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7"” YIO OWL [So IQE'VL” FCD 70A “Er—I EMA Problem 1.62. (The heat equation.) 30 Chapter 1 Energy in Thermal Physics Consider the segment of the rod between the two dashed lines, centered on point 3: with
width A27. The rate of heat ﬂow into this segment from the left is Q1__ 05”
E k ktAdm Am
1:2 (according to the Fourier heat conduction law), where A is the crossvsectional area of the
rod. Similarly, the rate of heat ﬂow into the segment from the right is Q2 _ dT
At _ +k‘A d2: 7
we with a plus sign because I’m deﬁning Q2 to be positive when heat ﬂows into the segment.
Thus the total rate of heat ﬂow into the segment is 2
Q C” )zktAAcci—T— _ :ktA dr At (12:2 ‘ 1+3;ﬂ d3 $“%
On the other hand, the heat input Q can be written as 0 AT, where C is the heat capacity
of the segment. The heat capacity, in turn, can be written as mc, where m is the mass of the segment and c is the speciﬁc heat. Therefore the previous equation becomes mc AT (PT But m/(A Am) is just the density of the rod, ,0, while AT/At is the derivative of the
segment’s temperature with respect to time. Finally, therefore, we can write g_ kt 62T_K62T
8t — pc 63:2 _ 8332’ with partial derivatives to denote the idea that T is now considered a function of both a:
and t. To check the trial solution T(ﬂ$,t)=To+£€_rz/4Kz ﬂ (where A now represents an arbitrary constant), we just evaluate its derivatives: 3T 1 3 2 A $2 2 A 1122 1 2 __ = __ — f2 7:1: MK: _ —x MK: : __ _ _ —a: /4Kt_
6t 2“” ‘3 +\/E(4Kt2)e ﬂ(4Kt2 2t)e ’
ﬂ : ﬁ(__§_)e_Iz/m. 83 ﬂ 2Kt ’ aZT A 1 2 a: 2 2
— = m __ 1 I41“ _ —ac MK:
62:2 Vii e + ( ) 6 J Problem 1.63 Multiplying the last line through by K and comparing to the ﬁrst line, we see that they
agree and so this trial solution does indeed work. What does it look like? Plotted as a
function of 50, it is a Gaussian “bell curve” that gets shorter and broader as It increases. To
plot it for four different t values I used the Mathematica instruction Plot [Evaluate [Table [Exp [x"2/ (410] /Sth [t] , {t , 1 ,4}]] , {x, ~10 , 10}] ; This produced the following graph: —lO —5 10
(I’m using units of 1', t, and T such that A and K are both numerically equal to 1.)
Physically, this solution corresponds to a “pulse” of excess energy that is placed at a: = 0
at time t = 0. At t = 0 the excess energy is all at the same place, so the pulse is inﬁnitely
high and narrow. It then spreads out according Fourier’s law, quickly at ﬁrst (when the
temperature gradients are large) and slowly later on (when the temperature gradients are small). As t a» 00 the pulse becomes inﬁnitely wide, as the rod’s temperature becomes
uniform. Problem 2.2. (Flipping 20 coins.) a Each coin has two ossible states, and the coins are independent so the total number
P ,
of microstates is 22” = 1048576, or a little over a million. (b) The sequence given corresponds to just one particular microstate. If the coins are fair
every microstate is equally probable, so the probability of any one of them, including
this, is 1/220 or a little less than one in a million. (And yet, amazingly, I got exactly
that sequence, on the ﬁrst try, when I was writing the problem!) (0) The number of ways of getting exactly 8 heads is 20 . .1. . . .' .
()22019 81716151413:125970. 8 87654 321 So the probability of getting exactly 8 heads is 125970 / 1048576 x: 12.0%. Problem 2.5. (Microstates of a small Einstein solid.) To represent each microstate I‘ll
use a sequence of digits, for the number of energy units in the ﬁrst, second, and third oscillators, respectively. (a) N = 3, q = 4:
400 310 031 220 211 040 301 103 202 121
004 130 013 022 1 12 1 count 15 microstates. And according to the formula, there should be 4+3—1 6! 65
=_m=~_—ﬁ 5.
( 4 ) 4121 2 1 (b) N=3, q=5: 500 410 041 320 032 311 221
050 401 104 302 203 131 212
005 140 014 230 023 113 122 I count 21 microstates. And according to the formula, there should be 5+3~1 71 76
( 5 )—ﬁ"__21‘ (0) N23, q=6: 600 501 015 042 141 033 132
060 150 420 204 114 321 213
006 051 402 024 330 312 123
510 105 240 411 303 231 222 I count 28 microstates. And according to the formula, there should be 6+3~1 8! 87
( 6 yams—48. Problem 2.13. (Fun with logarithms.)
(a) ealnb : (elnb)a : ba (b) ln[a + b) = ln [a(1 — 2)] = 111 a + 111(1 — 22) The second logarithm can be simpliﬁed using the approximation ln(1 + 59) z 3:, which is valid when Ix] << 1. In this case the
logarithm is just b/a, so ln(a + b) m 1na + (Ia/a). Problem 2.14. We want to solve the equation
81023 = 10‘” for 3:. A good ﬁrst step is to take the natural log of both sides. Then the left—hand side
becomes simply 1023, while the righthand side becomes In 103 = a: In 10. Therefore, 1023 _ 3 22
— 11110 4.34X 10 . 35' So 81023 2 104.34x1022 Problem 2.16. First note that for 1000 coins, the total number of possible outcomes
(microstates) is 21000. (a) The number of ways of getting exactly 500 heads and 500 tails is 1000) 1000! N 10001000€_1000\/27T  1000 R 21000 500 E (500!)2 N (5005008—500‘/27T.500)2 L 1/5007r' The probability is this divided by 21000, or simply 1/x/5007r = .025. So the chance of
getting exactly 500 heads is about 2.5%, or 1 in 40. (b) The number of ways of getting exactly 600 heads and 400 tails is 9(500) : ( 9(600)_ 1000 _ 1000! N 1000100064000w/27r1000
_ 600 _ 6001400! N 6005005*500\/27r  600 4004DDet4°°\/27r  400
10001000 : 60060040040w4807r‘ Again, the probability is this divided by 21000: H600): = 9) (a) 480W. 600600400400¢4807r 6005004004oox/4807r _ 6 4 Now all the factors can be evaluated on a calculator. The result is P(600) = 4.6x 10‘“,
much smaller than P(500). Problem 2.18. First note that [N — 1)! z: N !/N , since dividing by N cancels the ﬁnal
factor in NI, leaving just the ﬁrst N— 1 factors. Similarly, (q+N~ 1)! : (q+N)!/(q+N).
Thus, as the hint says, mm®=G+ZUZM+NW=M+MP N_ Now apply Stirling’s approximation to each of the factorials and cancel as many factors as
possible: mN)[email protected]+NVW€WMV%@+N) N [email protected]+NWW N
,q N qqe“91/27quNe‘N\/2NN (IiN Q"? NN 27TQ(9+N)I Finally, write (q + N)9+N as (q + N)q  (q + N)N, to obtain Hen/m Problem 2. 22. (Estimating the width of the multiplicity peak. ) (a) The number of energy units in Solid A can be 0, l, 2, etc ,up to 2N, so the total
number of macrostates is 2N + 1. (b) According to Problem 2.18, the multiplicity of any large Einstein solid is “(W : (g‘Z—N)q(iiv—N)N MINE?) Here we want to substitute 2N for both N and q, so the multiplicity reduces to Qtotal : (MYNU 2N+2N)2 2N _ _22N22N
271' 2N2; (2N+2N) STN (c) The multiplicity of just Solid A, when it has exactly half of the energy (q— — N), 18 N+N>N (N_N_+N)N _ N N /122NN
QA“(N V 271' N(N+N) 2 2 4TN Since the multiplicity of Solid B is the same, the multiplicity of the composite system
in this macrostate is just the square of 9A, 241V ”most likely = 47TN  ((1) Since the height of the peak is Qmostnkely While the area under the peak is (Item, the
width of the peak must be roughly Qtotal i 24N/V SWN _ 47TN _
Qmostlikely Lﬁ 24N/4TTN — V 87TN This is fairly large, but the width of the entire graph of 9 vs. 9,, is 2N + 1, so the fractional width of the peak is \/ 27rN/ (2N) = «MW/(2N) ~ 1/\/IV. This is quite small:
when N = 1023, the fractional width is less than 10““, or one part in 100 billion. 27TN. Problem 2.24 49 Problem 2.24. (Multiplicity of a large two—state paramagnet.)
(a) The most likely macrostate is NT 2 N1 = N /2. At this point the multiplicity is _ N! N! N NNe “Nx/EWN _2N 2
max — NleL! =(N02 N ((% )N/ze —N/2 27rN/2)2 — WN'
(b) By Stirling’s approximation, the multiplicity is
(2—— N! N NNe"N\/277N NN N NTlNl!Nvale‘Nu/Q—erT‘NF‘e—N1/21TN'1=NNTNNL 27rN1N1' If we set N7 = (N/2) + :3, then N1 = (JV/2) H 9:, and the multiplicity can be written At this point it becomes simpler to work with the logarithm of the multiplicity: 1119 =NlnN— gangY —$2] —~:cln(% +33) +xln(% — 2:) wiser—$21 So far I haven’ t assumed anything about the size of 3 relative to N. But if :1: < N
we can expand each of the logarithms containing two terms. For example Item =1n(%)+1n[1(3$)l ~2me1— (t1 111C} i =3) = Inez) +111 [1 a: 33"] ”111(g) 3: 2% With these approximations, the logarithm of the multiplicity becomes Similarly, N 23:2 N 25:2 N 2:122
1119 NlnN—Nln§+ WHxlng—F+x1n_2___l\7
1 {IV—1N 2:1:2
+n 271' n2+N3
2
=N1n2_2i+1n 2 _%2_1"_ (C)
(d) The last term is much smaller than the others and can be neglected. Exponentiating
the remaining terms gives 922N1/;2Ne‘212/N (fora:<<N). This is a Gaussian function, peaked at :c = 0, where its value agrees with the result
of part (a). The Gaussian function falls off to Us of its peak value when 2232/N : 1 or a: = N / 2.
So the full width of the peak would be twice this, or m. For N = 106, the half—width of the peak in the multiplicity function would be m
or about 700. So an excess of 1000 heads puts us only a little beyond the point where
the Gaussian has fallen off to Us of its maximum value. I Wouldn’t be surprised to
obtain approximately this many heads, though I might be surprised to get an excess of
exactly 1000. On the other hand, an excess of 10,000 heads lies far outside of the peak
in the multiplicity function. At this point the Gaussian has fallen off to (3—200 m 10‘8"r
of its maximum value. If I got a result anywhere close to this, I would be quite certain
that the coins were not fair. ...
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 Spring '10
 N.Darnton

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