Reading Quiz

Question 1:

Sound waves are much like light waves. We can use Bose-Einstein statistics in a similar fashion to what we used for photons and apply them to phonons. What are three important differences between light waves and sound waves?

Answer:

See the bulleted list on p. 308. Your responses below.
  1. Sound waves travel (much) slower than light, and speed depends on stiffness and density. Sound waves can be longitudinally or transversley polarized, while light is only Trnasversely polarized. Sound waves cannot have wavelengths shorter than twice the atomic spacing.
  2. 1. Sound waves are much slower than light. 2. Sound has longitudional polarization as well as transverse. 3. Sound wavelengths can only be 2x the size of the atomic spacing of their medium or larger.
  3. The differences in between light waves and sound waves are: sound waves travel faster than light waves; sound waves can be transversely and longitudinally polarized, whereas light waves can only be transversely polarized; sound waves have wavelengths no shorter than twice the atomic spacing in a solid, while light waves have arbitrarily short wavelengths in a solid.
  4. 1. Sound waves travel at less than the speed of light. 2. Sound waves have three polarization states as opposed to two. 3. Sound waves must have a wavelength of at least twice the atomic spacing.
  5. Sound waves are slower than light waves and their speed depends on the material. Sound waves have three polarizations instead of two (the third is longitudinal polarization). Thirdly, the lower limit of sound wavelength is double the distance between atoms.
  6. speed of sound << speed of light; sound waves have 3, not 2, polarizations; sound waves can not have arbitrarily small wavelengths
  7. 1) sound waves are slower then light waves 2) sound waves can be longitudinally polarized, light waves can't. 3) sound waves cannot have wavelengths shorter then twice the atomic spacing.
  8. Sound waves are MUCH slower Sound can be both transverse and longitudinal Sound waves have a lower limit on wavelength
  9. 1 > sound waves are much slower than light waves 2 > sound waves have three polarization axes, since they are longitudinal in addition to transverse 3 > there is a lower limit on the wavelength of a sound wave in a solid, that depends on the atomic spacing of the solid
  10. Sound waves travel much slower than light waves. Sound waves can be longitudinally polarized as well as transversely polarized while light waves are only transversely polarized. Light waves can have short wavelengths, sound waves in solids cannot have wavelengths shorter than twice the atomic spacing.
  11. 1. speed of waves 2. polarization 3. the relation of wavelength to atomic spacing

Question 2:

Debye's clever idea was to approximate a cube as an eighth of a sphere. According to the last paragraph on p. 309, "You can easily show that the radius of the sphere has to be" given by eq. (7.107). So show me. Note that for a solid with N atoms, each with three degrees of freedom, the total number of modes is 3N. So while the text states "To preserve the total number of degrees of freedom, he chose a sphere whose total volume is N", I might correct that to "total volume is 3N". Note that 3 is separate from any 3 that might come in from polarization arguments.

Answer:

We went over this in class. Your answers below.
  1. Since V= 4/3 * pi * r^3, and V=3N, 3N=4/3*pi*r^3. But really we are looking at a sphere whose (1/8) portion is of volume 3N. Hence, 3N=1/6*pi*r^3. [ Maybe the total volume of the 1/8th portion is supposed to be N? ] That aside, there should be a factor of three for the different polarizations (in the volume) so that things become 3N=1/2*pi*r^3 which gives the correct radius.
  2. Volume of cube=N=Volume of 1/8 sphere=(1/8)*4/3*pi*nmax^3=nmav^3*pi/6=N nmax=(6N/pi)^(1/3)
  3. We start with the volume of a sphere: V = (4/3) * Pi * r^3 But we want the volume of an eighth of a sphere, V/8 = (1/6) * Pi * r^3 From Figure 7.27, we can see that the eighth of sphere has volume N, N = (1/6) * Pi * r^3 moving the (1/6) and Pi over to the left side, we have 6N/Pi = r^3 Taking the cube root and realizing that r = n_max, n_max = (6N/Pi)^(1/3)
  4. 1/8*4/3*pi*R^3 = 3*N R^3 = 18/pi*N R = (18*N/pi)^(1/3)
  5. V_total = N = (1/8 of a spere in n-space) = (1/8)*(4/3)*pi*n^3 N = (1/6)(pi)n^3 6N/pi = n^3 n_max = (6*N/pi)^1/3
  6. V = N = 1/8 4/3 Pi n_max^3 = 1/6 Pi n_max^3 ==> n_max = (6 N/Pi)^(1/3)
  7. 3N= 4/3 pi r^3 9/4 N/pi= r^3 r= ( 9/4 * N/pi)^(1/3) ? there should be another factor of three in there for the polarization, maybe?
  8. Area of sphere is 4/3 Pi nmax^3. So eight is 1/6 Pi nmax^3 Set this equal to N and solve for nmax = (6N/Pi)^1/3
  9. V = 3N = (4/3)pi*R^3 (3/4)3N = pi*R^3 (9/4)N = pi*R^3 9N/4pi = R^3 R = (9N/4pi)^(1/3)
  10. V=(4*Pi/3) r^3 for a sphere. If the volume of the sphere as stated by Debye is N, then N=(4*Pi/3)r^3. We're dealing with 1/8 of a sphere, so N=(Pi/6)r^3. r=(6N/Pi)^1/3.
  11. eek i don't get this

Question 3:

(based on Problem 7.58). The speed of sound in copper is 3560 m/s. Use this value to calculate its theoretical Debye temperature, using TD as defined in eq. (7.111). (In a previous homework, you found that for copper, N/V = 8.47 x 1028 m-3.)

Answer:

Plugging numbers in, I get 466 K.
  1. T_D=h*c_s/(2*k)*(6N/ pi*V)^1/3, if N/V=8.47*10^28 m^-3, then T_D is: ((h * (3 560 (m / s))) / (2 * k)) * (((8.47 * ((10^28) (m^-3)) * 6) / pi)^(1 / 3)) = = 465.465105 kelvin
  2. TD=491k
  3. T_D = 465 K
  4. T_D = 465.4 K
  5. 465.3 K
  6. T_D = 465.5 K
  7. T_D=(h c_s)/(2 k) * (6N/pi V)^(1/3) (the T cancels) which numerically gives me 465.342 thanks, calculator!!
  8. at room temperature TD = 9.3*10^-18 (this seems wrong)
  9. T_D = 465.5 K
  10. 930.7 Kelvin
  11. 465K

Question 4:

Lot of neat difficult stuff in this reading and recently. What material from this reading or previous classes would you like me to go over in more detail?

Answer:

Your responses below.
  3. In the text, on page 309, he says that he chose a sphere of total volume N. You said just before (2) of this reading quiz that you chose a sphere of total volume 3N. In the footnote of Figure 7.27, it is implied that the volume of 1/8 of the sphere is N. The third case is the only one that led me to the correct answer. Which interperetation of the volume of the sphere is correct?
  5. I'm not completely clear about why the volume of an eigth-sphere is a valid approximation for the cube volume. Isn't this trying to compare ~(1/2)r^3 to r^3? More of an application-type question: I once saw an energy level diagram for a fluorescence process in which vibrational levels of the electronic excited state decayed to the lowest vibrational level through emission of "phonons" (most diagrams just say "non-radiative relaxation"). Now it makes sense, but in thinking about it, does sonication function on the reverse process? basically phonon-induced vibrational excitation without increasing temperature because the water bath maintains isothermal conditions?
  7. I'm not really comfortable with the idea of phonons as "particles" of sound. But I do like the word.
  8. From graph 7.29 it doesn't look THAT much different than the Einstein model. It's obvious that Debye is better for Low Temp, but what about medium temp. Are there any models that work for the intermediate vaules of temperature.
  9. What subtlety am I missing in the calculation of the radius of the n-sphere? How does my 9/4 resolve to a 6?
  10. This reading wasn't too difficult. I had more problems with the last one. Especially with the derivation that we did in class the other day. ICE-31, part A. This is a problem I can take care of in office hours though.
  11. I don't get the volume part, how does this work?