What is the full form of POLS

Complete function systems: Fourier transformation and multipole expansion

tasks

Occasionally the tasks contain more information than is necessary for the solution. For some others, however, data from general knowledge, from other sources, or useful estimates are required.

easy tasks with just a few arithmetic steps

••

Moderately difficult tasks that require some thinking and possibly a combination of different concepts

•••

Demanding tasks that require advanced concepts (possibly also from later chapters) or your own mathematical modeling

13.1 • Spherically symmetric Fourier transformation

Prove: If a function \ (f ({\ varvec {r}}) \) is spherically symmetric, then this also applies to its Fourier transform \ (\ tilde {f} ({\ varvec {k}}) \ ).

13.2 ••• Discrete Fourier transform

The function \ (f_ {n} (x) \) (n a natural number) have the value 1 in \ (2n + 1 \) intervals of width x0 and the distance d otherwise the value 0, whereby the function graph is symmetrical to yAxis (the figure represents this function especially for n = 3 dar).

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  1. (a)

    Calculate the Fourier transform \ (\ tilde {f} _ {n} (k) \) and represent the function \ (f_ {n} (x) \) with it.

  2. (b)

    The function \ (f (x): = \ lim_ {n \ to \ infty} f_ {n} (x) \) is periodic ("grid"), so it can be represented by a Fourier series. Determine this representation (with exponential functions as basis functions).

  3. (c)
    Use your results from subtasks (a) and (b) to justify that the following applies:
    $$ \ sum_ {j = - \ infty} ^ {\ infty} \ mathrm {e} ^ {\ mathrm {i} jx} = 2 \ uppi \ sum_ {m = - \ infty} ^ {\ infty} \ delta (x-2 \ uppi m). $$
    This is a discrete analogue of (13.18.)

Solution hint:

(a) Use the behavior of the Fourier transform under displacements from the box “Overview: Important properties of the Fourier transform” in Sect. 13.1.

13.3 • Cauchy's integral theorem

Calculate the integral
$$ \ oint z \, \ mathrm {d} z, $$
  1. (a)

    by choosing a square of side length \ (2a \) as the path, which is symmetrical to the origin and whose sides are parallel to the axes, and

  2. (b)

    by means of Cauchy's integral theorem.

13.4 • Residual theorem

Calculate the integral
$$ \ oint \ frac {1} {z} \, \ mathrm {d} z, $$
  1. (a)

    by choosing a square of side length \ (2a \) as the path, which is symmetrical to the origin and whose sides are parallel to the axes, and

  2. (b)

    using the residual theorem.

13.5 •• Closed path of integration in infinity

Calculate the integral
$$ \ int _ {- \ infty} ^ {\ infty} \ frac {1} {1 + x ^ {4}} \, \ mathrm {d} x $$
with the help of an "infinitely" closed integration path in the complex level.

13.6 •• Complete system of functions: Hermite polynomials

The set of functions is given
$$ f_ {n} (x) = x ^ {n} \ mathrm {e} ^ {- x ^ {2} / 2}, $$
in which n run through all natural numbers. Determine for n = 0,1,2,3 from this orthonormal functions \ (h_ {n} (x) \) with respect to the scalar product
$$ \ langle f, g \ rangle = \ int _ {- \ infty} ^ {\ infty} f (x) g (x) \, \ mathrm {d} x. $$
annotation: Instead, the scalar product can also be defined more generally with a so-called weight function \ (\ rho (x) \):
$$ \ langle f, g \ rangle = \ int _ {- \ infty} ^ {\ infty} f (x) g (x) \ rho (x) \, \ mathrm {d} x. $$
With \ (\ rho (x) = \ mathrm {e} ^ {- x ^ {2}} \) we are looking for the polynomials that are orthonormal with respect to this scalar product (“Hermite polynomials”; these are important in quantum mechanics ).

13.7 • Multipole moments and choice of the coordinate system

Prove:
  1. (a)

    If the total charge of a charge distribution is zero, its (Cartesian) dipole moment is independent of a shift in the coordinate system.

  2. (b)

    If the total charge and the dipole moment of a charge distribution are zero, then its (Cartesian) quadrupole moment is independent of a shift in the coordinate system.

13.8 • Cartesian multipole moments: charges squared

A charge distribution from four point charges of the amount is given q0, which are located at the corners of a square of side length \ (2a \), the center of which is the origin and whose sides are parallel to the coordinate axes. Let the sign of the point charge be positive at the point \ ((a, a) \) and alternate from corner to corner. Determine the electrical potential of this charge distribution up to and including the (Cartesian) quadrupole moment.

13.9 • Cartesian multipole moments: cuboid

In a cuboid with side lengths \ (2a \), \ (2b \) and \ (2c \), whose center is the origin and whose edges are parallel to the axes, the charge density is proportional to zn (n be a non-negative integer) and have the value \ (\ rho_ {0} \) at \ (z = c \), but be independent of x and y. Determine the (Cartesian) multipole moments of this charge distribution up to and including the quadrupole moment.

Solution hint:

Distinguish the two cases that n is an even or an odd number.

13.10 •• Cartesian multipole moments: pyramid

A homogeneous charge distribution (charge density \ (\ rho_ {0} \)) has the shape of a three-sided pyramid with the vertices \ (O (0; 0; 0) \), \ (A (a; 0; 0) \), \ (B (0; b; 0) \) and \ (C (0; 0; c) \) (see figure).

Open image in new window

Determine the (Cartesian) multipole moments of this charge distribution up to and including the quadrupole moment. What is the result especially for \ (a = b = c \)?

13.11 • Spherical multipole moments and potential

Prove: The electrical potential has the angle dependence of one of the spherical surface functions if and only if this also applies to the charge distribution (with the same spherical surface function).

Solution hint:

“Exactly then” means equivalence of both statements, so both directions of the assertion have to be shown.

13.12 •• Vector-valued surface integral

Prove the following formula, where over the surface of a sphere with radius R. is integrated around the origin, whereby the location \ ({\ boldsymbol {r}} ^ {\ prime} \) recorded during the integration can be inside or outside the sphere:
$$ \ frac {3} {4 \ uppi R ^ {3}} \ oint_ {r = R} \ mathbf {d} \ boldsymbol {f} \ frac {1} {| {\ boldsymbol {r}} - { \ varvec {r}} ^ {\ prime} |} = \ frac {{\ varvec {r}} ^ {\ prime}} {[{\ rm max} (r ^ {\ prime}, R)] ^ { 3}}. $$

Solution hint:

Use spherical coordinates where the zAxis in the direction of \ ({\ varvec {r}} ^ {\ prime} \), and the development of \ (1 / | {\ varvec {r}} - {\ varvec {r}} ^ { \ prime} | \) according to spherical surface functions or Legendre polynomials.

13.13 • Spherical and Cartesian multipole moments

Determine the relationships between the Cartesian and spherical multipole moments up to and including the quadrupole moment.

Solution hint:

Use the explicit expressions for the spherical surface functions Ylm up to l = 2.

13.14 ••• Spherical multipole moments: circular disk

In the case of a charge distribution, the charge density is only unequal to zero in a circular disk with a radius r0 in the x-y-Plane around the origin. In this circular disk the surface charge density has the value \ (\ sigma_ {0} \). Determine the spherical multipole moments of this charge distribution.

Solution hint:

Use the properties of the spherical surface functions or the (assigned) Legendre functions from this chapter and the mathematical literature.

13.15 ••• Spherical multipole moments: circular line

A charge distribution is only unequal to zero on a circular line with a radius r0 in the x-z-Plane around the origin. The circular line has the line charge density \ (\ lambda_ {0} \). Explain: The spherical multipole moments of this charge distribution do not disappear only if l and m both are even numbers.

Solution hint:

Use the properties of the spherical surface functions or the (assigned) Legendre functions from this chapter and the mathematical literature.

Detailed solutions to the tasks

13.1

The spatial Fourier transform is
$$ \ begin {aligned} \ tilde {f} ({\ varvec {k}}) & = \ frac {1} {\ left (\ sqrt {2 \ uppi} \ right) ^ {3}} \ int \ mathrm {d} ^ {3} x \, f ({\ varvec {r}}) \ mathrm {e} ^ {\ mathrm {i} {\ varvec {k}} \ cdot {\ varvec {r}}} \ & = \ frac {1} {\ left (\ sqrt {2 \ uppi} \ right) ^ {3}} \ int \ mathrm {d} ^ {3} x \, f (r) \ mathrm {e } ^ {\ mathrm {i} kr \ cos \ alpha} \ end {aligned} $$
for a spherically symmetrical function f, where \ (k = | {\ varvec {k}} | \) and \ (r = | {\ varvec {r}} | \) have been set and α is the angle between these two vectors. Because of the symmetry, one can always place the coordinate system for integration in such a way that α equals the angle \ (\ vartheta \) of \ ({\ boldsymbol {r}} \) to zAxis is. So is
$$ \ tilde {f} ({\ boldsymbol {k}}) = \ frac {1} {\ left (\ sqrt {2 \ uppi} \ right) ^ {3}} \ int_ {0} ^ {\ infty } r ^ {2} \, \ mathrm {d} r \, f (r) \ int_ {0} ^ {2 \ uppi} \ mathrm {d} \ varphi \ int _ {- 1} ^ {1} \ mathrm {d} \ cos \ vartheta \, \ mathrm {e} ^ {\ mathrm {i} kr \ cos \ vartheta}. $$
One can see directly from this expression that it only depends on \ (k = | {\ varvec {k}} | \), but not on the orientation of \ ({\ varvec {k}} \), i.e. spherical is symmetrical. However, you can evaluate it further:
$$ \ begin {aligned} \ tilde {f} (k) & = \ frac {1} {\ sqrt {2 \ uppi}} \ int_ {0} ^ {\ infty} r ^ {2} \, \ mathrm {d} r \, f (r) \ frac {\ mathrm {e} ^ {\ mathrm {i} kr} - \ mathrm {e} ^ {- \ mathrm {i} kr}} {\ mathrm {i} kr} \ & = \ frac {2} {\ sqrt {2 \ uppi} k ^ {3}} \ int_ {0} ^ {\ infty} x \ kern 1.0ptf \ left (\ frac {x} {k } \ right) \ sin x \, \ mathrm {d} x, \ end {aligned} $$
where \ (x = kr \) was substituted.

13.2

  1. (a)
    Defined as an abbreviation
    $$ g (x) = \ left \ {\ begin {array} [] {ll} 1 & \ text {f \ "{u} r \} -x_ {0} / 2 \ leq x \ leq x_ {0} / 2 \ 0 & \ text {otherwise} \ end {array} \ right., $$
    so you can write:
    $$ f (x) = \ sum_ {j = -n} ^ {n} g (x-j (x_ {0} + d)). $$
    Let us first calculate the Fourier transform of G:
    $$ \ tilde {g} (k) = \ frac {1} {\ sqrt {2 \ uppi}} \ int _ {- x_ {0} / 2} ^ {x_ {0} / 2} \ mathrm {e} ^ {- \ mathrm {i} kx} \, \ mathrm {d} x = - \ sqrt {\ frac {2} {\ uppi}} \ frac {\ sin \ left (\ frac {kx_ {0}} { 2} \ right)} {k}. $$
    Using the rule for the Fourier transform of a shifted function, one obtains
    $$ \ tilde {f} _ {n} (k) = - \ sum_ {j = -n} ^ {n} \ sqrt {\ frac {2} {\ uppi}} \ frac {\ sin \ left (\ frac {kx_ {0}} {2} \ right)} {k} \ mathrm {e} ^ {- \ mathrm {i} j (x_ {0} + d) k}. $$
    You can put the constant factor in front of the sum and swap the sign of the summation variable without changing anything in the sum:
    $$ \ tilde {f} _ {n} (k) = - \ sqrt {\ frac {2} {\ uppi}} \ frac {\ sin \ left (\ frac {kx_ {0}} {2} \ right )} {k} \ sum_ {j = -n} ^ {n} \ mathrm {e} ^ {\ mathrm {i} j (x_ {0} + d) k}. $$
    So with that
    $$ f_ {n} (x) = - \ int _ {- \ infty} ^ {\ infty} \ frac {\ sin \ left (\ frac {kx_ {0}} {2} \ right)} {\ uppi k } \ sum_ {j = -n} ^ {n} \ mathrm {e} ^ {\ mathrm {i} j (x_ {0} + d) k} \ mathrm {e} ^ {\ mathrm {i} kx} \ mathrm {d} k. $$
  2. (b)
    In the exponential representation of the Fourier series, the coefficients result from (8.84) as follows:
    $$ c_ {m} = \ frac {1} {x_ {0} + d} \ int _ {- x_ {0} / 2} ^ {x_ {0} / 2 + d} f (x) \ mathrm {e } ^ {- 2 \ uppi \ mathrm {i} mx / (x_ {0} + d)} \ mathrm {d} x, $$
    where \ (c = -x_ {0} / 2 \) was chosen and
    $$ \ omega = \ frac {2 \ uppi} {T} = \ frac {2 \ uppi} {x_ {0} + d} $$
    is. Since in the selected integration area \ (f (x) \) only between \ (- x_ {0} / 2 \) and \ (x_ {0} / 2 \) is different from zero and has the value 1 there, we have
    $$ \ begin {aligned} c_ {m} & = \ frac {1} {x_ {0} + d} \ int _ {- x_ {0} / 2} ^ {x_ {0} / 2} \ mathrm {e } ^ {- 2 \ uppi \ mathrm {i} mx / (x_ {0} + d)} \ mathrm {d} x \ & = - \ frac {1} {\ uppi m} \ sin \ left (\ frac {2 \ uppi mx_ {0}} {2 (x_ {0} + d)} \ right) \ end {aligned} $$
    and with (8.83) the Fourier series representation
    $$ f (x) = - \ sum_ {m = - \ infty} ^ {\ infty} \ frac {1} {\ uppi m} \ sin \ left (\ frac {2 \ uppi mx_ {0}} {2 (x_ {0} + d)} \ right) \ mathrm {e} ^ {2 \ uppi \ mathrm {i} mx / (x_ {0} + d)}. $$
  3. (c)
    With the two representations for \ (f (x) \) from subtasks (a) and (b), the following must apply:
    $$ \ begin {aligned} & \ quad- \ int _ {- \ infty} ^ {\ infty} \ frac {\ sin \ left (\ frac {kx_ {0}} {2} \ right)} {\ uppi k } \ sum_ {j = - \ infty} ^ {\ infty} \ mathrm {e} ^ {\ mathrm {i} j (x_ {0} + d) k} \ mathrm {e} ^ {\ mathrm {i} kx} \ mathrm {d} k \ & = - \ sum_ {m = - \ infty} ^ {\ infty} \ frac {1} {\ uppi m} \ sin \ left (\ frac {2 \ uppi mX_ { 0}} {2} \ right) \ mathrm {e} ^ {2 \ uppi \ mathrm {i} mX}, \ end {aligned} $$
    where \ (X = x / (x_ {0} + d) \) was introduced for the abbreviation.
    Let us now multiply both sides by \ (- \ uppi \) and replace \ (k = k ^ {\ prime} / (x_ {0} + d) \). This is listing
    $$ \ begin {aligned} & \ int _ {- \ infty} ^ {\ infty} \ frac {1} {k ^ {\ prime}} \ sin \ left (\ frac {k ^ {\ prime} X_ {0 }} {2} \ right) \ mathrm {e} ^ {\ mathrm {i} k ^ {\ prime} X} \ left (\ sum_ {j = - \ infty} ^ {\ infty} \ mathrm {e} ^ {\ mathrm {i} jk ^ {\ prime}} \ right) \ mathrm {d} k ^ {\ prime} \ & = 2 \ uppi \ sum_ {m = - \ infty} ^ {\ infty} \ frac {1} {2 \ uppi m} \ sin \ left (\ frac {2 \ uppi mX_ {0}} {2} \ right) \ mathrm {e} ^ {2 \ uppi \ mathrm {i} mX}. \ end {aligned} $$
    On the right side of the equation, a factor \ (2 \ uppi \) was added in front of the sum. If one compares the two sides, one sees that of the integral over \ (k ^ {\ prime} \) only a sum over the discrete values ​​\ (2 \ uppi m \) remains. So it applies
    $$ \ sum_ {j = - \ infty} ^ {\ infty} \ mathrm {e} ^ {\ mathrm {i} jk ^ {\ prime}} = 2 \ uppi \ sum_ {m = - \ infty} ^ { \ infty} \ delta \ left (k ^ {\ prime} -2 \ uppi m \ right). $$
    This was to be shown.

13.3

  1. (a)
    The four sides of the square (starting at the point \ ((a, a) \) of the complex plane, in the mathematically positive sense) are described by
    $$ \ begin {aligned} z & = x + \ mathrm {i} a \ quad \ text {with \} -a \ leq x \ leq a, \ z & = - a + \ mathrm {i} y \ quad \ text { with \} -a \ leq y \ leq a, \ z & = x- \ mathrm {i} a \ quad \ text {with \} -a \ leq x \ leq a, \ z & = a + \ mathrm {i } y \ quad \ text {with \} -a \ leq y \ leq a. \ end {aligned} $$
    So is
    $$ \ begin {aligned} \ oint z \, \ mathrm {d} z & = \ int_ {a} ^ {- a} (x + \ mathrm {i} a) \, \ mathrm {d} x + \ int_ {a } ^ {- a} (- a + \ mathrm {i} y) \ mathrm {i} \, \ mathrm {d} y \ & \ quad + \ int _ {- a} ^ {a} (x- \ mathrm { i} a) \, \ mathrm {d} x + \ int _ {- a} ^ {a} (a + \ mathrm {i} y) \ mathrm {i} \, \ mathrm {d} y \ & = - \ int _ {- a} ^ {a} (x + \ mathrm {i} a) \, \ mathrm {d} x- \ int _ {- a} ^ {a} (- a + \ mathrm {i} y) \ mathrm { i} \, \ mathrm {d} y \ & \ quad + \ int _ {- a} ^ {a} (x- \ mathrm {i} a) \, \ mathrm {d} x + \ int _ {- a} ^ {a} (a + \ mathrm {i} y) \ mathrm {i} \, \ mathrm {d} y \ & = - 2 \ mathrm {i} a \ int _ {- a} ^ {a} \, \ mathrm {d} x + 2 \ mathrm {i} a \ int _ {- a} ^ {a} \, \ mathrm {d} y = 0. \ end {aligned} $$
  2. (b)
    Cauchy's integral theorem gives directly
    $$ \ oint z \, \ mathrm {d} z = 0, $$
    since the function \ (f (z) = z \) is analytic (within the square, but of course also in the whole complex plane).

13.4

  1. (a)
    With the same parameterization of the integral as in the previous exercise, the result is
    $$ \ begin {aligned} \ oint \ frac {1} {z} \, \ mathrm {d} z & = - \ int _ {- a} ^ {a} \ frac {1} {x + \ mathrm {i} a } \, \ mathrm {d} x- \ int _ {- a} ^ {a} \ frac {1} {- a + \ mathrm {i} y} \ mathrm {i} \, \ mathrm {d} y \ & \ quad + \ int _ {- a} ^ {a} \ frac {1} {x- \ mathrm {i} a} \, \ mathrm {d} x + \ int _ {- a} ^ {a} \ frac {1 } {a + \ mathrm {i} y} \ mathrm {i} \, \ mathrm {d} y \ & = \ int _ {- a} ^ {a} \ frac {2 \ mathrm {i} a} {x ^ {2} + a ^ {2}} \, \ mathrm {d} x + \ int _ {- a} ^ {a} \ frac {2 \ mathrm {i} a} {y ^ {2} + a ^ { 2}} \, \ mathrm {d} y \ & = 4 \ mathrm {i} \ int _ {- 1} ^ {1} \ frac {\ mathrm {d} u} {u ^ {2} +1} , \ end {aligned} $$
    where \ (x = au \) or \ (y = au \) was substituted. This integral can be evaluated elementarily:
    $$ 4 \ mathrm {i} \ int _ {- 1} ^ {1} \ frac {\ mathrm {d} u} {u ^ {2} +1} = 4 \ mathrm {i} \ left [\ arctan u \ right] _ {- 1} ^ {1} = 2 \ uppi \ mathrm {i}. $$
  2. (b)
    The residual theorem gives
    $$ \ oint \ frac {1} {z} \, \ mathrm {d} z = 2 \ uppi \ mathrm {i} \, \ text {Res} _ {0} \ frac {1} {z} = 2 \ uppi \ mathrm {i}, $$
    since the residual at the only pole \ (z_ {1} = 0 \) within the square is equal to 1.

13.5

As in the example in Sect. 13.2, the integral along the real axis of -R. to R. and the integration path is closed with a semicircle in the upper half-plane around the origin (alternatively, a semicircle can be used in the lower half-plane, but the opposite sign must be observed)
$$ \ int _ {- R} ^ {R} \ frac {1} {1 + z ^ {4}} \, \ mathrm {d} z _ {\ text {rA}} = \ int \ frac {1} { 1 + z ^ {4}} \, \ mathrm {d} z _ {\ text {Hk}} + \ oint \ frac {1} {1 + z ^ {4}} \, \ mathrm {d} z _ {\ text {gW}}. $$
Again, as in the example, one can easily argue that the contribution of the semicircle for \ (R \ to \ infty \) approaches 0. So it stays again
$$ \ int _ {- \ infty} ^ {\ infty} \ frac {1} {1 + x ^ {4}} \, \ mathrm {d} x = \ oint \ frac {1} {1 + z ^ { 4}} \, \ mathrm {d} z _ {\ text {gW}}. $$
The integral over the closed path is evaluated with the residual theorem. The integrand has the poles
$$ \ begin {aligned} z_ {1} & = \ mathrm {e} ^ {\ uppi \ mathrm {i} / 4} = \ frac {1} {\ sqrt {2}} (1+ \ mathrm {i }), \ z_ {2} & = \ mathrm {e} ^ {3 \ uppi \ mathrm {i} / 4} = \ frac {1} {\ sqrt {2}} (- 1+ \ mathrm {i }), \ z_ {3} & = \ mathrm {e} ^ {5 \ uppi \ mathrm {i} / 4} = \ frac {1} {\ sqrt {2}} (- 1- \ mathrm {i }), \ z_ {4} & = \ mathrm {e} ^ {7 \ uppi \ mathrm {i} / 4} = \ frac {1} {\ sqrt {2}} (1- \ mathrm {i} ) \ end {aligned} $$
and can thus be written as
$$ \ frac {1} {1 + z ^ {4}} = \ frac {1} {(z-z_ {1}) (z-z_ {2}) (z-z_ {3}) (z- z_ {4})}. $$
The residuals of the two poles in the upper half-plane are then
$$ \ begin {aligned} \ text {Res} _ {z_ {1}} \ frac {1} {1 + z ^ {4}} & = \ lim_ {z \ to z_ {1}} \ frac {z -z_ {1}} {(z-z_ {1}) (z-z_ {2}) (z-z_ {3}) (z-z_ {4})} \ & = \ frac {1} { (z_ {1} -z_ {2}) (z_ {1} -z_ {3}) (z_ {1} -z_ {4})} \ & = \ frac {\ sqrt {2}} {4 \ mathrm {i} (1+ \ mathrm {i})}, \ \ text {Res} _ {z_ {2}} \ frac {1} {1 + z ^ {4}} & = \ frac {1} {(z_ {2} -z_ {1}) (z_ {2} -z_ {3}) (z_ {2} -z_ {4})} \ & = \ frac {\ sqrt {2}} {4 \ mathrm {i} (1- \ mathrm {i})}. \ end {aligned} $$
So you have
$$ \ begin {aligned} \ int _ {- \ infty} ^ {\ infty} \ frac {1} {1 + x ^ {4}} \, \ mathrm {d} x & = 2 \ uppi \ mathrm {i} \ left (\ text {Res} _ {z_ {1}} \ frac {1} {1 + z ^ {4}} + \ text {Res} _ {z_ {2}} \ frac {1} {1+ z ^ {4}} \ right) \ & = \ frac {\ sqrt {2} \ uppi} {2} \ left (\ frac {1} {1+ \ mathrm {i}} + \ frac {1} {1- \ mathrm {i}} \ right) = \ frac {\ sqrt {2} \ uppi} {2}. \ End {aligned} $$

13.6

The general formulas are required to calculate the integrals
$$ \ int _ {- \ infty} ^ {\ infty} x ^ {2n + 1} \ mathrm {e} ^ {- x ^ {2}} \, \ mathrm {d} x = 0 $$
(because of symmetry) and
$$ \ int _ {- \ infty} ^ {\ infty} x ^ {2n} \ mathrm {e} ^ {- x ^ {2}} \, \ mathrm {d} x = \ frac {(2n-1) !!} {2 ^ {n}} \ sqrt {\ uppi}, $$
which can easily be found in formulas. First is the function f0 normalized:
$$ \ langle f_ {0}, f_ {0} \ rangle = \ int _ {- \ infty} ^ {\ infty} \ mathrm {e} ^ {- x ^ {2}} \, \ mathrm {d} x = \ sqrt {\ uppi}, $$
so is
$$ h_ {0} (x) = \ sqrt {\ frac {1} {\ sqrt {\ uppi}}} \ mathrm {e} ^ {- x ^ {2} / 2}. $$
Around H1 to get is from f1 first of all the part that is parallel to it is removed again H0 is:
$$ \ bar {f} _ {1} (x) = f_ {1} (x) - \ frac {\ langle h_ {0}, f_ {1} \ rangle} {\ langle h_ {0}, h_ { 0} \ rangle} h_ {0} (x) = f_ {1} (x) - \ langle h_ {0}, f_ {1} \ rangle h_ {0} (x), $$
there H0 is already normalized. With
$$ \ langle h_ {0}, f_ {1} \ rangle \ propto \ int _ {- \ infty} ^ {\ infty} x \ mathrm {e} ^ {- x ^ {2}} \, \ mathrm {d } x = 0 $$
surrendered
$$ \ bar {f} _ {1} (x) = f_ {1} (x) = x \ mathrm {e} ^ {- x ^ {2} / 2}. $$
This function is also standardized:
$$ \ langle \ bar {f} _ {1}, \ bar {f} _ {1} \ rangle = \ int _ {- \ infty} ^ {\ infty} x ^ {2} \ mathrm {e} ^ { -x ^ {2}} \, \ mathrm {d} x = \ frac {\ sqrt {\ uppi}} {2}, $$
so
$$ h_ {1} (x) = \ sqrt {\ frac {2} {\ sqrt {\ uppi}}} x \ mathrm {e} ^ {- x ^ {2} / 2}. $$
With
$$ \ langle h_ {0}, f_ {2} \ rangle = \ frac {1} {\ sqrt [4] {\ uppi}} \ frac {\ sqrt {\ uppi}} {2} = \ frac {\ sqrt [4] {\ uppi}} {2} \ quad \ text {and} \ quad \ langle h_ {1}, f_ {2} \ rangle = 0 $$
you get
$$ \ bar {f} _ {2} (x) = \ left (x ^ {2} - \ frac {1} {2} \ right) \ mathrm {e} ^ {- x ^ {2} / 2 } $$
and with
$$ \ langle \ bar {f} _ {2}, \ bar {f} _ {2} \ rangle = \ int _ {- \ infty} ^ {\ infty} \ left (x ^ {4} -x ^ { 2} + \ frac {1} {4} \ right) \ mathrm {e} ^ {- x ^ {2}} \, \ mathrm {d} x = \ frac {\ sqrt {\ uppi}} {2} $$
then
$$ h_ {2} (x) = \ sqrt {\ frac {2} {\ sqrt {\ uppi}}} \ left (x ^ {2} - \ frac {1} {2} \ right) \ mathrm { e} ^ {- x ^ {2} / 2}. $$
Finally it results from
$$ \ begin {aligned} \ bar {f} _ {3} (x) & = f_ {3} (x) - \ langle h_ {0}, f_ {3} \ rangle h_ {0} (x) \ \ & \ quad- \ langle h_ {1}, f_ {3} \ rangle h_ {1} (x) - \ langle h_ {2}, f_ {3} \ rangle h_ {2} (x) \ end {aligned } $$
first
$$ \ bar {f} _ {3} (x) = \ left (x ^ {3} - \ frac {3} {2} x \ right) \ mathrm {e} ^ {- x ^ {2} / 2} $$
and then
$$ h_ {3} (x) = \ frac {2} {\ sqrt {3 \ sqrt {\ uppi}}} \ left (x ^ {3} - \ frac {3} {2} x \ right) \ mathrm {e} ^ {- x ^ {2} / 2}. $$

13.7

  1. (a)
    The dipole moment is
    $$ {\ varvec {p}} = \ int \ mathrm {d} V \, {\ varvec {r}} \ rho ({\ varvec {r}}). $$
    If the coordinate system is shifted by \ ({\ boldsymbol {r}} _ {0} \) one has
    $$ \ rho ^ {\ prime} ({\ varvec {r}}) = \ rho ({\ varvec {r}} - {\ varvec {r}} _ {0}) $$
    and thus
    $$ \ begin {aligned} {\ varvec {p}} ^ {\ prime} & = \ int \ mathrm {d} V \, {\ varvec {r}} \ rho ^ {\ prime} ({\ varvec { r}}) = \ int \ mathrm {d} V \, {\ varvec {r}} \ rho ({\ varvec {r}} - {\ varvec {r}} _ {0}) \ & = \ int \ mathrm {d} V ^ {\ prime} \, ({\ varvec {r}} ^ {\ prime} + {\ varvec {r}} _ {0}) \ rho ({\ varvec {r}} ^ {\ prime}), \ end {aligned} $$
    where \ ({\ varvec {r}} ^ {\ prime} = {\ varvec {r}} - {\ varvec {r}} _ {0} \) set and \ (\ mathrm {d} V = \ mathrm {d} V ^ {\ prime} \) was exploited. But this results
    $$ {\ varvec {p}} ^ {\ prime} = \ int \ mathrm {d} V ^ {\ prime} \, {\ varvec {r}} ^ {\ prime} \ rho ({\ varvec {r }} ^ {\ prime}) + {\ varvec {r}} _ {0} \ int \ mathrm {d} V ^ {\ prime} \, \ rho ({\ varvec {r}} ^ {\ prime} ) = {\ varvec {p}} + q {\ varvec {r}} _ {0}. $$
    Is the total charge q = 0, so it is indeed \ ({\ varvec {p}} = {\ varvec {p}} ^ {\ prime} \).
  2. (b)
    The components of the quadrupole moment are
    $$ Q_ {ij} = \ int \ rho ({\ boldsymbol {r}}) \ left (3x_ {i} x_ {j} -r ^ {2} \ delta_ {ij} \ right) \, \ mathrm { d} V. $$
    By shifting the coordinate system, we get as above
    $$ \ begin {aligned} Q ^ {\ prime} _ {ij} & = \ int \ rho ({\ varvec {r}} ^ {\ prime}) \ Big (3 (x ^ {\ prime} _ { i} + x_ {0, i}) (x ^ {\ prime} _ {j} + x_ {0, j}) \ & \ quad - ({\ varvec {r}} ^ {\ prime} + { \ boldsymbol {r}} _ {0}) ^ {2} \ delta_ {ij} \ Big) \, \ mathrm {d} V ^ {\ prime}. \ end {aligned} $$
    If one breaks the brackets and integrates the summands individually, this leads to
    $$ \ begin {aligned} Q ^ {\ prime} _ {ij} & = Q_ {ij} + (3x_ {0, i} p_ {j} + 3p_ {i} x_ {0, j} -2 {\ boldsymbol {p}} \ cdot {\ boldsymbol {r}} _ {0} \ delta_ {ij}) \ & \ quad + (3x_ {0, i} x_ {0, j} - {\ boldsymbol {r}} _ {0} ^ {2} \ delta_ {ij}) q. \ End {aligned} $$
    Disappearance of total charge q and dipole moment \ ({\ varvec {p}} \), so it is indeed \ ({\ varvec {Q}} ^ {\ prime} = {\ varvec {Q}} \).

13.8

The charge density is
$$ \ begin {aligned} \ rho ({\ varvec {r}}) & = \! q_ {0} \ delta (z) \ left [\ delta (xa) \ delta (ya) - \ delta (x + a) \ delta (ya) \ right. \ & \ quad + \ left. \ delta (x + a) \ delta (y + a) - \ delta (xa) \ delta (y + a) \ right]. \ end {aligned} $$
For the monopole moment, i.e. the total charge, the obvious result is immediately obtained
$$ q = \ int \ rho ({\ varvec {r}}) \, \ mathrm {d} V = q_ {0} -q_ {0} + q_ {0} -q_ {0} = 0. $$
Also the x-Component of the dipole moment vanishes:
$$ p_ {1} = \ int x \ rho ({\ varvec {r}}) \, \ mathrm {d} V = aq_ {0} - (- a) q_ {0} + (- a) q_ { 0} -aq_ {0} = 0, $$
also \ (p_ {2} = 0 \). Because of the delta function in z is also \ (p_ {3} = 0 \). For the diagonal elements of the quadrupole tensor one has
$$ \ begin {aligned} Q_ {11} & = \ int \ left (3x ^ {2} -r ^ {2} \ right) \ rho ({\ varvec {r}}) \, \ mathrm {d} V \ & = \ int \ left (2x ^ {2} -y ^ {2} -z ^ {2} \ right) \ rho ({\ varvec {r}}) \, \ mathrm {d} V, \ Q_ {22} & = \ int \ left (2y ^ {2} -x ^ {2} -z ^ {2} \ right) \ rho ({\ varvec {r}}) \, \ mathrm {d } V, \ Q_ {33} & = \ int \ left (2z ^ {2} -x ^ {2} -y ^ {2} \ right) \ rho ({\ varvec {r}}) \, \ mathrm {d} V. \ end {aligned} $$
We calculate first
$$ \ int x ^ {2} \ rho ({\ varvec {r}}) \, \ mathrm {d} V = \ int y ^ {2} \ rho ({\ varvec {r}}) \, \ mathrm {d} V = \ int z ^ {2} \ rho ({\ varvec {r}}) \, \ mathrm {d} V = 0 $$
(the last integral results again immediately because of the delta function in z) and thus
$$ Q_ {11} = Q_ {22} = Q_ {33} = 0. $$
For the same reason, you also get it instantly
$$ Q_ {12} = Q_ {21} = \ int 3xy \ rho ({\ varvec {r}}) \, \ mathrm {d} V = 12a ^ {2} q_ {0}. $$
This is the electrical potential up to and including the quadrupole component
$$ \ phi ({\ varvec {r}}) = \ frac {1} {2r ^ {5}} (x, y, z) ^ {\ top} \ left (\ begin {array} [] {ccc } 0 & 12a ^ {2} q_ {0} & 0 \ 12a ^ {2} q_ {0} & 0 & 0 \ 0 & 0 & 0 \ \ end {array} \ right) \ left (\ begin {array} [] {c} x \ y \ z \ \ end {array} \ right) $$
$$ = 12a ^ {2} q_ {0} \ frac {xy} {r ^ {5}} = 6a ^ {2} q_ {0} \ frac {\ sin ^ {2} \ vartheta \, \ sin ( 2 \ varphi)} {r ^ {3}}. $$

13.9

The charge density is
$$ \ rho ({\ boldsymbol {r}}) = \ begin {cases} \ rho_ {0} \ left (\ frac {z} {c} \ right) ^ {n} & \ text {f \ "{ u} r \} -a \ leq x \ leq a, \, - b \ leq y \ leq b, \ & \ quad-c \ leq z \ leq c \ 0 & \ text {otherwise} \ end {cases }. $$
For the total charge you get first
$$ \ begin {aligned} q & = \ int \ rho ({\ boldsymbol {r}}) \, \ mathrm {d} V = 2a \ cdot 2b \ cdot \ rho_ {0} \ cdot \ frac {1} { c ^ {n}} \ int _ {- c} ^ {c} z ^ {n} \, \ mathrm {d} z \ & = \ begin {cases} \ frac {8abc \ rho_ {0}} {n +1} = \ frac {V \ rho_ {0}} {n + 1} & \ text {f \ "{u} r even \} n \ 0 & \ text {f \" {u} r odd \} n \ end {cases}, \ end {aligned} $$
in which V. is the volume of the cuboid.
Because of the independence of the charge density from x and y are at p1 and p2 the integrands are odd functions in x or. y, and you immediately get Bei p3 a case distinction is again necessary:
$$ \ begin {aligned} p_ {3} & = \ int z \ rho ({\ boldsymbol {r}}) \, \ mathrm {d} V = 2a \ cdot 2b \ cdot \ rho_ {0} \ cdot \ frac {1} {c ^ {n}} \ int _ {- c} ^ {c} z ^ {n + 1} \, \ mathrm {d} z \ & = \ begin {cases} 0 & \ text {f \ "{u} r even \} n \ \ frac {8abc ^ {2} \ rho_ {0}} {n + 2} = \ frac {V \ rho_ {0} c} {n + 2} & \ text {f \ "{u} r odd \} n \ end {cases}. \ end {aligned} $$
For the components of the quadrupole tensor one has initially at straight lines n
$$ \ begin {aligned} \ int x ^ {2} \ rho ({\ boldsymbol {r}}) \, \ mathrm {d} V & = \ frac {V \ rho_ {0} a ^ {2}} { 3 (n + 1)} = \ frac {qa ^ {2}} {3}, \ \ int y ^ {2} \ rho ({\ varvec {r}}) \, \ mathrm {d} V & = \ frac {qb ^ {2}} {3}, \ \ int z ^ {2} \ rho ({\ varvec {r}}) \, \ mathrm {d} V & = \ frac {V \ rho_ {0 } c ^ {2}} {n + 3} = \ frac {n + 1} {n + 3} qc ^ {2} \ end {aligned} $$
and thus
$$ \ begin {aligned} Q_ {11} & = q \ left [\ frac {2a ^ {2} -b ^ {2}} {3} - \ frac {n + 1} {n + 3} c ^ {2} \ right], \ Q_ {22} & = q \ left [\ frac {2b ^ {2} -a ^ {2}} {3} - \ frac {n + 1} {n + 3} c ^ {2} \ right], \ Q_ {33} & = q \ left [2 \ frac {n + 1} {n + 3} c ^ {2} - \ frac {a ^ {2} + b ^ {2}} {3} \ right]. \ End {aligned} $$
With odd n is against
$$ Q_ {11} = Q_ {22} = Q_ {33} = 0. $$
For the non-diagonal elements it results
$$ Q_ {12} = Q_ {13} = Q_ {23} = 0 $$
for all n, since integrands always occur which are odd in x or in y are.

The electrical potential has for just n i.e. a monopole and a quadrupole component, for odd n on the other hand only a dipole part.

13.10

The integrals appearing here can all be written as follows:
$$ \ begin {aligned} \ int \ mathrm {d} V & = \ int_ {0} ^ {a} \ mathrm {d} x \ int_ {0} ^ {b- \ frac {b} {a} x} \ mathrm {d} y \ int_ {0} ^ {c- \ frac {c} {a} x- \ frac {c} {b} y} \ mathrm {d} z \ & = abc \ int_ {0 } ^ {1} \ mathrm {d} x ^ {\ prime} \ int_ {0} ^ {1-x ^ {\ prime}} \ mathrm {d} y ^ {\ prime} \ int_ {0} ^ { 1-x ^ {\ prime} -y ^ {\ prime}} \ mathrm {d} z ^ {\ prime} \ & = abc \ int_ {0} ^ {1} \ mathrm {d} y ^ {\ prime} \ int_ {0} ^ {1-y ^ {\ prime}} \ mathrm {d} z ^ {\ prime} \ int_ {0} ^ {1-y ^ {\ prime} -z ^ {\ prime }} \ mathrm {d} x ^ {\ prime} \ & = abc \ int_ {0} ^ {1} \ mathrm {d} z ^ {\ prime} \ int_ {0} ^ {1-z ^ { \ prime}} \ mathrm {d} x ^ {\ prime} \ int_ {0} ^ {1-z ^ {\ prime} -x ^ {\ prime}} \ mathrm {d} y ^ {\ prime}, \ end {aligned} $$
where \ (x ^ {\ prime} = x / a \), \ (y ^ {\ prime} = y / b \) and \ (z ^ {\ prime} = z / c \) have been substituted and the last both relationships apply for reasons of symmetry. So you first have for the total charge
$$ \ begin {aligned} q & = abc \ int_ {0} ^ {1} \ mathrm {d} x ^ {\ prime} \ int_ {0} ^ {1-x ^ {\ prime}} \ mathrm {d } y ^ {\ prime} \ int_ {0} ^ {1-x ^ {\ prime} -y ^ {\ prime}} \ mathrm {d} z ^ {\ prime} \, \ rho_ {0} \ & = abc \ rho_ {0} \ int_ {0} ^ {1} \ mathrm {d} x ^ {\ prime} \ int_ {0} ^ {1-x ^ {\ prime}} \ mathrm {d} y ^ {\ prime} \, (1-x ^ {\ prime} -y ^ {\ prime}) \ & = abc \ rho_ {0} \ int_ {0} ^ {1} \ mathrm {d} x ^ {\ prime} \, \ left ((1-x ^ {\ prime}) (1-x ^ {\ prime}) - \ frac {1} {2} (1-x ^ {\ prime}) ^ { 2} \ right) \ & = abc \ rho_ {0} \ int_ {0} ^ {1} \ mathrm {d} x ^ {\ prime} \, \ frac {1} {2} (1-x ^ {\ prime}) ^ {2} = abc \ rho_ {0} \ cdot \ frac {1} {6} = V \ rho_ {0}, \ end {aligned} $$
where the volume \ (V = abc / 6 \) of the Pyramida can also be calculated elementary-geometrically.
The x- The component of the dipole moment is corresponding
$$ \ begin {aligned} p_ {1} & = abc \ int_ {0} ^ {1} \ mathrm {d} x ^ {\ prime} \ int_ {0} ^ {1-x ^ {\ prime}} \ mathrm {d} y ^ {\ prime} \ int_ {0} ^ {1-x ^ {\ prime} -y ^ {\ prime}} \ mathrm {d} z ^ {\ prime} \, ax ^ { \ prime} \ rho_ {0} \ & = \ frac {a ^ {2} bc \ rho_ {0}} {2} \ int_ {0} ^ {1} \ mathrm {d} x ^ {\ prime} x ^ {\ prime} (1-x ^ {\ prime}) ^ {2} = \ frac {a ^ {2} bc \ rho_ {0}} {24} = \ frac {qa} {4} \ end {aligned} $$
and for reasons of symmetry
$$ p_ {2} = \ frac {qb} {4}, \ quad p_ {3} = \ frac {qc} {4}. $$
For the quadrupole tensor we first calculate
$$ \ begin {aligned} \ int x ^ {2} \ rho ({\ varvec {r}}) \, \ mathrm {d} V & = \ frac {qa ^ {2}} {10}, \ \ int y ^ {2} \ rho ({\ boldsymbol {r}}) \, \ mathrm {d} V & = \ frac {qb ^ {2}} {10}, \ \ int z ^ {2} \ rho ({\ boldsymbol {r}}) \, \ mathrm {d} V & = \ frac {qc ^ {2}} {10} \ end {aligned} $$
and thus
$$ \ begin {aligned} Q_ {11} & = \ frac {q (2a ^ {2} -b ^ {2} -c ^ {2})} {10}, \ Q_ {22} & = \ frac {q (2b ^ {2} -c ^ {2} -a ^ {2})} {10}, \ Q_ {33} & = \ frac {q (2c ^ {2} -a ^ {2 } -b ^ {2})} {10}. \ end {aligned} $$
The non-diagonal elements are somewhat more complex to calculate, here as an example
$$ \ begin {aligned} Q_ {12} & = 3a ^ {2} b ^ {2} c \ rho_ {0} \ int_ {0} ^ {1} \ mathrm {d} x ^ {\ prime} \ , x ^ {\ prime} \ int_ {0} ^ {1-x ^ {\ prime}} \ mathrm {d} y ^ {\ prime} \, y ^ {\ prime} (1-x ^ {\ prime } -y ^ {\ prime}) \ & = 3a ^ {2} b ^ {2} c \ rho_ {0} \ int_ {0} ^ {1} \ mathrm {d} x ^ {\ prime} \ , x ^ {\ prime} \ left [(1-x ^ {\ prime}) \ frac {1} {2} (1-x ^ {\ prime}) ^ {2} \ right. \ & \ left .- \ frac {1} {3} (1-x ^ {\ prime}) ^ {3} \ right] \ & = \ frac {a ^ {2} b ^ {2} c \ rho_ {0} } {2} \ int_ {0} ^ {1} \ mathrm {d} x ^ {\ prime} \, x ^ {\ prime} (1-x ^ {\ prime}) ^ {3} \ & = \ frac {a ^ {2} b ^ {2} c \ rho_ {0}} {2} \ cdot \ frac {1} {20} = \ frac {3qab} {20} \ end {aligned} $$
and for reasons of symmetry
$$ Q_ {13} = \ frac {qac} {20}, \ quad Q_ {23} = \ frac {qbc} {20}. $$
For \ (a = b = c \) one has special
$$ Q_ {11} = Q_ {22} = Q_ {33} = 0, $$
$$ Q_ {12} = Q_ {13} = Q_ {23} = \ frac {3qa ^ {2}} {20}. $$
In this case, the quadrupole tensor only has components outside the diagonal.

13.11

First we show: If the potential has the angle dependence of one of the spherical surface functions,
$$ \ phi ({\ boldsymbol {r}}) = \ bar {\ phi} (r) Y_ {lm} (\ vartheta, \ varphi), $$
this also applies to the distribution of the charge. This follows directly from the Poisson equation and from the fact that the spherical surface functions are eigenfunctions of the Laplace operator:
$$ \ begin {aligned} \ rho ({\ varvec {r}}) & = - \ frac {{\ Updelta}} {4 \ uppi} \ phi ({\ varvec {r}}) \ & = - \ frac {{\ Updelta} _ {r} \ bar {\ phi} (r)} {4 \ uppi} Y_ {lm} (\ vartheta, \ varphi) - \ frac {{\ Updelta} _ {\ Upomega} Y_ {lm} (\ vartheta, \ varphi)} {4 \ uppi r ^ {2}} \ bar {\ phi} (r) \ & = \ frac {-r ^ {2} {\ Updelta} _ { r} \ bar {\ phi} (r) + l (l + 1) \ bar {\ phi} (r)} {4 \ uppi r ^ {2}} Y_ {lm} (\ vartheta, \ varphi) \ \ & = \ bar {\ rho} (r) Y_ {lm} (\ vartheta, \ varphi). \ end {aligned} $$
Now we show the other direction: If the charge distribution has the angle dependence of one of the spherical surface functions,
$$ \ rho ({\ boldsymbol {r}}) = \ bar {\ rho} (r) Y_ {lm} (\ vartheta, \ varphi), $$
this also applies to potential. This follows from the multipole expansion and the orthogonality of the spherical surface functions:
$$ \ begin {aligned} q_ {l ^ {\ prime} m ^ {\ prime}} & = \ frac {4 \ uppi} {2l ^ {\ prime} +1} \ int_ {0} ^ {\ infty } r ^ {\ prime 2} \, \ mathrm {d} r ^ {\ prime} \, \ bar {\ rho} (r ^ {\ prime}) r ^ {\ prime l ^ {\ prime}} \ \ & \ cdot \ int \ mathrm {d} \ Upomega ^ {\ prime} \, Y_ {lm} (\ vartheta ^ {\ prime}, \ varphi ^ {\ prime}) Y ^ {*} _ {l ^ {\ prime} m ^ {\ prime}} (\ vartheta ^ {\ prime}, \ varphi ^ {\ prime}) \ & = \ frac {4 \ uppi} {2l ^ {\ prime} +1} \ int_ {0} ^ {\ infty} \ mathrm {d} r ^ {\ prime} \, \ bar {\ rho} (r ^ {\ prime}) r ^ {\ prime l ^ {\ prime} +2} \, \ delta_ {ll ^ {\ prime}} \ delta_ {mm ^ {\ prime}}, \ end {aligned} $$
so
$$ \ begin {aligned} \ phi ({\ varvec {r}}) & = \ sum_ {l ^ {\ prime} = 0} ^ {\ infty} \ frac {1} {r ^ {l ^ {\ prime} +1}} \ sum_ {m ^ {\ prime} = - l ^ {\ prime}} ^ {l ^ {\ prime}} \ frac {4 \ uppi} {2l ^ {\ prime} +1} \ delta_ {ll ^ {\ prime}} \ delta_ {mm ^ {\ prime}} Y_ {l ^ {\ prime} m ^ {\ prime}} (\ vartheta, \ varphi) \ & \ cdot \ int \ mathrm {d} r ^ {\ prime} \, \ bar {\ rho} (r ^ {\ prime}) r ^ {\ prime l ^ {\ prime} +2} \ & = \ frac {4 \ uppi } {2l + 1} \ frac {\ int \ mathrm {d} r ^ {\ prime} \, \ bar {\ rho} (r ^ {\ prime}) r ^ {\ prime l + 2}} {r ^ {l + 1}} Y_ {lm} (\ vartheta, \ varphi) \ & = \ bar {\ phi} (r) Y_ {lm} (\ vartheta, \ varphi). \ end {aligned} $$

13.12

In spherical coordinates the surface element in (13.213) is through
$$ \ mathbf {d} \ varvec {f} = R ^ {2} \ mathrm {d} \ Upomega \, {\ varvec {\ hat {e}}} _ {r} $$
$$ = R ^ {2} \ mathrm {d} \ Upomega \ left (\ sin \ vartheta \, \ cos \ varphi \, {\ boldsymbol {\ hat {e}}} _ {x} + \ sin \ vartheta \, \ sin \ varphi \, {\ varvec {\ hat {e}}} _ {y} + \ cos \ vartheta \, {\ varvec {\ hat {e}}} _ {z} \ right) $$
given. If \ ({\ varvec {r}} ^ {\ prime} \) on the zAxis, the expansion (13.193) with \ (\ alpha = \ vartheta \) can be used:
$$ \ frac {1} {| {\ varvec {r}} - r ^ {\ prime} {\ varvec {\ hat {e}}} _ {z} |} = \ sum_ {l = 0} ^ { \ infty} \ frac {r _ {<} ^ {l}} {r _ {>} ^ {l + 1}} P_ {l} (\ cos \ vartheta). $$
The \ (\ varphi \) integration vanishes for the contributions proportionally to \ ({\ varvec {\ hat {e}}} _ {x} \) and \ ({\ varphi {\ hat {e}}} _ { y} \) because of \ (\ int_ {0} ^ {2 \ uppi} \ cos \ varphi \, \ mathrm {d} \ varphi = \ int_ {0} ^ {2 \ uppi} \ sin \ varphi \, \ mathrm {d} \ varphi = 0 \). It remains to be evaluated
$$ \ begin {aligned} & \ oint_ {r = R} \ mathbf {d} \ varvec {f} \ frac {1} {| {\ varvec {r}} - r ^ {\ prime} {\ varvec { \ hat {e}}} _ {z} |} \ & = 2 \ uppi R ^ {2} {\ boldsymbol {\ hat {e}}} _ {z} \ sum_ {l = 0} ^ {\ infty} \ frac {r _ {<} ^ {l}} {r _ {>} ^ {l + 1}} \ int _ {- 1} ^ {1} \ cos \ vartheta \, P_ {l} (\ cos \ vartheta) \ mathrm {d} (\ cos \ vartheta). \ end {aligned} $$
Since \ (\ cos \ vartheta \ equiv P_ {1} (\ cos \ vartheta) \), the remaining integral with (13.91) yields
$$ 2 \ uppi R ^ {2} {\ boldsymbol {\ hat {e}}} _ {z} \ sum_ {l = 0} ^ {\ infty} \ frac {r _ {<} ^ {l}} {r_ {>} ^ {l + 1}} \ frac {2} {2l + 1} \ delta_ {l1} = \ frac {4 \ uppi} {3} R ^ {2} {\ boldsymbol {\ hat {e} }} _ {z} \ frac {r _ {<}} {r _ {>} ^ {2}}, $$
where \ (r _ {<} = {\ rm min} (R, r ^ {\ prime}) \) and \ (r _ {>} = {\ rm max} (R, r ^ {\ prime}) \) is. If \ ({\ varvec {\ hat {e}}} _ {z} \) is written more generally as \ ({\ varvec {r}} ^ {\ prime} / r ^ {\ prime} \), then obtained we
$$ \ oint_ {r = R} \ mathbf {d} \ varvec {f} \ frac {1} {| {\ varvec {r}} - {\ varvec {r}} ^ {\ prime} |} = \ frac {4 \ uppi} {3} R ^ {2} \ frac {{\ boldsymbol {r}} ^ {\ prime}} {r ^ {\ prime}} \ begin {cases} \ displaystyle \ frac {R} {r ^ {\ prime 2}} & \ mbox {f {\ "u} r $ r ^ {\ prime}> R $} \ \ displaystyle \ frac {r ^ {\ prime}} {R ^ {2 }} & \ mbox {f {\ "u} r $ r ^ {\ prime} This gives the result given (13.213).

13.13

First of all, one has for the monopoly moment
$$ \ begin {aligned} q_ {0 \, 0} & = \ frac {4 \ uppi} {2 \ cdot 0 + 1} \ int \ mathrm {d} V ^ {\ prime} \, \ rho ({ \ boldsymbol {r}} ^ {\ prime}) r ^ {\ prime 0} Y ^ {*} _ {00} (\ vartheta ^ {\ prime}, \ varphi ^ {\ prime}) \ & = 4 \ uppi \ int \ mathrm {d} V ^ {\ prime} \, \ rho ({\ varvec {r}} ^ {\ prime}) \ sqrt {\ frac {1} {4 \ uppi}} = \ sqrt {4 \ uppi} q. \ End {aligned} $$
At the dipole moments is
$$ \ begin {aligned} q_ {1 \, 0} & = \ frac {4 \ uppi} {2 \ cdot 1 + 1} \ int \ mathrm {d} V ^ {\ prime} \, \ rho ({ \ boldsymbol {r}} ^ {\ prime}) r ^ {\ prime 1} Y ^ {*} _ {10} (\ vartheta ^ {\ prime}, \ varphi ^ {\ prime}) \ & = \ frac {4 \ uppi} {3} \ int \ mathrm {d} V ^ {\ prime} \, \ rho ({\ varvec {r}} ^ {\ prime}) r ^ {\ prime} \ sqrt {\ frac {3} {4 \ uppi}} \ cos \ vartheta ^ {\ prime} \ & = \ sqrt {\ frac {4 \ uppi} {3}} \ int \ mathrm {d} V ^ {\ prime} \, \ rho ({\ varvec {r}} ^ {\ prime}) z ^ {\ prime} = \ sqrt {\ frac {4 \ uppi} {3}} p_ {3} \ end {aligned} $$
and
$$ q_ {1 \, 1} = - \ frac {4 \ uppi} {3} \ int \ mathrm {d} V ^ {\ prime} \, \ rho ({\ varvec {r}} ^ {\ prime }) r ^ {\ prime} \ sqrt {\ frac {3} {8 \ uppi}} \ sin \ vartheta ^ {\ prime} \ mathrm {e} ^ {- \ mathrm {i} \ varphi ^ {\ prime }} $$
$$ = - \ sqrt {\ frac {2 \ uppi} {3}} \ int \ mathrm {d} V ^ {\ prime} \, \ rho ({\ varvec {r}} ^ {\ prime}) r ^ {\ prime} \ sin \ vartheta ^ {\ prime} (\ cos \ varphi ^ {\ prime} - \ mathrm {i} \ sin \ varphi ^ {\ prime}) $$
$$ = - \ sqrt {\ frac {2 \ uppi} {3}} \ int \ mathrm {d} V ^ {\ prime} \, \ rho ({\ varvec {r}} ^ {\ prime}) ( x ^ {\ prime} - \ mathrm {i} y ^ {\ prime}) $$
$$ = - \ sqrt {\ frac {2 \ uppi} {3}} (p_ {1} - \ mathrm {i} p_ {2}); $$
corresponding
$$ q_ {1, -1} = \ sqrt {\ frac {2 \ uppi} {3}} (p_ {1} + \ mathrm {i} p_ {2}). $$
In a similar way you get then too
$$ \ begin {aligned} q_ {20} & = \ sqrt {\ frac {\ uppi} {5}} Q_ {33}, \ q_ {21} & = - \ sqrt {\ frac {2 \ uppi} {15}} (Q_ {13} - \ mathrm {i} Q_ {23}), \ q_ {2, -1} & = \ sqrt {\ frac {2 \ uppi} {15}} (Q_ {13 } + \ mathrm {i} Q_ {23}), \ q_ {22} & = \ sqrt {\ frac {\ uppi} {30}} (Q_ {11} -Q_ {22} -2 \ mathrm {i } Q_ {13}), \ q_ {2, -2} & = \ sqrt {\ frac {\ uppi} {30}} (Q_ {11} -Q_ {22} +2 \ mathrm {i} Q_ { 13}). \ End {aligned} $$

13.14

The charge density is
$$ \ rho ({\ boldsymbol {r}}) = \ sigma_ {0} \ frac {\ delta (\ cos \ vartheta)} {r} $$
for \ (r \ leq r_ {0} \) and otherwise zero. So you have
$$ q_ {lm} = \ frac {4 \ uppi \ sigma_ {0}} {2l + 1} \ int_ {0} ^ {r_ {0}} r ^ {2} \, \ mathrm {d} r \ , r ^ {l-1} \ int_ {0} ^ {2 \ uppi} \ mathrm {d} \ varphi \ int _ {- 1} ^ {1} \ mathrm {d} \ cos \ vartheta $$
$$ \ quad \, \, \ cdot \ delta (\ cos \ vartheta) Y ^ {*} _ {lm} (\ vartheta, \ varphi). $$
The integrals over r and \ (\ vartheta \) can be executed immediately:
$$ q_ {lm} = \ frac {4 \ uppi \ sigma_ {0} r_ {0} ^ {l + 2}} {(2l + 1) (l + 2)} \ int_ {0} ^ {2 \ uppi} \ mathrm {d} \ varphi \, Y ^ {*} _ {lm} (\ uppi / 2, \ varphi). $$
For the integral over \ (\ varphi \) note that the \ (\ varphi \) dependence on Ylm is only contained in the factor \ (\ mathrm {e} ^ {\ mathrm {i} m \ varphi} \). Because of
$$ \ int_ {0} ^ {2 \ uppi} \ mathrm {e} ^ {\ mathrm {i} m \ varphi} \, \ mathrm {d} \ varphi = 2 \ uppi \ delta_ {m \, 0} $$
it immediately follows that \ (q_ {lm} = 0 \) for all \ (m \ neq 0 \) and
$$ \ begin {aligned} q_ {l0} & = \ frac {8 \ uppi ^ {2} \ sigma_ {0} r_ {0} ^ {l + 2}} {(2l + 1) (l + 2) } Y ^ {*} _ {l0} (\ uppi / 2.0) \ & = \ frac {2 \ uppi \ sigma_ {0} r_ {0} ^ {l + 2}} {l + 2} \ sqrt {\ frac {4 \ uppi} {2l + 1}} P_ {l} (0). \ end {aligned} $$
For the further evaluation one uses first that for odd l the Legendre functions P.l are odd functions, so \ (P_ {l} (0) = 0 \) holds. For straight l on the other hand, one can use the following formula, for example, which can be found in the mathematical literature:
$$ P_ {l} (x) = \ frac {1} {2 ^ {l}} \ sum_ {k = 0} ^ {l / 2} (- 1) ^ {k} \ left (\ begin {array } [] {c} l \ k \ \ end {array} \ right) \ left (\ begin {array} [] {c} 2l-2k \ l \ \ end {array} \ right) x ^ {l-2k}. $$
For x = 0 only the summand with \ (k = l / 2 \) remains, so
$$ P_ {l} (0) = \ frac {(- 1) ^ {l / 2}} {2 ^ {l}} \ left (\ begin {array} [] {c} l \ l / 2 \ \ end {array} \ right). $$
Finally stays for just l:
$$ q_ {l0} = \ frac {2 \ uppi \ sigma_ {0} r_ {0} ^ {l + 2}} {l + 2} \ sqrt {\ frac {4 \ uppi} {2l + 1}} \ frac {(- 1) ^ {l / 2}} {2 ^ {l}} \ left (\ begin {array} [] {c} l \ l / 2 \ \ end {array} \ right) . $$
All other multipole moments disappear.

13.15

The charge density is
$$ \ rho ({\ varvec {r}}) = \ lambda_ {0} \ frac {\ delta (r-r_ {0}) \ left (\ delta (\ varphi) + \ delta (\ varphi- \ uppi ) \ right)} {r \ sin \ vartheta}. $$
Inserted into the general equation for the spherical multipole moments (13.196) is initially
$$ \ begin {aligned} q_ {lm} & = \ frac {4 \ uppi \ lambda_ {0}} {2l + 1} \ int_ {0} ^ {\ infty} r ^ {2} \, \ mathrm { d} r \, r ^ {l-1} \ delta (r-r_ {0}) \ int_ {0} ^ {2 \ uppi} \ mathrm {d} \ varphi \ & \ cdot \ int _ {- 1 } ^ {1} \ frac {\ mathrm {d} \ cos \ vartheta} {\ sin \ vartheta} \ left (\ delta (\ varphi) + \ delta (\ varphi- \ uppi) \ right) Y ^ {* } _ {lm} (\ vartheta, \ varphi). \ end {aligned} $$
The integrals over r and \ (\ varphi \) can be executed immediately:
$$ q_ {lm} = \ frac {4 \ uppi \ lambda_ {0} r_ {0} ^ {l + 1}} {2l + 1} \ int _ {- 1} ^ {1} \ frac {\ mathrm { d} \ cos \ vartheta} {\ sin \ vartheta} \ left (Y ^ {*} _ {lm} (\ vartheta, 0) + Y ^ {*} _ {lm} (\ vartheta, \ uppi) \ right ). $$
For the further evaluation of this expression one should first note that the part of Ylm is given by \ (\ mathrm {e} ^ {\ mathrm {i} m \ varphi} \). Here but
$$ \ mathrm {e} ^ {\ mathrm {i} m \ cdot 0} + \ mathrm {e} ^ {\ mathrm {i} m \ cdot \ uppi} = 1 + (- 1) ^ {m} = \ left \ {\ begin {array} [] {ll} 2 & \ text {f \ "{u} r even \} m \ 0 & \ text {f \" {u} r odd \} m \ \ end {array} \ right. $$
holds, it immediately follows that \ (q_ {lm} = 0 \) for all odd m and that one for straight m the sum \ (Y ^ {*} _ {lm} (\ vartheta, 0) + Y ^ {*} _ {lm} (\ vartheta, \ uppi) \) through
$$ 2Y ^ {*} _ {lm} (\ vartheta, 0) = 2 \ sqrt {\ frac {2l + 1} {4 \ uppi} \ frac {(lm)!} {(L + m)!} } P_ {l} ^ {\ kern 1.0ptm} (\ cos \ vartheta) $$
can replace. In the following we limit ourselves to straight m and thus have
$$ \ begin {aligned} q_ {lm} & = 2 \ lambda_ {0} r_ {0} ^ {l + 1} \ sqrt {\ frac {4 \ uppi} {2l + 1} \ frac {(lm) !} {(l + m)!}} \ int _ {- 1} ^ {1} \ frac {\ mathrm {d} \ cos \ vartheta} {\ sin \ vartheta} P_ {l} ^ {\ kern 1.0ptm } (\ cos \ vartheta) \ & = 2 \ lambda_ {0} r_ {0} ^ {l + 1} \ sqrt {\ frac {4 \ uppi} {2l + 1} \ frac {(lm)!} {(l + m)!}} \ int _ {- 1} ^ {1} \ mathrm {d} x \ frac {P_ {l} ^ {\ kern 1.0ptm} (x)} {\ sqrt {1-x ^ {2}}}. \ End {aligned} $$

For odd l are the Legendre polynomials P.l odd functions. Because of (13.161) this is true for even m also for the assigned Legendre functions. For odd l so the integral disappears and with it qlm.

Note: at least for m = 0 one can also evaluate the integral exactly; namely (Gradshteyn & Ryzhik 1994):
$$ \ int _ {- 1} ^ {1} (1-x ^ {2}) ^ {- 1/2} P_ {2n} (x) \ mathrm {d} x = \ left (\ frac {\ Upgamma (1/2 + n)} {n!} \ Right) ^ {2} $$
for all natural numbers n. (The gamma function \ (\ Upgamma (x) \) is discussed in more detail in the "Mathematical Background" 32.4.6.) If we also use that for the gamma function
$$ \ Upgamma (1/2 + n) = \ frac {(2n + 1) !!} {2 ^ {n}} \ Upgamma (1/2) = \ frac {(2n + 1) !!} { 2 ^ {n}} \ sqrt {\ uppi}, $$
holds, one finally obtains
$$ q_ {l \, 0} = 2 ^ {1-l} \ lambda_ {0} r_ {0} ^ {l + 1} \ sqrt {\ frac {4 \ uppi ^ {3}} {2l + 1 }} \ left (\ frac {(l + 1) !!} {(l / 2)!} \ right) ^ {2} $$
for all straight l.