electrodynamics / Classical Electrodynamics for Undergraduates - H. Norbury
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CHAPTER 2. VECTORS |
Let dlr and dlµ be inØnitessimal increments of length obtained when moving in the e^r and e^µ directions respectively. From Fig. 2.9 we have
dlr = dr |
(2.64) |
and |
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dlµ = rdµ |
(2.65) |
as shown in Fig. 2.11. The inØnitessimal displacement vector is then |
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dl = dlre^r + dlµe^µ = dre^r + rdµe^µ |
(2.66) |
The circumference of a circle (for which plane polar coordinates are eminently appropriate) is given by R dlµ = R 20ºrdµ = 2ºr. The inØnitessimal area element (magnitude only) is
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dA ¥ dlrdlµ = rdrdµ |
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(2.67) |
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so that the area of a circle is |
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R |
2º |
1 2 |
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R dlrdlµ = R |
0 rdrR |
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ºR2. |
R dA = |
0 |
dµ = (2 R )(2º) = |
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2.8.4Spherical (3-dimensional) Polar Coordinates
This coordinate system is speciØed by (r; µ; ¡) and unit vectors (^er; e^µ; e^¡) shown in Fig. 2.12. In order to sweep out a sphere ¡ varies from 0 to 2º, but µ now only varies from 0 to º. (Note that e^µ points 'down' whereas for plane polar coordinates it pointed 'up'.) Again the basis vectors move as point P moves. The relation between polar and rectangular coordinates is obtained from Fig. 2.13 as
x = r sin µ cos ¡ |
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y = r sin µ sin ¡ |
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z = r cos µ |
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(2.68) |
where r2 = x2 + y2 + z2. The unit vectors are related by |
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e^r = sin µ cos ¡i + sin µ sin ¡j + cos µk |
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e^µ = cos µ cos ¡i + cos µ sin ¡j |
° sin µk |
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(2.69) |
e^¡ = ° sin ¡i + cos ¡j |
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2.8. CURVILINEAR COORDINATES
or
01 0
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e^r |
C |
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B |
cos µ cos ¡ |
cos µ sin ¡ |
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B e^¡ |
= |
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sin ¡ |
cos ¡ |
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@ |
e^µ |
A |
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sin µ cos ¡ |
sin µ sin ¡ |
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° |
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° sin µ |
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C B |
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cos µ |
^j |
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(2.70) |
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1 0 |
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A @ |
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A |
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k |
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(do Problem 2.13) Any vector A is written as A = Are^r + Aµe^µ + A¡e^¡. Let dlr; dlµ; dl¡ be inØnitessimal increments of length obtained when moving in the e^r; e^µ and e^¡ directions respectively. Clearly
dlr = dr |
(2.71) |
and |
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dlµ = rdµ |
(2.72) |
but |
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dl¡ = r sin µd¡ |
(2.73) |
which can be seen from Fig. 2.14. The inØnitessimal displacement vector is then
dl = dlre^r + dlµe^µ + dl¡e^¡ = dre^r + rdµe^µ + r sin µd¡e^¡ |
(2.74) |
There are three inØnitessimal area elements that we can form, namely dA0 = dlrdlµ = rdrdµ (variable r) or dA00 = dlrdl¡ = r2 sin µdrd¡ (variable r) and
the one which has Øxed r and gives the area patch on the surface of a sphere
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dA = dlµdl¡ = r2 sin µdµd¡: |
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(2.75) |
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Thus the surface area of a sphere is dA = |
dlµdl¡ = R2 |
0º sin µdµ |
02ºd¡ = |
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4ºR2 |
. The inØnitessimal volume |
element is |
R |
R |
R |
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R |
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dø = dlrdlµdl¡ = r2 sin µdrdµd¡: |
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(2.76) |
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= 34 ºR3. |
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R |
R |
R |
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so that the volume of a sphere is |
dø = dlrdlµdl¡ = 0Rr2dr 0º sin µdµ 02ºd¡ |
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2.8.5Cylindrical (3-dimensional) Polar Coordinates
^
These coordinates speciØed by ( ; ¡; z) and (^e ; e^¡; k) are shown in Fig 2.15. It is worthwhile to note that for a Øxed slice in z, cylindrical polar coordinates are identical to plane polar coordinates. The angle µ and radius r used in plane polar coordinates is replaced by angle ¡ and radius in cylindrical
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CHAPTER 2. VECTORS |
polar coordinates. In other words cylindrical polar coordinates are just plane polar coordinates with a z axis tacked on. The relation between cylindrical polar coordinates and rectangular coordinates is thus
x = cos ¡ |
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y = sin ¡ |
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z = z |
(2.77) |
the Ørst two of which are analagous to equation (2.57). Unit vectors are related by
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e^ = cos ¡i + sin ¡j |
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e^¡ = ° sin ¡i + cos ¡j |
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(2.78) |
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e^z = k |
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again the Ørst two of which are just equation (2.63). Any vector is |
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A = A e^ + A¡e^¡ + Aze^z: |
(2.79) |
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Note that for spherical polar coordinates the position vector of point P is r = re^r and for plane polar coordinates r = re^r also. However in cylindrical polar coordinates we have
r = e^ + ze^z |
(2.80) |
as can be seen from Fig. 2.15. The inØnitessimal elements of length are |
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dl = d |
(2.81) |
dl¡ = d¡ |
(2.82) |
and |
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dlz = dz |
(2.83) |
where (2.81) and (2.82) are the same as (2.64) and (2.65). The inØnitessimal displacement vector is
dl = dl e^ + dl¡e^¡ + dlze^z = d e^ + d¡e^¡ + dze^z |
(2.84) |
to be compared to (2.66). Exercise: derive the formula for the volume of a cylinder.
2.9. SUMMARY |
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2.8.6Div, Grad and Curl in Curvilinear Coordinates
See Gri±ths.
(to be written up later)
2.9Summary
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CHAPTER 2. VECTORS |
2.10Problems
2.1a) Vector C points out of the page and D points to the right in the same plane. What is the direction of C £ D and D £ C ?
b) B points to the left and A points down the page. What is the direction of B £ A and A £ B ?
2.2 a) Write down the values of the following Kronecker delta symbols: ±11; ±12; ±33; ±13.
b) Write down the values of the following Levi-Civita symbols:
111, 121, 312, 132.
2.3Show that (A £ B)z = AxBy ° AyBx.
2.4Show that the determinant formula (2.23) gives the same results as (2.22).
2.5Show that (2.24) is true for the following values of indices:
a)i = 1; j = 1; l = 1; m = 1,
b)i = 3; j = 1; l = 1; m = 3.
2.6 Prove the following vector identities:
a)A:(B £ C) = B:(C £ A) = C:(A £ B)
b)A:(B £ C) = °B:(A £ C)
c)A £ (B £ C) = B(A:C) ° C(A:B)
d)(A £ B):(C £ D) = (A:C)(B:D) ° (A:D)(B:C)
e)A £ [B £ (C £ D)] = B[A:(C £ D)] ° (A:B)(C £ D).
2.7Calculate the gradient of f(x; y; z) = x + yz2.
° ^ ^
2.8 Calculate the divergence of C = yi + xj and interpret your result.
2.10. PROBLEMS |
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2.9 Calculate the curl of B = j and interpret your result.
2.10 Let f = xy2 and a = (0; 0; 0) and b = (1; 1; 0). Evaluate R ba5f:dl along two diÆerent integration paths and show that the results are the same.
2.11 Check the fundamental theorm of gradients using the function and end points of Problem 2.10.
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2^ |
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2.12 Check Gauss' divergence theorem using C = xzi+y |
j +yzk |
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using the unit cube with a corner at the origin as shown in Fig. 2.6.
2.13 Prove equation (2.69) or (2.70).
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CHAPTER 2. VECTORS |
2.11Answers
2.1a) C £ D points up the page and D £ C points down the page.
b) B£A points out of the page and A£B points into the page.
2.2 a) 1, 0, 1, 0. b) 0, 0, +1, -1.
^2^ ^
2.7i + z j + 2yzk.
2.80
2.90
2.101
2.12. SOLUTIONS |
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2.12Solutions
2.1a) C £ D points up the page and D £ C points down the page.
b) B£A points out of the page and A£B points into the page.
2.2 a) 1, 0, 1, 0.
b) 0, 0, +1, -1.
2.3 From (2.20) A £ B = ijkAiBje^k. Therefore the kth component is (A £ B)k = ijkAiBj. Thus (A £ B)3 = ij3AiBj =1j3A1Bj + 2j3A2Bj + 3j3A3Bj. Now 3j3 = 0 for all values of j and in the Ørst two terms the only non-zero values
will be j = 2 and j = |
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respectively. |
Realizing |
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us from writing out all |
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terms in the |
sum over |
j. |
Thus |
(A £ B)3 = 123A1B2 + 213A2B1 = A1B2 ° A2B1 meaning that
(A £ B)z = AxBy ° AyBx.
2.4
A |
£ |
B = |
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A1 |
A2 |
A3 |
Ø |
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e^1 |
e^2 |
e^3 |
Ø |
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B1 |
B2 |
B3 |
Ø |
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Ø |
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Ø |
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Ø |
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= e^1A2B3 + e^2A3B1 + e^3A1B2 °e^3A1B2 ° e^2A1B3 ° e^1A3B2 = (A2B3 ° A3B2)^e1 + (A3B1 ° A1B3)^e2 +(A1B2 ° A2B1)^e3:
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CHAPTER 2. VECTORS |
2.5 kij klm = ±il±jm ° ±im±jl
a) The left hand side is
kij klm = k11 k11
=111 111 + 211 211 + 311 311
=(0)(0) + (0)(0) + (0)(0)
=0:
The right hand side is ±11±11 °±11±11 = (1)(1)°(1)(1) = 1°1 = 0 b) The left hand side is
kij klm = k31 k13
=131 113 + 231 213 + 331 313
=(0)(0) + (+1)(°1) + (0)(0)
=0 ° 1 + 0 = °1:
The right hand side is ±31±13 ° ±33±11 = (0)(0) ° (1)(1) = °1
2.12. SOLUTIONS |
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2.6
a)
A:(B £ C) = Ak(B £ C)k = Ak ijkBiCj = Bi ijkCjAk = Bi jkiCjAk = Bi(C £ A)i = B:(C £ A)
Also this is
= Cj ijkAkBi = Cj kijAkBi = Cj(A £ B)j = C:(A £ B):
b)
A:(B £ C) = Ak(B £ C)k = Ak ijkBiCj
=Bi ijkAkCj = Bi jkiAkCj = °Bi kjiAkCj
=°Bi(A £ C)i = °B:(A £ C):
c)
A£ (B £ C) = ijkAi(B £ C)je^k = ijkAi jlmBlCme^k
=ijk jlmAiBlCme^k = ° ikj jlmAiBlCme^k
=°(±il±km ° ±im±kl)AiBlCme^k
=°AiBiCke^k + AiBkCie^k
= °C(A:B) + B(A:C):
d)
(A £ B):(C £ D) = (A £ B)k(C £ D)k
=ijkAiBj lmkClDm
=ijk lmkAiBjClDm
=kij klmAiBjClDm
=(±il±jm ° ±im±jl)AiBjClDm
=AiBjCiDj ° AiBjCjDi
=(A:C)(B:D) ° (A:D)(B:C):
e)
A £ [B £ (C £ D)] = ijkAi[B £ (C £ D)]je^k