Mathematical Physics I (Fall 2021): Midterm Exam Solution

Oct. 23, 2021

[total 20 pts, closed book/cellphone, no calculator, 90 minutes]

1. (a) [1 pt] Find the interval of convergence of the following power series (x is a real number).Be sure to investigate the endpoints of the interval.

∞∑n=2

(−1)n(x+ 1)n

2n lnn

(b) [2 pt] The Doppler frequency shift formula is written as ν ′ = ν(1∓ v

c

)−1when the source

is moving at speed v relative to the stationary receiver (negative sign when the source movestowards the receiver). Meanwhile, the relativistic version of the formula can be written as

ν ′ = ν(1± v

c

) (1− v2

c2

)− 12

(positive sign when the source moves towards the receiver). Show

that the two formulae agree in the classical limit — i.e., if terms of order v2

c2can be neglected.

• (a) ρ = limn→∞

∣∣∣ (x+1) lnn2 ln(n+1)

∣∣∣ = |x+1|2 < 1. The endpoints could be investigated in the same manner

as in Examples of Boas Chapter 1, Section 10, yielding the interval of convergence −3 < x ≤ 1.

Note that the series with x = −3 is divergent by the comparison test with∞∑n=1

1n .

• (b) Expanding the two functions in binomial power series up to order v2

c2, ν(1∓ v

c

)−1 'ν(

1± vc + (−1)(−2)

2!v2

c2

), while ν

(1± v

c

) (1− v2

c2

)− 12 ' ν

(1± v

c

) (1 + v2

2c2

)' ν

(1± v

c + 12v2

c2

).

2. (a) [1 pt] Find the disk of convergence of the following power series (z is a complex number).

∞∑n=0

(n!)3(z − 1 + i)n

(3n)!

1

(b) [2 pt] Verify that sinh−1z = ln (z ±√z2 + 1).

• (a) ρ = limn→∞

∣∣∣ (n+1)3(z−1+i)(3n+3)(3n+2)(3n+1)

∣∣∣ = |z−1+i|27 < 1.

• (b) sinh−1 z = ω → With u ≡ eω, you find z = sinhω = eω−e−ω

2 = u−u−1

2 → u2−2zu−1 = 0

→ ω = lnu = ln(z ±√z2 + 1) = sinh−1z.

3. (a) [2 pt] Solve the set of equations below by at least two out of three methods listed here: (i)by row reducing the augmented matrix, (ii) by using Cramer’s rule, (iii) by finding the inverseof the coefficient matrix. Clearly indicate which of the three methods you are using so the gradercould follow each of your answers easily.

x− 2y = 4

5x + z = 7

x+ 2y − z = 3

(b) [2 pt] The Pauli spin matrices used in quantum mechanics are

σ1 =

(0 11 0

), σ2 =

(0 −ii 0

), σ3 =

(1 00 −1

).

For σ2, show that eiθσ2 becomes a rotation matrix of an angle θ. Note that functions of matricesare defined by their power series expansions, e.g., eM = I +M + 1

2!M2 + 1

3!M3 + · · · .

(c) [2 pt] Rotate the quadric surface 7x2 + 4y2 + z2 − 8xz = 36 to principal axes (x′, y′, z′).What is the shortest distance from the origin to the surface? What is the name of this surface?Examples of quadric surfaces in 3 dimension include, but are not limited to, the following.

• (a-1) As in Example 1 of Boas Chapter 3, Section 2,1 −2 0 45 0 1 71 2 −1 3

→

1 −2 0 40 10 1 −130 4 −1 −1

→

1 −2 0 40 10 1 −130 0 −14

104210

→

1 −2 0 40 10 1 −130 0 1 −3

→

1 −2 0 40 1 0 −10 0 1 −3

→

1 0 0 20 1 0 −10 0 1 −3

2

• (a-2) As in Example 5 of Boas Chapter 3, Section 3,

x =

∣∣∣∣∣∣4 −2 07 0 13 2 −1

∣∣∣∣∣∣ /∣∣∣∣∣∣1 −2 05 0 11 2 −1

∣∣∣∣∣∣ =

∣∣∣∣∣∣4 −2 07 0 110 2 0

∣∣∣∣∣∣ /∣∣∣∣∣∣1 −2 05 0 16 2 0

∣∣∣∣∣∣ = −∣∣∣∣ 4 −210 2

∣∣∣∣ /(− ∣∣∣∣1 −26 2

∣∣∣∣) = 2, etc.

• (a-3) As in Example 3 of Boas Chapter 3, Section 6,

M =

1 −2 05 0 11 2 −1

→ det(M) = −14 and C =

−2 6 10−2 −1 −4−2 −1 10

→ M−1 =1

14

2 2 2−6 1 1−10 4 −10

• (b) As in Example 4 of Boas Chapter 3, Section 6, with σ22 = I and (iσ2)

2 = −I, you find

eiθσ2 = (cos θ)I + i(sin θ)σ2 = (cos θ)

(1 00 1

)+ i(sin θ)

(0 −ii 0

)=

(cos θ sin θ−sin θ cos θ

).

• (c) As in Example 2 of Boas Chapter 3, Section 12,

(x, y, z)

7 0 −40 4 0−4 0 1

xyz

= 36 →

∣∣∣∣∣∣7− λ 0 −4

0 4− λ 0−4 0 1− λ

∣∣∣∣∣∣ = 0 = (−1)2

∣∣∣∣∣∣4− λ 0 0

0 7− λ −40 −4 1− λ

∣∣∣∣∣∣ ,from which λ = 9, 4,−1 are acquired. Therefore, the new quadric surface equation relative tothe principal axes becomes

(x′, y′, z′)

9 0 00 4 00 0 −1

x′y′z′

= 36 → x′2

22+y′2

32− z′2

62= 1

that represents a hyperboloid of one sheet (일엽쌍곡면). At (x′, y′, z′) = (±2, 0, 0), the surfacehas the shortest distance to the origin, d = 2.

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4. (a) [1 pt] Show that

d

dx

∫ 2/x

1/x

coshxt

tdt = 0.

(b) [2 pt] Using the Lagrange multiplier method, find the shortest distance from the origin tothe line of intersection of the two planes 2x+ y − z = 1 and x− y + z = 2.

• (a) Using Leibniz’ rule in Eq.(12.13) of Boas Chapter 4,

d

dx

∫ 2/x

1/x

coshxt

tdt =

cosh 2

2/x

(−2

x2

)− cosh 1

1/x

(−1

x2

)+

∫ 2/x

1/xsinhxt dt

= −cosh 2

x+

cosh 1

x+

[coshxt

x

]2/x1/x

= 0.

• (b) From Thm.(9.20) of Boas Chapter 4, we write F = x2+y2+z2+λ1(2x+y−z)+λ2(x−y+z).Then, ∂F

∂x = 2x+ 2λ1 + λ2 = 0, ∂F∂y = 2y+ λ1 − λ2 = 0, and ∂F

∂z = 2z − λ1 + λ2 = 0. Combining

the equations with the constraints 2x+ y − z = 1 and x− y + z = 2 gives (x, y, z) = (1,−12 ,

12).

Thus, d =√x2 + y2 + z2 =

√62 .

5. [2 pt] In the integral

I =

∫ 1

0dy

∫ 1−y

0e(x−y)/(x+y)dx,

make the change of variables u = x− y and v = x+ y to show that I = 12 sinh(1).

• The new intervals of integration for u and v are found to be [−v, v] and [0, 1], respectively.With x = 1

2(v + u), y = 12(v − u), and

J =

∣∣∣∣∣∣∂x∂u

∂x∂v

∂y∂u

∂y∂v

∣∣∣∣∣∣ =

∣∣∣∣∣∣12

12

−12

12

∣∣∣∣∣∣ =1

2,

the integral I becomes∫ 1

0dv

∫ v

−veu/v|J |du =

1

2

∫ 1

0dv[veu/v]

∣∣∣∣v−v

=1

2

∫ 1

0v(e− e−1)dv =

1

4(e− e−1) =

1

2sinh(1).

6. (a) [1 pt] Prove that (A×B)× (C×D) = [A · (B×D)]C− [A · (B×C)]D. You are welcomedto tackle this problem in two different ways for an additional +1 point — that is, writing andproving in ordinary vector notation and in tensor notation.

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(b) [2 pt] A certain force field is given in spherical coordinates (r, θ, φ) by

F = er2P cos θ

r3+ eθ

P sin θ

r3(r ≥ P/2).

Calculate∮F · ds for a unit circle in the plane of θ = π/2. What does this indicate about the

force F being conservative or nonconservative? Then, examine ∇ × F. If you believe F maybe described by F = −∇ψ, find ψ(r, θ, φ); otherwise, simply state that no acceptable potentialexists. You may use the following formula to acquire the gradient and the curl in spherical

coordinates when the element of arc length is given by ds =3∑i=1

eihidxi:

∇u =

3∑i=1

ei1

hi

∂u

∂xiand ∇×V =

1

h1h2h3

∣∣∣∣∣∣∣∣h1e1 h2e2 h3e3∂∂x1

∂∂x2

∂∂x3

h1V1 h2V2 h3V3

∣∣∣∣∣∣∣∣ .

• (a-1) Using Thm.(3.9) and Eq.(3.3) of Boas Chapter 6, (A×B)× (C×D) = [(A×B) ·D]C−[(A×B) ·C]D = [A · (B×D)]C− [A · (B×C)]D = (ABD)C− (ABC)D.

• (a-2) Alternatively, writing in tensor notation and then using Eqs. (5.8) and (5.11) of BoasChapter 10, the ith component of the vector is [(A×B)× (C×D)]i = εijk(A×B)j(C×D)k =εijk · εjmnAmBn · εkpqCpDq = εjmn(εkijεkpq) ·AmBnCpDq = εjmn(δipδjq − δiqδjp)AmBnCpDq =εjmn(AmBnCiDj−AmBnCjDi) = [(A×B)jDj ]Ci− [(A×B)jCj ]Di = [(ABD)C− (ABC)D]i.

• (b-1) For a unit circle in the plane of θ = π/2 (i.e., dr = dθ = 0), you can easily find∮F · ds =

∫ 2π0 Fφrdφ = 0, which suggests — but does not prove — that F is conservative.

• (b-2) From the ∇×V formula given in the problem, you get

∇× F =1

r2sin θ

∣∣∣∣∣∣∣∣er reθ rsinθ eφ∂∂r

∂∂θ

∂∂φ

2P cos θr3

P sin θr2

0

∣∣∣∣∣∣∣∣ =1

r2sin θ· rsin θ eφ

[−2P sin θ

r3− −2P sin θ

r3

]= 0.

Thus Thm.(11.10) of Boas Chapter 6 states that F can be written as a gradient of a single-valued scalar potential, −∇ψ. From −∂ψ

∂r = 2P cos θr3

, −1r∂ψ∂θ = P sin θ

r3and − 1

rsinθ∂ψ∂φ = 0, you can

acquire ψ(r, θ, φ) = P cos θr2

.

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