Reading Quiz

Question 1:

Consider rolling two fair six-sided dice, and adding the result of the numbers that turn up. What would be a microstate description of this system? What would be a macrostate description of this system?

Answer:

The microstate could be what numbers came up on each die. The macrostate could be the sum of those numbers.
  1. microstate description= the value of die 1, and the value of die 2 macroastate description= the sum of the the two values
  2. Say one die is purple and the other is red. A microstate description tells you the number you rolled on each die (say, purple - 4, red - 1). A macrostate description tells you the total number rolled (in the above microstate, 5).
  3. A microstate description of this system would be, say, the first die was a 3 and the second die was a 6. A macrostate description of this same system would be that a 3 and a 6 were rolled, with no statement about which die had which number on it.
  4. A microstate description would give the which dice gave which number. The macrostate description would just be the two numbers that resulted from the rolls without specifying which dice they came up on.
  5. A microstate description is a description of each possible outcome including constituent die rolls. A macrostate is a total kind of description, such as the sum of the die rolls without the results of each die.
  6. a macrostate description would be the sum of the numbers of the die (2 through 12). A micro state system is the combination of numbers used to get a numerical sum (for example a 4 and a 3 as opposed to a 2 and a 5).
  7. microstate description: rolled a 3 and a 5 macrostate description: got an 8
  8. A microstate description lists each possible outcome of the roll. For instance, two and three is a microstate because the state of each die is given. A macrostate description tells how many of each number are rolled but does not tell which die has that number.
  9. A microstate would be {outcome of die 1, outcome of die 2}, i.e. {1,3} or {6,7}. A macrostate would be something like "rolling a 4" or "rolling two 3's."
  10. A microstate description of the system would be to list in detail all of the possible combinations for the results of both die. ie 1,6. 2,4. 6,1. etc. A macrostate would be to describe them by the number that turns up for the total. Such as 5, 19, 4 or 12.
  11. Microstate description would describe one of the possible outcomes by rolling both dice. Examples: 5 and 2; 3 and 4; etc... Macrostate description would just be the sum of the two numbers. Example, and integer between 2 and 12, such as 7, or perhaps 9.

Question 2:

Flip 10 coins. What is the multiplicity of the macrostate with 8 heads?

Answer:

This is "8 choose 2", which from eq. (2.6) is 10!/(8!*(10-8)!) = 45.
  1. 45
  2. omega = 10 choose 8 = 45
  3. The multiplicity of the macrostate with 8 heads is 45.
  4. The multiplicity of the macrostate with 8 heads is 45.
  5. Using [2.6], 10 choose 8 is => 10!/(8!*(10-8)!) = 10*9 / 2 = 45.
  6. the multiplicity is 45 based on eq 2.6
  7. choosing 8 heads out of 10 coins --> 45 possible microstates multiplicity - 45
  8. 10factorial/(8factorial * 2) = 45
  9. The multiplicity is 10 choose 8 = 10 x 9 / 5 = 90 / 5 = 18.
  10. Using equaiton 2.6 I get a multiplicity of 45.
  11. The total number of microstates that would give you 8 heads. 10!/8!(10-8)! 45

Question 3:

How many oscillators are there in an Eintein solid with 2000 atoms?

Answer:

Each atom is really 3 oscillators, so there are 6000 oscillators total. And while we could study any number N of oscillators, on N's divisible by 3 are physically meaningful.
  1. 3*2000=6000 6000 oscillators
  2. Each atom corresponds to 3 oscillators (one for each dimension), so 6000.
  3. An Einstein solid with 2000 atoms has 6000 oscillators.
  4. There are 6000 oscillators in an Einstein solid with 2000 atoms.
  5. As on page 54, for N oscillators there are N/3 atoms. Since we have 2000 atoms, 2000=N/3, N=6000.
  6. 6000 oscillators
  7. Each atom corresponds to three oscillators (one in each dimension), so 6000 oscillators.
  8. 6000 because each atom oscillates in three independent directions.
  9. 6000 oscillators
  10. If there are N oscillators, there is only N/3 atoms. Thus 2000 atoms corresponds to 6000 oscillators.
  11. 6000

Question 4:

What is the multiplicity for 1 atom with 4 units of energy?

Answer:

Using q = 4 (energy units) and N = 3 (3 oscillators for each atom) in eq. (2.9), we obtain a multiplicity of 15.
  1. (4+1-1)!/(4!0!)=4!/4!=1
  2. omega(1,4) = 4 choose 4 = 1
  3. The multiplicity for 1 atom with 4 units of energy is 15.
  4. The multiplicity for 1 atom with 4 units of energy is 15.
  5. Using [2.9], (q+N-1) choose (q) [where q is units of energy, N is oscillators, our case q=4, N=3]. Then (4+3-1) choose (4) = (6) choose (4) = 6!/(4!*(6-4)!)=6*5/2= 15.
  6. one, there is only one way to distribute 4 units of energy among 1 particle (all of the energy on that particle.
  7. 1 atom --> 3 oscillators 4 energy units multiplicity = 15
  8. 15
  9. N = 3, q = 4, multiplicy = 15
  10. using equation 2.9 I get a multiplicity of 15
  11. (4+1-1)!/4!*(1-1)! 1

Question 5:

What material from the reading or previous classes would you like me to go over in more detail?

Answer:

Your responses below.
  1. none
  3. I can't think of anything right now.
  4. Most of this material seems to be pretty straight forward.
  5. Nope
  6. not particulary,
  7. The 'dot and line' graphical representation as a proof of the formula for the multiplicity of an Einstein solid was a little confusing.
  8. the dot/line proof of the multiplicity formula for an Einstein solid
  11. This Einstein model stuff is a little confusing, i'm just having trouble visualizing what this is physically.