\subsection{Thermal conductivity versus radiative transfer} 
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\subsubsection{The heat equation}
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In many climatological texts it seems 
to be implicated that 
thermal radiation needs not be taken 
into account when dealing with 
heat conduction, which is incorrect\TschSpace% 
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\cite{Weise1966}.
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Rather, always the entire heat flow 
density $\textbf{q}$ must be taken into account.
This is given by the equation
%
\begin{equation}
\textbf{q} = - \lambda \cdot \mbox{\textbf{grad}} \, T
\end{equation}
%
in terms of the gradient 
of the temperature $T$.
It is inadmissible to separate 
the radiation transfer from 
the heat conduction, when
balances are computed.

In the following, a quasi one-dimensional
experimental situation for the determination 
of the thermal conductivity is considered
(Fig.~\ref{Fig:heattransport}).
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\begin{figure}[htbp]
\ifthenelse{\equal{IJMPB}{\TschStyle}}{%
\centerline{\psfig{file=PictureLarge_heat_conduction_.eps,width=2.00in}}
}{}
\ifthenelse{\equal{arXiv}{\TschStyle}}{%
\centerline{\includegraphics[scale=1.00]{PictureSmall_heat_conduction_.png}}
\vspace*{-16pt}
}{}
\ifthenelse{\equal{TeX4ht}{\TschStyle}}{%
\centerline{\includegraphics[scale=1.00]{PictureLarge_heat_conduction_.png}}
\vspace*{-16pt}
}{}
\vspace*{8pt}
\caption{A simple heat transport problem.}
\label{Fig:heattransport}
\end{figure}
%
With $F$ being the cross section,
$d$ the distance between the two walls,
and $Q$ being the heat per time 
transported
% 
from 1 to 2, such that,
%
\begin{equation}
q_{x} 
= 
\frac{Q}{F} 
\end{equation}
%
we have
%
\begin{equation}
Q
= 
F \cdot q_x 
= 
- \lambda  \cdot F \cdot \frac{\partial T}{\partial x}
=
- \lambda \cdot F \cdot \frac{T_2 - T_1}{d}
=
\lambda \cdot F \cdot \frac{T_1 - T_2}{d}
\end{equation}
% 
in case of a stationary temperature distribution. 

$Q$ is produced and measured 
for the stationary situation 
by Joule heat 
(i.e.\ electric heat) 
at the higher temperature. 
The heat transfer by radiation 
cannot be separated from the heat 
transfer of kinetic energy. 
Of course, one tries to avoid 
the heat convection by 
the experimental arrangement. 
Hence any effects 
of the thermal radiation 
(long wave atmospheric radiation to Earth) 
are  simply contained in the stationary 
temperatures and the measured Joule heat.

In the non-stationary case the divergence
of the heat flow no longer vanishes,
and we have for constant thermal conductivity 
%
\begin{equation}
\mbox{\textbf{div}} \, \textbf{q} 
= 
- \lambda \cdot 
\mbox{\textbf{div}} \, \mbox{\textbf{grad}} \, T
=
- \lambda \cdot
\Delta T
=
- \varrho\,c_{\rm v} \cdot \frac{\partial T}{\partial t}
\label{Eq:preheatequation}
\end{equation}
%
where 
$\Delta T$ 
is the Laplacean of the temperature 
and 
$\varrho\,c_{\rm v}$
the specific heat of unit volume. We finally obtain 


%
\begin{equation}
\frac{\partial T}{\partial t}
=
\frac{\lambda}{\varrho\,c_{\rm v}}\,
\Delta T
\end{equation}
%
It is important to note, 
that the thermal conductivity 
is divided by 
$\varrho\,c_{\rm v}$,
which means that 
the isochoric thermal diffusivity
%
\begin{equation}
a_{\rm v} = \frac{\lambda}{\varrho\,c_{\rm v}}
\end{equation}
% 
of gases and metals 
can be of the the same order of magnitude, 
even if 
the thermal conductivities $\lambda$ 
are completely different. 

Unfortunately, the work on even
the simplest examples of heat conduction
problems needs techniques of mathematical
physics, which are far beyond the 
undergraduate level.
Because a concise treatment of the 
partial differential equations 
lies even outside the scope of this paper,
the following statements should 
suffice: Under certain circumstances 
it is possible to calculate the 
space-time dependent temperature distribution 
with given initial values and boundary conditions. 
If the temperature changes have 
the characteristic length 
$L_{\mbox{\scriptsize\rm char}}$, 
the characteristic time for 
the heat compensation process is
%
\begin{equation}
\frac{
1
}
{
t_{\mbox{\scriptsize\rm char}}
}
=
\frac{ 
\lambda 
}
{ 
\varrho \, c_{\rm v} 
}
\cdot
\frac{
1
}
{
{\rm L_{\mbox{\scriptsize\rm char}}^2}
}
\end{equation}
%
If the radius of the Moon were used 
as the characteristic length and typical values 
for the other variables, the 
relaxation time would be equivalent 
\textit{to many times the age of the universe}.
Therefore, an average ground temperature 
(over hundreds of years) is 
\textbf{no indicator at all} 
that the total irradiated solar 
energy is emitted. If there were 
a difference, it would be 
impossible to measure it, due to the 
large relaxation times. 
At long relaxation times, 
the heat flow from the Earth's 
core is an important factor 
for the long term reactions 
of the average ground temperature; 
after all, according to certain hypotheses 
the surfaces of the planetary bodies 
are supposed to have been very hot 
and to have cooled down. These 
temperature changes can never 
be separated experimentally 
from those, which were caused 
by solar radiation.
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%
\subsubsection{Heat transfer across and near interfaces}
%
In the real world things become even more complex 
through the existence of interfaces, namely 
%
\begin{itemize}
\item solid-gas interfaces
\item solid-liquid interfaces
\item liquid-gas interfaces
\end{itemize}
%
for which a general theory of heat transport does not exist yet.
The mechanisms of air cooling and water cooling and the influence 
of radiation have been studied in engineering thermodynamics\TschSpace% 
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\cite{AlfredSchackBook,Kreith1999,Baukal2000}
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% 
and are of practical interest e.g.\ in solar collectors, 
fire research, chemistry, nuclear engineering, electronic 
cooling, and in constructing reliable computer hardware\TschSpace%
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\cite{Safran1994,Bouali2006}.
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Obviously, there are of utmost importance
in geophysics and atmospheric physics as well. 
Since they add an additional degree of complexity 
to the problem discussed
here, they are not discussed further in this context.    
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\subsubsection{In the kitchen: Physics-obsessed housewife versus IPCC}
%
In Section~\ref{Sec:WaterPot} 
it was indicated how simple it is 
to falsify 
the atmospheric greenhouse hypotheses,
namely by observing a water pot
on the stove:
Without water filled in, the bottom of the pot
will soon become glowing red.
However, with water filled in, 
the bottom of the pot will be substantially 
colder.

In particular, such an experiment can be performed on
a glass-ceramic stove. The role of the Sun is
played by the electrical heating coils or by 
infrared halogen lamps that are used as heating 
elements. Glas-ceramic has a very low 
heat conduction coefficient, but lets infrared 
radiation pass very well. The dihydrogen monoxide in the pot,
which not only plays the role of the 
\lq\lq greenhouse gas\rq\rq\
but also realizes a very dense phase 
of such a magic substance,
absorbs the infrared extremely well. 
Nevertheless,
there is no additional \lq\lq backwarming\rq\rq\ effect of the
bottom of the pot. In the opposite, the ground becomes colder.

There are countless similar experiments possible 
that immediately show that the atmospheric
greenhouse picture is absolutely ridiculous 
from an educated physicist's point of view
or 
from the perspective
of a well-trained salesman offering
high performance tinted glass that reduces
solar heat gain mainly in the infrared\TschSpace%
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\cite{EfficientWindowsOrg}:
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\begin{quote}
\lq\lq 
Daylight and view are two of the fundamental 
attributes of a window. Unfortunately, windows 
are also the source of significant solar heat 
gain during times when it is unwanted. 
Traditional solutions to reducing solar 
heat gain such as tinted glazing or shades 
mean that the amount of light is reduced 
as well. New glazings with low-solar-gain 
Low-E (spectrally selective) coatings 
can provide better solar heat gain 
reduction than tinted glass, with 
a minimal loss of visible light. 
This also means that views can be 
clearer and unobstructed.\rq\rq
\end{quote}

Ironically, this works already in the case of 
dihydrogen monoxide.
Such experiments can be performed easily 
on every overhead projector, showing that 
the absorption of the infrared portion of the 
incoming radiation by water is a non-neglible
and leads to a drop of the temperature of
the illuminated surface dressed by an infrared 
absorbing layer that is transparent to visible
light. 
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