PHZ3400 Midterm Two Solution: Difference between revisions

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


This is the 3rd chance for the retards who failed the 2nd.
To those who did well: ''Shoo!''
==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 <math>T < T_c</math>.'''
===Canonical partition function===
Given a thermodynamically large system that is in constant thermal contact with the environment, where both the volume <math>V</math> and the number of constituent particles <math>\eta</math> are fixed. <math>\eta</math> (<math>\eta</math> = 1, 2, 3, ...) are the exact states (microstates) that the system can occupy, generally, these can be regarded as discrete quantum states of the system.
Therefore, the canonical partition function is defined as:
<math>Z = \sum_{n}^{\infty} e^{-\beta E_n} \;</math> 
where
<math>Z \;</math> = statistical function of Temperature
<math>e \;</math> = base of the natural logarithm
<math>\beta = \tfrac{1}{k_{B}T} \;</math> = the inverse temperature where <math>k_B</math> is the Boltzmann const
<math>E_{n} \;</math> = total energy of the system
<math>n \;</math> = the number of particles. 
===Thermodynamic Limit===
Spontaneous symmetry breaking (i.e. phase transitions) cannot occur for a 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>. There must be an infinite number of spins. Thus systems must be sufficiently large, <math> \eta \rightarrow \infty \;</math>, in order to reach the ''thermodynamic limit'' and for spontaneous symmetry breaking to occur. 
<font color="blue">I still don't know what this is. What is it defined as? The limit of the number of spins as it goes to infinity:<math>\lim_{\eta \rightarrow \infty } \;</math> ? The temperature  at which spontaneous symmetry breaking can occur? The point where the domain wall begins? [[User:KimberlyWynne|KimberlyWynne]] 00:25, 16 April 2009 (EDT) </font>
===Domain Wall Argument===
A domain wall is an interface transition of 180° angular displacements on magnetic spin.
[[Image:Domainwall1.jpg|center]]
Given a ferromagnetic system where all of the spins are up at ''t'' = 0, and we want to find the time ''t'' it takes to reverse them to all spin down. The energetically easiest way to do this would be to "push" a domain wall through the system. <ref>[http://www.physics.fsu.edu/Users/Dobrosavljevic/Phase%20Transitions/ssb.pdf]</ref>
[[Image:DomainWall.jpeg|center|thumb|300px|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.
[[Image:Probability.jpeg|center|thumb|300px|This illustrates the probability of states.  <math>\Delta E</math> is the energy of the Domain Wall ]]
====Probability====
The probability of reaching the all-spin-down configuration
<math>P(E)\propto e^{-\tfrac{E}{k_{B}T}} \;</math>
When there is random motion, the probability is proportional to time
<math>P(E)\propto T \;</math>.
Since we are dealing with a domain wall, we can replace E with <math>\Delta E</math> from the illustration, which is also defined as
<math>\Delta E = \delta L^{d-1} \;</math>
where
<math>\delta\;</math> = surface tension <math>\propto J</math> - the interaction between spins
<math>L\; </math> = system size
With this substitution, we get
<math>P(E) \propto  e^{-E/kT}  \propto  \tau \propto e^{-A \tfrac{L^{d-1}}{T}}</math>.
where
<math>A \;</math> = dummy variable representing some constant
<math>d \;</math> = dimension, so for a 3-dimensional case we get <math>L^{3-1} = L^{2}</math>.
<math>\tau</math> = relaxation time, therefore the equation gives a relationship between the relaxation time and the size of the system. 
====Relaxation time====
The relaxation time <math>\tau \;</math> and the size of the system have a direct correlation. As the size of the system increases the relaxation 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.
<references/>
==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 (<math>T_c</math>) 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 (<math>T_c</math>). The <math>T_c</math> will be unique based on the ferromagnet.
[[Image:problem2.jpg|center|thumb|300px|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 [http://www.physics.fsu.edu/Users/Dobrosavljevic/Phase%20Transitions/ssb.pdf]]]
===(B) Finite External Magnetic Field===
[[Image:Tc.gif|center|thumb|300px|Ignore the temperature but observe the trend as the magnetization approaches 0 at the <math>T_c</math>. Image is from [http://www.irm.umn.edu/hg2m/hg2m_b/Image9.gif] ]]
===Behavior around the Curie Temperature===
==Problem 3==
'''Consider a solid with vibrations well described by the "Einstein Model", where a ''single vibrational frequency'' <math>\omega_0</math> is considered. [Note: Phonons obey Bose-Einstein statistics.]'''
Bose-Einstein Distribution
<math>\eta _{phonon} = \frac{1}{e^{\tfrac{\hbar \omega}{kT}}+1}\;</math>
===(A)===
'''What is the density of phonons with this frequency as a function of the temperature <math>T</math>?  Sketch this function. '''
===(B)===
''' What is the phonon density at very low temperatures? '''
<math>\lim_{T \to 0} \eta (T) = \lim_{T \to 0} \frac{1}{e^{\tfrac{\hbar \omega}{kT}}+1}\;</math>
===(C)===
'''What is the phonon density at very high temperatures?'''
<math>\lim_{T \to \infty } \eta (T) = \lim_{T \to \infty } \frac{1}{e^{\tfrac{\hbar \omega}{kT}}+1} \;</math>
===(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.'''
<math>\tau ^{-1} \sim  \eta(T) \sim T = \frac{1}{\delta } = \frac{m}{\eta e^2 \tau} \;</math>
==Problem 4==
'''Consider a neutron star of radius R = 10km, and mass M = 3x10<sup>30</sup> kg (approximately 1.5 solar masses). Given that the mass of a single neutron is m=1.7*10<sup>-27</sup> kg, that h-bar = 1x10<sup>-34</sup> m<sup>2</sup> 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.
:<math>V =\frac{4}{3} \pi R^3 \,</math>
===(B)===
'''Derive in detail (as we did in class; show all the steps) the relation between Fermi energy and the number density.'''
:<math>E_f = \frac{\hbar^2 \pi^2}{2 m L^2} n^2 \,</math>.
For a 3-D object, in this case a sphere,
:<math>N =2\times\frac{1}{8}\times\frac{4}{3} \pi n_f^3 \,</math>
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 <math>\frac{4}{3} \pi n_f^3 \,</math> comes from the volume of the sphere with radius <math>n_f</math>
Solving for <math>n_f</math> and then substituting into <math>E_f = \frac{\hbar^2 \pi^2}{2 m L^2} n^2 \,</math> yields
:<math>n_f=\left(\frac{3 N}{\pi}\right)^{1/3} </math>
and then finally
:<math>E_f =\frac{\hbar^2 \pi^2}{2m L^2} \left( \frac{3 N}{\pi} \right)^{2/3}</math>
In order to get this in terms of number density, we substitute L<sup>2</sup> with V<sup>2/3</sup>.
===(C)===
'''Estimate the Fermi energy of this neutron star in Joules'''
===(D)===
'''Using the fact that 1eV = <math>1.60217646 \times 10^{19} </math>Joules, and that 1meV ~ 10K, estimate the Fermi temperature of this neutron star. How does it compare to that of a typical nucleus, where <math>E_F ~ 30MeV</math>?'''
===(E)===
'''Given that the typical temperature of this neutron star is T ~ <math>10^6</math> K, is this neutron star "cool" or is it "hot"(as compared to the Fermi temperature)?'''

Latest revision as of 20:43, 17 March 2011

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