PHZ3400 Midterm Two Solution: Difference between revisions

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'''Explain the concept of the "Thermodynamic Limit", and present the corresponding domain wall argument (''derivation'' of a formula) to estimate the relaxation time as a function of system size, at <math>T < T_c</math>.'''
'''Explain the concept of the "Thermodynamic Limit", and present the corresponding domain wall argument (''derivation'' of a formula) to estimate the relaxation time as a function of system size, at <math>T < T_c</math>.'''


The partition function
===Canonical partition function===
Suppose we have a thermodynamically large system that is in constant thermal contact with the environment, which has temperature T, with both the volume <math>V</math> of the system and the number of constituent particles <math>\eta</math> fixed. This kind of system is called a canonical ensemble. Let us label with <math>\eta</math> (<math>\eta</math> = 1, 2, 3, ...) the exact states (microstates) that the system can occupy, and denote the total energy of the system when it is in microstate <math>\eta</math> as <math>E_{\eta}</math>. Generally, these microstates can be regarded as discrete quantum states of the system.
 
The canonical partition function is defined as:


<math>Z = \sum_{n}^{\infty} e^{-\beta E_n} \;</math>   
<math>Z = \sum_{n}^{\infty} e^{-\beta E_n} \;</math>   
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where  
where  


<math>Z \;</math> =  
<math>Z \;</math> = statistical function of Temperature


<math>e \;</math> =
<math>e \;</math> = base of the natural logarithm


<math>\beta \;</math> =
<math>\beta = \tfrac{1}{k_{B}T} \;</math> = the inverse temperature where <math>k_B</math> is the Boltzmann const


<math>E_{n} \;</math> =
<math>E_{n} \;</math> = total energy of the system


<math>n \;</math> = the number of particles.   
<math>n \;</math> = the number of particles.   
===Thermodynamic Limit===
<font color="blue">I still don't know what this is. [[User:KimberlyWynne|KimberlyWynne]] 00:25, 16 April 2009 (EDT) </font>


However, symmetry breaking, including phase transitions, cannot occur for any finite system<ref>http://prola.aps.org/abstract/PR/v87/i3/p404_1  
However, symmetry breaking, including phase transitions, cannot occur for any finite system<ref>http://prola.aps.org/abstract/PR/v87/i3/p404_1  
C. N. Yang and T. D. Lee, ''Statistical Theory of Equations of State and Phase Transitions. I. Theory of Condensation'' Phys. Rev. 87, 404 - 409 (1952)</ref>.  Thus systems must be sufficiently large, <math> n\rightarrow \infty</math>, in order to reach the thermodynamic limit and for spontaneous symmetry breaking to occur.   
C. N. Yang and T. D. Lee, ''Statistical Theory of Equations of State and Phase Transitions. I. Theory of Condensation'' Phys. Rev. 87, 404 - 409 (1952)</ref>.  Thus systems must be sufficiently large, <math> n\rightarrow \infty</math>, in order to reach the thermodynamic limit and for spontaneous symmetry breaking to occur.   


[[Image:DomainWall.jpeg|left|thumb|300px|This illustrates the most energetically favorable configuration of the system.  The green line represents the Domain Wall.]]
===Domain Wall Argument===
[[Image:Probability.jpeg|right|thumb|300px|This illustrates the probability of states<math>\Delta E</math> is the energy of the Domain Wall ]]
Suppose we start in a ferromagnetic system where all of the spins are up and we would like to make them spin down. This can be done by applying a domain wall to the system.  The image shows the most energetically favorable configuration of the system.   
 
[[Image:DomainWall.jpeg|center|thumb|300px|This illustrates the most energetically favorable configuration of the systemThe green line represents the Domain Wall.]]


Suppose we start in a ferromagnetic system where all of the spins are up and we would like to make them spin down.  This can be done by applying a domain wall to the system.  The image to the left shows the most energetically favorable configuration of the system.
===Derivation===
This can be further described by this image, which shows the probability of the system being in those states.  Notice that the image shows the system is most likely to be found in a state where there is a mix of up and down spinsThe domain wall will, by its nature, try to find the most energetically favorable state and it will thus lie where the energy of the probability of the states of the system is at a maximum.


This can be further described by the image to the right, which shows the probability of the system being in those states.  Notice that the image shows the system is most likely to be found in a state where there is a mix of up and down spins.  The domain wall will, by its nature, try to find the most energetically favorable state and it will thus lie where the energy of the probability of the states of the system is at a maximum.
[[Image:Probability.jpeg|center|thumb|300px|This illustrates the probability of states.  <math>\Delta E</math> is the energy of the Domain Wall ]]


===Probability ===
====Probability====


<math>P(E)\propto e^{-\tfrac{E}{kT}}</math>  
<math>P(E)\propto e^{-\tfrac{E}{kT}}</math>  

Revision as of 00:25, 16 April 2009

PHZ 3400 – Midterm Two Exam (with solutions) – April 10, 2009

Problem 1

Explain the concept of the "Thermodynamic Limit", and present the corresponding domain wall argument (derivation of a formula) to estimate the relaxation time as a function of system size, at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T < T_c} .

Canonical partition function

Suppose we have a thermodynamically large system that is in constant thermal contact with the environment, which has temperature T, with both the volume Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V} of the system and the number of constituent particles Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta} fixed. This kind of system is called a canonical ensemble. Let us label with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta} (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta} = 1, 2, 3, ...) the exact states (microstates) that the system can occupy, and denote the total energy of the system when it is in microstate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta} as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{\eta}} . Generally, these microstates can be regarded as discrete quantum states of the system.

The canonical partition function is defined as:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z = \sum_{n}^{\infty} e^{-\beta E_n} \;}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z \;} = statistical function of Temperature

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e \;} = base of the natural logarithm

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta = \tfrac{1}{k_{B}T} \;} = the inverse temperature where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_B} is the Boltzmann const

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{n} \;} = total energy of the system

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \;} = the number of particles.

Thermodynamic Limit

I still don't know what this is. KimberlyWynne 00:25, 16 April 2009 (EDT)

However, symmetry breaking, including phase transitions, cannot occur for any finite system[1]. Thus systems must be sufficiently large, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\rightarrow \infty} , in order to reach the thermodynamic limit and for spontaneous symmetry breaking to occur.

Domain Wall Argument

Suppose we start in a ferromagnetic system where all of the spins are up and we would like to make them spin down. This can be done by applying a domain wall to the system. The image shows the most energetically favorable configuration of the system.

This illustrates the most energetically favorable configuration of the system. The green line represents the Domain Wall.

Derivation

This can be further described by this image, which shows the probability of the system being in those states. Notice that the image shows the system is most likely to be found in a state where there is a mix of up and down spins. The domain wall will, by its nature, try to find the most energetically favorable state and it will thus lie where the energy of the probability of the states of the system is at a maximum.

This illustrates the probability of states. Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta E} is the energy of the Domain Wall

Probability

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(E)\propto e^{-\tfrac{E}{kT}}}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E \;} = Energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T \;} = Temperature.

When there is random motion

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(E)\propto T} .

Since we are dealing with a domain wall, we can replace E with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta E} , which is also defined as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta E = \delta L^{d-1} \;}

where

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \delta\;} = surface tension Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \propto J} - the interaction between spins

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L\; } = the size of the system.

With this substitution, we get

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(E) \propto e^{-E/kT} \propto \tau ^{-1} \propto e^{-AL^{d-1}/T}} .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d \;} = dimension, so for a three dimensional case we get Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L^2} .

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau} = relaxation time, therefore and the equation gives a relationship between the relaxation time and the size of the system.

The relaxation time and the size of the system have a direct correlation, as the size of the system increases the orelaxation time also increases, albeit at a greater rate. Thus if the size of the system is very large (macroscopic objects) then the relaxation time becomes larger than the age of the universe, and the system will stay in its current state.

Another interesting relation is between the relaxation time and temperature. As the temperature gets lower, the relaxation time gets larger. This fits in well with superconductors since they are only conducting at low temperatures. This means that as the semiconductors cool down the spins align and exhibit symmetry breaking.

  1. http://prola.aps.org/abstract/PR/v87/i3/p404_1 C. N. Yang and T. D. Lee, Statistical Theory of Equations of State and Phase Transitions. I. Theory of Condensation Phys. Rev. 87, 404 - 409 (1952)

Problem 2

Sketch the magnetization of a ferromagnet as a function of temperature T, for (A) Zero external magnetic field and (B) Finite external magnetic field. How is the behavior around the Curie Temperature (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} ) affected by the field?

(A) Zero External Magnetic field

For zero external magnetic field, the magnetization of a ferromagnet as a function of temperature T should go to the Curie Temperature (Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} ). The Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} will be unique based on the ferromagnet.

This image shows the results for zero external magnetic field(blue) overlayed with the results with an external magnetic field(teal). Image from Dr. Dobrosavljevic's notes [1]

(B) Finite External Magnetic Field

Ignore the temperature but observe the trend as the magnetization approaches 0 at the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} . Image is from [2]

Behavior around the Curie Temperature

Problem 3

Consider a solid with vibrations well described by the "Einstein Model", where a single vibrational frequency Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega_0} is considered. [Note: Phonons obey Bose-Einstein statistics.]

Bose-Einstein Distribution

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \eta _{phonon} = \frac{1}{e^{\tfrac{\hbar \omega}{kT}}+1}\;}

(A)

What is the density of phonons with this frequency as a function of the temperature Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T} ? Sketch this function.

(B)

What is the phonon density at very low temperatures?

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{T \to 0} \eta (T) = \lim_{T \to 0} \frac{1}{e^{\tfrac{\hbar \omega}{kT}}+1}\;}

(C)

What is the phonon density at very high temperatures?

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{T \to \infty } \eta (T) = \lim_{T \to \infty } \frac{1}{e^{\tfrac{\hbar \omega}{kT}}+1} \;}

(D)

Assuming that the scattering rate of electrons in such a solid is simply proportional to the density of phonons, determine the temperature dependence of resistivity at high temperatures.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \tau ^{-1} \sim \eta(T) \sim T = \frac{1}{\delta } = \frac{m}{\eta e^2 \tau} \;}

Problem 4

Consider a neutron star of radius R = 10km, and mass M = 3x1030 kg (approximately 1.5 solar masses). Given that the mass of a single neutron is m=1.7*10-27 kg, that h-bar = 1x10-34 m2 kg/s, and that neutrons obey Fermi-Dirac Statistics, calculate:

(A)

The number density n=N/V of this star. N = total number of objects in volume V. Using this, we first can find the volume of the neutron star.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V =\frac{4}{3} \pi R^3 \,}

(B)

Derive in detail (as we did in class; show all the steps) the relation between Fermi energy and the number density.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_f = \frac{\hbar^2 \pi^2}{2 m L^2} n^2 \,} .

For a 3-D object, in this case a sphere,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N =2\times\frac{1}{8}\times\frac{4}{3} \pi n_f^3 \,}

where the factor 2 comes from the fact that there are 2 spin states, 1/8 comes from the fact that only 1/8 of the sphere is positive, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4}{3} \pi n_f^3 \,} comes from the volume of the sphere with radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_f}

Solving for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_f} and then substituting into Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_f = \frac{\hbar^2 \pi^2}{2 m L^2} n^2 \,} yields

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_f=\left(\frac{3 N}{\pi}\right)^{1/3} }

and then finally

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_f =\frac{\hbar^2 \pi^2}{2m L^2} \left( \frac{3 N}{\pi} \right)^{2/3}}

In order to get this in terms of number density, we substitute L2 with V2/3.

(C)

Estimate the Fermi energy of this neutron star in Joules

(D)

Using the fact that 1eV = Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1.60217646 \times 10^{19} } Joules, and that 1meV ~ 10K, estimate the Fermi temperature of this neutron star. How does it compare to that of a typical nucleus, where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_F ~ 30MeV} ?

(E)

Given that the typical temperature of this neutron star is T ~ K, is this neutron star "cool" or is it "hot"(as compared to the Fermi temperature)?