Last Time
Last time it was derived that the probability of being in a state in a microcanonical ensemble is
[ P_v = \frac
]
. Boundary conditions are
[ N, V, ]
and
[ E ]
constant. In an isolated system, the energy is constant. There is degeneracy in the number of states available to acess. Any state is not more probable than another. The probability to be in any of the states is inversely proportional to the number of states.
[ P_v = \frac{e^{\beta E_v}}{\sum_v e^{ \beta E_v}} ]
Consider a canonical ensemble. The variables of
[ N, V, ]
and
[ T ]
are fixed. There is an ensemble of systems with heat-conducting boundaries. All systems exchange energy, and the whole ensemble is at a fixed energy. It could be treated as one large many-body function. Consider distribution with most permutations. Maximize permutation number.
Consider the canonical ensemble probability function. Assume that all energies are the same. The partition function is a summation over states all with the same energy. Sum
[ \Omega ]
times.
[ P_v^{\mbox{can}}= \frac{e^{\beta E}}{\sum_K e^{\beta E}} ]
[ P_v^{\mbox{can}}= \frac{e^{\beta E}}{\sum_E \Omega e^{\beta E}} ]
[ P_v^{\mbox{can}}= \frac
]
Determining thermodynamic properties with the probability function
It is possible to find mechanical thermodynamic properties with the probability function,
[ P_v ]
. Below is a general case and calculations of the average energy and pressure. With the determination of\betafor a canonical ensemble, it is possible to calculate many things.
[ \overline M = \sum_v M_v P_v ]
[ \overline E = \frac{\sum_v E_v e^{-\beta E_v}}
]
[ \overline p = \frac{\sum_v P_v e^{-\beta E_v}}
]
[ \overline p = \frac{\sum_v \left ( \frac{-\partial E_v}
\right )_
e^{-\beta E_v}}
]
Summary
Microcanonical Ensemble
- Unknown macro: {latex}constant
[ (N, V, E) ]
- Unknown macro: {latex}
[ P_v = \frac
Unknown macro: {1}Unknown macro: {Omega (E)}]
- Degeneracy
- Unknown macro: {latex}
[ E \rightarrow \Omega (E) ]
- No energy state is more likely than another
Canonical Ensemble
- Unknown macro: {latex}constant
[ (N, V, T) ]
- Unknown macro: {latex}
[ P_v = \frac{e^{\beta E_v}}{\sum_v e^{ \beta E_v}} ]
- Whole ensemble is isolated (fixed energy)
- Little systems within the whole ensemble can exchange energy
- Since the whole ensemble is at constant energy, it can be expressed as the many body (i.e. identical system) non-interacting wavefunctions, which obey the postulate ofequal a priori probability. i.e. every state of the ensemble has an equal probalbility.
- Some distributions have more permutations than others
- As we increase the number, we'll put more weight on some states [EXPLAIN]
- Consider when a special case when energy is constant
- Unknown macro: {latex}
[ E_v = E ]
- Unknown macro: {latex}
[ P_v^{\mbox{can}}= \frac
Unknown macro: {1}Unknown macro: {Omega}]
Probability function
The probability function,
[ P_v ]
, is used to derive mechanical thermodynamic properties.
[ \overline M = \sum_v M_v P_v ]
Energy
[ \overline E = \frac{\sum_v E_v e^{-\beta E_v}}
]
-
- Pressure
- Unknown macro: {latex}
[ \overline p = \frac{\sum_v P_v e^{-\beta E_v}}
Unknown macro: {Q}]
Unknown macro: {latex}[ \overline p = \frac{\sum_v \left ( \frac{-\partial E_v}
Unknown macro: {partial V}\right )_
Unknown macro: {T,N}e^{-\beta E_v}}
Unknown macro: {Q}]
- Pressure
Today
The terms
[ \beta ]
and
[ S ]
are calculated today. These are non-mechanical properties, and there is an insight into what entropy means. It is not possible to express these in terms of one particle. There is a need for a whole system to be defined.
Determination of beta
What is the destination? Consider the equation below. It is an equation of state, and it is used to find
[ \beta ]
.
[ \frac
_
- T(\frac
)_
= -P ]
Consider the energy function of volume. The dimension of the box is relevant when considering the energy, and the partition function depends on volume.
[ \left ( \frac
\right )_
=\frac
\left ( \frac{\sum_v E_v e^{-\beta E_v}}
\right ) ]
[ \left ( \frac
\right )_
=\frac{ \sum_v \frac
e^{-\beta E_v} }
+\frac{\sum_v \ E_v (- \beta \frac
) e^{-\beta E_v}}
]
[ \left ( \frac
\right )_
= -\frac
\left ( \sum_v E_v e^{-\beta E_v} \right ) \left( -\sum_v \beta \frac
e^{-\beta E_v} \right ) ]
[ \left ( \frac
\right )_
= - \overline
+ \beta \overline
+ \beta \overline
\overline
]
A relation below is used in the derivation. Plug this into an above expression.
[ \left ( \frac
\right )_
= \overline E \cdot \overline p - \overline
]
[ \left ( \frac
\right )_
+ \beta \left ( \frac
\right )_
= - \overline p ]
The thermodynamic relations need to be equivalent. Compare a result with the equation of state from thermodynamics.
[ \left (\frac
\right )_
- T \left (\frac
\right )_
= -P ]
[ \left ( \frac
\right )_
+ \beta \left ( \frac
\right )_
= - \overline p ]
[ \beta \frac
= -T \frac
]
[ \frac{-\partial T}
= \frac
]
[ -\ln T =\ln \beta + C ]
[ \beta = \frac
]
[ \beta = \frac
]
The relation derived above is true for both the microcanonical and canonical ensemble; it is not dependent on the type of ensemble. The termkork_Bis the universal constant that is independent of system. There is now a complete expression of probability.
[ Q = \mbox
]
[ Q = \sum_v e^{-\beta E_v} ]
[ Q = \sum_v e^{\frac{-E_v}{k_B T}} ]
The partition function is a measure of the number of states that are thermally accessible. When the temperature is very small, only a few states will contribute in the summation of the partition function.
Determination of entropy, S
The partition function,
[ Q ]
, is a function of number of moles, volume, and temperature. The term
[ \beta ]
is dependent on temperature, and energy,
[ E_v ]
is a function of the number of moles and volume. The partition function is a state function, which means it is not path dependent.
[ Q=\sum_v e^{-\beta E_v} ]
Consider a function,
[ f = \ln Q ]
. A function is written below. Consider the whole derivative.
[ df = \left ( \frac
\right )_
d \beta + \sum_v \left ( \frac
\right ){\beta, E{K \ne V}} d E_v ]
Consider
[ \left ( \frac
\right )_
]
[ \left ( \frac
\right )_
= \frac
\cdot \frac
]
[ \left ( \frac
\right )_
= \frac{\sum_v E_v e^{\beta E_v}}
]
[ \left ( \frac
\right )_
= - \overline E ]
Consider
[ \left ( \frac
\right ){\beta, E{K \ne V}} ]
[ \left ( \frac
\right ){\beta, E{K \ne V}} = \frac{\beta e^{\beta E_v}}
]
[ \left ( \frac
\right ){\beta, E{K \ne V}} = \beta P_v ]
Add
[ (-\beta d \overline E + \beta d \overline E) ]
to both sides of the equation and rearrange terms.
[ d(f+\beta \overline E) = \beta (d \overline E - \sum_v P_v d E_v ) ]
Since the terms
[ E ]
and
[ f ]
are state functions, the expression
[ d(f+\beta \overline E) ]
is also a state function. The term
[ d \overline E ]
is the change in internal energy.
Cause the system to undergo a small change in volume. Let the system undergo an infinitesimal change of state,
[ dV ]
along a reversible path, while keeping the system in contact with a heat bath and at constant temperature.
[ \sum_v P_v d E_v = \sum_v P_v \left ( \frac
\right ) dV ]
[ \sum_v P_v d E_v = \sum_v P_v (-pv) dV ]
[ \sum_v P_v d E_v = -\overline p dV ]
[ \sum_v P_v d E_v = dW_{\mbox{reversible}} ]
Substitute this relation into a previous equation and apply the first and second law.
[ d(f+\beta \overline E) = \beta (d \overline E - \sum_v P_v d E_v ) ]
[ d(f+\beta \overline E) = \beta ( d \overline E - dW_{\mbox{reversible}}) ]
[ \mbox
[
[ \beta ( d \overline E - dW_{\mbox{reversible}}) = \beta d Q_{\mbox{reversible}} ]
[ \mbox
]
[ \beta d Q_{\mbox{reversible}} = \frac
]
The function
[ f = \ln Q ]
is known. Rewrite the above expression in terms of
[ dS ]
. Integrate and disregard the integration constant.
[ dS=k_B d(f+\beta \overline
) ]
[ S=k_B f + \frac
+ \mbox
]
[ S=k_B f + \frac
]
What does this expression mean?
Summary
Determine non-mechanical properties from
[ \beta ]
and
[ S ]
- Find mathematical relations from thermodynamics between average mechanical properties and compare to similar thermodynamic relations.
Determination of
[ \beta ]
- Consider equation of state,\frac
Unknown macro: {partial E}Unknown macro: {partial V}\right )_
_
Unknown macro: {N,T}- T(\frac{partial P}}
Unknown macro: {partial T})_
Unknown macro: {N,V}= -P
- Find expression of\left ( \frac
Unknown macro: {partial overline E}
Unknown macro: {N,B} - Find expression of\left ( \frac
- Find\left ( \frac
Unknown macro: {partial overline E}Unknown macro: {partial V}\right )_Unknown macro: {N,beta}+ \beta \left ( \fracUnknown macro: {partial overline p}Unknown macro: {partial beta}\right )_
\right )_
Unknown macro: {N,V}= - \overline p</math>Compare result with thermodynamic equation of state and find <math> \beta = \frac
Unknown macro: {1}Unknown macro: {k_B T}- Plug into partition function:Q = \sum_v e^{\frac{-E_v}
Unknown macro: {k_B T}}
Determination of entropy,S
- Consider the whole derivative of functionf = \ln Q
**df = \left ( \fracUnknown macro: {partial f}
Unknown macro: {E_v}= - \overline Ed \beta + \sum_v \left ( \frac
Unknown macro: {partial f}Unknown macro: {partial E_v}\right ){\beta, E{K \ne V}} d E_v
***\left ( \fracUnknown macro: {partial beta}\right )_
***\left ( \fracUnknown macro: {partial f}Unknown macro: {partial E_v}\right ){\beta, E{K \ne V}} = \beta P_v- Add(-\beta \d \overline E + \beta \d \overline E)to both sides of the equation and rearrange terms.
***d(f+\beta \overline E) = \beta (d \overline E - \sum_v P_v d E_v )
***\sum_v P_v d E_v = dW_{\mbox{reversible}}
****d(f+\beta \overline E) = \beta (d \overline E - \sum_v P_v d E_v )
****\beta ( d \overline E - dW_{\mbox{reversible}}) = \beta d Q_{\mbox{reversible}}
*****S=k_B f + \fracUnknown macro: {overline E}Unknown macro: {T}
- Plug into partition function:Q = \sum_v e^{\frac{-E_v}
Physical Interpretation of\partial Qand\partial W(our derivation)
Begin with an expression of energy. Components of the expression can be equated with work and heat.
\overline E = \sum_v P_v E_v
d \overline E = \sum_v E_v d P_v + \sum_v P_v d E_v
\delta Q = \sum_v E_v d P_v
\delta W = \sum_v P_v d E_v
Consider the physical meaning of these expressions. Work is associated with changing energy levels slightly while changing probabilities of being at a particular energy level are fixed. Heating a system results in keeping the energy levels fixed while chaning the probability of being at a particular energy level.
A problem from this class that is asked to be done as additinoal homework is to show that entropy can be expressed asS = -k_B \sum_v P_v \ln P_v. A tip is to plug in the expression for probability below.
P_v = \frac{e^{- \beta E_v}}{\sum_v e^{- \beta E_v} }
Entropy is not a mechanical property. There is a need to sum over all probabilities. There are no microscopic counterparts to temperature and entropy. They are properties of the whole probability distribution.
Values of energy levels are needed to solve for probabilities. Given a system with knownN, V,andT, solve energy eigenvalues,E_vand eigenfunctions,\Psi_vof the Hamiltonian. Plug these values into the expression of probability.
\hat H \Psi = E \Psi
P_v = \frac{e^{-\beta E_v}}
Summary
Differential of\overline E = \sum_v P_v E_v
*d \overline E = \sum_v E_v d P_v + \sum_v P_v d E_v
**\sum_v E_v d P_v = \delta Qrelates to changing probability or changing the population
-
-
- Heat absorption: Energy levels stay fixed, but change population of those levels.
**\sum_v P_v d E_v = \delta Wrelates to changing energy levels - Work: Change energy levels slightly, but keep population of those levels fixed.
- Heat absorption: Energy levels stay fixed, but change population of those levels.
-
Homework
- Show that S can be expressed asS = -k_B \sum_v P_v \ln P_v(problem 2.5)
- TipP_v = \frac{e^{- \beta E_v}}{\sum_v e^{- \beta E_v} }
- Need to sum over all probabilities
Temperature and entropy have no microscopic counterparts, hence they are properties of the whole probability distribution.
System withN, V, andTconstant
- FindE_vand\Psi_v
- Use Hamiltonian\hat H \Psi = E \Psi
- Plug into probability expression,P_v = \frac{e^{-\beta E_v}}
Eight sided dice
Entropy is proportional to the number of optimal binary questions that must be asked to find what state a system is in. A more complex system requires more questions be asked. Consider an eigth-sided dice. What face is up on average? A dumb question could be regarded as a question wherein the probability of an answer of "yes" is\frac
. Below is a calculation of the average number of dumb questions necessary to determine what side of the dice is up. The number seven appears twice in the calculation; if the side of the dice is known not to be seven after the first seven questions, the side of the top of the dice must be eight.
\frac
(1+2+3+4+5+6+7+7)=4.375</math><br></center>Smart questions function approach the problem with the lens of a binary tree. Each question divides the field equally. The calculation is below of the average number of smart questions required to calculate what side of an eight-sided is up. A general expression of the number of optimal questions is also below. The last expression is of similar form to entropy, <math>S</math> in Problem 2.5.<center><br><math>\mbox
</math><br><math>\log_2 (8) = 3</math><br><math>\mbox
</math><br><math> - \sum_i P_i \log_2 (P_i)</math><br></center>Fewer questions are required in a biased system. See theclass slides.SummaryConsider an eight-sided diceDumb questionsprobability of positive response is <math>\frac
-
- average number of questions to determine face:\frac
Unknown macro: {1}Unknown macro: {8}
(1+2+3+4+5+6+7+7)=4.375</math>Smart questions: divide and conquerforming binary tree<math>\log_2 (8) = 3</math><math> - \sum_i P_i \log_2 (P_i)</math>Looks like <math>S</math> from Problem 2.5Relative EnergiesConsider energies of states and determine whether the probability, <math>P_v</math>, depends on the chosen zero energy. Consider <math>N, V,</math> and <math>T</math> constant. A plot of relative energies is below, and the values can be found by Hamilton's equations.<center><br><math>\hat H \Psi = E \Psi </math></center><br>There is degeneracy associated with each energy level.<center><br><math>E_0: \lbrace \Psi_
Unknown macro: {0,1},\Psi_
Unknown macro: {0,2},.....,\Psi_
Unknown macro: {0,Omega(E_0)}\rbrace
E_2: \lbrace \Psi_
Unknown macro: {2,1},\Psi_
Unknown macro: {2,2},.....,\Psi_
Unknown macro: {2,Omega(E_2)}\rbrace
The degeneracy ofE_Nis\Omega(E_N), and below is a plot of energy and degeneracy.
Unable to render embedded object: File (Degeneracy_and_energy.PNG) not found.
Look at the probability function. The energy is denoted byiand the state byj.
\mbox{wavefunction
\Psi_
Unknown macro: {ij}\mbox{probability function
P_v = P_
P_
Unknown macro: {ij}= \frac{e^{-\beta E_i}}
Unknown macro: {Q}Consider a relative energy scale. When calculating the partition function, sum over all the states. Consider\Delta E_i \ge 0. The calculation below demonstrates thatP_vdoes not depend on the chosen energy level. In the exam,E_1can be chosen to be zero if givenE_1, E_2,andE_3. Physics is blind to choosing that value to be zero.
Q = \sum_
Unknown macro: {v=i,j}e^{-\beta E_i}
Q = e^{-\beta E_0} \sum_
e^{-\beta (E_1 - E_0)}
Q = e^{-\beta E_0} \sum_
Unknown macro: {ij}e^{-\beta (\Delta E_i)}
P_
= \frac{e^{\beta E_0} \cdot e^{\beta \Delta E_i}}{e^{-\beta E_0} \cdot \sum_
Unknown macro: {ij}e^{-\beta \Delta E_i}}
P_
= \frac{ e^{-\beta \Delta E_i}}{\sum_
Unknown macro: {ij}e^{-\beta (\Delta E_i)}
Summary
Consider a system withN, V,andTfixed
- Find the energy values with\hat H \Psi = E \Psi
- The wavefunction is\Psi_
Unknown macro: {i,j}
- There is a degeneracy associated with each energy
**E_0: \lbrace \Psi_
,\Psi_
Unknown macro: {0,2},.....,\Psi_
Unknown macro: {0,Omega(E_0)}\rbrace </math><math>E_2: \lbrace \Psi_
Unknown macro: {2,1},\Psi_
Unknown macro: {2,2},.....,\Psi_
Unknown macro: {2,Omega(E_2)}\rbrace </math><math>\Omega(E_N)</math> is the degeneracy of <math>E_N</math><p>
</p>Probability with relative energyExpression of probability with energy <math>E_i</math>: <math> P_Unknown macro: {ij}= \frac{e^{-\beta E_i}}
Unknown macro: {Q}- Expression of probability with relative energy\Delta E_i
_
= \frac{ e^{-\beta \Delta E_i}}{\sum_
Unknown macro: {ij}e^{-\beta (\Delta E_i)}
-
- This is good; the probability is independent of the chosen zero of energy
- Physics is blind to choosingE_0=0
What happends toP_
</math> with T?Consider the dependence of probability with temperature. As temperature approaches zero, <math>\beta</math> approaches infinity. An expression of probability is below. Consider the ground state wherein <math>\Delta E_0 = 0</math>, and <math>\Omega(E_0)</math> is the degeneracy at groundstate.<center><br><math> P_v = P_
Unknown macro: {ij}= \frac{ e^{-\beta \Delta E_i}}{\sum_
e^{-\beta (\Delta E_i)}</math><br><math>P_
Unknown macro: {ij}= \frac{ e^{-\beta \Delta E_i}}{\sum_
e^{-\beta (\Delta E_i)}</math><br><math> \lim_
Unknown macro: {beta to infty}\frac{ e^{-\beta \Delta E_i}}{\sum_
Unknown macro: {ij}e^{-\beta (\Delta E_i)}} = \frac
Unknown macro: {Omega(E_0)}**\Omega(E_0)is the degeneracy at groundstateConsider the probability of not being in the ground state.
\lim_
Unknown macro: {beta to infty}ty} \frac{ e^{-\beta \Delta E_i}}{\sum_
Unknown macro: {ij}e^{-\beta (\Delta E_i)}} = 0
Summary
T \to 0(\beta \to > \infty)
- Probability
**P_v = P_
</math><math>P_
Unknown macro: {ij}= \frac{ e^{-\beta \Delta E_i}}{\sum_
e^{-\beta (\Delta E_i)}</math>Probaility of being in groundstate<math>\Delta E_0 = 0 </math><math> \lim_
\frac{ e^{-\beta \Delta E_i}}{\sum_
Unknown macro: {ij}e^{-\beta (\Delta E_i)}} = \frac
Unknown macro: {1}
- average number of questions to determine face:\frac
- Probability of not being in groundstate
- All\lim_
Unknown macro: {beta to infty}ty} \frac{ e^{-\beta \Delta E_1}}{\sum_Unknown macro: {ij}P_
e^{-\beta (\Delta E_i)}} \to 0
Assume\Omega (E_0) 1
Assume that the degeneracy of the ground state is one,\Omega(E_0)=1. Consider the limit as temperature approaches zero.
\lim_
Unknown macro: {T to 0}P_
Unknown macro: {GS}= 1
S = -k_B \sum_
Unknown macro: {ij}</math><br><math>S = 0</math><br></center>There is no question what state the system is in. There is no ignorance. The state of the system is completely determined.Summary<math>\lim_\ln P_
Unknown macro: {T to 0}P_Unknown macro: {GS}= 1 </math><math>S = -k_B \sum_Unknown macro: {ij}\ln P_P_
Unknown macro: {ij}</math><math>S = 0</math>State of system is completely determinedNo ignoranceSystem is orderedA more ordered system means less entropy, which means fewer quesions need to ask to determine state<center>
10/25/06
</center>
- All\lim_