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# SOLID STATE I PHYS 845

Clemson

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This 15 page Class Notes was uploaded by Joshuah Kunde on Saturday September 26, 2015. The Class Notes belongs to PHYS 845 at Clemson University taught by Staff in Fall. Since its upload, it has received 38 views. For similar materials see /class/214241/phys-845-clemson-university in Physics 2 at Clemson University.

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Date Created: 09/26/15

First Test Phys 845 Prof Daw due in class 28 Oct 2004 You may use Marder and your class notes and no other reference 1 33 points Given a 2D rectangular lattice With c a 2 see Marder7s Figure 13 You7re also given that the average electron density is 1 elec tronunit cell Calculate the Fermi wavevector assuming a nearly free elec tron gas Try to make an accurate drawing of the Fermi surface 2 33 points Repeat Marder7s Problem 6 from Chapter 87 but set 7393 12 In this case7 calculate and plot the bands without using a cal culator or computer 3 34 points Assume that an electron moves in a 1D periodic potential of this form V V0 cos 27rxa Construct a solution using a planewave basis consisting of just the three smallest reciprocal vectors GL1 727Ta G0 Girl 27ra With this basis7 set up the reduced hamiltonian Find the energies at the F pt Chapter 1 Dimensional Analysis The rst step in modeling any physical phenomena is the identi cation of the relevant vari ables and then relating these variables via known physical laws For suf ciently simple phenomena we can usually construct a quantitative relationship among these variables from rst principles however for many complex phenomena which often occur in engineering applications such an ab initio theory is often dif cult if not impossible In these situations modeling methods are indispensable and one of the most powerful modeling methods is dimensional analysis You have probably encountered dimensional analysis in your previous physics courses when you were admonished to check your units77 to ensure that the left and right hand sides of an equation had the same units so that your calculation of a force had the units of kg ms2 In a sense this is all there is to dimensional analysis although checking units77 is certainly the most trivial example of dimensional analysis incidentally if you arent in the habit of checking units do itl Here we will use dimensional analysis to actually solve problems or at least infer some information about the solution Much of this material is taken from Refs 1 and 2 Ref 3 provides many interesting applications of dimensional analysis and scaling to biological systems the science of allornetry The basic idea is the following physical laws do not depend upon arbitrariness in the choice of the basic units of measurement In other words Newton7s second law F ma is true whether we choose to measure mass in kilograms acceleration in meters per second squared and force in newtons or whether we measure mass in slugs acceleration in feet per second squared and force in pounds As a concrete example consider the angular frequency of small oscillations of a point pendulum of length l and mass m w 11 where g is the acceleration due to gravity which is 98 ms2 on earth in the 81 system of units see below To derive Eq 11 one usually needs to solve the differential equation which results from applying Newton7s second law to the pendulum do itl Let7s instead deduce 11 from dimensional considerations alone What can u depend upon It is reasonable to assume that the relevant variables are m l and 9 it is hard to imagine others at least for a point pendulum Now suppose that we change the system of units so that the unit of mass is changed by a factor of M the unit of length is changed by a factor of L and the unit of time is changed by a factor of T With this change of units the units of frequency will change by a factor of T l the units of velocity will change by a factor of LT l and the units of acceleration by a factor of LT Z Therefore the units of the quantity gl will change by T4 and those of gl12 will change by T l Consequently the ratio n L 12 9l is invariant under a change of units H is called a dimensionless number Since it doesnt depend upon the variables mgl it is in fact a constant Therefore from dimensional considerations alone we nd that i 9 w 7 const gtlt 13 A few comments are in order 1 the frequency is independent of the mass of the pen dulum bob a somewhat surprising conclusion to the uninitiated 2 the constant cannot be determined from dimensional analysis alone These results are typical of dimensional analysisiuncovering often unexpected relations among the variables while at the same time failing to pin down numerical constants lndeed to x the numerical constants we need a real theory of the phenomena in question which goes beyond dimensional considerations 11 Units Before proceeding further with dimensional considerations we rst need to discuss units of measurement 111 The SI system of units In this course we will adopt the SI system of units1 which is described in some detail in the Physicist s Desk Reference 4 which I will abbreviate as PDR from now on pp 4710 In the SI system the base or de ned units are the meter m the kilogram kg the second s the kelvin K and the ampere A The de nitions of these units in terms of fundamental physical processes are given in the PDR All other units are derived For instance the SI unit of energy the joule J is equal to 1 kg mZsz The derived units are also listed in the PDR The Sl system is referred to as a LMT class since the de ned units are length L mass M and time T if we add thermal and electrical phenomena then we have a LMTQI class in the SI system 1In some older texts this is referred to as the MKS systemi 2We should also add the mole mol and the candela cd but these will seldom enter into our models 112 Atomic units The Sl system of units is not the only possible choice however and often the choice of units is dictated by the scale of the physical phenomena under consideration For instance in studying phenomena at the atomic and molecular level it is useful to choose a unit of length which is comparable to the size of an atom We do this by using the electron charge e as the unit of charge and the electron mass me as the unit of mass At the atomic level the forces are all electromagnetic and therefore the energies are always proportional to 624713960 which has dimensions MLgT Z Another quantity that appears in quantum physics is h Planck7s constant divided by 27139 which has dimensions MLZT l Dimensional analysis then tells us that the atomic unit of length is hZ 0529 10 10 14 7715624713960 X m QB This is called the Bohr radius or simply the bohr because in the Bohr model it is the radius of the smallest orbit for an electron circling a xed proton One can similarly nd the unit of time by dimensional analysis Rather than do this directly we nd rst the atomic unit of energy which is e2 1 e2 2 m h mg H60 436 gtlt10 18 J 272 eV 15 where eV electron volt 16 gtlt 10 19 J this is the kinetic energy acquired by an electron which is accelerated from rest through a potential difference of one Volt The energy Eh is called the hartree and is twice the ground state energy of an electron circling a xed proton The unit of time is then hEh It is worth noting that in atomic units the unit of velocity is aBhEh eZ47T60h the ratio of this velocity to the speed of light is a dimensionless number called the ne structure constant a 2 6 MN m i 16 hc 137 The fact that this number is small is important when treating the interaction of radiation with matter 12 Dimensions dimensional homogeneity and inde pendent dimensions Returning to the discussion above recall that if the units of length are changed by a factor of L and the units of time are changed by a factor of T then the units of velocity change by a factor of LT l We call LT l the dimensions of the velocity it tells us the factor by which the numerical value of the velocity changes under a change in the units within the LMT class Following a convention suggested by Maxwell we denote the dimensions of a 3 Table 11 Dimensions of some commonly encountered physical quantities in the LMTQI c ass L Length L M Mass M T Time T 1 Velocity LT l a Acceleration LT Z F Force MLT Z p Mass density ML S p Pressure ML lT Z a Angle 1 E Energy MLZT Z 6 Temperature 6 S Entropy MLZT ZQ l I Electric current I Q Electric charge IT E Electric eld MLT SI l B Magnetic eld MT ZI l physical quantity b by ab thus7 1 LT l A dimensionless quantity would have a5 1 1 e7 its numerical value is the same in all systems of units within a given class What about more complicated quantities such as force From Newton7s second law7 F ma7 so that F a MLT Z Proceeding in this way7 we can easily construct the dimensions of any physical quantity some of the more commonly encountered quantities are included in Table 11 We see that all of the dimensions in the Table are power law nwnomz39als7 of the form in the LMT class as OL ZMZ TC 17 where C and abc are constants ln fact7 this is a general result which can be proven mathematically see Sec 14 of Barenblatt7s book7 Ref This property is often called dimensional homogeneity and is really the key to dimensional analysis To see why this is useful7 consider again the determination of the period of a point pendulum7 in a more abstract form We have for the dimensions to T l7 g LT Z7 l L7 and M If u is a function of glm7 then its dimensions must be a power law monomial of the dimensions of these quantities We then have M T l glalllblmlc LT72aLch L W T Z ZMC 18 with a b and c constants which are determined by comparing the dimensions on both sides of the equation We see that ab0 72a71 00 19 The solution is then a 12 b 712 c 0 and we recover Eq 12 A set of quantities a1ak is said to have independent dimensions if none of these quantities have dimensions which can be represented as a product of powers of the dimensions of the remaining quantities As an example the density ML g the velocity LT l and the force MLT Z have independent dimensions so that there is no product of powers of these quantities which is dimensionless3 On the other hand the density velocity and pressure ML lT Z are not independent for we can write p pHng ie ppvz is a dimensionless quantity Now suppose we have a relationship between a quantity a which is being determined in some experiment which we will refer to as the governed parameter and a set of quantities a1an which are under experimental control the governing parameters which is of the form afa1akak1an 110 where a1 ak have independent dimensions For example this would mean that the dimensions of the governed parameter a is determined by the dimensions of a1ak while all of the ass with s gt k can be written as products of powers of of the dimensions of a1 ak eg ak11afa2 would be dimensionless with p r an appropriately chosen set of constants With this set of de nitions it is possible to prove that Eq 110 can be written as a a aafa2 lta kmgt 111 with ltlgt some function of dimensionless quantities only The great simpli cation is that while the function f in Eq 110 was a function of n variables the function ltlgt in Eq 111 is only afunetion ofn 7 k variables Eq 111 is a 1 t t t t of B quot L s 11 Theorem which is the central result of dimensional analysis The formal proof can again be found in Barenblatt7s book Dimensional analysis cannot supply us with the dimensionless function ltlgtiwe need a real theory for that As a simple example of how this works let7s return to the pendulum but this time well assume that the mass can be distributed so that we relax the condition of the mass being concentrated at a point The governed parameter is the frequency v the governing parameters are g l which we can interpret as the distance between the pivot point and the center of mass m and the moment of inertia about the pivot point I Since I MLZ the set g m l I is not independent we can choose as our independent parameters gm l as before with Iml2 a dimensionless parameter In the notation developed above n 4 3Prove this formally by writing Flambwc 1 and then show that the only solution is a b c 01 Alternatively show that it is impossible to write F plumb for any a bi and k 3 Therefore7 dimensional analysis tells us that w o 112 with lt1gt some function which cannot be determined from dimensional analysis alone we need a theory in order to determine it 13 Examples Some of the preceding discussion may have seemed a little abstract 1 will try to esh it out a bit with several examples 131 Oscillations of a star A star undergoes some mode of oscillation How does the frequency w of oscillation depend upon the properties of the star The rst step is the identi cation of the physically relevant variables Certainly the density p and the radius R are important well also need the grav itational constant G which appears in Newton7s law of universal gravitation We could add the mass m to the list7 but if we assume that the density is constant as a rst approximation7 then m p47TR33 and the mass is redundant Therefore7 w is the governed parameter7 with dimensions w T l7 and pRG are the governing parameters7 with dimensions p ML fl7 R L7 and G M lLST Z check the last one You can easily check that AR G have independent dimensions therefore7 n 37 k 37 so the function lt1gt is simply a constant in this case Next7 determine the exponents w T l PlalRlblGlc MaicL73ab30T72o 113 Equating exponents on both sides7 we have 1700 73ab300 72071 114 Solving7 we nd a c 12 b 07 so that w C Gp 115 with C a constant We see that the frequency of oscillation is proportional to the square root of the density7 and independent of the radius Once again7 the determination of 0 requires a real theory of stellar oscillation7 but the interesting dependence upon the physical parameters has been obtained from dimensional considerations alone 132 Gravity waves on water Next consider waves on the surface of water which are called gravity waves or sometimes capillary waves How does the frequency to depend upon the wavenurnber4 k of the wave The relationship to wk is known as the dispersion relation for the wave The relevant variables would appear to be pg k which have dirnensions p ML B g LT 2 and k L lg these quantities have independent dirnensions so it 3 k 3 Now we can determine the exponents lwl T l Plalglblklc MaL73abicT72b7 so that a0 i3abic0 72b71 117 with the solution a 0 b c 12 Therefore to 0 118 with 0 another undeterrnined constant We see that the frequency of water waves is propor tional to the square root of the wavenurnber in contrast to sound or light waves for which the frequency is proportional to the wavenurnber This has the interesting consequence that the group velocity of these waves is 119 awak 02M while the phase velocity is at wk C gk so that 119 v 2 Recall that the group velocity describes the large scale lurnps77 which would occur when we superirnpose two waves while the phase velocity describes the short scale wavelets inside the lurnps For water waves these wavelets travel twice as fast as the lurnps You might worry about the effects of surface tension 039 on the dispersion relationship We can include these in our dimensional analysis by recalling that the surface tension is the energy per unit area of the surface of the water so it has dimensions a MT The dimensions of the surface tension are not independent of the dimensions of pg k in fact it is easy to show that a pHgHsl z so that UkZpg is dirnensionless Then using the same arguments as before we have w g Up 1 119 with ltIgt sorne undeterrnined function A calculation of the dispersion relation for gravity waves starting from the fundamental equations of uid mechanics 5 gives u g k 1Uk2pg 120 4Recall that k 27rA with A the wavelength of the wave 7 so that our function s is ltIgtx 1 121 Dimensional analysis enabled us to deduce the correct form of the solution7 ie7 the possible combinations of the variables Of course7 only a complete theory could provide us with the function What have we gained We originally started with to being a function of the four variables p97k70 what dimensional analysis tells us is that it is really only a function of the combination UkZpg7 even though we don7t know the function Notice that this is an important fact if you are trying to measure the dependence of w on the physical parameters p97k70 If you needed to make say 10 separate measurements on each variable while holding the others xed7 then without dimensional analysis you would naively need to make 104 separate measurements Dimensional analysis tells you that you only really need to measure the combinations gk and UkZpg7 so only need to make 102 measurements to characterize w Dimensional analysis can be a labor saving devicel 133 Energy in a nuclear explosion We next turn to a famous example worked out by the eminent British uid dynamicist G l Taylor5 In a nuclear explosion there is an essentially instantaneous release of energy E in a small region of space This produces a spherical shock wave7 with the pressure inside the shock wave thousands of times greater that the initial air pressure7 which may be neglected How does the radius R of this shock wave grow with time t The relevant governing variables are E7 t7 and the initial air density p0 with dimensions MLZT Z7 t T7 and p0 ML g This set of variables has independent dimensions7 so 71 37 k 3 We next determine the exponents lRl L Elalpolbltlc MabL2a73bT72ac so that ab07 2173b17 72ac0 123 with the solution a 157 b 715 c 25 Therefore R OE15p315t25 124 with C an undetermined constant If we could plot the radius R of the shock as a function of time t on a log log plot7 the slope of the line should be 25 The intercept of the graph would provide information about the energy E released in the explosion7 if the constant 0 could be determined By solving a model shock wave problem Taylor estimated 0 to be 5Taylor s name is associated with many phenomena in uid mechanics the RayleighTaylor instability7 SaffmanTaylor fingering7 Taylor cells7 Taylor columns7 etc 8 about 1 he was able to take de classi ed movies of nuclear tests and using his model infer the yield of the bombs This data of course was strictly classi ed it came as a surprise to the American intelligence community that this data was essentially publicly available to those well versed in dimensional analysis 134 Solution of the diffusion equation Dimensional analysis can also be used to solve certain types of partial differential equations If this seems too good to be true it isn7t Here we will concentrate on the solution of the diffusion equation This material is optional but you might nd it interesting if you have some background in differential equations We7ll start by deriving the one dimensional diffusion or heat equation6 Let Tt represent the temperature of a metal bar at a point x at time t 1711 use 739 to avoid confusion with the symbol for the dimension of time T The rst step is the derivation of a continuity equation for the heat ow in the bar Let the bar have a cross sectional area A so that the in nitesimal volume of the bar between a and xA is A Am The quantity of heat contained in this volume is cpTA An with C17 the speci c heat at constant pressure per unit volume it has dimensions cpl ML lT ZQ l In a time interval dt this heat changes by an amount cp676tA Azdt due to the change in temperature This change in the heat must come from somewhere and is the result of a ux of heat q t through the area A q is the heat owing through a unit area per unit time Into the left side of the volume an amount of heat qut ows in a time dt on the right hand side of the volume a quantity Aq 6q6xAxldt ows out in a time dt so that the net accumulation of heat in the volume is 7A6q6xAxdt Equating the two expressions for the rate of change of the heat in the volume A Ax we nd 3739 3q cpa 7 a 125 which is the equation of continuity It is a mathematical expression of the conservation of heat in the in nitesimal volume A An We supplement this with a phenomenological law of heat conduction known as Fourier s law the heat ux is proportional to the negative of the local temperature gradient heat ows from a hot reservoir to a cold reservoir 3739 q fag 126 with n the thermal conductivity of the metal bar The thermal conductivity is usually measured in units of W cm 1 K l and has dimensions a MLT 30 1 See the Table on p 38 of the PDR for the thermal conductivities of some materials Combining Eqs 125 and 126 we obtain the di usion equation often called the heat equation 3739 32739 E Dy 127 51 prefer the term diffusion equation since we are just describing the diffusion of heat where D Hcp is the thermal dz usz39m39ty of the metal bar it has dimensions D KCp LZT l as it should Eq 127 is the diffusion equation for heat The diffusion equation will appear in many other contexts during this course It usually results from combining a continuity equation with an empirical law which expresses a current or ux in terms of some local gradient Suppose that the bar is very long so that we can consider the idealized case of an in nite bar At an initial time t 0 we add an amount of heat Q0 with dimensions Q0 MLZT Z at some point of the bar which we will arbitrarily call x 0 We could do this for instance by brie y holding a match to the bar The heat is conserved at all times so that 00 CpA Txtdx Q0 128 00 How does this heat diffuse away from x 0 as a function of time t ie what is Tzt7 We rst identify the important parameters The temperature 739 certainly depends upon 2 t and the di fusivity D we see from Eq 128 that it also depends upon the initial conditions through the combination Q E QoCPA What are the dimensions We have L t T D LZT l and Q L0 so that n 4 These dimensions are not independent for the quantity xDt12 is dimensionless so that k 3 We will choose as our independent quantities t D Q Now express 739 in terms of these variables Tl 9 lilallelQlc LZWTHQC 129 We nd 2bc0 aib0 01 130 which has the solution a 712 b 712 c 1 Therefore dimensional analysis tells us that the solution of the diffusion equation is of the form 1 Q 95 with ltIgt a function which we still need to determine The important point is that ltIgt is only a function of the combination mDt12 and not z and t separately To determine ltIgt let7s introduce the dimensionless variable 2 mDt12 Now use the chain rule to calculate various derivatives of 739 3739 7 Q 62dltIgtz amp Dal2 d2 7 Qdlt1gt2 7 E d2 132 627id2lt1gt2 133 Dt32 d2 10 g 1 Q Q 32 dlt1gtz 3 6t 2D12t32 2Dt1Zot dz dlt1gtz Z 1 Q Toms211 d2 13934 Substituting Eqs 133 and 134 into the diffusion equation 1277 and canceling various factors7 we obtain a differential equation for lt17 d2lt1gt2 z dlt1gtz 1 lt1gt 0 135 dz 2 dz 2 Dimensional analysis has reduced the problem from the solution of a partial di erential equa tion in two variables to the solution of an ordinary di erential equation in one variable The normalization condition7 Eq 1287 becomes in these variables f lt1gt2d2 1 136 foo You might think that Eq 135 is hard to solve however7 it turns out that it is an exact differential7 d dlt1gt 2 lt1gt 0 137 dz dz 2 l 7 which we can integrate once to obtain dlt1gt g 3113 const 138 However7 since any physically reasonable solution would have both lt1 a 0 and dlt1gtdz a 0 as x a 007 the integration constant must be zero We now need to solve a rst order differential equation7 which we do by dividing Eq 138 by lt17 multiplying by dz7 and integrating7 with the result that lnlt1gt 7224 const7 or ltIgtz 05124 139 with C a constant To determine 0 we use the normalization condition7 E017 136 0 e ZZ ldz C47r12 1 140 where the integral known as a Gaussian integral can be found in integral tables Therefore 0 147T12 Returning to our original variables7 we have Q zz 4D T7t we t This is the complete solution for the temperature distribution in a one dimensional bar due to a point source of heat7 7For the mathematically sophisticated7 llll mention that the same solution can be obtained using the method of Fourier transforms applied to the diffusion equation 11 14 Similarity modeling and estimating The notion of similarity is familiar from geometry Two triangles are said to be similar if all of their angles are equal7 even if the sides of the two triangles are of different lengths The two triangles have the same shape the larger one is simply a scaled up version of the smaller one This notion can be generalized to include physical phenomena This is important when modeling physical phenomenaifor instance7 testing a prototype of a plane with a scale model in a wind tunnel The design of the model is dictated by dimensional analysis Return to the mathematical statement of the l l Theorem7 Eq 111 We can identify the following dimensionless parameters Hazl H1 142 and so on7 such that Eq 111 can be written as H ltlgtll1 l39likk 143 The parameters 11111 7Hwy are known as similarity parameters Now if two physical phenomena are similar7 they will be described by the same function ltlgt Denote the similarity parameters of the model and the prototype by the superscripts m and p7 respectively Then if the two are similar7 their similarity parameters are equal H HYquot 115ka HEEL 144 so that HQ HM 145 Therefore7 in order to have an accurate physical model of a prototype7 we must rst identify all of the similarity parameters7 and then insure that they are equal for the model and the prototype Finally7 we come to estimating In this course we will often make order of magnitude estimates7 where we try to obtain an estimate to within a factor of ten sometimes better7 sometimes worse This means that we often drop factors of two etc7 although one should exercise some caution in doing this Estimating in this fashion is often aided by rst doing some dimensional analysis Once we know how the governed parameter which we are trying to estimate scales with other quantities7 we can often use our own personal experience as a guide in making the estimate More examples of this later in the course especially elastic similarity Bibliography 1 G l Barenblatt7 Dimensional Analysis New York Gordon and Breach Science Pub lishers7 1987 2 H L Langhaar7 Dimensional Analysis and the Theory of Models New York Wiley7 1951 3 T A McMahon and J T Bonner7 On Size and Life New York Scienti c American Library7 1983 4 H L Anderson edl7 A Physicist s Desk Reference New York American Institute of Physics7 1989 5 L D Landau and E M Lifshitz7 Fluid Mechanics New York Pergamon Press7 19597 61 6 G l Taylor7 The formation of a blast wave by a very intense explosion H The atomic explosion of 19457 Proc Roy Soc London A2017 159 1950

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