PHZ3400 Midterm Two Solution: Difference between revisions

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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 <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}} = e^{-\beta E} \;</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^{-\beta \Delta E}  \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 zero at the Curie Temperature (<math>T_c</math>). The <math>T_c</math> will be unique based on the ferromagnet.  The image below illustrates this behavior.
[[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] ]]
===(B) Finite External Magnetic Field===
When there is an external magnetic field, the magnetism does not drop to zero at the Curie temperature.  Rather, the magnetization of the ferromagnet will trail off as shown by the teal line in the image below.
[[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]]]
===Behavior around the Curie Temperature===
Even though the curie temperature is unique to each magnet, the behavior at the Curie point is similar for all of them.  When the magnet reaches its Curie temperature, the spins lose their alignment and become randomly oriented.  This prevents a magnetic field from forming.  For superconductors, the curie temperature is very low, which is why they are non-conducting at room 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) Density as a function of T===
'''What is the density of phonons with this frequency as a function of the temperature <math>T</math>?  Sketch this function. '''
[[Image:PhotonD.jpg|center|thumb|300px|This image represents the photon density n as a function of the temperature <math>T</math>]]
===(B) Density at Low Temperatures===
''' 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>
However, since as <math>T\to0 </math>  'e' becomes very large, we can disreguard the 1, thus the equation becomes:
<math>n(T) = e^{- \hbar \omega / kT} </math>
===(C) Density at High Temperatures===
'''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>
As <math>T\to\infty</math> 'e' becomes very small and we can make a substitution such that
<math> e^{- \hbar \omega / kT} \sim \tfrac{1}{1 + {\tfrac{\hbar \omega}{kT}}} </math>
Therfore we can say that <math>n(T) \sim \tfrac{kT}{\hbar \omega} \sim T. </math>
===(D) Temperature dependence of Resistivity===
'''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.'''
The Drude formula can be used to relate the resistivity and the phonon density through a few simple mathematical manipulations.  The Drude formula is as follows:
<math> \sigma = \tfrac{ne^2 \tau}{m} </math>
where <math>\sigma</math> = conductivity, <math>n</math> = phonon density, <math>m</math> = mass, and <math>\tau = \tfrac{l}{v}</math> where <math>l</math> = mean free path and <math>v</math> = velocity.
Since resitivity (<math>\delta</math>) and conductivity (<math>\sigma</math>) are inversely proportional, we can write that
<math>\tfrac{1}{\sigma} = \delta = \tfrac{m}{ne^2\tau}</math>
Now when <math>n(T) \sim \tfrac{1}{\tau}</math>, it is also proportional to T.  <math>n(T) \propto  T</math>
Thus when written together we have:
<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) Number Density===
'''Calculate the number density <math>n = \frac{N}{V}</math> of this star.'''
N = total number of objects and V = volume.  We can begin by calculating the volume of the neutron star.
Assuming that the neutron star is spherical:
<math>V =\frac{4}{3} \pi R^3 </math>
We are given that the radius is 10km which is equal to <math>1x10^4</math>m.  Plugging this in:
<math>V =\frac{4}{3} \pi (1x10^4m)^3 = 4.189x10^{12}m^3</math>
Now we can calculate N:
<math>N = \frac{M}{m}</math> where <math>M = 3x10^{30}kg</math> = mass of the neutron star and <math>m = 1.7x10^{-27}kg</math> = mass of a single neutron.  Plugging in we get the following:
<math>N = \frac{M}{m} = \frac{3x10^{30}kg}{1.7x10^{-27}kg} = 2.5x10^{57}</math>
Having both of these numbers, we can now calculate the number density (<math>n</math>)
<math>n = \frac{N}{V} = \frac{2.5x10^{57}}{4.189x10^{12}m^3} = 5.968x10^{44}\frac{1}{m^3}</math>
===(B) Fermi Energy and Number Density===
'''Derive in detail (as we did in class; show all the steps) the relation between Fermi energy and the number density.'''
We begin this derivation by saying that
<math>E(k) = \frac{\hbar ^2k_f^2}{2m}</math>
We can solve this for <math>k_f</math> such that
<math>k_f = \sqrt{ \frac{2mE_f}{\hbar ^2}}</math>
Keeping in mind that
<math>n = \frac {N}{V}</math>
<math>n = \frac{2}{L} \sum_{k < |k_F |} = \frac{2}{2 \pi } \sum_{k < |k_F |}\frac{2 \pi}{L} = \frac{2}{2 \pi} 2\int_0^{k_F} dk = \frac{2k_F}{\pi}</math>
Therefore Fermi Energy is
<math>E_f = \frac{\hbar^2 \pi^2 n^2}{8m}.</math>
===(C) Value of Fermi Energy ===
'''Estimate the Fermi energy of this neutron star in Joules.'''
We begin with the given equation:  <math>E_f = \tfrac{\hbar ^2}{2m}(\tfrac{3\pi ^2N}{V})^{2/3}</math>.
Since in part (A) we solved for the number density <math>n = \frac{N}{V}</math> we can replace <math>\frac{N}{V}</math> with <math>n</math> to get:
<math>E_f = \tfrac{\hbar ^2}{2m}(3\pi ^2n)^{2/3}</math>
Now we can plug in all the variables such that
<math>E_f = \tfrac{(1x10^{-34}\tfrac{m^2kg}{s}) ^2}{2(1.7x10^{-27}kg)}(3  \pi ^2  5.968x10^{44}\tfrac{1}{m^2})^{2/3}</math> .
By completing the calculations we can find that the solution is
<math> E_f = 1.131x10^{-68} \tfrac {m^2 kg}{s^2} = 1.995x10^{-11} J</math> .
===(D) Fermi Temperature===
'''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>?'''
Using what we calculated in part (D), we can now compute the unit change from J to eV and from eV to meV.
<math> \frac{1.13x10^{-11}J}{}  \frac{1eV}{1.60217646x10^{-19}J} \frac {1000meV}{1eV} = 1.245x10^{11}meV </math>
Since we know that 1meV ~ 10K, we can then multiply our <math>E_f</math> value by 10 and we get that the Fermi temperature of this neutron star is
<math> 1.245x10^{12} K </math>.
In MeV, the the Fermi Energy is
<math>E_f = 124.5MeV</math>.
This is very much greater than that of a typical nucleus which has a Fermi Energy of
<math>E_f ~ 30MeV</math>.
===(E) Temperature Comparison===
'''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)?'''
It is hot.
This neutron star has a <math>T_f</math> ~ <math>10^{12}</math> K which is several orders of magnitude greater than the temperature of a typical neutron star which has a temperature of T ~ <math>10^6</math> K.

Latest revision as of 20:43, 17 March 2011

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