If one considers an exponential distribution such as
e^(-KE-PE)
where PE is a function of elevation
then at ground level one would have
e^(-KE)
and at a given elevation h
e^(-KE-PE) = e^(-PE)e^(-KE)
Renormalizing for the lower density the distribution at elevation becomes
e^(-KE)
which is identical with the original distribution at ground level and
indicates that the gas is isothermal. This only works with exponential
distributions.
The Maxwell Boltzmann distribution can be written in several versions.
See http://homepage.univie.ac.at/franz.vesely/sp_english/sp/node8.html
The one that describes the velocity distribution is exponential:
\begin{displaymath} f(\vec{v}) = \left[ \frac{m}{2 \pi kT} \right]^{3/2}
e^{-mv^{2}/2 kT} \end{displaymath}
but a Boltzmann term needs to be added to describe the effect of
potential energy
\begin{displaymath} f(\vec{v}) = \left[ \frac{m}{2 \pi kT} \right]^{3/2}
e^{-mv^{2}/2 kT} \end{displaymath} e^-(PE/kT)
This density in velocity space is commonly called*Maxwell-Boltzmann
distribution density*. The same name is also used for a slightly
different object, namely the distribution density of the*modulus*of the
particle velocity (the ``speed'')
\begin{displaymath} f(\vert \vec{v} \vert) = 4 \pi v^{2} f(\vec{v}) = 4
\pi \left[ \frac{m}{2 \pi k T} \right]^{3/2} v^{2}e^{-mv^{2}/2kT}
\end{displaymath}
(Similarly a Boltzmann term needs to be added. This distribution is not
exponential as it has the v^2 factor in front. )
My problem is to justify using the exponential distribution, obviously
without having to invoke the Second Law which is being challenged.
On 11/21/2014 11:19 AM, meekerdb wrote:
On 11/20/2014 9:07 PM, George wrote:
Brent you are right.
Maxwell distribution is not exponential with energy. For the purpose
of comparing the different distributions, I was attempting to give
the same form to all distributions Maxwell, Fermi-Dirac and
Bose-Einstein independently of the scaling factor in front of the
exponential. i.e.,
The trouble is that it's not just a scaling factor in front, it's a
normalization and the normalization has to produce the right
dimensions. The functions you right below are all dimensionless, so
they can only be density functions relative to a dimensionless
variable, e.g. x=(E/kT)
Maxwell: 1/e^x
Fermi-Dirac 1/(e^x + 1)
Bose-Einstein: 1/(e^x - 1)
I may not have been correct in doing this.
I agree, Maxwell distribution is not exponential with _energy_.
If we assume that the distribution is also not exponential with
_elevation_ then the renormalized distribution after vertical
translation does not overlap the original distribution. Therefore
there is a spontaneous atmospheric temperature lapse and Loschmidt
was right after all!
There is a "spontaneous" atmospheric lapse rate which in the standard
atmosphere model is linear, -6.5degK/km, from sea level to 10km. And
you could run a heat engine using the temperature difference - just as
people have proposed running a heat engine between warm surface water
and cold deep ocean water. But why would that violate the 2nd law?
The atmosphere is heated by the surface where sunlight is absorbed and
it's lost by radiation to space in the upper atmosphere - so there's a
gradient and free energy which can be turned into work.
Loschmidt assumed that the temperature gradient would be self
regenerating independently of the sun and wind.
Breaking the Second Law does not require QM. All that is required is
a Maxwellian gas in a force field.
The question therefore is whether Maxwell distribution is exponential
with _elevation_.
What does it mean for the M-B distribution to be exponential with
elevation?
See my comment on top of post.
As a density function over energy it has one parameter, kT. Are you
asking whether T=T_0*exp(-h/h_0) where T_0 is the surface temperature,
h is the altitude, and h_0 is some altitude scale. If so, the answer
is no. The function is T=T_0 - 6.5h for T in degK and h in km. But
that only works up to 10km. Changes in molecular species (different
masses) become important at higher altitude.
If it is then Loschmidt falls on Maxwellian gases. If it is not, then
Loschmidt is completely vindicated for any kind of gas. I need to
think about this. Any idea?
Loschmidt considered just a gas or other substance in an isolated
column (no solar heating, no radiative cooling), so the atmosphere
isn't a good example.
This is the example he used. As good physicist do,we can always run a
thought experiment in which a column of gas in a gravitational field is
isolated from its surrounding.
Brent
George
On 11/20/2014 6:41 PM, meekerdb wrote:
On 11/20/2014 6:28 PM, George wrote:
Maxwell's distribution
f = e^(-E/kT) where E = (1/2) mv^2
?? Distribution with respect to energy is:
Note the sqrt(E) factor.
http://en.wikipedia.org/wiki/Maxwell-Boltzmann_distribution
Brent
can be looked at in different ways. It is a Chi Square distribution
with respect to velocity v, and exponential with respect to kinetic
energy E.
The _most likely (mode)_ kinetic energy is zero
Not for a Maxwell-Boltzmann distribution.
Brent
but the _mean_ kinetic energy is not zero . The distribution decays
exponentially with higher energies.
George
On 11/20/2014 6:13 PM, meekerdb wrote:
If it were the momentum or velocity the mean would be zero, but it
wouldn't be exponential. If you just considered the speed
(absolute magnitude of velocity) in a particular direction you get
an exponential distribution. Is that what the graph represents?
Brent
On 11/20/2014 5:03 PM, LizR wrote:
The average kinetic energy of an air molecule is zero, I imagine,
because they're all travelling in different directions and cancel
out? Or doesn't it work like that?
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