Chen C.J. Physics of Solar Energy

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290  Spherical Trigonometry

Figure B.1 The spherical trian-

gle. A great circle is deﬁned by the

intersection of a plane that cuts the

sphere in two equal halves with the

sphere. Given three points A, B, and

C on the sphere, three great circles can

be deﬁned. The three arcs AB, BC,

and CA, each less than 180

◦

or π in

rad, form a spherical triangle.

B.2 Cosine Formula

In planar trigonometry, there is a cosine formula

= b

+ c

− 2bc cos A. (B.1)

In spherical trigonometry, a similar formula exists:

cos a =cosb cos c +sinb sin c cos A. (B.2)

When the arcs are short and the spherical triangle approaches a planar triangle, Eq.

B.2 reduces to Eq. B.1. In fact, for small arcs,

cos b ≈ 1 −

, (B.3)

and

sin b ≈ b, (B.4)

and so on. Substituting Eqs B.3 and B.4 into Eq. B.2 reduces it to Eq. B.1.

Here we give a simple proof of the cosine formula in spherical trigonometry by an

analogy to that in planar trigonometry. To simplify notation, we set the radius of the

sphere OA = OB = OC = 1. By extending line OB to intersect a line tangential to

AB at a point D,wehave

AD =tanc; OD = sec c. (B.5)

Similarly, by extending line OC to intersect a line tengential to AC at at a point E,

we have

AE =tanb; OE = sec b. (B.6)

From the planar triangle DAE, using the planar cosine formula,

= AD

+ AE

− 2 AD · AE cos



DAE

=tan

c +tan

b − 2tanb tan c cos A.

(B.7)

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B.3 Sine Formula  291

Figure B.2 Derivation of cosine for-

mula. The derivation is based on the pro-

jection of a spherical triangle onto a plane

and the cosine formula in planar trigonome-

try. A straight line tangential to arc AB inter-

sects the extension of line OB at D. Another

straight line tangential to arc AC intersects the

extension of line OC at E. By applying the co-

sine formula in planar trigonometry on trian-

gles ODE and ADE, after some brief algebra,

the corresponding cosine formula is obtained.

Similarly, for the planar triangle DOE,

= OD

+ OE

− 2 OD · OE cos



DOE

= sec

c + sec

b − 2 sec b sec c cos a.

(B.8)

Subtrating Eq. B.7 from B.8, and notice that

sec

b =1+tan

b, sec

c =1+tan

c, (B.9)

we obtain

2 − 2 sec b sec c cos a = −2tanb tan c cos A. (B.10)

Multiplying both sides with cos b cos c yields

cos a =cosb cos c +sinb sin c cos A. (B.11)

Similarly, for vertices B and C,

cos b =cosc cos a +sinc sin a cos B; (B.12)

cos c =cosa cos b +sina sin b cos C. (B.13)

B.3 Sine Formula

In planar trigonometry, there is the sine formula

sin A

sin B

sin C

. (B.14)

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292  Spherical Trigonometry

In spherical trigonometry, a similar formula exists,

sin a

sin A

sin b

sin B

sin c

sin C

. (B.15)

Obviously, for small arcs, Eq. B.15 reduces to Eq. B.14.

To prove Eq. B.15, we rewrite Eq. B.2 as

sin b sin c cos A =cosa − cos b cos c. (B.16)

Squaring, we obtain

sin

b sin

c cos

A =cos

a − 2cosa cos b cos c +cos

b cos

c. (B.17)

The left-hand side can be written as

sin

b sin

c − sin

b sin

c sin

A, (B.18)

1 − cos

b − cos

c +cos

b cos

c − sin

b sin

c sin

A. (B.19)

Hence,

sin

b sin

c sin

A =1− cos

a − cos

b − cos

c +2cos

a cos

b cos

c. (B.20)

Deﬁne

Z =

sin

a sin

b sin

1 − cos

a − cos

b − cos

c +2cos

a cos

b cos

. (B.21)

Equation B.20 can be written as

sin a

sin A

= ±

√

Z. (B.22)

By deﬁnition, in a spherical triangle, the sides and the vertical angles are always smaller

than 180

◦

. Therefore, in Eq. B.22, only a positive sign is admissible. Because Z is

symmetric to A, B and C, we obtain

sin a

sin A

sin b

sin B

sin c

sin C

. (B.23)

B.4 Formula C

We rewrite the cosine formula B.11 in the following form and use Eq. B.13:

sin b sin c cos A =cosa − cos b cos c

=cosa − cos b (cos a cos b +sina sin b cos C)

=cosa sin

b − cos b sin a sin b cos C.

(B.24)

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B.4 Formula C  293

Dividing both sides by sin b, one obtains formula C

sin c cos A =cosa sin b − sin a cos b cos C. (B.25)

Similarly,

sin a cos B =cosb sin c − sin b cos c cos A, (B.26)

sin a cos C =cosc sin b −sin c cos b cos A, (B.27)

and so on.

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294  Spherical Trigonometry

Problems

B.1. Show that if one of the arcs is 180

◦

, then no spherical triangle can be constructed.

B.2. If one of the vertex angles of a spherical triangle, for example, C, is a right angle,

show that for small arcs the cosine formula leads to the Pythagorean theorem.

B.3. Using the cosine and sine formulas, show that

cot a sin b =cotA sin C +cosb cos C, (B.28)

cot c sin a =cotC sin B +cosa cos B, (B.29)

and so on.

B.4. For a rectangular spherical triangle, where C=90

◦

, show that

sin a =sinc sin A, (B.30)

sin b =sinc sin B, (B.31)

tan a =tanc cos B, (B.32)

tan b =tanc cos A, (B.33)

tan a =sinb tan A, (B.34)

tan b =sina tan B. (B.35)

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Appendix C

Quantum Mechanics Primer

Usually, introductory quantum mechanics starts with Schr¨odinger’s equation and using

partial diﬀerential equations as the mathematical tools. For example, the hydrogen

atom problem is resolved with spherical harmonics and Laguerre polynomials. A possi-

ble shortcoming with this approach is that the readers become submerged in pages and

pages of mathematical formulas and lose the conceptual understanding of the physics.

Historically, before Erwin Schr¨odinger discovered the partial diﬀerential equation for-

mat, Heisenberg and Pauli developed the algebraic approach of quantum mechanics,

and resolved several basic problems in quantum mechanics, including harmonic oscil-

lator, angular momentum, and the hydrogen atom. From a pedagogic point of view,

the succinct notation of the algebraic approach, especially the Dirac notation, could be

conceptually more directly related to the underlying physics. From a practical point of

view, to handle the problems with the utilization of solar energy, analytic approach for

partial diﬀerential equations is not useful because numerical calculations and pertur-

bation methods are the norm. Furthermore, the more advanced methods of quantum

mechanics, such as quantum electrodynamics, rely on the algebraic method rather than

the partial diﬀerential equation method.

This Appendix is a brief summary of the algebraic approach to quantum mechanics,

exempliﬁed by the problems of the harmonic oscillator, angular momentum, and hy-

drogen atom. For clarity, we use the Dirac notation, and adding a hat on an operator

to distinguish it from a (in general complex) number.

C.1 Harmonic Oscillator

The Hamiltonian of a one-dimensional harmonic oscillator is

H =

ˆp

mω

ˆq

, (C.1)

where the momentum ˆp and coordinate ˆq satisfy the commutation relation

[ˆp, ˆq] ≡ ˆpˆq − ˆqˆp = i. (C.2)

We introduce a pair of operators, the annihilation operator,

295

Physics of Solar Energy C. Julian Chen

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296  Quantum Mechanics Primer

ˆa =

√

2mω

ˆp + i



mω

2

ˆq, (C.3)

and the creation operator

ˆa

†

√

2mω

ˆp − i



mω

2

ˆq, (C.4)

the meanings of these terms well be clariﬁed soon. Using the commutation relation C.2,

we have

H = ωˆa

†

ˆa +

ω, (C.5)

and the commutation relation

[ˆa, ˆa

†

] ≡ ˆaˆa

†

− ˆa

†

ˆa =1. (C.6)

To ﬁnd the eigenstates and energy levels of the Hamiltonian C.1,

H|n = E

|n, (C.7)

it is suﬃcient to ﬁnd the eigenstates and eigenvalues of the operator ˆa

†

ˆa,

ˆa

†

ˆa|n = u

|n, (C.8)

and E

=(u

)ω. As a consequence of Eq. C.6, if |n is an eigenstate with

eigenvalue u

,thenˆa|n is also an eigenstate

ˆa

†

ˆa ˆa|n =(u

− 1) ˆa|n (C.9)

with eigenvalue u

− 1. Because n|ˆa

†

ˆa|n must not be negative, there must be an

eigenstate with minimum value, u

= 0. For such a state,

ˆa

†

ˆa|0 =0. (C.10)

On the other hand, if |n is an eigenstate with eigenvalue u

,thenˆa

†

|n is also an

eigenstate

ˆa

†

ˆa(ˆa

†

|n)=(u

+1)(ˆa

†

|n) (C.11)

with eigenvalue u

+ 1. Starting with |0, by applying ˆa

†

many times, we have

ˆa

†

ˆa



ˆa

†



|0 = n



ˆa

†



|0. (C.12)

Therefore, we conclude that the eigenvalues of the operator ˆa

†

ˆa are non-negative

integers, and the eigenstates can be constructed from the state with a zero eigenvalue,

|n = C



ˆa

†



|0. (C.13)

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C.2 Angular Momentum  297

The ﬁnal step is to ﬁnd the normalization constant C

. The zeroth-order state is

by deﬁnition normalized,

0|0 =1. (C.14)

By applying ˆa

†

many times, we have

0|(ˆa)



ˆa

†



|0 = n!. (C.15)

Therefore, C

=(n!)

−1/2

, and ﬁnally

|n =

√



ˆa

†



|0. (C.16)

Those are also eigenstates of the harmonic oscillator problem,

H|n =



n +



ω |n. (C.17)

The energy level of the harmonic oscillator is thus quantized, with energy quanta ω.

The operator ˆa

†

adds an energy quanta to the oscillator, thus the name creation op-

erator; the operator ˆa removes an energy quanta from the oscillator, thus the name

annihilation operator. These operators play essential roles in quantum electrodynamics,

or the bona ﬁde quantum theroy of radiation.

C.2 Angular Momentum

The deﬁnition of angular momentum in quantum mechanics is similar to that in classical

mechanics, except that the order of coordinate r and momentum p is ﬁxed,

m =

r ×

p, (C.18)

or in tensor notation,

ˆm

= 

ijk

ˆx

ˆp

, (C.19)

where the unit axial tensor 

ijk

is a tensor antisymmetric to all three suﬃxes, with



123

= 1, and changes sign by exchanging two identical indices, where the value is zero.

Asumoverj and k is implied.

To simplify notation, a dimensionless version of the angular momentum is deﬁned,

= 

−1

ˆm

= 

−1



ijk

ˆx

ˆp

. (C.20)

The commutation relations can be obtained from the commutation relations of mo-

mentum and coordinate,

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298  Quantum Mechanics Primer

[ˆp

, ˆx

]=iδ

, (C.21)

which are

−

= i

−

= i

−

= i

(C.22)

or in tensor form,

[

]=i

ijk

. (C.23)

From the components we can form an operator as the square of the modulus of the

angular momentum vector,

. (C.24)

As a result of commutation relations C.22,

commutes with a component, for example,

[

]=0. (C.25)

Therefore, we can ﬁnd states |l, m which are simultaneously eigenstate of

and

|l, m= λ|l, m,

|l, m= μ|l, m.

(C.26)

In the following, we introduce a pair of operators which are similar to the creation

and annihilation operators in the problem of harmonic oscillators,

+ i

−

− i

(C.27)

which have the commutation relations

[

, (C.28)

[

−

]=−

−

(C.29)

and the identity

−

. (C.30)

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C.2 Angular Momentum  299

Using arguments similar to those in the harmonic oscillator problem, we ﬁnd that

|l, m is also an eigenstate of

with eigenvalue μ + 1 and

−

|l, m is an eigenstate of

with eigenvalue μ −1. Applying those operators many times, we have





|l, m =(μ + n)





|l, m (C.31)

and



−



|l, m =(μ − n)



−



|l, m. (C.32)

However, because of Eq. C.24, the eigenvalue of

cannot grow indeﬁnitely. Its absolute

value must have a maximum. Because of the symmetry, the absolute value of the

positive maximum and the absolute value of the negative maximum must beequal.

Also, because the diﬀerence between the positive maximum and the negative maximum

must be an integer, both must be one-half of an integer. Assigning this number as l,

the possible eigenvalues of

, often assigned as m,mustbe

m = −l, −l +1, −l +2, ...l − 2,l− 1,l. (C.33)

with

l =0,

, 1,

, 2,

, 3 ... (C.34)

It is obvious that l, the maximum absolute value of

, is also a quantum number

for the total angular momentum

. In the following, we will ﬁnd the eigenvalue of the

operator

.Becausem = l is the maximum eigenvalue of

, one must have

|l, l =0. (C.35)

In view of Eq. C.30,

|l, l =



−



|l, l = l(l +1)|l, l. (C.36)

Because l and m are independent, ﬁnally we have the eigenvalues and eigenstates for

the angular momentum operator,

|l, m= l(l +1) |l, m,

ˆm

|l, m= m |l, m.

(C.37)