Motion SchwarzchildGeom

8/4/2019 Motion SchwarzchildGeom

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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

=


)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~ >> .


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

=


)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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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+

=


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 +

+


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 > .


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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35 Page

=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


)]r(BJr[)r(B

r

dr

dt

222

22

=


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37 Page

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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40 Page

)(#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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41 Page

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

=+


=

+=+ 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

=

+

+


=

+

+

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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44 Page

=

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


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


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)


sinaa

sinaaaax2

2131

2213312

= (F.3)



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51 Page

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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53 Page

)MG2r)(hr(

)MG2h)(rr(q

=+

+ ( 1q0


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54 Page

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)


sinaa

sinaaaax

24142

2412421

= (I.3)


sink1

d

)x(G

dx

22

= (I.4)

where

42

32

31

412

aa

aa

aa

aak

= (I.5)

and


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55 Page

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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56 Page

)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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57 Page

= )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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58 Page

+

=

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


59/61


60/61

60 Page

+

= )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 )


61/61

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

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