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.
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? 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.
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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