4|ThreeDimensionalMotion
127
Allthesesayisthataparticularcomponentoftheforceis(minus)thederivativeofthepotentialenergy
withrespecttothedistanceinthatdirection. Itisjustastrueof
r
:
F
r
@U
@r
whethersphericalorcylindrical. The e gravitationalforcebytheEarthon amass
m
isradialandis a
functionofthesinglecoordinate
r
(atleasttoagoodapproximation). Inthisapproximationsithas
thepotentialenergy:
F
r
GMm
r
2
dU
dr
!
U
(
r
)= 
GMm
r
(4
:
20)
Example
Twodimensionsareenoughtoseewhat’shappeninghere.Take
U
(
x;y
)=
1
2
k
(
x
2
+
y
2
)
;
then
F
x
@U
@x
kx
and
F
y
@U
@y
ky
Assembletheforcevectorfromthesecomponentsand
~
F
(
x;y
)= 
kx
^
x
ky
^
y
k~r
Fig.4.3
Theforceisdirectedradially towardtheorigin,pointingfromhighertoward lowerpotentialenergies,
andtheforce’smagnitudeisproportionaltothedistancetotheorigin:
F
=
k
x
2
+
y
2
1
=
2
.Thisisjust
aspringforce. Thedrawingshowsasetofequipotentials|curveswith
U
=aconstant|forequal
incrementsinthevalueof
U
. Thearrowsrepresenttheforcevectorsatvariousstartingpoints,with
theirmagnitudesproportionaltothedistancetotheorigin.
Aretheseequationsaseasytoderivein threedimensionsasin one? ? If f you’refamiliarenough
withvectormanipulationstheyare.
~
F
=
m~a
!
~
F
.
~v
=
m
d~v
dt
.
~v
=
d
mv
2
=
2
dt
!
Z
t
f
t
i
~
F
.
~vdt
=
Z
t
f
t
i
d
mv
2
=
2
dt
dt
=
1
2
mv
2
f
1
2
mv
2
i
(4
:
21)
Arethesemanipulationslegal? Yes,butuntilyou u are comfortablewiththem you shouldwritethem
outin
x
-
y
-
z
componentstoseewhat’shappening,especiallythe
m~a
.
~v
equation.
d
dt
(
v
2
x
+
v
2
y
+
v
2
z
)=2
v
x
dv
x
dt
+2
v
y
dv
y
dt
+2
v
z
dv
z
dt
=2
~v
.
~a
andyouseethatthemanipulationswerevalid.
The rstof the integrals in this equation is not necessarily y well-dened withoutmore e eort.
R
t
f
t
i
~
F
.
~vdt
willalmost always dependon thepath you takefrom thebeginningtothe end. . Simply
givingthelimits
t
i
and
t
f
isnotenough.Inthreedimensionstherearemanywaystogetfromonepoint
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4|ThreeDimensionalMotion
128
toanother;inonedimension,therearenotsomany,buteventhereyoucanlookbackattheanalysis
atgure4.2andleadingtoEq.(4.16)toseethattheworkintegralfromzerotozerodependsonthe
path.
W
=
Z
0
0
F
x
dx
=0
ifthemassdoesn’tmoveatall
W
=
Z
x
top
0
F
x
dx
+
Z
0
x
top
F
x
dx
=
mg
sin
k
mg
cos
x
top
mg
sin
+
k
mg
cos
x
top
= 2
k
mg
cos
x
top
6=0
ifitgoesupandback
Insection2.3agraphofpotentialenergy
U
(
x
)becameausefultooltogainaqualitative(and
quantitative)understandingofthemotionofamass.Evenwhenyoucan’tsolvetheequationsexplicitly
youcanseehowthemassbehaves,bouncingbetweenstoppingpoints. Thesamesortofpictureshelp
inthreedimensionstoo.
Justastheequation(4.13)showsthatpowerinonedimensionis
dW=dt
=
F
x
v
x
,theequation
forworkinthreedimensionsgives
W
=
Z
~
F
.
d
~
=
Z
~
F
.
~vdt
!
dW
dt
=
~
F
.
~v
(4
:
22)
Thisistruewhethertheintegraldependsonthepath ornot becausepoweris aderivativeandyou
needtogoonlytheshortdistance
~r
=
~v
t
toevaluateit.
W
=
Z
~r
+
~r
~r
~
F
.
d
~
=
~
F
(
~r
)
.
~r
!
dW
dt
=
~
F
.
d~r
dt
=
~
F
.
~v
Iftheworkintegral,
R
~
F
.
~vdt
=
R
~
F
.
d
~
,doesinfactdependononlythestartingandending
points,andnotonhowyougofromonetotheother,thenpotentialenergy exists,otherwiseitdoes
not. Theequation(4.14)denes
U
asafunctionoftheupperlimit,
x
f
;whathappenstotheintegral
astheupperlimitischanged?Rewritethatequation,takingthezeroofthepotentialtobeat
x
0
.
U
(
x
)= 
Z
x
x
0
F
x
(
x
0
)
dx
0
then
U
(
x
+
x
)= 
Z
x
+
x
x
0
F
x
(
x
0
)
dx
0
Subtractthese,andthepartsoftheintegralsfrom
x
0
to
x
cancel.
U
(
x
+
x
U
(
x
)= 
Z
x
+
x
x
F
x
(
x
0
)
dx
0
Forsmallenough
x
theintegralisjust
F
x
(
x
)
x
,soyoucanwritethisasadierential
dU
F
x
(
x
)
dx
Whathappens in three dimensions? ? Much h the samething. . If f theintegraldoes dependon only the
endpointsandnotonhowyougetfromonetotheother,
U
(
~r
)= 
Z
~r
~r
0
~
F
(
~r
0
)
.
d
~
0
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4|ThreeDimensionalMotion
129
Here,
d
~
isthelittlebitofpathalongwhichyouareintegrating.Makealittlechangeintheendpoint
and
U
(
~r
+
~r
U
(
~r
)= 
Z
~r
+
~r
~r
0
~
F
(
~r
0
)
.
d
~
0
+
Z
~r
~r
0
~
F
(
~r
0
)
.
d
~
0
Z
~r
+
~r
~r
~
F
(
~r
0
)
.
d
~
0
and justas inonedimension,whentheincrement
~r
issmallenoughtheintegral isjustaproduct,
andthisis
dU
~
F
.
d~r
(4
:
23)
Howdoyoupicturethis? Asyoumovearound,thepotentialenergywillchange.Ifyoumovein
adirectionperpendiculartotheforce,thedotproductiszeroandthepotentialenergydoesnotchange
inthatdirection. Moveinthedirectionof
~
F
and
U
decreases. Moveinagainst
~
F
andthepotential
energyincreases. ForthegravitationaleldoftheEarth,theforceisclosetoradial(intheprettygood
approximationthattheEarthisspherical). Thatimpliesthatthesurfacesofconstant
U
arespheres.
Theequationfor
U
isEq.(4.20),
U
GMm=r
,andifthisisaconstantthen
r
isconstant. The
equipotentialsarespheres.
Fig.4.4
ThisisamapoftheEarth’sgravitationalpotentialenergy
mgh
overthesurfaceoftheEarthin
amountainousregion.Thefactor
mg
isomittedfromthecontours,sothattheyrepresenttheheight
abovemeansealevelasyouwanderovertheground.Thecomponentofforceyoumustcontendwithon
yourhikeisdownhill,andEq.(4.23)saysthatifyoumove(
d
~
)perpendiculartothedownhilldirection
then
U
doesn’tchange.Youarewalkingalongacontour.Comparethesimplersetofcontoursinthe
gure4.3.
Example
Ifintwodimensions
U
=
y
x
2
where
and
arepositiveconstants,theequipotentialsare
y
x
2
=
C
(4
:
24)
andEach
C
denesadierentparabolaasanequipotential. Theforcecorrespondingtothispotential
is everywhere perpendicularto theseparabolas. . Whatis theforce? ? Equation n (4.23) determinesthe
answer. Ifyoutake
d~r
tobeinthedirectionsuchthat
y
isaconstant,then
d
~
=^
xdx;
so
dU
~
F
.
d
~
~
F
.
^
xdx
F
x
dx
Thatjustgivestherstoftheequations (4.19),
F
x
@U=@x
=2
x
. Fortheothercomponent,
justmoveinthe
y
-directiontoget
F
y
.Drawsomeequipotentialsanddrawsomeforcevectors.
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4|ThreeDimensionalMotion
130
4.3EandBelds
Howdoesachargebehaveinanelectromagneticeld?Theforcelawis
~
F
=
q
~
E
+
~v
~
B
)=
m
d
2
~r
dt
2
(4
:
25)
The values ofthe electric andmagneticeldshaveto becomputed somewhereelse,because that’s
anothersubject. Suppose e that someone else derives them ormeasures them,theproblemnowis to
determinetheireectoncharges.Ifitsoundsstraight-forward,that’sprobablybecauseyou’velookedat
thecasesofuniformeldswhenyoutookabeginningcourseonthesubject. Eventhere,youprobably
sawonlythetwoeldsseparately. Whenyouputthemtogetheryoualreadygetsomeinterestingand
surprising results. . When n you allowthe elds to be functions of positionthe mathematics escalates
dramatically,butthere’soneimportantsimplicationthatappliestoallmagneticelds,uniformornot:
amagneticelddoesnowork. Recalltheexpression(4.22)forpower:
~
F
.
~v
. Iftheforceismagnetic
then
power=
~
F
.
~v
=
q~v
~
B
.
~v
=0
(4
:
26)
It’stheelectriceldalonethatdoeswork.*
Startwiththecaseofuniformelds
~
E
=0and
~
B
=
B
0
^
z
. You’veseentheresult. Thecharge
willgoinacircle. Wellno,ahelixismorelikely.Theequation(4.26)saysthatthekineticenergyand
hencethespeedofthechargewillbeconstant. Yourecallfromapreviouscoursethattheexpression
fortheradialaccelerationofamassmovinginacircleis
v
2
=r
,sousethathere. Themagneticforce
istowardthecenterofthecircle,so
F
=
qvB
0
=
mv
2
=r
!
v=r
=
qB
0
=m
=
!
Theperiodofthecircularorbitisthecircumferenceoverthespeed: 2
r=v
=2
=!
,andthis
!
isthe
(constant)angularfrequency oftheorbit,(radians/second). . Wheredoes s ahelixenter? ? Don’tforget
thatthereisnoforcealongthe
z
-axis,sothecomponentofvelocity inthatdirection is aconstant.
Combinethatwiththecircularmotionaroundthe
z
-axisandyouhaveahelix.
HereIwillgetthesameresultusingmethodsthatarenotsoelementary,butthatallowforlater
generalization.
m
d
2
~r
dt
2
=
m
d~v
dt
=
q~v
B
0
^
z
=
q
^
x
_
x
+^
y
_
y
+^
z
_
z
B
0
^
z
=
qB
0
_
y
^
x
_
x
^
y
(4
:
27)
Let
qB
0
=m
=
!
,thenthisbreaksintocomponents
d
_
x
dt
=
!
_
y;
d
_
y
dt
!
_
x;
d
_
z
dt
=0
(4
:
28)
Thethirdoftheseequationsstandsbyitselfandsaysthatthe
z
-componentofvelocityisaconstant.
Asfortheothertwo,theyaresimultaneousequationsfor _
x
and _
y
. Howdoyousolvethem? Several
ways. (1)Justas s withalgebraicequationsyou caneliminateoneofthevariables between themand
getasingleequationinonedependentvariable. (2)Followthemethodsofsection3.9andassumean
exponentialsolution. (3)Usematrixmethods. (4)Useoperatormethods.(5)Usecomplexalgebra.
Therstmethodisnotthemostgeneral,butitisaperfectlygoodwaytogettothesolution.
Morepowerfulmethodsarenotneeded,sotheycanwaitforapageortwo. Toeliminateoneofthe
* Thenhowdoesitappearthatinanelectricmotorthemagneticeldispushingonthecurrents
anddoingwork? ThisisasubtleprobleminvolvingtheHalleect,andIwon’ttrytoresolveithere.
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4|ThreeDimensionalMotion
131
coordinates between the equations (4.28), , take _
y
from the rst equation and substitute itintothe
secondequation.
_
y
=
1
!
d
_
x
dt
!
d
_
y
dt
=
1
!
d
2
_
x
dt
2
!
_
x
!
d
2
_
x
dt
2
!
2
_
x
Thisisasimpleharmonicoscillatorequationfor _
x
,sothesolutionis
_
x
=
A
cos(
!t
+
)
;
thentherstequationof(4.28)says
d
_
x
dt
A!
sin(
!t
+
)=
!
_
y
(4
:
29)
Younowhavethesolutionsfor _
x
and for _
y
. Oneintegraleachandyouhavethefunctions
x
(
t
) and
y
(
t
)themselves(plus
z
).
x
(
t
)=
1
!
A
sin(
!t
+
)+
C;
y
(
t
)=
1
!
A
cos(
!t
+
)+
D;
z
(
t
)=
z
0
+
v
z
t
(4
:
30)
Theconstant
A
isarbitrary,so
A=!
istoo. Callthecombinationanotherarbitraryconstant,andIwill
picktheletter
R
.
x
(
t
)=
R
sin(
!t
+
)+
C;
y
(
t
)=
R
cos(
!t
+
)+
D
(4
:
31)
Verifythatthesereallydosatisfytheequations(4.28). Theydescribeacircleofradius
R
andwitha
centeratthe
x
-
y
coordinates
C
and
D
.
(
x
C
)
2
+(
y
D
)
2
=
R
2
~v
Isitgoinginthecorrectdirection?Set
=0,
C
=0,
D
=0,andtake
q>
0tocheck.
x
(
t
)=
R
sin(
!t
)
;
y
(
t
)=
R
cos(
!t
)
(4
:
32)
Equation(4.32)startsat(
x;y
)=(0
;R
),andalittletimelaterthe
x
-coordinateisalittlebitpositive.
Thissays
~v
/^
x
,with
~
B
/^
z
.Theaccelerationisalong
~v
~
B
,andthisisproportionalto^
x
^
z
^
y
.
Thisisexactlywhatthepictureshows|accelerationtowardthecenterofthecircle.(WheredidIuse
assumptionthat
q>
0? Andwhatifitisn’t?)
With
v
z
<
0inthepicture,youhave
Fig.4.5
Thequantity
!
=
qB=m
iscalledthecyclotronfrequency. Thenamereferstoitsapplication
in the rstparticleacceleratortoachievehighenergyatlowvoltages. . Itwasinventedin1932,and
despiteitsantiquity,thisacceleratorstillhassomeapplications,mostlyinmedicine.
OperatorSolution
Thisisanexcusetoshowyouatotallydierentwaytoattackproblems.Itwilllooklikenothingyou’ve
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4|ThreeDimensionalMotion
132
seen before andyoumay wellaskif itis legitimate. . Itis. Theequationof f motion foracharge ina
magneticeldis
m
d~v
dt
=
q~v
~
B;
or
d~v
dt
q
m
~
B
~v
Thishasthesameformasthesimpleequation
dx=dt
=
x
,andyouknowthatthesolutiontothat
isanexponential. Startingfromtheinitialposition
x
0
theresultis
dx
dt
=
x
!
x
=
e
t
x
0
Thenwhycan’tIsolvethepreviouslineas
d~v
dt
q
m
~
B
~v
!
~v
=
e
qt
m
~
B
~v
0
(4
:
33)
NowifIcanjustgureoutwhattheexponentialof \
~
B
"is.
Icandeneitbyitspowerseriesrepresentation.Tokeepthealgebraicfactorsfromaccumulating,
let
~!
q
~
B=m
,then
e
x
=1+
x
+
x
2
2!
+
x
3
3!
+
with
x
=
t~!
gives
~v
(
t
)=
e
t~!
~v
0
=
~v
0
+
t~!
~v
0
+
t
2
2!
~!
(
~!
~v
0
)+
t
3
3!
~!
(
~!
(
~!
~v
0
))+
(4
:
34)
Thereisstillsomemanipulation toputthis seriesintoamanageableform (lessthanyoumaythink
though).Eachtermintheseriesisawell-denedproductofvectors,soIcanatleastaskifitsatises
thedierentialequation.Youdothattheusualway: Youplugin. Dierentiatetheseries(4.34)term
byterm.
d
dt
~v
(
t
)=
~!
~v
0
+
2
t
2!
~!
(
~!
~v
0
)+
3
t
2
3!
~!
(
~!
(
~!
~v
0
))+
andthisis
~!
(4.34)itself. Itworks!
Nowforsomevectormanipulation.Theinitialvector
~v
0
canbeinanydirection,butintheseries
for
~v
allthetermsaftertherstareperpendiculartothedirectionof
~!
(or
~
B
). Call
~
B
the^
z
-direction,
then
v
z
staysconstantandonlytheothercomponentschange. Justset
v
z
=0fornow,andyoucan
additbacklater.
~!
~v
0
rotates
~v
0
by90
about
~!
andeachsuccessivefactorof
~!
eectsarotationbyanother90
aboutthe+
~!
direction. Looking
downalongthe
z
-axis
~
B
(
^
out)and
~!
q
~
B=m
(
^
in)
,therstfeware(andyes,thedirectionsare
right;lookagain)
~v
0
~!
~v
0
(
~!
)
2
~v
0
(
~!
)
3
~v
0
(
~!
)
4
~v
0
Fig.4.6
Now the seriesof termsinEq.(4.34) are easytosum. . The1
st
, 3
rd
,5
th
,etc.arein ()one
directionandthe2
nd
,4
th
,etc.arein()theperpendiculardirection.Indrawingthepicturethelengths
4|ThreeDimensionalMotion
133
ofthevectorschangebecauseofthefactor
!
. Ifyoudrawthepictureusingtheunitvector^
!
asthe
factor,thenallthelengthswouldbethesame. Thecrossproductbyaunitvector^
z
appliedtothe^
x
-^
y
planeispurelya90
rotationaboutthe
z
-axis.
Writetheinitialvelocityastwoterms. Onealongthe
z
-axisandtheotherperpendiculartoit:
~v
0
=^
zv
0
z
+
~v
1
,andonlythe
~v
1
isrotated.Equation(4.34)is
~v
(
t
)=^
zv
0
z
+
~v
1
+
!t
^
!
~v
1
+
!
2
t
2
2!
~v
1
+
!
3
t
3
3!
^
!
~v
1
+
!
4
t
4
4!
~v
1
+
:::
=^
zv
0
z
+
~v
1
!
2
t
2
2!
+
!
4
t
4
4!

+^
!
~v
1
!t
!
3
t
3
3!
+
!
5
t
5
5!

=^
zv
0
z
+
~v
1
cos
!t
+^
!
~v
1
sin
!t
(4
:
35)
Thepositionasafunctionoftimeisnow
~r
0
+
Z
t
0
dt
0
~v
(
t
0
)=
~r
0
+^
zv
0
z
t
+
1
!
^
!
~v
1
+
1
!
~v
1
sin
!t
^
!
~v
1
cos
!t
(4
:
36)
DoesthisagreewithEq.(4.31)?Itrepresentsmotionalongahelix:aconstantvelocityinonedirection
anduniformcircularmotionaboutthatdirectionasanaxis.Seeproblem4.20.
BothEandB
Withbotheldspresent(butstilluniform)theequationsare
~
F
=
q
~
E
+
~v
~
B
)=
m
d~v
dt
(4
:
37)
Thisisalinear,inhomogeneousdierentialequationfor
~v
.Assuchyoucandividetheproblemintwo,
ndingthegeneralsolutiontothehomogeneouspartandndinganyonesolutiontotheinhomogeneous
part. ThehomogeneouspartisEq.(4.31),sothat’sdone. Forasolutiontotheinhomogeneouspart,
noticethat
~
E
and
~
B
areconstants,soaconstant
~v
isaplausibleguess. Itisn’trightbecause
~v
~
B
isperpendicularto
~
B
,andthatisnotnecessarilyinthedirectionof
~
E
. Afunctionthatislinearin
t
works.
x
y
~
E
~
B
out
Insteadofpursuingageneralsolutiontotheseequations,takeaspecialcaseand
solveitbythesimplestpossiblemethods.Afterthatgoforthegeneralities.Take
~
B
along
the^
z
-directionand
~
E
alongthe^
y
-direction,bothuniform. (
q>
0.) Startthechargeat
theoriginwithzerovelocity. Therstthingtohappenisthatthechargeacceleratesalong
^
y
becauseoftheelectriceld,andthemagneticelddoesnothing(initially)becausethe
velocity iszero. . Inalittletimethevelocityhasa ^
y
-componentandnowthemagnetic
eldwillact. Themagneticforce
q~v
~
B
isinthe^
y
^
z
=^
x
-direction,soitpushesthechargealong
the
x
-axis,perpendiculartoboth
~
E
and
~
B
.
m
d~v
dt
=
q
E
^
y
+
~v
B
^
z
!
m
_
v
x
=
qBv
y
m
_
v
y
=
qE
qBv
x
m
_
v
z
=0
(4
:
38)
Thethirdequationsaysthat
v
z
startedatzeroandsoitstaysatzero. Let
!
=
qB=m
,thenwithout
the
E
-termthesearetheequations(4.28)withsolutions
v
x
(
t
) and
v
y
(
t
)in(4.29),soyouknowthe
solutionstothehomogeneouspartoftheseequations. Nowtryguessingaparticularsolutionforthe
inhomogeneouspart,andstartwiththesimplestguesses.Tryaconstant,andifthatdoesn’twork,add
4|ThreeDimensionalMotion
134
atermproportionalto
t
. If
v
y
isaconstant,therstimpliesthat
v
x
/
t
,thenthesecondequation
can’tberight. If
v
x
isaconstant,therstimpliesthat
v
y
iszero,andtheseconddoesworkforthe
rightvalueoftheconstant
v
x
.
v
x;
inh
=
E
B
; v
y;
inh
=0
;
v
x;
hom
=
A
cos(
!t
+
)
; v
y;
hom
A
sin(
!t
+
)
Thegeneralsolutionisthesumofthese.
v
x
(
t
)=
E
B
+
A
cos(
!t
+
)
;
v
y
(
t
)= 
A
sin(
!t
+
)
(4
:
39)
The velocityat timezeroiszero, sothisdetermines
=0and
A
E=B
. Nowintegratethese,
startingattheorigin,
x
(0)=0=
y
(0),andgetting
x
(
t
)=
E
B
t
E
B!
sin(
!t
)
;
y
(
t
)=
E
B!
1 cos(
!t
)
(4
:
40)
x
y
2
E=B!
~
E
x
?
?
~
B
J
Fig.4.7
Thiscurveiscalledacycloid.* Ifapebbleiscaughtinthetreadofyourcar’stire,thenasyoudrive
downtheroadthatpebblewilltracethiscycloidalcurve.
Whatistheinitialbehaviorofthismotion?Expandthesolutionforsmalltime.
x
(
t
)
E
B
t
E
B!
!t
!
3
t
3
=
6
=
E!
2
6
B
t
3
and
y
(
t
)
E
B!
1 1+
!
2
t
2
=
2
=
E!
2
B
t
2
=
qE
2
m
t
2
Forthesesmalltimes,the
t
2
termismuchbiggerthan
t
3
,sothemotionstartsalongthe
y
-directionas
expected. Theshapeofthecurveneartheoriginbehavesas
y
/
x
2
=
3
,consistentwiththegraphjust
beforeEq.(4.38).
TheaveragevelocityisfromEq.(4.40^
xE=B
,perpendiculartobothelds,andyoucanwrite
itas theproduct
~
E
~
B=B
2
. Is s itpossibleforthisresulttobebiggerthanthespeedoflight? ? Of
course,justmakethemagneticeldsmallenough,butthenyouhavetousetherelativisticversionof
theseequations. Theresultwhenyoudoitwiththecorrectrelativisticequationsandsmallenough
B
isthatthereisnooscillation,andthemasswillcontinuetoaccelerate,approachingthespeedoflight
astimegoeson.Nomorecycloid. Theconditionforthissortofoscillatorymotionissimply
E<cB
,
thoughwellbelowthispointtheequationswillnolongerbeexactlythoseofacycloid.
Insolvingtheequationsforcrossedelectricandmagneticelds,alltheequationsusedcompo-
nents. Ididn’ttrytoshowmanipulationsusingthewholevectorformalismwith
~
E
’sand
~
B
’s.Canyou
doitthatway?Yes,butitisalotofeortwithoutenoughbenet.
OtherInitialConditions
Theequations(4.40)assumethatthechargestartedfromrest.Amoregeneralinitialconditioniseasy
enoughtoimplementsimplybyintegratingtheequations(4.39).
x
(
t
)=
+
E
B
t
+
sin(
!t
+
)
;
y
(
t
)=
cos(
!t
+
)
(4
:
41)
* seetheFamousCurvesIndexmentionedattheendofsection0.3.
4|ThreeDimensionalMotion
135
Dependingontheinitialconditions,orequivalentlyonyourchoicesofthefourarbitraryconstantsthis
equation,youwillgetothershapesforthecurves.Theylooklike
y
x
y
x
Fig.4.8
Thecurvesarecalledtrochoidsinsteadofcycloids,andthesetwoexamplessimplyhaverespec-
tivelypositiveandnegativeinitialcomponentsofvelocityinthe
x
-direction.
Non-uniformB-elds
Youmaythinkthatanon-uniformmagneticeldwillbealittlemoredicultthanauniformone. You
wouldbewrong; evenwithout
~
E
in the mixit’s alot more dicult. . The e simplestsuch
B
-eld has
spacedependencelinearin
x
,
y
,and
z
,anditissomethinglike
~
B
=
B
0
^
z
+
z
^
z
1
2
x
^
x
1
2
y
^
y
ThefactorsarenecessaryinordertosatisfyMaxwell’sequations. Theanalogoftheequations(4.27)
and(4.28)involveproductsoffunctions,suchas
x
_
y
and
y
_
z
. Thismakesthedierentialequationsnon-
linearandfarmorediculttosolve.Infacttheycan’tbesolvedexactlyintermsofstandardfunctions.
They requireeither r nding purely y numericalsolutions orusingfarmoresophisticatedapproximation
methods.
4.4MagneticMirrors
ThereisonecasehoweveronwhichIwanttospendsometime. That’snotbecauseitisanyeasier,
butbecause theresult is bothinteresting and important, so that it is worththe time tolookatit.
TheEarth’smagneticeldisnotuniform,beinglargernearthemagneticpolesandweakerinbetween.
Thesolarwindconsistsofchargedparticles,electronsandions,sentathighspeedfromthesun,and
in the absenceofthe Earth’smagneticeld its abrasiveeectwouldgraduallystriptheEarthofits
atmosphere. ThismaybewhathashappenedonMarsbecausethatplanet’smagneticeldisfartoo
weaktode ecttheseparticles.
When charges reach the Earth,many arede ected, butsomearetrapped in theEarth’s eld
andformbeltsofchargedparticlesthatbouncearoundinmagneticmirrors. ThesearetheVanAllen
belts.Inauniformmagneticeld,achargemovesaccordingtoEq.(4.30)|ahelix.TheEarth’seld
isnotuniform,buttoagoodapproximationanelectronwillspiralaroundaB-eldline,withtheaxis
ofthehelixbendingaroundtofollowtheeld. Furthermore,astheeldgetsstrongertheradiusofthe
electron’sorbitwillgetsmalleranditsmotionalongtheeldlinewillslow. Eventuallytheelectron’s
orbitwillstopandreverse,sendingitbacktowardtheothermagneticpole. Themirroreectiswhat
trapsthechargesintheVanAllenbelts.
TheequationsdescribingtheEarth’smagneticeldaresucientlycomplicatedthatyoudon’t
wanttostartwiththem. Instead,takeasimpliedmodeloftheeldthatrequireslessgeometrybut
thatstillcapturestheessenceoftheproblem.
+
z
B-eldlines
z
=0
~
B
=
B
0
^
z
+
B
0
z
2
^
z
z
(
x
^
x
+
y
^
y
)
=‘
2
(4
:
42)
4|ThreeDimensionalMotion
136
The
z
-axis is left and right, , and the
x
-axis isup. . Look k athowthe
~
B
-eld varies along the
z
-axis.
When
x
=
y
=0all youhaveis
~
B
=
B
0
1+
z
2
=‘
2
^
z
,sothattheeldgets strongernearthetwo
ends.Also,asyoumovetoward
z
youpickupacomponentof
~
B
thatpointstowardthe
z
-axis.This
isliketheEarth’seld,whichislargestneartheNorthandSouthmagneticpoles,thoughitdiersin
thatitdoesn’thavetocurvearoundtheEarth. Thismathematicalmodelsimplystraightensoutthe
eldlineswhilestillmaintainingtheessentialstructureoftheeld. Theparameter
isthescaleover
whichtheeldchangesalot. Ontheaxis,ifyougofrom
z
=0to
z
=
theeldstrengthdoubles
andtheeldlinescomeclosertoeachother. WithoutderivingitIwillsimplystatethattheequation
forthemagneticeldlinesis
r
?
=
p
x
2+
y
=
=
p
2+
z
2,where
isaparameterthatspecies
whicheldlineyouarereferringto,andthelineliesinaplanewiththe
z
-axis,whichisleft-to-right.
(Seeproblem4.36forthederivation. Ifyouknowtherighttrick,it’sreallyeasy.) ) Thepictureshows
manyeldlines,withthemagneticeldstrengthlargestontheleftandright.
Theequationsdescribingthemotionofanelectroninamagneticeldare
d~r
dt
=
~v;
d~v
dt
=
q
m
~v
~
B
Withinitialdata
~r
(0)=
~r
0
and
~v
(0)=
~v
0
youcan,usingalotofarithmetic,marchtheelectronalittle
bitatatimethroughitsmotion.Callthelittlebitoftimebythetraditionalname: 
t
.Thestepfrom
timezeroisapproximatedby
~v
(
t
)=
~v
(0)+(
q=m
)
~v
(0)
~
B
~r
(0)
t;
~r
(
t
)=
~r
(0)+
~v
(
t
)
t
(4
:
43)
Aftertherststepcomesthesecondstep,from
t
to2
t
.It’sthesameastherststepexceptthat
theinitialconditionsarenowthevaluesof
~r
(
t
)and
~v
(
t
)thatyoujustcomputed. Andkeepdoing
thisone
t
atatime. (Computersaregoodatblindlyrepetitivearithmetic.)* * Repeatthisfor
n
=0
onupto
n
=theendofyourpatience,
~v
(
n
+1)
t
=
~v
(
n
t
)+(
q=m
)
~v
(
n
t
)
~
B
~r
(
n
t
)
t
~r
(
n
+1)
t
=
~r
(
n
t
)+
~v
(
n
+1)
t
t
Isyourresultaccurateenough? Ifnotyoucanuseasmaller
t
.Ifstillnot,thenyoucanlearnoneof
themoresophisticatedmethodstodothissortofcomputation.Runge-KuttaorRunge-Kutta-Fehlberg
arepopulartechniques. Lookupchapter11of\MathematicalToolsforPhysics"foranintroduction
tothesubject. TheprimitiveformthatIwroteaboveisrarelyusedinpracticebecauseitisnotvery
accurate,andthebettermethodsarenotmuchhardertoimplement.
Inmakingthisplot,Iusedafourth-orderRunge-Kuttamethodtostepthrough themotion of
thechargeasittraversestheeld.
Fig.4.9
* Why y isn’t the lastterm in Eq.(4.43)
~v
(0)
t
? Both h forms arecorrect,butthe one Igaveis
slightlymorestable.
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