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Astronomy 9620a / Physics 9302a - 1st Problem List and Assignment

Your solution to problems 4, 7, 13, and 16 has to be handed in on Thursday, Oct. 17, 2013.

1. Prove the following relations (going from the left- to the right-hand side of the equality):

a) a × b( ) ⋅ c × d( ) = a ⋅ c( ) b ⋅d( ) − a ⋅d( ) b ⋅ c( ) b) ∇ × ψa( ) = ∇ψ × a +ψ ∇ × a c) ∇ a ⋅b( ) = a ⋅∇( )b + b ⋅∇( )a + a × ∇ × b( ) + b × ∇ × a( ) d) ∇ × a × b( ) = a ∇ ⋅b( ) − b ∇ ⋅a( ) + b ⋅∇( )a − a ⋅∇( )b.

Solution. a) We use index calculus to get (Einstein summation convention assumed)

a × b( ) ⋅ c × d( ) = εijka jbk( ) εimncmdn( )= εijkεimnajbkcmdn= δ jmδkn − δ jnδkm( )ajbkcmdn= ajbkcjdk − ajbkckd j

= a ⋅ c( ) b ⋅d( ) − a ⋅d( ) b ⋅ c( ).

(1.1)

b) The second relation (with ∂i = ∂ ∂xi )

∇ × ψa( )⎡⎣ ⎤⎦i = εijk∂ j ψak( )= εijk ak∂ jψ +ψ ∂ jak⎡⎣ ⎤⎦= ∇ψ × a +ψ ∇ × a[ ]i .

(1.2)

c) The third relation

∇ a ⋅b( )⎡⎣ ⎤⎦i = ∂i a jbj( )= bj∂ia j + aj∂ibj= bj∂ia j + aj∂ibj + bj∂ jai + aj∂ jbi − bj∂ jai − aj∂ jbi= aj∂ jbi + bj∂ jai + aj ∂ibj − ∂ jbi( ) + bj ∂ia j − ∂ jai( )= aj∂ jbi + bj∂ jai + ajεijkεmnk∂mbn + bjεijkεmnk∂man= aj∂ jbi + bj∂ jai + εijka jεkmn∂mbn + εijkbjεkmn∂man= a ⋅∇( )b + b ⋅∇( )a + a × ∇ × b( ) + b × ∇ × a( )⎡⎣ ⎤⎦i .

(1.3)

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d) Finally, the fourth relation

∇ × a × b( )⎡⎣ ⎤⎦i = εijk∂ jεkmnambn= δ imδ jn − δ inδ jm( ) bn∂ jam + am∂ jbn( )= bj∂ jai + ai∂ jbj − bi∂ ja j − aj∂ jbi= a ∇ ⋅b( ) − b ∇ ⋅a( ) + b ⋅∇( )a − a ⋅∇( )b⎡⎣ ⎤⎦i .

(1.4)

2. The time-average potential of a neutral hydrogen atom in its ground state is given by

Φ =q4πε0

e−αr

r1+ αr

2⎛⎝⎜

⎞⎠⎟, (2.1)

where q is the magnitude of the electronic charge, and α−1 = a0 2 with a0 the Bohr radius. Find the distribution of charge (both continuous and discrete) that will give this potential and interpret your result physically (for example, calculate the total charge).

Solution. Let us first rewrite equation (2.1) as follows

Φ =q4πε0

e−αr α2+1r

⎛⎝⎜

⎞⎠⎟. (2.2)

We now need to apply the Poisson equation to equation (2.2) remembering the following relation for the product of two functions

∇2 fg( ) = ∇ ⋅∇ fg( ) = ∇ ⋅ g∇f + f∇g( )

= g∇2 f + 2∇f ⋅∇g + f∇2g. (2.3)

Setting

f = e−αr

g = α2+1r, (2.4)

with

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∇f = ∂f∂r

= −αe−αrer

∇2 f = 1r∂2

∂r2rf( ) = 1

r∂∂r

1−αr( )e−αr⎡⎣ ⎤⎦ = α −2r

⎛⎝⎜

⎞⎠⎟αe−αr

∇g = ∂g∂r

= −1r2er

∇2g = ∇2 1r

⎛⎝⎜

⎞⎠⎟= −4πδ x( ).

(2.5)

If we now apply these equations to the Poisson equation, we find

∇2Φ = −ρε0

=q4πε0

e−αr α2+1r

⎛⎝⎜

⎞⎠⎟

α −2r

⎛⎝⎜

⎞⎠⎟α + 2 α

r2− 4πδ x( )⎡

⎣⎢⎤⎦⎥

= −q4πε0

4πδ x( ) − α 3

2e−αr

⎡

⎣⎢

⎤

⎦⎥.

(2.6)

Then, the charge distribution is

ρ = q δ x( ) − 1πa0

3 e−2ra0

⎡

⎣⎢⎢

⎤

⎦⎥⎥. (2.7)

We see that the first term corresponds to the proton charge, while the second term is the square of the norm of the radial wave function of the electron in the ground state (multiplied by its charge). Integrating ρ over all space yields a total charge of zero, as expected.

3. A simple capacitor is a device formed by two insulated conductors adjacent to each other. If equal and opposite charges are placed on the conductors, there will be a certain difference of potential between them. The ratio of the magnitude of the charge on one conductor to the magnitude of the potential difference is called the capacitance (in SI units it is measured in farads). Using Gauss’ Law, calculate the capacitance of

a) two large, flat, conducting sheets of area A , separated by a small distance d ; b) two concentric conducting spheres with radii a, b b > a( ) ;

c) two concentric conducting cylinders of length L , large compared to their radii a, b b > a( ) .

Solution.

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a) It is clear by the symmetry of the problem that if the sheets are large enough (we neglect the edge effects when d A1 2 ), then the electric field between them will be oriented normal to their surface. If we now consider an imaginary closed, rectangular surface that straddles one conducting sheet, extending some distance ξ < d( ) away on either side, then using Gauss’ Law for this surface we find

2EΔa = qε0A

Δa, (3.1)

with E and Δa the electric field at, and the area of, the imaginary Gauss surface inside the capacitor (the electric field in the conducting sheet is zero). Alternatively, we can write

E =q

2ε0A. (3.2)

Obviously, the surface charge density of the other sheet will contribute equally to the electric field Ec within the capacitor. It is therefore clear that this electric field is constant with

Ec =q

ε0A, (3.3)

and that the potential difference between the two plates is

Φ = Ecd =qdε0A

. (3.4)

The capacitance is therefore

C =qΦ

=ε0Ad. (3.5)

b) We consider a closed spherical surface of radius r located in the space between the two conducting spheres (it is also concentric to the two conducting spheres). Since the outer sphere does not contribute to the electric filed within its volume, the electric field between the spheres is entirely due to the charge distribution on the inner sphere with

E =q

4πε0r2 er , (3.6)

where q is the total charge on a sphere. The potential difference between the two spheres that form the capacitor is therefore

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Φ = −q4πε0

drr2b

a

∫ =q4πε0

1a−1b

⎛⎝⎜

⎞⎠⎟. (3.7)

The capacitance is then

C =qΦ

= 4πε01a−1b

⎛⎝⎜

⎞⎠⎟−1

. (3.8)

c) Because the cylinders are very long compared to their radii, we are justified in neglecting edge effects. Then, we see from the symmetry of the problem that electric field will only have a radial component. Just as in the preceding case, the electric field in the space between the two conducting surfaces is entirely due to the charge located on the smaller surface. Considering a cylindrical surface of radius r extending within the space between the two conducting cylinders and using Gauss’ Law, we find that the electric field in that space is due to the charge q located on the surface of the smaller cylinder is

E =q

2πε0rL. (3.9)

The potential difference between the two conducting cylinders is therefore

Φ = −q

2πε0Ldrra

b

∫ = −q

2πε0Lln b

a⎛⎝⎜

⎞⎠⎟. (3.10)

And the capacitance is

C =qΦ

=2πε0L

ln ba

⎛⎝⎜

⎞⎠⎟. (3.11)

4. a) Find the electric field a distance z along the symmetry axis of a uniformly charged ring of total charge Q and radius R .

b) Use the field for a ring calculated in a) to find the electric field a distance z along the symmetry axis of a uniformly charged disk of total charge Q and radius R .

c) Use the results of b) to find the electric field due to the charged disk at z = 0 and z R . For the latter, would it have been possible to derive this result based simply on physical arguments? Explain your answer.

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Solution. a) The field from the elemental charge dq = Q 2πR( )Rdϕ = Q 2π( )dϕ , where dϕ is an infinitesimal angle in the plane of the ring is given by

dE = dq4πε0r

2 er

= Qdϕ8π 2ε0 R2 + z2( ) er ,

(4.1)

where er is the unit vector linking the elemental charge to the point where the field is evaluated. Since that location is on the symmetry axis, all the elemental charges from the ring will contribute to yield a total field oriented along the z-axis , the field component in direction perpendicular to the z-axis will cancel out. We then have

Ez = ez ⋅dE0

2π

∫=

Qcos θ( )4πε0 R2 + z2( )

= Qz4πε0 R2 + z2( )3 2

,

(4.2)

where ez ⋅er = cos θ( ) . b) The disk can be thought of as being made of rings of radius ρ and thickness dρ , with 0 ≤ ρ ≤ R . We then write

Ez z( ) = 14πε0

Q 2πρπR2

⎛⎝⎜

⎞⎠⎟

zρ2 + z2( )3 2

⎛

⎝⎜⎜

⎞

⎠⎟⎟dρ

0

R

∫

= Qz2πε0R

2ρdρ

ρ2 + z2( )3 20

R

∫

= Q2πε0R

2 1−z

z2 + R2⎛

⎝⎜⎞

⎠⎟.

(4.3)

c) It is straightforward to calculate that

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Ez 0( ) = Q2πε0R

2

Ez z R( ) = Q2πε0R

2 1− 1+ R2

z2⎛⎝⎜

⎞⎠⎟

−1 2⎡

⎣⎢⎢

⎤

⎦⎥⎥

Q

2πε0R2 1− 1− R2

2z2+

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

Q

4πε0z2 .

(4.4)

This last equation could have be easily guessed from the fact that at z R the disk appears as a point charge Q , resulting in the calculated approximation for the electric field.

5. Electrostatic Screening and the Debye Length. Let’s assume that we have a system consisting of distributions (i.e., number densities in m−3 ) ne and ni of electrons and positive ions, respectively. The distributions are initially in equilibrium with homogeneous densities ne = ni (independent of position), such that the electrostatic potential Φ is zero everywhere. We now introduce a charge Q that locally perturbs the distributions of electrons and ions, and destroys their homogeneity in its vicinity. If we assume that far away from the charge both densities still have the same value n , then we can expect from thermodynamic equilibrium considerations that the following relations will hold in the neighborhood of Q

ni = ne

−qΦκBT

ne = neqΦκBT ,

(5.1)

where q, κ B, and T are the charge of the ions (−q for the electrons), the Boltzmann constant, and the temperature of the system, respectively. a) Use equations (5.1) and the Poisson equation to show that the electrostatic potential around Q is given by

Φ r( ) = Q4πε0r

e−rλD , (5.2)

where r is the distance (radius) away from Q and

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λD =ε0κ BT2nq2

⎛⎝⎜

⎞⎠⎟

1 2

(5.3)

is the so-called Debye length. To obtain equation (5.2) you can safely assume that qΦ κ BT( ) is a small quantity, such that it is appropriate to linearize the exponentials of equations (5.1). Also, assume that the electrostatic potential is zero at infinity. b) Using just a few sentences, explain the physical implication of equation (5.2). That is, what is the spatial extent over which the presence of Q is felt? Over what length scale can this plasma be considered neutral? [Hints: Poisson’s equation is

∇2Φ x( ) = −ρ x( )ε0

, (5.4)

and the Laplacian operator in spherical coordinates is given by

∇2Φ =1r∂2

∂r2rΦ( ) + 1

r2 sin θ( )∂∂θ

sin θ( ) ∂Φ∂θ

⎛⎝⎜

⎞⎠⎟+

1r2 sin2 θ( )

∂2Φ∂ϕ 2 . (5.5)

Also, the following second order differential equation

∂2

∂r2ψ[ ] = ψ

R2, (5.6)

has for solution

ψ r( ) = Ae−rR + Be

rR + C, (5.7)

with R a constant.] Solution. The Laplace equation for this system in the vicinity of, but not at the position of, Q is

∇2Φ = −qε0

ni − ne( )

= −qnε0

e−qΦκBT − e

qΦκBT

⎛

⎝⎜

⎞

⎠⎟

2nq2

ε0κ BTΦ,

(5.8)

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where we have used e

±qΦ κBT( ) 1± qΦ κ BT( ) for the last step. Using the equations (5.5) and (5.8), and the fact that the system is spherically symmetric, we can write

1r∂2

∂r2rΦ r( )⎡⎣ ⎤⎦ =

Φ r( )λD2 , (5.9)

or, if we define a new function ψ r( ) ≡ rΦ r( ) ,

∂2ψ∂r2

=ψλD2 . (5.10)

Equation (5.10) is easily solved to yield

Φ r( ) ≡ ψ r( )r

=1rAe

−rλD + Be

rλD + C

⎛

⎝⎜

⎞

⎠⎟ , (5.11)

where A, B, and C are some constants. However, we require that the electrostatic potential tends towards zero as r→∞ , and that

limr→0

Φ r( ) = Q4πε0r

, (5.12)

then B = C = 0 and A = Q 4πε0( ) , which finally leaves

Φ r( ) = Q4πε0r

e−rλD . (5.13)

b) Equation (5.13) implies that the presence of the charge Q introduced in the system will only affect the electron and ion distributions within a volume of radius approximately equal to the Debye length. It is seen that beyond r λD the potential caused by the charge becomes negligible. It follows that a plasma that is globally neutral can be also considered neutral everywhere when the relevant physical quantities are averaged on scales large compared to the Debye length.

6. Consider a hollow sphere of radius R , on which the potential V ϕ( ) = V cos ϕ( ) (6.1)

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is applied (ϕ is the usual azimuth angle for spherical coordinates). Find the potential Φ x( ) everywhere inside the sphere.

[Hints: To ensure that the potential is unambiguously defined everywhere, you can assume that there is a tiny hole at both poles on the sphere. Use the most general (and physically plausible) expansion for the potential. Don’t forget that any function can be expressed as a series of functions belonging to a complete set of basis functions; choose the right basis. You may use the table of spherical harmonics given at the end, but limit yourself to l ≤ 3 .]

Solution.

Using spherical coordinates, and spherical harmonics, we can expand the potential Φ x( ) as follows

Φ r,θ,ϕ( ) = AlmrlYlm θ,ϕ( )

l=−m

m

∑l=0

∞

∑ , (6.2)

where we omitted any terms in r− l+1( ) since the potential must be finite at the origin. On the surface of the sphere, the potential must satisfy the boundary condition

Φ R,θ,ϕ( ) = AlmRlYlm θ,ϕ( )

l=−m

m

∑l=0

∞

∑ = V cos ϕ( ). (6.3)

It follows from equation (6.3) that the coefficients AlmR

l are simply given by AlmR

l = V dϕ cos ϕ( )Ylm* θ,ϕ( )sin θ( ) dθ0

π

∫0

2π

∫ . (6.4)

However, because Ylm θ,ϕ( ) = Plm θ( )eimϕ , and cos ϕ( )∝ e± iϕ , then AlmR

l will be zero unless m = ±1. Therefore,

Al ,±1 =

πVRl Pl ,±1

* θ( )sin θ( ) dθ0

π

∫=πVRl Pl ,±1 θ( )sin θ( ) dθ

0

π

∫ , (6.5)

where Pl ,±1 θ( ) = Pl ,±1* θ( ) can be obtained from the table at the end. More precisely,

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P1,±1 θ( ) = 38πsin θ( )

P2,±1 θ( ) = 158πsin θ( )cos θ( )

P3,±1 θ( ) = 2164π

sin θ( ) 5cos2 θ( ) −1⎡⎣ ⎤⎦.

(6.6)

Upon inserting the different functions in equation (6.5), it is easy to see that terms in l = 2 will vanish since their integrand is odd. Now, from

sin2 θ( ) dθ

0

π

∫ =π2

sin2 θ( )cos2 θ( ) dθ0

π

∫ =π8, (6.7)

we find

A1,±1 = VRπ 2

238π

A3,±1 = VR3

π 2

82164π

, (6.8)

and

Φ r,θ,ϕ( ) Vπ 2 12

38π

rRY1,−1 θ,ϕ( ) −Y11 θ,ϕ( )⎡⎣ ⎤⎦

⎧⎨⎪

⎩⎪

+18

2164π

r3

R3Y1,−3 θ,ϕ( ) −Y13 θ,ϕ( )⎡⎣ ⎤⎦

⎫⎬⎪

⎭⎪.

(6.9)

Rearranging the different terms, we finally get

Φ r,θ,ϕ( ) V 3π

8sin θ( )cos ϕ( ) r

R⎧⎨⎩

+732

rR

⎛⎝⎜

⎞⎠⎟3

5cos2 θ( ) −1⎡⎣ ⎤⎦⎫⎬⎪

⎭⎪. (6.10)

7. An arbitrary function Φ that is a solution to the Laplace equation can always be expressed in cylindrical coordinates using a Bessel-Fourier expansion with

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Φ ρ,φ, z( ) = AmnJm kmnρ( ) + BmnNm kmnρ( )⎡⎣ ⎤⎦{n=1

∞

∑m=0

∞

∑× Cmn cos kmnz( ) + Dmn sin kmnz( )⎡⎣ ⎤⎦× Em cos mφ( ) + Fm sin mφ( )⎡⎣ ⎤⎦}

(7.1)

where Jm x( ) and Nm x( ) are the Bessel functions of, respectively, the first and second kind of order m , and it is implicit that F0 = 0 and E0 agrees with the usual definition for the corresponding Fourier coefficient (see Equation (2.2) of the Lecture Notes). The functions’ behaviors at the origin are such Jm 0( ) is finite and well behaved whereas Nm 0( ) is not. The validity of equation (7.1) stems, in part, from the fact that the set of Bessel functions is complete and forms a basis, such that any function f ρ( ) can be expanded over a given interval for ρ of length a with

f ρ( ) = αmnJm kmnρa

⎛⎝⎜

⎞⎠⎟+ βmnNm kmn

ρa

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥n=1

∞

∑m=0

∞

∑ , (7.2)

where

αmn =

2a2Jm+1

2 kmn( ) ρ f ρ( )Jm kmnρa

⎛⎝⎜

⎞⎠⎟dρ∫

βmn =2

a2Nm+12 kmn( ) ρ f ρ( )Nm kmn

ρa

⎛⎝⎜

⎞⎠⎟dρ∫

(7.3)

and the integrals are performed over the aforementioned interval.

Now consider a hollow cylindrical tube of radius a , with its axis of symmetry coincident with the z-axis , and its ends located at z = 0 and z = L . The potential on its end faces is zero, while the potential on the cylinder is given as

V φ, z( ) = V for − π 2 < φ < π 2−V for π 2 < φ < 3π 2.

⎧⎨⎪

⎩⎪ (7.4)

Find the potential anywhere inside the cylinder. Solution. Since we require the potential to be finite at the origin ρ = 0 , equation (7.1) reduces to

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Φ ρ,φ, z( ) = AmnJm kmnρ( ) Cmn cos kmnz( ) + Dmn sin kmnz( )⎡⎣ ⎤⎦

n=1

∞

∑m=0

∞

∑× Em cos mφ( ) + Fm sin mφ( )⎡⎣ ⎤⎦.

(7.5)

Applying the boundary conditions specified for the end faces we also find that Cmn = 0 . Furthermore, the symmetry of the potential V φ, z( ) on the cylinder implies that Fm = 0 . We then have

Φ ρ,φ, z( ) = AmnJm kmnρ( )sin kmnz( )cos mφ( )n=1

∞

∑m=0

∞

∑ , (7.6)

where the Bessel-Fourier coefficient Amn has been accordingly redefined, and

V φ, z( ) = AmnJm kmna( )sin kmnz( )cos mφ( )n=1

∞

∑m=0

∞

∑ . (7.7)

Clearly, for the potential to vanish at the end faces it must be that

kmn =

nπL, n = 1, 2, 3,… (7.8)

and kmn is independent of m . It follows from the definition for Fourier coefficients that when m ≠ 0

AmnJm kmna( ) = 2

πLdφ cos mφ( ) dzV φ, z( )

0

L

∫ sin kmnz( )0

2π

∫=2VπL

cos mφ( )dφ−π 2

π 2

∫ − cos mφ( )dφπ 2

3π 2

∫⎡⎣⎢

⎤⎦⎥

sin kmnz( )dz0

L

∫ , (7.9)

which further reduces to

AmnJm kmna( ) =8Vmnπ 2 1− −1( )n⎡⎣ ⎤⎦, m = 1,5,9,…

−8Vmnπ 2 1− −1( )n⎡⎣ ⎤⎦ m = 3,7,11,…

⎧

⎨⎪⎪

⎩⎪⎪

(7.10)

It is also understood that A0nJ0 k0na( ) = 0 , which according to equation (7.8) means that A0n = 0 . We therefore find that

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Φ ρ,φ, z( ) =

8Vπ 2

1− −1( )n⎡⎣ ⎤⎦mn

Jm nπ ρ L( )Jm nπ a L( ) sin nπ z

L⎛⎝⎜

⎞⎠⎟cos mφ( )

n=1

∞

∑m

∞

∑ , m = 1,5,9,…

−8Vπ 2

1− −1( )n⎡⎣ ⎤⎦mn

Jm nπ ρ L( )Jm nπ a L( ) sin nπ z

L⎛⎝⎜

⎞⎠⎟cos mφ( )

n=1

∞

∑m

∞

∑ m = 3,7,11,…

⎧

⎨

⎪⎪⎪

⎩

⎪⎪⎪

(7.11)

for 0 ≤ ρ ≤ a and 0 ≤ z ≤ L .

8. Prove the following theorem: For an arbitrary charge distribution ρ x( ) the values of the 2l +1( ) moments of the first non-vanishing multipole are independent of the origin of the coordinate axes, but the value of all higher multipole moments do in general depend on the choice of origin. (The different moments qlm for fixed l depend, of course, on the orientation of the axes.)

Solution. For a given l and m the corresponding multipole moment is given by qlm = Ylm

* θ,ϕ( )rlρ x( )d 3x∫ . (8.1) Before we can prove the theorem, we first have to show that the spherical harmonics obey the following relation Ylm θ,ϕ( )rl = Ca,b,cx

aybzca+b+c= l∑ , (8.2)

where Ca,b,c is some constant . To do so we consider a function Φl x( ) of order l , which we can express with the two following series

Φl x( ) = AlmrlYlm θ,ϕ( )

m=− l

l

∑ = Ba,b,cxaybzc

a+b+c= l∑ , (8.3)

with the constraint that a, b, and c are integer numbers. If we next consider one specific term of the first series, say for m = ′m , we can write Al ′m r

lYl ′m θ,ϕ( ) = Φl x( ) − AlmrlYlm θ,ϕ( )

m≠ ′m∑ . (8.4)

It is obvious, however, that the right hand side of equation (8.4) is just another function of the coordinates, say ′Φl x( ) , that can also be expanded as follows

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Al ′m r

lYl ′m θ,ϕ( ) = ′Φl x( )= ′Ba,b,cx

aybzca+b+c= l∑ , (8.5)

which proves equation (8.2). We are now ready to proceed, and we replace equation (8.1) with qlm = Ca,b,c xaybzcρ x( )d 3x∫

a+b+c= l∑ . (8.6)

If we now change the origin of the coordinate axes and make the following substitution

′x = x − x0′y = y − y0′z = z − z0 ,

(8.7)

then equation (8.6) becomes qlm = Ca,b,c ′x + x0( )a ′y + y0( )b ′z + z0( )c ρ ′x( )d 3 ′x∫

a+b+c= l∑ . (8.8)

Now each term within parentheses in equation (8.8) can be transformed with the binomial expansion as follows

′x + x0( )a = an

⎛⎝⎜

⎞⎠⎟

′x a−nx0n

n=0

a

∑ , (8.9)

with similar relations for the terms in ′y and ′z . Inserting equation (8.9) in equation (8.8) we have

qlm = Da,b,cn, p,r ′x a−n ′y b− p ′z c− rρ x( )d 3x∫

r=0

c

∑p=0

b

∑n=0

a

∑a+b+c= l∑ , (8.10)

with Da,b,c

n, p,r a new constant. Now, it should be clear that for every combination of n, p, and r with n + p + r > 0 corresponds a value ′l such that

′l = a − n( ) + b − p( ) + c − r( ) = a + b + c( ) − (n + p + r)= l − (n + p + r) < l.

(8.11)

If l is for the first non-vanishing multipole (i.e., q ′l m = 0 for ′l < l ), then ′x a−n ′y b− p ′z c− rρ ′x( )d 3 ′x∫ = 0, for ′l = l − (n + p + r) < l, (8.12)

16

since we can always write

′x e ′y f ′z g = De, f ,g ′r ′l Y ′l m ′θ , ′ϕ( )m=− ′l

′l

∑ , for e + f + g = ′l , (8.13)

while it is assumed here that q ′l m = Y ′l m

* ′θ , ′ϕ( ) ′r ′l ρ ′x( )d 3 ′x∫ = 0, for ′l < l. (8.14) Inserting equation (8.12) into equation (8.10) we are left with

qlm = Ca,b,c ′x a ′y b ′z cρ ′x( )d 3 ′x∫a+b+c= l∑

= Ylm* ′θ , ′ϕ( ) ′r lρ ′x( )d 3 ′x∫

= Ylm* θ,ϕ( )rlρ x( )d 3x∫ ,

(8.15)

and qlm is therefore independent of the origin of the coordinates axes. Finally, if we consider the case for ′′l = l +1 , because (contrary to the case where ′l < l ) we have ′x a ′y b ′z cρ ′x( )d 3 ′x∫ ≠ 0, when a + b + c = l < ′′l , (8.16) and we cannot simplify equation (8.10) as before, implying that equation (8.8) applies. The value of these multipole moments (i.e., for ′′l > l ) depends on the choice of origin. Obviously, the same is true in general for all values greater than l .

9. Given that the potential Φ x( ) due to a charge distribution ρ ′x( ) can be evaluated with the following relation

Φ x( ) = 14πε0

ρ ′x( )x − ′x

d 3 ′x∫ , (9.1)

expand 1 x − ′x with a Taylor series (see equation (1.84) of the lecture notes) to show that

Φ x( ) 1

4πε0

qr+p ⋅xr3

+12Qij

xix jr5

+⎡⎣⎢

⎤⎦⎥, (9.2)

where a summation on a repeated index is assumed, and

17

q = ρ ′x( )∫ d 3 ′x

p = ′x ρ ′x( )∫ d 3 ′x

Qij = 3 ′xi ′x j − ′r 2δ ij( )ρ ′x( )∫ d 3 ′x .

(9.3)

Solution.

Using a Taylor series expansion for 1 x − ′x we can write

1x − ′x

1r− ′x ⋅∇

1x − ′x

⎛⎝⎜

⎞⎠⎟ ′x =0

+12

′x ⋅∇( )2 1x − ′x

⎛⎝⎜

⎞⎠⎟ ′x =0

, (9.4)

with r = x . We, therefore, need the following derivatives

∂∂xi

1x − ′x

⎛⎝⎜

⎞⎠⎟ ′x =0

=∂∂xi

1

x j − ′x j( ) x j − ′x j( )⎛

⎝⎜⎜

⎞

⎠⎟⎟

′x =0

= −xi − ′xi( )x − ′x 3

′x =0

= −xir3,

(9.5)

and

∂2

∂xi∂x j

1x − ′x

⎛⎝⎜

⎞⎠⎟ ′x =0

= −∂∂x j

xi − ′xi( )x − ′x 3

⎡

⎣⎢⎢

⎤

⎦⎥⎥

′x =0

= −δ ijr3

+ 3xix jr5.

(9.6)

Now, from equation (9.6) and the last term in the right-hand side of equation (9.4), we have (summations on repeated indices implied) 1r53xix j − δ ijr

2( ) ′xi ′x jρ ′x( ) d 3 ′x∫ =1r53xix j − δ ijδmnxmxn( ) ′xi ′x jρ ′x( ) d 3 ′x∫

=1r5xix j 3 ′xi ′x jρ ′x( ) d 3 ′x∫ − δmnxmxn ′x j ′x jρ ′x( ) d 3 ′x∫

=1r5xix j 3 ′xi ′x j − δ ij ′r

2( ) j ρ ′x( ) d 3 ′x∫=1r5Qijxix j ,

(9.7)

18

where the quadrupole moment Qij is defined in the last of equations (9.3). Inserting equations (9.5) and (9.7) in equation (9.1), we find

Φ x( ) 14πε0

qr+xr3

⋅ ′x ρ ′x( )∫ d 3 ′x +12Qij

xix jr5

⎡⎣⎢

⎤⎦⎥

14πε0

qr+x ⋅pr3

+12Qij

xix jr5

⎡⎣⎢

⎤⎦⎥.

(9.8)

The dipole moment p is defined in the second of equations (9.3).

10. A perfectly conducting object has a hollow cavity in its interior. If a point charge is introduced in the cavity, what is the total charge induced on the surface of the cavity.

Solution. The electric field inside a perfect conductor is zero. So, if we choose an imaginary surface S that encloses the hollow cavity but is located within the conductor, then we must have (with n is the unit vector normal toS ) E ⋅n da

S∫ = 0. (10.1)

This implies, from equation (1.53) of the lecture notes, that the total charge enclosed by S must be zero. Therefore, if a point charge q is introduced anywhere within the cavity, then a charge −q must be induced on its surface to keep the total charge inside S to zero.

11. An electric dipole is pointing in the z direction and is placed at the origin of the coordinate system. Find the value of the electric field at any point.

Solution. We will use the usual definitions for the spherical coordinates r, θ, and ϕ

x = r sin θ( )cos ϕ( )y = r sin θ( )sin ϕ( )z = r cos θ( ).

(11.1)

The potential due to the dipole, given its orientation and that it is located at the origin, is

Φ x( ) = 14πε0

x ⋅pr3

=14πε0

pzr3

=14πε0

pcos θ( )r2

. (11.2)

Staying with spherical coordinates

19

Er = −∂Φ∂r

=14πε0

2pcos θ( )r3

Eθ = −1r∂Φ∂θ

=14πε0

psin θ( )r3

Eϕ = −1

r sin θ( )∂Φ∂ϕ

= 0.

(11.3)

12. We know that the force F acting on a charge q is given by F = qE , where E is the electric field at the position of the charge.

a) Now, consider a charge distribution ρ x( ) and write down the general expression for the total force acting on it. b) We choose a position x0 as the “origin” for calculating the different multipole moments of the distribution. Show that the force acting on the electric dipole moment is Fd = p ⋅∇( )E, (12.1) where p is the dipole moment, and it is understood that the gradient is evaluated at x0 .

c) We also know that the electrostatic energy of the dipole moment is Ud = −p ⋅E . Starting from equation (12.1) show that the total force on the dipole can also be expressed as Fd = −∇U . That is, Fd = ∇ p ⋅E( ). (12.2) [Hint. You may need the following relations

∇ ⋅ a × b( ) = b ⋅ ∇ × a( ) − a ⋅ ∇ × b( )∇ × a × b( ) = a ∇ ⋅b( ) − b ∇ ⋅a( ) + b ⋅∇( )a − a ⋅∇( )b

∇ a ⋅b( ) = a ⋅∇( )b + b ⋅∇( )a + a × ∇ × b( ) + b × ∇ × a( ). (12.3)

] Solution. a) The total force acting on the charge distribution is

F = ρ x( )E x( ) d 3x∫ . (12.4)

20

b) Let’s expand the electric field with a Taylor series about x0 , then

E x( ) E x0( ) + x − x0( ) ⋅∇x0

⎡⎣ ⎤⎦E x( ). (12.5) Inserting equation (12.5) into equation (12.4) yields

F ρ x( ) E x0( ) + x − x0( ) ⋅∇⎡⎣ ⎤⎦E⎡⎣ ⎤⎦ d3x∫

E x0( ) ρ x( ) d 3x∫ + ρ x( ) x − x0( ) d 3x∫⎡⎣ ⎤⎦ ⋅∇{ }E

qE x0( ) + p ⋅∇( )E, (12.6)

where it is understood that the gradient is evaluated at x0 , q is the total charge, and p is the dipole moment at x0 ,. That is, p = ρ x( ) x − x0( ) d 3x∫ . (12.7) Then the force acting on the dipole moment is Fd = p ⋅∇( )E. (12.8) c) Using the last of equations (12.3), we have ∇ p ⋅E( ) = p ⋅∇( )E + E ⋅∇( )p + p × ∇ × E( ) + E × ∇ × p( ). (12.9) But since the dipole moment is independent of x , then ∂i p j = 0 and the second and fourth terms on the right-hand side of equation (12.9) vanish. Moreover, since ∇ × E = 0 in electrostatics, then we are left with ∇ p ⋅E( ) = p ⋅∇( )E. (12.10)

13. We know that the potential energy of a dipole p subjected to an electric field E is given by W = −p ⋅E , where the electric is evaluated at the position occupied by the dipole. Assume that the dipole changes by an amount dp following a rotation dθ caused by the torque resulting from the electric field. a) Consider the corresponding change in potential energy dW stemming from the previous equation and compare it to the general relation dW = −τ ⋅dθ , which links it to the applied torque τ , and derive the equation for the torque acting on the dipole. b) Given that the force F acting on a charge q is expressed with F x( ) = qE x( ) , consider a charge distribution ρ x( ) and write down the general expression for the total torque

21

acting on it. Show that the lowest order term for the torque, when expanding the electric in a Taylor series, is the same as that obtained in a). c) Determine the torque acting on a dipole p2 due to the presence of another dipole p1 located some distance r away. d) Find the torque on p1 due to p2 .

e) Determine the total torque on the system composed of p1 and p2 (make sure to consider the contributions from the interaction force each dipole applies on the other). Solution. a) Using a simple differential we write

dW = −E ⋅dp

= −τ ⋅dθ . (13.1)

Since, like any vector field, the dipole p transforms as dp = dθ × p under a rotation dθ , we then have

dW = −E ⋅ dθ × p( )

= −dθ ⋅ p ×E( ), (13.2)

which under comparison with the last of equations (13.1) yields τ = p ×E. (13.3) b) Generalizing to a charge density we write for the torque acting on it τ = x ×E x( )ρ x( )d 3x∫ , (13.4)

where we recognize in E x( )ρ x( )d 3x the electric force acting on the elemental charge contained with the volume d 3x . Assuming the dipole to be located at x0 we expand the electric field with

E x( ) E x0( ) + x − x0( ) ⋅∇x0

⎡⎣ ⎤⎦E x( ), (13.5) which upon keeping only the leading term in the expansion yields

τ = x ×E x( )ρ x( )d 3x∫ xρ x( )d 3x∫⎡⎣ ⎤

⎦ ×E x0( ) p ×E x0( ).

(13.6)

22

This is the same expression as derived in equation (13.3). c) We know from equation (2.111) of the Lecture Notes that the electric field due to a dipole p1 some distance r away from it is

E1 r( ) = 3n n ⋅p1( )− p14πε0r

3 , (13.7)

with n = r r . Therefore the torque on p2 stemming from this field is

τ 2 = p2 ×E1 r( )

=3 p2 × n( ) n ⋅p1( )− p2 × p1

4πε0r3 .

(13.8)

d) To determine the torque τ 1 on p1 due to p2 we simply need to make the changes p1 ↔ p2 and n→−n in equation (13.8). This results in

τ 1 =3 p1 × n( ) n ⋅p2( ) + p2 × p1

4πε0r3 . (13.9)

e) To evaluate the total torque on the system we may be tempted to simply add together the two torques previously calculated to get

τ 1 + τ 2 =3

4πε0r3 p1 × n( ) n ⋅p2( ) + p2 × n( ) n ⋅p1( )⎡⎣ ⎤⎦. (13.10)

But this result cannot be complete since the conservation of angular momentum requires that the total torque must be zero (i.e., there is no external force or torque acting on the system composed of the two dipoles). We must also consider the force of interaction between the two torques. For example, we know from equation (12.10) that the force F21 on p2 due to p1 is F21 = ∇ p2 ⋅E1( ) = p2 ⋅∇( )E1 . Applying the needed derivatives to equation (13.7) yields

F21 =−3 n ⋅p2( )4πε0r

4 3n n ⋅p1( )− p1⎡⎣ ⎤⎦

+ 14πε0r

3

3 n ⋅p1( )r

p2 − n n ⋅p2( )⎡⎣ ⎤⎦ +3nrp1 ⋅p2 − n ⋅p1( ) n ⋅p2( )⎡⎣ ⎤⎦

⎧⎨⎩

⎫⎬⎭

= 34πε0r

4 n ⋅p1( )p2 + n ⋅p2( )p1 + p1 ⋅p2( )n− 5 n ⋅p1( ) n ⋅p2( )n⎡⎣ ⎤⎦,

(13.11)

23

where we have used ∂n ∂r = 0 , ∂n ∂θ = eθ , and ∂n ∂ϕ = sin θ( )eϕ (see equations (2.95) in the Lecture Notes when n = er ). Since according to Newton’s Third Law F12 = −F21 (it is straightforward to verify this by changing n→−n in equation (13.11)) we find that the torque due to this force of interaction, calculated about the mid-point between the two dipoles, is

τ 21 =r2× F21 − F12( )

= r × F21

= −34πε0r

3 p1 × n( ) n ⋅p2( ) + p2 × n( ) n ⋅p1( )⎡⎣ ⎤⎦.

(13.12)

Adding equations (13.10) and (13.12) we find that the total torque on the system is zero, as expected.

14. Two dipoles lie in the x, z( )-plane . p1 is placed at the origin and is pointing along the z-axis ; p2 is placed at the point x, z( ) and makes an angle α with p1 . Calculate the force on p2 in the general case (limit yourself to the x and z components of the force).

Solution.

If E is the electric due to the first dipole p1 at the position of the second dipole p2 , then the force acting on p2 is F = p2 ⋅∇( )E. (14.1) (You should try to prove this.) Using equations (11.3) with ϕ = 0 (to work in the x, z( )-plane ), and transforming to Cartesian coordinates, we find

24

Ex =14πε0

3p1xz

x2 + z2( )5 2

Ez =14πε0

p1 2z2 − x2( )

x2 + z2( )5 2. (14.2)

From these equations we can evaluate the following derivatives

∂Ex

∂x=

14πε0

3z z2 − 4x2( )x2 + z2( )7 2

p1,∂Ex

∂z=

14πε0

3x x2 − 4z2( )x2 + z2( )7 2

p1

∂Ez

∂x=

14πε0

3x x2 − 4z2( )x2 + z2( )7 2

p1,∂Ez

∂z=

14πε0

3z 3x2 − 2z2( )x2 + z2( )7 2

p1

(14.3)

and the force components are

Fx =p1p24πε0

sin α( ) 3z z2 − 4x2( )

x2 + z2( )7 2+ cos α( ) 3x x2 − 4z2( )

x2 + z2( )7 2⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

Fz =p1p24πε0

sin α( ) 3x x2 − 4z2( )x2 + z2( )7 2

+ cos α( ) 3z 3x2 − 2z2( )

x2 + z2( )7 2⎡

⎣

⎢⎢

⎤

⎦

⎥⎥.

(14.4)

15. Find the B field along the symmetry axis of a circular loop of radius a carrying a current I .

Solution. We will use the cylindrical coordinates r, θ, and z . The Biot and Savard law states that

dB =µ04π

I dl × x( )x 3 , (15.1)

where x = zez − aer is the vector going from a given point on the loop to the observation point on the symmetry axis, and dl = adθeθ . Therefore,

dB =µ04π

Ix 3 adθeθ × zez − aer( )⎡⎣ ⎤⎦

=µ04π

Ix 3 azdθer + a

2dθez( ). (15.2)

25

If we integrate over θ from 0 to 2π , we find that the first term vanishes since er = cos θ( )ex + sin θ( )ey , and finally

B =µ0Ia

2

2 z2 + a2( )32ez. (15.3)

16. A small current loop of radius R lies in the x, y( )-plane . A current J passes through the loop. a) Find the magnetic dipole moment m of the loop. b) Find the asymptotic (i.e., r R ) magnetic induction B from the magnetic potential due to m . c) Write the equation of motion for a particle of mass M and charge q in the asymptotic B field of the loop. Show that the particle stays in the x, y( )-plane if its original velocity is restricted to that plane. Solution. a) The magnetic moment is located in a plane; therefore, it is simply expressed as the product of the current and the area of the loop. That is, m = JπR2ez . (16.1) b) We know from equation (3.48) of the lecture notes that the magnetic field away from the magnetic dipole is given by

B x( ) = µ04π

3n n ⋅m( ) −mx 3

⎡

⎣⎢⎢

⎤

⎦⎥⎥, (16.2)

with n = x x = x r . If the vector x makes an angle θ with the z-axis , then n ⋅m = mcos θ( )ez , and

B x( ) = µ04π

mr3

3cos θ( )er − ez⎡⎣ ⎤⎦. (16.3)

However, we know that Cartesian and spherical coordinates are related in such a way that (see Problem 18) ez = cos θ( )er − sin θ( )eθ , and

B x( ) = µ04π

mr3

2cos θ( )er + sin θ( )eθ⎡⎣ ⎤⎦. (16.4)

26

c) Using the equation of the Lorentz force for a charged particle subjected to a magnetic induction, we can write

M dvdt

= qv × B. (16.5)

Breaking up the velocity of the particle into two components, one parallel and another perpendicular to B (i.e., v = ve + v⊥e⊥ ), then equation (16.5) transforms to

d 2rdt 2

= 0

d 2r⊥dt 2

=qBMv⊥ ,

(16.6)

with B =µ04π

mr31+ 3cos2 θ( )⎡⎣ ⎤⎦

12 . But if the particle has an initial motion restricted to the

x, y( )-plane (i.e., z0 = vz 0( ) = 0 , and θ = π 2 ), then

B = −µ04π

m

x2 + y2( )32ez , (16.7)

and e = ez , and e⊥ is in the x, y( )-plane . The solution to the first of equations (16.6) is easily shown to be z = z0 + vz 0( )t = z0 = 0. (16.8) The particle’s motion will therefore remain in the x, y( )-plane .

17. Magnetostatic problems can be treated in a way similar to electrostatic ones. Show that for the calculation of the fields, a material of magnetization M x( ) can be replaced by a volume polarization charge density ρM = −∇ ⋅M and a surface polarization charge densityσM = n ⋅M .

Solution. Consider Ampère’s law in cases when a magnetization M is present ∇ × B = µ0 J +∇ ×M[ ], (17.1) and let us define an equivalent magnetization MJ that is due to the current J . That is, we have

27

J = ∇ ×MJ . (17.2) Furthermore, we introduce the total effective magnetization MT =MJ +M . Limiting ourselves to cases where µ0 is constant, we can transform equation (17.1) to

∇ ×1µ0B −MT

⎛⎝⎜

⎞⎠⎟= 0, (17.3)

or, equivalently, if we define the magnetic field H as

H =1µ0B −MT , (17.4)

then we have ∇ ×H = 0. (17.5) Turning our attention to the equation for the divergence of the magnetic induction, we can write ∇ ⋅B = µ0∇ ⋅ H +MT( ) = 0, (17.6) or alternatively ∇ ⋅H = −∇ ⋅MT . (17.7) If we further define a volume polarization charge density ρM = −∇ ⋅MT , equation (17.7) becomes ∇ ⋅H = ρM . (17.8) Equations (17.5) and (17.8) are similar in form to the equations of electrostatics

∇ × E = 0

∇ ⋅E =ρTε0, (17.9)

where ρT = ρ − ∇ ⋅P is the total effective charge that takes into account the polarization of the medium. So, in analogy to electrostatics we can postulate that for magnetostatics H = −∇ΦM , (17.10)

28

with ΦM the magnetic scalar potential. Finally, again by analogy with electrostatics, if we assume that our medium of effective magnetization MT has an outer bound beyond which the magnetization suddenly falls to zero, then, considering an infinitesimally small pillbox straddling the boundary, we obtain the following boundary condition (see equation (1.65)) −n ⋅H = σM , (17.11) where σM is the effective magnetic surface-charge density, and n the outwardly directed normal. But since the total effective magnetic charge qM inside the pillbox is given by qM = ρM d 3x

V∫ = σM daS∫ , (17.12)

and

ρM d 3xV∫ = σM daS∫

− ∇ ⋅MT d3x

V∫= − MT ⋅ndaS∫ ,

(17.13)

then σM = n ⋅MT . (17.14) Although our treatment was limited to media of constant permeability, it can also be extended to ferromagnetic material, since in this case ρ = J = 0 .

18. When making a change from a system of Cartesian coordinates x, y, z( ) to another system α,β,γ( ) we find that the volume integral is expressed as dxdydz = J dαdβdγ∫∫∫∫∫∫ , (18.1) where J is the determinant of the Jacobian matrix that expresses the transformation of infinitesimal changes between the two systems of coordinates. More precisely, we have

dxdydz

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ = J

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

dαdβdγ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ . (18.2)

29

We assume that the α,β,γ( ) system is actually a spherical coordinate system r,θ,ϕ( ) , with the usual definitions

x = r sin θ( )cos ϕ( )y = r sin θ( )sin ϕ( )z = r cos θ( ).

(18.3)

a) Write down equations for dx, dy, and dz as functions of r, θ, ϕ, dr, dθ, and dϕ, and using equations (18.2) and (18.1) show that dxdydz = r2 sin θ( )drdθdϕ∫∫∫∫∫∫ . (18.4) b) Consider the two corresponding sets of unit basis vectors ex ,ey ,ez( ) and er ,eθ ,eϕ( ) . Express the vectors of the Cartesian basis as functions of the spherical basis, and write down the transformation matrix equation that links them as

ex ,ey ,ez( ) = er ,eθ ,eϕ( ) T⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥. (18.5)

Give an expression for T . c) Just as the infinitesimal vector dr can be written as

dr = ex ,ey ,ez( )dxdydz

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , (18.6)

or dr = dx ex + dyey + dzez , use the matrices J and T of a) and b) to transform equation (18.6) to express dr as

dr = er ,eθ ,eϕ( ) M⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

drdθdϕ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , (18.7)

or dr = aer + beθ + ceϕ . Give expressions for M , a, b, and c . d) An equivalent way of expressing the infinitesimal Cartesian volume element of equation (18.1) is to use the components of the vector dr = dx ex + dyey + dzez to

30

calculate dxdydz = dzez ⋅ dx ex × dyey( ) . Similarly, use dr = aer + beθ + ceϕ to verify

that the infinitesimal spherical volume element equals r2 sin θ( )dr dθ dϕ . e) Finally, invert equation (18.5) to obtain

er ,eθ ,eϕ( ) = ex ,ey ,ez( ) T −1

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥, (18.8)

then starting with r = r er verify that dr = aer + beθ + ceϕ .

Solution. a) From equation (18.3) we calculate

dx = dr sin θ( )cos ϕ( ) + r cos θ( )cos ϕ( )dθ − r sin θ( )sin ϕ( )dϕdy = dr sin θ( )sin ϕ( ) + r cos θ( )sin ϕ( )dθ + r sin θ( )cos ϕ( )dϕdz = dr cos θ( ) − r sin θ( )dθ.

(18.9)

We then have

dxdydz

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ =

sin θ( )cos ϕ( ) r cos θ( )cos ϕ( ) −r sin θ( )sin ϕ( )sin θ( )sin ϕ( ) r cos θ( )sin ϕ( ) r sin θ( )cos ϕ( )cos θ( ) −r sin θ( ) 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

drdθdϕ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ , (18.10)

where the matrix in bracket is J , the Jacobian matrix. Using the third row as the anchor, we can calculate its determinant as

J = cos θ( ) r2 sin θ( )cos θ( ) cos2 ϕ( ) + sin2 ϕ( )⎡⎣ ⎤⎦{ }−r sin θ( ) −r sin2 θ( ) sin2 ϕ( ) + cos2 ϕ( )⎡⎣ ⎤⎦{ }= r2 sin θ( ).

(18.11)

Using equation (18.1), we find that dxdydz = r2 sin θ( )drdθdϕ∫∫∫∫∫∫ . b) We can also write for the basis vectors

ex = sin θ( )cos ϕ( )er + cos θ( )cos ϕ( )eθ − sin ϕ( )eϕey = sin θ( )sin ϕ( )er + cos θ( )sin ϕ( )eθ + cos ϕ( )eϕez = cos θ( )er − sin θ( )eθ .

(18.12)

31

We then have

ex ,ey ,ez( ) = er ,eθ ,eϕ( )sin θ( )cos ϕ( ) sin θ( )sin ϕ( ) cos θ( )cos θ( )cos ϕ( ) cos θ( )sin ϕ( ) − sin θ( )

− sin ϕ( ) cos ϕ( ) 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥. (18.13)

c) Inserting equations (18.2) and (18.5) into equation (18.6) we get

dr = er ,eθ ,eϕ( ) T⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

J⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

drdθdϕ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

= er ,eθ ,eϕ( ) M⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

drdθdϕ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ ,

(18.14)

where

M[ ] = T[ ] J[ ] =1 0 00 r 00 0 r sin θ( )

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥. (18.15)

Thus we have

dr = er ,eθ ,eϕ( )1 0 00 r 00 0 r sin θ( )

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

drdθdϕ

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟ . (18.16)

Using equation (18.16), we can write dr = dr er + r dθ eθ + r sin θ( )dϕ eϕ , (18.17) and we have a = dr, b = r dθ, and c = r sin θ( )dϕ . d) With equation (18.17) we can calculate the infinitesimal spherical volume element with

32

dr er ⋅ r dθ eθ × r sin θ( )dϕ eϕ( ) = dr er ⋅ r

2 sin θ( )dθ dϕ er= r2 sin θ( )dr dθ dϕ.

(18.18)

e) It can easily be verified that the matrix T is orthonormal, so T −1 = T T (T T is the transpose of T ), and

er ,eθ ,eϕ( ) = ex ,ey ,ez( )sin θ( )cos ϕ( ) cos θ( )cos ϕ( ) − sin ϕ( )sin θ( )sin ϕ( ) cos θ( )sin ϕ( ) cos ϕ( )cos θ( ) − sin θ( ) 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥. (18.19)

We can now write r er = r sin θ( )cos ϕ( )ex + sin θ( )sin ϕ( )ey + cos θ( )ez⎡⎣ ⎤⎦, (18.20) from which we calculate the differential

dr = ∂r∂rdr +

∂r∂θ

dθ +∂r∂ϕ

dϕ

= er + r∂er∂r

⎛⎝⎜

⎞⎠⎟dr + r

∂er∂θ

dθ + r∂er∂ϕ

dϕ

= dr er+r dθ cos θ( )cos ϕ( )ex + cos θ( )sin ϕ( )ey − sin θ( )ez( )+r dϕ − sin θ( )sin ϕ( )ex + sin θ( )cos ϕ( )ey( )= dr er + r dθ eθ + r sin θ( )dϕ eϕ ,

(18.21)

where we used equation (18.19) for the last step. We, therefore, get the same result than was obtained in a different manner in d) (cf., equation (18.17))

19. Consider the gradient vector operator ∇ in Cartesian coordinates ∇ = ex∂x + ey∂y + ez∂z , (19.1)

where we used the following short hand notation for the partial derivatives ∂x ≡∂∂x

, etc.

We are interested in investigating the form that this operator takes when we make the change from a Cartesian to a spherical coordinate system. We use the usual definitions

33

x = r sin θ( )cos ϕ( )y = r sin θ( )sin ϕ( )z = r cos θ( ),

(19.2)

or, alternatively,

r = x2 + y2 + z2

tan θ( ) = x2 + y2

z

tan ϕ( ) = yx.

(19.3)

a) Using the chain rule we can write

∂x =∂r∂x

∂r +∂θ∂x

∂θ +∂ϕ∂x

∂ϕ . (19.4)

Write similar equations for ∂y and ∂z , and then write the matrix equation relating the components of the gradient operator from the two coordinate systems, that is

∂x∂y∂z

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= S⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

∂r∂θ∂ϕ

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟. (19.5)

(You have to express the matrix S using the different partial derivatives as its components, but do not evaluate the derivatives yet.) b) Using, when needed, the following equation for the derivative of a tangent function

ddutan α u( )⎡⎣ ⎤⎦ =

1cos2 α( )

dαdu, (19.6)

or, alternatively,

dαdu

= cos2 α( ) ⋅ ddutan α u( )⎡⎣ ⎤⎦, (19.7)

use equations (19.3) to show that

34

S =

sin θ( )cos ϕ( ) 1rcos θ( )cos ϕ( ) −

1r sin θ( ) sin ϕ( )

sin θ( )sin ϕ( ) 1rcos θ( )sin ϕ( ) 1

r sin θ( ) cos ϕ( )

cos θ( ) −1rsin θ( ) 0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

. (19.8)

(Here is an example as how to use equation (19.7)

∂ϕ∂y

= cos2 ϕ( )∂y tan ϕ( )⎡⎣ ⎤⎦ = cos2 ϕ( )∂y

yx

⎡⎣⎢

⎤⎦⎥

=cos2 ϕ( )

x=

cos2 ϕ( )r sin θ( )cos ϕ( ) =

cos ϕ( )r sin θ( ) ,

(19.9)

where the first of equations (19.2) was used for the second line.) c) Just as equation (19.1) can be written as

∇ = ex ,ey ,ez( )∂x∂y∂z

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟, (19.10)

use equation (19.5) and (19.8), along with the following result relating the two unit bases obtained in part b) of Problem 18

ex ,ey ,ez( ) = er ,eθ ,eϕ( ) T⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= er ,eθ ,eϕ( )sin θ( )cos ϕ( ) sin θ( )sin ϕ( ) cos θ( )cos θ( )cos ϕ( ) cos θ( )sin ϕ( ) − sin θ( )

− sin ϕ( ) cos ϕ( ) 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

(19.11)

to prove that

∇ = er∂r + eθ1r∂θ + eϕ

1r sin θ( ) ∂ϕ . (19.12)

d) Given a vector A = Axex + Ayey + Azez = Arer + Aθeθ + Aϕeϕ , (19.13)

35

we would like to find an expression for its divergence. To do so start with

∇ ⋅A = er∂r + eθ1r∂θ + eϕ

1r sin θ( ) ∂ϕ

⎛⎝⎜

⎞⎠⎟⋅ Arer + Aθeθ + Aϕeϕ( ), (19.14)

and use the inverse of equation (19.11) (which was calculated in part e) of Problem 18) to prove that

∇ ⋅A = ∂rAr + 2

Arr+1r∂θAθ +

cos θ( )r sin θ( ) Aθ +

1r sin θ( ) ∂ϕAϕ

=1r2

∂r r2Ar( ) + 1

r sin θ( ) ∂θ Aθ sin θ( )( ) + 1r sin θ( ) ∂ϕAϕ .

(19.15)

Solution. a) Using the chain rule we have

∂x =∂r∂x

∂r +∂θ∂x

∂θ +∂ϕ∂x

∂ϕ

∂y =∂r∂y

∂r +∂θ∂y

∂θ +∂ϕ∂y

∂ϕ

∂z =∂r∂z

∂r +∂θ∂z

∂θ +∂ϕ∂z

∂ϕ .

(19.16)

The system of equations (19.16) can be written in a matrix form as

∂x∂y∂z

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟= S⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

∂r∂θ∂ϕ

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟, (19.17)

with

S =

∂r∂x

∂θ∂x

∂ϕ∂x

∂r∂y

∂θ∂y

∂ϕ∂y

∂r∂z

∂θ∂z

∂ϕ∂z

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

. (19.18)

b) From equations (19.3) we calculate

36

∂r∂x

=x

x2 + y2 + z2= sin θ( )cos ϕ( )

∂r∂y

=y

x2 + y2 + z2= sin θ( )sin ϕ( )

∂r∂z

=z

x2 + y2 + z2= cos θ( ),

(19.19)

and by further using equation (19.7)

∂θ∂x

= cos2 θ( ) xz x2 + y2

=1rcos θ( )cos ϕ( )

∂θ∂y

= cos2 θ( ) yz x2 + y2

=1rcos θ( )sin ϕ( )

∂θ∂z

= − cos2 θ( ) x2 + y2

z2= −

1rsin θ( )

∂ϕ∂x

= − cos2 ϕ( ) yx2

= −1

r sin θ( ) sin ϕ( )

∂ϕ∂y

= cos2 ϕ( ) 1x=

1r sin θ( ) cos ϕ( )

∂ϕ∂z

= 0.

(19.20)

Inserting equations (19.19) and (19.20) into equation (19.18) we get

S =

sin θ( )cos ϕ( ) 1rcos θ( )cos ϕ( ) −

1r sin θ( ) sin ϕ( )

sin θ( )sin ϕ( ) 1rcos θ( )sin ϕ( ) 1

r sin θ( ) cos ϕ( )

cos θ( ) −1rsin θ( ) 0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

. (19.21)

c) We know that

∇ = ex ,ey ,ez( )∂x∂y∂z

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟, (19.22)

and upon using equations (19.11) and (19.17) we can express the gradient operator as

37

∇ = er ,eθ ,eϕ( ) N⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

∂r∂θ∂ϕ

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟, (19.23)

with N = TS. (19.24) Inserting equations (19.11) and (19.21) for the matricesT and S , respectively, into equation (19.24) we get

N =

1 0 0

0 1r

0

0 0 1r sin θ( )

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

, (19.25)

and

∇ = er∂r + eθ1r∂θ + eϕ

1r sin θ( ) ∂ϕ . (19.26)

c) Starting with part e) of Problem 18 we know that

er ,eθ ,eϕ( ) = ex ,ey ,ez( )sin θ( )cos ϕ( ) cos θ( )cos ϕ( ) − sin ϕ( )sin θ( )sin ϕ( ) cos θ( )sin ϕ( ) cos ϕ( )cos θ( ) − sin θ( ) 0

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥, (19.27)

from which we can easily calculate the following derivatives

∂rer = ∂reθ = ∂reϕ = 0

∂θer = cos θ( )cos ϕ( )ex + cos θ( )sin ϕ( )ey − sin θ( )ez = eθ∂θeθ = − sin θ( )cos ϕ( )ex − sin θ( )sin ϕ( )ey − cos θ( )ez = −er∂θeϕ = 0

∂ϕer = − sin θ( )sin ϕ( )ex + sin θ( )cos ϕ( )ey = sin θ( )eϕ∂ϕeθ = − cos θ( )sin ϕ( )ex + cos θ( )cos ϕ( )ey = cos θ( )eϕ∂ϕeϕ = − cos ϕ( )ex − sin ϕ( )ey = − sin θ( )er − cos θ( )eθ .

(19.28)

38

Using equation (19.14)

∇ ⋅A = er∂r + eθ1r∂θ + eϕ

1r sin θ( ) ∂ϕ

⎛⎝⎜

⎞⎠⎟i Arer + Aθeθ + Aϕeϕ( )

= ∂rAr( ) + 1rAr + ∂θAθ( ) + 1

r sin θ( ) Ar sin θ( ) + Aθ cos θ( ) + ∂ϕAϕ( ), (19.29)

and finally

∇ ⋅A = ∂rAr +

2rAr +

1r∂θAθ +

1r sin θ( ) Aθ cos θ( ) + 1

r sin θ( ) ∂ϕAϕ

=1r2

∂r r2Ar( ) + 1

r sin θ( ) ∂θ Aθ sin θ( )( ) + 1r sin θ( ) ∂ϕAϕ .

(19.30)

39

Table of spherical harmonics

Y00 =14π

Y10 =34πcos θ( )

Y1,±1 = 38πsin θ( )e± iϕ

Y20 =516π

3cos2 θ( ) −1⎡⎣ ⎤⎦

Y2,±1 = 158πsin θ( )cos θ( )e± iϕ

Y2,±2 =1532π

sin2 θ( )e± i2ϕ

Y30 =716π

5cos3 θ( ) − 3cos θ( )⎡⎣ ⎤⎦

Y3,±1 = 2164π

sin θ( ) 5cos2 θ( ) −1⎡⎣ ⎤⎦e± iϕ

Y3,±2 =10532π

sin2 θ( )cos θ( )e± i2ϕ

Y3,±3 = 3564π

sin3 θ( )e± i3ϕ