Question for theoretical physics about gases
 

Black body radiation is only relevant at red hot temperatures or in a
vacuum. In a fluid media like air or water heat is transported by
convection. You must be a newbie to these groups. Don't you remember Dr.
Convection who was the nemesis of the Jim Hansen crowd?


wrote in message
...
It is a most basic fact that objects in a room will all reach the same
temperature as the air in a room.

So how does the air transfer the energy to the solid substances such
as metal?

It is not denied that a piece of metal radiates according to Planck's
Blackbody Radiation Law, which describes the distribution of the
energy of the radiation. The total energy radiated from the surface
conforms with the Boltzman Stefan equation.

If the air brings the metal to it's temperature at equilibrium, the
metal will be radiating from it's surface the infrared energy
according to the Boltzman Stefan equation, which for 300K, (81F) is
460Wm-2. So a steel ball with 1 meter surface area will be radiating
this quantity of energy per second. If this quantity of energy is
leaving the surface of the metal at c and in this quantity per second,
the metal must be absorbing this quantity of energy at this rate from
the gases in the air.

A situation could be devised wherein it could be clearly shown that
the energy is not received as radiation from other solid surfaces.

This energy cannot be transmitted to the metal merely by the
collisions or conduction through the air molecules. A quantification
of the energy of the gas molecules and the transfer of energy to the
metal must be achieved in order to have valid theoretical physics of
this most basic of phenomena of physics.

This means the gas molecules are not transparent to the infrared as is
often believed, but are absorbing and radiating the full continous
spectrum of the infrared in the distribution similar to the Planck
Radiation Law. It is only this absorption and emission through the
full spectrum that the quantity of energy being transfered per second
can be achieved which is equivalent to the energy leaving the surface
of the metal.

It is ABSOLUTELY ESSENTIAL to the theory of greenhouse gases, that N2
and O2 are absolutely transparent and non-reactive to infrared
radiation energy. Otherwise, there is no rational that the trace gases
or the very minute changes in concentrations of trace gases can have
such theorized impact upon atmospheric temperatures.

If the theory of greenhouse gases were correct, very great differences
in the temperature of a piece of the metal would occur depending upon
which gases or combination of gases are used in experimentation.

This is certainly not the case. As it stands, there is no valid
theoretical theory of gases which can be shown by experiment to
support the widely held belief in the existence of the property of
'greenhouse' gases.

KD

Question for theoretical physics about gases
 

kiloVolts wrote:

Black body radiation is only relevant at red hot temperatures or in a
vacuum. In a fluid media like air or water heat is transported by
convection. You must be a newbie to these groups. Don't you remember Dr.
Convection who was the nemesis of the Jim Hansen crowd?

Your confidence is not proportional to your knowledge. Study some
thermodynamics.


wrote in message
...

It is a most basic fact that objects in a room will all reach the same
temperature as the air in a room.

So how does the air transfer the energy to the solid substances such
as metal?

It is not denied that a piece of metal radiates according to Planck's
Blackbody Radiation Law, which describes the distribution of the
energy of the radiation. The total energy radiated from the surface
conforms with the Boltzman Stefan equation.

If the air brings the metal to it's temperature at equilibrium, the
metal will be radiating from it's surface the infrared energy
according to the Boltzman Stefan equation, which for 300K, (81F) is
460Wm-2. So a steel ball with 1 meter surface area will be radiating
this quantity of energy per second. If this quantity of energy is
leaving the surface of the metal at c and in this quantity per second,
the metal must be absorbing this quantity of energy at this rate from
the gases in the air.

A situation could be devised wherein it could be clearly shown that
the energy is not received as radiation from other solid surfaces.

This energy cannot be transmitted to the metal merely by the
collisions or conduction through the air molecules. A quantification
of the energy of the gas molecules and the transfer of energy to the
metal must be achieved in order to have valid theoretical physics of
this most basic of phenomena of physics.

This means the gas molecules are not transparent to the infrared as is
often believed, but are absorbing and radiating the full continous
spectrum of the infrared in the distribution similar to the Planck
Radiation Law. It is only this absorption and emission through the
full spectrum that the quantity of energy being transfered per second
can be achieved which is equivalent to the energy leaving the surface
of the metal.

It is ABSOLUTELY ESSENTIAL to the theory of greenhouse gases, that N2
and O2 are absolutely transparent and non-reactive to infrared
radiation energy. Otherwise, there is no rational that the trace gases
or the very minute changes in concentrations of trace gases can have
such theorized impact upon atmospheric temperatures.

If the theory of greenhouse gases were correct, very great differences
in the temperature of a piece of the metal would occur depending upon
which gases or combination of gases are used in experimentation.

This is certainly not the case. As it stands, there is no valid
theoretical theory of gases which can be shown by experiment to
support the widely held belief in the existence of the property of
'greenhouse' gases.

KD

Question for theoretical physics about gases
 

On Feb 19, 8:00 pm, "kiloVolts"
mant...@u6t3s0s3A6s1e4Z4V7rH3h5b7q3a3s2aX7A2D9y6j 1z.com wrote:
Black body radiation is only relevant at red hot temperatures or in a
vacuum. In a fluid media like air or water heat is transported by
convection. You must be a newbie to these groups. Don't you remember Dr.
Convection who was the nemesis of the Jim Hansen crowd?


Convection only exist in gases because they are absobing and emitting
the infrared of all frequencies and maintain this radiation field.

Obviously you have no mathematics for the energy of which you are
speaking or the mathematics for the motion of the molecules in a gas.

Thanks for stating clearly this false idea, which is accepted by the
schools of theoretical physics which shows clearly that they have
entirely departed from actual laboratory analysis and any claim at all
to mathematical capability.

The energy of 1 mole of gas per deg K is R.
The mean kinetic energy of a molecule for it's velocity is therefore
kT.
Kinetic energy is derived by Newtons equation,
1/2mv^2

Since the average kinetic energy is equal for different molecules, the
average value of the square of the velocity is seen to be inversely
proportional to the mass of the molecule, and hence the average
velocity (root mean square average), is inveresely proportional to the
square root of the molecular weight.

Therefore at 0degC, hydrogen molecules have the velocity of,
1.84 x 10^3 m s-1
at 0degC oxygen molecules have the average speed of,
..46 x 10^3 m s-1

Botzman Stefan applies to all temperatures.
T^4 x 5.67E-8 = Wm-2.
This is the derivative for the overall energy, So a steel ball with
1sq. meter of surface area, has this overall quantity of energy, (460
Joules per second), leaving it's surface per second which is in
Joules.

The Planck Radiation Law only defines how this energy is distributed
at each frequency, and is entirely complaint and derived directly and
encompasses the Boltzman-Stefan equation.

The term, BLACKBODY, is actually not so important.The distribution
defined by Planck with the curve for each temperature, actually only
is valid when temperature is at equilibrium or not changing. Only then
does Planck's mathematics define the probability for the energy of the
oscillator at the time of emission of a photon. In Planck's actual
theory before it is abused by the theoretical schools, he meant to
define emissions as the product of electron oscillators, and to define
the probability of the energy of the oscillator at each specific
moment of emisison. An oscillator emits a quantity of energy and then
must recompense it's energy in a time interval which is called the
time of continuance.
The distribution curve is defined merely by the probabilities or
statistics of a system at thermodynamic equilibrium, such as the sun.

This radiation law also applies to gases and low temperatures. This
can be proved by the simple analysis of the steel ball which reaches
the same temperature as the air or gas in which it exists.

If the ball were placed in pure nitrogen, and conditions were devised
that radiation from other solid surfaces could not be affecting
temperature, there is no valid application of the theory that the gas
molecules are transfering energy to the metal by collisions, as is
absoutely required if it is considered that the N2 is transparent to
infrared radiation.

The metal ball radiates according to Boltzman Stefan, which is 460
Joules per second per sq meter at 300K. The energy of 1 mole of gas of
RT at 300K is
R = 8.31 Joules,
RT for 300K is 2493Joules.

Although the heat capacity of monatomic gases is 3/2 R, the 2493Joules
is actually the total kinetic energy for the sum of the velocities of
the molecules for 1 mole of gas. Therefore, it can be considered that
this is the energy of the mole of gas above absolute zero.

This is the sum energy which can be delivered by direct contact with
all of the molecules of the mole of gas at 300K. Of course there is
the total energy of the heat capacity, which is not more than 4.5R at
lower temperatures for most gases.

A steel ball with 1 sq meter surface area has a radius of,
28.209cm
and occupies 94,024 cubic centimeters of volume.
1 mole of gas @stp occupies 22,400cc of volume.

Adding volume of steel ball to volume of 1 mole of gas, =
116,429cc
Radius of sphere with this volume, =
30.29cm
30.29 radius minus 28.209 (radius of steel ball with surface area of 1
sq meter), =
2.08 centimeters.

So 1 mole of gas surrounding a steel ball with surface area of 1 sq
meter will encompass 2.08 cm of radius in it's volume around the ball.

The steel is at temperature equilibrium with the air. It radiates from
it's surface 460 Joules per second of infrared radiation energy
regardless of the specific distribution at each frequency.

If the gas around the ball is 'transparent' to the radiation, this
quantity of energy LEAVES THE VICINITY AT C.

At this rate of energy transfer per second and with the kinetic energy
of the velocities of the molecules being 2493 Joules, THE ENTIRE
ENERGY ABOVE ABSOLUTE ZERO FOR THE VELOCITIES OF THE MOLE OF GAS IS
ACHIEVED IN 5.4 SECONDS.

Incorporate the increase of radiated energy for increasing temperature
(which increases as a fourth power), and the increase of kinetic
energy for the molecular velocities (which increases as a direct
proportion to temperature), and it is even more clear that the
contemporary idea is clearly false, that it is merely the collisions
and kinetic energy of the molecules of gas which holds and transfers
energy. The energy simply cannot be transfered through the motions of
the molecules at the rate which experimentaion proves the energy IS
being transfered.

The gases trap a radiation field, although the absorption and emission
of these photons occurs at the velocity of light.

At equilibrium, if the gases were transparent to the infrared, there
is no means that this energy is being supplanted to the metal at the
rate it is leaving the surface.

For the metal and the 'non-greenhouse' gas to be at or near
temperature equilibrium, THE GAS IS ABSORBING AND RADIATING EQUIVALENT
ENERGY OF THE ENERGY RADIATION FROM THE METAL..

When a gas molecule absorbs an infrared photon of lower energies, the
direction of the subsequent emission is random. At lower temperatures,
individual emission of photons is normally at or near the energy level
of an absorbed photon. Infrared thus can 'appear' to travel unabsorbed
through low density gases, but do not traverse through the gas as do
high energy visible frequencies to which many gases ARE mostly
transparent.

Very simple experiments with pure nitrogen very simply PROVE IT TO BE
INVALID, the concept that this gas is transparent or non-reactive to
infrared.

Your denial of Stefans Law being applicable at low temperatures in
which energy increases as a fourth power to the temperature shows your
lack of education due to the entire lack of valid theoretical physics
in present schools and entire disregard for actual direct science on
which theory should be based.

The steel ball radiates according to Stefans law. You can forgoe
quoting the dogma of the eternal need of the schools to abuse Planck
and do their normal redefintion of the term, 'blackbody', as their
approach to doing physics with no mathematics at all. And be a part of
the hoax of theoretical physics to have a valid theory of gases, in
which they merely use semantics and redefinition of the terminology to
encourage the mere superstition and psuedo religous belief in the
false concept of 'greenhouse gases'.

Otherwise submit some Gdamn direct science with some form of
mathematics which in can some way demonstrate some direct science for
this widely used term.

As it is now, you only have theoretical calculations from a FALSE
Planck curve which entirely is an invalid application of his
mathematic for probability at temperature equilibrium for the
oscillators energy at each subsequent time of emission according to
temperature.

Energy is not specified or 'quantized' to be in this distribution, and
the dark bands of CO2 have NO EFFECT AT ALL in retaining outgoing
infrared energy as is supposed by this theoretical application which
has NO DIRECT LABORATORY science to support it's assertions. CO2
radiates absorbed energy at other frequencies than it's dark bands and
has no trouble at all doing this, as the CO2 laser can demonstrate.

Or explain this most simple and basic phenomena of physics and
chemistry is some terms that make sense on how the energy is
supplanted to the surface of the metal which supposedly leaves through
the damn, TRANSPARENT gas.

KD

Question for theoretical physics about gases
 

On Feb 21, 7:25*am, wrote:


Therefore at 0degC, hydrogen molecules have the velocity of,
1.84 x 10^3 m s-1
at 0degC oxygen molecules have the average speed of,
.46 x 10^3 *m s-1

An important note to this to keep in perspective the idea of the
velocities and collisions and supposed transfer of kinetic energy, the
mean free path or distance a molecule moves between collisions @stp
is only 500 Angstroms, or 200 times it's own diameter


So 1 mole of gas surrounding a steel ball with surface area of 1 sq
meter will encompass 2.08 cm of radius in it's volume around the ball.

Actually at 300K, the volume of the gas is 300/273 = 1.0989 x 22,400
=
24,615cc
92,024+ 24,615 =
30.31 cm radius of 1 mole of air around sphere of 1sq meter surface
area.
30.31 - 28.209 =
2.1 centimeter of added radius for the volume of 1 mole of gas around
the sphere of 1 sq meter surface area @300degK, or 81F.

Question for theoretical physics about gases
 

On Feb 19, 6:00*pm, "kiloVolts"
mant...@u6t3s0s3A6s1e4Z4V7rH3h5b7q3a3s2aX7A2D9y6j 1z.com wrote:
Black body radiation is only relevant at red hot temperatures or in a
vacuum. In a fluid media like air or water heat is transported by
convection. You must be a newbie to these groups. Don't you remember Dr.
Convection who was the nemesis of the Jim Hansen crowd?

wrote in message

...

It is a most basic fact that objects in a room will all reach the same
temperature as the air in a room.

So how does the air transfer the energy to the solid substances such
as metal?

It is not denied that a piece of metal radiates according to Planck's
Blackbody Radiation Law, which describes the distribution of the
energy of the radiation. The total energy radiated from the surface
conforms with the Boltzman Stefan equation.

If the air brings the metal to it's temperature at equilibrium, the
metal will be radiating from it's surface the infrared energy
according to the Boltzman Stefan equation, which for 300K, (81F) is
460Wm-2. So a steel ball with 1 meter surface area will be radiating
this quantity of energy per second. If this quantity of energy is
leaving the surface of the metal at c and in this quantity per second,
the metal must be absorbing this quantity of energy at this rate from
the gases in the air.

A situation could be devised wherein it could be clearly shown that
the energy is not received as radiation from other solid surfaces.

This energy cannot be transmitted to the metal merely by the
collisions or conduction through the air molecules. A quantification
of the energy of the gas molecules and the transfer of energy to the
metal must be achieved in order to have valid theoretical physics of
this most basic of phenomena of physics.

This means the gas molecules are not transparent to the infrared as is
often believed, but are absorbing and radiating the full continous
spectrum of the infrared in the distribution similar to the Planck
Radiation Law. It is only this absorption and emission through the
full spectrum that the quantity of energy being transfered per second
can be achieved which is equivalent to the energy leaving the surface
of the metal.

It is ABSOLUTELY ESSENTIAL to the theory of greenhouse gases, that N2
and O2 are absolutely transparent and non-reactive to infrared
radiation energy. Otherwise, there is no rational that the trace gases
or the very minute changes in concentrations of trace gases can have
such theorized impact upon atmospheric temperatures.

If the theory of greenhouse gases were correct, very great differences
in the temperature of a piece of the metal would occur depending upon
which gases or combination of gases are used in experimentation.

This is certainly not the case. As it stands, there is no valid
theoretical theory of gases which can be shown by experiment to
support the widely held belief in the existence of the property of
'greenhouse' gases.

KD

You are plain flat wrong

Question for theoretical physics about gases
 

On Feb 21, 5:45*pm, rich wrote:
On Feb 19, 6:00*pm, "kiloVolts"





mant...@u6t3s0s3A6s1e4Z4V7rH3h5b7q3a3s2aX7A2D9y6j 1z.com wrote:
Black body radiation is only relevant at red hot temperatures or in a
vacuum. In a fluid media like air or water heat is transported by
convection. You must be a newbie to these groups. Don't you remember Dr..
Convection who was the nemesis of the Jim Hansen crowd?

wrote in message

....

It is a most basic fact that objects in a room will all reach the same
temperature as the air in a room.

So how does the air transfer the energy to the solid substances such
as metal?

It is not denied that a piece of metal radiates according to Planck's
Blackbody Radiation Law, which describes the distribution of the
energy of the radiation. The total energy radiated from the surface
conforms with the Boltzman Stefan equation.

If the air brings the metal to it's temperature at equilibrium, the
metal will be radiating from it's surface the infrared energy
according to the Boltzman Stefan equation, which for 300K, (81F) is
460Wm-2. So a steel ball with 1 meter surface area will be radiating
this quantity of energy per second. If this quantity of energy is
leaving the surface of the metal at c and in this quantity per second,
the metal must be absorbing this quantity of energy at this rate from
the gases in the air.

A situation could be devised wherein it could be clearly shown that
the energy is not received as radiation from other solid surfaces.

This energy cannot be transmitted to the metal merely by the
collisions or conduction through the air molecules. A quantification
of the energy of the gas molecules and the transfer of energy to the
metal must be achieved in order to have valid theoretical physics of
this most basic of phenomena of physics.

This means the gas molecules are not transparent to the infrared as is
often believed, but are absorbing and radiating the full continous
spectrum of the infrared in the distribution similar to the Planck
Radiation Law. It is only this absorption and emission through the
full spectrum that the quantity of energy being transfered per second
can be achieved which is equivalent to the energy leaving the surface
of the metal.

It is ABSOLUTELY ESSENTIAL to the theory of greenhouse gases, that N2
and O2 are absolutely transparent and non-reactive to infrared
radiation energy. Otherwise, there is no rational that the trace gases
or the very minute changes in concentrations of trace gases can have
such theorized impact upon atmospheric temperatures.

If the theory of greenhouse gases were correct, very great differences
in the temperature of a piece of the metal would occur depending upon
which gases or combination of gases are used in experimentation.

This is certainly not the case. As it stands, there is no valid
theoretical theory of gases which can be shown by experiment to
support the widely held belief in the existence of the property of
'greenhouse' gases.

KD

' You are plain flat wrong- Hide quoted text -

Proof is in the pudding. Although most of greenie weenie science is
based upon psychological evaluation of anyone who does not likewise
repeat the invalid dogma as they, there are times in life when one can
devise and obtain an actual SCIENTIFIC PROOF.

But since you do not rely on PROOF for your beliefs, we can understand
that this concept is foreign to you, and we should not expect you to
trouble your little mind to attempt to understand what a PROOF is or
how this concept relates to science.

So right and wrong for you is a subjective concept and when it is your
communal insanity, you are assured that you are right and have no need
to apply science or mathematics.

KD


'http://groups.google.com/group/alt.global-warming/tree/browse_frm/
thread/f6cd38eff8d1777b/bc868a4e83400a37?hl=en&rnum=1&_done=%2Fgroup
%2Falt.global-warming%2Fbrowse_frm%2Fthread%2Ff6cd38eff8d1777b%3 Fhl
%3Den%26scoring%3Dd%26&scoring=d#doc_bc868a4e83400 a37