This paper is my work on Einstein's Relativity theory for speed greater than light.The paper mainly focus on to generalize the Lorentz factor , equation of negative energy and negative mass thus proving the existence of wormholes and the derivation of new space metric.

1. International Refereed Journal of Engineering & Technology (IRJET) – Volume I Issue I
27
Advanced Theory of Relativity
Mihir Kumar Jha , EEE Department ,Global academy of technology , Bangalore , India
ABSTRACT- The motive of this paper is to derive the
generalized equation for relativistic mass and rest mass
equation, equation for negative mass , relativistic kinetic
energy followed by the derivation of new metric of space -
time for speed greater than light.
INTRODUCTION
At the dawn of 20th
century, physicist believed that they
possessed enough knowledge to deduce the mystery of the
universe. During this golden era, a patent clerk in Switzerland
named Albert Einstein proposed two theories of relativity
including the famous equation E=MC². He used the relativistic
mass and moving mass equation given as follows
m = m₀/ (√ (1-v²/c²))--------1
Above equation was used to derive the energy mass equation.
If a close look is taken on the equation, there are two major
drawbacks described as below:
If speed is equal to light speed C, the equation is undefined.
And, If
speed is greater than light speed C, the mass becomes complex.
In order to overcome these drawbacks, the equation has to be
generalized which is valid for greater then equal to light speed
and thus determining the equation for negative energy and
negative mass.
DELAY FACTOR (Ф)
Consider a particle of mass m, initially moving with light speed
c. Let the time taken by the ball to cover a distance L₀ is Δt.
Now, let us consider the ball speed increases from c to (c+v)
then the time required to cover the same distance is Δt’.
Mathematically,
L₀= c*Δt -------------- 2 at v=c
And, L₀ =(c+v)* Δt’ ---------3 at v= (v+c)
From the above two equations, we can say
c*Δt = (c+v)* Δt’ --------------------4
Δt’ = Δt / (1+v/c)
Δt’ = Δt*Ф -----------------------5
Where, Ф is delay factor given by,
Ф = 1/(1+v/c)
LENGTH CONTRACTION
Consider the above explanation for delay factor, in which the
ball travels the distance L₀ at time Δt with speed c and covers
distance L at time Δt’ with the same speed. Then from equation
4 we have
c*Δt = (c+v)*Δt’ ----------------------6
c*Δt = c(1+v/c)*Δt’
c*Δt= (c*Δt’)*(1+v/c)
L₀ = L(1+v/c)
Hence, L = L₀/(1+v/c) ---------------7
Or, L= L₀*Ф
RELATEVISTIC MASS AND REST MASS
Consider a particle having a rest mass m₀ and moving mass m,
which is moving initially with a velocity v₁ and attains a
velocity v after time t. The initial and final momentum are
given by the following equations
P(i) = m₁v₁ for (v<c)------------------------8
P(f) = m₂v for(v>= c)------------------------ 9
The total momentum is the sum of initial momentum and final
momentum given by
P(t) = P(i) + P(f) -------------- 10
The total kinetic energy of the body is given by the equation
E= ∫ 𝐹. 𝑑𝑠 ----------------------------11
= ∫(dP/dt). ds
= ∫(ds/dt). dP
= ∫ 𝑣. 𝑑𝑃
= ∫ 𝑣. [𝑚₂𝑑𝑣 + 𝑣𝑑𝑚₂]
= ∫ 𝑚₂𝑣. 𝑑𝑣 + ∫ 𝑣². 𝑑𝑚₂
= ∫ 𝑚₂𝑣. 𝑑𝑣 + ∫(𝑣². 𝑑𝑚₂/dv ) .dv
Differentiating the above equation with respect to v, we get
P(t) = m₂v + v²dm₂/dv
P(t) = P(f) + v²dm₂/dv
P(t) – P(f) = v²dm₂/dv
P(i) = v²dm₂/dv
P(i) ∫ (
1
v2) dv
𝑐
0
= ∫ 𝑑𝑚
𝑚₂
𝑚₁
₂
-P(i)[1/c – 1/0] = m₁ - m₂ ------------A
Since [1/v] for v=0 is undefined value we need to calculate the
value through limit
I₁ = lim
𝑥→0
1/𝑥
= lim
𝑥→0
(𝑠𝑖𝑛²x + cos²x)/𝑥
= lim
𝑥→0
(𝑠𝑖𝑛2
x)/x + lim
𝑥→0
(cos²x)/𝑥 =0 (B)
Now, let I = lim
𝑥→0
(cos²x)/𝑥
Multiplying and dividing with x we have
I = lim
𝑥→0
(xcos²x)/𝑥²
Applying L- Hospital rule we get
I= lim
𝑥→0
(−2x(cosx ∗ sinx) + cos²x)/2𝑥
= lim(
𝑥→0
− cosx ∗ sinx) +1/2 lim
𝑥→0
cos²x/𝑥
= 0 + ( ½) I
I – ½ I = 0
Or, I = 0 ----------------C
Substituting the value of lim
𝑥→0
(cos²x)/𝑥 in equation B we have
I₁ = lim
𝑥→0
𝑠𝑖𝑛2x
x
+ 0
Again applying L- Hospital rule we get
I₁ = lim
𝑥→0
2𝑠𝑖𝑛x∗cosx
1
= 0
Hence the value of lim
𝑣→0
1/𝑣 is zero.

2. International Refereed Journal of Engineering & Technology (IRJET) – Volume I Issue I
28
Coming back to our equation A we have
-P(i)*[1/c – 1/0] = m₁ - m₂
- P(i)*1/c = m₁ - m₂ ----------D
Now Substituting for P(i) in above equation D we have
-m₁v₁/c = m₁ - m₂
Or, m₂ = m₁*(1-v₁/c) --------------13
In general,
m = m₀*(1+v/c) ------------------14
m = m₀/Ф [-v represents velocity in opposite direction]
RELATEVISTIC FORCE
Consider a mass kept on a surface of the earth. When the mass
is stationary (i:e v = 0) the only acceleration acting on the
particle is gravity (g). When particle starts gaining velocity the
force on the particle is given by
F = dp/dt ----------------------------15
= d(m.v)/dt
= d(m₀.v)*(1+v/c))/dt
= (m₀*a₀)*(1+v/c)²
= (m₀*(1+v/c))*(a₀*(1+v/c))
=(m₀*(1+v/c))*(g*(1+v/c))------16
[at v= 0 ; a₀ = g ]
F = F’/ Ф²
NEGATIVE ENERGY AND NEGATIVE MASS
Negative mass is a concept of matter whose mass is of opposite
sign to the mass of normal matter and the energy produced by
these matter is called negative energy.
In order to find the equation for negative mass let us consider
the total energy of a particle, given by the equation
E= ∫ 𝐹. 𝑑𝑠
𝑐
0
= ∫(𝑑𝑝/𝑑𝑡). 𝑑𝑠 = ∫ 𝑣. 𝑑𝑝
= ∫ 𝑑𝑝𝑣 – ∫ 𝑝. 𝑑𝑣
= mc² - ∫ mvdv
𝑐
0
= mc² - m₀ ∫ v ∗ (1 +
v
c
)dv
𝑐
0
= mc² - m₀[
v2
2
+
v³
3c
]0
𝑐
= mc² - m [c²/2 +c²/3 ]
= mc² - 5m₀c²/6
= mc² - 5mc²/6(1+v/c)
= m [c²- 5c²/6(1+v/c)]
= mc² [1-5c/6(c-v)]
Now we consider the value of v as v₁ such that [5c/6(c-v)] is
equal to [5c/6(c-v₁)].Thus
E = mc²[1-c/(c-v₁)]
= mc² [(c-v₁+c)/(c-v)]
= mc²(-v₁/(c-v)) = -mv₁c²/(c-v) ----------------------17
Hence ,
E = - mvc/(1-v/c)
E = 𝑀−
𝑣𝑐 ∗ Ф
Were M¯ is negative mass and E is the negative energy’
The above equation clearly states that the negative energy and
negative mass is generated at velocities equal to or greater than
light speed.
RELATEVISTIC ENERGY
From equation 17 we have
E = mc²[1-c/(c-v)]
= m₀c²(1-v/c)[1-1/(1-v/c)]
= m₀c²[(1-v/c)-1 ]
E = m₀c²[(1/ Ф)-1] -------------------------------19
SPACE-TIME INTERVAL
Space time is defined as any mathematical model that
combines space and time into a single interwoven continuum.
The mathematical form is a series of differential equation for
speed less than light is given by
Δs² = - (cΔt)² + Δx² + Δy² + Δz²--------20
And, the mathematical form of space time for speed greater or
equal to light speed is given by
Δs = cΔt + Δx + Δy + Δz ----------------21
In order to derive the above differential equation, we need to
find the equations for energy dependence on distance (s) and
time (t).
The energy of a particle having a velocity v is given by
E = - ∫ F. ds
𝑣
0
------------------------22
Substituting for force F in above equation
E = ∫ [m₀g/(1 + v/c)²] ds
𝑣
0
= m₀g ∫ [(1 + v/c)²] ds
𝑣
0
-------------23
= m₀g∫ [(1 + v/c)²]v dt
𝑡
0
--------------24
ENERGY DEPENDENCE ON DISTANCE
Since, S = S₀/(1+v/c)
Or, S₀/S = (1+v/c)
Substituting the value of Ф in equation 23
E = m₀g ∫ [S₀²/S²] ds
𝑆
0
= m₀gS₀²∫ (1/S² )ds
𝑆
0
= -[m₀g*(S₀/S)]
= -[m₀g*(S₀/S²)]*S
Substituting for S₀/S in above equation we get
E = -m*a*S -----------------------25
ENERGY DEPENDENCE ON TIME
Since t = t₀/(1+v/c)
Or, t₀/t = (1+v/c) and
v = c*(t₀/t -1)
Substituting t’/t and v in equation 24 we get
E = m₀g∫ [c ∗ (
t₀
t
– 1) ∗
t₀2
t²
] dt
𝑡
0
------26
=-[m₀gc(
t0
t
)²(-t/2) + (t₀/t²)t] ---27
= m₀a*c(t/2)
Let t/2 = T
Then the above equation can be written as
E = m₀a*c*T---------------28
Comparing both the energy equations we get
S = cT---------------------29

3. International Refereed Journal of Engineering & Technology (IRJET) – Volume I Issue I
29
The above equation says that the mechanics formulas hold
good for quantum level also. Since in mechanics the speed
distance and time is related as s = v*t
FOR VELOCITY LESS THAN LIGHT SPEED
From equation 29 if the speed decreases by c to (c-v) then the
Above equation is written as
S= (c-v)*T
Differentiating the above equation we get
dS = c*dT - d(V*T) ----------------------30
dS=c*dT-dS’ [ S = V*T ]
dS = c*dT - √(dx² + dy² + dz²) ---------31
dS – c*dT = -√(dx² + dy² + dz²)
squaring both sides we get
dS² + (c*dT)² - 2*dS*c*dT = (dx² + dy² + dz²)
------32
dS²+ (c*dT)² -dS*c*dt = (dx² + dy² + dz²)
-------------33 [since T=t/2]
since,
-2(dS*c*dT) = -dS²+dS√(dx² + dy² +dz²)**
----------------34 [from eq 31]
-2(dS*c*dt)/2 =-dS²- dS√(dx²+ dy²+ dz²)
-(dS*c*dt) = - dS² - dS√(dx² + dy² + dz²) ------------34
Substituting equation 31 in equation 29 we get
dS² +(c*dT)² -dS² - dS*√(dx² + dy² + dz²)
= (dx² + dy² + dz²) ---------------35
-dS*√(dx² + dy² + dz²) = - (c*dT)² +(dx² + dy² + dz²)
-dS*dS’ = - (c*dT)² +(dx² + dy² + dz²)
[ dS = √(dx² + dy² + dz²)]
dS₁² = - (c*dT)² +(dx² + dy² + dz²) ------36
Therefore we can write the above equation in general,
ΔS² = - (c*Δt)² + Δx² + Δy² + Δz² ------37
**[dS = c*dT + √(dx² + dy² + dz²)
Multiplying both side by dS
dS² = c*dT*dS + dS*√(dx² + dy² + dz²) -c*dT*dS = - ds² +
dS*√(dx² + dy² + dz²)]
FOR VELOCITY GREATER THAN LIGHT SPEED
From equation 29 we have
S = cT*(1+ V/c)
Differentiating the above equation with respect to t we get
dS/dt = c* dT/dt + d(VT)
dS = c*dT + dS
dS = c*dT + √( dx² + dy² + dz²)] ----- 38
Now consider two reference frame R(a) and R(b) where R(a) is
in three dimensional reference frame and R(b) is in four
dimensional reference frame given that both reference frame
coincide at t=0. Then the co-ordinate of three dimension given
by √( dx² + dy² + dz²)] can be written as dx’ + dy’ + dz’ in four
dimension.
For example if the co-ordinate in 3-D is s(x,y,z)=(3i,4j,0k) then
s = √ (3² + 4² + 0²) = 5 then in 4-D co-ordinate it can be
written as simply the algebraic sum of 4-D co-ordinate i.e. s =
(1i ,4y,0z,0t) where 4+1=5 or (1i,2j,1k,1t) where 1+2+1+1=5
and any other value which satisfies the above condition. Thus
using this concept in equation 35, it can be written in 4-D as
the simple algebraic sum of the 4-D co-ordinate whose sum is
equal to the value of 3-D co-ordinates
Thus,
dS = c*dT + dx’ + dy’ + dz’ -------------39
in general equation 36 can be written as
dS = c*dt + dx + dy + dz -------------40
The interesting thing about the above equation is the
Pythagoras theorem ceases as velocity increases greater than
the light speed and the particle can be seen in multiple co-
ordinates at the same time in four dimension in it’s line of
motion. The above equation is called mihir space time
equation.
MIHIR METRIC
In this section, we will find a solution to Einstein’s Field
equations that describes a
gravitational field exterior to an isolated sphere of mass M
assumed to be at rest.
Let us place this sphere at the origin of our coordinate system.
For simplicity, we will use spherical coordinates ρ, φ, θ:
x = ρ sin φ cos θ
y = ρ sin φ sin θ
z = ρ cos φ
Starting from the flat Mihir space time of advanced relativity
and changing to
spherical coordinates, we have the invariant interval from
equation 40
dS = c*dt + dx + dy + dz
= dt + dρ + ρ dφ + ρ sin φdθ ------41
The mihir space time tensor mμν is given by
Mμν= (
g₀₀ g₀₁ g₀₂ g₀₃
g₁₀ g₁₁ g₁₂ g₁₃
g₂₀ g₂₁ g₂₂ g₂₃
g₃₀ g₃₁ g₃₂ g₃₃
)
= (
𝑔𝑡𝑡 𝑔𝑡𝑟 𝑔𝑡𝜑 𝑔𝑡𝜃
𝑔𝑟𝑡 𝑔𝑟𝑟 𝑔𝑟𝜑 𝑔𝑟𝜃
𝑔𝜑𝑡 𝑔𝜑𝑟 𝑔𝜑𝜑 𝑔𝜑𝜃
𝑔𝜃𝑡 𝑔𝜃𝑟 𝑔𝜃𝜑 𝑔𝜃𝜃
)
In addition to having a spherically symmetric solution to
Einstein’s field equations
we also want it to be static. That is, the gravitational field is
unchanging with time
and independent of φ and θ. If it were not independent of φ and
θ then we would be
able to define a preferred direction in the space, but since this
is not so
grθ = grφ = gθr = gφr = gθφ = gφθ = 0
Similarly, to prevent a preferred direction in space time, we can
further restrict the
Coefficients
gtθ = gtφ = gθt = gφt = 0
Also, with a static and unchanging gravitational field, all of the
metric coefficients must be independent of t and the metric
should remain unchanged if we were to reverse time, that is,
apply the transformation t → −t. With that constraint, only dt
leaves ds unchanged and thus implies 𝑔𝑡𝑟 =𝑔 𝑟𝑡 = 0. At this
point we have reduced mμν to the following

4. International Refereed Journal of Engineering & Technology (IRJET) – Volume I Issue I
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mμν = (
𝑔𝑡𝑡 0 0 0
0 𝑔𝑟𝑟 0 0
0 0 𝑔𝜑𝜑 0
0 0 0 𝑔𝜃𝜃
)
We may then write the general form of the static spherically
symmetric space time as
𝑔𝑡𝑡 dt + 𝑔𝑟𝑟 dρ + 𝑔𝜑𝜑 dφ + 𝑔𝜃𝜃 sin φdθ
Now let us solve for the gμν coefficients. We start with a
generalization of the invariant interval in flat space time from
equation
.ds = U(ρ) dt + V (ρ) dρ +W(ρ) (ρ dφ + ρ sin φdθ)
where U, V,W are functions of ρ only. Since we can redefine
our radial coordinate, so let r = ρW(ρ), and then we can define
some A(r) and B(r) so that the above becomes
ds = A(r) dt + B(r) dr + r dφ + r sin φ---42
We next define functions m = m(r) and n = n(r) so that
A(r) = 𝑒 𝑚(𝑟)
= 𝑒2𝑚
and B(r) = 𝑒 𝑛(𝑟)
= 𝑒2𝑛
We set A(r) and B(r) equal to exponentials because we know
that they must be strictly positive, and with a little foresight, it
will make calculations easier later on. Then substituting into
equation (13) we have
ds = 𝑒2𝑚
dt + 𝑒2𝑛
dr + r dφ + r sin φdθ -----43
Thus mihir space time can be written as
Mμν = (
𝑒2𝑚
0 0 0
0 𝑒2𝑛
0 0
0 0 𝑟 0
0 0 0 𝑟 sin 𝜑
) -----44
Since Mμν is a diagonal matrix, g = det(𝑔𝑖𝑗) = 𝑒2𝑚+2𝑛
r² sinφ.In
order to solve for a static, spherically symmetric solution we
must find m(r) and n(r). However, to solve for them we must
use the Ricci Tensor given by
which relies on the Christoffel symbols given by
EVALUATION OF CHRISTOFFEL SYMBOLS
From equation 44 , we see that 𝑔 𝜇𝜈 = 0 for μ ≠ ν, and so 𝑔 𝜇𝜇 =
1/𝑔 𝜇𝜇 and 𝑔 𝜇𝜈 = 0
if μ ≠ ν. Thus, the coefficient 𝑔 𝛽𝜆 is 0 unless β = λ and
substituting this into the above we have
Notice that Γλ
𝜇𝜈 = Γλ
𝜈𝜇 so we have three cases: λ = ν, μ = ν
≠ λ, and μ, ν, λ distinct.
Recall that 𝑔 𝜇𝜈 = 0 for μ ≠ ν as we will use that to simplify
several terms.
Case 1. For λ = ν
Case 2. For μ = ν ≠ λ:
Case 3. For μ, ν, λ distinct:
Using the values for 𝑔 𝜇𝜈 from equation (44) we can calculate
the nonzero Christoffel symbols (in terms of m, n, r, and φ )
where ‘ = d/dr
Γ°₁₀ = Γ°₀₁ = m’ Γ¹₀₀ = -m’𝑒2𝑚−2𝑛
Γ¹₁₁ = n’
Γ¹₂₂ = -(1/2) 𝑒−2𝑛
Γ²₁₂ = Γ²₂₁ = 1/2r
Γ¹₃₃ = (1/2) 𝑒−2𝑛
sin 𝜑
Γ³₁₃ = Γ³₃₁ = 1/2r
Γ³₂₃ = Γ³₃₂ = (½) cot 𝜑
Γ²₃₃ = -(cos 𝜑)/2
And We know that ,
½ log 𝑔 = ½ log[𝑒2𝑚+2𝑛
𝑟² sin𝜑] = m + n + log(r) + ½ log
[sin𝜑] ----------------45
RICCI TENSOR COMPONENT
However, 𝑅 𝜇𝜈 for some values of μ and ν will be in terms of m,
n, r, and φ which we
must set equal to zero. Let us calculate those terms now and
show that the other
terms reduce to zero afterwards. Recall x° = t, x¹ = r, x² = φ, x³
= θ, equation (22),
and the nonzero values of the Christoffel symbols presented
previously
= 𝑒2𝑚−2𝑛
[ 2m’² + m’’ - m’n’ + m’/r ]
= [ m’’ + m’² - m’n’ –n’/r – 1/2r² ]
= [ -
½- ½ cot² 𝜑 + 3/2 (n’𝑒−2𝑛
) + ½(𝑒−2𝑛
) ]
And,

5. International Refereed Journal of Engineering & Technology (IRJET) – Volume I Issue I
31
= sin 𝜑 [1/2𝑛′(𝑒−2𝑛
) + ½ -½ (m’𝑒−2𝑛
)
+½r(𝑒−2𝑛
) ]
And the value of ricci component at 𝜇, 𝜈 = 1,2 is (0,0,-(cot
φ)/4r , (cot φ)/4r)
Thus, 𝑅 𝜇𝜈= {
0 + 0 +
cot 𝜑
4𝑟
−
cot 𝜑
4𝑟
= 0 , 𝑓𝑜𝑟 𝜇, 𝜈 = {1,2}
0 + 0 + 0 + 0 = 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝜇 𝜈
----------46
[For detailed explanation of above value refer the derivation of
Schwarzschild Metric]
Now, we have
R₀₀ = 𝑒2𝑚−2𝑛
[ 2m’² + m’’ - m’n’ + m’/r ]
---------47
R₁₁ = [ m’’ + m’² - m’n’ –n’/r – 1/2r² ]
----------48
R₂₂ = [ -½- ½ cot² 𝜑 + 3/2 (n’𝑒−2𝑛
) + ½(𝑒−2𝑛
) ] --------------
-------------49
R₃₃ = sin 𝜑 [1/2𝑛′(𝑒−2𝑛
) + ½ -½ (m’𝑒−2𝑛
) + ½r(𝑒−2𝑛
) ] ------
-----------------50
SOLVING FOR THE COEFFICIENTS
Since we are finding the solution for an isolated sphere of
mass, M, the Field Equations from equation (46) imply that
outside this mass, all components of the Ricci tensor are zero.
We can then set the above equations equal to zero and solve the
system for m and n so we can find an exact metric that
describes the gravitational field of a static spherically
symmetric space-time. Hence, we need
2m’² + m’’ - m’n’ + m’/r = 0 ------------51
m’’ + m’² - m’n’ –n’/r – 1/2r² = 0 ------52
-½- ½ cot² 𝜑 + 3/2 (n’𝑒−2𝑛
) + ½(𝑒−2𝑛
) = 0
-----------------------53 1/2𝑛′(𝑒−2𝑛
) + ½ -½
(m’𝑒−2𝑛
) + ½r(𝑒−2𝑛
) = 0
-------------------54
Subtracting equation 51 from equation 52 we get
m’² + m’/r + n’/r + 1/2r² = 0 ----------55
Now partially differentiating the above equation with respect to
m’ we get
2m’ + 1/r = 0
Or, m’ = -1/(2r) ----------------------56
Substituting m’ in equation we get
n’ = 1/(2r) ---------------------57
Therefore from equation 56 and 57 we have
m’ + n’ = 0 ----------------58
or, m + n = constant
and m = -n -----------------59
Now from equation 54 we have
½𝑛′(𝑒−2𝑛
) + ½ -½ (m’𝑒−2𝑛
) + ½r(𝑒−2𝑛
) = 0
Or, (2𝑒−2𝑛
)/r = -1
= 𝑒2𝑚
= -r /2 -----60 [ since , m = -n ]
We will need to solve for r in terms of M to find the exact
metric. With this in mind, suppose we release a ‘test’ particle
with so little mass that it does not disturb the space-time
metric. Also, suppose we release it from rest so that initially
𝑑𝑥 𝜇
= 0 for μ = 1, 2, 3
Now from equation we have
ds = 𝑒2𝑚
dt + 𝑒2𝑛
dr + r dφ + r sin φdθ
= 𝑒2𝑚
dt + 0 + 0 +0
However, for time like intervals, we can relate proper time τ
between two events as
dτ = ds/c
The second equality follows from our use of geometrized units;
we set c = 1 in the
formulation of Mihir space time. We then have
dτ = ds =𝑒2𝑚
𝑑𝑡
dt/dτ = 1/𝑒2𝑚
--------------------61
and, (dt/dτ) ² = (1/𝑒2𝑚
)² ----------62
From the geodesic equation we have
Notice that it relies on 𝑑𝑥 𝜇
/ dτ which motivates our previous
change from ds to dτ.
Recall that we are attempting to solve for r and that we want
our metric to reduce
to Newtonian gravitation at the weak field limit. Towards that
end, and having the
benefit of a little foresight, let us find the geodesic equation for
𝑥 𝜆
= 𝑥1
= r at the instant that we release our ‘test’ particle.
Remember that we are releasing this particle from rest and so
𝑑𝑥 𝜆
/𝑑𝜏= 0 for λ ≠ 0 and so we are left with only the following
in our summation
Now we see more of the motivation for switching to proper
time τ as we can easily substitute in from equation (61) and
move the right hand term to the other side to have
d²r/ dτ² = -m’(𝑒2𝑚−2𝑛
) 𝑒4𝑚
= - m’ 𝑒(2𝑚+2𝑚)−4𝑚
[as m= -n]
= -m’
= - ( -1/2r) = 1/2r ---------63
[since m’ = -1/2r]
Conveniently, we know that this must reduce to the predictions
of Newtonian gravity for the limit of a weak gravitational field
and so
d²r/ dτ² = −GM/r² --------------------64
Thus from equation 63 and 64 we have
1/2r = −GM/r²
or, r = -2MG = -2M --------------------65
Substituting for r in equation 60 we get
𝑒2𝑚
= -r/2 = 2MG/2 = M
And, 𝑒2𝑚
= 𝑒−2𝑛
= -2/r = 2/2MG = 1/M
[since G=1 for geometrized units ]

6. International Refereed Journal of Engineering & Technology (IRJET) – Volume I Issue I
32
Using this to substitute back into equation (60) and recalling
that m = −n we can
find an exact solution for equation (43)
ds = (-r/2) dt + (-2/r) dr +rdφ + r sin φ dθ
------------------66
And since from equation 65 we have r= - 2M substituting in
equation 66 we get
ds = Mdt + (1/M)d(-2M) + (-2M) dφ +
(- 2M) sin φ dθ -----------------67
The above equation is called MIHIR metric.