Transcript PART TWO Statistical Physics Chapter III:Statistic Distributions for ideal gases • 32 Statistics Regularities.

PART TWO Statistical Physics
Chapter III:Statistic Distributions for ideal gases
• 32 Statistics Regularities. Distributions,
Most Probable Distributions
(统计调节。分布，最大概率分布)
• The main objective of statistical physics:
• 1) to establish the behavior laws for
macroscopic quantities of a substance.
• 2) to offer a theoretical substantiation(证实) of
thermodynamic laws on the basis of atomic and
molecular ideas.
Basic Methods
• Condition : a system consisting of a large
number N of molecules(由微观极大数目的粒子构
成).
• Classical: using Newton’s classical
mechanics(经典力学) to describe the state of
the system；
• Quantum: using quantum-mechanical
description, the ideal of wave mechanics.
Newton’s classical mechanics
(经典力学)
• Ignoring: intramolecular(分子内) structure，
to visualize a molecule as a point or particle.
• The equation of Newton’s motion for each of
the N particles.
dv i
mi
  Fih
ih
dt
• Fih：i’th与h’th分子的作用力；vi: velocity.
• 求和存在的问题：1)要知道作用力或空间相关的作用
势；
2)要知道6N个初始条件：每个分子的三维坐标与动量。
3)假设上述条件已知，求和计算分子的路径。
困难与解决方法
• 数学计算上的求和的难度，使其几乎不可能。
因为系统的粒子数达到1025m-3。
• 即使知道了粒子的路径和运动方程，也未必能
提供以系统作为一个整体有用的信息。
• In a system consisting of a great number of
particles new purely statistical or probability laws
take effect that are foreign to (不适合于) a system
containing a small number of particles.
Statistical Method
• Assumption: It is possible
to measure rapidly the
energy of each molecule of
a gas. The results of such
measurement are present
graphically in the Figure.
• The axis of the abscissa(横坐标轴) is subdivided into equal
sections each 0 long, and 0 is sufficiently small enough.
All energies in l0~(l+1)0 is assumed to be equal to l0.
Energy Distribution Function and
“ Boxes”
• The relative number of molecules in the
range l0~(l+1)0 is denoted by n(l):
n(l ) 
N (l , l  1)
N
• N is the total number. n(l) is “energy distribution
function for molecules” or “energy distribution”
• To divide the x-axis into longer unequal segments
“Boxes”
• The number of molecules with lower or higher
energies is very small.(能量差别不大的分子属一个盒)
Cell
• A box is a larger unit which contains several
cells: molecules have the fully same energy.
• A box which energy l0~(l+m)0,
• m0 is the length of a box, which varies little.
• Actually, all the histograms(矩形高度) will be
close to some averaged histogram and large
deviations from it will be rare.
Microstate 微观状态
• To describe the state of a gas at some moment of time:
Microstate
• Classical mechanics: “coordinates(坐标) and velocity”
• 设粒子的坐标与势能相关，而动能与速度相关：
mi (vxi2  v yi2  vzi2 )
2
 U ( xi  yi  zi )  
•经典力学用坐标和动量描述粒子的“微观运动状态”。某一
时刻，整个系统的N个粒子构成了一个“微观状态”，如前
述的能量-粒子柱形图为能量分布图，或“状态分布”图。下
一个时刻，N个粒子的能量(坐标及动量)发生了变化，微观状
态也发生了变化。
• 一个气体分子平移运动的平均动量为
mv 2
3
 kT
2
2
• 整个系统气体分子的平均能量为
 
l
N (l , l  1) l
  n(l ) l
l
N
• It is possible to find the mean values of any energy
function if the “momentum distribution” are know.
• 一个宏观的热力学系统用P、V、T、S描述。在一个
状态下(平衡或不平衡)，均可用统计的微观状态描
述。即一个不变的热力学状态对应着许多的微观状
态，其数目被称之为“微观状态数”。
Basic physical postulate of
statistical physics
• “the greatest number of microstates of the
most probable distribution and is
equivalent to the equilibrium state of
thermodynamics”.------两者相同点
• Thermodynamics assumes that a system
remains in a state of equilibrium
indefinitely long, but statistical physics
predicts there existence of fluctuations(涨落)
spontaneous(自发) and rare deviations from
the equilibrium state. ------两者不同点
统计物理学关心的问题
• The problem of finding the most
probable distribution for ensembles
of non-interacting particles or for
ideal gases.
33. -Space. Boxes and Cells
• 借助于相-Space(六个坐标的空间)的概念导出统计分布。
• -Space : x,y,z, (ksai),(eta),(zita). System has N points.
• This six-dimensional surface is specified by the equation:

 2 2   2
2m
 U ( x, y , z )
•The concept of the phase volume(相体积) in the -Space is
introduced by the expression:
d  dxdydzddd
•Subdivided into(细分) the volume in the configurational
space and in the momentum space. (构型空间和动量空间)
dV  dxdydz
dVp  ddd
• It might be convenient to select the spherical
layers(球壳层): dV=4rdr2, dVp=4pdp2.
• 若简化为一维的运动：
 
2
2m
 m gx
•相空间用代表点dxd 表
示.
一个抛物线上的代表点能量相同。两个分子碰撞，
改变了各自的抛物线轨迹，但总能量不变。
Boxes in the -Space
• The qi and pi are applied to represent coordinate
and momentum. It is not homogeneous (均匀的)
in all the space.
• In phase volume d, the number of
representative points is dN. The density is
(qi,pi) = dN/d.
• A postulate is introduced: the distribution
function for the -Space, (qi,pi), depends only on
the particle energy  and not on qi and pi
individually.
• The -Space is subdivided into “boxes”
by carefully drawing the hypersurfaces of
“constant energy”. This energy layer is
sufficiently thin that the representative
points代表点confined in the layer have the
same energy  .
统计物理解决问题举例
• 一个三能级系统，0, 20, 30中，每个能级cells
(原胞)有6个空位，共有6个完全相同的粒子，总
能量为120,每个空位只能放一个，粒子如何分布？
• 粒子可以采取的分布方式为：
上图为粒子微观状态的表现，下图为分布函数，四种
微观状态出现的数目分别是1，6×15×6, 153, 202。
34 Bose-Einstein and Fermi-Dirac
Distributions
• Subdivision non-equidimensional energy boxes
and equidimensional cells.
• The ith energy box: having an energy i ,gi cells,
Ni representative points.
• How do these representative points distribute
among the cells.?
• Principle: any arrangement of representative
points in the cells to be equiprobable. 等概率的
• The distribution is realized by the most
probable distribution,--- the equilibrium state.
Two Hypotheses 两个假设
• 1. All particles of one kind are absolutely
identical to one another (所有粒子为全同).
• 2. These particles differ slightly just as
producting-line(生产线) identical parts
produced in a factory differ from one
another.
• Both of above:
“ particles of one kind are identical ”
微观状态的经典和量子描述
• N个全同粒子构成的体系，任意交换两个粒子的坐标和
动量时，经典力学认为其微观状态不同。因为经典力
学认为其运动轨迹是可以被跟踪的、每个粒子原则上
是可以被识别的。
• 量子力学认为，任意交换两个粒子的，其微观状态相
同。因为量子力学不可跟踪粒子的运动轨迹，运用的
是测不准原理和几率分布。
• 一个柱形图在量子力学条件下为一个微观状态，---此
为量子力学的“全同性原理”，全同粒子不可分辨；
而在经典条件下为多个微观状态，粒子可以分辨。
粒子的量子性
• 自然界中有两类量子粒子：fermion
and boson
• Fermions follow an important law:
the Pauli exclusion principle
• in a system of N identical fermions one
cell in the -space can contain no more
than one representative point.
• in a system of N identical bosons one cell
in the -space can contain any number
representative points from zero to N.
Statistical properties of the
different particles
• To illustrate the difference in the statistical
properties of the different particles by a simple
example:
“Arrange two particles on three cells 1, 2, 3”
For the classic particles, they are distinguishable
经典
玻色子
费米子
• Classical : 9 arrangements;
• Bosons: 6 arrangements;Fermion: 3 arrangements.
• How about Ni particles in gi cells?
For the Boson:
How to express?
The ith box :
Analyses
Calculate Wi
( Ni  gi  1)!
Wi 
Ni !( gi  1)!
• Wi: denoted as the number of different ways of
arranging Ni particles in gi cells.
• Two classes of objects: Particles & partitions
•
粒子： Ni 和隔离物：gi-1
• 不同的排列方式可以分为两种交换：
• (1)粒子和隔离物；(2)粒子和粒子。
• 因此，粒子和隔离物排列在一条线上，总的排列方式：
• (Ni +gi-1)! ：包含了全同粒子交换Ni !
和隔离物交换(gi-1)!
Boson系统的物理
量：
N   Ni
i
( Ni  gi  1)!
W  Wi  
Ni !( gi  1)!
• 要确定系统的最大几率分布，即确定W的最大值。
从数学上看，确定lnW较为方便。定义 = lnW
  ln W   [ln(N i  g i  1)! ln N i ! ln(g i  1)!]
i
  [(N i  g i ) ln(N i  g i )  N i ln N i  g i ln(g i )]
i
利用了近似等式：
ln Ni ! Ni (ln Ni 1)
求 的极大值
• 两个必要条件是：the total number of gas particles and
the total energy of the gas are fixed.
• 气体的分子总数一定；气体的总能量一定。用公式表示为
 Ni  N ,
i
 N i i  U
i
使用拉格朗日多项式变分的原理，使函数 = +N -  U
的变分为零。得到

 ln(N i  g i )  ln N i     i  0
N i
The most probable number of particles in a cell is:
Ni 
gi
e
   i
1
(34.6)
Fermi-Dirac distribution,
Fermion (费米子)
• Bose-Einstein distribution are specified by
gi
g i i
 N,
U
   
   
i e
i e
1
1
•Fermions are considered. For the ith energy box
with a number of cells gi, and a number of
particles Ni (Ni < gi), the different ways of
distribution differ from each other only in that
some cells are occupied by one particle and some
cells are empty---permutations(置换) of empty cells
i
i
• 图中，一种分布为一个微观状态，此分布的微观状态数
目为空胞的置换次数。空胞数为(gi-Ni)， 总置换数gi!，
总的粒子数置换数Ni!，总空胞置换数(gi-Ni)!微观状态为
gi !
Wi 
N i !( g i  N i )!
N个粒子的排列方式是：
gi !
W 
i N i !( g i  N i )!
微观状态数W的极大等价于的极大
   [ gi ln(gi ) Ni ln Ni  ( gi  Ni ) ln(gi  Ni )]
i
求的极大得到
• 同理利用Lagrangian方法，使函数的微分为零：

  ln Ni  ln(gi  Ni )    i  0
Ni
可以得到在Box上的粒子数：
Ni 
gi
e     1
l
并由此可以得到费米总粒子数的分布和能量公式：

i

i
gi
e
   l
1
g i i
e
   l
1
 N,
U
35. The Boltzmann Principle
• Two statistical distributions are known, but
the meaning to be imparted(赋予) .
• The important is to know the meaning of two
parameters  and . Make physical postulate:
•1) W=W1·W2······Wn
•
 =1+2+······+n
• is a extensive quantity
• 2) In an isolated system, and in “平衡态”
• Thermodynamics:
• Statistical Physics:
•极大时的熵S对应着热力学系统的平衡态。
• 极为自然的问题: 和S 之间是否存在联系？
•These arguments make it reasonable to postulate
that with a degree of accuracy up to a constant
multiplier thermodynamically defined entropy
coincides(一致) with the quantity .
意义
Boltzmann consider this situation and thought
that there must be some internal connection
between them. He applied a multiple constant to
establish an equation:
S  k lnW ,
23
k  1.3810 J / deg
------The Boltzmann Principle
• Boltzmann endue(赋予) the entropy an statistical
meaning . 宏观的熵其微观含义如何？
• It is convenient to use another definition form.
• 在统计物理中，常常使用更为方便的表达方式：
S  lnW
(35.1)
•Here , the entropy is assumed to be a
dimensionless quantity. Since the product TdS
must have the dimension of energy, then the
temperature must be in the energy units.
•即：熵S变成无维度的，温度变成能量单位的。
•The entropy of a system in a state of
equilibrium is
Smax  lnWmax
熵的物理意义
• Boltzmann: the increase in entropy in an
equalization process is the result of the
system passing from a less probable
states to the most probable state.
同一个物理系统，当它从完全混乱状态有序状态时，例如
无序排列的液体的水在温度不变的条件下分子结晶成冰，
熵如何变化？
例：一个红色墨滴滴入白色清水中，红色分子逐渐
扩散至均匀状态，其过程为熵增加。可以认为分子熵减小。
从有序变为无序。熵增加，反之，则为
因此，熵的统计意义就是分子的“混乱程度”.
等温分子的扩散，熵增加，有“虚的”吸热。(对应S1)
About an irreversible process:
• What is the fundamental difference between
the
statistical
interpretation
and
the
thermodynamic interpretation?
• From thermodynamics: a reverse process is
impossible by definition.
• From statistical physics: a transition from the
most probable distribution to the less probable.
• (除非由于涨落的影响，且涨落会很大。)
The discuss of distribution
• The Boltzmann principle can be used to find the
meaning of two parameters. By the formulae:

i
gi
e
   l
1
 N,

i
g i i
e
   l
1
U
The upper sign pertains to the Bose distribution,
and the lower to the Fermi distribution.
The entropy S is related with both N and U.
•同一个BOX内的所有CELLS都是等价的，每个粒子占
据不同cell的概率完全相同。Cells 越多，粒子占据该box
的概率越大。为了表示粒子占据Cells的概率，定义了：
Occupation Number
“占有数”函数
• 粒子在一个cell上占据
的概率可以表示为：
ni 
1
e     1
i
•由于该函数对费米分布来说是一个分布函数，但对玻色
分布来说有可能是大于1 的粒子数的函数：ni = Ni / g i ，
故将其称之为粒子的“占有数”函数。作用？
•首先，利用占有数函数可以对熵函数进行化简。在玻色
系统中，熵的表达式是：
S   [(N i  g i ) ln(N i  g i )  N i ln N i  g i ln(g i )]
i
  g i [(ni  1) ln(ni  1)  ni ln ni ],        (35.5)
i
her,
N i  g i ni
熵的化简形式(玻色分布)
• 将玻色分布的ni代入得到：
     i
  
S   g i    
 ln(1  e
i
1
e
i
•其中： ni 
1
e
, ni  1 
   i
   i
e
1
e   
将两个参量看作常数，得到：
   i
S  N  U   gi ln(1  e
i
i
i

)

1

 1 1  e  
i
)        (35.7)
熵的化简形式(费米分布)
S   [ g i ln(g i )  N i ln N i  ( g i  N i ) ln(g i  N i )]
i
  g i [ni ln(ni )  (1  ni ) ln(1  ni )]       (35.8)
i
• 将费米色分布的ni代入得到
     i
S   g i    
 ln(1  e  
i
1
e
i
i

)

得到
   i
S  N  U   gi ln(1  e
i
)        (35.10)
熵的统一表达式
   i
S  N  U   gi ln(1  e
)
(35.11)
i
• Here, minus  bosons; positive  fermions.
• Important emphasize:
• 1) 式(35.5)和式(35.8) 适用于平衡与非平衡态。
• 2)式(35.7) 和式(35.10)仅适用于平衡态，因为
在公式中包含了仅适用于平衡态时的极大熵
和极大占有数n1,n2,…,ni…以及确定平衡时熵
的两个参量与------热平衡的基本值。
The meaning of the parameters  and 
• Compare Eq(35.11) with the expression dS
in thermodynamics.
• 分析方法：定gi, i为常数。
• 1) 不变，S对求导；
• 2) 不变，S对 求导。
• gi和I将与气体的容器尺寸相关，即与V相关。
• 假设无任何外场的作用, 即式(35.11)不再包含
其他项，求导设定V不变，则
 and  的偏导数
   i
S  N  U   gi ln(1  e
)
(35.11)
i
 S 
 N 
 U 
g i i
    
  U   
   (
)
  
   ,V
   ,V
   ,V i 1  e
i
 S

 

 N

  
 ,V
 

 U

  
 ,V
 


 ,V
(35.12)
gi
 S 
 N 
 U 
 N 
 
  (   
)




i
e
1
    ,V
    ,V
    ,V
i
 N 
 U 
  
 


    ,V
    ,V
(35.13)
 and  的值
S
S
dS  (
) ,V d  (
)  ,V d


• 将式(35.12)和式(35.13)代入得到：
 N

 U

N
U
dS   ( ) ,V d  ( )  ,V d    (
)  ,V d  (
) ,V d 


 

 

dSV   dN   dU
已知： dU  TdS  PdV  dN
1
P

1

dSV  dU  dV  dN  dU  dN
T
T
T
T
T
1
1


   ( ),      ( )
T
kT
T
kT
• 关于单位：玻耳兹曼常数k是微观量，粒子数N是
宏观量，可以将此微观量转化为宏观量：
• Nak = 6.023×1023×1.38×10 –23 = 8.31 (J K-1mol-1)
• 在附录中R = 8.314 (J K-1mol-1); k = R/NA.
• 理想气体的物态方程：PV = nRT = NkT.
• 其中，N表示系统中的“粒子数”；n表示“摩尔
数”。
• Accordingly, in all following sections the chemical
potential  does not refer to one mole of substance,
as in thermodynamics, but to one particle, so that
therm = NA stat is true. ( s ,T)代替(，)在公式
统计物理学所表示的公式
• From the Eq.(35.11), the entropy is
   i
S  N  U   g i ln(1  e
)
(35.11)
i
TS   N  U  T  g i ln(1  e
(   i ) / T
)
i
此时的i为一个粒子的能量，表示某个粒子所处的能
级。而熵是对所有这种能级求和。虽然我们已知了
是化学势，也知道在热力学中它是相变的推动力，然
而，在统计物理学中是否还有新的含义？长远的应用？
•G =  N；(为一个分子的化学势)。
•F = U – TS = G – PV，F – G=U – TS – G= U -TS-  N
• 得到公式： PV  T  g ln(1  e(   ) / T )
i
i
i
•最终，统计公式可以写为：
Ni 

gi
( l   ) / T
gi
e
( i   ) / T
 N,

1
g i i
( l   ) / T
(35.20)
U
(35.21)
i e
e
1
1
参量和一旦确定，上式的物理意义会发生变化：
i
仅仅是温度的函数， 已经确定；上左式中，化学势
是粒子数N和温度T的函数，而上右式中，内能是化
学势和温度的函数。可以推断：内能是粒子数N和温
度T的函数。 ------理想气体的焦耳定律。
• 进一步分析，每个粒子的内能U/N将仅仅是温
度的函数，从而成为强度量。V/N也将如此。
• 如果外加其他场强的话，系统如何变化？
i is the function of intensive parameters, i.e.
field intensities.
In Conclusion
• The B.E. distribution and F.D. distribution are
derived by the box-cell method presupposing (预
先假定) that thermodynamic equilibrium state
sets in.
• The initial non-equilibrium particle distribution
 equilibrium distribution  particles change
their boxes to a equilibrium state.
• Reason: N=cons. and particles interact with
surrounding walls (thermostat).
• Indeviation: both N and U are fixed, only ~T.
36. The Maxwell-Boltzmann
Distribution
• Question: How does classical particles distribute?
• Let Box1 for N1, Box2 for N2, …
• Box n for Nn, … .
•
•
•
•
任意两个粒子两两交换：N! 次
不考虑N1个粒子在BOX1中的交换；
不考虑N2个粒子在BOX2中的交换；
因此，将N个粒子填充到BOX 中的数目是：
N!
W'
 Ni !
微观状态数
• 在一个BOX内，有gi个cells，Ni个粒子。由于
每个粒子均是可以被跟踪的，即可以被识别
的。因此，任意交换两个粒子为不同的微观状
态。
giNi
• 在一个BOX内总的交换次数是多少？
由此可以算出总的微观状态数目：
Ni
gi
N!
Ni
W 
 N !
 gi
i
Ni !
 Ni ! i
i
运用Stirling公式，得到：
  N ln N   [ N i ln gi  N i ln N i ]
i
• 使用拉格朗日变分公式：
    (  1) N  U
对Ni求导得到最后结果：
N i  g i e   i
   i
 gi e
N
i
   i
 g i i e
U
i
而量子的分布是：
ni 
1
e
   i
1
• If for any i the condition exp(i -)/kT >>1 is
satisfied, the unity in the denominator can be
ignored and we obtain the Maxwell-Boltzmann
distribution:
Ni  gi e
(   i ) / kT
ni  N i / g i  e (   i ) / kT
•对于理想气体近似为实际气体的条件，就是:
ni  Ni / gi  e
(   i ) / kT
 1
In this rarefied gas, the average interparticle distances
are large, so they cannot be confused---distinguishable.
在稀薄气体条件下，粒子之间的距离较远，不可被分辨的
“量子”与可以被分辨的经典粒子是完全一致的。
结论：稀薄浓度的量子粒子可
以近似为经典粒子处理
• The Boso-Einstein and the Fermi-Dirac
distributions are valid for all particles,
thile the Maxwell-Boltzmann distribution
is approximately true in the limiting case
of small occupation number.
• The entropy of a gas in an arbitrary
equilibrium or non-equilibrium state can
be obtained in two ways:
熵的计算
S   g i ni ln ni  N
i
• If the Boltzmann formula S=lnW is used,
the classical gas (36.2) would follow:
S   gi ni ln ni  N ln N
*
i
It is not true, otherwise, S* will be not an
extensive quantity.------ we return to the Gibbs
paradox.
Gibbs 的预言
• Gibbs’ foresight is worthy of admiration, for as far
back as the end of the nineteenth century he
anticipated the present-day concept of the
indistinguishability of particles.
• 值得注意的是，玻耳兹曼分布应用的范围是有效的，
其不等式是成立的。即便当gi很大时，Ni也会很小or
close to unity.
• In these conditions, the Stirling’s formula becomes
incorrect for Ni and gi.
• An general Gibbs’ method can be applicable to ideal
gses but also the systems of interacting particles.
• Problem: Page 190
• 中文教材内容补充：
• 在一个长为L，粒子数为N的容器内,粒子以波
动的形式运动。运动方式为：
• 其运动方式为驻波。
驻波的波长为 = L/n. n为正整
数。定义波矢为k=2/ .
波矢具有两个传播方向，定义
 2
kx 
nx ,
L
nx  0,1,2,...
动量与能量
2
p  k 
nx
L
2
px
2 2  2 2
 nx 

nx
2
2m
mL
1
2  nx  n y  nz
2
2
2
2
 
( px  p y  pz ) 
2m
m
L2
L
dnx 
dpx , ......
2
L 3
V
dnx dny dnz  (
) dpx dpy dpz  3 dpx dpy dpz
2
h
2
2
2
2
2
作业：P228： 6.2, 6.4, 6.5
• 上式表示在一定动量空间内代表点(量子态)的
数量。在一维空间，一个代表点的体积(Ldp)是
h。三维空间一个代表点的体积是h3。量子态
与动量之间有直接的对应关系。
• 利用能量-动量关系 =p2/2m.
• 动量有正负，能量是简并的。
• 相同能量而动量不同，为同一状
态。(能量---BOX,动量---Cell)
能量或状态是分立的，即在单位能量长度内其状
态数目用D(E)表示。故
D( )d  V3 4p 2 dP
h
D( ) 
2V
h3
(2m)3 / 2  1 / 2
什么是波矢 k ?
• 从量子的角度看，波矢对应着速度：
•
p = hk = mv
• 电子从低能级跳到高能级
能量和动量均发生了变化。
横坐标表示动量的变化，
纵坐标表示能量的变化。
当一个粒子(电子)碰到一个高速振动的粒子
(离子)时，会获取能量和动量。
What is the concept of “Boxes”
D( )d 
V
h3
4p 2 dP
D( ) 
(2m) 
• Here, D() is defined as the density of state.
dN 
D ( ) 
d
dN  is the number of energy in
the range of  ~  + d. How about
D() ~ V, ~m, ~ ?
What is the “box”? One box is one state(能量状态，不是
量子态), or one line in the figure, about one value of
.
2V
h3
3/ 2
1/ 2
•
•
•
•
37. Transition to continuously
Varying Energy. Degeneracy
Conditions for Ideal Gases
Three Statistical Distributions
1) the Bose-Einstein distribution
2) the Fermi-Dirac distribution
3) the Maxwell-Boltzmann distribution
1) N   N i  
gi
,
1
U 
e ( i   ) / kT  1
gi
 i gi
2) N   N i   (   ) / kT
, U   (   ) / kT
i
i
i e
i e i
1
1
3) N   N i   g i e (   i ) / kT , U    i g i e (   i ) / kT
i
i
i
i
e
( i   ) / kT
 i gi
i
i
Discussion
• In deriving the statistical distributions, the energy
was a “discretely varying quantity” ------“Box”.
• If it is suitable? In what degree? Size of the cell?
• If the energy layers(boxes) are sufficiently thin, we
can even replace above summation by integration.
• How do we integrate?
• By a new concept “phase volume” ----d=dqidpi
• In this volume the particle number is dN.
• If the volume of one cell is “a”, “g” weight factor
dqi dpi
d
gi  g
g
a
a
The meaning of g
• For instance, the spin of a particle is s, the
projection of the spin in any direction have
2s+1 different values(-s, -s+1, … s-1,s). In
this case g = 2s+1.
• The light quantum, photon, has not spin, but
has two vibrational directions, g = 2.
• The photon is Boson.
• The electron, the Fermion, g = 2s+1 = 2.
Three distributions
g
d
g
d
1) N   (   ) / kT
, U   (   ) / kT
a e
a e
1
1
g
d
g
d
2) N   (   ) / kT
, U   (   ) / kT
a e
a e
1
1
g (   ) / kT
g
3) N   e
d, U   e(   ) / kT d
a
a
• The important distinction(差异)。
• For the boson and fermion, a can be solved by
the comparison with the experiment results of
Cv.
U
Cv  (
)V
T
The discussion of MaxwellBoltzmann distribution
• The situation is quite different for the
Maxwell-Boltzmann distribution. The
chemical potential and the cell volume are
presented in the same form: exp(/T)/a.
UN
But two others
U N
 / T

e
d

 / T
e
d

(   ) / T
1

(
e

1
)
d

(   ) / T
1
(
e

1
)
d

Distinction
• The energy depends on  and,
consequently, (从而) on a。For the exact
statistical Fermi-Dirac and Bose-Einstein
distribution, the volume of a cell is not
arbitrary: exactly by laws of nature and
by experiment.
• Impossible in the case of small occupation
numbers, because of the phase volume of a
cell acquires arbitrary value.
在此情况下，可以考虑用经典分布替代量子分
布。在什么情况下可以替代，标准如何？
The criterion of validity of the
Maxwell-Boltzmann distribution
• In the case of mono-atomic ideal gas
g
d
N   (   ) / kT
a e
1
p2  2  2   2


,
2m
2m
能量依赖于动量，细分球形动量空间得到：
d  V  4p dp  V  2 (2m)  d
2
3/ 2
1/ 2
已知Maxwell-Boltzmann 分布的有效性为：
(   ) / T
 /T
e
 1, i.e. e  1 (including  0)
1
Na
Na
 /T
 / T
e

 e d 
g
gV (2m T)3 / 2




1
Na
Na
 / T
e

 e d 
3/ 2
g
gV (2m T)
• 对于一个单原子气体，一个元胞的体积为
a = h3,(Planck’s constant), and g = 1
• The criterion of validity of the MaxwellBoltzmann distribution is
 /T
N
V
 h 


 8m T 
2
3/ 2
 h 

 1, i.e. n
 8m T 
2
3/ 2
 1
The criterion are low density, high temperature,
and large molecular masses m.
简并条件
• The reverse criterion
3/ 2
2
N h 

  1 i.e.
V  8m T 
3/ 2
 h 
3/ 2


n
 T
(37.9)

 8m 
• The MB distribution is inapplicable;
• The gas obeys the BE or FD distribution.
• The gas is then said to be “degenerate”(简并)
2
• Eq.(37.9) is known as the “degeneracy criterion”
• For a gas: at low temperatures shows “quantum”
and at high temperatures shows “classical”.
Examples
• For the ordinary atoms, N/V~1019cm-3, m ~
( 10-23 to 10-24 )g, T<<10-1K as quantum.
• For ordinary gases, the normal MB distribution
is a good approximation down to rather low
temperatures.
• 随着低温技术的提高，最近几年统计物理学的主
要研究内容是极低温下气体所表现出的量子效应--“玻色凝聚”。
• 由于电子的质量仅为10-27g, 临界温度高达104K。