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Equations of Orbits, Deflection of Light,
Mercurys Perihelion Shift, Period of Revolution and
Time Delay of Signals
in Schwarzchilds Geometry.
Some Exact Formulas.
Solomon M. Antoniou
SKEMSYS
S cientific Knowledge Engineering
and Management Systems
37 oliatsou Street, Corinthos 20100, [email protected]
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Abstract
We present some exact solutions of the differential equations of motion connected
to Schwarzchilds solution of Einstein field equations in General Relativity. The
exact solutions are expressed in terms of Jacobis or Weirstrasss Elliptic
Functions or in terms of Elliptic Integrals.
The topics which are covered are the following:
I) Deflection of light in the Suns gravitational field
II) Equation of a closed orbit
III) Mercurys perihelion shift
IV) Trajectory of a light signal
V) Period of revolution
VI) Time delay of radio signals in the Suns gravitational field
Although the subject itself is rather old, some of the results of this paper appear for
the first time in the literature. In particular the formulae (10.29) for the period and
(11.19) for the time delay, to the best of our knowledge, appear for the first time in
the literature.
Keywords: General Relativity, Einsteins Equations, Schwarzchild Solution,
Equations of Orbits, Mercurys Perihelion Shift, Deflection of Light, Period of
Revolution, Time Delay of Signals, Weirstrass Elliptic Functions, Jacobi Elliptic
Functions, Elliptic Integrals, Exact Solutions.
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Contents
1. Introduction
2. The Schwarzchild Solution
3. Equations of the Geodesic Curves
4. Simplification of the Equations of Motion
5. Differential Equations. Elliptic Functions
5.1 Differential Equation of the Orbit
5.2 Differential Equation for the Calculation of the Time
6. Deflection of Light in the Suns Gravitational Field7. Equation of Closed Orbit
7.1 Equation of Closed Orbit using Weirstrass Elliptic
Function
7.2 Equation of Closed Orbit using Jacobis Elliptic
Function
7.3 Equation of Closed Orbit. Third Method
8. Mercurys Perihelion Shift
9. Trajectory of Light Signal
9.1 Trajectory of Light Signal using Weirstrass Elliptic
Function
9.2 Trajectory of Light Signal using Jacobis Elliptic
Function
10. Period of Revolution
11. Time Delay of Radio Signals in the Suns Gravitational Field
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1. Introduction
We analyze the problem of motion of test particles in the gravitational field
created by a heavy spherically symmetric object of mass M, which produces a
Schwarzchild geometry. We study different types of orbits, which mainly fall into
two categories: bound and unbound. We also consider the period of a test body
describing a cycle around a heavy object (say the Sun) and the time delay of radio
signals in the Suns gravitational field.
In all the cases we express the final solutions in terms of Elliptic Functions
(Jacobis or Weirstrasss) or Elliptic Integrals. We use a number of different
approaches, which are then proved to lead to equivalent solutions.
2. The Schwarzchild Solution.
Einsteins field equations in empty space are (Ref. [1])
0R =
where R is Riccis tensor.
By solving Einsteins equations for the general static isotropic metric
22222222 dsinrdrdr)r(Adt)r(Bd = (2.1)
we can determine the functions )r(B and )r(A .
These functions can be determined by imposing the boundary condition that for
r the metric tensor must approach the Minkowski tensor in spherical
coordinates:
1)r(Blim)r(Alimrr
==
A constant of integration which appears in the solution, can be fixed considering
that at great distances from a central mass M, the component Bg tt = must
approach to the quantity 21 , where is the Newtonian potentialr
GM
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where M is the mass of the object producing the field and G is Newtons
constant. Under these circumstances, we find that the functions )r(B and )r(A are
given by
r
MG21)r(B = (2.2)
and
1
r
MG21)r(A
= (2.3)
respectively.
The solution of Einsteins equations for the metric (2.1) with )r(B and )r(A given
by (2.2) and (2.3) respectively, is the very-well known Schwarzchildsolution.
It is obvious that
1)r(B)r(A = (2.4)
The non-vanishing Christoffel symbols associated to the metric (2.1) are given by
)r(A2
)r(A
rr = )r(A
r
r = )r(A
sinr
2
r =
)r(A2
)r(B tt
r = r
1 r
r == cossin = (2.5)
r
1 r
r == cot ==
)r(B2
)r(B rt
ttr
t ==
where the prime denotes differentiation with respect to r.
3. Equations of the Geodesic Curves
The differential equations of the geodesic curves in a curved space have the form
0dp
dx
dp
dx
dp
xd
2
2
=+ (3.1)
where p is an affine parameter and summation over repeated indices is always
understood.
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Using the non-vanishing components of the Christoffel symbols (relations (2.5)),
we find from equation (3.1) the equations
+
+
2222
2
2
dp
d
)r(A
sinr
dp
d
)r(A
r
dp
dr
)r(A2
)r(A
dp
rd
0dp
dt
)r(A2
)r(B2
=
+ (3.2)
0dp
dcossin
dp
dr
dp
d
r
2
dp
d2
2
2
=
+ (3.3)
0
dp
d
dp
dcot2
dp
dr
dp
d
r
2
dp
d
2
2
=++ (3.4)
0dp
dr
dp
dt
)r(B
)r(B
dp
td
2
2
=
+ (3.5)
Since the field is isotropic, we may consider the orbit of our particle to be confined
to the equatorial plane, that is
2
= (3.6)
Equation (3.3) is then immediately satisfied and we can forget about as a
dynamical variable.
Under this assumption, the system of equations (3.2)-(3.5) is equivalent to the
system
0dp
dt
)r(A2
)r(B
dp
d
)r(A
r
dp
dr
)r(A2
)r(A
dp
rd222
2
2
=
+
+ (3.7)
0dp
dr
dp
d
r
2
dp
d
2
2=+ (3.8)
0dp
dr
dp
dt
)r(B
)r(B
dp
td
2
2
=
+ (3.9)
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4. Simplification of the Equations of Motion
We can further simplify the equations of motion (3.7)-(3.9) we have derived. Full
details are given in Appendix A. The new equations are the following:
)r(B
1dpdt = (4.1)
Jdp
dr2 = (4.2)
E)r(B
1
r
J
d
dr)r(A
2
22
=+
(4.3)
where J is a constant having dimensions of angular momentum per unit mass and
E is a constant.
The relation between proper time and affine parameter p is given by
22 dpEd = (4.4)
where
0E > for particles
0E = for photons
The equation of the orbit )(rr = is obtained by eliminating dp betweenequations (4.2) and (4.3). We find
222
2
4 J
E
)r(BJ
1
r
1
d
dr
r
)r(A=+
(4.5)
which can be put in one of the following equivalent forms:
=
222
42
r
1
J
E
)r(BJ
1
)r(A
r
d
dr(4.6)
and
1r
JE
)r(A
1
d
dr
r
J
2
22
2=
++
(4.7)
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We can also determine the value of the constant J as follows:
If the orbit is an unbounded curve, at the closest distance from the object which
produces the gravitational field, we have 0rr = and so 0d
dr= , and equation (4.5)
gives
= E
)r(B
1rJ
0
20
2 (4.8)
In case the path is a closed curve, we find another expression for J (Appendix D):
)rr)(MG2r)(MG2r(
)rr(MG2J
22
++
++
= (4.9)
where +r and r are the aphelion and perihelion respectively.
Finally, since dt)r(Bdp = , as follows from equation (4.1), we can eliminate dp
from equation (4.3) and we get
E)r(B
1
r
J
dt
dr
)r(B
)r(A
2
22
2=+
(4.10)
5. Differential Equations. Elliptic Functions
5.1. Differential Equation of the Orbit.
The differential equation of the orbit can be expressed in the form
)r(fd
dr2
=
(5.1)
where )r(f is a fourth degree polynomial, which can easily be integrated.
Under the substitutionr
1u = , equation (5.1) is being converted into an equation of
the form
)u(gd
du2
=
(5.2)
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where )u(g is a third degree polynomial. The resulting equation can then be
integrated. The resulting integral is then converted to an elliptic integral or to a
linear combination of elliptic integrals of various types.
We can also transform the differential equation (5.2) into a Weirstrass or Jacobi
differential equation. In fact, under a linear transformation Uu += , the
differential equation is converted to
323
2
gUgU4d
dU=
(5.3)
The above differential equation has the general solution
)g,g;c(U 32+= (5.4)
where is Weirstrass elliptic function and c is a constant which can be
determined from the initial conditions.
Since, on the other hand,
)eU)(eU)(eU(4d
dU321
2
=
(5.5)
where 321 eee >> , under the substitution
32
32
ee
eUz
= ,31
322
ee
eek
= ( 1k0 2
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5.2. Differential equation for the calculation of time.
The differential equation for the calculation of time in terms of the distance is
given by (Appendix A, equation (A.19))
)r(Rdr
dt= (5.9)
where
2/1323 )]x(BxE)x(BxJx[
xx
MG2x
x)x(R
= (5.10)
The above differential equation can be integrated by separation of variables and its
solution can be expressed as a combination of Elliptic Integrals of various types.
6. Deflection of Light in the Suns Gravitational Field.
We start with equation (4.6) where 2J is given by equation (A.11) of Appendix A.
Since
2/1
222
2
r
1
J
E
)r(BJ
1
)r(A
r
d
dr
=
we obtain, by separation of variables
dr
r
1
J
E
)r(BJ
1r
)r(Ad
2/1
222
2
= (6.1)
We also have
=
=
22222 r
1E
)r(B
1
J
1
r
1
J
E
)r(BJ
1
2
1
020 r
1E
)r(B
1E
)r(B
1
r
1
=
(6.2)
and thus equation (6.1) can be written as
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dr
r
1E
)r(B
1E
)r(B
1
r
1r
)r(Ad
2/1
2
1
020
2
=
(6.3)
Integrating the previous equation, we obtain the equation of the orbit
+=r
2/1
2
1
020
2
dr
r
1E
)r(B
1E
)r(B
1
r
1r
)r(A)()r( (6.4)
This is the general equation of the orbit of a particle, which approaches the
gravitational field produced by an object of mass M.
If the particle is a photon, then 0E = and equation (6.4) takes the form
=
+=
r2/1
220
02
dr
r
1
)r(Br
)r(Br
)r(A)()r(
=
+=
r 2/1
220
200
22
dr
)r(Brr
)r(Br)r(Brr
)r(A)(
+=
r2/12
002
0dr
)]r(Br)r(Br[r
)r(B)r(Ar)(
and since 1)r(B)r(A = , we get finally
=
r2/12
0020
dx)]x(Br)r(Bx[x
1r)()r( (6.5)
The integral which appears in (6.5) can be transformed as follows. First of all we
have to transform the quantity under the integration. In Appendix B we find that
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= 2/12002 )]x(Br)r(Bx[x
)rx)(rx)(rx(xr
MG2r210
0
0
= (6.6)
where
++=
MG2r
MG6r1r
2
1r
0
001 and
+=
MG2r
MG6r1r
2
1r
0
002 (6.7)
Collecting everything together, we have
=
r 2100
00 dx
)rx)(rx)(rx(x
1
MG2r
rr)()r( (6.8)
We put
)rx)(rx)(rx(x)x(G 210 = (6.9)
We have to calculate the difference
=
0r0
000 dx
)x(G
1
MG2r
rr)()r( (6.10)
In Appendix C it is proved that
= a
022120r sink1
d
)rr(r
2dx
)x(G
1
0
where
= 10
121
rr
rrsina and
12
10
0
22
rr
rr
r
rk
=
Equation (6.10) is then equivalent to
=
= a
0221200
000
sink1
d
)rr(r
2
MG2r
rr)()r(
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)k;a(F)rr)(MG2r(
r2
120
0
= (6.11)
where )k;a(F is the Elliptic Integral of the first kind.
We also have
MG2r
MG6rrrr
0
0012
+= (6.12)
and thus equation (6.11) is equivalent to
)k;a(F)rr)(MG2r(
r2)()r(
4/1
120
20
0
= (6.13)
The deflection of the light is given by the equation
|)()r(|2 0 = (6.14)
and using (6.13), we get the expression
)k;a(F)rr)(MG2r(
r4
4/1
120
20
= (6.15)
7. Equation of Closed Orbits.
7.1 Equation of Closed Orbit using Weirstrass Elliptic Function
We start with equation (4.6):
)r(A
r
)r(A
r
J
Er
J
1
d
dr24
2
4
2
2
=
(7.1)
which, using (7.30), (7.31) and the expression for )r(A , can be converted to the
equation
rMG2rrb
GMa2r
b
ac
d
dr 2342
+
=
(7.2)
The quantities a, b and c are defined by (7.27), (7.28) and (7.29) respectively.
Under the substitution
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r
1u = ,
d
du
u
1
d
dr
2= (7.3)
equation (7.2) can be transformed to the equation
322
uMG2uub
GMa2
b
ac
d
du ++= (7.4)
We try a linear transformation
Uu += (7.5)
so as to convert (7.4) into the standard form:
323
2
gUgU4d
dU=
(7.6)
Under (7.5), equation (7.4) is transformed to
322
2 )U(MG2)U()U(b
GMa2
b
ac
d
dU +++++
=
which is equivalent to
+
++++=
UMG62b
aMG2
1U)MG61(UMG2
d
dU 2232
++
+ MG2
b
MGa2
b
ac
1 322
(7.7)
The coefficients and are to be determined so as the coefficient of3U
should be 4 and the coefficient of 2U should be zero:
4MG2 = and 0MG61 =+
Solving the previous system we find the values of the constants and :
MG
2= (7.8)
and
MG6
1 = (7.9)
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Thus the linear transformation (7.5) has the form
MG6
1U
MG
2u += (7.10)
After substituting the values of and into (7.7) we arrive at the equation
323
2
gUgU4d
dU=
(7.11)
where
b
GMa
12
1g
22
2 = (7.12)
and
4
GM
b
ac
12
GM
b
a
216
1g
2222
3
= (7.13)
The solution to (7.11) is given by
)g,g;c(U 32+= (7.14)
where is the Weirstrass elliptic function and c is a constant which can be
determined from the initial conditions.
We thus arrive at the following solution of equation (7.1):
MG6
1)g,g;c(
MG
2
r
132 ++= (7.15)
We can determine the constant c by imposing the initial condition:
0 = , = rr
MG6
1)g,g;c(
MG
2
r
132 +=
from which there follows:
=r12
rMG6)g,g;c( 32 (7.16)
On the other hand, it is a very well-known fact (Ref. [6]) that if
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)g,g;z( 32=
then
=
323 gxgx4
dxz
which means that
=r12
rMG6)g,g;c( 32
implies that
=2e~ 32
3gxgx4
dxc
where 2e~ is one of the roots of the polynomial 32
3 gxgx4 where 2g and 3g
are given by (7.12) and (7.13) respectively.
We also have
)e~x)(e~x)(e~x(4gxgx4 321323 =
where
+
++ +=rr
)rr(MG3rre~1 ,
=r12
rMG6e~2 ,+
+=r12
rMG6e~3
are the three roots of the polynomial with ordering
321 e~e~e~ >>
Using (17.4.65) of ref. [5], we get
)m;(Fe~e~
1
)e~x)(e~x)(e~x(4
dxc
21e~ 321
2
=
=
(7.17)
where
=
022 sinm1
d)m;(F (7.18)
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with
31
322
e~e~e~e~
m
= and
= 32
311
e~e~e~e~
sin (7.19)
7.2 Equation of Closed Orbit using Jacobi Elliptic Function.
First Case. Solution of the Equation (7.4):
We have that (7.4) is equivalent to
)eu)(eu)(eu(MG2d
du321
2
=
where 321 e,e,e are the three roots of the equation
0uMG2uub
GMa2
b
ac32 =++
We can find that these roots are given by
h
1
rrMG2
)rr(MG2rre1 =
+=
+
++ ,
=r
1e2 ,
+=
r
1e3
We remind the reader the assumption
MG2hrr >>> + and MG6r >
from which there follows the ordering 321 eee >> .Under the substitution
32
32
ee
euz
=
31
322
ee
eek
= ( 1k0 2
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)z1()ee(eu2
322 =
2323 z)ee(eu =
dzz)ee(2du 32 =
d
dzz)ee(2
d
du32 =
The lhs of equation (7.4) is transformed into
222
32
2
d
dzz)ee(4
d
du
=
The rhs of equation (7.4) is transformed into
= )eu)(eu)(eu(MG2 321
== ]z)ee[()]z1()ee([)]zk1([MG2 2322
32222
)zk1()z1(z)ee(MG2 22222322 =
Therefore the original equation (7.4) is being transformed into the equation
)zk1()z1(MG2d
dz4 2222
2
=
which is equivalent to
)zk1()z1(d
dz 2222
=
or
)zk1()z1(dx
dz 2222
=
(7.4a)
where
x = and2
)ee(MG
2
MG 31
==
The general solution of the equation (7.4a) is given by
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)k;x(snz += , = constant
and so
)k;(sn)ee(eu 2323 ++=
The equation of the orbit is thus written as
)k;(sn)ee(e
1r
2323 ++
=
We may calculate the constant in the previous formula by imposing some
initial conditions:
i) If we impose 0 = for += rr then from the equation of the orbit we have
)k;(sn)ee(e
1e1r
23233 +
==+
from which we get
0)k;(sn2 =
and then
0 =
In this case the equation of the orbit is
)k;(sn)ee(e
1r
2323 +
=
We can now transform this equation in a more standard form.
It is known that if )k;a(F = then asin)k;(sn = . We then have
2
a2cos1asin)k;(sn
22 ==
Introducing the angle by |a2| = , we find a2coscos = and so
2
cos1)k;(sn2
+=
We obtain in this way
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2
cos1)ee(e
1r
323+
+=
from which there follows that
cose1
Rr
+=
where
32 ee
2R
+= and
32
32
ee
eee
+
=
Since
++
=+
=
r
1
r
1
2
ee
2R
32
and
+=
+ r
1
r
1
2
1
R
1,
R may be identified with the semi-latus rectum.
We also have
+
++
=+
=rr
rr
ee
eee
32
32
If a is the semi-major axis, then
a)e1(r +=+ and a)e1(r =
ii) If we impose 0 = for = rr , then from the equation of the orbit we have
)k;(sn)ee(e
1
e
1r
23232 +
==
from which we get
2
sin1)k;(sn1)k;(sn
2 ===
and then
)k(K)k,2/(F ==
In this case the equation of the orbit is
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)k;K(sn)ee(e
1r
2323 ++
=
Since
)u(dn)u(cn)k;)k(Ku(sn =+
the previous equation of the orbit can be expressed as
)k;(dn
)k;(cn)ee(e
1r
2
2
323 +
=
or in equivalent form
)k;(cn)ee()k;(dne
)k;(dnr
232
23
2
+=
Second Case: Solution of the equation (7.11)
We have that (7.11) is equivalent to
)e~U)(e~U)(e~U(4d
dU321
2
=
where 321 e~,e~,e~ are the three roots of the equation 0gUgu4 323 =We can find that these roots are given by
+
++ +=rr
)rr(MG3rre~1 ,
=r12
rMG6e~2 ,
+
+=r12
rMG6e~3
with the ordering 321 e~e~e~ >> .
Under the substitution
3232
e~e~e~U
z
=
31
322
e~e~e~e~
k
= ( 1k0 2
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we get
2323 z)e~e~(e~U +=
)zk1(e~U222
1 =
)z1()e~e~(e~U 2322 =
2323 z)e~e~(e~U =
dzz)e~e~(2dU 32 =
d
dzz)e~e~(2
d
dU32 =
The lhs of equation (7.11) is transformed into
222
32
2
d
dzz)e~e~(4
d
dU
=
The rhs of equation (7.11) is transformed into
= )e~U)(e~U)(e~U(4 321
== ]z)e~e~[()]z1()e~e~([)]zk1([4 2322
32222
)zk1()z1(z)e~
e~
(422222
32
2
=Therefore the original equation (7.11) is being transformed into the equation
)zk1()z1(d
dz 22222
=
which is equivalent to
)zk1()z1(d
dz 2222
=
or
)zk1()z1(dx
dz 2222
=
(7.11a)
where
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x = and 31 e~e~ =
The general solution of the equation (7.11a) is given by
)k;x(snz += , = constant
and then
)k;(sn)e~e~(e~U2
323 ++=
Using (7.10), we arrive at the following equation of the orbit:
MG6
1)]k;(sn)e~e~(e~[
MG
2
r
1 2323 +++=
7.3 Equation of Closed Orbit. Third Method.
Since the path is a closed curve, there are only two points at which 0d
dr = .
Let us call r these two points. Using equation (4.5) in the form
0E)r(B
1
r
J
d
dr
r
)r(AJ
2
22
4
2
=++
we get
0E)r(B
1
)r(
J
2
2
=+ (7.20)
In other words
0E)r(B
1
)r(
J
2
2
=+++
(7.21a)
0E)r(B
1
)r(
J
2
2
=+
(7.21b)
Equations (7.21) is a system of two equations with two unknowns:2J and E.
This system is solved in Appendix D. We find
)rr)(MG2r)(MG2r(
)rr(MG2J
22
++
++
= (7.22)
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and
)rr)(MG2r)(MG2r(
rrMG2)MG2r()r()MG2r()r(E
22
++
+++++
= (7.23)
The angle )r( through which the radius vector of the particle rotates can be
calculated using (6.1), by the formula
= r
r2/1
222
2
dr
r
1
J
E
)r(BJ
1r
)r(A)r()r( (7.24)
With += rr we find
+
= +r
r2/1
222
2
dr
r
1
J
E
)r(BJ
1r
)r(A)r()r( (7.25)
We now have simplify the expression under the integral sign of (7.24). We have
=
=
2/1
2
2
2
22/1
222
2 1
J
xE
)x(BJ
xx
)x(A
x
1
J
E
)x(BJ
1x
)x(A
=
=2/12244 )x)x(BJx)x(BEx(
)x(B)x(AJ
2/1233 )x)x(BJx)x(BEx(x
J
= (7.26)
Using the substitutions
)rr(MG2)MG2r()r()MG2r()r(:a22
+++ += (7.27)
2)rr(MG2:b += (7.28)
)rr)(MG2r)(MG2r(:c ++ += (7.29)
expressions (7.22) and (7.23) become
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c
bJ2 = (7.30)
and
c
aE = (7.31)
respectively.
The expression x)x(BJx)x(BEx233 which appears in (7.26) can be further
transformed in Appendix E. We find
= x)x(BJx)x(BEx 233
)}hx)(rx)(rx({)MG2r)(MG2r)(rr(
)rMG2rMG2rr(MG2
+
= +++++
(7.32)
where
)rr(MG2rr
)rr(MG2h
++
++
= (7.33)
Using now (7.26) and (7.32), we get
=
2/1
2222
x
1
J
E
)x(BJ
1x
)x(A
)hx)(rx)(rx(
1
rMG2rMG2rr
)rr(
=
+++
+ (7.34)
Therefore equation (7.24) takes on the final form
= )r()r(
=
+++
+r
r)hx)(rx)(rx(
dx
rMG2rMG2rr
)rr((7.35)
The conversion of the integral in (7.35) into an elliptic integral is performed in
Appendix F. We have found that
= a
022sinm1
d)(#T)r()r( (7.36)
where
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+++
+
=r)hr(
2
rMG2rMG2rr
)rr()(#T (7.37)
2/11
hr
rr
rr
hrsina
=
+
+ (7.38)
+
+
=r)hr(
h)rr(m2 (7.39)
8. Mercurys Perihelion Shift
Mercurys perihelion shift (per revolution) is given by (ref. [1])
2|)r()r(|2 = (8.1)
From (7.36) we get, since 2
a = (from (7.38))
)m(K)(#T
sinm1
d)(#T)r()r(
2/
022
=
= + (8.2)
We thus find
2)m(K)(#T22|)r()r(|2 == (8.3)
9. Trajectory of a Light Signal.
9.1 Trajectory of Light Signal using Weirstrass Elliptic Function.
For a light signal we have 0E = . We get from equation (4.6)
)r(A
r
J
r
d
dr 2
2
42
=
(9.1)
where
)r(B
r
J 0
202
= , 00 rMG2
1)r(B = ,
1
r
MG2
1)r(A
= (9.2)
Using relations (9.2), we get from (9.1)
2420
02
rr
MG21r
r
)r(B
d
dr
=
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which is equivalent to
rMG2rrr
MG2r
d
dr 2430
02
+
=
(9.3)
The change of variables
r
1u = ,
d
du
u
1
d
dr
2= (9.4)
transforms equation (9.3) into
u
MG2
u
1
u
1
r
MG2r
d
du
u
1
2430
02
2+
=
which is equivalent to
30
0232
r
MG2ruuMG2
d
du +=
(9.5)
Using the linear transformation
MG6
1U
MG
2u += ,
d
dU
MG
2
d
du= (9.6)
equation (9.5) becomes
+=
32
MG6
1U
MG
2MG2
d
dU
MG
2
30
02
r
MG2r
MG6
1U
MG
2 +
+
which is equivalent to
3232 gUgU4
d
dU =
(9.7)
where
12
1g2 = (9.8)
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30
022
3r4
)MG2r(GM
216
1g
= (9.9)
The solution of (9.7) is given by
)g,g;c(U 32+= (9.10)
where is the Weirstrass elliptic function and c is a constant. We thus arrive at
the following solution of equation (9.1):
MG6
1)g,g;c(
MG
2
r
132 ++= (9.11)
9.2 Trajectory of Light Signal using Jacobis Elliptic Function.
First Case: Solution of the equation (9.5)Equation (9.5) is equivalent to
)eu)(eu)(eu(MG2d
du321
2
=
(9.12)
where 1e , 2e , 3e are the three roots of the equation
0r
MG2ruuMG2
3
0
023 =
+
given by
0
0001
rMG4
)MG6r)(MG2r(MG2re
++=
0
2r
1e =
0
000
3 rMG4
)MG6r)(MG2r(MG2re
+
=
with the ordering
321 eee >>
Using the substitution
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32
32
ee
euz
=
31
322
ee
ee
h
= ( 1h02
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0
0001
r24
)MG6r)(MG2r(3MG6re~
++=
0
02
r12
rMG6e~
=
0
0003
r24
)MG6r)(MG2r(3MG6re~
+=
with the ordering 321 e~e~e~ >> under the further assumption MG6r0 > .
Under the substitution
32
32
e~e~e~U
z
=
31
322
e~e~e~e~
g
= ( 1g0 2
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10. Period of Revolution
We shall now calculate the period of revolution of a test body describing a closed
orbit around a heavy object of mass M producing the gravitational field. We start
with equation (A.14) of Appendix A, written in equivalent form
)x(Rdr
dt= (10.1)
where
2/1323 ]x)x(BEx)x(BJx[
xx
MG2x
x)x(R
= (10.2)
The period of revolution is been expressed as
+
=
r
r
r
dx)x(Rdx)x(R2T (10.3)
where
)2(r == (10.4)
We have, according to (7.32):
= 323 x)x(BEx)x(BJx
)}hx)(rx)(rx({)MG2r)(MG2r)(rr(
)rMG2rMG2rr(MG2
+
= +++
++ (10.5)
Because of (10.5), we get from (10.2) that
)hx)(rx)(rx()MG2x(
xx)(#P)x(R
2
=
+(10.6)
where
)rMG2rMG2rr(MG2)MG2r)(MG2r)(rr()(#P
++++
+= (10.7)
We observe that
)x(H)MG2x(
x)(#P)x(R
3
= (10.8)
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where
)hx)(rx)(rx(x)x(H = + (10.9)
According to Appendix F, under the substitution
hxrx
rrhrsin2
=
++ (10.10)
sin)rr()hr(
sin)rr(h)hr(rx
2
2
++
++
= (10.11)
we have
sinm1
d
r)hr(
2
)x(H
dx
22
=
+(10.12)
where
+
+
=r)hr(
h)rr(m2 (10.13)
In Appendix G, under the same substitution (10.11) and introducing further the
notation
hr
rrk2
=
+
+ (10.14)
and
)MG2r)(hr(
)MG2h)(rr(q
=+
+ ( 1q0
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where
)rMG2rMG2rr)(hr)(MG2r(MG2
r)MG2r)(rr()r(2)(#Q 2
+++
++
+= (10.18)
d
sinm1)sinq1()sink1(
)sinm1(d)(S222222
322
= (10.19)
We now find the new limits of the integrals. Since
hx
rx
rr
hrsin2
= +
+ andsin)rr()hr(
sin)rr(h)hr(rx
2
2
++
++
=
we find
+= rx :21sin2 ==
= rx : 00sin2 ==
and
x = : ah
r
rr
hrsin
h
r
rr
hrsin
2/112
=
= +
+
+
+
= rx : 0 =
Using (10.3) and (10.17) we get the following formula for the period:
=
a
0
2/
0
d)(Sd)(S2)(#QT (10.20)
In order to calculate the period, we have to calculate the two integrals appearing in
(10.20). For this purpose, we first have to use the identity (Appendix H)
=
)sinq1()sink1(
)sinm1(
2222
322
+
+=
sink1
1
)kq(k
kqk2mk2qm)km(
qk
m
22224
42222222
4
6
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)sinq1()kq(q
)qm(
sink1
1
k)kq(k
)km(
222
32
22222
322
+ (10.21)
Using the previous partial fraction decomposition we find, using equation (10.20)
that the period is given by
=
sinm1
d2)(#QT
22
a
0
2/
0
1
+
sinm1)sink1(
d2)(#Q
2222
a
0
2/
0
2
+ sinm1)sink1(
d2
k)(#Q
2222
a
0
2/
023
sinm1)sinq1(
d2)(#Q
222
a
0
2/
0
4
(10.22)
where
)(#Q
qk
m)(#Q
4
6
1 =
)(#Q)kq(k
kqk2mk2qm)km()(#Q
224
42222222
2
+=
)(#Q)kq(k
)km()(#Q
22
322
3
=
)(#Q)kq(q
)qm()(#Q
22
32
4
=
We make further use of the substitutions
=2/
022sinm1
d)m(K (10.23)
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=a
022sinm1
d)m;a(F (10.24)
=2/
02222
2
sinm1)sink1(
d
)m;k,2
( (10.25)
=a
02222
2
sinm1)sink1(
d)m;k,a( (10.26)
=2/
0222 sinm1)sinq1(
d)m;q,
2
( (10.27)
=a
0222 sinm1)sinq1(
d)m;q,a( (10.28)
to convert equation (10.22) into a more compact form.
= )}m;a(F)m(K2{)(#QT 1
+ )}m;k,a()m;k,2
(2{)(#Q 222
+)}m;k,a()m;k,
2(2{
k)(#Q 2223
)}m;q,a()m;q,2
(2{)(#Q4 (10.29)
We have now to find the quantity )2(r == . In section (7.2) we have found
that the equation of the orbit is given by
)k;(sn)ee(e
1r
2323 +
=
where
h
1
)rr(MG2
)rr(MG2rre1 =
+=
+
++ ,
=r
1e2 ,
+=
r
1e3 ( 321 eee >> )
are the three roots of the equation
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0uMG2uub
aMG2
b
ac 32 =++
and
2)ee(MG 31 =
In obtaining the equation of the orbit we have considered the initial condition
0 = for += rr
We thus find that
)k;2(sn)ee(e
1)2(r
2323 +
=== (10.30)
11. Time Delay of Radio Signals in the Suns
Gravitational Field.
The time taken for the light to reach a star P whose distance from the Sun is Pr
and received back to Earth by the emitting device, is given by
)}r,r(t)r,r(t{2T 0P0 += (11.1)
where r is Earths distance from the Sun and 0r is the closest approach of the
orbit of light from the Sun (Ref. [4]).
We are going now to calculate the quantity )r,r(t 0 which represents the time
taken by the light to go from a point whose distance is r to a point whose distance
is 0r from a massive point of mass M, which produces the gravitational field.
Since for radio signals 0E = , we get from equation (4.10)
=
2
222
r
J
)r(B
1
)r(A
)r(B
dt
dr
which is equivalent to
)]r(BJr[)r(B
r
dr
dt
222
22
=
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from which we get
2/122 )]r(BJr[)r(B
r
dr
dt
= (11.2)
Since
)r(B
rJ
0
202 = (11.3)
we finally get from (11.2):
2/12
002
0
)]r(Br)r(Br[)r(B
)r(Br
dr
dt
= (11.4)
Separating the variables in the previous equation and integrating, we get
=r
r2/12
002
00
0
dx)]x(Br)r(Bx[)x(B
)r(Bx)r,r(t (11.5)
Since, as follows from (6.6)
= 2/12002 )]x(Br)r(Bx[x
)rx)(rx)(rx(x
r
MG2r210
0
0
= (11.6)
and
MG2x
x
r
MG2r
x
MG21
r
MG21
)x(B
)r(B
0
000
=
= (11.7)
we get from (11.5)
=r
r 210
3
0
0
dx)rx)(rx)(rx(x)MG2x(
x)r,r(t (11.8)
In Appendix I we find, under the substitution
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sin)rr()rr(
sin)rr(r)rr(rx
21012
2102120
=
that
= )rx)(rx)(rx(x)MG2x(
x
210
3
d
sink1)sinp1()sinm1(
)sink1()(#U
222222
322
+
= (11.9)
where
)rr(r
)rr(rk
120
1022
= (11.10)
12
102
rr
rrm
= (11.11)
12
10
0
2
rr
rr
MG2r
MG2rp
= (11.12)
and
4/1
00
0
20
120
0
20
MG6r
MG2r
MG2r
r2
rr
r
MG2r
r2
)(#U
+
== (11.13)
We calculate now the new limits of integration from (I.12) (Appendix I)
00sinrx2
0 ===
arr
rr
rr
rrsin
rr
rr
rr
rrsinrx
2/1
2
0
10
121
2
0
10
122
=
== (11.14)
The integral in (11.8) is converted into
=
= r
r 210
3
0
0
dx)rx)(rx)(rx(x)MG2x(
x)r,r(t
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=a
0222222
322
d
sink1)sinp1()sinm1(
)sink1()(#U (11.15)
We can convert the integral in (11.15) into a combination of Elliptic
Integrals. For this purpose we use the identity (J.4), Appendix J.
=
)sinp1()sinm1(
)sink1(
2222
322
++=
sinm1
1
)pm(m
)pm2kpmk2m()mk(
pm
k
22224
22224222
4
6
sinp1
1
)pm(p
)pk(
sinm1
1
m)pm(m
)mk(
222
32
22222
322
(11.16)
We thus find the following expression
+
=a
022
10sink1
d)(#U)r,r(t
+
+ a
02222
2sink1)sinm1(
d)(#U
+
+
a
022222
3sink1)sinm1(
d
m)(#U
+
+a
0222
4sink1)sinp1(
d)(#U (11.17)
where
)(#Upm
k)(#U4
6
1 =
)(#U)pm(m
pm2pkmk2m)mk()(#U
224
22224222
2
+=
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)(#U)pm(
)km(
m
1)(#U
2
322
23
=
)(#U)pm(p
)kp()(#U
22
32
+=
Introducing the notation
=a
022sink1
d)k;a(F
=a
02222
2
sink1)sinm1(
d)k;m,a(
=a
0222 sink1)sinp1(
d)k;p,a(
we obtain the expression
++= )k;m,a()(#U)k;a(F)(#U)r,r(t 2210
)k;p,a()(#U)k;m,a(m
)(#U 42
23+
+ (11.18)
Using (11.18), we obtain the following formula for the time delay:
=+= )}r,r(t)r,r(t{2T 0P0
++= )}k;(F)k;(F){(#U2 1
+++ )}k;m,()k;m,({)(#U2 222
++
+ )}k;m,()k;m,({
m)(#U2 22
23
)}k;p,()k;p,({)(#U2 4 ++ (11.19)
where
2/1
2
0
10
121
rr
rr
rr
rrsin
=
(11.20)
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and
2/1
2P
0P
10
121
rr
rr
rr
rrsin
= (11.21)
Appendix A. Simplification of the Equations of Motion.
Step 1. We divide equation (3.9) bydp
dtand we find
0dp
dr
)r(B
)r(B
dp
td
dp
dt
1
2
2
=
+
which is successively equivalent to
=
+=+ 0Bln
dp
dtln
dp
d0Bln
dp
d
dp
dtln
dp
d
=
=
1CB
dp
dtln0B
dp
dtln
dp
d
CBdp
dt=
We choose to normalize p so that the solution of the last equation should lookas
)r(B
1
dp
dt= (A.1)
Step 2. We divide equation (3.8) bydp
dand we find
0
dp
dr
r
2
dp
d
dpd
1
2
2
=+
which is successively equivalent to
=
+=+ 0rln
dp
dln
dp
d0rln
dp
d
dp
dln
dp
d 22
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=
=
2
22 Crdp
dln0r
dp
dln
dp
d
J
dp
dr2 = (A.2)
where J is a constant having dimensions of angular momentum per unit mass.
Step 3. We substitutedp
dtand
dp
das given by equations (A.1) and (A.2) into
equation (3.7) and we find
0)r(B
1
)r(A2
)r(B
r
J
)r(A
r
dp
dr
)r(A2
)r(A
dp
rd22
2
2
2
2
=
+
+
and multiplying bydp
dr)r(A2 we find
0dp
dr
)r(B
)r(B
dp
dr
r
J2
dp
dr)r(A
dp
rd
dp
dr)r(A2
23
23
2
2
=
+
+
which is successively equivalent to
=
+
+
0
)r(B
1
dp
d
r
J
dp
d
dp
dr)r(A
dp
d
2
22
=
+
0
)r(B
1
r
J
dp
dr)r(A
dp
d
2
22
E)r(B
1
r
J
dp
dr)r(A
2
22
=+
(A.3)
where E is a constant.
Step 4. Relation between proper time and affine parameter p.
We start from
22222222 dsinrdrdr)r(Adt)r(Bd = (2.1)
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and use the fact that2
= so as equation (2.1) takes the form
22222
drdr)r(Adt)r(Bd =
which can be written as
22
22
2
2
dp
dr
dp
dr)r(A
dp
dt)r(B
dp
d
= (A.4)
We substitute in the above equation the derivativesdp
dtand
dp
das given by
equations (A.1) and (A.2) respectively. We obtain the equation
2
22
2
2
rJ
dpdr)r(A
)r(B1
dpd
= (A.5)
We easily recognize that the rhs of the above equation is E, because of equation
(A.3). This means that
22 dpEd = (A.6)
where
0E > for particles
0E = for photons
Step 5. Equation of the orbit )(rr = .
The equation of the orbit is obtained by eliminating dp from the equations (A.2)
and (A.3). In fact using dJ
rdp
2
= , equation (A.3) takes one of the following
equivalent forms
222
2
4 J
E
)r(BJ
1
r
1
d
dr
r
)r(A=+
(A.7)
or
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=
222
42
r
1
J
E
)r(BJ
1
)r(A
r
d
dr(A.8)
or
1r
JE
)r(A
1
d
dr
r
J
2
22
2=
++
(A.9)
Step 6. The value of J.
6.1 Suppose the orbit is an unbounded curve. At the closest distance from the
object which produces the gravitational field, we have 0rr = and so 0d
dr= .
Therefore equation (A.7) gives
2
022
0 J
E
)r(BJ
1
r
1=
from which, since
2V1E = (A.10)
we get
+=2
0
2
0
2
V1)r(B
1
rJ (A.11)
6.2 When the path is a closed curve, the value of J is not longer given by
equation (A.11), because in this case there are two points for which 0d
dr= .
Let us call r these points. Using equation (A.7) we get
0E
)r(B
1
r
J
2
2
=+
(A.12)
In other words we have the system
0E)r(B
1
r
J
2
2
=+++
(A.13a)
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0E)r(B
1
r
J
2
2
=+
(A.13b)
which is solved in Appendix D.
Step 7.
Since dt)r(Bdp = (equation (A.1)), equation (A.3) gives us
E)r(B
1
r
J
)r(B
1
dt
dr)r(A
2
22
=+
(A.14)
We may simplify further equation (A.14). We find
)]r(BrE)r(BJr[)r(B
)r(Ar
dr
dt
222
22
=
from which we obtain
2/1323 )]r(BrE)r(BrJr[
rr
)r(B
)r(A
dr
dt
= (A.15)
We put
2/1323 )]x(BxE)x(BxJx[
xx
)x(B
)x(A)x(R
= (A.16)
We also have
2
2
22 )MG2x(
x
x
MG21
1
)x(B
)x(B)x(A
)x(B
)x(A
=
==
from which we get
MG2x
x
)x(B
)x(A
= (A.17)
Therefore the expression for )x(R is given by
2/1323 )]x(BxE)x(BxJx[
xx
MG2x
x)x(R
= (A.18)
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Equation (A.15) is equivalent to
)r(Rdr
dt= (A.19)
with )x(R given by (A.18).
Appendix B.
In this Appendix we prove identity (6.6). We have
== )]x(Br)r(Bx[x)]x(Br)r(Bx[x 200222/12
002
=
=
x
MG21xr
r
MG21xx 20
0
3
=
+
= 20
20
3
0
0 rMG2xrxr
MG2rx
+
=
MG2r
rMG2x
MG2r
rxx
r
MG2r
0
30
0
303
0
0 (B.1)
One of the roots of the cubic polynomial which appears in the bracket under the
square root of (B.1) is 0r and so
+=
+
MG2r
rMG2xrx)rx(
MG2r
rMG2x
MG2r
rx
0
20
02
00
30
0
303
The discriminant of the quadratic trinomial inside the parenthesis is
MG2r
)MG6r(r
MG2r
rMG8rD
0
020
0
202
0 +
=
+=
Suppose that MG2r0 > and the real roots are given by
+
+=MG2r
MG6r1r
2
1r
0
001 and
+
=MG2r
MG6r1r
2
1r
0
002
Therefore we have
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= 2/12002
)]x(Br)r(Bx[x
)rx()rx()rx(xr
MG2r210
0
0
= (B.2)
We have to mention that 20 rr > if MG3r0 > (provided that MG2r0 > ).
So instead of MG2r0 > we may impose the stronger condition MG3r0 > and
under this assumption we have
120 rrr >>
and under the assumption332
0 GM32r > we have the ordering
120 rrMG2MG3r >>>> (B.3)
Appendix C. In this Appendix we shall express the integral in (6.10) in terms
of elliptic integrals. We put
01 ra = , 22 ra = , 0a3 = and 14 ra = (C.1)
Since 20 rr > and 1r0 > , we have
21 aa > and 43 aa > (C.2)
Under these assumptions, the substitution (Ref. [3])
sinaa
sinaaaax
24142
2412421
= (C.3)
implies the relation
sink1
d
)x(G
dx
22= (C.4)
where
42
32
31
412
aa
aa
aa
aak
= (C.5)
and
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4231 aa
2 = (C.6)
Since
124242 rraaa == , 104141 rraaa == (C.7)and
03131 raaa == (C.8)
we find
)rr(r
2
120 = (C.9)
and
12
10
0
22
rr
rr
r
rk
= (C.10)
sin)rr()rr(
sin)rr(r)rr(rx
21012
2102120
= (C.11)
We have thus found that
sink1
d
)rr(r
2
)x(G
dx
22120
=
We also have the following relation
2
0
10
12
2
1
41
422
rx
rx
rr
rr
ax
ax
a
asin
=
= (C.12)
From the previous equation we can determine the new limits of the integral:
00sinrx2
0 ===
arr
rrsin
rr
rrsinx
2/1
10
121
10
122
=
== (C.13)
In this way we have found that
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= a
022120r sink1
d
)rr(r
2
)x(G
dx
0
(C.14)
Appendix D. In this Appendix we solve the system of equations (7.21).
Subtracting the two equations (7.21) we get
)r(B
1
)r(B
1
r
1
r
1J
22
2
++=
from which we obtain
)rr)(rr(
)rr(
)r(B)r(B
)r(B)r(BJ
22
++
+
+
++
= (D.1)
Adding the two equations (7.21) we get
++=
++22
2
r
1
r
1J
)r(B
1
)r(B
1E2
from which, using the expression (D.1) for2
J , we find
)rr)(r(B)r(B
r)r(Br)r(BE
22
22
++
++
= (D.2)
We have now to simplify further the two expressions we have derived above for
2J and E. Using the expression (2.2) for )r(B we obtain
)rr)(MG2r)(MG2r(
)rr(MG2J
22
++
++
= (D.3)
and
)rr)(MG2r)(MG2r(
)rr(MG2)MG2r()r()MG2r()r(
E
22
++
+++
+
+
= (D.4)
Appendix E. In this Appendix we prove (7.32).
Using (7.30) and (7.31), we have
= x)x(BJx)x(BEx 233
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=
= x
x
MG21
c
bx
x
MG21
c
ax 33
c
bMG2x
c
bx
c
MGa2x
c
a1 23 ++
= (E.1)
The roots of the cubic polynomial which appears in (E.1) are
+r , r and++
+
=rMG2rMG2)rr(
)rr(MG2:h (E.2)
We also make the assumption
MG2hrr >>> + and MG6r >
Since
)rMG2rMG2rr(MG2ac ++ =using (E.1) and (E.2) we obtain
= x)x(BJx)x(BEx 233
)}hx)(rx)(rx({)MG2r()MG2r()rr(
)rMG2rMG2rr(MG2
+
= +++
++ (E.3)
Appendix F.
We shall express the integral in (7.35) in terms of an elliptic integral.
We put
+= ra1 , = ra2 , ha3 = and 0a4 = (F.1)
Since + > rr and 0h > , we have
21 aa > and 43 aa > (F.2)
Under these assumptions, the substitution (Ref. [3])
sinaa
sinaaaax2
2131
2213312
= (F.3)
implies the relation
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sinm1
d
)x(H
dx
22= (F.4)
where
)hx()rx()rx(x)x(H = + (F.5)
42
12
13
432
aa
aa
aa
aam
= (F.6)
4231 aa
2 = (F.7)
We also have
3
2
21
312
ax
ax
a
a
sin
= (F.8)
Since
hraaa 3131 == +
+ == rraaa 2121 (F.9)
== raaa 4242
we find
+
=r)hr(
2 (F.10)
hx
rx
rr
hrsin
2
= +
+ (F.11)
+
+
=r)hr(
h)rr(m2 (F.12)
sin)rr()hr(
sin)rr(hr)hr(x 2
2
++++
= (F.13)
We thus arrive at
sinm1
d
r)hr(
2
)hx()rx()rx(x
dx
22=
++(F.14)
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For the limits we find
00sinrx2 ===
ahr
rr
rr
hr
sinhr
rr
rr
hr
sinrx
2/112
=
==
+
+
+
+
We thus have found
= ++
a
022
r
r sinm1
d
r)hr(
2
)hx()rx()rx(x
dx(F.15)
Appendix G. Introducing the notation
hr
rr
k2
= ++
(G.1)
we get from (10.11) that
sink1
r)sinm1(x
22
22
= (G.2)
and then
sink1
sin)kMG2rm()MG2r(MG2x
22
222
= (G.3)
We have further
=
=+
+
+
+
hr
rrMG2r
r)hr(
h)rr(kMG2rm 22
hr
)MG2h()rr(
=+
+ (G.4)
and then we get from (G.3) that
sink1
sinq1)MG2r(MG2x
22
2
= (G.5)
where
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)MG2r)(hr(
)MG2h)(rr(q
=+
+ ( 1q0
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Using the identity
+
=
xk1
1
kk
xk1
1
)xk1(
1
22
2
222(H.2)
relation (H.1) becomes
=
)xq1()xk1(
)xm1(
22
32
+
++=
xk1
1
)kq(k
)qk2qmmk2k()km(
qk
m
2224
22224222
4
6
xq1
1
)kq(q
)qm(
xk1
1
k)kq(k
)km(
22
32
2222
322
+ (H.3)
Substituting sinx 2= in (H.3) we arrive at the identity (10.21).
Appendix I.
We put
01 ra = , 22 ra = , 0a3 = and 14 ra = (I.1)
Since 20 rr > and 1r0 > , we have
21 aa > and 43 aa > (I.2)
Under these assumptions, the substitution (Ref. [3])
sinaa
sinaaaax
24142
2412421
= (I.3)
implies the relation
sink1
d
)x(G
dx
22
= (I.4)
where
42
32
31
412
aa
aa
aa
aak
= (I.5)
and
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4231 aa
2 = (I.6)
Since
124242 rraaa == , 104141 rraaa == (I.7)and
03131 raaa == (I.8)
we find
)rr(r
2
120 = (I.9)
12
10
0
22
rr
rr
r
r
k
= (I.10)
sin)rr()rr(
sin)rr(r)rr(rx
21012
2102120
= (I.11)
We also have the following relation
2
0
10
12
2
1
41
422
rx
rx
rr
rr
ax
ax
a
asin
=
= (I.12)
Relation (I.4) takes on the final form
sink1
d
)rr(r
2
)x(G
dx
22120 = (I.13)
We make the substitutions
== sin)rr(r)rr(r)(G 2102120
)sink1)(rr(r22
120 = (I.14)
and
== sin)rr()rr()(F 21012
)sinm1)(rr(22
12 = (I.15)
where
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)rr(r
)rr(rk
120
1022
= (I.16)
12
102
rr
rrm
= (I.17)
Because of the previous substitutions (I.14) and (I.15), we get, using (I.14)-(I.17)
that (I.11) becomes
sinm1
)sink1(rx
22
220
= (I.18)
Under the substitution (I.18), we find
=
=MG2
sinm1
)sink1(r)sinm1(
)sink1(r
MG2x
x
22
220322
32230
3
)]sinm1(MG2)sink1(r[)sinm1(
)sink1(r
22220
222
32230
= (I.19)
We have further
= )sinm1(MG2)sink1(r 22220
== sin)mMG2kr()MG2r( 22200
= sin
MG2r
mMG2kr1)MG2r(
2
0
220
0 (I.20)
and since
=
=
12
10
12
102220
rr
rrMG2
rr
)rr(rmMG2kr
)MG2r(rr
rr2
12
10
=
we get from (I.20)
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= )sinm1(MG2)sink1(r 22220
)sinp1)(MG2r( 20 = (I.21)
where
12
10
0
2
rr
rr
MG2r
MG2rp
= (I.22)
Using (I.21), we get from (I.19):
)sinp1()sinm1(
)sink1(
MG2r
r
MG2x
x
2322
322
0
30
3
=
Finally, collecting everything together, we have
= )rx)(rx)(rx(x)MG2x(
x
210
3
d
sink1)sinp1()sinm1(
)sink1()(#U
222222
322
= (I.23)
where
12
0
0
20
rr
r
MG2r
r2)(#U
= (I.24)
Appendix J. In this Appendix we prove identity (11.16).
For this purpose we use the partial fraction decomposition
=
)xp1()xm1(
)xk1(
22
32
+=
xm1
1
)pm(m
)pmkp2mk3()mk(
pm
k
2224
2222222
4
6
xp1
1
)pm(p
)pk(
)xm1(
1
)pm(m
)mk(
22
32
2224
322
(J.1)
Using the identity
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+
=
xm1
1
mm
xm1
1
)xm1(
1
22
2
222(J.2)
relation (J.1) takes the form
=
)xp1()xm1(
)xk1(22
32
++=
xm1
1
)pm(m
)pm2pkmk2m()mk(
pm
k
2224
22224222
4
6
xp1
1
)pm(p
)pk(
xm1
1
m)pm(m
)mk(
22
32
2222
322
(J.3)
After the substitution sinx 2= we arrive at the identity
=
)sinp1()sinm1(
)sink1(
2222
322
++=
sinm1
1
)pm(m
)pm2pkmk2m()mk(
pm
k
22224
22224222
4
6
sinp1
1
)pm(p
)pk(
sinm1
1
m)pm(m
)mk(
222
32
22222
322
(J.4)
Appendix K. Elliptic Integrals and their Series Expansions.
The Elliptic Integral of the First Kind, denoted by )m,a(F , is defined by
=a
022 sinm1
d)m,a(F
and has the power series expansion
=
+=
0nn2
n2)a(Sm
)2/1(!n
)2
1n(
)m,a(F
where
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+
= )k;m,a(
km
m)m1(
mn
1)k;n,a(
2
+++
2
12
kn1
k2sin
2
nm
tan)k,a(Fnm
k
mn
1
where
n1
knm
2
++
= ( 1m > )
Appendix L. The Weirstrass and Jacobi Elliptic Functions.
The relation between Weirstrass and Jacobi elliptic functions is given by
)m;x(sn
eee)x(
231
3+=
where
31 ee =
Using the above relation, the reader can easily establish the equivalence between
the solutions expressed in terms of Weirstrass elliptic functions and those
expressed in terms of Jacobi elliptic functions.We also have the following expansions:
++++=!5
x)mm141(
!3
x)m1(x)m;x(sn
542
32
and
=
+=2k
2k2k2
xcx
1)x(
where
20
gc 22 = ,
28
gc 33 = and
=+
=2k
2mmkmk cc
)1k2)(3k(
3c ( 4k )
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References[1] S. Weinberg: Gravitation and Cosmology
Wiley 1972
[2] L. D. Landau and E. M. Lifshitz: The Classical Theory of FieldsPergamon 1971
[3] C. A. Korn and T. M. Korn: Mathematical Handbook
Second Edition, McGraw-Hill 1968
[4] M. V. Berry: Principles of Cosmology and Gravitation
Institute of Physics Publishing,
Bristol and Philadelphia, 1993
[5] M. Abramowitz and I. Stegun: Handbook of Mathematical Functions
Dover 1965
[6] E. T. Whittaker and G. N. Watson: A Course of Modern Analysis
Cambridge University Press, 4th Edition 1927
[7] H. T. Davis: Introduction to Nonlinear Differential and Integral Equations
Dover 1962