On 09/08/15 23:30, Alastair wrote:
On Sunday, 9 August 2015 21:41:27 UTC+1, Dawlish wrote:

So if we took the cold object away, the hot one would still cool?
You are saying it is not the cold object that is making it cold.

The warmer object would cool by emitting more radiation than it
gained, if the surroundings were cooler.

And if we took the cold object away and the surrounding were not
cooler what would happen?


If the 'surroundings' are a uniform background of black-body radiation
at a temperature greater than the body in question then the body will
heat up till it is at the same temperature as the surroundings.

On 09/08/2015 14:13, Metman2012 wrote:
I've been following this thread with fascination. I'm not a physicist or
even a scientist, but I have a question that perhaps someone can answer.

Let's ask it with an example. There are three bodies, one at -50, one at
0 and one at 50 degrees. It's obvious that the one at 50 degrees is hot
radiating and the one at -50 is cold radiating. What is the one in the
middle doing? Unless I've completely missed the point, it's cold
radiating to the one at 50 and hot radiating to the one at -50. How can
it be both?

No you have pretty much understood it at a handwaving level.

"Cold radiation" is an invention of dodgy double glazing salesmen.
It has as much relevance today as the phlogiston theory of fire.

The question you pose above of -50, 0 and 50 C is what you might expect
if the dominant mechanism of heat transfer was by conduction. The mid
point being half way in temperature between the two ends.

Taking the end set points for temperatures as t=300K and T=400K for
arithmetic convenience and considering only radiative transfer and
taking each surface as a perfect black body radiator.


t=300K u= unknown T=400K
| | |
| -- kt^4 ku^4 --|--ku^4 kT^4--|
| | |

We can say with certainty that at equilibrium without the central
radiation shield the system would have a net flux F of

F = kT^4-kt^4

And with the intermediate radiation shield at equilibrium the net flux f
on either side of the divide must be equal so that

f = kT^4-ku^4 = ku^4 - kt^4

Hence 2u^4 = T^4 + t^4 or u^4 = (T^4+t^4)/2

Hence f = F/2

Putting the numbers in u^4 = (300^4 + 400^4)/2 ~= 360^4

In practice this technique is used in cryogenics where the radiation
shields are mirror finished so that instead of being a perfect black
body they mostly reflect incident photons and behave as a 5% black body.


Now let's add another body, say at 100 degrees. This one is now the hot
radiating one, and the one at 50 degrees now becomes a body which does
both. Now the reality of the universe is that there are many bodies, all
busily radiating. And we can't know which is the hottest and which the
coldest, so everything is radiating both hot and cold.

So am I being simplistic? Am I not understanding what all this is about?

Can someone answer in simple terms (one syllable or less) to explain
this please?

No you have it right. And because radiation loss is such a steep power
of temperature objects more than 10% above ambient on the Kelvin scale
start to radiate at a level which becomes increasingly significant.

Everything at a temperature above absolute zero is radiating energy at
everything else. The balance of an isolated object between incoming and
outgoing energy fluxes determines its final equilibrium temperature.

--
Regards,
Martin Brown

On 09/08/2015 15:29, Alastair wrote:
On Sunday, 9 August 2015 09:20:15 UTC+1, Martin Brown wrote:
On 07/08/2015 21:26, Col wrote:
Dawlish wrote:
On Friday, August 7, 2015 at 5:56:34 PM UTC+1, Alastair wrote:

I find your posts so insulting I have difficulty reading them.

I know. It's because they tell you that you are clearly and
unambiguously wrong.

Didn't you say that there was no 'proof' in science?

But there is "disproof" - a subtle difference.

https://en.wikipedia.org/wiki/Scient...tific_evidence

There is no proof of correctness of a theory. Every independent
experiment consistent with a theory merely improves confidence in it
until you find a novel *experiment* that breaks the status quo. We can
never be sure we have a complete description but we get a successively
better approximation to describing our universe as time passes.

However, a scientific theory must be capable of being *refuted* and one
clear refutation is more than enough to show that a widely held theory
is invalid or at the very least incomplete. You can prove that some
theory is wrong because it does not describe the universe we live in.

"Cold radiation" doesn't even get over the first hurdle it is complete
and utter ********(TM) in the same vein as N-rays and polywater.

Prove that!

It is self evident to anyone with even a basic understanding of modern
thermodynamics and twentieth century physics. I rest my case.

You are unable or unwilling to learn so the only thing that remains now
is to ensure that no-one is misled by your inane ramblings.

There is only one sort of thermal radiation from any body determined by
its absolute temperature, geometry and surface emissivity. The changes
in temperature of thermally isolated bodies is determined by the balance
of emitted radiation and received radiation.

There is no need to invoke your magyck of "cold radiation" here.

--
Regards,
Martin Brown

On 09/08/2015 16:17, Alastair wrote:
On Sunday, 9 August 2015 09:35:20 UTC+1, Martin Brown wrote:
On 07/08/2015 23:21, Alan LeHun wrote:
In article ,
says...

Exactly, if the radiation a body receives is from a cooler body,
then the first body will cool. So it is possible to cool a body with
radiation, and it only makes sense to call it cold radiation.

You are confusing and conflating the net flux of energy with
temperature.

No, you are.

You *really* don't understand your mistake at all.

I thought initially that you were trolling but it is now clear that you
do not understand the subject of radiative heat transfer at all.

I thought you were smart, but I see you are just as incapable of revising
your preconceived ideas as Dawlish. But he's got an excuse.

I understand the physics and want to make sure that no-one here is
misled by your incoherent ramblings about "cold radiation".

He isn't interested in the truth, only in making me appear foolish.

You have excelled in quest that by your original post and inability to
comprehend explanations given to you by several posters now. Your
attempt to demean Dawlish has backfired and made you look ignorant,
stubborn, unwilling to leanr and stupid in roughly equal measure.

I had thought you were better than that :-(

I am only interested in the truth.

--
Regards,
Martin Brown

"RedAcer" wrote in message
...
On 09/08/15 23:54, Alastair McDonald wrote:
"RedAcer" wrote in message
...

How many do you know?

Scores. I've got a degree in Physics and spent several years working on
a PhD in low temperature solid state physics.

When the photon reaches a body and is absorbed by an atom how does is it
'know' if that body is hotter or colder than the one it was emitted
from. How does the absorbing atom 'know' the temperature of the body the
photon came from?

I am tempted to reply: "The photon and atom ask a passing PhD student."
:-)

The photon and the atom only "know" the photon's frequency and hence its
energy. It is the difference between the absorbed photons' and emitted
photons'
energies which determines the objects change in temperature.

That is drivel. You need to learn some physics.
?
I didn't mention any emitted photons. Lets do one thing at a time.

"When the photon reaches a body and is absorbed by an atom how does is
it 'know' if that body is hotter or colder than the one it was emitted
from"

I think YOU need to learn some physics.

The photon doesn't know anything. It does not have a brain. And blackbody
radiation is not absorbed by atoms.

"RedAcer" wrote in message

The hotter body will cool anyway whether the colder body is there or
not. The radiation from the colder body means that the hot body will
cool more slowly that it would have done if the cold body weren't there.
It is not 'cold' radiation as you keep insisting. It does not cool the
hot body, it warms it up.(I'm assuming no background bodies or source of
radiation)

There is always background radiation of one type or another. Think about
it. You are either surrounded by walls, or by the earth and sky, or by
cosmic background radiation. Describe a real situation where it does not
exist.

Cheers, Alastair.

On Monday, August 10, 2015 at 3:12:41 PM UTC+1, Alastair wrote:
"RedAcer" wrote in message
...
On 09/08/15 23:54, Alastair McDonald wrote:
"RedAcer" wrote in message
...

How many do you know?

Scores. I've got a degree in Physics and spent several years working on
a PhD in low temperature solid state physics.

When the photon reaches a body and is absorbed by an atom how does is it
'know' if that body is hotter or colder than the one it was emitted
from. How does the absorbing atom 'know' the temperature of the body the
photon came from?

I am tempted to reply: "The photon and atom ask a passing PhD student."
:-)

The photon and the atom only "know" the photon's frequency and hence its
energy. It is the difference between the absorbed photons' and emitted
photons'
energies which determines the objects change in temperature.

That is drivel. You need to learn some physics.
?
I didn't mention any emitted photons. Lets do one thing at a time.

"When the photon reaches a body and is absorbed by an atom how does is
it 'know' if that body is hotter or colder than the one it was emitted
from"

I think YOU need to learn some physics.

Oh the irony......

On 10/08/15 15:09, Alastair McDonald wrote:
"RedAcer" wrote in message
...
On 09/08/15 23:54, Alastair McDonald wrote:
"RedAcer" wrote in message
...

How many do you know?

Scores. I've got a degree in Physics and spent several years working on
a PhD in low temperature solid state physics.

When the photon reaches a body and is absorbed by an atom how does is it
'know' if that body is hotter or colder than the one it was emitted
from. How does the absorbing atom 'know' the temperature of the body the
photon came from?

I am tempted to reply: "The photon and atom ask a passing PhD student."
:-)

The photon and the atom only "know" the photon's frequency and hence its
energy. It is the difference between the absorbed photons' and emitted
photons'
energies which determines the objects change in temperature.

That is drivel. You need to learn some physics.
?
I didn't mention any emitted photons. Lets do one thing at a time.

"When the photon reaches a body and is absorbed by an atom how does is
it 'know' if that body is hotter or colder than the one it was emitted
from"

I think YOU need to learn some physics.

The photon doesn't know anything. It does not have a brain.

Excellent. So you agree it is not a hot photon or a cold photon; just a
photon.

And blackbody
radiation is not absorbed by atoms.

Curious - what do you think absorbs/interacts with the photon when it
enters the 'body' in question?

On 10/08/15 15:12, Alastair McDonald wrote:
"RedAcer" wrote in message

The hotter body will cool anyway whether the colder body is there or
not. The radiation from the colder body means that the hot body will
cool more slowly that it would have done if the cold body weren't there.
It is not 'cold' radiation as you keep insisting. It does not cool the
hot body, it warms it up.(I'm assuming no background bodies or source of
radiation)

There is always background radiation of one type or another. Think about
it. You are either surrounded by walls, or by the earth and sky, or by
cosmic background radiation. Describe a real situation where it does not
exist.

We are all trying to explain some physics to you. The way that it's done
(in any physics class/book) is to concentrate on the salient features of
interest in the system and ignore/minimise other 'smaller' effects.
Assume we doing the experiment out in space where the CMBR is at 2.3K.
Let the cold body be at 200K and the hot at 300K. OK.
Terms of in the SB equation are proportional to T^4 and so can easily be
ignored in a first approximation.


Cheers, Alastair.

On 10/08/15 16:53, RedAcer wrote:
On 10/08/15 15:12, Alastair McDonald wrote:
"RedAcer" wrote in message

The hotter body will cool anyway whether the colder body is there or
not. The radiation from the colder body means that the hot body will
cool more slowly that it would have done if the cold body weren't there.
It is not 'cold' radiation as you keep insisting. It does not cool the
hot body, it warms it up.(I'm assuming no background bodies or source of
radiation)

There is always background radiation of one type or another. Think about
it. You are either surrounded by walls, or by the earth and sky, or by
cosmic background radiation. Describe a real situation where it does not
exist.

We are all trying to explain some physics to you. The way that it's done
(in any physics class/book) is to concentrate on the salient features of
interest in the system and ignore/minimise other 'smaller' effects.
Assume we doing the experiment out in space where the CMBR is at 2.3K.
Let the cold body be at 200K and the hot at 300K. OK.
Terms of in the SB equation are proportional to T^4 and so can easily be
ignored in a first approximation.

Last sentence not very clear, should be:-
"Terms in the SB equation are proportional to T^4 an so the CMBR can
ignored in a first approximation."