NO water vapor.
>>
>> So you've never seen anvil clouds? Towering cumulus can billow to
>> above 40000 ft, well into the stratosphere.
>
> Fortunately there isn't one over my head at the moment, so I'm not too
> concerned.
>
> But I can see from your reaction that you've absolutely nothing to
> offer, water vapor does simply not explain global warming,
Your ignorance is absolutely amazing. Does it take a lot of practice?
BDR-529
[snip]
> WV has already radiated its LWIR below the stratosphere. It doesn't need
> to compete.
That is complete bloody bugger, if something radiates then it radiates
still in all directions, and not just your direction you knucklehead.
[snip]
> Your ignorance is absolutely amazing. Does it take a lot of practice?
Perhaps I spot a little bit of anxiety? Could that be true when you're
continuously shown to be wrong by your opponents who just cite the facts?
How painful that must be, Billobwab.
Q
Bill Ward
> Bill Ward wrote:
>
> [snip]
>
>> WV has already radiated its LWIR below the stratosphere. It doesn't
>> need to compete.
>
> That is complete bloody bugger, if something radiates then it radiates
> still in all directions, and not just your direction you knucklehead.
Because of LTE, net energy begins to be radiated upward as the LWIR
optical depth approaches 1. That's up near the tropopause. Don't worry
if you don't understand, it's just over your head. That's what happens
when you're unable to think for yourself.
> [snip]
>
>> Your ignorance is absolutely amazing. Does it take a lot of practice?
>
> Perhaps I spot a little bit of anxiety? Could that be true when you're
> continuously shown to be wrong by your opponents who just cite the
> facts?
Nah, you're just projecting. Probably your desperation showing through.
> How painful that must be, Billobwab.
Tell us about it. It must be awful.
columbiaaccidentinvestigation
Tell us about it."
multiple personalities?
BDR-529
> On Tue, 17 Nov 2009 22:12:36 +0100, BDR-529 wrote:
>
>> Bill Ward wrote:
>>
>> [snip]
>>
>>> WV has already radiated its LWIR below the stratosphere. It doesn't
>>> need to compete.
>> That is complete bloody bugger, if something radiates then it radiates
>> still in all directions, and not just your direction you knucklehead.
>
> Because of LTE, net energy begins to be radiated upward as the LWIR
> optical depth approaches 1. That's up near the tropopause. Don't worry
> if you don't understand, it's just over your head. That's what happens
> when you're unable to think for yourself.
Energy is still radiated in ALL directions, there are no directional
antenna's in the problem.
The one trying to sell you otherwise is laughing his butt off.
>
>> [snip]
>>
>>> Your ignorance is absolutely amazing. Does it take a lot of practice?
>> Perhaps I spot a little bit of anxiety? Could that be true when you're
>> continuously shown to be wrong by your opponents who just cite the
>> facts?
>
> Nah, you're just projecting. Probably your desperation showing through.
>
>> How painful that must be, Billobwab.
>
> Tell us about it. It must be awful.
The real painful part is that you have no clue what your talking about.
Q
Rob Dekker
"I M @ good guy" <I...@good.guy> wrote in message news:a8u4g5dqhn3vjt8hq...@4ax.com...
Good Guy. It almost feels like we are talking a different language.
I really do not know where to begin.
I'm not even sure where you are in your understanding of radiative transfer and other basic physics concepts.
In short : What climate scientists call "back radiation" is absolutely there. It's the IR radiation from the atmosphere back to the
surface. Without that, Earth's surface would be about 33 C lower in temperature. This is also called the 'greenhouse effect' in
popular literature.
Back-radiation is also quite easy to measure :
You can measure this radiation if you look up into the sky (at a clear night preferable) with an IR detector.
If this detector would include a spectrum analyser, then you will see the CO2 lines and water-vapor lines light up nicely on an
envelope of the sky temperature. Incidentally, if you measure towards the ground, you will see a much more 'black-body' shape IR
spectrum, and the envelope will match the envelope of the ground.
Are you with me so far ?
My point was that none of this is relevant for the radiation budget of the planet as a whole.
Only the radiation that escapes to space is important.
Next question :
Since you read Colose's paper, did you fully understand it, or do you have any questions about it ?
Rob
>
>
>
>
Bill Ward
> Bill Ward wrote:
>> On Tue, 17 Nov 2009 22:12:36 +0100, BDR-529 wrote:
>>
>>> Bill Ward wrote:
>>>
>>> [snip]
>>>
>>>> WV has already radiated its LWIR below the stratosphere. It doesn't
>>>> need to compete.
>>> That is complete bloody bugger, if something radiates then it radiates
>>> still in all directions, and not just your direction you knucklehead.
>>
>> Because of LTE, net energy begins to be radiated upward as the LWIR
>> optical depth approaches 1. That's up near the tropopause. Don't
>> worry if you don't understand, it's just over your head. That's what
>> happens when you're unable to think for yourself.
>
> Energy is still radiated in ALL directions, there are no directional
> antenna's in the problem.
Whoosh... I warned you. Try thinking about the word "net".
> The one trying to sell you otherwise is laughing his butt off.
>
>
>>> [snip]
>>>
>>>> Your ignorance is absolutely amazing. Does it take a lot of
>>>> practice?
>>> Perhaps I spot a little bit of anxiety? Could that be true when you're
>>> continuously shown to be wrong by your opponents who just cite the
>>> facts?
>>
>> Nah, you're just projecting. Probably your desperation showing
>> through.
>>
>>> How painful that must be, Billobwab.
>>
>> Tell us about it. It must be awful.
>
> The real painful part is that you have no clue what your talking about.
Your projection is still showing.
columbiaaccidentinvestigation
Whoosh... "
bill at his best, missing as usual.....
I M @ good guy
wrote:
Just what is your first language?
>I really do not know where to begin.
That has become abundantly clear.
>I'm not even sure where you are in your understanding of radiative transfer and other basic physics concepts.
You think there is somebody that doesn't know em?
>In short : What climate scientists call "back radiation" is absolutely there. It's the IR radiation from the atmosphere >back to thesurface.
So they have measured the downward LWIR flux at
various altitudes, or do they just imagine back radiation
exists, and guess at the flux value?
>Without that, Earth's surface would be about 33 C lower in temperature.
Obviously you do not understand the science and
either remember the talking point phrases or look up
the gossip on a search engine.
Back radiation would only be possible with GHGs,
and the "saying" about 33 C warmer is not about GHGs,
it is about an Earth without an atmosphere.
If you don't know the difference, go back to
truck driving.
>This is also called the 'greenhouse effect' in
>popular literature.
By parrots. Lets see what the web says about the
mean temperature of the moon:
http://stars5.netfirms.com/datmoon.htm#specs
Minus 4 F, now that is a surprise, as the gossip
about GreenHouse Theory you mention causing the
mean temperature of Earth to be 33 C warmer means
the moon should be a lot colder than that.
http://www.asi.org/adb/02/05/01/surface-temperature.html
shows a 38 C difference between the Earth and moon.
Note the moon has no atmosphere.
You do know the difference between no atmosphere
and an atmosphere with no GHGs, don't you?
There is no good information about an Earth with an
atmosphere but no GHGs. I find many instances where
the writer assumes the moon reflects more light than the
Earth, causing other wrong assumptions.
>Back-radiation is also quite easy to measure :
So help me find the data measured at different
altitudes, surely this is not another crystal ball data.
>You can measure this radiation if you look up into the sky (at a clear night preferable) with an IR detector.
And you have done that, or know of somebody that has?
>If this detector would include a spectrum analyser, then you will see the CO2 lines and water-vapor lines light up nicely on an
>envelope of the sky temperature.
What would that tell me, I need back radiation in
watts per square meter to compare with the radiation
budget graphs.
>Incidentally, if you measure towards the ground, you will see a much more 'black-body' shape IR
>spectrum, and the envelope will match the envelope of the ground.
Why would I care?
>Are you with me so far ?
You are running around in circles, is that to get away from
the subject of measured back radiation at various altitudes?
>My point was that none of this is relevant for the radiation budget of the planet as a whole.
BS, if 342 watts were radiated back to the surface it
would mean the surface would have to radiate a lot more
than it absorbs from the sun.
More than 70 percent of the outgoing radiation
is from the atmosphere, and that doesn't jibe with a
large back radiation being absorbed by the surface.
>Only the radiation that escapes to space is important.
So GHGs radiating to space are important, because
less than 30 percent of the energy radiated to space
is direct from the surface.
>Next question :
>Since you read Colose's paper, did you fully understand it, or do you have any questions about it ?
>
>Rob
I did not find it to apply to my interest in measured
back radiation at various altitudes.
Rob Dekker
> On Wed, 18 Nov 2009 01:30:00 +0100, BDR-529 wrote:
>
>> Bill Ward wrote:
>>> On Tue, 17 Nov 2009 22:12:36 +0100, BDR-529 wrote:
>>>
>>>> Bill Ward wrote:
>>>>
>>>> [snip]
>>>>
>>>>> WV has already radiated its LWIR below the stratosphere. It doesn't
>>>>> need to compete.
>
>>>> That is complete bloody bugger, if something radiates then it radiates
>>>> still in all directions, and not just your direction you knucklehead.
>>>
>>> Because of LTE, net energy begins to be radiated upward as the LWIR
>>> optical depth approaches 1. That's up near the tropopause. Don't
>>> worry if you don't understand, it's just over your head. That's what
>>> happens when you're unable to think for yourself.
>>
>> Energy is still radiated in ALL directions, there are no directional
>> antenna's in the problem.
>
> Whoosh... I warned you. Try thinking about the word "net".
Bill, what are you trying to say ?
BDR is right that thermal radiation from atoms of the atmosphere is in all directions.
For most atmospheric analysis purposes, you are right that LTE is a reasonable assumption for every altitude in the troposphere.
Radiation levels (measured up or down) reduce by altitude, since it's simply colder up there.
These statements are not mutual exclusiove and in fact are built into all radiative transfer models.
So I fail to see any disagreement here.
Bill Ward
I thought the word "net" would give it away. Apparently not. The point
is that the "downward LWIR" is virtual in the sense that it can't warm
anything warmer than the source, or be detected by any thermal means
warmer than the source.
The only apparent purpose in mentioning it is to confuse people like Ima
into thinking it can heat the surface. It can't. The surface is cooling
slower than it would if the sink were colder, but absolutely no heating
of the surface can occur unless the surface is colder than the LWIR
source. The second law rules. Surely you know that.
BDR-529
[snip]
> I thought the word "net" would give it away. Apparently not. The point
> is that the "downward LWIR" is virtual in the sense that it can't warm
> anything warmer than the source, or be detected by any thermal means
> warmer than the source.
Please stop using terms like virtual downward LWIR while you already
know that radiation is omnidirectional. Don't try to fool us with your
goobligoop copied from the Miskolczi discussion pages, because that's
what you are trying to do.
> The only apparent purpose in mentioning it is to confuse people like Ima
> into thinking it can heat the surface. It can't. The surface is cooling
What the Hell it this supposed to mean? Such statements will effectively
kill any chance of getting results in a science journal, essentially
because reviewers don't understand you.
> slower than it would if the sink were colder, but absolutely no heating
> of the surface can occur unless the surface is colder than the LWIR
> source. The second law rules. Surely you know that.
What the hell are you talking about? Same remark.
Good, download a program called MODTRAN now, and stop this ridiculous
discussion. MODTRAN will solve the radiative transfer problem for you,
and it assumes that radiation, re-radiation, absorption and emission are
omidirectional. MODTRAN solves the following problem:
> http://en.wikipedia.org/wiki/Radiative_transfer
It is all about solving the following problem (cite wiki):
> The equation of radiative transfer simply says that as a beam of radiation travels, it loses energy to absorption, gains energy by emission, and redistributes energy by scattering. The differential form of the equation for radiative transfer is:
>
>
> \frac{1}{c}\frac{\partial}{\partial t}I_\nu + \hat{\Omega} \cdot \nabla I_\nu + (k_{\nu, s}+k_{\nu, a}) I_\nu = j_\nu + \frac{1}{4\pi c}k_{\nu, s} \int_\Omega I_\nu d\Omega
>
> where jν is the emission coefficient, kν,s is the scattering cross section,
See the wiki page for the rest of the explanation on how to solve the
resulting Eddington equations, and, apply it to a simple example. If you
are unable to do this for whatever reason then here is an example for
the Earth's atmosphere, it is all based on modtran, or better COART
which is founded on modtran where gas concentration are modified to see
what its effect is on IR absorption throughout the atmosphere:
Programs like COART have been successfully applied to spectra observed
by satellites and aircrafts. If you don't like what COART does then I
strongly suggest that you write your own program, write an article about
it, get it past the reviewers, and publish it in a journal. This is the
normal way of doing business in science.
Maybe take a look at figure 9 in http://tinyurl.com/ykbzofj
This shows the fundamental problem of the -3.15 W/m^2 at the tropopause
by a hypothetic doubling of the present day CO2 (380 ppmv). Water vapor
and methane I though is included in the discussion.
Hope this helps.
Q
I M @ good guy
Why does the construction of the detector need mentioned,
surely anybody attempting to measure LWIR is aware of what the
detector has to be,
Your comments suggest that the net flux flow must
be upward, which "back radiation" seems to be a conflicting
term.
>The only apparent purpose in mentioning it is to confuse people like Ima
>into thinking it can heat the surface. It can't. The surface is cooling
>slower than it would if the sink were colder, but absolutely no heating
>of the surface can occur unless the surface is colder than the LWIR
>source. The second law rules. Surely you know that.
Which has nothing to do with the measurement
of back radiation or downward LWIR, the upward LWIR
is a separate flow, it is not disrupted as if em was like
a liquid, water can't do much two way flowing, but em
can flow in all directions at once.
[ - Incidently, a magazine had an April 1st article - ]
[ - on "contra-polar light" about 54 years ago - ]
http://www.swtpc.com/mholley/PopularElectronics/Apr1955/PE_Apr_1955_pg04.jpg
[ - Page 27 Contra-Polar Energy - ]
If back radiation has not been measured, I really
feel disgusted with the whole issue. The atmosphere
can still absorb and retain thermal energy according to
the lapse whether or not there is downward LWIR more
than a few meters, and at this point I am beginning to
think that is the case.
Bill Ward
> Bill Ward wrote:
>
> [snip]
>
>> I thought the word "net" would give it away. Apparently not. The point
>> is that the "downward LWIR" is virtual in the sense that it can't warm
>> anything warmer than the source, or be detected by any thermal means
>> warmer than the source.
>
> Please stop using terms like virtual downward LWIR while you already
> know that radiation is omnidirectional. Don't try to fool us with your
> goobligoop copied from the Miskolczi discussion pages, because that's
> what you are trying to do.
Please stop snipping context. It shows your desperation.
>> The only apparent purpose in mentioning it is to confuse people like
>> Ima into thinking it can heat the surface. It can't. The surface is
>> cooling
>
> What the Hell it this supposed to mean? Such statements will effectively
> kill any chance of getting results in a science journal, essentially
> because reviewers don't understand you.
I'll try not to lose any sleep over that.
>> slower than it would if the sink were colder, but absolutely no heating
>> of the surface can occur unless the surface is colder than the LWIR
>> source. The second law rules. Surely you know that.
>
> What the hell are you talking about? Same remark.
The remark was directed at Dekker. I don't expect you to know anything.
>
> Good, download a program called MODTRAN now, and stop this ridiculous
> discussion. MODTRAN will solve the radiative transfer problem for you,
> and it assumes that radiation, re-radiation, absorption and emission are
> omidirectional. MODTRAN solves the following problem:
>
>> http://en.wikipedia.org/wiki/Radiative_transfer
>
> It is all about solving the following problem (cite wiki):
>
>> The equation of radiative transfer simply says that as a beam of
>> radiation travels, it loses energy to absorption, gains energy by
>> emission, and redistributes energy by scattering. The differential form
>> of the equation for radiative transfer is:
>>
>>
>> \frac{1}{c}\frac{\partial}{\partial t}I_\nu + \hat{\Omega} \cdot
>> \nabla I_\nu + (k_{\nu, s}+k_{\nu, a}) I_\nu = j_\nu +
>> \frac{1}{4\pi c}k_{\nu, s} \int_\Omega I_\nu d\Omega
>>
>> where jν is the emission coefficient, kν,s is the scattering cross
>> section,
>
> See the wiki page for the rest of the explanation on how to solve the
> resulting Eddington equations, and, apply it to a simple example. If you
> are unable to do this for whatever reason then here is an example for
> the Earth's atmosphere, it is all based on modtran, or better COART
> which is founded on modtran where gas concentration are modified to see
> what its effect is on IR absorption throughout the atmosphere:
That way you don't have to understand anything - you just put the numbers
in and accept whatever comes out as gospel. I can see the attraction
that would have for Q. It's wrong, but requires no thought, thus is much
easier than understanding how Miskolczi approaches the problem.
>> http://tinyurl.com/ykbzofj
>
> Programs like COART have been successfully applied to spectra observed
> by satellites and aircrafts. If you don't like what COART does then I
> strongly suggest that you write your own program, write an article about
> it, get it past the reviewers, and publish it in a journal. This is the
> normal way of doing business in science.
Really? Show us some of your published papers so we can see how that
works.
> Maybe take a look at figure 9 in http://tinyurl.com/ykbzofj
>
> This shows the fundamental problem of the -3.15 W/m^2 at the tropopause
> by a hypothetic doubling of the present day CO2 (380 ppmv). Water vapor
> and methane I though is included in the discussion.
The paper also says this:
"A weakness of the radiative transfer model approach is that it can not
easily insert a feedback mechanism to the hydrologic cycle on Earth.
Global warming would change the hydrologic cycle because a temperature
increase would affect terms like the evaporation and transpiration, and
cloud formation. An increment in the cloud coverage would be a
negative feed-back in the climate system."
That link corroborates what I've been saying all along - that the
radiation models don't work in the real world. The Earth has lots of
water, and a hydrologic cycle that rules the climate.
It looks like Q doesn't even read his links. It's a good thing he's
immune to embarrassment.
Rob Dekker
"Bill Ward" <bw...@ix.REMOVETHISnetcom.com> wrote in message news:p9udnT5lh53s5J7W...@giganews.com...
I think this whole discussion is one big misunderstanding created by different terminology.
For one, your terminology of "virtual" downward LWIR is very confusing, since downward LWIR is very real.
Your argument makes sense for nightime.
You are right that at night a cold sky cannot increase the temperature of the surface, so it that sense the surface it cannot "heat
up" from the sky.
It's good that you mention that it can slow down the cooling (cooling at night that is), otherwise I would have been very concerned
with your line of thiking.
So much for night time.
But your argument (that LWIR from a cold sky cannot 'heat' the warmer surface) starts to break down during the day.
Suppose that during the day, there is 700 W/m^2 SW radiation coming in.
Suppose that the surface heats up say from 20 C to 35 C during the day if there were only that radiation.
Now suppose LWIR is pretty constant day/night, and that it is at 300 W/m^2 (just giving it some number here) from a fairly 'cold'
sky.
Then, during the day, the surface receives 1000 W/m^2 rather than the 700 W/m^2 it received without the downward LWIR.
So sure enough, the surface will heat up to MORE than 35 C.
And this is caused by a sky that is colder than 20 C.
So, you cannot claim any more that downward LWIR cannot heat up the surface.
QED
Bill Ward
The net LWIR flux is upward, from warm to cold, because the 2nd law won't
allow anything else. "Back radiation" is a confusing term. The
troposphere is in LTE, and the flux is the same in all directions until
it gets near an edge, where it can radiate to space. Worrying about
"back radiation" is simply a distraction from the fact it's simply a
warm, semi-opaque gas. Radiative transport (except for that escaping
through the 10u "window") is not important till you get to an altitude
where the GHG can "see" out the top side to space.
>
>
>>The only apparent purpose in mentioning it is to confuse people like Ima
>>into thinking it can heat the surface. It can't. The surface is cooling
>>slower than it would if the sink were colder, but absolutely no heating
>>of the surface can occur unless the surface is colder than the LWIR
>>source. The second law rules. Surely you know that.
>
>
> Which has nothing to do with the measurement
> of back radiation or downward LWIR, the upward LWIR is a separate flow,
> it is not disrupted as if em was like a liquid, water can't do much two
> way flowing, but em can flow in all directions at once.
>
> [ - Incidently, a magazine had an April 1st article - ] [ - on
> "contra-polar light" about 54 years ago - ]
> http://www.swtpc.com/mholley/PopularElectronics/Apr1955/
PE_Apr_1955_pg04.jpg
> [ - Page 27 Contra-Polar Energy - ]
>
>
> If back radiation has not been measured, I really
> feel disgusted with the whole issue. The atmosphere can still
> absorb and retain thermal energy according to the lapse whether or not
> there is downward LWIR more than a few meters, and at this point I am
> beginning to think that is the case.
You don't actually have to measure it, you can determine it from the
temperature of the gas in LTE. It's in equilibrium, remember. What do
you think it will tell you?
BDR-529
> On Wed, 18 Nov 2009 06:18:30 +0100, BDR-529 wrote:
>
>> Bill Ward wrote:
>>
>> [snip]
>>
>>> I thought the word "net" would give it away. Apparently not. The point
>>> is that the "downward LWIR" is virtual in the sense that it can't warm
>>> anything warmer than the source, or be detected by any thermal means
>>> warmer than the source.
>> Please stop using terms like virtual downward LWIR while you already
>> know that radiation is omnidirectional. Don't try to fool us with your
>> goobligoop copied from the Miskolczi discussion pages, because that's
>> what you are trying to do.
>
> Please stop snipping context. It shows your desperation.
No, it makes postings more readable.
>
>>> The only apparent purpose in mentioning it is to confuse people like
>>> Ima into thinking it can heat the surface. It can't. The surface is
>>> cooling
>> What the Hell it this supposed to mean? Such statements will effectively
>> kill any chance of getting results in a science journal, essentially
>> because reviewers don't understand you.
>
> I'll try not to lose any sleep over that.
Because you don't publish.
>
>>> slower than it would if the sink were colder, but absolutely no heating
>>> of the surface can occur unless the surface is colder than the LWIR
>>> source. The second law rules. Surely you know that.
>> What the hell are you talking about? Same remark.
>
> The remark was directed at Dekker. I don't expect you to know anything.
You sound so agitated every time, if you want to say to somebody
something then say:
@dekker: bla bla bla
in a post
>> Good, download a program called MODTRAN now, and stop this ridiculous
>> discussion. MODTRAN will solve the radiative transfer problem for you,
>> and it assumes that radiation, re-radiation, absorption and emission are
>> omidirectional. MODTRAN solves the following problem:
>>
>>> http://en.wikipedia.org/wiki/Radiative_transfer
>> It is all about solving the following problem (cite wiki):
>>
>>> The equation of radiative transfer simply says that as a beam of
>>> radiation travels, it loses energy to absorption, gains energy by
>>> emission, and redistributes energy by scattering. The differential form
>>> of the equation for radiative transfer is:
>>>
>>>
>>> \frac{1}{c}\frac{\partial}{\partial t}I_\nu + \hat{\Omega} \cdot
>>> \nabla I_\nu + (k_{\nu, s}+k_{\nu, a}) I_\nu = j_\nu +
>>> \frac{1}{4\pi c}k_{\nu, s} \int_\Omega I_\nu d\Omega
>>>
>>> where jν is the emission coefficient, kν,s is the scattering cross
>>> section,
>> See the wiki page for the rest of the explanation on how to solve the
>> resulting Eddington equations, and, apply it to a simple example. If you
>> are unable to do this for whatever reason then here is an example for
>> the Earth's atmosphere, it is all based on modtran, or better COART
>> which is founded on modtran where gas concentration are modified to see
>> what its effect is on IR absorption throughout the atmosphere:
>
> That way you don't have to understand anything - you just put the numbers
No, not really, you can still code the thing yourself and verify the
outcome. I did.
> in and accept whatever comes out as gospel. I can see the attraction
This is not how it works. Have you ever tried to test methods / software
etc?
> that would have for Q. It's wrong, but requires no thought, thus is much
> easier than understanding how Miskolczi approaches the problem.
WHAT is wrong, Please please please, tell it. What specifically is wrong
in the radiative transfer theory. Yell it out, proclaim it, become
famous. Just do it. I didn't see anything from you that even proves an
error in the standard theory.
>
>>> http://tinyurl.com/ykbzofj
>> Programs like COART have been successfully applied to spectra observed
>> by satellites and aircrafts. If you don't like what COART does then I
>> strongly suggest that you write your own program, write an article about
>> it, get it past the reviewers, and publish it in a journal. This is the
>> normal way of doing business in science.
>
> Really? Show us some of your published papers so we can see how that
> works.
:-)
>
>> Maybe take a look at figure 9 in http://tinyurl.com/ykbzofj
>>
>> This shows the fundamental problem of the -3.15 W/m^2 at the tropopause
>> by a hypothetic doubling of the present day CO2 (380 ppmv). Water vapor
>> and methane I though is included in the discussion.
>
> The paper also says this:
>
> "A weakness of the radiative transfer model approach is that it can not
> easily insert a feedback mechanism to the hydrologic cycle on Earth.
> Global warming would change the hydrologic cycle because a temperature
> increase would affect terms like the evaporation and transpiration, and
> cloud formation. An increment in the cloud coverage would be a
> negative feed-back in the climate system."
>
> That link corroborates what I've been saying all along - that the
> radiation models don't work in the real world. The Earth has lots of
> water, and a hydrologic cycle that rules the climate.
Yet, modtran is able to explain what satellites and aircrafts have
measured, why did you conveniently not react to that part. It should
have rang a very large bell in front of you.
You did not even understand the statement you quote, the true meaning
is, if you want to run a climate / weather model than radiative modeling
is just a part of it. To do the rest means, add more complication, and
suddenly you talk about a GCM.
>
> It looks like Q doesn't even read his links. It's a good thing he's
> immune to embarrassment.
:-)
Authoritative bill, aka the big loser, speaks again, firmly stuck his
head in the sand, no way to convince him.
Q
BDR-529
What you are talking about is maybe a variation of solving the Eddington
equations with different boundary conditions, but no more. Get modtran,
and do it yourself.
Q
>
>
>>>>> The one trying to sell you otherwise is laughing his butt off.
>>>>>
>>>>>
>>>>>>> [snip]
>>>>>>>
>>>>>>>> Your ignorance is absolutely amazing. Does it take a lot of
>>>>>>>> practice?
>>>>>>> Perhaps I spot a little bit of anxiety? Could that be true when
>>>>>>> you're continuously shown to be wrong by your opponents who just
>>>>>>> cite the facts?
>>>>>> Nah, you're just projecting. Probably your desperation showing
>>>>>> through.
>>>>>>
>>>>>>> How painful that must be, Billobwab.
>>>>>> Tell us about it. It must be awful.
>>>>> The real painful part is that you have no clue what your talking
>>>>> about.
>>>> Your projection is still showing.
>>>>
>>>>
>>>>
>
>
--
Type-2 diabetes since July 2009, 1000 mg/day metformin, BMI 26.3
Rob Dekker
"I M @ good guy" <I...@good.guy> wrote in message news:vep6g554tl50u9i9p...@4ax.com...
Dutch.
>
>>I really do not know where to begin.
>
>
> That has become abundantly clear.
>
See below.
>
>>I'm not even sure where you are in your understanding of radiative transfer and other basic physics concepts.
>
>
> You think there is somebody that doesn't know em?
Yes.
>
>
>>In short : What climate scientists call "back radiation" is absolutely there. It's the IR radiation from the atmosphere >back to
>>thesurface.
>
>
> So they have measured the downward LWIR flux at
> various altitudes,
I think you already posted a link to a NASA paper where they did exactly that.
> or do they just imagine back radiation
> exists, and guess at the flux value?
Imagine ? guess ? Radiative transfer theory is directly derived from the Maxwell Equations.
This is rock-solid physics, undisputed and uncontested.
There is very little guess work involved there.
>
>>Without that, Earth's surface would be about 33 C lower in temperature.
>
>
> Obviously you do not understand the science and
> either remember the talking point phrases or look up
> the gossip on a search engine.
>
> Back radiation would only be possible with GHGs,
Close enough.
> and the "saying" about 33 C warmer is not about GHGs,
> it is about an Earth without an atmosphere.
Please tell me you are kidding here.
Because if not, then we are done talking.
>
> If you don't know the difference, go back to
> truck driving.
>
First, I would appreciate if you would tone your attitude down a bit. You have a big mouth and very little substance in your
arguments.
Second : there is nothing wrong with truck driving. It's a great profession.
>>This is also called the 'greenhouse effect' in
>>popular literature.
>
>
> By parrots. Lets see what the web says about the
> mean temperature of the moon:
>
> http://stars5.netfirms.com/datmoon.htm#specs
>
> Minus 4 F, now that is a surprise, as the gossip
> about GreenHouse Theory you mention causing the
> mean temperature of Earth to be 33 C warmer means
> the moon should be a lot colder than that.
Wow. Did you actually prove AGW theory wrong with this amazing deduction ? Mmm. Maybe not.
Let's try that again :
How did they determine the average temperature of the moon again ?
Oh ! Yes, they took the AVERAGE temperature of the lit side (Tmin) and the dark side (Tmax) of the moon. That is Tave =
(Tmin+Tmax)/2.
And emitted radiation goes with T^4 remember ?
So can you now calculate the EFFECTIVE radiating temperature of the moon (The temperature that corresponds to much radiation it
looses) ?
Tmin and Tmax vary much greater on the moon than they vary here on Earth remember ?
So feel it coming ?
Yes, and do you understand that that would be colder than the average temperature ?
In fact, it will be very close to 255 K.
Mmmm. Very much the same as the EFFECTIVE radiating temperature of planet Earth.
It seems you still have problems understanding Stefan Boltzmann equation and how it applies.
We did not even need radiative transfer theory for this.
>
> http://www.asi.org/adb/02/05/01/surface-temperature.html
>
> shows a 38 C difference between the Earth and moon.
>
> Note the moon has no atmosphere.
>
> You do know the difference between no atmosphere
> and an atmosphere with no GHGs, don't you?
Sigh.
I'd suggest you read it any way, and this time try to understand it, even if it does not immediately "apply to your interest".
Good guy, you don't even begin to understand the basic principles.
And when I explain something (or try to explain) you react with contempt and arrogance.
So good luck discussing climate science with other people.
But I am done talking with you.
>
>
>
>
>
>
Bill Ward
The calculator at:
<http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c2>
says it would be have to be ~270K to emit 300W/m^2.
The surface would have to be lower than that to transfer net energy to
the surface. At 20C the net LWIR would still be upward at ~117W/m^2.
Heat does not flow from cold to hot.
> Then, during the day, the surface receives 1000 W/m^2 rather than the
> 700 W/m^2 it received without the downward LWIR. So sure enough, the
> surface will heat up to MORE than 35 C. And this is caused by a sky that
> is colder than 20 C. So, you cannot claim any more that downward LWIR
> cannot heat up the surface.
Net energy never flows spontaneously from cold to hot. It's the law.
Thus "downward LWIR" can never transfer energy to a surface that's warmer
than the source of the LWIR. The net upward radiation will be dependent
on the target T, which will slow the cooling, but no energy will be
transferred against the temperature gradient.
In reality, the surface would attain LTE with the atmosphere, and the
emission temperature would depend on the emission altitude and local
lapse rate.
But you already knew that.
> QED
Not until you repeal the 2nd law.
Rob Dekker
Mmm. The 2nd law is not broken with my example above, since NET heat flow is
never from cold to warm.
But it does show that a cold sky can heat a warm surface.
If you don't believe that, then please tell me what's wrong with my
calculation.
Rob
>
Bill Ward
> Bill Ward wrote:
>> On Wed, 18 Nov 2009 06:18:30 +0100, BDR-529 wrote:
>>
>>> Bill Ward wrote:
>>>
>>> [snip]
>>>
>>>> I thought the word "net" would give it away. Apparently not. The
>>>> point is that the "downward LWIR" is virtual in the sense that it
>>>> can't warm anything warmer than the source, or be detected by any
>>>> thermal means warmer than the source.
>>> Please stop using terms like virtual downward LWIR while you already
>>> know that radiation is omnidirectional. Don't try to fool us with your
>>> goobligoop copied from the Miskolczi discussion pages, because that's
>>> what you are trying to do.
>>
>> Please stop snipping context. It shows your desperation.
>
> No, it makes postings more readable.
Not when it's misleading. I can see why you'd rather readers didn't see
your sorry performance, but having to snip just makes you look more
incompetent.
>>>> The only apparent purpose in mentioning it is to confuse people like
>>>> Ima into thinking it can heat the surface. It can't. The surface is
>>>> cooling
>>> What the Hell it this supposed to mean? Such statements will
>>> effectively kill any chance of getting results in a science journal,
>>> essentially because reviewers don't understand you.
>>
>> I'll try not to lose any sleep over that.
>
> Because you don't publish.
Almost true.
>>>> slower than it would if the sink were colder, but absolutely no
>>>> heating of the surface can occur unless the surface is colder than
>>>> the LWIR source. The second law rules. Surely you know that.
>>> What the hell are you talking about? Same remark.
>>
>> The remark was directed at Dekker. I don't expect you to know
>> anything.
>
> You sound so agitated every time, if you want to say to somebody
> something then say:
>
> @dekker: bla bla bla
>
> in a post
I was responding to his post, not yours.
For one thing the rad models have a non-physical temperature
discontinuity at the surface. Were you able to read Miskolczi, you'd
know that.
<http://www.met.hu/doc/idojaras/vol111001_01.pdf>
Here's a sample from page 13. Maybe you can understand part of it:
<begin excerpt>
There were several attempts to resolve the above deficiencies by
developing simple semi-empirical spectral models, see for example Weaver
and Ramanathan (1995), but the fundamental theoretical problem was never
resolved. The source of this inconsistency can be traced back to several
decades ago, when the semi-infinite solution was first used to solve
bounded atmosphere problems. About 80 years ago Milne stated: "Assumption
of infinite thickness involves little or no loss of generality", and
later, in the same paper, he created the concept of a secondary
(internal) boundary (Milne, 1922). He did not realize that the classic
Eddington solution is not the general solution of the bounded atmosphere
problem and he did not re-compute the appropriate integration constant.
This is the reason why scientists have problems with a mysterious surface
temperature discontinuity and unphysical solutions, as in Lorenz and
McKay (2003). To accommodate the finite flux optical depth of the
atmosphere and the existence of the transmitted radiative flux from the
surface, the proper equations must be derived.
<end excerpt>
There's more above this, but the equations don't translate well to ASCII.
>
>>>> http://tinyurl.com/ykbzofj
>>> Programs like COART have been successfully applied to spectra observed
>>> by satellites and aircrafts. If you don't like what COART does then I
>>> strongly suggest that you write your own program, write an article
>>> about it, get it past the reviewers, and publish it in a journal. This
>>> is the normal way of doing business in science.
>>
>> Really? Show us some of your published papers so we can see how that
>> works.
>
> :-)
None, eh? Big hat, no cattle.
>>> Maybe take a look at figure 9 in http://tinyurl.com/ykbzofj
>>>
>>> This shows the fundamental problem of the -3.15 W/m^2 at the
>>> tropopause by a hypothetic doubling of the present day CO2 (380 ppmv).
>>> Water vapor and methane I though is included in the discussion.
>>
>> The paper also says this:
>>
>> "A weakness of the radiative transfer model approach is that it can not
>> easily insert a feedback mechanism to the hydrologic cycle on Earth.
>> Global warming would change the hydrologic cycle because a temperature
>> increase would affect terms like the evaporation and transpiration, and
>> cloud formation. An increment in the cloud coverage would be a negative
>> feed-back in the climate system."
>>
>> That link corroborates what I've been saying all along - that the
>> radiation models don't work in the real world. The Earth has lots of
>> water, and a hydrologic cycle that rules the climate.
>
> Yet, modtran is able to explain what satellites and aircrafts have
> measured, why did you conveniently not react to that part. It should
> have rang a very large bell in front of you.
>
> You did not even understand the statement you quote, the true meaning
> is, if you want to run a climate / weather model than radiative modeling
> is just a part of it. To do the rest means, add more complication, and
> suddenly you talk about a GCM.
Why would I want to run a known inapplicable model? I'd rather
understand what's actually going on.
>> It looks like Q doesn't even read his links. It's a good thing he's
>> immune to embarrassment.
>
> :-)
>
> Authoritative bill, aka the big loser, speaks again, firmly stuck his
> head in the sand, no way to convince him.
When you can discuss Miskolczi, let me know. Right now, you're a useless
waste of my time.
BDR-529
[snip] (essentially, software allows me to view previous parts of the
discussion).
> For one thing the rad models have a non-physical temperature
> discontinuity at the surface. Were you able to read Miskolczi, you'd
> know that.
>
> <http://www.met.hu/doc/idojaras/vol111001_01.pdf>
>
> Here's a sample from page 13. Maybe you can understand part of it:
>
> <begin excerpt>
>
> There were several attempts to resolve the above deficiencies by
> developing simple semi-empirical spectral models, see for example Weaver
> and Ramanathan (1995), but the fundamental theoretical problem was never
> resolved. The source of this inconsistency can be traced back to several
> decades ago, when the semi-infinite solution was first used to solve
> bounded atmosphere problems. About 80 years ago Milne stated: "Assumption
> of infinite thickness involves little or no loss of generality", and
> later, in the same paper, he created the concept of a secondary
> (internal) boundary (Milne, 1922). He did not realize that the classic
> Eddington solution is not the general solution of the bounded atmosphere
> problem and he did not re-compute the appropriate integration constant.
> This is the reason why scientists have problems with a mysterious surface
> temperature discontinuity and unphysical solutions, as in Lorenz and
> McKay (2003). To accommodate the finite flux optical depth of the
> atmosphere and the existence of the transmitted radiative flux from the
> surface, the proper equations must be derived.
>
> <end excerpt>
>
> There's more above this, but the equations don't translate well to ASCII.
Ok, this is at least something that cuts meat.
But, would solving this discontinuity at the surface avoid the problem
of IR absorption in the stratosphere, probably not, but lets see.
In COART you define the surface temperature, this is the way Jin
implemented the boundary condition.
So, what you could tell me first is the magnitude of the error in the
boundary condition at the surface, and then to see what its effect it on
the upwelling flux calculation
Q
[snip]
> When you can discuss Miskolczi, let me know. Right now, you're a useless
> waste of my time.
In case you're running out of time, consider to buy a watch. Please
answer the above question, and lets see what MODTRAN says when you tell
what the size of the discontinuity at the boundary is.
Q
Bill Ward
Good. You're on the right track.
> But it does show that a cold sky can heat a warm surface. If you don't
> believe that, then please tell me what's wrong with my calculation.
It's your definition of "heat". Slowing the rate of cooling is not
heating. An insulating blanket is not a heater.
I M @ good guy
wrote:
While I agree with your description, I feel the numbers may
be off quite a bit.
And rather than me trying to apologize to Kd, I hope
he will read what you wrote above and see how a black night
sky can cause a windshield (or a leaf) to frost.
My concern here is that in too many places, some of
the AGW arguments use spectra or temperature as a proxy
of flux.
My last sentence is very clear and complete, otherwise
the physics of radiative transfer has to be as thought, only
the exact imbalance in the energy budget is the big issue.
The only thing worth mentioning otherwise is that
the atmospheric radiation by GHGs in daytime is a result
of SW heating, not from upward LWIR from the surface,
the surface radiating broadband at all times can radiate
directly to space in wavelengths not absorbed by GHGs.
And it has to be that clouds radiate broadband
downward that can cause the rather large difference
in temperature of my windshield on cloudy nights
compared to that on clear nights with the same
ambient temperature.
Neither dew or frost forms under a cloudy sky,
which seems to show the difference that broadband
LWIR can make.
So Kd is right about a surface not radiating more
to space with a black sky, the surface receives a lot
more energy from broadband radiation from the
water droplets in the clouds.
This thread is about putting numbers on all
that, flux is more a measure of energy transfer
rates than of temperature or wavelength.
And it is net energy transfer that is the
essence of the Earth's energy budget.
Bill Ward
Sorry I didn't make myself clear. You snipped again, and I'm done with
you. Read Miskolczi yourself.
I M @ good guy
<bw...@ix.REMOVETHISnetcom.com> wrote:
The flux can perhaps be inferred, but not "determined",
and the broadband from clouds should show a big difference
in flux (directional energy transfer), which may actually
diminish the emphasis on GHG LWIR, depending on how
much greater the LWIR broadband flux is compared to
the individual GHG LWIR combined fluxes.
Why would I be out of line in wanting numbers
put on all this? About 2 or 3 watts per square meter.
erschro...@gmail.com
The second law does NOT forbid a photon coming from a cold source to
be absorbed by a warm source. You continue to confuse conduction and
radiation. The earth absorbs the 3 K microwave background radiation
that hits it, for example.
erschro...@gmail.com
Tell you what, you tell us the physicists that accept him. The
physics journals he's been published in. The physics textbooks that
include his ideas.
BDR529
Who says I did not, does it matter. What I think is that you're running
away with your tail between your legs once we start asking questions
that generate sufficient detail to do an independent verification. If
you can't handle the heat, then get out of the kitchen.
Q
BDR529
>
> Reduced to the max.
>
> :-)
Do you mind not insinuate some text that I did not write, Mulehead?
Q
Rob Dekker
"Bill Ward" <bw...@ix.REMOVETHISnetcom.com> wrote in message news:n_WdnfOD0fL_KZ7W...@giganews.com...
Somehow this sounds a bit patronising.
You could simply state : "so far we are in agreement".
>
>> But it does show that a cold sky can heat a warm surface. If you don't
>> believe that, then please tell me what's wrong with my calculation.
>
> It's your definition of "heat". Slowing the rate of cooling is not
> heating.
Short of reverting to exactly what the definition of "heating" is, my example above shows that during the day (when there is no
'cooling' going on), downward LWIR clearly heats the surface.
> An insulating blanket is not a heater.
>
I revert back to my earlier statement that I think this thread is plagued by confusions in terminology.
What are we really trying to say ?
This thread started with the questing of downward LWIR radiation, and if it really existed or not.
I trust that that question is answered now with a resounding YES.
Not only does radiative tranfer theory predict it very explicitly, but, as Q it also comes out as a result of radiative transfer
modeling programs like MODTRAN.
Also, if somebody still questions the validity of radiative transfer theory, please note that you are up against one of the most
solid physics theories around. It is derived from the Maxwell Equations.
Besides that, it's effect is also very observable :
With reduced downward LWIR (on a clear and still night) water in an insulated cup (only open from the top) freezes quickly even if
the air around it is above freezing. No such thing happens if clouds are above.
If you still think downward LWIR does not exist or is "virtual" then please state that very explicitly.
Rob
I M @ good guy
wrote:
Chances are if the sun is not shining on the
surface, the surface is not heating.
It is more likely to stay about the same or
slowly drift toward the temperature of the air.
>> An insulating blanket is not a heater.
>>
>
>I revert back to my earlier statement that I think this thread is plagued by confusions in terminology.
>
>What are we really trying to say ?
>This thread started with the questing of downward LWIR radiation, and if it really existed or not.
>I trust that that question is answered now with a resounding YES.
But how much?
>Not only does radiative tranfer theory predict it very explicitly, but, as Q it also comes out as a result of radiative transfer
>modeling programs like MODTRAN.
What does that have to do with measuring downward LWIR?
>Also, if somebody still questions the validity of radiative transfer theory,
>please note that you are up against one of the most
>solid physics theories around.
Who questions em radiation, it is what cools the
atmosphere, and the atmosphere and radiation cool
the surface.
>It is derived from the Maxwell Equations.
>Besides that, it's effect is also very observable :
>With reduced downward LWIR (on a clear and still night)
>water in an insulated cup (only open from the top)
>freezes quickly even if the air around it is above freezing.
Not in Kd's yard. :-)
>No such thing happens if clouds are above.
But clouds radiate broadband, what percentage
of broadband is CO2 LWIR?
>If you still think downward LWIR does not exist or is "virtual" then please state that very explicitly.
>
>Rob
How much exists at each kilometer above the
surface, why am I having trouble finding that?
Bill Ward
I claim that no net energy can be transferred by "downward", or "back"
radiation unless the source is warmer than the sink. You agreed with
that. I further claim that there is no way to detect any "virtual" heat
transfer from cold to hot, because any thermal detector must be colder
than the radiation, and any coherent receiver must add energy from an
outside source, effectively cooling the receiver.
I also understand the concept that all bodies omnidirectionally radiate
and absorb thermal energy, and that one can mathematically add the signed
fluxes without introducing error. But the experiments showing that
always depend on a source and a cooler sink, because there has to be a
material detector to absorb the radiation. Until someone figures out how
to measure the energy of a photon without affecting it, the hypothesis
remains unproven.
The reasons I'm making an issue of this are that I'm trying to get people
to think things through for themselves instead of simply accepting dogma,
and also to point out the danger of misleading those who are not familiar
with radiative transfer by referring to "back radiation" as though it
represented an actual net energy flow.
Is that explicit enough?
I M @ good guy
When he mentioned "heating" of the surface
he was talking about two sources of radiation, in
daytime, with sun warmed atmosphere and/or
surface plus back radiation.
My interest is in having a data set, a measured
data set, for each of the arrows on an energy budget
graph.
BDR-529
This shows your complete ignorance towards the subject I guess.
>
>
>> Also, if somebody still questions the validity of radiative transfer theory,
>> please note that you are up against one of the most
>> solid physics theories around.
>
>
> Who questions em radiation, it is what cools the
> atmosphere, and the atmosphere and radiation cool
> the surface.
>
>
>> It is derived from the Maxwell Equations.
>> Besides that, it's effect is also very observable :
>> With reduced downward LWIR (on a clear and still night)
>> water in an insulated cup (only open from the top)
>> freezes quickly even if the air around it is above freezing.
>
>
> Not in Kd's yard. :-)
>
>
>> No such thing happens if clouds are above.
>
>
> But clouds radiate broadband, what percentage
> of broadband is CO2 LWIR?
Why don't you google for it?
>
>
>> If you still think downward LWIR does not exist or is "virtual" then please state that very explicitly.
>>
>> Rob
>
>
> How much exists at each kilometer above the
> surface, why am I having trouble finding that?
A lot of questions you can answer yourself.
Q
BDR-529
Bugger, this is not how it works in physics. Put a hot source next to a
cold source (still warmer than 0K though), and they will both emit
radiation.
> that. I further claim that there is no way to detect any "virtual" heat
> transfer from cold to hot, because any thermal detector must be colder
Try thinking of the real world, rather than your virtual game world.
> than the radiation, and any coherent receiver must add energy from an
> outside source, effectively cooling the receiver.
Bugger! Major bugger!
>
> I also understand the concept that all bodies omnidirectionally radiate
> and absorb thermal energy, and that one can mathematically add the signed
> fluxes without introducing error. But the experiments showing that
> always depend on a source and a cooler sink, because there has to be a
> material detector to absorb the radiation. Until someone figures out how
> to measure the energy of a photon without affecting it, the hypothesis
> remains unproven.
Bugger major bugger, nothing needs to be proven, check a physics book.
>
> The reasons I'm making an issue of this are that I'm trying to get people
> to think things through for themselves instead of simply accepting dogma,
> and also to point out the danger of misleading those who are not familiar
> with radiative transfer by referring to "back radiation" as though it
> represented an actual net energy flow.
>
> Is that explicit enough?
Explicit enough to explain that you a complete nitwit.
Q
I M @ good guy
No, it shows you are too stupid to know the difference
between a measurement and a calculation.
>>> Also, if somebody still questions the validity of radiative transfer theory,
>>> please note that you are up against one of the most
>>> solid physics theories around.
>>
>>
>> Who questions em radiation, it is what cools the
>> atmosphere, and the atmosphere and radiation cool
>> the surface.
>>
>>
>>> It is derived from the Maxwell Equations.
>>> Besides that, it's effect is also very observable :
>>> With reduced downward LWIR (on a clear and still night)
>>> water in an insulated cup (only open from the top)
>>> freezes quickly even if the air around it is above freezing.
>>
>>
>> Not in Kd's yard. :-)
>>
>>
>>> No such thing happens if clouds are above.
>>
>>
>> But clouds radiate broadband, what percentage
>> of broadband is CO2 LWIR?
>
>Why don't you google for it?
Why don't you respond to your inner circle AGW nutfriends
messages only, maybe they understand your stupid remarks.
>>> If you still think downward LWIR does not exist or is "virtual" then please state that very explicitly.
>>>
>>> Rob
>>
>>
>> How much exists at each kilometer above the
>> surface, why am I having trouble finding that?
>
>A lot of questions you can answer yourself.
>
>Q
Not unless I have a CO2 sensor and an airplane,
I am beginning to wonder if you know what measurement
means.
columbiaaccidentinvestigation
 Until someone figures out how
to measure the energy of a photon without affecting it, the
hypothesis remains unproven"
laughing.....
columbiaaccidentinvestigation
to measure the energy of a photon without affecting it, the
hypothesis remains unproven."
So how does the human eye work bill? I guess at this point you dont
believe what your eyes tell you, and those rods and cones mess it all
up because they effect the photons....
Rob Dekker
"Bill Ward" <bw...@ix.REMOVETHISnetcom.com> wrote in message news:_Lidne24BaHzAZnW...@giganews.com...
Since you mention "net" energy flow (counter-directional energy flow subtracted), your statement is a liberal interpretation of the
second law of thermodynamics.
Therefore, I am in agreement.
> I further claim that there is no way to detect any "virtual" heat
> transfer from cold to hot,
Can you remove the word "virtual" without changing the meaning of sentence for you ?
If not, can you define exactly what you mean with "virtual" ?
Below, I'll assume that with "virtual" heat transfer, you mean 'directional' heat transfer, means any energy transfer from one point
to another.
> because any thermal detector must be colder
> than the radiation, and any coherent receiver must add energy from an
> outside source, effectively cooling the receiver.
This statement is not correct.
We build receivers with less than 20 K noise temperature, while they operate at room temperature.
And if you think that the 20 K is as low as it goes, remember that the resolution (discriminate between two different temperatures)
of a detector can be much better than the noise temperature of the system. How else could Wilson have detected the CMBR in 1965 and
determined it to be 3.5 K (pretty close to the actual 2.7 K).
All this determined with a system that operated at room temperature.
>
> I also understand the concept that all bodies omnidirectionally radiate
> and absorb thermal energy, and that one can mathematically add the signed
> fluxes without introducing error.
Good.
> But the experiments showing that
> always depend on a source and a cooler sink, because there has to be a
> material detector to absorb the radiation. Until someone figures out how
> to measure the energy of a photon without affecting it, the hypothesis
> remains unproven.
No need to overrule Quantum Physics.
We can measure energy flow by radiation VERY accurately, both in direction and quantity.
Now that this is settled, can you then take away the word "virtual" and accept that radiation is a form of transfer of energy (heat)
from a cold to a warm object ?
And that this is actually happening (due to clouds and GHGs) and that it contributes to warming of the Earth's surface ?
Or do you need some additional evidence ?
>
> The reasons I'm making an issue of this are that I'm trying to get people
> to think things through for themselves instead of simply accepting dogma,
Thinking these things through for yourself is always a good thing.
> and also to point out the danger of misleading those who are not familiar
> with radiative transfer by referring to "back radiation" as though it
> represented an actual net energy flow.
Not a "net" energy flow, but an energy flow alright.
In graphic representations of the Earth's energy balance, it is important to put the numbers with each of the energy flow arrows,
both up and down. Because these flows are all real (and measurable).
>
> Is that explicit enough?
>
>
I hope so. What I understood from your argument is that you did not see that 'downward' radiation (back-radiation) actually exists,
or that it would always be the same as upward radiation.
What I tried to show is that this is not the case. These energy flows exist in reality, are directional, and they also change from
moment to moment. Just like convection and conduction heat flow changes from moment to moment. When you add up all the heat flow
numbers for one location and moment, then you will know if that location will warm up or cool down.
So it is incorrect to assume an everlasting equilibrium or to assume that heat can only flow from warm to cold.
BDR-529
ROFL!!
Maybe Bill Ward's eyes (and ears) just work differently, who knows?
Q
Bill Ward
I'm using the word "virtual in the sense of, "... the virtual is not what
something is as a matter of fact, but what it is in principle." You can
find that at
<http://en.wikipedia.org/wiki/Virtual>
Virtual particles are, in theory, constantly forming and disappearing all
though space, but can't be detected unless there is enough energy to
release them. Pair production when a gamma greater than 1.02MeV nears a
nucleus is one example.
>
> Below, I'll assume that with "virtual" heat transfer, you mean
> 'directional' heat transfer, means any energy transfer from one point to
> another.
Sorry, that's not what I mean by "virtual". I mean energy that can be
assumed to be there, but is not observable, like the electron and
positron before pair production occurs. If I meant directional, I would
say directional.
>> because any thermal detector must be colder than the radiation, and any
>> coherent receiver must add energy from an outside source, effectively
>> cooling the receiver.
>
> This statement is not correct.
> We build receivers with less than 20 K noise temperature, while they
> operate at room temperature. And if you think that the 20 K is as low as
> it goes, remember that the resolution (discriminate between two
> different temperatures) of a detector can be much better than the noise
> temperature of the system. How else could Wilson have detected the CMBR
> in 1965 and determined it to be 3.5 K (pretty close to the actual 2.7
> K). All this determined with a system that operated at room temperature.
There are basically two types of devices used to measure EMR. One is a
bolometer, which detects the EMR by transforming the energy directly to
heat, the same way the surface of the Earth converts solar radiation.
They obviously must be cooled below the source temperature.
<http://en.wikipedia.org/wiki/Bolometer>
The other is a coherent receiver, which does not convert the incoming
radiation to heat, but excites a tuned circuit, which involves an
amplifier and an external power supply. You can think of the external
power as the way around the 2nd law, as the energy is not spontaneously
flowing from cold to hot, but is given a boost, in the same way the heat
inside a refrigerator is boosted by the externally powered compression
and expansion cycle.
<http://trs-new.jpl.nasa.gov/dspace/bitstream/2014/10504/1/02-2516.pdf>
The coherent receiver is not detecting heat, but the direct EMR. That's
more like a radio transmitter and receiver. The AM band would represent
a very low temperature, but is easily received by a room temperature
radio. It's not heat, because heat is random, and coherent radiation is
not.
There's no free lunch. Heat is not transferred from the cold source to
the hot sink in either case.
>> I also understand the concept that all bodies omnidirectionally radiate
>> and absorb thermal energy, and that one can mathematically add the
>> signed fluxes without introducing error.
>
> Good.
>
>> But the experiments showing that
>> always depend on a source and a cooler sink, because there has to be a
>> material detector to absorb the radiation. Until someone figures out
>> how to measure the energy of a photon without affecting it, the
>> hypothesis remains unproven.
>
> No need to overrule Quantum Physics.
> We can measure energy flow by radiation VERY accurately, both in
> direction and quantity.
Thermal radiation with a detector warmer than the source? Explain how.
> Now that this is settled, can you then take away the word "virtual" and
> accept that radiation is a form of transfer of energy (heat) from a cold
> to a warm object ?
Your initial assumption was wrong about what I mean by "virtual", so no,
I don't accept that real heat can flow from cold to hot.
> And that this is actually happening (due to clouds and GHGs) and that it
> contributes to warming of the Earth's surface ?
Not while you claim that reducing cooling is equivalent to warming.
Or do you need some additional evidence ?
Whatcha got?
>> The reasons I'm making an issue of this are that I'm trying to get
>> people to think things through for themselves instead of simply
>> accepting dogma,
>
> Thinking these things through for yourself is always a good thing.
>
>> and also to point out the danger of misleading those who are not
>> familiar with radiative transfer by referring to "back radiation" as
>> though it represented an actual net energy flow.
>
> Not a "net" energy flow, but an energy flow alright. In graphic
> representations of the Earth's energy balance, it is important to put
> the numbers with each of the energy flow arrows, both up and down.
Not if you work with net flow. That's the point - it's too easy to
forget that the flux from hot to cold is always more than the flux from
cold to hot. The net is the only "real" quantity, as that's all that can
be actually measured.
> Because these flows are all real (and measurable).
Not directly. They are in the end calculated from the difference in T
between the source and some sink,
>> Is that explicit enough?
>>
>>
>>
> I hope so. What I understood from your argument is that you did not see
> that 'downward' radiation (back-radiation) actually exists, or that it
> would always be the same as upward radiation.
That would be LTE, with no net flux.
> What I tried to show is that this is not the case. These energy flows
> exist in reality, are directional, and they also change from moment to
> moment. Just like convection and conduction heat flow changes from
> moment to moment. When you add up all the heat flow numbers for one
> location and moment, then you will know if that location will warm up or
> cool down. So it is incorrect to assume an everlasting equilibrium or to
> assume that heat can only flow from warm to cold.
All you have to do is show a case where heat can be proven to flow from
cold to hot. I maintain that's impossible because of the 2nd law. If net
heat is spontaneously transferred, the source must be hotter than the
sink. The detector has to be a sink to convert it back to heat.
That doesn't mean the method of assuming there is a real transfer form
hot to cold and an opposite real transfer from cold to hot gives the
wrong answer, it's just that it can be misleading to those more familiar
with dealing in net flux.
columbiaaccidentinvestigation
he does not seem motivated to correct his post, rather it appears he
would rather hold others to higher standards than he does himself...
hda
See, the two gravity wave investigators met and had fun in
sustaining half-truths about radiation and name calling.
Why are you behaving like a frustrated professor, Ernst
Schrama ?
Why do you corrupt science, promote a virtual world with
computermodels ?
Why do you teach like an hegelian idiot ?
hda
Your indicated cold to hot radiation _has_no_ power.
I am now looking to a screen with a color temperature of
6000 K, but it is not as hot as the sun.
columbiaaccidentinvestigation
looking to a screen with a color temperature of  6000 K, but it is
not as hot as the sun."
The 6000 K refers to the correlated color temp. The correlated color
temp. can be broken down into the tri-stimulus values, for the short
(blue), medium (green) and long (red) light receptors in the human
eye. The differences in white points can be thought of as moving from
a lower temp to a higher temp, with the lower temps resulting in a
general (warmness) to the white point (appx 2300-5000K), and beyond
5000K the white tends to produce a blue (or more cool) white. Now
turn off your monitor, and thats the blackest black your eye will see
from the monitor, and the white being set at 6000K. You then have the
anchors of the luminosity scale for that device, with the gamma
setting changing the relationship of the grays. Now that luminosity
scale can be represented as the z axis of a 3 dimensional plane. The
(x) axis of that plane goes from green (-x) to red (+x), and blue (-y)
to yellow (+y), with luminosity moving from black to white on the (z)
axis. All the colors presented by your monitor can be viewed in this
3d representation, known as the color gamut of the device (monitor in
your case). This color gamut can be device specific, so when you
install your monitor or printers, you are installing the instructions
for that devices color gamut, this is known as the ICC profile.
Understanding the profile for monitors are easy, its printing where
things get tricky, as the white point is set by the paper.
columbiaaccidentinvestigation
>
> - Show quoted text -
see below....
Bill Ward
Exactly. It's not thermal radiation, but three narrowband sources
combined. Unless you're using incandescent backlight, which I doubt.
Maybe that has something to do with the efficiency difference between
incandescent and fluorescent lighting? ;-)
BDR-529
Why don't you send him an e-mail?
Q
Rob Dekker
"Bill Ward" <bw...@ix.REMOVETHISnetcom.com> wrote in message news:7a2dnclcHNi9pJvW...@giganews.com...
This seems like a red herring to me.
There is no way to detect a photon 'in flight' so to say (while it is still 'virtual').
But once it materializes, it constitutes energy transfer from the location where it originated to the point where it was detected.
At the point where it is detected, it consitutes heat.
So that should be detectable, even the source was colder than the sink.
>
>>> because any thermal detector must be colder than the radiation, and any
>>> coherent receiver must add energy from an outside source, effectively
>>> cooling the receiver.
>>
>> This statement is not correct.
>> We build receivers with less than 20 K noise temperature, while they
>> operate at room temperature. And if you think that the 20 K is as low as
>> it goes, remember that the resolution (discriminate between two
>> different temperatures) of a detector can be much better than the noise
>> temperature of the system. How else could Wilson have detected the CMBR
>> in 1965 and determined it to be 3.5 K (pretty close to the actual 2.7
>> K). All this determined with a system that operated at room temperature.
>
> There are basically two types of devices used to measure EMR. One is a
> bolometer, which detects the EMR by transforming the energy directly to
> heat, the same way the surface of the Earth converts solar radiation.
OK.
> They obviously must be cooled below the source temperature.
I don't think so.
Since the low temperature heat source is simply adding energy to the detector, it should be possible
to detect a source that has a lower temperature than the (system) temperature of the detector,
by looking at the DIFFERENCE in detector temperature that the low temperator source creates.
Even for a bolometer.
>
> <http://en.wikipedia.org/wiki/Bolometer>
>
> The other is a coherent receiver, which does not convert the incoming
> radiation to heat, but excites a tuned circuit, which involves an
> amplifier and an external power supply. You can think of the external
> power as the way around the 2nd law, as the energy is not spontaneously
> flowing from cold to hot, but is given a boost, in the same way the heat
> inside a refrigerator is boosted by the externally powered compression
> and expansion cycle.
>
> <http://trs-new.jpl.nasa.gov/dspace/bitstream/2014/10504/1/02-2516.pdf>
>
> The coherent receiver is not detecting heat, but the direct EMR. That's
> more like a radio transmitter and receiver. The AM band would represent
> a very low temperature, but is easily received by a room temperature
> radio. It's not heat, because heat is random, and coherent radiation is
> not.
Another way of looking at coherent receivers is that the photon energy gets 'amplified' before they are 'detected'.
They are still 'virtual' inside the receiver, until we actually 'decohere' them by measuring their energy (with a diode or other
non-linear device).
This gets quite a bit off-topic, but it's actually very interesting to do this though experiment in the context of causality and
quantum physics.
But for this discussion, I do not see much of a difference between a coherent receiver or a bolometer.
They both receive the same radiation energy, it's just that a coherent receiver first amplifies the photon (energy).
>
> There's no free lunch. Heat is not transferred from the cold source to
> the hot sink in either case.
I think it is, but see below.
>
>>> I also understand the concept that all bodies omnidirectionally radiate
>>> and absorb thermal energy, and that one can mathematically add the
>>> signed fluxes without introducing error.
>>
>> Good.
>>
>>> But the experiments showing that
>>> always depend on a source and a cooler sink, because there has to be a
>>> material detector to absorb the radiation. Until someone figures out
>>> how to measure the energy of a photon without affecting it, the
>>> hypothesis remains unproven.
>>
>> No need to overrule Quantum Physics.
>> We can measure energy flow by radiation VERY accurately, both in
>> direction and quantity.
>
> Thermal radiation with a detector warmer than the source? Explain how.
Let's try that (below).
>
>> Now that this is settled, can you then take away the word "virtual" and
>> accept that radiation is a form of transfer of energy (heat) from a cold
>> to a warm object ?
>
> Your initial assumption was wrong about what I mean by "virtual", so no,
> I don't accept that real heat can flow from cold to hot.
>
>> And that this is actually happening (due to clouds and GHGs) and that it
>> contributes to warming of the Earth's surface ?
>
> Not while you claim that reducing cooling is equivalent to warming.
>
> Or do you need some additional evidence ?
>
> Whatcha got?
>
See below..
OK. Now we are getting somewhere close to a definition of the problem and hypothesis that we discuss.
In physics, it is important to very precisely define the problem, or else the answer can be ambiguous or irrelevant.
So, let's start with a definition :
Do you accept the definition of heat as given by Wikipedia ? :
"heat is the process of energy transfer from one body or system due to thermal contact, which in turn is defined as an energy
transfer to a body in any other way than due to work performed on the body"
or in short :
"heat is the process of energy transfer from one body to a body in any other way than due to work performed on the body"
If so, radiation IS heat.
You should be OK with that, since I think you are mostly concerned with the detection of that heat, right ?
If so, then is this a good definition of your hypothesis (you write this, above) :
Bill's hypothesis 1 : "heat is not transferred from a cold source to a hot sink"
You OK with that ?
Or would you prefer something like this :
Bill's hypothesis 2 : "the heat transfer from a cold to a hot body is not physically detectable"
Or can you state your one hypothesis
Note that I did not say 'net' heat transfer. Only 'heat' transfer, which is essentially 'energy' transfer.
And you also put in TWO constraints on disproving these hypothesis :
In the process of experimentally disproving hypothesis 1, two constraints apply :
Bill's constraint 1 : " no coherent receivers can be used"
Bill's constraint 2 : "the (non-coherent) receiver at the sink is not allowed to be cooled".
I think these two constraints are completely irrelevant, and are simply an experimental physics form of sado masochism.
But I put them there any way, since I do not think there is any conceptual difference.
It just makes it harder to think of an experiment that will disprove the hypothesis.
>
> That doesn't mean the method of assuming there is a real transfer form
> hot to cold and an opposite real transfer from cold to hot gives the
> wrong answer, it's just that it can be misleading to those more familiar
> with dealing in net flux.
>
Please let me know if you accept the hypothesis, and the constraints.
Just promise me one thing : If I can think of such an experiment (I'm already thinking) and it honestly complies with your
hypothesis, then please accept it. OK ?
Rob
Rob Dekker
"hda" <age...@xs4all.nl.invalid> wrote in message news:v00dg59r9gcu1fbu4...@4ax.com...
I think that was Bill.
I stated that cold to hot radiation has power.
It's just less power than the hot to cold radiation that goes the other way.
I also claim that both energy flows are indepenent and are independently observable.
>
> I am now looking to a screen with a color temperature of
> 6000 K, but it is not as hot as the sun.
As Bill mentioned, that's because EM radiation (light and IR etc) can come from a variety of sources.
Thermal sources (such as the sun or an incandecent light bulb wire) generate light because their atoms get 'excited' (their
electrons change energy state) because of wild thermal movement of the atoms. If the material is 'black' (absorb all light) at low
temperature, thermal light sources radiate a 'black-body' spectrum. This changes 'color' depending on the temperature of the
material.
Other light sources (LEDs, fluorecents, lasers etc) generate light by excitation specific atoms (a semiconductor, neon gas, CO2 gas
etc) in non-thermal ways (often by an electric field).
So these do not need to be hot to emit light of various frequencies (color).
These 'cold' sources often emit only a single (set of) spectral lines, and not a 'black-body' spectrum like thermal sources do.
CO2 (and many other gases) atoms are excited by only certain EM frequencies (CO2 spectral lines).
Thus, these do not radiate 'black-body' spectrum, but only absorb and emit radiation of a certain frequency (color), called the CO2
absoption/emission spectrum.
This effect is at the basis of GHG theory.
That's it for now.
hda
<r...@verific.com> wrote:
YOUR indicated cold to hot radiation _has_no_ power.
>I stated that cold to hot radiation has power.
To stimulate your coherent receivers.
>It's just less power than the hot to cold radiation that goes the other way.
And thus meaningless in heattransfer wheater or climate.
>I also claim that both energy flows are indepenent and are independently observable.
One with help and the other not necessarily.
>
>>
>> I am now looking to a screen with a color temperature of
>> 6000 K, but it is not as hot as the sun.
But your theoretical academic suggests the color yellow is
actually radiating 3200 K, which of course it is not.
Rob Dekker
"hda" <age...@xs4all.nl.invalid> wrote in message news:hp5eg55qc9sfvb3s4...@4ax.com...
> On Fri, 20 Nov 2009 14:05:27 -0800, "Rob Dekker"
>>>>I hope so. What I understood from your argument is that you did not see that 'downward' radiation (back-radiation) actually
>>>>exists,
>>>>or that it would always be the same as upward radiation.
>>>>
>>>>What I tried to show is that this is not the case. These energy flows exist in reality, are directional, and they also change
>>>>from
>>>>moment to moment. Just like convection and conduction heat flow changes from moment to moment. When you add up all the heat flow
>>>>numbers for one location and moment, then you will know if that location will warm up or cool down.
>>>>So it is incorrect to assume an everlasting equilibrium or to assume that heat can only flow from warm to cold.
>>>>
>>>
>>> Your indicated cold to hot radiation _has_no_ power.
>>
>>I think that was Bill.
>
> YOUR indicated cold to hot radiation _has_no_ power.
Are you saying this as a statement of fact, or do you have evidence that 'my' indicated "cold to hot" radiation does not exist ?
>
>>I stated that cold to hot radiation has power.
>
> To stimulate your coherent receivers.
Yes, or anything else that it hits.
>
>>It's just less power than the hot to cold radiation that goes the other way.
>
> And thus meaningless in heattransfer wheater or climate.
>
Only meaningless is you can prove that it does not exist or does make no difference for wheater or climate.
>>I also claim that both energy flows are indepenent and are independently observable.
>
> One with help and the other not necessarily.
>
You are on thin ice here, buddy. This "cold to hot" (as well as "hot to cold") radiation, is very accurately predicted by Planck's
law.
So you would have to dis-prove Planck's law to make it go away.
http://en.wikipedia.org/wiki/Planck's_law
Some people think that the Stefan Bolzmann constant is falsified by Al Gore to serve AGW theory.
Do you think that is true ?
>>
>>>
>>> I am now looking to a screen with a color temperature of
>>> 6000 K, but it is not as hot as the sun.
>
> But your theoretical academic suggests the color yellow is
> actually radiating 3200 K, which of course it is not.
Correction, by stating what you just said about your computer screen, YOU suggested that the color yellow is radiation 3200 K.
Reality is the other way around.
Bill Ward
Please explain how my definition of "virtual" is a red herring. It's in
the wiki, and its what I meant. How much more on topic can I get?
> There is no way to detect a photon 'in flight' so to say (while it is
> still 'virtual'). But once it materializes, it constitutes energy
> transfer from the location where it originated to the point where it was
> detected. At the point where it is detected, it consitutes heat.
Minor point - the energy in the photon becomes heat (random translational
motion) in the absorber. Photons are not heat.
> So that should be detectable, even the source was colder than the sink.
All you need to do is find a way to detect it.
>>>> because any thermal detector must be colder than the radiation, and
>>>> any coherent receiver must add energy from an outside source,
>>>> effectively cooling the receiver.
>>>
>>> This statement is not correct.
>>> We build receivers with less than 20 K noise temperature, while they
>>> operate at room temperature. And if you think that the 20 K is as low
>>> as it goes, remember that the resolution (discriminate between two
>>> different temperatures) of a detector can be much better than the
>>> noise temperature of the system. How else could Wilson have detected
>>> the CMBR in 1965 and determined it to be 3.5 K (pretty close to the
>>> actual 2.7 K). All this determined with a system that operated at room
>>> temperature.
>>
>> There are basically two types of devices used to measure EMR. One is a
>> bolometer, which detects the EMR by transforming the energy directly to
>> heat, the same way the surface of the Earth converts solar radiation.
>
> OK.
>
>> They obviously must be cooled below the source temperature.
>
> I don't think so.
> Since the low temperature heat source is simply adding energy to the
> detector, it should be possible to detect a source that has a lower
> temperature than the (system) temperature of the detector, by looking at
> the DIFFERENCE in detector temperature that the low temperator source
> creates. Even for a bolometer.
Ah, but if the bolometer is warmer than the "source", the reading will be
negative, indicating energy was still transferred from hot to cold. It's
not nice to fool Mother Nature.
>> <http://en.wikipedia.org/wiki/Bolometer>
>>
>> The other is a coherent receiver, which does not convert the incoming
>> radiation to heat, but excites a tuned circuit, which involves an
>> amplifier and an external power supply. You can think of the external
>> power as the way around the 2nd law, as the energy is not spontaneously
>> flowing from cold to hot, but is given a boost, in the same way the
>> heat inside a refrigerator is boosted by the externally powered
>> compression and expansion cycle.
>>
>> <http://trs-new.jpl.nasa.gov/dspace/bitstream/2014/10504/1/02-2516.pdf>
>>
>> The coherent receiver is not detecting heat, but the direct EMR.
>> That's more like a radio transmitter and receiver. The AM band would
>> represent a very low temperature, but is easily received by a room
>> temperature radio. It's not heat, because heat is random, and coherent
>> radiation is not.
>
> Another way of looking at coherent receivers is that the photon energy
> gets 'amplified' before they are 'detected'. They are still 'virtual'
> inside the receiver, until we actually 'decohere' them by measuring
> their energy (with a diode or other non-linear device).
> This gets quite a bit off-topic, but it's actually very interesting to
> do this though experiment in the context of causality and quantum
> physics.
Indeed. I actually think it's on topic wrt the reality/virtuality of
"back radiation".
> But for this discussion, I do not see much of a difference between a
> coherent receiver or a bolometer. They both receive the same radiation
> energy, it's just that a coherent receiver first amplifies the photon
> (energy).
The required outside energy prevents the transfer from being
"spontaneous", in the same way a refrigerator "allows" energy to flow
from cold to hot.
Alternatively, one can say the photon was never converted to heat - it
never transferred its energy to a material body as heat, but ended up as
a current which was amplified. Coherent detection does not involve
conversion to heat.
No, actually I prefer the hyperphysics definition:
<http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/heat.html#c1>
"Heat may be defined as energy in transit from a high temperature object
to a lower temperature object. An object does not possess "heat"; the
appropriate term for the microscopic energy in an object is internal
energy. The internal energy may be increased by transferring energy to
the object from a higher temperature (hotter) object - this is properly
called heating."
>
> If so, radiation IS heat.
Not according to hyperphysics. It's the internal (kinetic) energy of a
body. Radiation is photons.
> You should be OK with that, since I think you are mostly concerned with
> the detection of that heat, right ?
Not OK, because the definition doesn't fit.
> If so, then is this a good definition of your hypothesis (you write
> this, above) :
>
> Bill's hypothesis 1 : "heat is not transferred from a cold source
> to a hot sink"
>
> You OK with that ?
Yup.
> Or would you prefer something like this :
>
> Bill's hypothesis 2 : "the heat transfer from a cold to a hot body is
> not physically detectable"
More complex, so not as good as the first. It assumes heat transfer from
cold to hot.
> Or can you state your one hypothesis
I'll stick with the one I explicitly posted a few paragraphs above.
> Note that I did not say 'net' heat transfer. Only 'heat' transfer, which
> is essentially 'energy' transfer.
No, they're not the same. Heat is always energy, but energy is not
always heat. Energy does always end up as heat, though. I'm sure you're
familiar with the concept of entropy.
> And you also put in TWO constraints on disproving these hypothesis :
>
> In the process of experimentally disproving hypothesis 1, two
> constraints apply :
>
> Bill's constraint 1 : " no coherent receivers can be used"
Right. Heat is not transferred under that condition. EM energy is.
> Bill's constraint 2 : "the (non-coherent) receiver at the sink is not
> allowed to be cooled".
Right. If the detector is warmer than the "source", heat still flows
from hot to cold.
> I think these two constraints are completely irrelevant, and are simply
> an experimental physics form of sado masochism. But I put them there any
> way, since I do not think there is any conceptual difference.
>
> It just makes it harder to think of an experiment that will disprove the
> hypothesis.
>> That doesn't mean the method of assuming there is a real transfer form
>> hot to cold and an opposite real transfer from cold to hot gives the
>> wrong answer, it's just that it can be misleading to those more
>> familiar with dealing in net flux.
>>
>>
> Please let me know if you accept the hypothesis, and the constraints.
If you keep in mind the difference between heat and EM radiation, I'm OK
with it.
> Just promise me one thing : If I can think of such an experiment (I'm
> already thinking) and it honestly complies with your hypothesis, then
> please accept it. OK ?
Sure. I've also tried to find a way around it. I haven't yet, but I'd
be happy (and surprised) to hear one. The 2nd law keeps getting in the
way.
Rob Dekker
"Bill Ward" <bw...@ix.REMOVETHISnetcom.com> wrote in message news:EoSdnbfMuOOUq5rW...@giganews.com...
> On Fri, 20 Nov 2009 12:57:13 -0800, Rob Dekker wrote:
>
>>>> Below, I'll assume that with "virtual" heat transfer, you mean
>>>> 'directional' heat transfer, means any energy transfer from one point
>>>> to another.
>>>
>>> Sorry, that's not what I mean by "virtual". I mean energy that can be
>>> assumed to be there, but is not observable, like the electron and
>>> positron before pair production occurs. If I meant directional, I
>>> would say directional.
>>
>>
>> This seems like a red herring to me.
>
> Please explain how my definition of "virtual" is a red herring. It's in
> the wiki, and its what I meant. How much more on topic can I get?
>
Your point on this definition of "virtual" back-radiation would have merit if the hot object (Earth) were a 'mirror'.
In that case, photons from the cold object do not manifest themselves, and thus do not transfer heat to the hot object.
In that case 'back-radiation' would be 'virtual'.
>> There is no way to detect a photon 'in flight' so to say (while it is
>> still 'virtual'). But once it materializes, it constitutes energy
>> transfer from the location where it originated to the point where it was
>> detected. At the point where it is detected, it consitutes heat.
>
> Minor point - the energy in the photon becomes heat (random translational
> motion) in the absorber.
Not a minor point, because we are not talking about a mirror. We are talking about a hot black-body object.
In that case, photons from cold object DO manifest themselves on the hot object, and thus transfer energy from cold to hot.
And this transfer of energy is real and detectable, thus not "virtual".
> Photons are not heat.
That is true.
As I stated in the beginning, I think this thread is plagued by misunderstanding created by different terminology.
It I can detect the effect by an increase in temperature, then that means there was energy transferred (from cold to hot).
I don't see how arguing about what the 'net' energy flow is makes any difference for this fact.
Also, the fact that bolometers are more sensitive when they are cooled does not exclude their ability to detect objects that are
colder then the bolometer itself.
And that is not fooling Mother Nature at all. It's simply caused by the fact that there is actual energy transfer happening.
>
>>> <http://en.wikipedia.org/wiki/Bolometer>
>>>
>>> The other is a coherent receiver, which does not convert the incoming
>>> radiation to heat, but excites a tuned circuit, which involves an
>>> amplifier and an external power supply. You can think of the external
>>> power as the way around the 2nd law, as the energy is not spontaneously
>>> flowing from cold to hot, but is given a boost, in the same way the
>>> heat inside a refrigerator is boosted by the externally powered
>>> compression and expansion cycle.
>>>
>>> <http://trs-new.jpl.nasa.gov/dspace/bitstream/2014/10504/1/02-2516.pdf>
>>>
>>> The coherent receiver is not detecting heat, but the direct EMR.
>>> That's more like a radio transmitter and receiver. The AM band would
>>> represent a very low temperature, but is easily received by a room
>>> temperature radio. It's not heat, because heat is random, and coherent
>>> radiation is not.
>>
>> Another way of looking at coherent receivers is that the photon energy
>> gets 'amplified' before they are 'detected'. They are still 'virtual'
>> inside the receiver, until we actually 'decohere' them by measuring
>> their energy (with a diode or other non-linear device).
>> This gets quite a bit off-topic, but it's actually very interesting to
>> do this though experiment in the context of causality and quantum
>> physics.
>
> Indeed. I actually think it's on topic wrt the reality/virtuality of
> "back radiation".
See above, for the "mirror" case you are right on track.
But for a black-body, energy transfer is real, since photons are no longer 'entangled'. They are now turned to energy.
>
>> But for this discussion, I do not see much of a difference between a
>> coherent receiver or a bolometer. They both receive the same radiation
>> energy, it's just that a coherent receiver first amplifies the photon
>> (energy).
>
> The required outside energy prevents the transfer from being
> "spontaneous", in the same way a refrigerator "allows" energy to flow
> from cold to hot.
Why does this exclude detecting the energy flow ?
>
> Alternatively, one can say the photon was never converted to heat - it
> never transferred its energy to a material body as heat, but ended up as
> a current which was amplified. Coherent detection does not involve
> conversion to heat.
Ah. Here we get into causality and the essence of what a 'photon' really is.
In coherent detection, the photon gets amplified (by 'modulating' a current or so), but one could claim that it still 'exists',
since all it's properties are still preserved.
Even phase-coupling of two receivers can be constructed, which allows the photon to take both paths simultaniously.
So one can argue that the photon still exists in both receivers.
Only after a non-linear detector (diode or so) does it's (now greatly amplified) energy turn to heat.
That's off-topic, but still fun huh ?
Well, here we have a problem.
You start with the DEFINITION that heat is energy in transit from a high temperature to a low temperature.
So any experiment to show actual energy transfer from low to high is now by DEFINITION no longer called heat.
Nothing I can do with that.
>>
>> If so, radiation IS heat.
>
> Not according to hyperphysics. It's the internal (kinetic) energy of a
> body. Radiation is photons.
>
>> You should be OK with that, since I think you are mostly concerned with
>> the detection of that heat, right ?
>
> Not OK, because the definition doesn't fit.
So then maybe you should change the definition (to the Wiki definition).
Yours sounds very restrictive any way, because it maked "a warm blanket" a contradiction in terms.
>
>> If so, then is this a good definition of your hypothesis (you write
>> this, above) :
>>
>> Bill's hypothesis 1 : "heat is not transferred from a cold source
>> to a hot sink"
>>
>> You OK with that ?
>
> Yup.
I guessed so, because that hypothesis is true by your own DEFINITION of heat transfer.
>
>> Or would you prefer something like this :
>>
>> Bill's hypothesis 2 : "the heat transfer from a cold to a hot body is
>> not physically detectable"
>
> More complex, so not as good as the first. It assumes heat transfer from
> cold to hot.
Since you ruled it our by DEFINITION it is no longer science.
So let's not call this energy transfer "heat". Lets call it "energy" and use the Wiki definition.
Or state another term for this energy transfer that goes from cold to hot.
What is the term that you want to use ?
>
>> Or can you state your one hypothesis
>
> I'll stick with the one I explicitly posted a few paragraphs above.
Can you re-state that ?
Here is the thing with science : It is exists, then it should be detectable.
If it is "virtual" or only there "in principle" (such as a photon 'in flight') then it is not detectable.
So let's Let's give it a name :
GHGs warm the Earth surface by :
Call it "back-radiation" (I think this is what most scientists prefer)
Call it "heat transfer from cold to hot" (this is what I throw out)
Call it "insolation, like a blanket" (I think this is what Al Gore uses).
Then, let's conceive of an experiment that detects this effect.
If we can't find such an experiment, then we can declare it "virtual"
OK ?
Bill Ward
> "Bill Ward" <bw...@ix.REMOVETHISnetcom.com> wrote in message
> news:EoSdnbfMuOOUq5rW...@giganews.com...
>> On Fri, 20 Nov 2009 12:57:13 -0800, Rob Dekker wrote:
>>
> [ snipped old text ]
>>>>> Below, I'll assume that with "virtual" heat transfer, you mean
>>>>> 'directional' heat transfer, means any energy transfer from one
>>>>> point to another.
>>>>
>>>> Sorry, that's not what I mean by "virtual". I mean energy that can
>>>> be assumed to be there, but is not observable, like the electron and
>>>> positron before pair production occurs. If I meant directional, I
>>>> would say directional.
>>>
>>>
>>> This seems like a red herring to me.
>>
>> Please explain how my definition of "virtual" is a red herring. It's
>> in the wiki, and its what I meant. How much more on topic can I get?
>>
>>
> Your point on this definition of "virtual" back-radiation would have
> merit if the hot object (Earth) were a 'mirror'. In that case, photons
> from the cold object do not manifest themselves, and thus do not
> transfer heat to the hot object. In that case 'back-radiation' would be
> 'virtual'.
Now you're bringing in absorptivity/emissivity for no apparent reason.
You still can't transfer heat to a warmer object, no matter how black it
is.
>>> There is no way to detect a photon 'in flight' so to say (while it is
>>> still 'virtual'). But once it materializes, it constitutes energy
>>> transfer from the location where it originated to the point where it
>>> was detected. At the point where it is detected, it consitutes heat.
>>
>> Minor point - the energy in the photon becomes heat (random
>> translational motion) in the absorber.
>
> Not a minor point, because we are not talking about a mirror. We are
> talking about a hot black-body object. In that case, photons from cold
> object DO manifest themselves on the hot object, and thus transfer
> energy from cold to hot. And this transfer of energy is real and
> detectable, thus not "virtual".
You say they manifest themselves on the warmer body as heat, yet can't
provide an experiment that proves that. Is that scientific? It seems to
me to be an unprovable hypothesis.
The whole point is the absence of heat flow from cold to hot. If the
bolometer is warmer than the "source", the energy flow is real, but from
the bolometer to the "source" (now actually a sink), so the bolometer is
cooled by the "source". That's transfer from hot to cold, which no one
is questioning. It seems kind of silly to assume "cold rays" from the
target are cooling the bolometer.
> Also, the fact that bolometers are more sensitive when they are cooled
> does not exclude their ability to detect objects that are colder then
> the bolometer itself.
The heat flow is still from hot to cold.
> And that is not fooling Mother Nature at all. It's simply caused by the
> fact that there is actual energy transfer happening.
Yeah, backwards. The bolometer cools by radiating energy to the cooler
target. Heat doesn't flow the other way.
So you say, but can't prove. All we can prove is that we never see heat
flow from cold to hot.
>>> But for this discussion, I do not see much of a difference between a
>>> coherent receiver or a bolometer. They both receive the same radiation
>>> energy, it's just that a coherent receiver first amplifies the photon
>>> (energy).
>>
>> The required outside energy prevents the transfer from being
>> "spontaneous", in the same way a refrigerator "allows" energy to flow
>> from cold to hot.
>
> Why does this exclude detecting the energy flow ?
We're not talking about energy flow, we're talking about heat flow. In a
closed system, entropy never decreases.
>> Alternatively, one can say the photon was never converted to heat - it
>> never transferred its energy to a material body as heat, but ended up
>> as a current which was amplified. Coherent detection does not involve
>> conversion to heat.
>
> Ah. Here we get into causality and the essence of what a 'photon' really
> is. In coherent detection, the photon gets amplified (by 'modulating' a
> current or so), but one could claim that it still 'exists', since all
> it's properties are still preserved. Even phase-coupling of two
> receivers can be constructed, which allows the photon to take both paths
> simultaniously. So one can argue that the photon still exists in both
> receivers. Only after a non-linear detector (diode or so) does it's (now
> greatly amplified) energy turn to heat. That's off-topic, but still fun
> huh ?
Whatever. Unless the photon is converted to heat (internal energy), heat
has not been transferred.
That's not just my definition, it's quoted from hyperphysics, and it
happens to be the one I learned long ago.
> Nothing I can do with that.
That's what I've been trying to tell you. Earlier in the thread, you
correctly nailed the issue as a misunderstanding, but I don't think it
was mine.
>>> If so, radiation IS heat.
>>
>> Not according to hyperphysics. It's the internal (kinetic) energy of a
>> body. Radiation is photons.
Silence is agreement?
>>> You should be OK with that, since I think you are mostly concerned
>>> with the detection of that heat, right ?
>>
>> Not OK, because the definition doesn't fit.
>
> So then maybe you should change the definition (to the Wiki definition).
> Yours sounds very restrictive any way, because it maked "a warm blanket"
> a contradiction in terms.
You would prefer to redefine the meaning of heat rather than accept my
argument? Why would you think a wiki entry would be preferable to
hyperphysics? Did you even look at the hyperphysics link?
>>> If so, then is this a good definition of your hypothesis (you write
>>> this, above) :
>>>
>>> Bill's hypothesis 1 : "heat is not transferred from a cold
>>> source to a hot sink"
>>>
>>> You OK with that ?
>>
>> Yup.
>
> I guessed so, because that hypothesis is true by your own DEFINITION of
> heat transfer.
Not just mine, the commonly used definition.
>>> Or would you prefer something like this :
>>>
>>> Bill's hypothesis 2 : "the heat transfer from a cold to a hot body
>>> is not physically detectable"
>>
>> More complex, so not as good as the first. It assumes heat transfer
>> from cold to hot.
>
> Since you ruled it our by DEFINITION it is no longer science. So let's
> not call this energy transfer "heat". Lets call it "energy" and use the
> Wiki definition.
OK, but you're now talking about a different issue. Energy can be
transmitted without regard to temperature, but not heat.
> Or state another term for this energy transfer that goes from cold to
> hot. What is the term that you want to use?
I used "virtual", because it's theoretically there, but unobservable. I
think I covered that earlier.
>>> Or can you state your one hypothesis
>>
>> I'll stick with the one I explicitly posted a few paragraphs above.
>
> Can you re-state that?
Sorry. You snipped it, you find it. It's right after you asked me to be
more explicit. It's not significantly different than the two you posted
above.
>>> Note that I did not say 'net' heat transfer. Only 'heat' transfer,
>>> which is essentially 'energy' transfer.
>>
>> No, they're not the same. Heat is always energy, but energy is not
>> always heat. Energy does always end up as heat, though. I'm sure
>> you're familiar with the concept of entropy.
Again, silence is agreement?
Emitting radiation to space from a higher, cooler altitude than the
surface. The lapse rate projects the equivalent temp back to the
surface. I thought we'd established that in a previous thread.
> Call it "back-radiation" (I think this is what most scientists prefer)
> Call it "heat transfer from cold to hot" (this is what I throw out) Call
> it "insolation, like a blanket" (I think this is what Al Gore uses).
>
> Then, let's conceive of an experiment that detects this effect. If we
> can't find such an experiment, then we can declare it "virtual"
>
> OK ?
Fine by me. Good luck. I looked, but couldn't find one that doesn't
violate the 2nd law. Entropy always manages to increase.
Shaun Carl
I was browsing through some discussion of global warming and I came
across this discussion on heat transfer.
I think someone asked how can heat flow from a cold body to a hot
body. Heat can't really flow because its something that describes
temperature and temperature can't flow either. If what was meant is
that photons cannot be passed from a colder body to a hotter one, then
my question is: how does the colder body know it's next to a hotter
one and vice versa. And if an even colder one was placed in its
vicinity how does the one having the middle temperature know just to
send photons to the coldest one? Is there some mysterious
communication between them? Not that I know of.
Bill Ward
Good question. I'm involved in such a discussion with Rob Dekker, and
neither of us can think of an experiment that shows heat being
transferred from a cold "source" to a hot "sink". The second law keeps
interfering.
In fact the classic definition of heat precludes such transfer:
<http://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.html>
My best speculation is that the cold (low energy) photons cannot transfer
their energy to the hot "receptor" (2nd law), so they can't materialize
there (no wave function collapse).
But QM is not my long suit. It still seems rather weird to me, so I'd
love to hear from someone more familiar with it. The issue is moot,
AFAIC, except for the fact it's easy for those not familiar with the
"back radiation" approach to get it confused with net energy transfer.
BTW, here's an experiment that confirms the "spooky action at a distance"
of Bell:
http://www.physorg.com/news132830327.html
In the quantum world, distance doesn't seem to matter.
I M @ good guy
If clouds can reduce the cooling of a windshield
very substantially, it has to mean some absorption of
the radiation from the clouds, albeit broadband.
So surfaces, and various levels of the atmosphere
do exchange radiant energy back and forth, but from
all indications, according to the inverse square rule,
with the exception of any window of surface broadband
to space.
I see no reason why back radiation is not
being measured at various levels, it would be
a big help in a better understanding of GHG
theory.
I thought Dekker said it has been, but
then said no more about direct measurements.
Rob Dekker
"I M @ good guy" <I...@good.guy> wrote in message
news:ubqgg5to4am2870mn...@4ax.com...
Compliments !
Only the main radiation from the clouds towards the surface (especially at
night) is in IR.
This IR radiation is indeed the reason that your windshield does not freeze
on still nights with overcast.
> So surfaces, and various levels of the atmosphere
> do exchange radiant energy back and forth, but from
> all indications, according to the inverse square rule,
> with the exception of any window of surface broadband
> to space.
That's right.
Once again, the broadband radiation (typically called short-wave (SW) is
only present during the day.
Of course, SW is the main heater of the planet, since it comes directly from
the sun.
>
>
> I see no reason why back radiation is not
> being measured at various levels, it would be
> a big help in a better understanding of GHG
> theory.
Radiation IS measured, and I'm sure there have been experiments measuring it
at different altitudes too.
The thing is that this radiation is varying from day to night, and even from
hour to hour or minute to minute, as the temperatures of the atmosphere and
the surface change rapidly throughout the 24 hour day.
Also, these radiation level change from location to location.
There is quite a big difference between radiation levels when there is
overcast, versus clear sky.
If you know the temperatures through the atmosphere and the surface, and you
know the composition and pressure of the atmopsphere at various altitudes,
then you can calculate the radiation levels extremely precise, using
radiation transfer theory.
Radiation transfer theory is undisputed (since it results directly from the
Maxwell Equations), and this may be another reason why not too many
researchers want to invest time (and money) to measure radiation levels at
various locations and altitude.
>
> I thought Dekker said it has been, but
> then said no more about direct measurements.
>
Bill did not give me a chance yet to define what the experiment should show.
So far, he ruled out all proposals by his own definition of what is an
acceptable experiment to prove that back-radiation (from the sky down to
Earth surface) is really happening.
But I think that the windshield experiment that you describe above is
already showing that there is radiation from clouds down to Earth that
prevents your windshield from freezing.
>
>
>
>
>
Rob Dekker
> On Sat, 21 Nov 2009 10:43:07 -0800, Shaun Carl wrote:
>
>> Hi,
>> I was browsing through some discussion of global warming and I came
>> across this discussion on heat transfer. I think someone asked how can
>> heat flow from a cold body to a hot body. Heat can't really flow because
>> its something that describes temperature and temperature can't flow
>> either. If what was meant is that photons cannot be passed from a colder
>> body to a hotter one, then my question is: how does the colder body know
>> it's next to a hotter one and vice versa. And if an even colder one was
>> placed in its vicinity how does the one having the middle temperature
>> know just to send photons to the coldest one? Is there some mysterious
>> communication between them? Not that I know of.
>
> Good question. I'm involved in such a discussion with Rob Dekker, and
> neither of us can think of an experiment that shows heat being
> transferred from a cold "source" to a hot "sink". The second law keeps
> interfering.
Bill, with all due respect, but this is not an accurate summary of what we
discussed.
There are multiple experiments that I can propose that there is actual
energy(back-radiation) from the atmosphere down to Earth surface.
But the problem is that so far, you have rules out any proposal for any
experiment, by your own arguments and definitions.
I can't use an IR detector, since you claim it's a coherent receiver (which
you claim is not proving anything).
I can't use a bolometer at sink temperature, since you claim it is not
showing anything.
I can't even use the definition of 'heat' since your definition defines that
heat flows only from hot to cold.
So the only thing left over after your boundary conditions are met is an
experiment that violates the 2nd law of thermodynamics.
So there are no experiments that I can propose that would convice you.
>
> In fact the classic definition of heat precludes such transfer:
>
> <http://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.html>
>
> My best speculation is that the cold (low energy) photons cannot transfer
> their energy to the hot "receptor" (2nd law), so they can't materialize
> there (no wave function collapse).
OK. Now we are getting somewhere.
Let me propose an experiment that shows that the low energy photons from the
cold source can transfer their energy to a hot source :
Infrared penetrates water only a few micrometers.
SW (broadband light) penetrates much deeper (several dozens of meters).
So, if we measure the temperature difference between the top of water and a
few centimeters down, then we should find that this difference is greater
when there is a cloud overhead, then when there is clear sky overhead. The
IR radiation from the cloud should make the top few milimeters of water
'warmer' than it would be without the cloud.
If I can show you that, would you then accept that 'back-radiation' from the
clouds is actually there, and that there is thus IR energy transferred from
(cold) cloud to (warm) water and that thus wave functions collapse on a warm
surface even if they come from a cold source ?
>
> But QM is not my long suit. It still seems rather weird to me, so I'd
> love to hear from someone more familiar with it.
I'm very familiar with QM.
> The issue is moot,
> AFAIC, except for the fact it's easy for those not familiar with the
> "back radiation" approach to get it confused with net energy transfer.
Agreed. The second law is never broken.
But it is a form of 'insulation', or 'slowing down cooling'.
And because the Earth is constantly warmed from the sun, any obstruction in
cooling will warm it.
>
> BTW, here's an experiment that confirms the "spooky action at a distance"
> of Bell:
>
> http://www.physorg.com/news132830327.html
>
> In the quantum world, distance doesn't seem to matter.
That's correct, but that is a completely different matter.
The photons emitted from a cold source have action when they materialize
(wave function collapse) on ANY surface (hot or cold). With that, they
transfer their energy, and emit heat.
Rob Dekker
"Shaun Carl" <shaun...@gmail.com> wrote in message
news:abc6f1c7-b8e2-4a0a...@g23g2000yqh.googlegroups.com...
Hi Shaun,
You are absolutely right.
If photons would not transfer energy from a cold object to a hot object, but
they would the other way around, then they would need to somehow retain the
infomation about the their 'temperature of origin'.
Of course, photons are not that 'intelligent'. They have only very limited
information : their frequency (energy) and their phase of origin.
In fact, they don't even know where they are in the Universe. They can only
'materialize' with a certain probability, which depends on their phase
accumulation of all the paths that they can take through the Universe. Their
phase changes with the spacetime difference between their start and end
points.
Any way, long story short : They do not know where they hit, and they don't
care if the end of their journey is warmer or colder than the temperature of
the object where they originated.
Rob
Bill Ward
Which is still not a measurement.
> Radiation transfer theory is undisputed (since it results directly from
> the Maxwell Equations), and this may be another reason why not too many
> researchers want to invest time (and money) to measure radiation levels
> at various locations and altitude.
>
>
>> I thought Dekker said it has been, but
>> then said no more about direct measurements.
>>
>>
> Bill did not give me a chance yet to define what the experiment should
> show. So far, he ruled out all proposals by his own definition of what
> is an acceptable experiment to prove that back-radiation (from the sky
> down to Earth surface) is really happening.
Not just my definition, hyperphysics and the rest of the world also uses
it.
<http://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.html>
<quote>
Heat may be defined as energy in transit from a high temperature object
to a lower temperature object. An object does not possess "heat"; the
appropriate term for the microscopic energy in an object is internal
energy. The internal energy may be increased by transferring energy to
the object from a higher temperature (hotter) object - this is properly
called heating.
</quote>
The definition is succinct, clear, unambiguous, and accommodates the 2nd
law. What's not to like about it?
> But I think that the windshield experiment that you describe above is
> already showing that there is radiation from clouds down to Earth that
> prevents your windshield from freezing.
That's the problem with the "back radiation" approach. Rob neglected to
mention that the windshield still must be heating the clouds (cooling
itself) if it's warmer than the clouds. It's just not cooling as much as
if it were radiating to deep space.
With the net radiation approach, you'd say the windshield radiates more
energy to deep space than it does to clouds. I believe that's less
likely to confuse those less familiar with radiative transfer.
If Rob thinks hyperphysics is wrong, he should notify them immediately,
as according to the site, it's being used to teach thousands of students
all over the world.
I'm sorry Rob gave up on inventing a way to actually measure heat
transferred by "back radiation" without violating the 2nd law. I was
hoping he would succeed where I've failed.
Bill Ward
Not just my definition, it's the one hyperphysics uses. It's also the
one I was taught, and has served me well for quite a while.
If you have an issue with it, I'd suggest you contact hyperphysics and
tell them about their mistake. The site is used to teach thousands of
students all over the globe, so I'm sure if you convince them their
definition is wrong, they'll change it. They're way more competent in
physics than I am.
Here - Ill repost it for your convenience:
"Heat may be defined as energy in transit from a high temperature object
to a lower temperature object. An object does not possess "heat"; the
appropriate term for the microscopic energy in an object is internal
energy. The internal energy may be increased by transferring energy to
the object from a higher temperature (hotter) object - this is properly
called heating."
> I can't use an IR
> detector, since you claim it's a coherent receiver (which you claim is
> not proving anything).
I'm assuming you mean a coherent IR detector. You agreed photons are not
heat, didn't you? Detecting photons is not the same as detecting heat.
If you have a method other than coherent detectors and bolometers,
there's a big market in radio astronomy waiting for you.
> I can't use a bolometer at sink temperature, since you claim it is not
> showing anything.
I don't exactly know what you mean by "at sink temperature", because
bolometers are normally the sink, so they're cooled to a lower T than
the source. If you have a "source" colder than the bolometer, you will
see energy flowing from hot to cold, from the bolometer to the source,
and the bolometer cools, showing a negative energy flow. That's not
"back radiation", because it's flowing from hot to cold.
> I can't even use the definition of 'heat' since your definition defines
> that heat flows only from hot to cold.
Not just mine - hyperphysics wrote it.
> So the only thing left over after your boundary conditions are met is an
> experiment that violates the 2nd law of thermodynamics.
That's exactly the point I've been trying to make. I was hoping you
could find a way around it.
> So there are no experiments that I can propose that would convice you.
Then quit complaining and accept my argument. Or convince hyperphysics
their definition is wrong.
>> In fact the classic definition of heat precludes such transfer:
>>
>> <http://hyperphysics.phy-astr.gsu.edu/hbase/HFrame.html>
>>
>> My best speculation is that the cold (low energy) photons cannot
>> transfer their energy to the hot "receptor" (2nd law), so they can't
>> materialize there (no wave function collapse).
>
> OK. Now we are getting somewhere.
> Let me propose an experiment that shows that the low energy photons from
> the cold source can transfer their energy to a hot source :
>
> Infrared penetrates water only a few micrometers. SW (broadband light)
> penetrates much deeper (several dozens of meters).
>
> So, if we measure the temperature difference between the top of water
> and a few centimeters down, then we should find that this difference is
> greater when there is a cloud overhead, then when there is clear sky
> overhead. The IR radiation from the cloud should make the top few
> milimeters of water 'warmer' than it would be without the cloud.
That's not explicit or direct. Maybe you can propose an experiment with
defined filters (ideal if necessary) and specific sources that would
demonstrate whatever it is you are trying to show. It looks to me like
you're trying to use the surface of the ocean as the world's cheesiest
bolometer.
> If I can show you that, would you then accept that 'back-radiation' from
> the clouds is actually there, and that there is thus IR energy
> transferred from (cold) cloud to (warm) water and that thus wave
> functions collapse on a warm surface even if they come from a cold
> source ?
No, because the alternative hypothesis, that the net energy transferred
is proportional to the difference of the 4th powers of the temperatures
would do the same thing. Remember, we're trying to find a way to
distinguish between the net flow approach and the "back radiation"
approach. I claim there is no way to measure a difference between them,
thus no way to actually measure "back radiation.
>> But QM is not my long suit. It still seems rather weird to me, so I'd
>> love to hear from someone more familiar with it.
>
> I'm very familiar with QM.
>
>> The issue is moot,
>> AFAIC, except for the fact it's easy for those not familiar with the
>> "back radiation" approach to get it confused with net energy transfer.
>
> Agreed. The second law is never broken. But it is a form of
> 'insulation', or 'slowing down cooling'. And because the Earth is
> constantly warmed from the sun, any obstruction in cooling will warm it.
The Sun is heating the Earth, the "insulation" can only slow the
cooling. The insulation can't heat the earth unless it's warmer than the
target. If the Earth is warmer than the insulation, it will heat the
insulation.
>> BTW, here's an experiment that confirms the "spooky action at a
>> distance" of Bell:
>>
>> http://www.physorg.com/news132830327.html
>>
>> In the quantum world, distance doesn't seem to matter.
>
> That's correct, but that is a completely different matter. The photons
> emitted from a cold source have action when they materialize (wave
> function collapse) on ANY surface (hot or cold). With that, they
> transfer their energy, and emit heat.
How do you know wfs of photons from a cold source collapse on a warmer
surface? What test shows that?
Bill Ward
> "Shaun Carl" <shaun...@gmail.com> wrote in message
> news:abc6f1c7-
b8e2-4a0a-9a5...@g23g2000yqh.googlegroups.com...
>> Hi,
>> I was browsing through some discussion of global warming and I came
>> across this discussion on heat transfer. I think someone asked how can
>> heat flow from a cold body to a hot body. Heat can't really flow
>> because its something that describes temperature and temperature can't
>> flow either. If what was meant is that photons cannot be passed from a
>> colder body to a hotter one, then my question is: how does the colder
>> body know it's next to a hotter one and vice versa. And if an even
>> colder one was placed in its vicinity how does the one having the
>> middle temperature know just to send photons to the coldest one? Is
>> there some mysterious communication between them? Not that I know of.
>>
>>
>>
> Hi Shaun,
>
> You are absolutely right.
> If photons would not transfer energy from a cold object to a hot object,
> but they would the other way around, then they would need to somehow
> retain the infomation about the their 'temperature of origin'. Of
> course, photons are not that 'intelligent'. They have only very limited
> information : their frequency (energy) and their phase of origin.
Wow. And here I thought the energy would carry the information about the
effective source temperature. But then, as I said, QM isn't my long suit.
Can you quote a source for that?
columbiaaccidentinvestigation
> On Sat, 21 Nov 2009 23:42:56 -0800, Rob Dekker wrote:
> > news:abc6f1c7-
>
> b8e2-4a0a-9a5a-7ee842589...@g23g2000yqh.googlegroups.com...
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
so how does bill explain the suns corona temp. being hotter than the
photosphere. The idea is there are examples in our universe, but he
does not want to venture into those areas....
columbiaaccidentinvestigation
ask bill to investigate the suns corona.
hda
<r...@verific.com> wrote:
>
>"hda" <age...@xs4all.nl.invalid> wrote in message news:hp5eg55qc9sfvb3s4...@4ax.com...
>> On Fri, 20 Nov 2009 14:05:27 -0800, "Rob Dekker"
>....
>>>>>I hope so. What I understood from your argument is that you did not see that 'downward' radiation (back-radiation) actually
>>>>>exists,
>>>>>or that it would always be the same as upward radiation.
>>>>>
>>>>>What I tried to show is that this is not the case. These energy flows exist in reality, are directional, and they also change
>>>>>from
>>>>>moment to moment. Just like convection and conduction heat flow changes from moment to moment. When you add up all the heat flow
>>>>>numbers for one location and moment, then you will know if that location will warm up or cool down.
>>>>>So it is incorrect to assume an everlasting equilibrium or to assume that heat can only flow from warm to cold.
>>>>>
>>>>
>>>> Your indicated cold to hot radiation _has_no_ power.
>>>
>>>I think that was Bill.
>>
>> YOUR indicated cold to hot radiation _has_no_ power.
>
>Are you saying this as a statement of fact, or do you have evidence that 'my' indicated "cold to hot" radiation does not exist ?
I want to clarify it as a latent situation, that in the
cold-to-hot pipe the cold-to-hot photons have no
manifestation of heatpower. In a [heatpipe] cold-to-hot pipe
you could introduce as many positive and negative
photogenetic, antimatter or mathematical components as you
like, but only the netresult is effectively meaningfull to
humans in weather. Bottomline-stuff.
Sure every measurement or manifestation is a net-result
conclusive collapse of waves.
[People only can communicate by the manifestion, not by the
mystery] One promotes belief/trust too much. Soon we have
science as another religion. QM as metaphysical recursive
woowoo ?
>
>>
>>>I stated that cold to hot radiation has power.
>>
>> To stimulate your coherent receivers.
>
>Yes, or anything else that it hits.
>
>>
>>>It's just less power than the hot to cold radiation that goes the other way.
>>
>> And thus meaningless in heattransfer wheater or climate.
>>
>
>Only meaningless is you can prove that it does not exist or does make no difference for wheater or climate.
>
>>>I also claim that both energy flows are indepenent and are independently observable.
>>
>> One with help and the other not necessarily.
>>
>
>You are on thin ice here, buddy. This "cold to hot" (as well as "hot to cold") radiation, is very accurately predicted by Planck's
>law.
>So you would have to dis-prove Planck's law to make it go away.
>http://en.wikipedia.org/wiki/Planck's_law
>
>Some people think that the Stefan Bolzmann constant is falsified by Al Gore to serve AGW theory.
>Do you think that is true ?
Corrupted ? No.
(scientifically use the word corrupted, the word falsified
triggers popperian approval)
I M @ good guy
While there is definitely things right about the ordinary
concepts of net energy flow by IR radiation, being absolutely
certain about anything removes the discussion from "science"
to some other status.
There are things that are assumed, and experiment may
support those assumptions, but there may be something else
involved.
Some of the things that have appeared in the last 100
years have been spectacular, and really surprised those who
were the established experts.
And other things that were so cut and dried turned
out to be duds or not worth the effort.
I thought solar sails would be in common use ny now,
but haven't seen any.
The simple toy "radiometer" might be an example of
something that could be used to test or even measure
back radiation, indirectly, like by comparison or some
logical way, but I have no faith in the claims of 200 plus
watts per square meter, and I do not believe that any
direct flux from CO2 in the stratosphere ever makes
it to the surface, what little there is may add a little
energy to the air at levels in between and a small
part of that might make it to the surface.
From what I have seen, visible light has much
more energy than any infra-Red, and flux is for sure
a function of frequency that makes IR different.
The fact that the atmosphere can radiate
a given flux greater than the solar flux through
a given cross section in the upper atmosphere
may be misleading because it probably originates
from a lot of different levels all at once.
I would like to see an experiment where a
wide plate is cooled down and my windshield
placed below it to see if the conditions of a
black sky can be replicated.
Rob Dekker
> On Sat, 21 Nov 2009 23:42:56 -0800, Rob Dekker wrote:
>
>> "Shaun Carl" <shaun...@gmail.com> wrote in message
>> news:abc6f1c7-
> b8e2-4a0a-9a5...@g23g2000yqh.googlegroups.com...
>>> Hi,
>>> I was browsing through some discussion of global warming and I came
>>> across this discussion on heat transfer. I think someone asked how can
>>> heat flow from a cold body to a hot body. Heat can't really flow
>>> because its something that describes temperature and temperature can't
>>> flow either. If what was meant is that photons cannot be passed from a
>>> colder body to a hotter one, then my question is: how does the colder
>>> body know it's next to a hotter one and vice versa. And if an even
>>> colder one was placed in its vicinity how does the one having the
>>> middle temperature know just to send photons to the coldest one? Is
>>> there some mysterious communication between them? Not that I know of.
>>>
>>>
>>>
>> Hi Shaun,
>>
>> You are absolutely right.
>> If photons would not transfer energy from a cold object to a hot object,
>> but they would the other way around, then they would need to somehow
>> retain the infomation about the their 'temperature of origin'. Of
>> course, photons are not that 'intelligent'. They have only very limited
>> information : their frequency (energy) and their phase of origin.
>
> Wow. And here I thought the energy would carry the information about the
> effective source temperature.
Are you serious, Bill ?
I really do not mean this in a bad way at all, but I am surprised.
I thought a lot about what it means if photons actually 'knew' what the
temperature of the body was when they were emitted.
I ran into a lot of tricky questions :
Since a photon gets emitted from a single atom, or atom collision, how would
the photon even know what the temperature is ?
After all, individual atoms don't really have a 'temperature'.
Next, how would they carry that temperature info with them ?
And then, how and why would they choose to only materialize on an atom with
a lower temperature than their origin ?
And what is the probability factor (of the photon materializing) when the
temperatures of souce and sink are very close ?
What happens when source and sink greatly vary in speed ? Does the
'materialization temperature' change ?
And what other properties does the photon carry with it ?
> But then, as I said, QM isn't my long suit.
I appreciate you being honest, but at the same time, you would hopefully
understand that there is no need for photons to have this property at all.
Simply the Stefan Boltzmann law already defines that radiation from a hot
black-body is always higher than radiation from a cold black-body. And that
pretty much makes sure that the second law will not be violated for
'experiments' like the ones we are looking for. No need for photons to
'know' their 'temperature' of origin.
> Can you quote a source for that?
I don't think you will find any article that explicitly states which
properties a photon does NOT have.
But read the "Physical properties" section here in Wiki :
http://en.wikipedia.org/wiki/Photon
A photon is massless, moves at c, has energy E=hc/lambda (with lambda being
the wavelength), and left or right circular spin momentum. That's it for
properties.
No mentioning of temperature.
Consequently, a photon that left from a cold black-body WILL materialize on
a hot black-body, and thus transfer energy.
Which name do you want to give this energy transfer ? In the context of cold
sky/clouds over a warmer surface, I call it back-radiation.
Rob Dekker
"I M @ good guy" <I...@good.guy> wrote in message
Actually, the essence of science (at least in physics) is that any proposed
theory has to hold under ALL circumstances.
That comes pretty close to being 'absolutely certain'.
Here is a good example of that :
Max Planck assumed that atoms that interact with each other exchange quanta
(photons) of energy E=hv (v being the frequency (c/lambda) of the photon).
That simple assumption explained the otherwise bizarre (not explainable with
classical physics) black-body radiation spectrum. Planck's law holds under
all circumstances. We have not found any exceptions to it. Not anywhere.
Stefan Boltzmann law (the integral of Planck's law over all frequencies)
thus also holds under all circumstances.
That means that it certainly holds for some cold clouds over a warm surface.
>
> There are things that are assumed, and experiment may
> support those assumptions, but there may be something else
> involved.
>
> Some of the things that have appeared in the last 100
> years have been spectacular, and really surprised those who
> were the established experts.
>
> And other things that were so cut and dried turned
> out to be duds or not worth the effort.
>
> I thought solar sails would be in common use ny now,
> but haven't seen any.
>
> The simple toy "radiometer" might be an example of
> something that could be used to test or even measure
> back radiation, indirectly, like by comparison or some
> logical way, but I have no faith in the claims of 200 plus
> watts per square meter, and I do not believe that any
> direct flux from CO2 in the stratosphere ever makes
> it to the surface, what little there is may add a little
> energy to the air at levels in between and a small
> part of that might make it to the surface.
>
I know that you have states questions about the radiation levels of cold sky
and clouds, but Stefan Boltzmann law is very clear on this :
http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
So a cloud (assuming it radiates as a black-body in IR) with a temperature
of 278 K (5 C, just above freezing) radiates some 338 W/m^2 of IR energy.
There is no question about this (back-radiation) at all. Not with any
scientist as far as I know.
> From what I have seen, visible light has much
> more energy than any infra-Red, and flux is for sure
> a function of frequency that makes IR different.
>
I think you confuse flux with frequency. They are completely separate
things.
A 40 W fashlight light incandecent light bulb radiates visible light (high
frequency), but the flux at 1 meter distance is low.
A 4000 W space heater radiates IR (low frequency) but the flux at 1 meter
distance is 100 times greater than the light bulb.
Flux is related to power (and distance if it is a point source), while
frequency is related to color.
> The fact that the atmosphere can radiate
> a given flux greater than the solar flux through
> a given cross section in the upper atmosphere
> may be misleading because it probably originates
> from a lot of different levels all at once.
>
That's true. Every spectral line (frequency) has a different 'optical depth'
(how far it penetrates through the atmosphere).
This means that every spectral line originates from a different altitude.
Not just as measured from the surface, but also as measured from space.
So this is how you can see that if CO2 concentration increases, that the
radiation energy in the CO2 spectral lines will change, because the optical
depth reduces. From the atmosphere to space, they radiate from a higher
altitude (where it is colder, thus less radiation), while from the
atmosphere to the surface they radiate from a lower altitude (where it is
warmer, thus more radiation) to the surface. Thus, Earth cools less, while
the surface warms. That's CO2 induced global warming in a nutshell. Nice huh
?
>
> I would like to see an experiment where a
> wide plate is cooled down and my windshield
> placed below it to see if the conditions of a
> black sky can be replicated.
>
Yep. That's a good idea !
You may also want to put aluminum foil on top of that cold plate. This will
prevent the plate from being warmed up by clouds overhead.
>
>
>
Bill Ward
Every radiated photon has an energy related to the source temperature.
Those emitted from a hot object have higher energy than those that were
emitted at a colder temperature, per Planck's law. That's why the Sun at
6000K emits short wave high energy photons, and the earth at 255K emits
LWIR low energy photons. So all the photon has to do is ask itself what
its energy is, and voila, it "knows" what its source temperature was/is.
How much QM have you really taken? That's pretty basic physics.
If you're really interested, you can look at:
<http://en.wikipedia.org/wiki/Planck's_law>
and
<http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/disene.html#c1>
I was basing my speculation around the idea that if the temperature of
the absorber is higher than the source, there may be no states available
to accommodate the energy of the low energy photon, preventing it from
being absorbed. I know I don't know enough about QM to be sure, but from
what little I do know, it seems plausible. Since you were apparently
unaware of Planck's law, even after I initially commented ,"...energy
would carry the information about the effective source temperature", I
suspect you know even less than I do.
>> Can you quote a source for that?
>
> I don't think you will find any article that explicitly states which
> properties a photon does NOT have.
> But read the "Physical properties" section here in Wiki :
> http://en.wikipedia.org/wiki/Photon
>
> A photon is massless, moves at c, has energy E=hc/lambda (with lambda
> being the wavelength), and left or right circular spin momentum. That's
> it for properties.
>
> No mentioning of temperature.
Wouldn't you guess energy might be a clue to temperature? You'd have to
be completely innocent of any knowledge of the origin of QM not to see
that, especially with a broad hint.
> Consequently, a photon that left from a cold black-body WILL materialize
> on a hot black-body, and thus transfer energy.
You have more confidence than knowledge, I'm afraid.
> Which name do you want to give this energy transfer ? In the context of
> cold sky/clouds over a warmer surface, I call it back-radiation.
I call it a lack of high-school physics.
columbiaaccidentinvestigation
to measure the energy of a photon without affecting it, the
hypothesis remains unproven"
thats a stupid statement.....
columbiaaccidentinvestigation
toy "radiometer" might be an example of something that could be used
to test or even measure back radiation, indirectly, like by comparison
or some logical way, but I have no faith in the claims of 200 plus
watts per square meter, and I do not believe that any direct flux from
CO2 in the stratosphere ever makes it to the surface, what little
there is may add a little
energy to the air at levels in between and a small part of that might
make it to the surface."
laughing, dont be so emotional when you find out you were wrong....
http://cat.inist.fr/?aModele=afficheN&cpsidt=13418307
"Titre du document / Document title
Atmospheric longwave irradiance uncertainty: Pyrgeometers compared to
an absolute sky-scanning radiometer, atmospheric emitted radiance
interferometer, and radiative transfer model calculations"
I M @ good guy
wrote:
Theories are a dime a dozen, I was not speaking of theories.
>That comes pretty close to being 'absolutely certain'.
Name one thing that is absolutely certain.
>Here is a good example of that :
>Max Planck assumed that atoms that interact with each other exchange quanta
>(photons) of energy E=hv (v being the frequency (c/lambda) of the photon).
>That simple assumption explained the otherwise bizarre (not explainable with
>classical physics) black-body radiation spectrum. Planck's law holds under
>all circumstances. We have not found any exceptions to it. Not anywhere.
>Stefan Boltzmann law (the integral of Planck's law over all frequencies)
>thus also holds under all circumstances.
So classical physics was no absolutely certain.
>That means that it certainly holds for some cold clouds over a warm surface.
This is where the AGW flavor of GHG theory is flawed,
talking about surface, and taking air temperature in the air.
>> There are things that are assumed, and experiment may
>> support those assumptions, but there may be something else
>> involved.
>>
>> Some of the things that have appeared in the last 100
>> years have been spectacular, and really surprised those who
>> were the established experts.
>>
>> And other things that were so cut and dried turned
>> out to be duds or not worth the effort.
>>
>> I thought solar sails would be in common use ny now,
>> but haven't seen any.
>>
>> The simple toy "radiometer" might be an example of
>> something that could be used to test or even measure
>> back radiation, indirectly, like by comparison or some
>> logical way, but I have no faith in the claims of 200 plus
>> watts per square meter, and I do not believe that any
>> direct flux from CO2 in the stratosphere ever makes
>> it to the surface, what little there is may add a little
>> energy to the air at levels in between and a small
>> part of that might make it to the surface.
>
>I know that you have states questions about the radiation levels of cold sky
>and clouds, but Stefan Boltzmann law is very clear on this :
>http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
>
>So a cloud (assuming it radiates as a black-body in IR) with a temperature
>of 278 K (5 C, just above freezing) radiates some 338 W/m^2 of IR energy.
I challenge that quantity per unit area.
Whats the matter, IR can't be focused with a
large diameter refractor or reflector collector apparatus?
It should be easy to boil water with one.
>There is no question about this (back-radiation) at all. Not with any
>scientist as far as I know.
That is not convincing.
>> From what I have seen, visible light has much
>> more energy than any infra-Red, and flux is for sure
>> a function of frequency that makes IR different.
>>
>
>I think you confuse flux with frequency. They are completely separate
>things.
I think you are missing something.
>A 40 W fashlight light incandecent light bulb radiates visible light (high
>frequency), but the flux at 1 meter distance is low.
>A 4000 W space heater radiates IR (low frequency) but the flux at 1 meter
>distance is 100 times greater than the light bulb.
>
>Flux is related to power (and distance if it is a point source), while
>frequency is related to color.
Check out some power equations and let me know.
>> The fact that the atmosphere can radiate
>> a given flux greater than the solar flux through
>> a given cross section in the upper atmosphere
>> may be misleading because it probably originates
>> from a lot of different levels all at once.
>>
>
>That's true. Every spectral line (frequency) has a different 'optical depth'
>(how far it penetrates through the atmosphere).
>This means that every spectral line originates from a different altitude.
>Not just as measured from the surface, but also as measured from space.
I am not interested in spectra at all, the bottom
line is flux, in fact, there should be measurements
made of upward and downward radiation at the same
location at the same instant to verify the net energy.
>So this is how you can see that if CO2 concentration increases, that the
>radiation energy in the CO2 spectral lines will change, because the optical
>depth reduces.
The CO2 concentration should not vary a lot with
altitude, what do you mean energy in the spectral lines?
>From the atmosphere to space, they radiate from a higher
>altitude (where it is colder, thus less radiation), while from the
>atmosphere to the surface they radiate from a lower altitude (where it is
>warmer, thus more radiation) to the surface. Thus, Earth cools less, while
>the surface warms. That's CO2 induced global warming in a nutshell. Nice huh
>?
Sounds like a fantasy a computer modeler might have.
>> I would like to see an experiment where a
>> wide plate is cooled down and my windshield
>> placed below it to see if the conditions of a
>> black sky can be replicated.
>>
>
>Yep. That's a good idea !
>You may also want to put aluminum foil on top of that cold plate. This will
>prevent the plate from being warmed up by clouds overhead.
Not for long, as soon as the thin foil reaches
equilibrium with the clouds (modified by ambient
air) it wouldn't help much, besides it could be done
indoors.
I M @ good guy
http://theory.uwinnipeg.ca/mod_tech/node151.html
I haven't seen any frequency graphs that attempted
to show the relative energy at each frequency, but with
electricity at a given voltage, I think power is directly
proportional to frequency.
And I am pretty sure I have seen literature where
photons carry some kind of information, there must
be more to x-rays and gamma rays and cosmic rays
than just frequency.
The absolute certainty in any discussion really
bothers me, AFAIK, one of the first lessons at the
beginning of graduate school is that there are
usually in-betweens or exceptions.
I really think any talk given on climate science
should begin with, "Assuming CO2 causes a warming
of the atmosphere....", it would seem to make any
statements made more credible.
hda
Rob states about absolute energy starting from T=0 Kelvin. A
one way energy stream of photons.
> Whats the matter, IR can't be focused with a
>large diameter refractor or reflector collector apparatus?
Probably you assume a 2-way stream of photons.
>
> It should be easy to boil water with one.
>
>>There is no question about this (back-radiation) at all. Not with any
>>scientist as far as I know.
>
>
> That is not convincing.
>
>
>>> From what I have seen, visible light has much
>>> more energy than any infra-Red, and flux is for sure
>>> a function of frequency that makes IR different.
>>>
>>
>>I think you confuse flux with frequency. They are completely separate
>>things.
E = k * Frequency = Joules per second = Watt = BTU/h
Flux is Watt per square meter = BTU/(sqft*h)
Bill Ward
Each photon has energy proportional to its frequency. If you know one,
you can derive the other. It may help if you reread my explanation above
and look at this:
<http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html#c3>
Poke around in hyperphysics and you'll find some interesting stuff,
nicely explained.
> And I am pretty sure I have seen literature where
> photons carry some kind of information, there must be more to x-rays and
> gamma rays and cosmic rays than just frequency.
No, not really. When the frequency is shifted, as in the astronomical
red shift, there's no difference in anything but energy.
Bill Ward
Then why did you post this, Rob? It appears you had no knowledge of
Planck when you wrote it. You must be a very quick study.
<quote from Message-ID: <1oKdnQfGPJhP1JfW...@giganews.com>
I ran into a lot of tricky questions : Since a photon gets emitted from a
single atom, or atom collision, how would the photon even know what the
temperature is ? After all, individual atoms don't really have a
'temperature'. Next, how would they carry that temperature info with them
? And then, how and why would they choose to only materialize on an atom
with a lower temperature than their origin ? And what is the probability
factor (of the photon materializing) when the temperatures of souce and
sink are very close ? What happens when source and sink greatly vary in
speed ? Does the 'materialization temperature' change ? And what other
properties does the photon carry with it?
<\quote>
Are you going to respond to my comments in that thread? I can repost it
if you didn't get it.
Only to absolute zero, which is hard to do and harder to measure. Even
the black sky is 3K.
> There is no question about this (back-radiation) at all. Not with any
> scientist as far as I know.
That's only an argument to authority. Can you explain any of it, or are
you simply taking it on faith?
>> From what I have seen, visible light has much
>> more energy than any infra-Red, and flux is for sure a function of
>> frequency that makes IR different.
>>
>>
> I think you confuse flux with frequency. They are completely separate
> things.
>
> A 40 W fashlight light incandecent light bulb radiates visible light
> (high frequency), but the flux at 1 meter distance is low. A 4000 W
> space heater radiates IR (low frequency) but the flux at 1 meter
> distance is 100 times greater than the light bulb.
>
> Flux is related to power (and distance if it is a point source), while
> frequency is related to color.
And source temperature.
>> The fact that the atmosphere can radiate
>> a given flux greater than the solar flux through a given cross section
>> in the upper atmosphere may be misleading because it probably
>> originates from a lot of different levels all at once.
>>
>>
> That's true. Every spectral line (frequency) has a different 'optical
> depth' (how far it penetrates through the atmosphere). This means that
> every spectral line originates from a different altitude. Not just as
> measured from the surface, but also as measured from space.
>
> So this is how you can see that if CO2 concentration increases, that the
> radiation energy in the CO2 spectral lines will change, because the
> optical depth reduces. From the atmosphere to space, they radiate from a
> higher altitude (where it is colder, thus less radiation), while from
> the atmosphere to the surface they radiate from a lower altitude (where
> it is warmer, thus more radiation) to the surface. Thus, Earth cools
> less, while the surface warms. That's CO2 induced global warming in a
> nutshell. Nice huh ?
Until you get into the details. Reading Miskolczi would help. Then you
notice CO2 is only a minor player compared to water.
I M @ good guy
<bw...@ix.REMOVETHISnetcom.com> wrote:
Do you mean "Cosmological Red Shift"?
Wouldn't your statement above make frequency
and energy the same, a red shift moves the lines.
The thing here is that the energy of the back
radiation would be related to the flux per unit of
area.
But all we seem to be getting is comments from
climate modelers.
I M @ good guy
<bw...@ix.REMOVETHISnetcom.com> wrote:
So my challenge will not be answered? :-)
This seems to present a problem, some of the
energy budget graphs show large downward values,
can somebody spot whether or not all of them are
net energy transfer or are some the calculate max
to a black sky?
Could there be two different ways to present
the IR radiation?