PHYSICS 165: First Long Exam H Quizzagan1 1 Bachelor of Science in Physics Department of Physical Sciences University of the Philippines Baguio Email:
[email protected] 1. A container of volume V contains N molecules of a gas. We assume that the gas is dilute so that the position of any one molecule is independent of all other molecules. Although the density is uniform on average, there are fluctuations in the density. Divide the volume into two parts V1 and V2 with V = V1 + V2 . Let us illustrate the said problem as follows: Figure 1: Gas molecules in volume V .
We can try to treat this as a Random Walk Problem such that a molecule is either on V1 or in V2 but the former location does not affect that later one. For this to work, we have to check the following definition of variables: p = probability that a molecule is in V1 (1) q = probability that a molecule is in V2
(2)
N = total number of molecules f ound in V
(3)
N1 = number of molecules f ound in V1
(4)
N2 = number of molecules f ound in V2
(5)
(a) (2 points) What is the probability p that a molecule is in the volume V1 ? SOLUTION: Note that in the problem, the position of any one molecule is described as statistically independent of the others which implies that the probability that a given molecule is located anywhere in the volume V is equally likely. From the figure, we know that V = V1 + V2
(6)
Hence, if we use the definitions described in (1) to (5), then we know that N = N1 + N2
1
(7)
Hence, by simple statistics, the probability p that a molecule can be found in V1 is simply the number of chances by which a molecule is in V1 divided by the total number of possibilities that it can be found anywhere in the volume V . We can represent this as: p=
V1 V1 = V1 + V2 V
(8)
(b) (2 points) What is the probability that N1 molecules are in V1 and N2 molecules are in V2 , where N = N1 + N2 ? SOLUTION: Let us declare that WN (N1 ) denotes the probability that there is an N1 number of molecules that can be found in V1 an that WN (N2 ) is the probability of finding N2 number of molecules in V2 . Note that we can treat this with a Binomial Distribution which we know to be denoted as WN (N1 ) =
N! pN1 q N2 N1 ! · N2 !
(9)
Furthermore, we should note that for the definition we stated on (1) and (2), we have the following condition: p+q =1 (10) but there is no assurance that p = q. Hence, we can’t utilize this assumption. Hence, for this case, we have the following: WN (N1 ) =
N! pN1 q (N −N1 ) N1 ! · (N − N1 )!
(11)
which describes the probability that N1 molecules are in V1 . WN (N2 ) =
N! q N2 p(N −N2 ) N2 ! · (N − N2 )!
(12)
which denotes the probability that N2 molecules are in V2 . We can rewrite (11) and (12) using the obtain expression from (8) which leads us to: N! WN (N1 ) = N1 ! · (N − N1 )!
N! WN (N2 ) = N2 ! · (N − N2 )!
V1 V
N1
V1 1− V
V1 1− V
N2
V1 V
(N −N1 ) (13) (N −N2 )
(c) (3 points) What is the average number of molecules in each part? SOLUTION: When solving for average or mean, we recall that, generally, PM f (ui )P (ui ) f (u) = i=1 PM i=1 P (ui )
(14)
(15)
but we know that the denominator, which is the sum of all probabilities for all possibilities (molecule is in either V1 or V2 ) is simply 1. Hence, we have, f (u) =
M X
f (ui )P (ui )
(16)
N1 W (N1 )
(17)
i=1
Using (13) on (16), we have N1 =
N X i=0
2
N X
N! N1 = N1 N ! · (N − N1 )! 1 i=0
V1 V
N1 (N −N1 ) V1 1− V
(18)
We know from Chapter One in Reif’s textbook that this leads to a simplified expression of the form: N1 = pN (p + q)N −1 (19) but we know from (10) that this simplifies further to N1 = pN
(20)
which describes the average number of molecules that can be found in V1 . Similarly, N2 = qN
(21)
We simply use (8) to rewrite (20) and (21):
V1 V
N1 = N2 =
N
(22)
V1 1− N V
(23)
(d) (3 points) What are the relative fluctuations of the number of molecules in each part? SOLUTION: In order to compute for the relative fluctuations of the number of molecules in each part, we simply 2 calculate for the dispersion (∆N1 )2 and divide it with N1 . Again, from Chapter One, we know that (∆x)2 = (x − x)2 = x2 − x2 (24) We reconfigure this for us to use (∆N1 )2 = N12 − N1
2
(25)
We have already computed in Chapter One that this leads us to (∆N1 )2 = N pq
(26)
Now, we simply rewrite (26) using (8) and (10). Hence, we have V1 V1 (∆N1 )2 = N 1− V V
(27)
Finally, to get the relative fluctation, we have (∆N1 )2 N1
2
N =
V1 V
V1 1− V N1
(28)
2
We evaluate (28) using (22). Hence, (∆N1 )2 N1
2
=
V1 N1 1 − V N1
2
1− =
V1 V
N1
(29)
Similarly, (∆N2 )2 N2
2
3
=
V1 1− V N2
(30)
(e) (2 points) What happens to the relative fluctuation in V1 when V1 V ? SOLUTION: V1 We know from (29) the relative fluctuation. If V1 V , then the term → 0. Hence, (29) V becomes (1 − 0) (∆N1 )2 1 = = (31) 2 N N 1 1 N1 (f) (2 points) What value should the (N1 − N1 )2 assume when V1 V ? SOLUTION: We know that V −1 V1 2 2 (N1 − N1 ) = (∆N1 ) = N pq = N 1− V V Hence, if V1 V , then the term
(32)
V1 → 0. Hence, we have: V (N1 − N1 )2 → N (0)(1 − 0)
(33)
(N1 − N1 )2 → 0
(34)
2. Consider an isolated system consisting of a large number N of every weakly interacting localized particles 1 of spin . Each particle has a magnetic moment µ which can point either parallel or antiparallel to an 2 applied field H. The energy E of the system is then E = −(n1 − n2 )µH, where n1 is the number of spins aligned parallel to H and n2 is the number of spins aligned antiparallel to H. SOLUTION: Before we start solving the problem, let me note that I adopted the recipe described in Chapter Two of Reif in analyzing systems similar to this. Firstly, we have to describe the state of the entire system. Note that for each particle in the system, the quantum number m which tells us about the spin orientation of each particle can only assume the values m=±
1 2
(35)
as for each particle, it has only two possible spin orientation which is either up or down. Hence, the state of the system is the quantum number m fir each of the N particles which we can denote as m1 , m2 , m3 , ..., mN
(36)
n1 = number of spins parallel to H
(37)
n2 = number of spins anti − parallel to H
(38)
N = n1 + n2
(39)
Let
(a) (5 points) Consider the energy range between E and E + δE where δE is very small compared to E but is microscopically large so that δE µH. What is the total number of states Ω(E) lying in this energy range? SOLUTION: We can treat this with the binomial distribution as we have two sets of possible values for our parameter, the quantum number m. Hence, we have the following number of distinct possibilities where you have n1 number of spins paralle and n2 number of spins anti-parallel to H: Φ(E) =
4
N! (n1 )!(n2 )!
(40)
Recall from Chapter Two in Reif that the total number of states Ω(E) in the range between E and E + δE is expressed as Ω(E) = Φ(E + δE) − Φ(E) =
∂Φ δE ∂E
(41)
wherein Φ(E) is the total number of possible sets of values of the m quantum numbers which we denoted in (40). Therefore, δE N! (42) Ω(E) = (n1 )!(n2 )! 2µH We note that E = −(n1 − n2 )µH
(43)
E = −[n1 − (N − n1 )]µH = −[2n1 − N ]µH
(44)
Using (39), we can rewrite (43) to be
By simple transpositions, we know from (44) that n1 =
N µH − E N E = − 2µH 2 2µH
(45)
Similarly, we can easily see that E = −[(N − n2 ) − n2 ]µH = −[N − 2n2 ]µH n2 =
(46)
N E E + N µH = + 2µH 2 2µH
(47)
We can therefore use (45) and (47) to rewrite (42) N! δE Ω(E) = E E N N 2µH − + ! ! 2 2µH 2 2µH
(48)
(b) (5 points) Write down an expression for ln Ω(E) as a function of E. Simplify this expression by applying Stirling’s formula in its simplest form. SOLUTION: We simply get the natural logarith of (48) and apply rules on natural logarithm. We have ln Ω(E) = ln
N! δE + ln (n1 )!(N − n1 )! 2µH
ln Ω(E) = ln N ! − ln n1 ! − ln (N − n1 )! + ln
(49) δE 2µH
(50)
Recall that Stirling’s formula gives us a way to approximate such a form of expression with ln n! = n ln n − n
(51)
Hence, we can rewrite (50) using (51) such that ln Ω(E) = N ln N − N − n1 ln n1 + n1 − (N − n1 ) ln (N − n1 ) + N − n1 + ln
δE 2µH
(52)
This gives us ln Ω(E) = N ln N − n1 ln n1 − (N − n1 ) ln (N − n1 ) + ln
5
δE 2µH
(53)
We can expand the third logarithmic term using Taylor’s series such that: y ln x + y = ln x + ln 1 + x
(54)
Now, we have h δE n1 i + ln ln Ω(E) = N ln N − n1 ln n1 − (N − n1 ) ln N + ln 1 − N 2µH
(55)
We now use DPMA on the third term on (55) to get n1 n1 δE ln Ω(E) = N ln N − n1 ln n1 − N ln N − N ln 1 − + n1 ln N + n1 ln 1 − + ln (56) N N 2µH The first and third term in (56) cancels out. We also use DPMA on the second and fifth term, as well as the fourth and sixth term. This leads us to n1 δE n1 + n1 ln N + n1 ln 1 − + ln (57) ln Ω(E) = −n1 ln n1 − N ln 1 − N N 2µH n1 δE ln Ω(E) = −n1 (ln n1 − ln N ) + (n1 − N ) ln 1 − + ln N 2µH
(58)
Using the rules of logarithm, we now have: ln Ω(E) = −n1 ln
n1 δE n1 + (n1 − N ) ln 1 − + ln N N 2µH
(59)
Recal from (45) that 1 2
E N− µH
(60)
Hence, we can use this to rewrite the terms in (59), such that, 1 E 1 N E N E 1 E δE ln Ω(E) = − N− ln − + − − N ln 1− +ln 2 µH 2 N N µH 2 2µH 2 N µH 2µH (61) This simplifies to 1 E 1 E 1 E 1 E δE ln Ω(E) = − N− ln 1− − N+ ln 1− + ln 2 µH 2 N µH 2 2µH 2 N µH 2µH (62) (c) (5 points) Assume that the energy E is in the region where Ω(E) is appreciable, i.e., that it is not close to the exteme possible value ±N µH which it can assume. In this case, apply a Gaussian approximation to part (a) to obtain a simple expression for Ω(E) as a function of E. SOLUTION: For this system, we note that N → ∞. For such a case, we know the the Gaussian distribution provides an approximate expression of the probability since this distribution has a pronounced maximum at a certain value. If we consider a region that is near this maximum point (near but not too close), then we can approximate the probability function as a continuous function of the continuous variable n1 with only the integral values having physical relevance. We recall from (4) that f (n1 ) =
N! (n1 )!(N − n1 )!
(63)
In this case, we expand the ln f (n1 ) because its slowly varying compared to just f (n1 ) and hence, we see that the power series of the natural logarithmic function converges more rapidly than for 6
the f (n1 ) which is what we want. Hence, we expand this around the maximum which we denote as n ˜ = n( max). We use Taylor series expansion to get 1 ln f (n1 ) = ln f (n1 ) + B1 η + B2 η 2 + ... 2
(64)
dk ln f (n1 ) Bk = dnk1 n1 =˜ n
(65)
such that
d f (n1 ) = 0 at this maximum dn1 point. We can also ignore the higher order derivatives in (64). Hence, we end up with Hence, we know then that the second term in (64) vanishes because
1 ln f (n1 ) = ln f (n˜1 ) + B2 η 2 , η = n1 − n ˜ 2
(66)
As we can see, we can just get the exponential value for each side to end up with a value of f (n1 ). e
ln f (n1 )
=e
ln f (n˜1 )+
1 B2 η 2 2
1 |B2 |η 2 f (n1 ) = f (n˜1 )e 2 −
(67)
(68)
You might be wondering why in (68) there is a negative exponent. As we can see from (65), B2 denotes the second derivative of f (n1 ) with respect to n1 evaluated at the maximum n˜1 . From Math 53, we know the the second derivative of a function tells us about the curvature at that point and in this case, the curvature at the maximum should be less than 0, hence B2 = −|B2 |
(69)
Note that f (n˜1 ) is integrated over all values of n1 and this should give us the sum total of all possibilities of spin direction. Therefore, we know that Z
+∞
1 |B2 |η 2 dη = 2N f (n˜1 )e 2 −
(70)
−∞
How do we determine the contant f (n˜1 )? Note that f (n1 ) and n1 could be treated as though they are quasicontinuous variables. Hence, the treatment we did above. Z f (n˜1 )
1 |B2 |η 2 2 dη = 2N e
+∞ −
(71)
−∞
How do we evaluate the integral above? We recall that Z +∞ 2 1√ e−ax xn dx = π, if a = 1 and x = 0 2 0
(72)
Hence, s
2π = 2N |B2 | r |B2 | n f (n˜1 ) = 2 2π
f (n˜1 )
7
(73)
(74)
We recall again from Math 53 that the first derivative of a function tells us about the slope of the tangent line at that point. In our case, the first derivative evaluated at the maximum should be equal to zero. We use the first and second term only of (59) to evaluate this. B1 =
d ln f (n1 ) d h n1 n1 i = −n1 ln + (n1 − N ) ln 1 − dn1 dn1 N N
Let us evaluate (75) term by term first term: d h n1 i d n1 n1 d −n1 ln = −n1 · ln + ln · (−n1 ) dn1 N dn1 N N dn1
(75)
(76)
Here, we used the multiplication rule in differentiation. 1 d h n1 i n1 = −n1 · · du − ln (−1) −n1 ln dn1 N u N where u=
1 n1 , du = N N
(77)
(78)
which implies that d d 1 N 1 n1 = ln u = du = ln · dn1 N du u n1 N
(79)
d h n1 i N 1 n1 n1 −n1 ln = −n1 − ln = −1 − ln dn1 N n1 N N N
(80)
Hence,
second term: d n1 d d n1 n1 = (n1 − N ) + ln 1 − (n1 − N ) ln 1 − ln 1 − (n1 − N ) dn1 N dn1 N N dn1 where we again used the multiplication rule of differentiation. let n1 du 1 u=1− , =− N dn1 N Hence, the first term in (80) can be rewritten and we have N n1 d n1 1 = (n1 − N ) + ln 1 − (1) (n1 − N ) ln 1 − − dn1 N N − n1 N N d n1 n1 (n1 − N ) ln 1 − = 1 + ln 1 − dn1 N N
(81)
(82)
(83)
(84)
We replace (80) and (84) into (75) and we get B1 = 0 = −1 − ln
n1 N − n1 + 1 + ln N N
(85)
We evaluate this at n1 = n˜1 . We rewrite (85) 0 = − ln n˜1 + ln N + ln (N − n˜1 ) − ln N = − ln n˜1 + ln (N − n˜1 )
(86)
This implies that ln n˜1 = ln (N − n˜1 )
(87)
Hence, n˜1 = N − n˜1 ⇒ n˜1 = 8
1 N 2
(88)
We use the expression obtained in (85) to get the second derivative which is simply dB1 1 1 B2 = =− − dn1 n1 N − n1 n1 =n˜1 = N 2 B2 = −
2 2 4 − =− N N N
(89)
(90)
Now we combine our results from (74) and (90) to finally rewrite (68): 1 |B2 | − |B2 |η2 2 ·e 2π
(91)
1 4 (n1 −n˜1 )2 4 1 − e 2N N 2π
(92)
r f (n1 ) = 2
N
r f (n1 ) = 2
N
2
r f (n1 ) = 2N Recall that Ω(E) = f (n1 )
2 e Nπ
−
N 2 n1 − N 2
(93)
δE N E , then using n1 = − , we can say that 2µH 2 2µH
r Ω(E) = 2N
2
2 E − 2 δE N 2µH e Nπ 2µH
(94)
3. Consider two spin systems A and A0 placed in an external field H. System A consists of N weakly 1 interacting localized particles of spin and a magnetic moment µ. Similarly, system A0 consists of N 0 2 1 and a magnetic moment of µ‘. The two systems are weakly interacting localized particles of spin 2 then placed in thermal contact with each other. Suppose that |b| 1 and |b0 | 1 so that ln Ω(E) = −
E2 2µ2 H 2 N
ln 2 +Nr N π 2
(95)
can be used for the densities of states of the two systems. (a) (3 points) Find the relation between the absolute temperature T and the total energy E of the initial state of system A. Do likewise for T 0 and E 0 of system A0 . SOLUTION: Recall that ∂ β= ln [Ω(E)] (96) ∂E Using (95), we have: 1 E β= − 2 2 (2E) + 0 = − 2 2 (97) 2µ H N µ H N and we know that β=
1 kT
(98)
Hence, 1 E µ2 H 2 N =− 2 2 ⇒E=− kT µ H N kT We also know that E0 = − 9
µ02 H 2 N 0 kT 0
(99)
(100)
(b) (3 points) In the most probable situation corresponding to the final thermal equilibrium, how is ˜ of system A related to the energy E˜0 of system A0 ? the energy E SOLUTION: In equilibrium, we know that: β = −β 0 ⇒ T = T 0 (101) Using our results from (99) and (100), we see that µ2 H 2 N − E kT = E0 µ02 H 2 N 0 − kT
(102)
µ2 N E = 02 0 0 E µ N
(103)
Simplified
(c) (3 points) What is the heat Q absorbed by system A in going from the initial situation to the final situation when it is in equilibrium with A0 ? SOLUTION: We know the energy is conserved which can lead us to the safe conclusion that E + E 0 = µN H + µ0 N 0 H
(104)
E = µN H + µN 0 H − E 0
(105)
From (103), we have E = µN H + µN 0 H −
Eµ02 N 0 µ2 N
(106)
Multiplying both sides of (106) by µ2 N , we obtain Eµ2 N + Eµ02 N 0 = (µN H + µ0 N 0 H)(µ2 N )
(107)
Applying DPMA then dividing both sides, we get E(µ2 N + µ02 N 0 ) = (µN H + µ0 N 0 H)(µ2 N ) E=
(µN H + µ0 N 0 H)(µ2 N ) (µ2 N + µ02 N 0 )
(108) (109)
Furthermore, we know that ∆E = Q − W ⇒ Q = ∆E
(110)
(µN H + µ0 N 0 H)(µ2 N ) − µN H (µ2 N + µ02 N 0 )
(111)
(µN H + µ0 N 0 H)(µ2 N ) − µN H(µ2 N + µ02 N 0 ) (µ2 N + µ02 N 0 )
(112)
µN Hµ2 N + µ0 N 0 Hµ2 N − µN Hµ2 N − µN Hµ02 N 0 ) (µ2 N + µ02 N 0 )
(113)
Q= We expand simplify (111) Q= Q=
As we can see, the first and third term of the numerator (113) cancels out and we are left with Q=
µN 0 Hµ2 N − µN Hµ02 N 0 ) (µ2 N + µ02 N 0 )
(114)
We can factor out terms from (114) such that Q=
(N 0 N H)(µ0 µ2 − µµ02 ) (µ2 N + µ02 N 0 ) 10
(115)
(d) (3 points) What is the probability P (E)dE that A has its energy in the range between E and E + dE? SOLUTION: Recall that P (E) α Ω(E)Ω0 (E0 − E), E0 = µN H + µ0 N 0 H (116) The total energy E0 is as we calculated (104) as E + E 0 . Now, if we get the exponential value of both sides of (95), then we get back to the function Ω(E). Let use the first term of that as our proportionality variable such that −
P (E) α e
E2 2µ2 H 2 N
−
e
(E0 − E)2 2µ2 H 2 N
(117)
For proportionalities, we introduce a proportionality constant to conver (117) to an equality such that E2 (E0 − E)2 − − 2µ2 H 2 N 2µ2 H 2 N P (E)dE = Ce (118) We then normalize using the condition to determine this constant such that
Z
+∞
−
Ce
(E0 − E)2 E2 − 2 2 2µ H N 2µ2 H 2 N
−∞
11
dE = 1
(119)
