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Solution Manual for Mathematical Methods and Physical Insights An Integrated Approach
Alec J. Schramm 1 / 4
- Coordinating Coordinates
2.1 Starting with~r=cos^{+sin^|:
^= @~r=@ j@~r=@j = cos^{+ sin^| p cos 2
- sin
2
= cos^{+ sin^|X and ^ = @~r=@ j@~r=@j = sin^{+cos^| p
2 sin 2 + 2 cos 2
=sin^{+ cos^|X Similar manipulations in spherical coordinates verify Eqn. (2.16).
2.2⇢1
⇢2 d@
ö⇢1
ö⇢2ö
@2 ö @1
@ö⇢
@# d# @ ö " !" d"
2.3 Cylindrical:d~r=
@~r @ d+ @~r @ d+ @~r @z
dz:
@~r @ = ^
@~r @
= ^jcos^{+ sin^|j= ^ @~r @ = ^
@~r @
= ^ jsin^{+cos^|j= ^
@~r @z = ^ k
@~r @z
= ^ k
Spherical:d~r=
@~r @~r dr+ @~r @ d+ @~r @
d:
@~r @r = ^r
@~r @r
= ^r
sincos^{+ sinsin^|+ cos ^ k
= ^r @~r @ = ^
@~r @
= ^
r
coscos^{+ cossin^|sin ^ k
= ^ r @~r @ = ^
@~r @
= ^ jr(sinsin^{+ sincos^|)j= ^ rsin ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. 2 / 4
2.4<> From Eqn. (2.18), the matrix mapping f^{;^|; ^ kgtof^; ^ ; ^ kgisM=
cossin0 sincos0
- 0 1
- So the matrix mapping
!.Similarly, Eqn. (2.19) gives the matrixNmappingf^{;^|; ^ kgtof^r; ^ ; ^
spherical coordinates into cylindrical coordinates isM N 1 . Since these are rotations, we can save a lot of work invokingN 1 =N T . Then that the mapping fromf^r; ^ ; ^ gtof^; ^ ; ^ kg multiplies out to be M N T =
sincos0
- 0 1
cossin0 !
:
The inverse transformation isN M T which is just the transpose ofM N T .
2.5<> Writing out the matrix equation
^r ^
^
!=
sincossinsincos coscoscossinsin sin cos 0 !^{ ^| ^ k !; it's straightforward to verify that^r^r= ^
^ = ^
^ = 1and that^r ^ = ^ , etc.
2.6<> (a)<> cartesian:x
2 +y 2 +z 2 = 1
cylindrical:
2 +z 2 = 1
spherical:r= 1
(b)<> cartesian:x
2 +y 2 = 1
cylindrical:= 1
spherical:rsin= 1
2.7<> The direction cosines are the cartesian components of a unit vector from the origin making angles; ; with the axes. Thus, given two dierent unit vectors ~ Aand ~ B, we see ~ A ~ B= cos gives the identity in (b); part (a) is just a special case of this result.
2.8<> Decomposing the vectors into cartesian components, but using spherical coordinates, ~r=rsincos^{+rsinsin^|+rcos ^ k ~r
=r
sin
cos
^{+r
sin
sin
^|+r
cos 0^ k Then ~r~r
rr
cos =rr
sinsin
coscos
+rr
sinsin
sinsin
+rr
coscos
:
Solving:
cos= = sinsin
coscos
- sinsin
- coscos
= sinsin
cos(
- + coscos
:
2©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. 3 / 4
2.9<> Executing the steps outlined in Example 2.2:
~a= d dt
_^+ _
^
= ^+ _
_ ^+ _ _
^ +
^ + _
_ ^
= ^+ _
_
^
- _
_
^ +
^
_
_ ^
= ^+ 2 _
_
^ +
^
_
_ ^ =
_
2
^+
- 2 _
_
^
=
!2
^+ (+ 2 _!)
^
:
2.10<> First,jJj 2 =jJj jJj=jJ T j jJj=jJ T Jj. Then jJj 2 =
@x=@u @y=@u @x=@v @y=@v
@x=@u @x=@v @y=@u @y=@v
=
( @x @u ) 2
- (
- @x
- (
@y @u )
@u @x @v + @y @u @y @v @x @v @x @u + @y @v @y @u ( @x @v ) 2
@y @v ) 2
Before trying to calculate this horric determinant, note that since^u@~r=@uand^v@~r=@v, the o-diagonal terms are just^u^v which vanishes for an orthogonal system. Moreover, the diagonal terms are just the scale factorsh 2 uandh 2
- Thus
jJj 2 =
h 2 u0 0h 2 v
=h 2 u h 2 v
:X
2.11<> Since ^ and ^ span the tangent plane to the sphere, then using little more than^{^|= ^ kdoes
the trick:
^n ^
^ =
^{coscos+ ^|cossin ^ ksin
(^{sin+ ^|cos) = ^ kcoscos 2 + ^ kcossin 2
- ^|sinsin+ ^{sincos
= ^{sincos+ ^|sinsin+ ^ kcos^rX 2.12<> Area elements (a)<> In cylindrical coordinates, the scale factors areh= 1,h=,hz= 1. Then i.<> on surface of constant,d~a= ^ d dz ii.<> on surface of constant,d~a= ^ d dz iii.<> on surface of constantz,d~a= ^ k d d (b)<> In spherical coordinates, the scale factors arehr= 1,h=r,h=rsin. Then i.<> on surface of constantr,d~a= ^r r 2 sin d d= ^r r 2 d ii.<> on surface of constant,d~a= ^ rsin dr d iii.<> on surface of constant,d~a= ^ r dr d 2.13<> (a)<> Directly leveraging Eqn. (2.12) immediately yields
radial equation:mrmr
_
2 = k r 2
angular equation:r
- 2 _r
_ = 0 (b)<> Simplifying the angular equation as 1 r d dt
r 2_
= 1 mr d dt
mr 2_
= 0reveals angular mo-
mentum conservation:`mr
2_ =constant ©Alec J. Schramm 2022. This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press.
- / 4
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