Tensor Analysis & Geometry
Spherical Coordinates
x1 = r sin¸ cosĆ x2 = r sin¸ sinĆ
x3 = r cos¸ x4 = ct
, , ,
2 2 2 2
3-dimensional line element: ds2 = dr + r (d¸ + sin2 ¸ dĆ )
Christoffel Symbols
1 " grm " grn " gmn
ëÅ‚ öÅ‚
Christoffel Symbol of the first kind: “ = [mn,r]= + - ÷Å‚
.
ìÅ‚
rmn
2 " xn " xm " xr Å‚Å‚
íÅ‚
z z rz
Å„Å‚ üÅ‚
Christoffel Symbol of the second kind: “ = = g “ .
òÅ‚
mn mnr
mnżł
þÅ‚
Derivation of the Riemann Curvature Tensor
In general, a second order differentiation on a covariant vector is independent of the order in which it is carried
out, i.e.:
2 2
" Vi " Vi
= .
j j
" x " xk " xk" x
However, the presence of Christoffel symbols can have an effect on this statement. We investigate this by first
finding the general second derivatives for both permutations of the differentiating parameters:
r r
(Vi; j ) = Vi, jk - “ Vr, j - “ Vi,r .
ik jk
;
k
s
But Vi; j = Vi, j - “ Vs ,
ij
s
" “
ij
s r s r s
4" (Vi: j ) = Vi, jk - Vs - “ Vs,k - “ [Vr, j - “ Vs]- “ [Vi,r - “ Vs]
ik rj jk ir
:
k
" xk ij
s
" “
ij
s r r s r r s
Ô! (Vi: j ) = Vi, jk - Vs - “ Vs,k - “ Vr, j + “ “ Vs - “ Vi,r + “ “ Vs
ik ik rj jk jk ir
:
k
" xk ij
Now we interchange j and k (which is the other possible way of determining this second derivative):
s
" “
ik s r r s r r s
(Vi:k ) = Vi,kj - Vs - “ Vs, j - “ Vr,k + “ “ Vs - “ Vi,r + “ “ Vs .
ik ij ij rk kj kj ir
: j j
" x
We now find the difference between these two. On the RHS, the first, third, fourth, sixth, and seventh terms
cancel out, thus giving the result:
s s
s s
îÅ‚ Å‚Å‚
" “ " “
" “ " “
ij r s r s ij r s r s
ik ik
Vi: jk - Vi:kj = - Vs + “ “ Vs + Vs - “ “ Vs = - + “ “ - “ “ .
ïÅ‚
ij rk
j j
" xk ik rj " x " x " xk ik rj ij rk śłVs
ïÅ‚ śł
ðÅ‚ ûÅ‚
We define the Riemann (or Riemann-Christoffel) Curvature tensor by:
s
s
" “
" “
ij
s ik r s r s
Rijk = - + “ “ - “ “ .
j
" x " xk ik rj ij rk
s
The difference between the covariant derivatives can thus be written as Vi: jk - Vi:kj = RijkVs . The Riemann
tensor used in this equation is called the Riemann curvature tensor of the second kind. The curvature tensor of
the first kind is defined as:
Rijkl = gir Rr .
jkl
Symmetry Properties:
First skew symmetry Rijkl = - Rjikl
Second skew symmetry Rijkl = - Rijlk
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Block symmetry Rijkl = Rklij
Bianchi s identity Rijkl + Riklj + Riljk = 0
The Ricci Tensor
The Ricci tensor of the first kind is simply a contraction of the Riemann tensor:
k
Rij = Rijk .
The last index can be raised to yield the Ricci tensor of the second kind:
ik
Rij = g Rki .
If this tensor is finally contracted by letting I = j, we get the Ricci curvature scalar. If it is zero, the space is flat.
From the first of the two equations above the Ricci tensor of the first kind can be calculated directly by:
k
k
" “
" “
ij
k ik r k r k
Rij = Rijk = - + “ “ - “ “ .
j
" x " xk ik rj ij rk
Transformation Of A Geodesic From Parameter u To v, Where v = f(u)
Given a particular geodesic in terms of a parameter u, in this section the geodesic equation will be transformed
so that it is in terms of a new parameter v.
2
ëÅ‚ öÅ‚
D dxa ÷Å‚ d xa a dxb dxc
ìÅ‚
Start with = + “ = 0 .
ìÅ‚ ÷Å‚
du du du2 bc du du
íÅ‚ Å‚Å‚
dxa dxa dv
Substitute = ,
du dv du
2
ëÅ‚ öÅ‚
D " xa dv " xa dv " xb dv dxc " xa d v
a
ìÅ‚ ÷Å‚
then = + “ + = 0
bc
ìÅ‚ ÷Å‚
du " v du " u" v du " v du du " v du2
íÅ‚ Å‚Å‚
2
" xa dv " xb dxc dv " xa d v
a
Ô! + “ = -
bc
" u" v du " v du du " v du2
2
" xa dv du du " xb dxc dv du du " xa d v du du
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
a
Ô! + “ =
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚ -
ìÅ‚ ÷Å‚
bc
" u" v du dv dv " v du du dv dv " v du2 dv dv
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
2
2
" xa a " xb dxc " xa d v du
ëÅ‚ öÅ‚
Ô! + “ = - .
ìÅ‚ ÷Å‚
" v2 bc " v dv " v du2 dv
íÅ‚ Å‚Å‚
Since the LHS is now in terms of v, partial differentiation can be replaced by normal differentiation:
2
2
dxa a dxb dxc dxa d v dv
ëÅ‚ öÅ‚
Ô! + “ = - .
ìÅ‚ ÷Å‚
dv2 bc dv dv dv du2 du
íÅ‚ Å‚Å‚
2
2
d v dv
ëÅ‚ öÅ‚
By letting  = - , we arrive at the final equation:
ìÅ‚ ÷Å‚
du2 du
íÅ‚ Å‚Å‚
dxa a dxb dxc dxa
+ “ =  .
dv2 bc dv dv dv
General Relativity
The Metric Tensor for Special Relativity
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1 0 0 0
îÅ‚ Å‚Å‚
0
ïÅ‚ - 1 0 0 śł
· = .
µ½
ïÅ‚ 0 0 - 1 0 śł
ïÅ‚ śł
0 0 0 - 1ûÅ‚
ðÅ‚
Einstein's Law of Gravitation
Simply stated, Einstein's law of gravitation is:
Rµ½ = 0 .
This condition holds when the local space is completely devoid of all forms of matter and energy.
Derivation of the Schwarzchild Solution
In this section the line-element solution to the field equations for a quasi-static gravitational field produced by a
spherical body will be derived.
We start by setting up a general line element employing spherical coordinates:
2 2 2 2 2 2
c2dÄ = Ac2dt - Bdr - Wr (d¸ + sin2 ¸ dĆ ).
Before we continue, a few assumptions need to be made:

The space is asymptotically flat. This means that A = B 1 as r " .

The gravitational field only affects time and radial distance, so W = 1.
We can immediately define the metric tensor:
2 2
g00 = Ac2 , g11 = - B , g22 = - r and g33 = - r sin2 ¸ .
Since we are dealing with the empty space surrounding the body, the Ricci tensor needs to equal zero. With this
in mind, the derivation begins. We first calculate the Christoffel symbols. Note that since a Christoffel symbol
of the second kind is defined as
z
z Å„Å‚ üÅ‚ rz
“ = = g “
òÅ‚
mn mnr
mnżł
þÅ‚
we need only calculate them for values when r = z.
0 00
1 1
2
“ = g “ = A- 1c- 2[g00,1 + g10,0 - g10,0]= A- 1A
10 100 2 2
1
1 1 1
2
“ = g11“ = - B- 1[g10,0 + g01,0 - g00,1]= B- 1g00,1 = B- 1A c2
00 001 2 2 2
1
1 1 1
2
“ = g11“ = - B- 1[g11,1 + g11,1 - g11,1]= - B- 1g11,1 = B- 1B
11 111 2 2 2
1
1 1 1
“ = g11“ = - B- 1[g12,2 + g21,2 - g22,1]= - B- 1g22,1 = - B- 1 Å"(2r)= - rB- 1
22 221 2 2 2
1 2
1 1 1
“ = g11“ = - B- 1[g13,3 + g31,3 - g33,1]= - B- 1(- g33,1)= - B- 1(2r sin ¸ )= - rB- 1 sin2 ¸
33 331 2 2 2
2 22 - 1
1 - 2
1 - 2
“ = g “ = - r [g21,2 + g22,1 - g21,2]= - r Å"(- 2r)= r
21 212 2 2
2 22
1 - 2
1 - 2 2
“ = g “ = - r [g23,3 + g32,3 - g33,2]= - r Å" 2r cos¸ sin¸ = - cos¸ sin¸
33 332 2 2
3 33 - 1
1 - 2 - 2
1 - 2
“ = g “ = - r sin ¸ [g33,1 + g13,3 - g13,3]= - r sin- 2 ¸ Å"(- 2r sin2 ¸ )= r
13 133 2 2
3 33 2
1 - 2 - 2
1 - 2
“ = g “ = - r sin ¸ [g33,2 + g23,3 - g23,3]= - r sin- 2 ¸ Å"(- 2r cos¸ sin¸ )= cot¸
23 233 2 2
We now solve the field equations:
0 0 0 0 0 0 1 0 1 0 2 0 2 0 3 0 3 0
R00 = [“ - “ + (“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )]
00,0 00,0 00 00 00 00 00 10 00 10 00 20 00 20 00 30 00 30
1 1 0 1 0 1 1 1 1 1 2 1 2 1 3 1 3 1
+[“ - “ +(“ “ - “ “ )+(“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )]
01,0 00,1 01 00 00 01 01 10 00 11 01 20 00 21 01 30 00 31
2 2 0 2 0 2 1 2 1 2 2 2 2 2 3 2 3 2
+[“ - “ + (“ “ - “ “ )+(“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )]
02,0 00,2 02 00 00 02 02 10 00 12 02 20 00 22 02 30 00 32
3 3 0 3 0 3 1 3 1 3 2 3 2 3 3 3 3 3
+[“ - “ + (“ “ - “ “ )+(“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )]
03,0 00,3 03 00 00 03 03 10 00 13 03 20 00 23 03 30 00 33
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0 0 0 0 0 0 1 0 1 0 2 0 2 0 3 0 3 0
îÅ‚
) )
R11 = “ - “ +(“ “ - “ “ +(“ “ - “ “ + (“ “ - “ “ )+ (“ “ - “ “ )]
10,1 11,0 10 01 11 00 10 11 11 10 10 21 11 20 10 31 11 30
ïÅ‚
ðÅ‚
1 1 0 1 0 1 1 1 1 1 2 1 2 1 3 1 3 1
+[“ - “ + (“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )]
11,1 11,1 11 01 11 01 11 11 11 11 11 21 11 21 11 31 11 31
2 2 0 2 0 2 1 2 1 2 2 2 2 2 3 2 3 2
îÅ‚
+ “ - “ + (“ “ - “ “ )+(“ “ - “ “ )+(“ “ - “ “ )+ (“ “ - “ “ )]
12,1 11,2 12 01 11 02 12 11 11 12 12 21 11 22 12 31 11 32
ïÅ‚
ðÅ‚
3 3 0 3 0 3 1 3 1 3 2 3 2 3 3 3 3 3
îÅ‚
+ “ - “ + (“ “ - “ “ )+(“ “ - “ “ )+ (“ “ - “ “ )+(“ “ - “ “ )]
13,1 11,3 13 01 11 03 13 11 11 13 13 21 11 23 13 31 11 33
ïÅ‚
ðÅ‚
0 0 0 0 0 0 1 0 1 0 2 0 2 0 3 0 3 0
)
R22 = [“ - “ + (“ “ - “ “ )+(“ “ - “ “ + (“ “ - “ “ )+ (“ “ - “ “ )]
20,2 22,0 20 02 22 00 20 12 22 10 20 22 22 20 20 32 22 30
1 1 0 1 0 1 1 1 1 1 2 1 2 1 3 1 3 1
) )
+[“ - “ + (“ “ - “ “ )+(“ “ - “ “ +(“ “ - “ “ + (“ “ - “ “ )]
21,2 22,1 21 02 22 01 21 12 22 11 21 22 22 21 21 32 22 31
2 2 0 2 0 2 1 2 1 2 2 2 2 2 3 2 3 2
+[“ - “ + (“ “ - “ “ )+(“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )]
22,2 22,2 22 02 22 02 22 12 22 12 22 22 22 22 22 32 22 32
3 3 0 3 0 3 1 3 1 3 2 3 2 3 3 3 3 3
+[“ - “ + (“ “ - “ “ )+(“ “ - “ “ )+ (“ “ - “ “ )+(“ “ - “ “ )]
23,2 22,3 23 02 22 03 23 12 22 13 23 22 22 23 23 32 22 33
0 0 0 0 0 0 1 0 1 0 2 0 2 0 3 0 3 0
R33 = [“ - “ + (“ “ - “ “ )+(“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )]
30,3 33,0 30 03 33 00 30 13 33 10 30 23 33 20 30 33 33 30
1 1 0 1 0 1 1 1 1 1 2 1 2 1 3 1 3 1
+[“ - “ + (“ “ - “ “ )+(“ “ - “ “ )+ (“ “ - “ “ )+(“ “ - “ “ )]
31,3 33,1 31 03 33 01 31 13 33 11 31 23 33 21 31 33 33 31
2 2 0 2 0 2 1 2 1 2 2 2 2 2 3 2 3 2
+[“ - “ + (“ “ - “ “ )+(“ “ - “ “ )+ (“ “ - “ “ )+(“ “ - “ “ )]
32,3 33,2 32 03 33 02 32 13 33 12 32 23 33 22 32 33 33 32
3 3 0 3 0 3 1 3 1 3 2 3 2 3 3 3 3 3
+[“ - “ + (“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )+ (“ “ - “ “ )]
33,3 33,3 33 03 33 03 33 13 33 13 33 23 33 23 33 33 33 33
We are left with
1 0 1 1 1 1 2 1 3
R00 = - “ + “ “ - “ “ - “ “ - “ “ ,
00,1 01 00 00 11 00 12 00 13
0 0 0 1 0 2 1 2 2 2 3 1 3 3 3
R11 = “ + “ “ - “ “ + “ - “ “ + “ “ + “ - “ “ + “ “ ,
10,1 10 01 11 10 12,1 11 12 12 21 13,1 11 13 13 31
1 0 1 1 1 2 1 3 1 3 3 3
R22 = - “ “ - “ - “ “ + “ “ + “ - “ “ + “ “ ,
22 10 22,1 22 11 21 22 23,2 22 13 23 32
1 0 1 1 1 3 1 2 1 2 3 2
R33 = - “ “ - “ - “ “ + “ “ - “ - “ “ + “ “ .
33 10 33,1 33 11 31 33 33,2 33 12 32 33
These equations must equal zero, thus after substitution, we have:
2 2 2 2 2 2 2 Å‚Å‚
B A A A A A 1
îÅ‚
R00 = - + - , (1)
ïÅ‚ śł
4B 2 4A r B
ðÅ‚ ûÅ‚
2 2 2 2 2 2 2
A A A B B A
R11 = - + - - , (2)
4A 2 4B Br
2 2
A r B r 1
R22 = - + - 1, (3)
2AB 2B2 B
2 2
A r B r 1
îÅ‚
R33 = - + - 1Å‚Å‚ sin2¸ = R22 sin2¸ . (4)
ïÅ‚ śł
2AB 2B2 B
ðÅ‚ ûÅ‚
Equation 2 becomes:
2 2 2 2 2 2 2 2 2 2 2 2 2 2
A A A B B A B A A A A B
0 = - + - - Ò! = - + - .
4A 2 4B Br Br 4A 2 4B
This can be substituted into equation 1 to give:
2 2 Å‚Å‚ 2 2 2 2
B A A 1 B A A B A
îÅ‚
0 = - - Ò! = - Ò! = - .
ïÅ‚ śł
Br r B Br r B A
ðÅ‚ ûÅ‚
This can also be written as:
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GENERAL RELATIVITY, TENSOR ANALYSIS AND GEOMETRY
dB dA dB dA 1
= - Ô! = - Ò! B = Ò! AB = 1,
Bdr Adr B A A
upon solving the simple differential equation. We use this solution to simplify equation 3:
2 2
A r A r
2 2 2
0 = + + A - 1 Ô! 1 = A r + A = A r + Ar = (Ar)2
2 2A
We now integrate:
d
(Ar)dr Ò! r + k = Ar ,
+"dr = +"
dr
where k is an integration constant. The equation can be rearranged to find A:
k
A = 1+ .
r
Since B = A- 1 ,
- 1
k
ëÅ‚ öÅ‚
B = 1+ .
ìÅ‚ ÷Å‚
r
íÅ‚ Å‚Å‚
The value of k is the last thing to obtain. In the next section on the approximation of Newtonian gravitation,
g00 = 1+ h00 . It can immediately be seen that the h00 is equivalent to k. In the Newtonian approximation, it
1
is required that the Newtonian gravitational potential V = c2h00 . Using the Newtonian potential
2
V = - GM r , this gives a value of k = - 2GM rc2 . By substituting A and B back into the original line-
element equation at the start of this section, we have the Schwarzchild solution:
- 1
2GM 2GM
ëÅ‚ öÅ‚ ëÅ‚ öÅ‚
2 2 2 2 2 2
c2dÄ = 1- ÷Å‚ - ìÅ‚
c2dt 1- ÷Å‚ - r (d¸ + sin2 ¸ dĆ ).
dr
ìÅ‚
rc2 Å‚Å‚ rc2 Å‚Å‚
íÅ‚ íÅ‚
Utilising Geodesic Equation to Find GR Approximation of Newtonian Gravity
A particle travels through spacetime along a geodesic, given by the equation:
dxµ µ dx½ dxÃ
+ “ = 0 .
½Ã
2
dÄ dÄ dÄ
Ä is the time experienced relative to the particle. The equation simply states that relative to a free particle, it
experiences no net acceleration (though other objects appear to accelerate if a gravitational field is present).
We wish to determine the motion of the particle relative to coordinate time, denoted by t. The equation would
give the path of the particle in accordance to what other observers would see if they thought they were in a
gravitational field.
With the above said, the following equation can be immediately written, transforming from proper time to
coordinate time:
2
2 2
öÅ‚
d xµ µ dx½ dxà ëÅ‚ d t dt dxµ
ëÅ‚ öÅ‚
ìÅ‚ ÷Å‚
+ “ = .
ìÅ‚ ÷Å‚
½Ã
2 2 2
ìÅ‚ ÷Å‚
dt dt dt dÄ dÄ dt
íÅ‚ Å‚Å‚
íÅ‚ Å‚Å‚
First, we expand and consider its spatial components:
2
2 j 2
öÅ‚
d xi i dx dxk i dx0 dxk i dx0 dx0 ëÅ‚ d t dt dxi
ëÅ‚ öÅ‚
ìÅ‚ ÷Å‚
+ “ + 2“ + “ = .
ìÅ‚ ÷Å‚
jk 0k 00
2 2 2
ìÅ‚ ÷Å‚
dt dt dt dt dt dt dt dÄ dÄ dt
íÅ‚ Å‚Å‚
íÅ‚ Å‚Å‚
j
dx dx0 i dx0 dxk
i
One of the Christoffel symbols has a coefficient of two since “ = “ . We simplify the
j 0 0k
dt dt dt dt
equation:
2
2 j 2
öÅ‚
d xi i dx dxk i dxk i ëÅ‚ d t dt dxi
ëÅ‚ öÅ‚
ìÅ‚ ÷Å‚
+ “ + 2“ c + c2“ = .
ìÅ‚ ÷Å‚
jk 0k 00
2 2 2
ìÅ‚ ÷Å‚
dt dt dt dt dÄ dÄ dt
íÅ‚ Å‚Å‚
íÅ‚ Å‚Å‚
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GENERAL RELATIVITY, TENSOR ANALYSIS AND GEOMETRY
We assume that the gravitational field is quasi-static, i.e. that it doesn't change with respect to time. Therefore,
any derivatives of the metric tensor with respect to time can be left out. Now we evaluate the connection
coefficients:
iÁ
" gÁ j " gÁ k " g
ëÅ‚ öÅ‚
1 g
jk
i iÁ
ìÅ‚
“ = g “ = + - ÷Å‚
. These derivatives are quite small, and can be neglected.1
jk Á jk
j
ìÅ‚
2 2 " xk " x " xÁ ÷Å‚
íÅ‚ Å‚Å‚
iÁ
" gÁ 0 " gÁ k " g0k iÁ " hÁ " h0k
ëÅ‚ öÅ‚ öÅ‚
g
0 k
i
ìÅ‚
2“ = 2 + - ÷Å‚ ìÅ‚
H" (· + hiÁ )ëÅ‚ ""hÁ + - ÷Å‚
0k
Á Á
ìÅ‚ ÷Å‚ ìÅ‚ ÷Å‚
2 " xk " x0 " x xk " x0 " x
íÅ‚ Å‚Å‚ íÅ‚ Å‚Å‚
" hÁ " h0k " hµ½
öÅ‚ " h0s " h0k
ëÅ‚ öÅ‚
Á k
iÁ is
ìÅ‚
H" (· + hiÁ )ëÅ‚ ""h0 + - ÷Å‚ - ´ on neglecting terms involving .
H"
ìÅ‚ - ÷Å‚
Á
ìÅ‚ ÷Å‚
xk " x0 " x " xk " xs Å‚Å‚ " x0
íÅ‚
íÅ‚ Å‚Å‚
iº iz
" gº 0 " g0º " g00 ´ " h00
g ëÅ‚ öÅ‚
i
“ = + - ÷Å‚
H" .
ìÅ‚
00
2 " x0 " x0 " xº Å‚Å‚ " xz
2
íÅ‚
Now we need to evaluate the RHS of the equation. We start by first looking at the following line element:
2
µ µ
dÄ 1 1
ëÅ‚ öÅ‚
2
c2dÄ = gµ½ dxµ dx½ Ô! = (· + hµ½ )dx dx½ = (1+ hµ½ )dx dx½ . Neglecting
ìÅ‚ ÷Å‚
dt c2 µ½ dt dt c2 dt dt
íÅ‚ Å‚Å‚
terms involving the spatial components, which are small in comparison to the temporal components, we get:
2
dÄ dÄ
ëÅ‚ öÅ‚ 1 2
1
= (1+ h00 )Ò! = (1+ h00 ) H" (1+ h00).
ìÅ‚ ÷Å‚
2
dt dt
íÅ‚ Å‚Å‚
2
d Ä dh00 dh00
Now = = c , so the RHS of our major equation becomes:
2
dt dt dx0
dh00
2 c
2 i
Å‚Å‚
dxi îÅ‚ d t dt
ëÅ‚ öÅ‚
dx0 dxi H" c dh00 h00 .
1
H" (1- )dx
ïÅ‚ ìÅ‚ ÷Å‚ śł
2 2
1
dt dÄ dÄ 1+ h00 dt dx0 2 dt
íÅ‚ Å‚Å‚
ïÅ‚ śł
2
ðÅ‚ ûÅ‚
This is negligible since it involves temporal derivatives of the gravitational field.
By plugging everything into our equation, we get:
2 iz
" h0s " h0k
d xi is ëÅ‚ öÅ‚ dxk ´ " h00
- ´ + c2 = 0 .
ìÅ‚ - ÷Å‚
2 z
dt " xk " xs Å‚Å‚ dt 2 " x
íÅ‚
Through multiplying by mass m and rearranging the equation, we get:
2 iz
d xi ´ " h00 is " h0s " h0k dxk
ëÅ‚ öÅ‚
m = - mc2 + m´ .
ìÅ‚ - ÷Å‚
2 z
dt 2 " x " xk " xs Å‚Å‚ dt
íÅ‚
The term on the left is the force that the particle appears to experience. The first term on the right is some kind
of potential of the field, since its temporal component is involved. The last term, which involves perpendicular
velocities, is indicative of some sort of Coriolis force. We are not interested in the Coriolis effects, so we shall
assume that we are in a non-rotating frame, so we get:
2 iz
d xi ´ " h00
m = - mc2 .
2 z
dt 2 " x
1
If we denote a potential by V = c2h00 , then the equation simply becomes F = - m" V , or
2
2
d xi iz " V
= - ´ . We want the metric tensor to be flat when there is no gravitational field present. The
2 z
dt " x
µ½ µ½ µ½
1
The metric tensor g H" · + hµ½ , where · is the familiar metric tensor of Special Relativity, and
hµ½ are small terms which include the action of any gravitational fields which may be present, and are small in
µ½
comparison to the · in weak gravitational fields, as opposed to the awesome sucking power of a black hole!
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GENERAL RELATIVITY, TENSOR ANALYSIS AND GEOMETRY
equation for the potential leads us to the expression g00 = · + h00 = 1+ 2V c2 . To finally get the actual
00
expression for the potential in terms of mass, we need to use the line element from the Schwarzchild solution.
The temporal component gives V = - GM r . With this expression, we can easily obtain an approximation of
the gravitational force:
2
F = - m" V = - GMm r .
Field Equations in the Presence of Matter: The Poisson Approximation
Let us write the equation:
µ½ µ½
1
Rµ½ - g = º T ,
2
or, more compactly as:
µ½
Gµ½ = º T ,
where G is the Einstein tensor.
As a test for General Relativity, at velocities which are small in comparison to the speed of light, there must be
2
an approximation to Poisson s equation: " V = 4Ä„ GÁ . To achieve this requires the weak field
approximation by leaving out negligible terms in the Ricci tensor.2
µ½
Gµ½ = º T
µ½ µ½
1
Rµ½ - g R = º T
2
µ½ µ½
1
gµ½ Rµ½ - gµ½ g R = º T gµ½
2
µ½
R - 2R = º T gµ½
µ½
4" R = - º T gµ½
We substitute this back into our original equation:
µ½ µ½ µ½ µ½ µ½
1 1
Rµ½ = º T + g R = º T - g º T gµ½
2 2
We are approximating that the material energy tensor has a negligible value in for all µ , ½ except when µ = ½ =
0, so we get:
00 00 00 00
1 1
R00 = º T - g º T g00 = º T .
2 2
We now make the Ricci tensor covariant:
00
1 1
g00 g00R00 = º T g00 g00 Ô! R00 = º T00
2 2
Weak field approximation.
1 1
Rµ½ H" (gÄ…Ä… ,µ½ - g½Ä… ,µÄ… - gµÄ… ,½Ä… + gµ½ ,Ä…Ä… )H" gµ½ ,Ä…Ä…
2 2
1
2 2
1 1
4" R00 H" g00,Ä…Ä… H" " g00 = " V
2 2
c2
1
2
1 1
4" " V = º T00 = ºÁ v0v0
2 2
c2
The particles are traveling at the travelling at the speed of light through time, so we get:
2
1
" V = ºÁ c4
2
This must equal Poisson s equation, stated earlier
8Ä„ G
This gives a value º = .
c4
2
Note that this particular calculation would be shorter if we took the Einstein tensor and the material energy
tensor to be covariant as opposed to contravariant, but due to the actual form of the material energy tensor, I
prefer it to be contravariant.
Courtney James Mewton Page 7 GR, Tensor Analysis & Geometry