I.
19.20 )
circular orbit
unstable
a)
Vmax
rmcxwrt.hn
✓
rmnx
>
rmax
=
,
d¥
speed
where
in
→ o
.
have
we
where
:
ftp.lm-ep
=
1-
12M¥
T
"
-
K
¥-1
"
3M
=
.
frame Ñ is
,
=
d¥T d¥
v=
.
where
d¥ has
no
radial component
circular
orbit )
Not
the
not and we
=
radius
In
Ui EI
components of
are
this orthonormal frame
Not
at
is
!÷µ 121^-11?
a
In observer 's
so
My
I → a.
As
greatest potential
In -4-11-121%19
=
plot at
A
has
4-
velocity
of the
.
Ut
=
-
Ui EE
eI.eu#--nqy--le-I.e-I=Ret---landeI.e-I--Nu---O
particle
.
-
.
lift [ 0.0.0.x]
v2.x =/
'
__
✗
=
f-
e¢✗=[ 0,010 I ]
,
eÉ=[ g.
0.0.0
-11-2%1.5=-1
]
y=( 1-
2¥)
-
±
eÉ=[ (1-215%40,0)
a.
eI=r:¥if=r¥
Hiei
=
-
(1-21) .it#.l1-2M-rj-i----l1-2M-rHddt--
And for circular orbits
µ¥T=Y→
,
v-r.ctzm-rl-t.de#- rltzY-1-EF splug ihgihr- rmnx- 3MV3M
b)
-
(1-5)
Same
=
±
-
NIV
radius
in § -15--1
-
.
photon
For circular
Wmax
-
-
-
orbit
3M )
as in
a)
,
.
F-
3M
.
f-
speed of light
.
2
.
Assume the observer
is
light my
the
,
boat
near
the
neutron star
:
-
-
let-1
¥7
observer
•
on
-
-
-
The farthest point
backside
on
farthest
from the axis
tangent
to the
no
circle
is seen
which
,
through
by symmetry
is
light ray
the curve
.
Sy
bending
za-184 =L
A- Ito
=
1¥
1=(11-84)/2
4¥)
=
4¥
.
mail.5kmk-wkm-i.cn
%)
-
=
A.
=
⇐ %)
-
E- if
°
=
Go
-1718
=
72.8°
.
3.
For
a)
the
light ray
light ray
distance
's
will be scattered
The
detection angle
So
some
,
b)
from
away
deflected
depends
ray
on
larger
than
527 M
,
.
the distance
back
turn
can
is
axis
from
axis
its origin
to
(b)
and M
.
.
<
-
?
I
•
-
-
-
-
→
angle of deflection when it is equal to E.
light ray will be directly reflected When it is
For the
be
the
,
.
d
{
^
b.
•
close
to IT
angle
deviated
¥
exceed
,
and this
so
,
,
the
emitting
area
beat #
A
and
is
but E)
=
go
has
lighting
a
back with
limit that it
impact parameter
-
.
Its latentf- a lbltu ¥1T
Tv 2b /
b ti 2¥
4h da ble b't
approximate
ble
to 1st order
for ¥
'
-
-
-
-
is
the
impact parameter
that
light
my deflects by
cannot
is still received in the
a.
=
extra
a
.
between trouble with
-
=
will
angle
that the
radius d receiver
So
light ray
The
,
exactly
IT
.
.
By
WI
19.78 )
a.
zfodw [1- will ¥wÑ
-
=
where
plug
The
number
-
solve for bl a. and
We can
And
1- will
equation fur
back into
received
N=t* A
¥w)
=
0
.
b' la numerically
area
A
.
.
is then
-
c)
By
analysis btn
Na f-* Mi .dz
dimensional
Then
the buser contains
mass
Unveil
of
to
M
?
M
f-*
Regardless of
to
"
and b' la should all be
,
multiples of M
.
In Box 2.1
The
.
black hole M
Mñ
.
,
1019 photons
is
per second
approximately
(number)/see
=
.
to 'tg
-
=
1×10
m
at 104 !
which is also
seconds to receive one
f-* to "
.
0.742×10-28×1015 cm
is in order
dz
.
photon
a
small
figure
reflected back
,
,
there
need
which is not
around
applicable
.
.
4. 46.1 )
di
gap
is
-_
dtit ( Hfltrz ) ) diet 11 total
-
=
-
(
Htttz
1- tttz
,
independent of
are
to 1
,
Also
So
,
,
,
o
along
killing
,
✗
(0
o)
t
-
Z
andy
0, I
,
-0
-
,
so
.
dy4 do
)
killing
vectors with these two
10)
the metric is also
vector with this
symmetry
is
unchanged
11
,
o
,
o,
since
1)
flt-2-1
.
is
unchanged
.
J
.
G)
spatial
so
coordinates of
particle
worldline should be
straight line
a
In
ok
is invariant
at rest
,
.
fitz > €
"
e-
origin
-
"-2440
>
±
C- zip > In 't)
ez )4ñ< link
>z
y zP< stench
-
-
b)
rÑt-z< Tdbnkl
¥*=Éh*ltio)=±tlt -01
=±a
e-¥
e-
the
fort >o
.
So
,
maximum
8L
>
ti
=
happens
In
.
at -1=0
Lt
6.
d5=
dti-llthxxlt-zydx.tl/thyylt-zddytdE
-
Measuring
distance from @ Not
to
Hit it
,
dtio
IX. Yit)
till
)µu,o
=
"
)
where
We
an
ds-jcthxxlt-zxdxl-llthyy.tt#dy4d-=
parametrize
it from
o
to
the
from
path
L*
.
Giro ) to 1×1%2-1
Use hi given
with
parameter
d.
.
L-xt-XYLL-xl-YZCL-x-Z.nl
1¥ ¥* ¥* )
Since
✗(
=
,
we can
write
.
✗
ill)
=
hid
ds becomes
Then
d5=( It hxxlt-h44.nxjdi.tl/-hyylt-nZN).lnYTdd-(nZTdii
Let
To
first
↳
dd-H-lhxxlt-nml.at/lthyylt-n7blnY4-ln4Y
=
=
order
±
approximation
↳ + I
fÉda[4th
*
it nm
-
)
.
Hit
/ It hyylt-nnln44-ln.TT
84-11=4-1 )
Slits
Since
we
=
Lt
-
E) Y
da
-48*-1 hxxlt-nZXHP-lsyythyylt-ntnllnptszil.MY]
sxy-ohxylt-h74-hyx-ohzz-ohxz-hzx-hyz-hz.ge
can
write
slot
8kt)
compactly
as
)=Éfo↳dd[sijthijlt
-
vital ] nini
0
,
7.
Without
gravitational
wave
1
I
→
light
Orient the
Along
the hull
"
d. ×
=
an
-
So ,
disxi
•*
To find
Assume
=
( 1.
toward
X
-
axis
the coordinates
,
geodesic
1,
is
are
oixi
⇐
along
So
.
d*=
+
x-axis
d=✗
,
solutions of
-
.
geodesic egs
:
TIP d¥dd¥
↳ -0 for hnek
ground spacetime
.
,
¥×=¥× -4 ¥m=d¥×
0,01
.
ftp.sluiuM
-0
-8T¥ -28T¥ -8T¥
8×+64=8×4×7
hap
=
µ
,
need
we
changes
0
fitz,
-
fttz )
O
to
Christoffel symbol
( fit
has
frequency
.
w
)
o
( gap
-
perturbation
in
-
.
giiTIp=±(°§¥•t%¥→ %¥-i )
Taking
ray
l
-8T¥ d¥ ¥5
=
sink null
=
my
TIP %Éd¥
dxz
along light
→✗
geodesic
ñsxi
d¥
a
d5=o
displacement
,
first
order
of
is
diagonal )
hap
ftp.t-hii-kptniisl-jp-zfohf! -134¥ -31¥
This -01
Mii -1
strip 1=101*+34,4 -01¥ )
scgii
-
.
for ii-xiy.tt
-
.
①
sT¥=±e¥÷t%÷°-%¥iI=o
For
②
i-x.y.it
8T¥
.
-0
-
STIX =±fh¥i+%¥-
%¥
i-tx.hn/-o.ST-Ix--o
For
i-X.hxx-flt-z-8-fx-I-tlt.it )
For
③
-
8T¥ =E(%¥ -101¥ 4¥71
For i=×
8T¥ =±(¥¥-)=±3h¥=±tlez)
-
,
For
i=y
8T¥
.
For i=z ,
Thus
,
8T¥
22 SX
=
-
-
I Coto -21¥)
=
=
-
I
2ft-24
az
=
Éflttzy
228g
0
=
2×2
azsz
=
Etty
-
an
z)
%¥And
-
-
=
we
①
=
c-
-
-
and
,
,
Z)
)
,
then
I awwscwx)
have
two
1-
'
1-1-2-1
¥¥=
endpoints
set
,
-
awwslwttz) )
=
=
Ea sinlwx) -1C
8×101--8×14=0
so
Edw
-
E. a-wlwscwxltx.lt ( 1- wslwll )) -11
-
82=-18
Sy
0
( wscwxltx.CI
-
awaswx
for
our
null
geodesic
.
S×= 3-za-w.wscwxl -10×+0
IFL 4- wslwll )
8x=
similarly
-
sin ( WH
a
=
If 't
-
on
-
I (0+0-0)=0
have
we
It fit
=
)
( 1- wslwll )) -11
)