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Unformatted text preview: 78 CHAPTER 7 THE DESCRIPTION OF CURVED SPACETIME 7—2. The following line element corresponds to ﬂat spacetime
d32 = ~dt2 + 2da: dt + dy2 + c122 . Find a coordinate transformation which puts the line element in the usual ﬂat
space form (7.1). Solution: Its hard to give a general prescription for solving this kind of
problem. Guesswork and trial and error are the main methods. We, therefore,
present some solutions without trying to explain exactly how they were arrived
at. ds2 = —alt2 + 2dr dt + dy2 + dz2
The transformation t:t’+x’, :rzct', yzy’, z=z’ leads to ds2 2 ~ (0315’ + dzr’)2 + 2 dx’ (dt’ + dm’) + dy'2 + dz?
= — (dt')2 + (dx’)2 + (dy')2 + (aiz')2 which is the standard form of the ﬂat space line element. 7—3. [GP] The Sagnac Eﬁect The Sagnac effect was worked out in an inertial
frame in Box 3.1 Two light waves propogate in opposite directions around a
rotating ring. The phase of a wave with frequency a) at time t a distance 5
around the ring is \11 E ~w(t — S) + const. (The speed 1) ofa light wave is 1.)
When there is a difference in phase of a multiple of 27r the waves constructively
interfere. It is also possible to work out the Sagnac effect in a frame in rotating with
the interferometer. The line element of ﬂat spacetime in that frame can be
found by deﬁning deﬁning a new coordinate (25 = qb’ + Qt. Derive the condition
for constructive interference in this frame. Solution: In the rotating frame the line element for ﬂat spacetime is d5? 2 —dt2 + d'r2 + r2 [(102 + sing a (da' + swag] . 78 PROBLEM 7.14 87 Evaluated on the world line 300(7), all functions become functions of 7'. For example
daa _ Baa daﬂ aaa _# _ __ : ___ 7.
dT 8x7 (17' 8x7 u
Thus
(1 __ 890;; a g ,y 8a“ 7 ﬂ Gab/3
8—;(3 abu +gaﬁagﬁul. The only possible difﬁculty with this problem is keeping the indices straight. 714. In a certain spacetime geometry the metric is
(132 = —(1 — Ar2)2dt2 + (1 — 14%)?er + 7‘2(d02 + sin2 aw?) . a) Calculate the proper distance along a radial line from the center 7" = 0 to
a coordinate radius 7‘ : R. b) Calculate the area of a sphere of coordinate radius r = R.
c) Calculate the three volume of a sphere of coordinate radius r = R. d) Calculate the four volume of a four dimensional tube bounded by a sphere
of coordinate radius R and two if = constant planes separated by a time T. Solution: a)
s=/OR<1—Ar2)dr=R(1§§2>
where g E Rx/Z.
b) A = Aphere(Rd6)(Rsin9d¢) 7r 27r
/ dQ/ d¢R2sin6=47rR2
0 0 87 88 CHAPTER 7. THE DESCRIPTION OF CURVED SPACETIME V = [sphere [(1—Ar2)dr] [rdﬁHrsiangt] z fanfare 022w r2(1—Ar2)sin6d0 = (42RB)<1—<2>e> where g is as above. d)
V4 = f [(1 — Ar?) at] [ (1 — Ar2) dr] [rd0][rsin was]
: font/ORdr (1—Ar2)2r2/01rd6 02wd¢sin6 Gm) (1 a $62 + 35“) ll 715. [S] Calculate the area of the peanut illustrated in Figure 2.7. Solution: From the line element (2.21) and (7.28), an element of area is
dA : (ad9) (af(t9)d¢). The area of the peanut is A=/07rd0/02"d¢ a2f(0) f(0) : sint9 (1 — sin2 0),
A : 27ra2. for 88 102 CHAPTER 8. GEODESICS for constants A, B, C, and D. Eliminating S gives a linear relation
y 2 mm + b for constants m and b. This is the equation of all straight lines. 82. In usual spherical coordinates the metric on a twodimensional sphere is [cf. (2.15)]
d32 = (12(d6‘2 + 51112 6d¢>2) Where a is a constant. (a) Calculate the Christoifel symbols “by hand”. (b) Show that a great circle is a solution of the geodesic equation. Make use of the freedom to orient the coordinates so the equation of a great circle is simple.) Solution: a) H R2 0
9” 0 R2 sin2 9 9 ' 0 (R sin 9)“2 Fig 2 ~sin9cos0 , IE2 :cot0 all the rest are zero. b) Orient coordinates so that the great circle lies along the equator 6 = 7r / 2.
The equation of the great circle is then 6 = 7r / 2, Q5 : S/a where S is the
distance around and de/dS : (0,1/a). The geodesic equation is then d%’4
0352 : Evidently the left hand side vanishes, and the right hand side vanishes
because the relevant Christoﬁel symbols vanish at 6 :— 7r / 2. 102 PROBLEM 8.5 105 c) For example, in the nonrelativistic limit, the 2: equation becomes alga: dy
—~ = —2o w n2 .
dt2 dt + m The second term on the right hand side is the xcomponent centrifugal
force 9 X (Q X when S2 = 952 The ﬁrst term is the Coriolis force 262’ x (da‘f/dt) 85. Derive the Christoﬁel symbols Pi; and F3”; for the wormhole metric (7.39)
directly from the general formula (8.19) and not starting from variational
principle of extremal proper time. Solution: There is not much to say about the solution to this problem except
to evaluate (8.19). The diagonal metric simpliﬁes the sums, 6.9. Empaw = 9” Zr We write out the two equations analogous to (8.20): 39¢ ag¢¢ 39m __ .2
(W5 + W 8gb —rsmo9 (59% + 89% _ z — (b2 + T2) sin0c080 1Q
%
f—j
S
‘6—
ll which gives
7" 0 __  4’ _.
F¢¢——Sln60086, 8—6. Show by direct calculation from (8.15) that the norm of the four velocity
u u is a constant along a geodesic. Solution: Let N E u  u 2 gaﬂuauﬁ, then dN du" agaﬂ “:2 ___ﬁ aﬁv
d7' gaﬁdTU—l—Baﬂuuu
39
m a 376 01/5 aﬁ'y
2gagF76uuu—laﬂuuu ...
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This note was uploaded on 12/01/2011 for the course PHYS 5523 taught by Professor Kennefick during the Fall '11 term at Arkansas.
 Fall '11
 Kennefick
 Theory Of Relativity

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