'intersections
Solutions
of that
sphcre with planes Ax+By+z=O going through the origin. Such
lintersections are givcn in terms of 0 nd g by substituting x=csinOcosg,
/=asin0sin
Exercises of
Chapter
g, and z=acos? in the equation of the plane:
A straightforward calculation now shows that this is a solution of (l). lVe, therefore,
conclude that the geodesics of a sphcrical surface are simply its great circles!
to
5.1。 l
L DerlnitiOn 5。 1.1 let ib>二 :c>‐ :a>to obtain
IC‖
lnterchanging
l
くlal+‖ c― al
lal―
‐
=〉
―lal≦ ‖
C― al
眸‖
α and c we get
lC‖
≦la― Cl=l― (C― a】 =卜 llC―
■ I LSI twO equations Ogetter imメ
l
5
y血 鑢
a‖
al
=ト ー
.
ー
llCIニ ト 11≦ ト al.chanJng
lld― lall=IICi― l-811≦ トー a鴻 〒膝+al
.
(―
Thじ sc are the inequalitiぃ
αЮ ‐
α we get
‐
.betWeen lthe LHS anl the middlO terms of the inequalities
wc are looking for.The inequalities‐
betwech the middle terIIIs and lthe RHS arc
‐
immediate consequences of part(iv)Of DerlnitiOn 5。 1.1.
te the"dittance"betw∝ nl(X)and t(X);Ⅲ en.
5,1.2 LctdJ de・ °
│
ィ三
レ 場fと
ニ
(χ )―
│
(・
‐
1,ll intO
and break up the intc■ d〔 ‐
Now,for the sake of deiniteness assume i●
‐
lrl,‐ 14]・ 1‐ 11,+111,[11,lrl],[1/1,1].In
【
subintervals[-1,-1/i],
Ⅲe rlrst subintewal
e.明
れ
∬ご 牌
∴∴ぶ蹴lTlぷ :L:蹴 Fts武 11:elaIT∝
Tch鳳
‐
1l and
vanish.In the sccond subil"Ⅳ al t=Ob,CauSe,fOr all this subinteⅢ
'
al,‐ 1くxく
by deinition.the function mutt vanish.I Similarly for“ e fOurth ttbiⅢ terval t=1
TherefOrO,dl can be EwHtten as
′
+1111。 )二 場←1犠 ■II脇 。)二 12と
、 イ =ぶ 脇(χ 12と
,
For the i“ t integral,wc have(dO nOt confusc i with√ 1)
と
=:畔 ←
十
畔(∵り
12+2■ ⇒
2牛
│
52
.
F:[:(―
+│):(十 二
ナ
十
)† :T:l
‐
l
│
警l∬ t熙:l畷lli職 脚
灘
I馳 ∫
l■ ‐
理 イ=O for iく
t∬
dtt Wtt VL“
6(i -
『
Js
il
= 6(r
- x')d(y
#
i.Indeed dij=dji and s宙
does hot affect the dittance.Hcnce,we can wdte,gen,rJly,`hm弓 =0・
y')6(z
“
ttt'1r'
(r-:')+f b-r't+4 k-"tt
41'47'4i'
wherc we have uscd the results
dimcnsions and write
of
〓
″ 9
ン
一
χ
為
δ
〓
″ 9
二
ゝガ
隅
二
わrar(Z:L`)=
5.2.2 Generalizing the result of the previous exercise to three dimensions, we
write.
' _.
p"(f) =a6(i - [) + qr6(f - fr)
so rhat now we have two infinities, one at
understood that
r, r',
and
can easily generalize this to n
.[d'kc*'r'-''t
k all have n
comlxtnents.
5.2.4 We note that very generally 5(y-yg)dy=6(x-xg)dx because a change in thc
dummy variable of integration does not change the integral. Obviously we have to
demand that y=y' whenever x=x' otherwisc one side wou.ld be zero while the other
side is infinite. There is anotler subtlety that has to be taken care of. The delta
function is never ncgativc (it is zero cverywhere except at one point where it must
be positive to givc a positive intcgral). Therefore, the two delta functions arE positive
(or zero). However, dy is arbiuarily related to dx. Thus it is possiblc to have a dy
which has opposite sign to dx. To prevent this, wc nccd to take the absolute value of
r,
and one at
c (whose midpoint is) located
direction, we have g2=-Q1=-Q, and rr=-rr=aezi
For-a dipole of length
p,(f) =q[6(i -
aA,)
follow.
can
5.2.5 In this case lhe
rr. The total
![[ o,VY',= Ilja,a(r - i,tp', + [[! a,6(f -f,\a3r
z
it is
fi$
tle
dy and dx, then we get 6(y-y6)ldyl=6(x-xo)ldxl. Now let y=f111 and yg-f(rg) and write
6(f(x)-f(xg))ldfl=6(x-x9)ldxl If we choose xg in such a way that f(x9)=0, the result will
as exPccted.
immediately
where
*
Example 5.2.2.
6(f -7')=
note that p.(x) and p"(x) are zenr everywhere except at x=x' where
they are infinite. This is to be expected, as a point mass (charge) has finite mass
(chargc) but zero volume, hence, infinite density. Also the deniity is expected to be
zero ev_erywhere except at the location of the point particle. ThiJ is reflicted in the
above functions as wcll. The third important prop€rty of the mass (charge) density is
tfat i_a integral (ovcr all space) must give the total mass (charge) of thJ parricle. -For
the above densities we have
z')
=#ffld'tu';1t-7l
hing i and j
5.2.1 First of all
-
=(*!__*.,o'r**t\*f _*,"*,'-'r\*[:_*,,"'u-r)
ノ
We can d,o sh° W that the same result holds for jく
-
= er
I
rezult of Exercise 5.2.4 is not valid because at the new zcro of
f(x) the left hand side is infinitc while &e RHS is zsro. We can follow the same line of
argument as before and write
charge is
6(/(r))=
ez
at the o.igin and pointing in
values of x close to r.1t ttc new zero of f(x). Combining the two equations,
noting that 6(x) is nonzerc only at a single point, wc obtain
for
- 6(f + ai,\
d(/(x)) =
6(i - { ) + . . . + 4,r6(i
"(f) =a,
5.?.3 The intcgral can be separated in
write
54
-
ir) = \!_,0
ol'-a;
la.,
the
For N point particles we obtain
p
pf
r6(i
-
ir
)
and in general
Cartesian coordinate system. Hence,
wc
if f(x)
has
n
#uV-',) * #6(r
w,
la..
roots
may
"l
55
- x,)
and
if f(x)=x2-az,
For example,
6(x'
5.2.6
Integrate
-
where wc have used(1)and the fact that all the
From(2)we can Obtain an‐ 1:
then ldf/dxl=2lxl and the last equation, gives
by parts, and
-
frtar
a) + 6(x +
a)l
If f(x) is well'behaved at t-, the first term will be zero because 6(*e)=s. The second
tenn is just the value of f(x)=dfl61 at x=0 by the delinition of the delia function.
Hence,
1#),_, = -r'(o)
ki*,--t?)r+a,k,
x)1:1xyax =
5.3.2
-f'(ro)
ao-t are two
2と
First note
山 at since
n,we can"五 te
Co(x) is orthogonal to all polynomials of degree less
n,= !lc^6\"r{:4ao.=
[!c,qt\*.r, +k?^ +...!n{ia,
= k,l'c,r'nd, + kiltc,x*twdx +..-= *,J!c,1x,1x,w(x)dx
The double integral runs over all thc x-y plane. We can change to polar coordinates r
and 0 , and integrate over all their possible values namely from 0 to 2* for 0 and from
0 to - for r. The Cartesian elemcnt of area dxdy turns into polar element of area
rdrd0. Wc then have
f2=寛 lη b∬
(l)
than
n.
Now
5‐
3.3
―
ち
frar=2く
Multiply and dividc M=2n(2n‐
=
(?)f*'
(rlrc. (r)d
x)*
Thc second integral above is simply
e"-rh-,=
+
a
^-,
\-r.
-[ft,[t(')r[&,'r'
[lc *,(x)lz w(x)dx
Thus, we have
+...1w@)dx
″
=´ ←:>二
二
:)‐
_:X″
=12■
―
1/n!。
r=ィ 万
Now take a
^″
・
:)・ :
57
obtain
2 out of
each term
―
…
り(″ 丼―
:
〕
1が
:」
"!
(“
56
I″
―
r2[)=π
・
1)(2n‐2)(2n‐ 3)“ 。
(n+1)
1)(2n‐ 2)(2n‐ 3).¨ (■ +1)n(n¨ 1).¨ 2・
'M=2n(2n‐
of the nulnerator to get
t
=.:″
∝
less
│′
Ю f
O
:
つ
毛ヤ り
り
轟
of x are all
_1→
y 鷺
b
all the remaining terms vanishing because the degrees
(5.33) by Co-1(x)w(x) and integrate to get
multiply
rhan
absolutely unrelated constants.
by I, note ,a" [-"-f al is also equal to [, and write
Denote the integral
ィ=膿 …xル
5.3.l
1 d=T(* t)
Idcntifying cro with h+t/ko, Fo with ao and Tn with an-I, gives (5.35) as desired.
The use of ao and ao_q for the two constants in (5.33) is admittedly confusing.
One may be tempted to raise the indices of (3) by one unit to obtain ao! This is, of
coufire, wrong because ao and
This can also be generalized to
Il_a'k -
(3)
To find fo, wc use the fact that (5.33) must hold for all x, and, therefore, coefficients
of equal F)wers of x must equal on both sides. Hence, for xn. we have
ソ0準 ―
二膨》←オ←凛=∝ χ
二∝χ
I岳 夕)ト
=
brackets
)
obtain
f t'1,y1,r*
in cudy
‐ ら
4-台 (≒ 絆
.1
a')=
other terurs
幼 κ
二
+十レト告
濃
→
←卜1れ げズ
'
r(/(,))=i#o(,-,*)
-LL.
″!
漱ぶド
「
帆ξ
躍 鴨T
t計 謬
ttWヽ 甜ぷ蹴』
Ψ 譜咄
|.p
♂
"\'
a, = 22 n(n - r)i
= 2'
+:)=(・ ―
―
―
=(2-:X"―
:)。
¨ ―″+:lr(″
(“
T″
+:)∀
0≦
″≦
“
and v7hen口 Fn,we get
rl“
す
:)=("―
「:)… :F皓 )
:X″
Jl,
and
│
(t
- r' )"*
=←
5。
二
…
:X卜 ∋:
D en
30_ギ 2ソ
〒“
Tり :lχ
...
- n + r)V
(n
intcgration
oft4f
by parts we
. r'- (r - r')*
^
&
obtain
Il_{"*
2
-
t\(zn
-
as
3).
..
5.3
. rl= (2n + t)2'
(' - ;X" - i)
it
f,(r-r')"ar=6ffi
5.3.5 First consider the case where k is odd. Ignoring the multiplicative factor in
the Rodriguez formula, we may write
11_11{ズ
{:χ
22ズ
= (2n + t)l(zn
2ソ
‐
No■ utt u=(1_12)・ 1,and dv=u二 12dx t。
=2“
:' )*'ar
Hence, we obtain
#
:“
fi#[t'
2←
'と
*]
Den=1Zn*t)2,ffi
カ
ird]
輩準ニ
_χ
と
=0+2“ ル
'「
―
χ
り
- z\(a x)(r- "' )"-'
Now use the result obtaincd in Exercisc 5.3.3 to write
3.4 1n"grate by parts using u=(1-x2)■ ,and dVttx to get
il←
r)
The denominator can be written
…
=尭 がⅨ―
亀
い⇒
υ
211≒
1尭
#=篭器
l10-χ l粋 =(lχ
J],x'(r -
"
=,r.-,
- 3.5.7--.(2n-l)2n+l
y=22●
2ズ
n(n-
1
需
So that
21fr
[,(t-r')'&=2'n(n-r)...2.
Now substitute this in,he expression for M to get
i
= 2-
in particular after m=n
Dividing both sides by r(1/2)we obtain
需
n(n- 11" -
[r
The trend is now clear. Thus, we can write a general zath step
:)=(・ :X"二 :)I("― :)
:)i(″
I“
t tt - "' )'-' |], - t] i ;
{i
3[F_lx-2χ
11_11:χ
4← _.2ド
Simllarty
Ⅲ
‐
2F2司
沖
「χ
=
I:*l#[t' - *I]]* =#[t' - of 1.
}
t; 11-x21k is a polynomial in x all of whose terms will vanish at x=() except the
constant term. Thercfore, we havc to find the constant term of 1Ak'1761k-t) {t-*z;t.
Using the binomial expansion, wc obtain
#lo
1
l
58
")']*
By Lemma.5.3.5, the upper limit gives zero. For the lower limit we note that 1dk-lldxk-
。
btain
■
-
-,' )' I-
=
#[*
(i]-" v ].- = I [;]'/
59
[#
('")'].=,
whose constant tcrm is obtained when 2j=k-1 or j=(k-l)/2. Since
worry about j not being an integer. Thus, we have
#[t
-,')'J.-. =f rr
-, ]-,1?
\2 )
θ
ご
θ
=券 ル=1
=券 ・
伺→
ル→
k is odd we need not
Hence,〈 ″│→
-{-"'-'
5。
in the multiplicative
-,
!ir,1;1*=+l
'ttf-
tl$;l)!
6(e
-
e,) = (e10,)= (0 l{
lV,,+,-*
I
and, as expected, the evenness
appearing in the expansion.
For 0'=2mr wJ ob,tain
=f,
where
=
(ol n'1fu1a')
of the
D-function prevents the sine terms from
is something wrong with this equation because the RHS is independent of
while the LHS depends on m! See Example 5.4.10 for clarification.
["-r,1-,)el=il1]nut* * [\r,{'oa,]
There
Il,r,rl,*=!rhl-,#ltr-,,)'l*=h#[(t-r)'il,=o
in the last st"p *L
I
5(0-2mttl= I irqe-2a*)- I irt'
Ztt *-2E *--
!ir,1,ya,=|[fi ar,l**[ir,1,y*l=]lltre)ax+l'r,(-,r)a(-,r)J
a,+
=
[email protected]')
\ o'\=
^L+Licosn(O' 2x
fr'or
The last equality follows from the fact that (-l)k=-l when k is odd.
For evcn k we distinguish two cases: k=0 in which case the integral is simply
and k-2 for which we havc
][fiao
1a(L_- z['rl) fl
E*__
′
卜
の
ノ
衝レ
衝)♂ 券Σ
=Σ l_← ノ
督
“
を
←
ffiffi'-L)' l-'-"',
=
0)
where ye used the completeness of the functions ent l^ffr in the interval [-n,+rl.
This equation can also be wriiten in terms of sines and cosines. We note that \=bo
+b -oL= | 12fr+ | l2rc= I I n, Bn=i (bn-b-n)=0. Hence,
constant, yields
=r(-rF
an ottonoma sd in卜 π
,+π ].
4.2
=[}rr?*-,),=11if,ffi(-l)?
which, after putting
θ
ワπおm
ノ
イ
=琺 ,and′ “
m
5.4.3 FЮ m 6.56)and the reality of【 0)we get
used Lemma 5.3.5.
r(θ )=Σ 4`―
“=∫
(θ)
Let n=‐ m in the above cquatiO■ . Then, we get
s.4.1
consider tm> and
tp
@l^l=(tt ^lfr)e* ,(eln) =(tt:ttnpil
wirh
(nlnl =(nfrln) = fu{$l,lelae(rild =
[.,{^le)(elfide
(3-alθ
′
θ
=券 ル
=券
-1'L=券
←
ⅢTど け
「
劇
満 ノ
石L下 レ
(・
=*#^fi=l)*'-(-1)"-'l=o
!.r m*
n
ど =Σ だ♂θ
ズの=Σ だ"ど =Σ だ議
and write
where
in the last
step we changed the dummy index back to n. Comparing this with
Eq.(5.56) gives bl" =b,or b-.=bi. In particular, for n={) we have bo={. ttris shows
that bo is a real number. Using the definition of Ao and Bo, we obtain \=bo+b-o=
bo+b*o= 2Rebo and Bo=i($o-b-o)=i(bo-bro)=-2lmbo. Hcnce, \ -d Bo are real. As a
particular case we see that if b-o=-bo, thcn \=g which means that bo has no real
part, i.e., it is a purely imaginary
For m=n, wc gct
60
number.
61
514141‐ I Sincci the
iS CVen, we cxpect only thc cosinc
lHCnCI:"e輌 te
5。
十
へ χ
)―
二
∞
S瞥 う
と
へ
毛脚→
v7ith l
Hcnce,
4.5 FЮ m(5。 72)we haVo
LC=岳 ー
=C衝 ∝
→
fO=岩 レ ー
レと
costtチ
I(χ )=:へ
and
terms to be present.
where wc used the resultl of Exalnple S.2.2. In particular the FouHer transfonn,
isダ (1)=√ :π δ
(た )。 Simllarly,the Fou甍 er transfom of
と
ズ
∝
D=轟 レー
り=轟
綾
│=:ル S厳 =務 血
.
十CO:(´ ―子 〉
ト
4二 :氏,臓 ∞Sザ駐と=券 二
lCOSll+子 〉
(タ
n(“
(ι +r′
1:券
=可
勉
Ⅲ
{赫 満
・
+薯
We can obtaln an expans10n for ■ lf we substitute x=a,
C° S筆
l
"(子
)}
.(勉 )2
2
_1(滋 )2_(“ →
in their arguments. This
in the next Exercise.
5.4.7 (a) Take the complex conjugate of
(5。
g・
o))In (5。
1下
‐
If g(■
gり
=岩 恩―
けり =轟 酔 ウ
よ
り
'り
C―
IIヒ
│
which shows that the Fouder transfom is even (Odd)when g(X)iS eVen (Odd)。
m
血 J
c曲
t
e
62
2
(2\
)and if it iS odd, 3(‐ y)=‐ 3(y)。 Hencc, we can write
)iS CVen then g(‐ y)=gも ′
封島 「
→
ズ
→
ヌlpJ3o=土 ズ
卜
封
■
47¥浮 π
Σ
詰 ‐′
二=1_■ _■ _2
4
15 63 143
72b)and uSe the fact that
ロ 0=ズ →
り=轟 Pg・ り〒
岩レ セ
3・
This ininite sum converges more rapidly than the ex,anSiOns de壺 ved in
becaus6 of the second powe'of h in the denominator.The rlrst few tems
both sides of
lS a
72b)make a change of vanables x=‐ y, then
N● w let ka・ ■/4 to to get
‐
}
Note that both g(x) and its Fouricr transform are even
gencral property of Fourier transform that is discusscd
ルa, and dividing by sinた α, we get
滋cotた =1122
ra―
{ニ
g(x) is real to get
蜆
勉
中
詳葛静こ
可赫:二 て
叫
Multiplying both sid,s by
3L―
}〒
:〕
We, therefore, have
cosl=2滋
δ
(x)is
=島 レー
篭0=島 レ とー
ズ
→
:→
ー
た
と
一
χ
と
と
が
菌
戸タと:1(― ル :∬ー
―
+ル L司 券券中
と
=島 性よЁ
考
:『 ン
5.4.6
二
血 → ζ・
ぴ
:毒
17f笠
││
二デ
+品島
た
滋
=品 血
ビ手
│
=義
免 of l
5.4.8 Lct
改→=。 →γ
2fffd嗚 Fr″
755
63
3ψ
2‖ ご ピ
│つ =(2→ γ
5.4.9
,
││││■
From Eq。
(π
■Ⅲ
Ⅲ血t thesc
(ら
2
=ニ
whiti c8uld be wHtten as
‐
‖ごl_t2■ 司+4d″ ルFた 0
3た
to
integral for O(i):
o(D=
0と =二
=ギ
find the Fourier transform of the potential from fte
differential equation. To find the potential itself, we substitute 6(f),in the Fourier
and
(5.81),we get
―
χ
/1岩 ニ
ヤ
ズ
Щ
ルー
中
"げ
ー
て
ルr司の
ギ
. =二 I島 二δ
│'1:]の
に
・‐
多
│ =ニ ズ
1轟 ←
‐
〒
り
レT二 綺
二わ轟レ的
'ン yか 侍
ダ
ャ
幾
界レ
O,ハ =",う =二
司⇒q準 響
刊Ⅲ4歌弓
it is
“
ち
ち
ち
ズ
Hl+ち L=ズ →
腸ル十り
卜=ニ ズ
5.4.10 From DerlnitiOn 5.4.3 and Eq。
Hence, the expression in brackets must be zero:
We note how simple
[!!a't$"o'
Therefore.
10二
substitute
F(E)=&"ffIrt,to17Y-nr
in the preceeding ,equation
to obtain
イ
タ
争昨¬″ ″
欧つ
た
=‖ イ
レ
気
耗わり♂
[島 丹 ∵
rab静 ∬ げぐ
〒
‖♂
ず
`3ら
=‖ 岬→
占 │‐
60篭
where we used Example 5.4.8 with
Example 5.4.E, that
lfff♂
f-7
(1)
Q=Q=l. In the last step we used the fact, again
the inversc transform
"f
Ur*(E) is jusr l/r-r,1. The last
in Eq.(l) above is, of course, the general form of
distribution p(r') given in Example 5.4.9.
64
electrostatic potential .for
:
ち
″
=轟 ニ
3レ 亀
昇ルLン嗚
ト
ャ
2‖ ご (π レ
3ゅ
)γ
81)we haVe
「(1)=7爾 二′・
ン喘亀
inithc Poislon equatiOn to obtain
2Φ
3る
2ノ F=_(2π
▽
.¢ γ
‐
→2fffご (Ix_た
│■ ‐
(5。
from
integral
a
charge
65
‐
岩レ
'
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Chapter 5 - Skeptical Educator