.MCAD 301010000 1 74
.CMD PLOTFORMAT
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 0 21 15
.CMD FORMAT  rd=d ct=10 im=i et=3 zt=50 pr=3 mass length time charge
.CMD SET ORIGIN 0
.CMD SET TOL 0.001000000000000
.CMD SET PRNCOLWIDTH 8
.CMD SET PRNPRECISION 4
.CMD PRINT_SETUP 1.200000 0
.CMD DEFINE_FONTSTYLE_NAME fontID=0 name=Variables
.CMD DEFINE_FONTSTYLE_NAME fontID=1 name=Constants
.CMD DEFINE_FONTSTYLE_NAME fontID=2 name=Text
.CMD DEFINE_FONTSTYLE_NAME fontID=4 name=User^1
.CMD DEFINE_FONTSTYLE_NAME fontID=5 name=User^2
.CMD DEFINE_FONTSTYLE_NAME fontID=6 name=User^3
.CMD DEFINE_FONTSTYLE_NAME fontID=7 name=User^4
.CMD DEFINE_FONTSTYLE_NAME fontID=8 name=User^5
.CMD DEFINE_FONTSTYLE_NAME fontID=9 name=User^6
.CMD DEFINE_FONTSTYLE_NAME fontID=10 name=User^7
.CMD DEFINE_FONTSTYLE fontID=0 family=Times^New^Roman points=10 bold=0 italic=0 underline=0
.CMD DEFINE_FONTSTYLE fontID=1 family=Times^New^Roman points=10 bold=0 italic=0 underline=0
.CMD DEFINE_FONTSTYLE fontID=2 family=Arial points=10 bold=0 italic=0 underline=0
.CMD DEFINE_FONTSTYLE fontID=4 family=Arial points=10 bold=0 italic=0 underline=0
.CMD DEFINE_FONTSTYLE fontID=5 family=Courier^New points=10 bold=0 italic=0 underline=0
.CMD DEFINE_FONTSTYLE fontID=6 family=System points=10 bold=0 italic=0 underline=0
.CMD DEFINE_FONTSTYLE fontID=7 family=Script points=10 bold=0 italic=0 underline=0
.CMD DEFINE_FONTSTYLE fontID=8 family=Roman points=10 bold=0 italic=0 underline=0
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.CMD DEFINE_FONTSTYLE fontID=10 family=Times^New^Roman points=10 bold=0 italic=0 underline=0
.CMD UNITS U=1
.TXT 5 3 0 0
Cg a66.500000,71.000000,104
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {\ul Loschmidt's Conjecture}{, from
 Problem 22 in Garrod's }{\i Stat Mech &
 Thermo}{ Book (Oxford '95)}}
}
.TXT 4 1 0 0
Cg a69.500000,70.000000,140
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {First imagine a gas of one particle
 in a uniform gravitational field.  If the
 atom has been dropped\par from a height
 h, it's velocity is...}}
}
.EQN 4 27 0 0
v(h,x):\(2*g*(h-x))
.TXT 5 -23 0 0
Cg a5.250000,69.000000,8
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {where}}
}
.EQN 0 8 0 0
g=?(m)/((sec)^(2))
.TXT 0 18 0 0
Cg a21.875000,35.000000,34
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Time spent per unit length is..}}
}
.EQN 0 24 0 0
dtdx(h,x):(1)/(v(h,x))
.TXT 5 -47 0 0
Cg a32.500000,71.000000,49
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {while time spend for the whole 
trip down is...}}
}
.EQN 0 35 0 0
t(h):\((2*h)/(g))
.TXT 5 -40 0 0
Cg a64.500000,69.000000,99
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {The probability for finding the
 particle in a height interval dx between
 0 and h is therefore...}}
}
.EQN 4 16 0 0
dpdx(h,x):(dtdx(h,x))/(t(h))
.TXT 0 21 0 0
Cg a2.750000,31.000000,7
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {e.g.}}
}
.EQN 0 5 0 0
dpdx(10*m,5*m)=?(m)^(-1)
.EQN 1 -38 0 0
x:0,.1;9.9
.EQN 2 -9 0 0
&0*(m)/(sec)&v(10*m,x*m)@&&x
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15
.EQN 0 37 0 0
&0*(m)^(-1)&dpdx(10*m,x*m)@&&x
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15
.TXT 24 -35 0 0
Cg a69.000000,69.000000,311
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Now let's imagine that our particle
 is rather massive, say 250,000 atomic mass
 units.  We choose this value so that the
 scale height of an atmosphere of these 
particles will be approximately a meter 
at room temperature, but we can only pick
 this number with hindsight which the reader
 does not have yet.  }}
}
.EQN 10 0 0 0
Avgdro:6.0221367*(10)^(23)
.EQN 0 19 0 0
amu:(gm)/(Avgdro)
.EQN 0 14 0 0
M:250000*amu
.EQN 0 16 0 0
M=?kg
.TXT 5 -51 0 0
Cg a30.000000,68.000000,47
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Kinetic energy as a function of
 height is,,,}}
}
.EQN 0 32 0 0
K(h,x):0.5*M*(v(h,x))^(2)
.EQN 0 21 0 0
eV:1.60217733*(10)^(-19)*joule
.TXT 4 -50 0 0
Cg a38.125000,68.000000,59
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Average kinetic energy per unit
 height interval dx is...}}
}
.EQN 0 41 0 0
dKdx(h,x):K(h,x)*dpdx(h,x)
.EQN 3 -43 0 0
&&K(10*m,x*m)@&&x
0 0 1 1 0
0 0 1 1 0
0 1 0 0 Kinetic Energy vs. Height x
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 0 21 15
.EQN 0 35 0 0
&0*(eV)/(m)&dKdx(10*m,x*m)@&&x
0 0 1 1 0
0 0 1 1 0
0 1 0 0 Energy Density vs. Height
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 0 21 15
.TXT 27 -35 0 0
Cg a72.000000,72.000000,458
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Thus even though the particle spends
 more time at large heights, the average
 kinetic energy per unit height interval
 (i.e. the kinetic energy density) is higher
 at lower heights.  Does this mean the temperature
 of a gas of such particles will also be
 higher at lower heights?  Since the Maxwell-Boltzmann
 distribution addresses how the velocity
 of particles in such a gas will be distributed,
 we turn to that to see wherein and if there
 is a conflict.}}
}
.TXT 20 -1 0 0
Cg a62.500000,72.000000,91
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {\ul The 1-Dimensional Maxwell-Boltzmann
 Isothermal Gas in a Uniform Gravitational
 Field.}}
}
.TXT 3 1 0 0
Cg a31.000000,72.000000,46
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {First defining some more units/constants...}}
}
.EQN 3 0 0 0
nat:1
.EQN 0 9 0 0
bit:ln(2)*nat
.EQN 0 14 0 0
byte:8*bit
.EQN 0 12 0 0
K:1.380658*(10)^(-23)*(joule)/(nat)
.TXT 0 19 0 0
Cg a12.875000,21.875000,21
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {<= redefine Kelvin}}
}
.EQN 5 -54 0 0
k:1
.EQN 0 8 0 0
k=?(joule)/(nat*K)
.TXT 0 18 0 0
Cg a43.875000,45.000000,63
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {<= Boltzmann's Constant is thus
 simply for units conversion.}}
}
.TXT 4 -27 0 0
Cg a18.125000,72.000000,26
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {At 0C, kT then becomes:}}
}
.EQN 3 4 0 0
kT(T):k*T
.EQN 0 12 0 0
kT(273.15*K)=?(eV)/(nat)
.EQN 0 23 0 0
{0:\b}NAME(T):(1)/(k*T)
.EQN 0 11 0 0
{0:\b}NAME(273.15*K)=?(nat)/(eV)
.TXT 4 -50 0 0
Cg a16.375000,19.000000,25
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {At room temperature...}}
}
.EQN 1 21 0 0
T:295.15*K
.EQN 0 14 0 0
k*T=?(eV)/(nat)
.EQN 0 16 0 0
(1)/(k*T)=?(nat)/(eV)
.TXT 5 -51 0 0
Cg a47.625000,71.000000,59
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Scale Height for our isothermal
 atmosphere then becomes:}}
}
.EQN 0 45 0 0
s:(k*T)/(M*g)
.EQN 0 10 0 0
s=?m
.TXT 6 -54 0 0
Cg a44.125000,72.000000,68
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {If we have one kg of particles,
 the total number of particles is:}}
}
.EQN 0 46 0 0
Ntot:(1*kg)/(M)
.EQN 0 11 0 0
Ntot=?_n_u_l_l_
.TXT 8 -55 0 0
Cg a18.125000,70.000000,32
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {The distribution itself is...}}
}
.EQN 0 27 0 0
C:(Ntot*sec)/(3.931*(m)^(2))
.EQN 0 12 0 0
dndxdv(x,v):C*(e)^(-((0.5*M*(v)^(2)+M*g*x)/(k*T)))
.EQN 3 15 0 0
3.926*(2.412)/(2.409)=?_n_u_l_l_
.EQN 5 -50 0 0
dndx(x):(0*(m)/(sec)&20*(m)/(sec)`dndxdv(x,v)&v)
.EQN 0 31 0 0
dndv(v):(0*m&10*m`dndxdv(x,v)&x)
.TXT 8 -30 0 0
Cg a11.500000,65.000000,18
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Examples are...}}
}
.EQN 0 28 0 0
dndxdv(.5*m,1*(m)/(sec))=?(sec)/((m)^(2))
.EQN 4 -28 0 0
dndx(1*m)=?(m)^(-1)
.EQN 0 29 0 0
dndx(10*m)=?(m)^(-1)
.EQN 4 -29 0 0
dndv(1*(m)/(sec))=?(sec)/(m)
.EQN 0 29 0 0
dndv(10*(m)/(sec))=?(sec)/(m)
.EQN 6 -32 0 0
x:0*m,.5*m;6*m
.EQN 1 41 0 0
v:0*(m)/(sec),.5*(m)/(sec);10*(m)/(sec)
.EQN 2 -46 0 0
&0*(m)^(-1)&dndx(x)@&&x
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 23 16
.EQN 1 38 0 0
&0*(sec)/(m)&dndv(v)@&&v
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15
.EQN 27 -38 0 0
n:(0*(m)/(sec)&20*(m)/(sec)`dndv(v)&v)
.EQN 0 21 0 0
n=?_n_u_l_l_
.EQN 0 14 0 0
dedv(v):0.5*M*(v)^(2)*dndv(v)
.EQN 7 -22 0 0
ke:(0*(m)/(sec)&20*(m)/(sec)`dedv(v)&v)
.EQN 0 23 0 0
ke=?joule
.EQN 17 -34 0 0
dedxdv(x,v):0.5*M*(v)^(2)*dndxdv(x,v)
.EQN 0 32 0 0
dedx(x):(0*(m)/(sec)&20*(m)/(sec)`dedxdv(x,v)&v)
.EQN 11 -23 0 0
v:0*(m)/(sec),0.5*(m)/(sec);20*(m)/(sec)
.EQN 1 27 0 0
&0*(joule)/(m)&dedx(x)@&&x
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15
.EQN 1 -36 0 0
&&dedv(v)@&&v
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15
.TXT 23 1 0 0
Cg a62.625000,66.000000,92
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Combining these distributions yields
 an average energy per particle as a function
 of x...}}
}
.EQN 4 -1 0 0
eav(x):(dedx(x))/(dndx(x))
.EQN 0 17 0 0
eav(1*m)=?eV
.EQN 0 17 0 0
eav(2*m)=?eV
.TXT 0 18 0 0
Cg a3.875000,21.000000,6
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain Note:}
}
.EQN 0 6 0 0
(1)/(2)*k*T=?eV
.EQN 3 -59 0 0
4*(10)^(-21)*joule&0*joule&eav(x)@&&x
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15
.TXT 1 36 0 0
Cg a34.000000,34.000000,324
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Hence the range of particle total
 energies has been chosen so that the average
 energy per particle (and hence dS/dE) is
 the same throughout the gas.  We examine
 total energy per particle (or max height
 h) for different x & v here, & on the next
 page look at its distribution directly 
with help of a variable change.}}
}
.EQN 16 9 0 0
etot(x,v):0.5*M*(v)^(2)+M*g*x
.EQN 8 -43 0 0
detotdxdv(x,v):etot(x,v)*dndxdv(x,v)
.EQN 0 38 0 0
detotdx(x):(0*(m)/(sec)&20*(m)/(sec)`detotdxdv(x,v)&v)
.EQN 10 -41 0 0
detotdv(v):(0*m&10*m`detotdxdv(x,v)&x)
.EQN 0 33 0 0
Etot:(0*m&10*m`detotdx(x)&x)
.EQN 0 23 0 0
Etot=?joule
.EQN 6 -55 0 0
&&detotdx(x)@&&x
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15
.EQN 0 35 0 0
&&detotdv(v)@&&v
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15
.TXT 33 -35 0 0
Cg a72.000000,72.000000,627
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {The variable change we seek is 
from the number of particles with coordinate
 x and velocity v, to the number of particles
 with maximum height h, and fractional height
 in their trajectory f=x/h.  These variables
 are chosen in such a way that the initial
 distribution may be created in its "noninteracting"
 equilibrium form simply by dropping our
 particles from the appropriate distribution
 of heights at random times.  In other words,
 the energetics are determined by the distribution
 of heights, while the f variable for a 
given particle depends simply on the temporal
 "phase", i.e. on the specific time it was
 dropped.}}
}
.TXT 17 0 0 0
Cg a30.875000,72.000000,47
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {The new variables are related to
 the old by:}}
}
.EQN 0 36 0 0
h2(x,v):x+((v)^(2))/(2*g)
.EQN 0 17 0 0
f2(x,v):(x)/(h2(x,v))
.TXT 5 -32 0 0
Cg a9.375000,72.000000,16
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Conversely...}}
}
.EQN 0 15 0 0
x1(h,f):f*h
.EQN 0 15 0 0
v1(h,f):\(2*g*h*(1-f))
.TXT 8 -52 0 0
Cg a9.500000,66.000000,17
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Starting with }}
}
.EQN 0 10 0 0
n÷(0*m&{0:\¥}NAME*m`(0*(m)/(sec)&{0:\¥}NAME*(m)/(sec)`dndxdv(x,v)&v)&x)
.TXT 0 26 0 0
Cg a11.750000,33.000000,18
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {we seek dndhdf:}}
}
.EQN 0 13 0 0
n÷(0*m&{0:\¥}NAME*m`(0&1`dndhdf(h,f)&f)&h)
.TXT 7 -49 0 0
Cg a72.875000,73.000000,212
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {This amounts to switching distribution
 variables as in Ch. 1 of Garrod.  This 
is neatly done with the Jacobian of the 
variable change, namely d(x,v)/d(h,f) = 
-\{gh/[2*(1-f)]\}^0.5.  As a result, we 
can write:}}
}
.EQN 8 4 0 0
dndhdf(h,f):\((g*h)/(2*(1-f)))*C*(e)^((-M*g*h)/(k*T))
.TXT 0 30 0 0
Cg a10.500000,39.000000,17
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {For example...}}
}
.EQN 0 12 0 0
dndhdf(1*m,.5)=?(m)^(-1)
.TXT 4 -45 0 0
Cg a34.125000,72.000000,51
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {We can then define the marginal
 distributions...}}
}
.EQN 5 3 0 0
dndh(h):(0&1`dndhdf(h,f)&f)
.EQN 0 34 0 0
dndf(f):(0*m&10*m`dndhdf(h,f)&h)
.EQN 5 -35 0 0
h:0*m,0.2*m;10*m
.EQN 0 21 0 0
f:0,.02;0.98
.TXT 0 24 0 0
Cg a2.750000,26.000000,7
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {e.g.}}
}
.EQN 0 5 0 0
dndf(.5)=?_n_u_l_l_
.EQN 1 -16 0 0
&0&dndf(f)@&&f
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15
.EQN 1 -36 0 0
&&dndh(h)@&&h
0 0 1 1 0
0 0 1 1 0
0 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 15
.TXT 23 0 0 0
Cg a72.000000,72.000000,510
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {The left graph here describes the
 distribution of heights h one must drop
 particles from if the distribution is to
 "begin" in its equilibrated or most uncertain
 form, given the amount of energy available
 to the gas.  We can check the self-consistency
 of these results in various ways.  First,
 notice the similarity of the graph on the
 right to the dpdx graph for a single particle
 gas on the first page of this worksheet.
  In fact, if we convert f in the graph 
above to x=hf for a specific value of h...}}
}
.TXT 11 0 0 0
Cg a27.000000,27.000000,55
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {...for example, using h=scale ht.
 s to be specific,}}
}
.EQN 0 30 0 0
x:0*m,0.02*s;.98*s
.EQN 2 6 0 0
&0*(m)^(-1)&dpdx(s,x),(dndf((x)/(s)))/(s*n)@&&x
0 0 1 1 0
0 0 1 1 0
4 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 16
.EQN 3 -33 0 0
h:s
.EQN 0 11 0 0
h=?m
.EQN 4 -8 0 0
dpdx(h,(h)/(2))=?(m)^(-1)
.EQN 6 0 0 0
(dndf(0.5))/(h*n)=?(m)^(-1)
.TXT 5 -6 0 0
Cg a31.375000,67.000000,50
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {They agree in spite of quite different
 origins!}}
}
.TXT 9 0 0 0
Cg a70.000000,72.000000,103
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Secondly, we can integrate the 
distribution over h to see if we get the
 correct number of particles:}}
}
.EQN 6 6 0 0
nh:(0*m&20*m`dndh(h)&h)
.EQN 0 25 0 0
nh=?_n_u_l_l_
.EQN 0 20 0 0
n=?_n_u_l_l_
.TXT 6 -49 0 0
Cg a14.125000,70.000000,21
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Not bad agreement!}}
}
.TXT 4 -2 0 0
Cg a71.875000,72.000000,124
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Lastly, we can examine the kinetic
 energy predicted by Maxwell-Boltzmann as
 a function of height x=hf in the trajectory.}}
}
.EQN 6 5 0 0
dKdf(h,f):dndf(f)*M*g*h*(1-f)
.TXT 0 31 0 0
Cg a2.750000,36.000000,7
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {e.g.}}
}
.EQN 0 8 0 0
dKdx(h,(h)/(2))=?(joule)/(m)
.EQN 4 -38 0 0
&&dKdx(h,x),(dKdf(h,(x)/(s)))/(h*n)@&&x
0 0 1 1 0
0 0 1 1 0
4 1 0 0 NO-TRACE-STRING
0 2 1 0 NO-TRACE-STRING
0 3 2 0 NO-TRACE-STRING
0 4 3 0 NO-TRACE-STRING
0 1 4 0 NO-TRACE-STRING
0 2 5 0 NO-TRACE-STRING
0 3 6 0 NO-TRACE-STRING
0 4 0 0 NO-TRACE-STRING
0 1 1 0 NO-TRACE-STRING
0 2 2 0 NO-TRACE-STRING
0 3 3 0 NO-TRACE-STRING
0 4 4 0 NO-TRACE-STRING
0 1 5 0 NO-TRACE-STRING
0 2 6 0 NO-TRACE-STRING
0 3 0 0 NO-TRACE-STRING
0 4 1 0 NO-TRACE-STRING
0 1 21 16
.EQN 4 38 0 0
(dKdf(h,0.5))/(h*n)=?(joule)/(m)
.TXT 7 -1 0 0
Cg a18.000000,29.000000,28
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Does this agree, or what!}}
}
.TXT 14 -43 0 0
Cg a72.000000,72.000000,507
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Thus we have the answer.  Loschmidt
 was correct in that any given particle 
will contribute most of its kinetic energy
 to motion at low altitudes, relative to
 its maximum height attainable with the 
kinetic energy it has.  However, the equilibration
 process, which makes the rearrangement 
of energy among particles maximally uncertain,
 results in a distribution of particle total
 energies which gives rise to uniform dS/dE
 and hence uniform average kinetic energy
 per particle as a function of height.}}
}
.EQN 13 7 0 0
i:0,1;20
.EQN 0 13 0 0
j:0,1;20
.EQN 0 17 0 0
(ndxdv)[(i,j):dndxdv((i)/(20)*5*m,(j)/(20)*10*(m)/(sec))
.EQN 5 -36 0 0
ndxdv{45 35 0 60 90    30    30 0 2 1 0}{57}
.EQN 0 37 0 0
(ndhdf)[(i,j):dndhdf((i)/(20)*5*m,(j)/(20)*.99)
.EQN 5 0 0 0
ndhdf{45 35 0 60 90    30    30 0 2 1 0}{57}
.TXT 32 -36 0 0
Cg a66.250000,70.000000,219
{\rtf1\ansi \deff0
{\fonttbl
{\f0\fnil Arial;}
}
{\plain {Above the x-axis goes along / and
 the v-axis\par along \\.  To the right,
 the h-axis goes along / \par and the f-axis
 goes along \\.  These plots illustrate 
the underlying differential distributions
 in\par both variable sets.}}
}