PHZ 3113 Fall 2010
Homework #3, Due Monday, September 20
1. (a) Let f (x, y) be a function of x and y, with y a function of x and r. Write (∂f /∂x)r in
terms of (∂f /∂x)y , (∂f /∂y)x , and (∂y/∂x)r . Let f (x, r) be a function of x and r, with r a
function of x and y. Write (∂f /∂x)y in terms of (∂f /∂x)r , (∂f /∂r)x , and (∂r/∂x)y .
Start with df (x, y) and dy(x, r),
df =
∂f ∂x y
use these to write
df =
dx +
∂f ∂y x
dy,
dy =
∂y ∂x r
dx +
∂y ∂r x
dr;
∂f
∂f ∂y ∂f ∂y dx +
+
dr;
∂x y
∂y x ∂x r
∂y x ∂r x
and thus conclude
∂f ∂f ∂f ∂y =
+
.
∂x r
∂x y
∂y x ∂x r
Or, start with df (x, r) and dr(x, y),
df =
∂f ∂x r
write
df =
and thus conclude
dx +
∂f ∂r x
dr,
dr =
∂r ∂x y
dx +
∂r ∂y x
dy;
∂f ∂r ∂f ∂r ∂f dx +
+
dy;
∂x r
∂r x ∂x y
∂r x ∂x x
∂f ∂r =
+
.
∂x y
∂x r
∂r x ∂x y
∂f ∂f The form of this result, in yet another set of variables, is also presented in Problem 27,
Section 4.7 in the text.
(b) Let
x2 − y 2
f= 2
,
a + x2 + y 2
where a is a constant. Compute (∂f /∂x)y . Let r 2 = x2 + y 2 . Write f as a function of x and
r. Compute (∂f /∂x)r . Compare your results with the expressions found in (a).
For f (x, y) as given, the derivatives are
∂f 2x(a2 + 2y 2 )
= 2
,
∂x y (a + x2 + y 2 )2
∂f 2y(a2 + 2x2 )
=− 2
.
∂y x
(a + x2 + y 2 )2
(In this case, I will allow us to skip the details of taking the derivatives.)
For y 2 = r 2 − x2 ,
f (x, r) =
2x2 − r 2
,
a2 + r 2
∂f ∂x r
=
4x
,
2
(a + r 2 )
∂f ∂r x
=−
2r(a2 + 2x2 )
.
(a2 + r 2 )2
Compared with (a),
∂f ∂f ∂y x
2x(a2 + 2y 2)
2y(a2 + 2x2 )
+
=
−
× −
∂x y
∂y x ∂x r (a2 + x2 + y 2)2 (a2 + x2 + y 2 )2
y
4x
4x
∂f
= 2
= 2
=
.
2
2
2
∂x r
a +x +y
a +r
∂f ∂x r
+
∂f ∂r 2r(a2 + 2x2 ) x 4x
−
= 2
×
∂r x ∂x y (a + r 2 )
r
(a2 + r 2 )2
∂f 2
2
2
2
2x(a + 2r − 2x )
2x(a + 2y 2)
=
= 2
=
.
∂x y
(a2 + r 2 )2
(a + x2 + y 2)2
They should match, and they do.
2. The variables x and y are related to the variables u and v by x =
Write the Laplacian operator ∇2 f
∇2 f =
√
2u cos v, y =
√
2u sin v.
∂2f
∂2f
+
,
∂x2
∂y 2
in the variables u and v.
Write
∂f du +
dv
∂u ∂v
∂f ∂u ∂u ∂f ∂v ∂v =
dx +
dy +
dx +
dy
∂u
∂x
∂y
∂v
∂x
∂y
∂f
∂u
∂f
∂v
∂f
∂u
∂f
∂v
=
+
du +
+
dy.
∂u ∂x
∂v ∂x
∂u ∂y
∂v ∂y
df =
∂f Thus,
∂f ∂f ∂u ∂f ∂v =
+
∂x
∂u ∂x
∂v ∂x
∂f ∂f ∂u ∂f ∂v =
+
.
∂y
∂u ∂y
∂v ∂y
Iterate to obtain second derivatives,
∂2f
∂ ∂f ∂u ∂ ∂f ∂v ∂ ∂f =
+
=
∂x ∂x
∂u ∂x ∂x
∂v ∂x ∂x
∂x2
2
2
2
∂ f ∂u
∂ f ∂u ∂v ∂ 2 f ∂v 2
=
+2
+ 2
∂u ∂v ∂x ∂x
∂u2 ∂x
∂v ∂x
2
∂ ∂f
∂u
∂ ∂f
∂v
∂ f
∂ ∂f
=
+
=
∂y ∂y
∂u ∂y
∂y
∂v ∂y ∂y
∂y 2
∂ 2 f ∂u 2
∂ 2 f ∂u ∂v ∂ 2 f ∂v 2
+ 2
=
+
2
∂u ∂v ∂y ∂y
∂u2 ∂y
∂v ∂y
The inverse relations are u = 12 (x2 + y 2), v = tan−1 y/x, and the coordinate derivatives are
∂u
= x,
∂x
∂u
= y,
∂y
∂v
y
,
=− 2
∂x
x + y2
∂v
x
;
= 2
∂y
x + y2
and so
xy
∂2f
∂2f
∂2f 2
∂2f
xy
2
2
∇ f=
− 2
+ 2 =
(x + y ) + 2
+
∂u ∂v
∂x2
∂y
∂u2
x + y 2 x2 + y 2
∂2f
y2
∂2f
x2
1 ∂2f
+ 2
=
2u
+
+
.
∂v (x2 + y 2)2 (x2 + y 2 )2
∂u2 2u ∂v 2
3. The free energy F (T, V ) of an ideal gas is
V mkT 3/2
F = −NkT ln
,
N 2πh̄
where N, k, and h̄ are constants.
(a) From dF = −S dT − p dV , Compute S and p. Do you recognize the pressure you obtain?
∂F V mkT 3/2
3
S=−
+ ln
= Nk
∂T V
2
N 2πh̄
∂F NkT
p=−
=
∂V T
V
pV = NkT is the ideal gas law; if N = νN0 , where N0 is Avogadro’s number 6 × 1023 , and
N0 k =, then this is pV = νRT .
(b) The internal energy is given by U = F + T S. Compute U. Your answer might appear to
be simple, but show from dU that U should be a function of S and V . Write U as a function
of S and V . Compute the pressure from U.
The easy calculation is
V mkT 3/2
3
3NkT
V mkT 3/2
U = F + T S = −NkT ln
+ T × Nk
=
+ ln
.
N 2πh̄
2
N 2πh̄
2
However, from
dU = dF + d(T S) = dF + T dS + S dT = −S dT − p dV + T dS + S dT = T dS − pdV,
the internal energy is formally a function of S and V , not of T , and so T must be written
as a function of S and V ,
T =
2S
2πh̄ N 2/3
exp
−1 ,
mk V
3Nk
U=
2S
3πh̄ N 5/3
exp
−1 .
m V 2/3
3Nk
∂U 2U
NkT
=
.
p=−
=
∂V
3V
V
This is still the ideal gas law.
4. Find the point on the curve defined by
5 2
5
x + 3xy + y 2 = 1
2
2
that is closest to the point (x, y) = (1, 1).
The (squared) distance from the point (1, 1) to a point (x, y) in the plane is d2 = (x − 1)2 +
(y − 1)2 = f (x, y). Minimization of f subject to constraint φ = 0 is found by the method of
Lagrange multipliers (9.7) from
5
5
F = f + λφ = (x − 1)2 + (y − 1)2 + λ ( x2 + 3 xy + y 2 − 1).
2
2
Extrema or stationary points of F occur where the derivatives vanish,
∂F
= 2(x − 1) + λ (5x + 3y) = 0,
∂x
∂F
= 2(y − 1) + λ (3x + 5y) = 0,
∂y
5
∂F
5
= x2 + 3xy + y 2 − 1 = 0.
∂λ
2
2
These must all be satisfied simultaneously. One approach is to start by solving the first two
equation for x and y in terms of λ,
x=
1
,
1 + 4λ
y=
1
.
1 + 4λ
Then, the constraint ∂F/∂λ = φ = 0 gives two solutions for λ,
√
8
2
1
±
.
−
1
=
0,
λ
=
−
a,b
4
2
(1 + 4λ)2
which leads back to
1
1 (x, y)a,b = ± √ , ± √ .
2 2 2 2
These are not the only solutions; when λ = −1 the first two equations are not independent,
but both give 3x + 3y + 2 = 0, or y = − 23 − x. Then, the constraint φ = 0 gives
1
1
1
1 4
1
2x2 + x − = 0,
(x, y)c,d = − ± √ , − ∓ √ .
3
9
3 3 2 3 3 2
These are “extrema,” which can be minima, maxima, or saddle points. The distances are
9 √
9 √
11
d2a = − 2 = 0.8357,
= 3.6666.
d2b = + 2 = 3.6642,
d2c = d2d =
4
4
3
The first of these is the smallest, by far. Thus, the closest point is
1
1 (x, y)a = √ , √ .
2 2 2 2
The figure shows the locus of the constraint, an ellipse with axes at ±45◦ , and the√four
stationary points. Point a is the solution, the minimum of distance, with d2 √
= 94 − 2 =
9
2
0.0.8357; points c, d are maxima, d2 = 11
3 = 3.6666; and point b, with d = 4 + 2 = 3.6642,
is a local but not global minimum.