Behaviour of electromagnetic radiation in gravitational fields

PHOTO-GEOMETRODYNAMICS (DRAFT)

JAMES SPEARS

Abstract. Hubble’s law is accounted for completely a result of the geometry of space-time. Introducing a point mass origin about which a spherically symmetric space-time isdefined, the instantaneous trajectory of particles traversing Eddington-Finkelstein radialnull geodesics are used to argue higher-dimensional locations of coordinate singularitiesin (u, r) and (v, r) space. Combining (u, r) or (v, r) into a single four-dimensional ra-dial parameter and introducing a fifth coordinate, five-dimensional space-time manifoldsare defined. It is then shown that Hubble’s redshift vs. distance relationship can bedescribed, with the scaled five-dimensional line element, as the gravitational redshift infive-dimensional space-time calculated using the projection of the path of the photonbetween four-dimensional radial shells, onto the path between three-dimensional radialshells.

Date: April 1, 2015.

1

2 JAMES SPEARS

1. Introduction

When a skilled experimentalist wields a knowledge of the natural sciences exacting theunknown, the task of the theorist is to explain how the laws of nature have always allowedfor such curiosities. The focus of this article is the relationship credited to Edwin P. Hubble[?]; who, observed a correlation that exists between a) the apparent distance, and b) theredshift of spectral data concerning electromagnetic radiation emitted from stellar objects.

2. Interpreting the Light-like Interval

For completeness and clarity of procedure begin with the special relativistic case. Theinvariant interval for the flat Minkowski space-time is given [?] in line element form as

(1) ds2 = −c2dt2 + dx2 + dy2 + dz2 .

Applying a coordinate transformation from cartesian to spherical polar and geometrizingthe unit system such that c = 1, the line element becomes

(2) ds2 = −dt2 + dr2 + r2dΩ2 ,

where

(3) dΩ2 := dθ2 + sin2 θdφ2 .

Consider intervals along the radial line dθ = dφ = 0. Combining this with the stipulationthat light-like intervals satisfy

(4) ds2 = 0 ,

the relation depicting the signature characteristic of world line of the photon in specialrelativity becomes evident

(5) dt = dr .

Simple integration of both the right and left hand side reveals

(6) t = r + const. .

The constant of integration in this case represents time information relating to the path ofthe photon.

In the presence of a point mass M , Schwarzschild’s solution provides the idealized meansto interpret light traversing space and time. The solution has the familiar line element form[?]

(7) ds2 = −(

1− rsr

)dt2 +

dr2(1− rs

r

) + r2dΩ2 ,

in spherical polar coordinates and geometrized units c = G = 1 where rs = 2M . Consider-ing intervals along the radial line dθ = dφ = 0 and applying the condition (4) for light-likeintervals presents the equality

(8)(

1− rsr

)dt2 =

dr2(1− rs

r

) .

PHOTO-GEOMETRODYNAMICS (DRAFT) 3

Separating like terms and taking the square root of the left and right hand side,

(9) dt =dr(

1− rsr

) .Straightforward integration reveals

(10) t = r + rs ln

∣∣∣∣r − rsrs

∣∣∣∣+ const. ,

Where once again the constant of integration represents information about the location ofthe photon at a specific point in time. Defining a term r∗ such that

(11) r∗ = r + rs ln

∣∣∣∣r − rsrs

∣∣∣∣ ,allows for the Eddington-Finkelstein coordinate transformations [?] u, and v in compactnotation where

(12) u := t− r∗,

and

(13) v := t+ r∗.

Substituting the coordinate u, the line element (7) becomes

(14) ds2 = −(

1− rsr

)du2 − 2dudr + r2dΩ2 ,

known as out-going Eddington-Finkelstein coordinates. Considering null geodesics ds2 = 0along the radial line dθ = dφ = 0, (14) has the solutions

(15) u = −2r∗ + const.,

and

(16) u = const. .

Making a similar substitution for v, the line element (7) is then given

(17) ds2 = −(

1− rsr

)dv2 + 2dvdr + r2dΩ2 ,

known as in-going Eddington-Finkelstein coordinates. Considering null geodesics ds2 = 0along the radial line dθ = dφ = 0, (17) has the solutions

(18) v = 2r∗ + const.,

and

(19) v = const. .

4 JAMES SPEARS

3. Instantaneous Trajectory of Light in the Presence of Mass

The trajectory of a particle travelling radial null geodesics (15) or (18) in (u, r) or (v, r)space for some point (uγ , rγ) or (vγ , rγ) on the path P are given by the equations of linestangent to (15) or (18) respectively. These are

(20) uT =−2r + 2rγ(

1− rsrγ

) − 2r∗|rγ + const. ,

and

(21) vT =2r − 2rγ(1− rs

rγ

) + 2r∗|rγ + const. .

Where the constants in equations line (36) and (37) are equal to the constants in equationsline (15) and (18) respectively. In the last section these constants were both set equal tozero. Thus the singularity, with respect to a particle travelling a path tangential to (15)or (18) at rγ , lies at coordinates (us, rs) or (vs, rs) where

(22) (us, rs) = (uT |rs , rs) ,

and

(23) (vs, rs) = (vT |rs , rs)

where again rs = 2M in units c = G = 1. Combining coordinates (u, r) or (v, r) intoa single four-dimensional radial parameter R (u, x, y, z) we have that the trajectory of aparticle travelling null geodesics (15) or (18) crosses the event horizon

(24) Rs = rs

√4 ln

∣∣∣∣rγ − rsrs

∣∣∣∣2 + 1 ,

from the origin in (u, r) and (v, r) space.

4. Dimensional Independence of Event, and Dependence of Measurement

Here it is supposed that for a photon at a constant radial distance rγ at a constanttime uγ or vγ (depending on the coordinate transformation used) has a trajectory equal tothe line tangent to the null geodesic in (u, r) or (v, r) space. In order for this event to beindependent of dimension it is claimed that there must exist a particle in higher dimensionsof space and time which our observation is a projection of. We claim that there must exista particle in five-dimensions of space and time which has a five-dimensional instantaneoustrajectory that projects onto the instantaneous trajectory in four-dimensions of space andtime.

When a measurement is made however, it is supposed there is complete dependanceon the dimension of the observer. Meaning that the when one takes a measurement ofphysical phenomenon involving light one must also consider the higher dimensions of spaceand time and the projection of particles traversing space and time though these spaces.


5. Implementing the (u, r), (v, r) Coordinate Singularity

Determining the metric tensor gµν describing the the five-dimensional space-time is inother words finding solutions to the Einstein field equations, which in this case are a setof 15 second-order partial differential equations. The complete form of the Einstein fieldequations in Einstein’s summation notation are written

(25) Gµν ≡ Rµν −1

2gµνR+ gµνΛ = 8πTµν .

Naming these components we have Gµν , called the Einstein tensor, Rµν is the Ricci tensor,gµν is the general metric tensor, R is the Ricci scalar curvature, Tµν is the stress energytensor, and Λ is the cosmological constant. We will take the stress energy tensor to bezero, that is Tµν = 0. We will also assume a cosmological constant of zero Λ = 0. The fieldequations take the form

(26) Rµν −1

2gµνR = 0

The reader will see that we begin assuming there exists a solution of a certain exponentialform of the general metric tensor. The Ricci tensor and the Ricci scalar curvature will bedefined shortly, but here the reader can note that these components are functions of thecoefficients of the general metric tensor. So it is sensible to begin by assuming there existsa solution of a general form in order to further manipulate the components before anexplicit solution is found. Using the mathematical definitions of the non-trivial, non-zerocomponents of the field equations we will substitute our supposed solution in it’s generalform and state the field equations in this way. There will come a point where given certaininitial conditions and restrictions we know the solution to have, we are able to deduce closedform relations characteristic of the physics behind these mathematical objects. The explicitform of this solution is supposed to be known using equation line (24) as the location ofthe coordinate singularity in (u, r) and (v, r) space as per our assumptions. The geodesicequation is given

d2xρ

ds2+ Γρµν

dxµ

ds

dxν

ds= 0 ,

where s is a scalar parameter of motion (for example we will implement the equation usingproper time τ), Γρµν is the Christoffel symbol defined shortly, and xµ is the µth coordinate.After our deductions we will be able to express this equation of motion in terms of theunknown constant of integration. Being the general relativistic equivalent of Newton’s lawsof motion, defining straight paths in curved spaces, will be used to deduce the sign of theconstant of integration found in the explicit form (gµν in terms of R, M , and rγ). Sincewe know the location of the coordinate singularity, or we should say, the four-dimensionallocation that the instantaneous path of a photon at (uγ , rγ) or (vγ , rγ) will cross the eventhorizon, the system of differential equations needs only to be shown to be satisfied by theproposed solution. It is in a way the converse argument used to deduce the exact form ofthe Schwarzschild solution by four-dimensional free fall. However we are jumping ahead,that realization is yet to be made.

6 JAMES SPEARS

We can now define the terms required to proceed. Rµν is the Ricci tensor and is defined

by summing the Riemann tensor Rλµνσ over one index

Rλµνσ ≡∂Γλσµ∂xν

−∂Γλνµ∂xσ

+ ΓλνρΓρσµ − ΓλσρΓ

ρνµ ,

⇒ Rλµνλ =∂Γλλµ∂xν

−∂Γλνµ∂xλ

+ ΓλνρΓρλµ − ΓλλρΓ

ρνµ = Rµν .

The Christoffel symbol Γλµν is defined

Γλµν ≡1

2gλσ

(∂gνσ∂xµ

+∂gµσ∂xν

− ∂gµν∂xσ

).

We note here the scalar curvature R ≡ gµνRµν . Where gµν are the coefficients of thematrix inverse of gµν .

5.1. Determining the general form of the line element. The invariant interval ds2

for the ‘flat’ five-dimensional Minkowski space-time representing the special relativisticcase is given in Einstein summation notation and line element form as

ds2 = ηµνdxµdxν = −c2dT 2 + du2 + dx2 + dy2 + dz2 = −c2dT 2 + dv2 + dx2 + dy2 + dz2 ,

with coordinates defined(x0, x1, x2, x3, x4

)= (T, x, y, z, u) or

(x0, x1, x2, x3, x4

)= (T, x, y, z, v)

in geometrized units c = 1. We see ηµν has the definition

ηµν =

−1 0 0 0 00 1 0 0 00 0 1 0 00 0 0 1 00 0 0 0 1

.

The tensor ηµν serves the same purpose as the general metric tensor gµν featured in theEinstein field equations of general relativity for which we wish to solve. In the ‘special ’case there are no energy densities to be considered when the measurements made by asystem of ideal observers are to be calculated. This concept will be useful when settingsome of the restrictions of the ‘general ’ case mentioned earlier. The supposition that thespace-time be asymptotically flat is made and means that at an infinite distance from anobject of mass the space-time will reduce to the ‘special ’ relativistic case.

We wish to define the curvature of space-time about a spherically symmetric, non-rotating object of mass in a vacuum. Thus it is sensible to define the origin of our coor-dinate system at the centre of the mass M , in question. It is also beneficial to change thecoordinate system from cartesian coordinates to four-dimensional spherical polar analogcoordinates using the transformation

x = R sin θ sinφ sinψ ,

y = R sin θ cosφ sinψ ,

z = R cos θ sinψ ,


u = R cosψ or v = R cosψ ,

By making such a transition we will see we have reduced the number of coefficients ofgµν we need to solve for in the general case. After taking the differential of the abovecoordinate transformation and substituting the values in to the equation for the line elementin cartesian coordinates, the line element of the Minkowski space-time in spherical polarcoordinates is shown to take the form

ds2 = −c2dT 2 + dR2 +R2dθ2 +R2 sin2 θdφ2 +R2 sin2 θ sin2 φdψ2 .

With coordinates defined(x0, x1, x2, x3

)= (T,R, θ, φ, ψ) in geometrized units c = 1 where

c in this instance is the speed of the five-dimensional photon. Notice that for constant Tand constant R the metric is that of a three-sphere.

We must now generalize the line element for introducing curvature while retaining ourinitial assumptions. We have mentioned that the solution is static. By definition thismeans two things are true. 1: The metric is unchanged if one applies a reversal of timeT → −T , and 2: the metric is independent of T . Spherical symmetry implies that under areversal of coordinates θ → −θ, φ→ −φ, and ψ → −ψ the metric is unchanged. Thus oursolution must remain independent of these coordinates as well. The conditions we have justoutlined have the consequence of diagonalizing the metric such that if we are to introducemetric coefficients the line element takes the form

ds2 = −g00dT2 + g11dR

2 + g22dθ2 + g33dφ

2 + g44dψ2 .

The definition of spherical symmetry may be made more precise by requiring that everypoint in the space-time where dr = dt = 0, lies on a two-surface which is a two-sphere thathas the line element [?].

dl2 = f (R, T )[dθ2 + sin2 θ

(dφ2 + sin2 φdψ2

)]We may state there exists a function f(R, T ) and define our radial coordinate as the root ofsuch a function defined as the radius of curvature of the three-sphere. Doing so makes ournotational reference to the radial coordinate some what arbitrary. We note the flexibilityof our notation and continue to use the coordinate R.

Our line element satisfying conditions for being static and spherically symmetric is thenof the form

ds2 = −α(R)dt2 + β(R)dR2 +R2 sin2 θdφ2 +R2 sin2 θ sin2 φdψ2 ,

where α(R) and β(R) are radially dependant functions. It is advantageous to elaborate onour dependance. As our functions α(R) and β(R) at this time are completely arbitrary,we further define such functions having exponential forms which is permitted so long asg00 < 0 and g11 > 0. This will also make our future calculations easier. So we define

α(R) = e2Λ(R) ,

and

β(R) = e2Φ(R) ,

8 JAMES SPEARS

making our line element

ds2 = −e2Λ(R)dT 2 + e2Φ(R)dR2 +R2dθ2 +R2 sin2 θdφ2 +R2 sin2 θ sin2 φdψ2 .

In matrix form the general metric tensor gµν may be written

gµν =

−e2Λ(R) 0 0 0 0

0 e2Φ(R) 0 0 00 0 R2 0 00 0 0 R2 sin2 θ 00 0 0 0 R2 sin2 θ sin2 φ

.

5.2. Solving for the Christoffel symbols. We are now in a position to solve the Einsteinfield equations. Let us begin by solving for the Christoffel symbols. Firstly we will deducesome of the non-zero elements. Recall

Γλµν =1

2gλσ

(∂gµσ∂xν

+∂gνσ∂xµ

− ∂gµν∂xσ

).

We notice that if we are to solve for every Christoffel symbol, that equates to 125 corre-sponding equations. Diagonalizing the metric has provided us a method by which we maydeduce some of the trivially zero symbols. Note that the metric gµν is a diagonal matrixi.e. gµν = 0 for µ 6= ν. Thus the Christoffel symbols for which λ 6= σ are zero. We restatethe Christoffel symbol

Γλµν =1

2gλλ

(∂gµλ∂xν

+∂gνλ∂xµ

− ∂gµν∂xλ

).

Immediately we notice for µ 6= ν 6= λ the term is trivially zero, again for reason that gµνis a diagonal matrix. We will refer to µ 6= ν 6= λ as case (1). The next case examined isλ = ν

Γλµλ =1

2gλλ

(∂gµλ∂xλ

+∂gλλ∂xµ

−∂gµλ∂xλ

)=

1

2gλλ

(∂gλλ∂xµ

),

which we will refer to as case (2). Similarly note if we were to swap λ and µ, the Christoffelsymbols take the form

Γλλµ =1

2gλλ

(∂gλλ∂xµ

+∂gµλ∂xλ

−∂gλµ∂xλ

)= Γλµλ .

For the next case which we will refer to as case (3), refer back to the definition of theChristoffel symbol and let µ = ν

Γλµµ =1

2gλλ

(∂gµλ∂xµ

+∂gµλ∂xµ

− ∂gµµ∂xλ

)= −1

2gλλ

(∂gµµ∂xλ

).

So far we have reduced the number of Christoffel symbols for which we need to solve from125 equations to 45 equations. Having significantly reduced the work required we may nowbegin solving the equations. Note that the ′ (prime) notation implies the partial derivativewith respect to x1 = R. All 45 Christoffel symbols and the corresponding equations aregiven and solved below. Note that any gαα is a function of at most three variables, either


x1 = R, x2 = θ, or x3 = φ and so any partial derivative of a metric coefficient with respectto variables x0 = T , x4 = ψ will be zero. Note too that in every case we will use thefact that the inverse of a diagonal matrix is also a diagonal matrix with the correspondingmetric coefficients inverted, that is gλλ = 1

gλλfor our calculations.

Γ000 =

(∂g00

∂x0

)2g00

=

(∂∂T e

2Λ(R))

2e2Λ(R)= 0

Γ001 = Γ0

10 =

(∂g00

∂x1

)2g00

=

(∂∂Re

2Λ(R))

2e2Λ(R)= Λ′(R)

Γ002 = Γ0

20 =

(∂g00

∂x2

)2g00

=

(∂∂θe

2Λ(R))

2e2Λ(R)= 0

Γ003 = Γ0

30 =

(∂g00

∂x3

)2g00

=

(∂∂φe

2Λ(R))

2e2Λ(R)= 0

Γ004 = Γ0

40 =

(∂g00

∂x4

)2g00

=

(∂∂ψe

2Λ(R))

2e2Λ(R)= 0

Γ011 = −

(∂g11

∂x0

)2g00

=

(∂∂T e

2Φ(R))

2e2Λ(R)= 0

Γ022 = −

(∂g22

∂x0

)2g00

=

(∂∂TR

2)

2e2Λ(R)= 0

Γ033 = −

(∂g33

∂x0

)2g00

=

(∂∂TR

2 sin2 θ)

2e2Λ(R)= 0

Γ044 = −

(∂g44

∂x0

)2g00

=

(∂∂TR

2 sin2 θ)

2e2Λ(R)= 0

Γ100 = −

(∂g00

∂x1

)2g11

=

(∂∂Re

2Λ(R))

2e2Φ(R)= Λ′(R) · e

2Λ(R)

e2Φ(R)

Γ101 = Γ1

10 =

(∂g11

∂x0

)2g11

=

(∂∂T e

2Φ(R))

2e2Φ(R)= 0

10 JAMES SPEARS

Γ111 =

(∂g11

∂x1

)2g11

=

(∂∂Re

2Φ(R))

2e2Φ(R)= Φ′(R)

Γ112 = Γ1

21 =

(∂g11

∂x2

)2g11

=

(∂∂θe

2Φ(R))

2e2Φ(R)= 0

Γ113 = Γ1

31 =

(∂g11

∂x3

)2g11

=

(∂∂φe

2Φ(R))

2e2Φ(R)= 0

Γ114 = Γ1

41 =

(∂g11

∂x4

)2g11

=

(∂∂ψe

2Φ(R))

2e2Φ(R)= 0

Γ122 = −

(∂g22

∂x1

)2g11

= −(∂∂RR

2)

2e2Φ(R)= − R

e2Φ(R)

Γ133 = −

(∂g33

∂x1

)2g11

= −(∂∂RR

2 sin2 θ)

2e2Φ(R)= −R sin2 θ

e2Φ(R)

Γ144 = −

(∂g44

∂x1

)2g11

= −(∂∂RR

2 sin2 θ sin2 φ)

2e2Φ(R)= −R sin2 θ sin2 φ

e2Φ(R)

Γ202 = Γ2

20 =

(∂g22

∂x0

)2g22

=

(∂∂TR

2)

2R2= 0

Γ221 = Γ2

12 =

(∂g22

∂x1

)2g22

=

(∂∂RR

2)

2R2=

1

R

Γ232 = Γ2

23 =

(∂g22

∂x3

)2g22

=

(∂∂φR

2)

2R2= 0

Γ242 = Γ2

24 =

(∂g22

∂x4

)2g22

=

(∂∂ψR

2)

2R2= 0

Γ233 = −

(∂g33

∂x2

)2g22

= −(∂∂θR

2 sin2 θ)

2R2= − sin θ cos θ

Γ244 = −

(∂g44

∂x2

)2g22

= −(∂∂θR

2 sin2 θ sin2 φ)

2R2= − sin θ cos θ sin2 φ


Γ200 = −

(∂g00

∂x2

)2g22

=

(∂∂θe

2Λ(R))

2R2= 0

Γ211 = −

(∂g11

∂x2

)2g22

= −(∂∂θe

2Φ(R))

2R2= 0

Γ222 =

(∂g22

∂x2

)2g22

=

(∂R2

∂θ

)2R2

= 0

Γ303 = Γ3

30 =

(∂g33

∂x0

)2g33

=

(∂∂TR

2 sin2 θ)

2R2 sin2 θ= 0

Γ300 = −

(∂g00

∂x3

)2g33

=

(∂∂φe

2Λ(R))

2R2 sin2 θ= 0

Γ311 = −

(∂g11

∂x3

)2g33

= −

(∂∂φe

2Φ(R))

2R2 sin2 θ= 0

Γ322 = −

(∂g22

∂x3

)2g33

=

(∂R2

∂φ

)2R2 sin2 θ

= 0

Γ333 =

(∂g33

∂x3

)2g33

=

(∂∂φR

2 sin2 θ)

2R2 sin2 θ= 0

Γ331 = Γ3

13 =

(∂g33

∂x1

)2g33

=

(∂∂RR

2 sin2 θ)

2R2 sin2 θ=

1

R

Γ332 = Γ3

23 =

(∂g33

∂x2

)2g33

=

(∂∂θR

2 sin2 θ)

2R2 sin2 θ= cot θ

Γ343 = Γ3

34 =

(∂g33

∂x4

)2g22

=

(∂∂ψR

2 sin2 θ)

2R2= 0

Γ344 = −

(∂g44

∂x3

)2g33

=

(∂∂φR

2 sin2 θ sin2 φ)

2R2 sin2 θ= − sinφ cosφ

Γ400 = −

(∂g00

∂x4

)2g44

=

(∂∂ψe

2Λ(R))

2R2 sin2 θ sin2 φ= 0

12 JAMES SPEARS

Γ411 = −

(∂g11

∂x4

)2g44

= −

(∂∂ψe

2Φ(R))


Γ422 = −

(∂g22

∂x4

)2g44

=

(∂∂ψR

2)


Γ433 = −

(∂g33

∂x4

)2g44

=

(∂∂ψR

2 sin2 θ sin2 φ)


Γ404 = Γ4

40 =

(∂g44

∂x0

)2g44

=

(∂∂TR

2 sin2 θ sin2 φ)


Γ414 = Γ4

41 =

(∂g44

∂x1

)2g44

=

(∂∂RR

2 sin2 θ sin2 φ)

2R2 sin2 θ sin2 φ=

1

R

Γ424 = Γ4

42 =

(∂g44

∂x2

)2g44

=

(∂∂θR

2 sin2 θ sin2 φ)

2R2 sin2 θ sin2 φ= cot θ

Γ434 = Γ4

43 =

(∂g44

∂x3

)2g44

=

(∂∂φR

2 sin2 θ sin2 φ)

2R2 sin2 θ sin2 φ= cotφ

Γ444 =

(∂g44

∂x4

)2g44

=

(∂∂ψR

2 sin2 θ sin2 φ)


We may now represent the Christoffel symbols in matrix form accounting for all 125equations.

Γ0µν =

0 Λ′(R) 0 0 0

Λ′(R) 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 0

Γ1µν =

Λ′(R)·e2Λ(R)

e2Φ(R) 0 0 0 0

0 Φ′(R) 0 0 00 0 − R

e2Φ(R) 0 0

0 0 0 −R sin2 θe2Φ(R) 0

0 0 0 0 −R sin2 θ sin2 φe2Φ(R)


Γ2µν =

0 0 0 0 00 0 1

R 0 00 1

R 0 0 00 0 0 − sin θ cos θ 00 0 0 0 − sin2 φ sin θ cos θ

Γ3µν =

0 0 0 0 00 0 0 1

R 00 0 0 cot θ 00 1

R cot θ 0 00 0 0 0 − sinφ cosφ

Γ4µν =

0 0 0 0 00 0 0 0 1

R0 0 0 0 cot θ0 0 0 0 cotφ0 1

R cot θ cotφ 0

5.3. Determining the Ricci tensor. Recall, the Ricci tensor is given by summing theRiemann tensor over one index. The Einstein vacuum field equations are the set of secondorder partial differential equations which satisfy

(27) Rµν =∂Γλλµ∂xν



ρνµ = 0 .

We begin constructing the Ricci tensor by solving for the non-trivial elements which are

R00 =∂Γλλ0

∂x0− ∂Γλ00

∂xλ+ Γλ0ρΓ

ρλ0 − ΓλλρΓ

ρ00

= −∂Γ100

∂x1+ Γ0

01Γ100 − Γ1

11Γ100 − Γ2

21Γ100 − Γ3

31Γ100 − Γ4

41Γ100 ,

⇒ e2Λ(R)−2Φ(R) ·(−Λ′(R)2 − Λ′′(R) + Λ′(R)Φ′(R)− 3Λ′(R)

R

)= 0 .

R11 =∂Γλλ1

∂x1− ∂Γλ11

∂xλ+ Γλ1ρΓ


ρ11

=∂Γ0

01

∂x1+ Γ0

10Γ001 − Γ0

01Γ111 − Γ2

21Γ111 − Γ3

31Γ111 − Γ4

41Γ111

⇒ Λ′′(R) + Λ′(R)2 − Λ′(R)Φ′(R)− 3Φ′(R)

R= 0

R22 =∂Γλλ2

∂x2− ∂Γλ22

∂xλ+ Γλ2ρΓ


ρ22

=∂Γ3

32

∂x2+∂Γ4

42

∂x2− ∂Γ1

22

∂x1+ Γ1

22Γ212 + Γ3

23Γ332 + Γ4

24Γ442 − Γ0

01Γ122 − Γ1

11Γ122 − Γ3

31Γ122 − Γ4

41Γ122

⇒ e−2Φ(R) ·(2 + Λ′(R) ·R− Φ′(R) ·R

)− 2 = 0

14 JAMES SPEARS

R33 =∂Γλλ3

∂x3− ∂Γλ33

∂xλ+ Γλ3ρΓ


ρ33

=∂Γ4

43

∂x3−∂Γ1

33

∂x1−∂Γ2

33

∂x2+Γ3

31Γ133+Γ3

32Γ233+Γ4

34Γ443−Γ0

01Γ133−Γ1

11Γ133−Γ2

21Γ133−Γ4

41Γ133−Γ4

42Γ233

⇒ sin2 θ ·[e−2Φ(R) ·

(2 + Λ′(R) ·R− Φ′(R) ·R

)− 2]

= sin2 θ ·R22 = 0

R44 =∂Γλλ4

∂x4− ∂Γλ44

∂xλ+ Γλ4ρΓ


ρ44

= −∂Γ144

∂x1−∂Γ2

44

∂x2−∂Γ3

44

∂x3+Γ1

44Γ414+Γ3

44Γ433+Γ4

22Γ244−Γ0

01Γ144−Γ1

11Γ144−Γ2

21Γ144−Γ3

31Γ144−Γ3

32Γ244

⇒ sin2 θ sin2 φ ·[e−2Φ(R) ·

(2 + Λ′(R) ·R− Φ′(R) ·R

)− 2]

= sin2 φ ·R33 = 0

For completeness we will show that the remaining values of the Ricci tensor are zero. Onceagain we note the Ricci tensor in summation notation

Rµν =∂Γλλµ∂xν



ρνµ = 0 ,

and notice that the tensor is represented as four individual terms also in summation nota-tion. So we will begin solving term by term for µ 6= ν. The first of four terms is written

∂Γλλµ∂xν

.

We may immediately eliminate all ν = 0 and ν = 4 as the Christoffel symbols havedependance on at most two variables x1, x2, and x3. Summing out the term excludingµ = ν

∂Γλλµ∂xν

=∂Γλλ0

∂x1+∂Γλλ0

∂x2+∂Γλλ0

∂x3+∂Γλλ1

∂x2+∂Γλλ1

∂x3+∂Γλλ2

∂x1+∂Γλλ2

∂x3+∂Γλλ3

∂x1+∂Γλλ3

∂x2+∂Γλλ4

∂x1+∂Γλλ4

∂x2+∂Γλλ4

∂x3.

We refer to pg. 12 and write the terms which are non-trivially zero (do not have corre-sponding Christoffel symbols equal to zero).

∂Γλλµ∂xν

=∂Γ0

01

∂x2+∂Γ1

11

∂x2+∂Γ2

21

∂x2+∂Γ3

31

∂x2+∂Γ3

31

∂x2+∂Γ0

01

∂x3+∂Γ1

11

∂x3+∂Γ2

21

∂x3+

∂Γ331

∂x3+∂Γ3

31

∂x3+∂Γ3

32

∂x1+∂Γ4

42

∂x1+∂Γ3

32

∂x3+∂Γ4

42

∂x3+∂Γ4

43

∂x1+∂Γ4

43

∂x2

Referring again to pg. 12 we see all the terms have partial derivatives with respect tonon-dependant variables and thus are all equal to zero. We have shown

∂Γλλµ∂xν

= 0 ,

for all µ 6= ν. The second term of the tensor equation is written

∂Γλνµ∂xλ

.


Once again we may immediately eliminate all terms where λ = 0 or λ = 4 as we know thatthere exists no Christoffel symbol with dependance on either x0 or x4 inclusive. We alsohave the initial condition that µ 6= ν and we expand out the summation as

∂Γλνµ∂xλ

=∂Γ1

νµ

∂x1+∂Γ2

νµ

∂x2+∂Γ3

νµ

∂x3.

Referring again to pg. 12 and find the Christoffel symbols Γ1µν = 0 for µ 6= ν. Keeping in

mind Γλµλ = Γλλµ, and eliminating λ 6= µ 6= ν

∂Γ2νµ

∂x2+∂Γ3

νµ

∂x3= 2

∂Γ212

∂x2+ 2

∂Γ313

∂x3+ 2

∂Γ323

∂x3.

We now refer again to pg. 12 and find the Christoffel symbols Γ212, Γ3

13 are not functionsof x2 and Γ3

23 is not function of x3. So

∂Γλνµ∂xλ

= 0 ,

for µ 6= ν. The third term in the tensor equation is written

ΓλνρΓρλµ .

Expanding out for λ and ρ

ΓλνρΓρλµ = Γ0

ν0Γ00µ + Γ0

ν1Γ10µ + Γ0

ν2Γ20µ + Γ0

ν3Γ30µ + Γ0

ν4Γ40µ+

Γ1ν0Γ0

1µ + Γ1ν1Γ1

1µ + Γ1ν2Γ2

1µ + Γ1ν3Γ3

1µ + Γ1ν4Γ4

1µ+

Γ2ν0Γ0

2µ + Γ2ν1Γ1

2µ + Γ2ν2Γ2

2µ + Γ2ν3Γ3

2µ + Γ2ν4Γ4

2µ+

Γ3ν0Γ0

3µ + Γ3ν1Γ1

3µ + Γ3ν2Γ2

3µ + Γ3ν3Γ3

3µ + Γ3ν4Γ4

3µ+

Γ4ν0Γ0

4µ + Γ4ν1Γ1

4µ + Γ4ν2Γ2

4µ + Γ4ν3Γ3

4µ + Γ4ν4Γ4

4µ .

We will continue examining the term case by case beginning with (µ, ν) = (0, 1). Oncethe summation is written explicitly for the given case we will refer to pg. 14 and state thenon-zero Christoffel symbols if any.


10Γ000 + Γ0

11Γ100 + Γ0

12Γ200 + Γ0

13Γ300 + Γ0

14Γ400+

Γ110Γ0

10 + Γ111Γ1

10 + Γ112Γ2

10 + Γ113Γ3

10 + Γ114Γ4

10+

Γ210Γ0

20 + Γ211Γ1

20 + Γ212Γ2

20 + Γ213Γ3

20 + Γ214Γ4

20+

Γ310Γ0

30 + Γ311Γ1

30 + Γ312Γ2

30 + Γ313Γ3

30 + Γ314Γ4

30+

Γ410Γ0

40 + Γ411Γ1

40 + Γ412Γ2

40 + Γ413Γ3

40 + Γ414Γ4

40 .

We find that every product in the summation has at least one Christoffel symbol equalto zero. For (µ, ν) = (0, 1) , ΓλνρΓ

ρλµ = 0.

16 JAMES SPEARS

For the case (µ, ν) = (0, 2).


20Γ000 + Γ0

21Γ100 + Γ0

22Γ200 + Γ0

23Γ300 + Γ0

24Γ400+

Γ120Γ0

10 + Γ121Γ1

10 + Γ122Γ2

10 + Γ123Γ3

10 + Γ124Γ4

10+

Γ220Γ0

20 + Γ221Γ1

20 + Γ222Γ2

20 + Γ223Γ3

20 + Γ224Γ4

20+

Γ320Γ0

30 + Γ321Γ1

30 + Γ322Γ2

30 + Γ323Γ3

30 + Γ324Γ4

30+

Γ420Γ0

40 + Γ421Γ1

40 + Γ422Γ2

40 + Γ423Γ3

40 + Γ424Γ4

40 .


ρλµ = 0.



30Γ000 + Γ0

31Γ100 + Γ0

32Γ200 + Γ0

33Γ300 + Γ0

34Γ400+

Γ130Γ0

10 + Γ131Γ1

10 + Γ132Γ2

10 + Γ133Γ3

10 + Γ134Γ4

10+

Γ230Γ0

20 + Γ231Γ1

20 + Γ232Γ2

20 + Γ233Γ3

20 + Γ234Γ4

20+

Γ330Γ0

30 + Γ331Γ1

30 + Γ332Γ2

30 + Γ333Γ3

30 + Γ334Γ4

30+

Γ430Γ0

40 + Γ431Γ1

40 + Γ432Γ2

40 + Γ433Γ3

40 + Γ434Γ4

40 .


ρλµ = 0.



40Γ000 + Γ0

41Γ100 + Γ0

42Γ200 + Γ0

43Γ300 + Γ0

44Γ400+

Γ140Γ0

10 + Γ141Γ1

10 + Γ142Γ2

10 + Γ143Γ3

10 + Γ144Γ4

10+

Γ240Γ0

20 + Γ241Γ1

20 + Γ242Γ2

20 + Γ243Γ3

20 + Γ244Γ4

20+

Γ340Γ0

30 + Γ341Γ1

30 + Γ342Γ2

30 + Γ343Γ3

30 + Γ344Γ4

30+

Γ440Γ0

40 + Γ441Γ1

40 + Γ442Γ2

40 + Γ443Γ3

40 + Γ444Γ4

40 .

We find that every product in the summation has at least one Christoffel symbol equal tozero. For (µ, ν) = (0, 4) , ΓλνρΓ

ρλµ = 0.


For the case (µ, ν) = (1, 0)


00Γ001 + Γ0

01Γ101 + Γ0

02Γ201 + Γ0

03Γ301 + Γ0

04Γ401+

Γ100Γ0

11 + Γ101Γ1

11 + Γ102Γ2

11 + Γ103Γ3

11 + Γ104Γ4

11+

Γ200Γ0

21 + Γ201Γ1

21 + Γ202Γ2

21 + Γ203Γ3

21 + Γ204Γ4

21+

Γ300Γ0

31 + Γ301Γ1

31 + Γ302Γ2

31 + Γ303Γ3

31 + Γ304Γ4

31+

Γ400Γ0

41 + Γ401Γ1

41 + Γ402Γ2

41 + Γ403Γ3

41 + Γ404Γ4

41 .


ρλµ = 0.



20Γ001 + Γ0

21Γ101 + Γ0

22Γ201 + Γ0

23Γ301 + Γ0

24Γ401+

Γ120Γ0

11 + Γ121Γ1

11 + Γ122Γ2

11 + Γ123Γ3

11 + Γ124Γ4

11+

Γ220Γ0

21 + Γ221Γ1

21 + Γ222Γ2

21 + Γ223Γ3

21 + Γ224Γ4

21+

Γ320Γ0

31 + Γ321Γ1

31 + Γ322Γ2

31 + Γ323Γ3

31 + Γ324Γ4

31+

Γ420Γ0

41 + Γ421Γ1

41 + Γ422Γ2

41 + Γ423Γ3

41 + Γ424Γ4

41 .

We find that every product in the summation has at least one Christoffel symbol equalto zero except for the last term of the summation. For (µ, ν) = (1, 2) , ΓλνρΓ

ρλµ = 2cot θR−1.



30Γ001 + Γ0

31Γ101 + Γ0

32Γ201 + Γ0

33Γ301 + Γ0

34Γ401+

Γ130Γ0

11 + Γ131Γ1

11 + Γ132Γ2

11 + Γ133Γ3

11 + Γ134Γ4

11+

Γ230Γ0

21 + Γ231Γ1

21 + Γ232Γ2

21 + Γ233Γ3

21 + Γ234Γ4

21+

Γ330Γ0

31 + Γ331Γ1

31 + Γ332Γ2

31 + Γ333Γ3

31 + Γ334Γ4

31+

Γ430Γ0

41 + Γ431Γ1

41 + Γ432Γ2

41 + Γ433Γ3

41 + Γ434Γ4

41 .


ρλµ = cotφR−1.

18 JAMES SPEARS



40Γ001 + Γ0

41Γ101 + Γ0

42Γ201 + Γ0

43Γ301 + Γ0

44Γ401+

Γ140Γ0

11 + Γ141Γ1

11 + Γ142Γ2

11 + Γ143Γ3

11 + Γ144Γ4

11+

Γ240Γ0

21 + Γ241Γ1

21 + Γ242Γ2

21 + Γ243Γ3

21 + Γ244Γ4

21+

Γ340Γ0

31 + Γ341Γ1

31 + Γ342Γ2

31 + Γ343Γ3

31 + Γ344Γ4

31+

Γ440Γ0

41 + Γ441Γ1

41 + Γ442Γ2

41 + Γ443Γ3

41 + Γ444Γ4

41 .


ρλµ = 0.



00Γ002 + Γ0

01Γ102 + Γ0

02Γ202 + Γ0

03Γ302 + Γ0

04Γ402+

Γ100Γ0

12 + Γ101Γ1

12 + Γ102Γ2

12 + Γ103Γ3

12 + Γ104Γ4

12+

Γ200Γ0

22 + Γ201Γ1

22 + Γ202Γ2

22 + Γ203Γ3

22 + Γ204Γ4

22+

Γ300Γ0

32 + Γ301Γ1

32 + Γ302Γ2

32 + Γ303Γ3

32 + Γ304Γ4

32+

Γ400Γ0

42 + Γ401Γ1

42 + Γ402Γ2

42 + Γ403Γ3

42 + Γ404Γ4

42 .


ρλµ = 0.



10Γ002 + Γ0

11Γ102 + Γ0

12Γ202 + Γ0

13Γ302 + Γ0

14Γ402+

Γ110Γ0

12 + Γ111Γ1

12 + Γ112Γ2

12 + Γ113Γ3

12 + Γ114Γ4

12+

Γ210Γ0

22 + Γ211Γ1

22 + Γ212Γ2

22 + Γ213Γ3

22 + Γ214Γ4

22+

Γ310Γ0

32 + Γ311Γ1

32 + Γ312Γ2

32 + Γ313Γ3

32 + Γ314Γ4

32+

Γ410Γ0

42 + Γ411Γ1

42 + Γ412Γ2

42 + Γ413Γ3

42 + Γ414Γ4

42 .

We find that every product in the summation has at least one Christoffel symbol equalto zero except for the last term of the summation. For (µ, ν) = (2, 1) , ΓλνρΓ

ρλµ = 2cot θR−1.




30Γ002 + Γ0

31Γ102 + Γ0

32Γ202 + Γ0

33Γ302 + Γ0

34Γ402+

Γ130Γ0

12 + Γ131Γ1

12 + Γ132Γ2

12 + Γ133Γ3

12 + Γ134Γ4

12+

Γ230Γ0

22 + Γ231Γ1

22 + Γ232Γ2

22 + Γ233Γ3

22 + Γ234Γ4

22+

Γ330Γ0

32 + Γ331Γ1

32 + Γ332Γ2

32 + Γ333Γ3

32 + Γ334Γ4

32+

Γ430Γ0

42 + Γ431Γ1

42 + Γ432Γ2

42 + Γ433Γ3

42 + Γ434Γ4

42 .


ρλµ = cot θ cotφ. For the case (µ, ν) = (2, 4)


40Γ002 + Γ0

41Γ102 + Γ0

42Γ202 + Γ0

43Γ302 + Γ0

44Γ402+

Γ140Γ0

12 + Γ141Γ1

12 + Γ142Γ2

12 + Γ143Γ3

12 + Γ144Γ4

12+

Γ240Γ0

22 + Γ241Γ1

22 + Γ242Γ2

22 + Γ243Γ3

22 + Γ244Γ4

22+

Γ340Γ0

32 + Γ341Γ1

32 + Γ342Γ2

32 + Γ343Γ3

32 + Γ344Γ4

32+

Γ440Γ0

42 + Γ441Γ1

42 + Γ442Γ2

42 + Γ443Γ3

42 + Γ444Γ4

42 .


ρλµ = 0. For the case (µ, ν) = (3, 0)


00Γ003 + Γ0

01Γ103 + Γ0

02Γ203 + Γ0

03Γ303 + Γ0

04Γ403+

Γ100Γ0

13 + Γ101Γ1

13 + Γ102Γ2

13 + Γ103Γ3

13 + Γ104Γ4

13+

Γ200Γ0

23 + Γ201Γ1

23 + Γ202Γ2

23 + Γ203Γ3

23 + Γ204Γ4

23+

Γ300Γ0

33 + Γ301Γ1

33 + Γ302Γ2

33 + Γ303Γ3

33 + Γ304Γ4

33+

Γ400Γ0

43 + Γ401Γ1

43 + Γ402Γ2

43 + Γ403Γ3

43 + Γ404Γ4

43 .


ρλµ = 0.

20 JAMES SPEARS



10Γ003 + Γ0

11Γ103 + Γ0

12Γ203 + Γ0

13Γ303 + Γ0

14Γ403+

Γ110Γ0

13 + Γ111Γ1

13 + Γ112Γ2

13 + Γ113Γ3

13 + Γ114Γ4

13+

Γ210Γ0

23 + Γ211Γ1

23 + Γ212Γ2

23 + Γ213Γ3

23 + Γ214Γ4

23+

Γ310Γ0

33 + Γ311Γ1

33 + Γ312Γ2

33 + Γ313Γ3

33 + Γ314Γ4

33+

Γ410Γ0

43 + Γ411Γ1

43 + Γ412Γ2

43 + Γ413Γ3

43 + Γ414Γ4

43 .


ρλµ = cotφR−1.



20Γ003 + Γ0

21Γ103 + Γ0

22Γ203 + Γ0

23Γ303 + Γ0

24Γ403+

Γ120Γ0

13 + Γ121Γ1

13 + Γ122Γ2

13 + Γ123Γ3

13 + Γ124Γ4

13+

Γ220Γ0

23 + Γ221Γ1

23 + Γ222Γ2

23 + Γ223Γ3

23 + Γ224Γ4

23+

Γ320Γ0

33 + Γ321Γ1

33 + Γ322Γ2

33 + Γ323Γ3

33 + Γ324Γ4

33+

Γ420Γ0

43 + Γ421Γ1

43 + Γ422Γ2

43 + Γ423Γ3

43 + Γ424Γ4

43 .


ρλµ = cot θ cotφ.



40Γ003 + Γ0

41Γ103 + Γ0

42Γ203 + Γ0

43Γ303 + Γ0

44Γ403+

Γ140Γ0

13 + Γ141Γ1

13 + Γ142Γ2

13 + Γ143Γ3

13 + Γ144Γ4

13+

Γ240Γ0

23 + Γ241Γ1

23 + Γ242Γ2

23 + Γ243Γ3

23 + Γ244Γ4

23+

Γ340Γ0

33 + Γ341Γ1

33 + Γ342Γ2

33 + Γ343Γ3

33 + Γ344Γ4

33+

Γ440Γ0

43 + Γ441Γ1

43 + Γ442Γ2

43 + Γ443Γ3

43 + Γ444Γ4

43 .


ρλµ = 0.




00Γ004 + Γ0

01Γ104 + Γ0

02Γ204 + Γ0

03Γ304 + Γ0

04Γ404+

Γ100Γ0

14 + Γ101Γ1

14 + Γ102Γ2

14 + Γ103Γ3

14 + Γ104Γ4

14+

Γ200Γ0

24 + Γ201Γ1

24 + Γ202Γ2

24 + Γ203Γ3

24 + Γ204Γ4

24+

Γ300Γ0

34 + Γ301Γ1

34 + Γ302Γ2

34 + Γ303Γ3

34 + Γ304Γ4

34+

Γ400Γ0

44 + Γ401Γ1

44 + Γ402Γ2

44 + Γ403Γ3

44 + Γ404Γ4

44 .


ρλµ = 0.



10Γ004 + Γ0

11Γ104 + Γ0

12Γ204 + Γ0

13Γ304 + Γ0

14Γ404+

Γ110Γ0

14 + Γ111Γ1

14 + Γ112Γ2

14 + Γ113Γ3

14 + Γ114Γ4

14+

Γ210Γ0

24 + Γ211Γ1

24 + Γ212Γ2

24 + Γ213Γ3

24 + Γ214Γ4

24+

Γ310Γ0

34 + Γ311Γ1

34 + Γ312Γ2

34 + Γ313Γ3

34 + Γ314Γ4

34+

Γ410Γ0

44 + Γ411Γ1

44 + Γ412Γ2

44 + Γ413Γ3

44 + Γ414Γ4

44 .


ρλµ = 0.



20Γ004 + Γ0

21Γ104 + Γ0

22Γ204 + Γ0

23Γ304 + Γ0

24Γ404+

Γ120Γ0

14 + Γ121Γ1

14 + Γ122Γ2

14 + Γ123Γ3

14 + Γ124Γ4

14+

Γ220Γ0

24 + Γ221Γ1

24 + Γ222Γ2

24 + Γ223Γ3

24 + Γ224Γ4

24+

Γ320Γ0

34 + Γ321Γ1

34 + Γ322Γ2

34 + Γ323Γ3

34 + Γ324Γ4

34+

Γ420Γ0

44 + Γ421Γ1

44 + Γ422Γ2

44 + Γ423Γ3

44 + Γ424Γ4

44 .


ρλµ = 0.

22 JAMES SPEARS



30Γ004 + Γ0

31Γ104 + Γ0

32Γ204 + Γ0

33Γ304 + Γ0

34Γ404+

Γ130Γ0

14 + Γ131Γ1

14 + Γ132Γ2

14 + Γ133Γ3

14 + Γ134Γ4

14+

Γ230Γ0

24 + Γ231Γ1

24 + Γ232Γ2

24 + Γ233Γ3

24 + Γ234Γ4

24+

Γ330Γ0

34 + Γ331Γ1

34 + Γ332Γ2

34 + Γ333Γ3

34 + Γ334Γ4

34+

Γ430Γ0

44 + Γ431Γ1

44 + Γ432Γ2

44 + Γ433Γ3

44 + Γ434Γ4

44 .


ρλµ = 0. After our analysis of the third term we find that

it is trivially zero in every instance except for R12 = R21, R13 = R31, and R23 = R32. Wehave

ΓλνρΓρλµ =

2 cot θR | µ, ν ∈ 1, 2

cotφR | µ, ν ∈ 1, 3

cot θ cotφ | µ, ν ∈ 2, 3

0 | µ, ν /∈ 1, 2 , 1, 3 , or 2, 3

.

We now examine the fourth and final term comprising the Ricci tensor equation that iswritten

ΓλλρΓρνµ .

Following our same techniques we begin by expanding out for λ and ρ

ΓλλρΓρνµ = Γ0

00Γ0νµ + Γ0

01Γ1νµ + Γ0

02Γ2νµ + Γ0

03Γ3νµ + Γ0

04Γ4νµ+

Γ110Γ0

νµ + Γ111Γ1

νµ + Γ112Γ2

νµ + Γ113Γ3

νµ + Γ114Γ4

νµ+

Γ220Γ0

νµ + Γ221Γ1

νµ + Γ222Γ2

νµ + Γ223Γ3

νµ + Γ224Γ4

νµ+

Γ330Γ0

νµ + Γ331Γ1

νµ + Γ332Γ2

νµ + Γ333Γ3

νµ + Γ334Γ4

νµ+

Γ440Γ0

νµ + Γ441Γ1

νµ + Γ442Γ2

νµ + Γ443Γ3

νµ + Γ444Γ4

νµ .

We will continue examining the term case by case beginning with (µ, ν) = (0, 1). Onceagain once we have stated the case explicitly we will refer to pg.12 and note any non-zero


Christoffel symbols.


00Γ010 + Γ0

01Γ110 + Γ0

02Γ210 + Γ0

03Γ310 + Γ0

04Γ410+

Γ110Γ0

10 + Γ111Γ1

10 + Γ112Γ2

10 + Γ113Γ3

10 + Γ114Γ4

10+

Γ220Γ0

10 + Γ221Γ1

10 + Γ222Γ2

10 + Γ223Γ3

10 + Γ224Γ4

10+

Γ330Γ0

10 + Γ331Γ1

10 + Γ332Γ2

10 + Γ333Γ3

10 + Γ334Γ4

10+

Γ440Γ0

10 + Γ441Γ1

10 + Γ442Γ2

10 + Γ443Γ3

10 + Γ444Γ4

10 .

We find that every product in the summation has at least one Christoffel symbol equalto zero. For (µ, ν) = (0, 1) , ΓλλρΓ

ρνµ = 0.



00Γ020 + Γ0

01Γ120 + Γ0

02Γ220 + Γ0

03Γ320 + Γ0

04Γ420+

Γ110Γ0

20 + Γ111Γ1

20 + Γ112Γ2

20 + Γ113Γ3

20 + Γ114Γ4

20+

Γ220Γ0

20 + Γ221Γ1

20 + Γ222Γ2

20 + Γ223Γ3

20 + Γ224Γ4

20+

Γ330Γ0

20 + Γ331Γ1

20 + Γ332Γ2

20 + Γ333Γ3

20 + Γ334Γ4

20+

Γ440Γ0

20 + Γ441Γ1

20 + Γ442Γ2

20 + Γ443Γ3

20 + Γ444Γ4

20 .


ρνµ = 0.



00Γ030 + Γ0

01Γ130 + Γ0

02Γ230 + Γ0

03Γ330 + Γ0

04Γ430+

Γ110Γ0

30 + Γ111Γ1

30 + Γ112Γ2

30 + Γ113Γ3

30 + Γ114Γ4

30+

Γ220Γ0

30 + Γ221Γ1

30 + Γ222Γ2

30 + Γ223Γ3

30 + Γ224Γ4

30+

Γ330Γ0

30 + Γ331Γ1

30 + Γ332Γ2

30 + Γ333Γ3

30 + Γ334Γ4

30+

Γ440Γ0

30 + Γ441Γ1

30 + Γ442Γ2

30 + Γ443Γ3

30 + Γ444Γ4

30 .


ρνµ = 0.



00Γ040 + Γ0

01Γ140 + Γ0

02Γ240 + Γ0

03Γ340 + Γ0

04Γ440+

Γ110Γ0

40 + Γ111Γ1

40 + Γ112Γ2

40 + Γ113Γ3

40 + Γ114Γ4

40+

Γ220Γ0

40 + Γ221Γ1

40 + Γ222Γ2

40 + Γ223Γ3

40 + Γ224Γ4

40+

Γ330Γ0

40 + Γ331Γ1

40 + Γ332Γ2

40 + Γ333Γ3

40 + Γ334Γ4

40+

Γ440Γ0

40 + Γ441Γ1

40 + Γ442Γ2

40 + Γ443Γ3

40 + Γ444Γ4

40 .


ρνµ = 0.

24 JAMES SPEARS



00Γ001 + Γ0

01Γ101 + Γ0

02Γ201 + Γ0

03Γ301 + Γ0

04Γ401+

Γ110Γ0

01 + Γ111Γ1

01 + Γ112Γ2

01 + Γ113Γ3

01 + Γ114Γ4

01+

Γ220Γ0

01 + Γ221Γ1

01 + Γ222Γ2

01 + Γ223Γ3

01 + Γ224Γ4

01+

Γ330Γ0

01 + Γ331Γ1

01 + Γ332Γ2

01 + Γ333Γ3

01 + Γ334Γ4

01+

Γ440Γ0

01 + Γ441Γ1

01 + Γ442Γ2

01 + Γ443Γ3

01 + Γ444Γ4

01 .


ρνµ = 0.



00Γ021 + Γ0

01Γ121 + Γ0

02Γ221 + Γ0

03Γ321 + Γ0

04Γ421+

Γ110Γ0

21 + Γ111Γ1

21 + Γ112Γ2

21 + Γ113Γ3

21 + Γ114Γ4

21+

Γ220Γ0

21 + Γ221Γ1

21 + Γ222Γ2

21 + Γ223Γ3

21 + Γ224Γ4

21+

Γ330Γ0

21 + Γ331Γ1

21 + Γ332Γ2

21 + Γ333Γ3

21 + Γ334Γ4

21+

Γ440Γ0

21 + Γ441Γ1

21 + Γ442Γ2

21 + Γ443Γ3

21 + Γ444Γ4

21 .

We find that every product in the summation has at least one Christoffel symbol equalto zero except the second last term of the summation. For (µ, ν) = (1, 2) , ΓλλρΓ

ρνµ =

2 cot θR−1.For the case (µ, ν) = (1, 3).


00Γ031 + Γ0

01Γ131 + Γ0

02Γ231 + Γ0

03Γ331 + Γ0

04Γ431+

Γ110Γ0

31 + Γ111Γ1

31 + Γ112Γ2

31 + Γ113Γ3

31 + Γ114Γ4

31+

Γ220Γ0

31 + Γ221Γ1

31 + Γ222Γ2

31 + Γ223Γ3

31 + Γ224Γ4

31+

Γ330Γ0

31 + Γ331Γ1

31 + Γ332Γ2

31 + Γ333Γ3

31 + Γ334Γ4

31+

Γ440Γ0

31 + Γ441Γ1

31 + Γ442Γ2

31 + Γ443Γ3

31 + Γ444Γ4

31 .


ρνµ = cotφR−1.



00Γ041 + Γ0

01Γ141 + Γ0

02Γ241 + Γ0

03Γ341 + Γ0

04Γ441+

Γ110Γ0

41 + Γ111Γ1

41 + Γ112Γ2

41 + Γ113Γ3

41 + Γ114Γ4

41+

Γ220Γ0

41 + Γ221Γ1

41 + Γ222Γ2

41 + Γ223Γ3

41 + Γ224Γ4

41+

Γ330Γ0

41 + Γ331Γ1

41 + Γ332Γ2

41 + Γ333Γ3

41 + Γ334Γ4

41+

Γ440Γ0

41 + Γ441Γ1

41 + Γ442Γ2

41 + Γ443Γ3

41 + Γ444Γ4

41 .



ρνµ = 0.



00Γ002 + Γ0

01Γ102 + Γ0

02Γ202 + Γ0

03Γ302 + Γ0

04Γ402+

Γ110Γ0

02 + Γ111Γ1

02 + Γ112Γ2

02 + Γ113Γ3

02 + Γ114Γ4

02+

Γ220Γ0

02 + Γ221Γ1

02 + Γ222Γ2

02 + Γ223Γ3

02 + Γ224Γ4

02+

Γ330Γ0

02 + Γ331Γ1

02 + Γ332Γ2

02 + Γ333Γ3

02 + Γ334Γ4

02+

Γ440Γ0

02 + Γ441Γ1

02 + Γ442Γ2

02 + Γ443Γ3

02 + Γ444Γ4

02 .


ρνµ = 0.



00Γ012 + Γ0

01Γ112 + Γ0

02Γ212 + Γ0

03Γ312 + Γ0

04Γ412+

Γ110Γ0

12 + Γ111Γ1

12 + Γ112Γ2

12 + Γ113Γ3

12 + Γ114Γ4

12+

Γ220Γ0

12 + Γ221Γ1

12 + Γ222Γ2

12 + Γ223Γ3

12 + Γ224Γ4

12+

Γ330Γ0

12 + Γ331Γ1

12 + Γ332Γ2

12 + Γ333Γ3

12 + Γ334Γ4

12+

Γ440Γ0

12 + Γ441Γ1

12 + Γ442Γ2

12 + Γ443Γ3

12 + Γ444Γ4

12 .

We find that every product in the summation has at least one Christoffel symbol equalto zero except for the second last term of the summation. For (µ, ν) = (2, 1) , ΓλλρΓ

ρνµ =

2 cot θR−1.For the case (µ, ν) = (2, 3).


00Γ032 + Γ0

01Γ132 + Γ0

02Γ232 + Γ0

03Γ332 + Γ0

04Γ432+

Γ110Γ0

32 + Γ111Γ1

32 + Γ112Γ2

32 + Γ113Γ3

32 + Γ114Γ4

32+

Γ220Γ0

32 + Γ221Γ1

32 + Γ222Γ2

32 + Γ223Γ3

32 + Γ224Γ4

32+

Γ330Γ0

32 + Γ331Γ1

32 + Γ332Γ2

32 + Γ333Γ3

32 + Γ334Γ4

32+

Γ440Γ0

32 + Γ441Γ1

32 + Γ442Γ2

32 + Γ443Γ3

32 + Γ444Γ4

32 .


ρνµ = cot θ cotφ.

26 JAMES SPEARS



00Γ042 + Γ0

01Γ142 + Γ0

02Γ242 + Γ0

03Γ342 + Γ0

04Γ442+

Γ110Γ0

42 + Γ111Γ1

42 + Γ112Γ2

42 + Γ113Γ3

42 + Γ114Γ4

42+

Γ220Γ0

42 + Γ221Γ1

42 + Γ222Γ2

42 + Γ223Γ3

42 + Γ224Γ4

42+

Γ330Γ0

42 + Γ331Γ1

42 + Γ332Γ2

42 + Γ333Γ3

42 + Γ334Γ4

42+

Γ440Γ0

42 + Γ441Γ1

42 + Γ442Γ2

42 + Γ443Γ3

42 + Γ444Γ4

42 .


ρνµ = 0.



00Γ003 + Γ0

01Γ103 + Γ0

02Γ203 + Γ0

03Γ303 + Γ0

04Γ403+

Γ110Γ0

03 + Γ111Γ1

03 + Γ112Γ2

03 + Γ113Γ3

03 + Γ114Γ4

03+

Γ220Γ0

03 + Γ221Γ1

03 + Γ222Γ2

03 + Γ223Γ3

03 + Γ224Γ4

03+

Γ330Γ0

03 + Γ331Γ1

03 + Γ332Γ2

03 + Γ333Γ3

03 + Γ334Γ4

03+

Γ440Γ0

03 + Γ441Γ1

03 + Γ442Γ2

03 + Γ443Γ3

03 + Γ444Γ4

03 .


ρνµ = 0.



00Γ013 + Γ0

01Γ113 + Γ0

02Γ213 + Γ0

03Γ313 + Γ0

04Γ413+

Γ110Γ0

13 + Γ111Γ1

13 + Γ112Γ2

13 + Γ113Γ3

13 + Γ114Γ4

13+

Γ220Γ0

13 + Γ221Γ1

13 + Γ222Γ2

13 + Γ223Γ3

13 + Γ224Γ4

13+

Γ330Γ0

13 + Γ331Γ1

13 + Γ332Γ2

13 + Γ333Γ3

13 + Γ334Γ4

13+

Γ440Γ0

13 + Γ441Γ1

13 + Γ442Γ2

13 + Γ443Γ3

13 + Γ444Γ4

13 .


ρνµ = cotφR−1.



00Γ023 + Γ0

01Γ123 + Γ0

02Γ223 + Γ0

03Γ323 + Γ0

04Γ423+

Γ110Γ0

23 + Γ111Γ1

23 + Γ112Γ2

23 + Γ113Γ3

23 + Γ114Γ4

23+

Γ220Γ0

23 + Γ221Γ1

23 + Γ222Γ2

23 + Γ223Γ3

23 + Γ224Γ4

23+

Γ330Γ0

23 + Γ331Γ1

23 + Γ332Γ2

23 + Γ333Γ3

23 + Γ334Γ4

23+

Γ440Γ0

23 + Γ441Γ1

23 + Γ442Γ2

23 + Γ443Γ3

23 + Γ444Γ4

23 .


ρνµ = cot θ cotφ.




00Γ043 + Γ0

01Γ143 + Γ0

02Γ243 + Γ0

03Γ343 + Γ0

04Γ443+

Γ110Γ0

43 + Γ111Γ1

43 + Γ112Γ2

43 + Γ113Γ3

43 + Γ114Γ4

43+

Γ220Γ0

43 + Γ221Γ1

43 + Γ222Γ2

43 + Γ223Γ3

43 + Γ224Γ4

43+

Γ330Γ0

43 + Γ331Γ1

43 + Γ332Γ2

43 + Γ333Γ3

43 + Γ334Γ4

43+

Γ440Γ0

43 + Γ441Γ1

43 + Γ442Γ2

43 + Γ443Γ3

43 + Γ444Γ4

43 .


ρνµ = 0.



00Γ004 + Γ0

01Γ104 + Γ0

02Γ204 + Γ0

03Γ304 + Γ0

04Γ404+

Γ110Γ0

04 + Γ111Γ1

04 + Γ112Γ2

04 + Γ113Γ3

04 + Γ114Γ4

04+

Γ220Γ0

04 + Γ221Γ1

04 + Γ222Γ2

04 + Γ223Γ3

04 + Γ224Γ4

04+

Γ330Γ0

04 + Γ331Γ1

04 + Γ332Γ2

04 + Γ333Γ3

04 + Γ334Γ4

04+

Γ440Γ0

04 + Γ441Γ1

04 + Γ442Γ2

04 + Γ443Γ3

04 + Γ444Γ4

04 .


ρνµ = 0.



00Γ014 + Γ0

01Γ114 + Γ0

02Γ214 + Γ0

03Γ314 + Γ0

04Γ414+

Γ110Γ0

14 + Γ111Γ1

14 + Γ112Γ2

14 + Γ113Γ3

14 + Γ114Γ4

14+

Γ220Γ0

14 + Γ221Γ1

14 + Γ222Γ2

14 + Γ223Γ3

14 + Γ224Γ4

14+

Γ330Γ0

14 + Γ331Γ1

14 + Γ332Γ2

14 + Γ333Γ3

14 + Γ334Γ4

14+

Γ440Γ0

14 + Γ441Γ1

14 + Γ442Γ2

14 + Γ443Γ3

14 + Γ444Γ4

14 .


ρνµ = 0.



00Γ024 + Γ0

01Γ124 + Γ0

02Γ224 + Γ0

03Γ324 + Γ0

04Γ424+

Γ110Γ0

24 + Γ111Γ1

24 + Γ112Γ2

24 + Γ113Γ3

24 + Γ114Γ4

24+

Γ220Γ0

24 + Γ221Γ1

24 + Γ222Γ2

24 + Γ223Γ3

24 + Γ224Γ4

24+

Γ330Γ0

24 + Γ331Γ1

24 + Γ332Γ2

24 + Γ333Γ3

24 + Γ334Γ4

24+

Γ440Γ0

24 + Γ441Γ1

24 + Γ442Γ2

24 + Γ443Γ3

24 + Γ444Γ4

24 .


ρνµ = 0.

28 JAMES SPEARS



00Γ034 + Γ0

01Γ134 + Γ0

02Γ234 + Γ0

03Γ334 + Γ0

04Γ434+

Γ110Γ0

34 + Γ111Γ1

34 + Γ112Γ2

34 + Γ113Γ3

34 + Γ114Γ4

34+

Γ220Γ0

34 + Γ221Γ1

34 + Γ222Γ2

34 + Γ223Γ3

34 + Γ224Γ4

34+

Γ330Γ0

34 + Γ331Γ1

34 + Γ332Γ2

34 + Γ333Γ3

34 + Γ334Γ4

34+

Γ440Γ0

34 + Γ441Γ1

34 + Γ442Γ2

34 + Γ443Γ3

34 + Γ444Γ4

34 .


ρνµ = 0.

Thus we may conclude for µ 6= ν where the Ricci tensor is given in summation notion

Rµν =∂Γλλµ∂xν



ρνµ = 0

Rµν =

0− 0 + 2 cot θR − 2 cot θ

R | µ, ν ∈ 1, 2

0− 0 + cotφR − cotφ

R | µ, ν ∈ 1, 3

0− 0 + cot θ cotφ− cot θ cotφ | µ, ν ∈ 2, 3

0− 0 + 0− 0 | µ, ν /∈ 1, 2 , 1, 3 , or 2, 3

.

5.4. Defining the Ricci scalar curvature. We may now contract the Ricci tensor Rµνto obtain the Ricci scalar curvature R. The scalar curvature is defined R ≡ gµνRµν [?].For our case we note

R = g00R00 + g11R11 + g22R22 + g33R33 + g44R44

5.5. The Einstein Tensor. Before we continue to solve for a more explicit form of theline element we will construct the Einstein tensor. Recall

Gµν ≡ Rµν −1

2gµνR+ gµνΛ = 8πTµν .

For our purposes the Einstein field equations reduce to the original vacuum field equationsof General Relativity [?].

Gµν ≡ Rµν −1

2gµνR = 0 ,


5.6. Explicit Form of the Line Element. We have a general form from which we caneasily construct the field equations using the functions e2Λ(R), and e2Φ(R), although we areable to solve for a more explicit form. We will do so by referencing the Ricci tensor andnoting the components of Rµν in terms of Λ(R), Φ(R), R, θ, and φ which are not triviallyequal to zero.

R00 = e2Λ′(R)−2Φ′(R) ·(−Λ′(R)2 − Λ′′(R) + Λ′(R)Φ′(R)− 3Λ′(R)

R

)= 0 .

R11 = Λ′′(R) + Λ′(R)2 − Λ′(R)Φ′(R)− 3Φ′(R)

R= 0

R22 = e−2Φ(R) ·(2 + Λ′(R) ·R− Φ′(R) ·R

)− 2 = 0

R33 = sin2 θ ·[e−2Φ(R) ·

(2 + Λ′(R) ·R− Φ′(R) ·R

)− 2]

= sin2 θ ·R22 = 0

R44 = sin2 θ sin2 φ ·[e−2Φ(R) ·

(2 + Λ′(R) ·R− Φ′(R) ·R

)− 2]

= sin2 φ ·R33 = 0

Adding the first and second equations,

R00 +R11 = −(

3Λ′(R)

R+

3Φ′(R)

R

)= 0 ,

provides the relationΛ′(R) + Φ′(R) = 0 ,

⇒ Λ(R) + Φ(R) = K | K is a constant .

We now are in a position to note some restrictions on the functions Λ(r), and Φ(r), andthus our solution. We know that at an infinite distance from the object of mass M , thespace-time must be that of a nearly Minkowski space. Mathematically,

limR−→∞

e2Λ(R) = 1 , and limR−→∞

e2Φ(R) = 1 ,

implieslim

R−→∞Λ(R) = 0, and lim

R−→∞Φ(R) = 0 .

Now that we have deduced Λ′(R) + Φ′(R) = 0, we may rewrite R22, as

R22 = e−2Φ(R) ·(2− 2Φ′(R) ·R

)− 2 = 0 ,

⇒ 1− Φ′(R) ·R = e2Φ(R) .

For which we can check has the solution

Φ(r) = −1

2ln

(1 +

C

R2

),

and so may write

e−2Φ(R) = e2Λ(R) =

(1 +

C

R2

)Setting the constant of integration C will be the final step in determining the explicit

form of the invariant interval ds2 of this five-dimensional static spherically symmetricsolution. We have already supposed to known the coordinate location of the singularity, or

30 JAMES SPEARS

as it should be more clearly stated the location the photon having trajectory tangential tothe radial null geodesic in (u, r) or (v, r) space cross the event horizon. Although the signof the constant of integration resulting from the solution of the Einstein field equationswill tell us the sign. The line element ds2 have the form

ds2 = −e2Λ(R)dT 2 + e2Φ(R)dR2 +R2dθ2 +R2 sin2 θdφ2 +R2 sin2 θ sin2 φdψ2 .

For the proper time interval dτ2 along the radial line dR = dθ = dφ = dψ = 0 metricsimplifies to

dτ2 = −ds2 = e2Λ(R)dT 2 ,(dT

dτ

)2

= e−2Λ(R) ,

⇒ dT

dτ= e−Λ(R) .

Recall the geodesic equation

d2xρ

dτ2= −Γρµν

dxµ

dτ

dxν

dτ,

which defines the path of matter in curved space. By supposing the release of a test particlefrom rest we are given dxµ = dxν = 0 for µ = ν 6= 0. Thus the only term in our summationis

d2x1

dτ2+ Γ1

00

(dx0

dτ

)2

=d2R

dτ2+ Γ1

00

(dT

dτ

)2

= 0 .

⇒ d2R

dτ2= −Λ′(R) · e−2Φ(R) = −Λ′(R) · e2Λ(R) = −1

2

d

dRe2Λ(R) ,

and so we haved2R

dτ2= −1

2

d

dR

(1 +

C

R2

)=

(C

R3

).

The second derivative of the radial distance of a particle released in a gravitational fieldin the classical Newtonian theory is given by Newton’s second law of motion which states

F = ma ,

where

F =−GMm

r2⇒ d2r

dτ2=−GMr2

and so supposing the direction of acceleration is independent of dimension we are able touse equation line (24) pg. 4 and state

C = −2M

√4 ln2

∣∣∣∣rγ − 2M

2M

∣∣∣∣ .


We recall e2Λ(R) = e−2Φ(R), and express the line element of the introduced geometry,which describes the space-time exterior (at radial distances R > R0 where R0 is the four-dimensional radius of the body) of a static, spherically symmetric four-dimensional objectof mass M .

(28) ds2 = −c2

1−2M

√4 ln2

∣∣∣ rγ−2M2M

∣∣∣R2

dT 2 +dR21−

2M

√4 ln2

∣∣∣ rγ−2M

2M

∣∣∣R2

+R2dθ2 +R2 sin2 θdφ2 +R2 sin2 θ sin2 φdψ2 .

6. Parameterizing u, and v as scalars of motion

Returning to the four-dimensional space-time and defining a coordinate transformationof a scaled line element, one may parameterize the coordinates u, and v as the scalars ofmotion in (u, r) and (v, r) space. Let

(29) u := n (t− r∗) ,

and

(30) v := n (t+ r∗) .

Substituting the coordinate transformations into the line elements

(31) ds2 = −(

1− rsr

)du2 − 2ndudr + n2r2dΩ2 ,

and

(32) ds2 = −(

1− rsr

)dv2 + 2ndvdr + n2r2dΩ2 .

respectively produces

(33) ds2 = −n2(

1− rsr

)dt2 +

n2dr2(1− rs

r

) + n2r2dΩ2 ,

the product of the Schwarzschild line element and a constant n2. Radial null geodesicsds2 = dθ = dφ = 0 of (22) and (23) are

(34) u = −2nr∗ + const.,

(35) u = const.

and

(36) v = 2nr∗ + const. ,

(37) v = const. .

32 JAMES SPEARS

respectively. Now, considering the space-like interval dσ where dt = dθ = dφ = 0 in thespace-time defined by line element (33), the definition for proper distance is given

(38) σ = n

∫P

dr√1− rs

r

,

for some path P . In order to parameterize u and v as scalars of motion it must be shownthat both u and v are proportional to σ, given by equation line (38), to within a constant.Suppose this is the case.

The condition of simultaneous reception implies an intersection of null geodesics at aminimum of one point, the reception point (uρ, rρ) or (vρ, rρ). This is achieved by settingthe constants of integration. Suppose constants abbreviated “const.” in lines (15) and (29)or (18) and (31) are initially set to zero. The difference between two in-going null curves(15) and (29), and out-going null curves (18) and (31) respectively at the reception pointrρ is given by the expression

(39) ∆uρ = 2r∗|rρ (n− 1) ,

and

(40) ∆vρ = 2r∗|rρ (1− n) .

And so, the in-going null curves (20) of the out-going transformation (22) and the out-going null curves (21) of the in-going transformation (23) which intersect the null curves(15) and (18) (having constants abbreviated “const.” equal to zero in both cases) at thereception point rρ are given

(41) u = −2nr∗ − 2r∗|rρ (n− 1) ,

and

(42) v = 2nr∗ + 2r∗|rρ (1− n)

respectively. We can now solve for the interval between emissions at rγ . This difference isgiven again for the in-going null curves of the out-going transformation and the out-goingnull curvres of the in-going transformation respectively

(43) ∆uγ = 2r∗|rγ (n− 1)− 2r∗|rρ (n− 1) ,

and

(44) ∆vγ = 2r∗|rγ (1− n)− 2r∗|rρ (1− n) .

Combing relations (43) and (44) with the definition of proper distance (33); for each ofthe transformations u and v, the difference between emission and reception ∆urγ→rρ or∆vrγ→rρ increase and decrease 1:1 with proper distance of null geodesics (15) and (25) or(18) and (27) over the path P for u, v respectively. Thus u and v are indeed scalars ofmotion for the scaled null geodesics (34), (35), and (36), (37) respectively.

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Behaviour of electromagnetic radiation in gravitational fields