Parallel tansport sssqrd

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Refresher Drills
Equations of Parallel Transport
Roa, F.J.P.
In this draft I will no longer go into the conceptual details concerning parallel transport of a
vector along parametrized paths. This draft mainly presents few review drills related to parallel
transport.
Simply quoting the result, the equation for parallel transport of a vector is given by
(1)
𝑑𝑉 𝜇
𝑑𝜆
+ Γ𝜎𝜌
𝜇 𝑑𝑥 𝜎
𝑑𝜆
𝑉 𝜌
= 0
where the Christoffel connections Γ as specific to the given metric solutions to Einstein’s field
equation. In this draft we will be dealing with Schwarzschild metric solutions.
The vector that we parallel transport has the contravariant ((1,0) tensors) components 𝑉 𝜇
that we
parallel transport along coordinate paths 𝑥 𝜇
(𝜆) that we parametrize in terms of an independent
parameter 𝜆. These parametrized paths have their corresponding tangent vectors
(2)
𝑊 𝜎
=
𝑑𝑥 𝜎
𝑑𝜆
As previously said we will be dealing with the Schwarzschild solutions whose fundamental
spacetime line element has the form
(3.1)
𝑑𝑆2
= −𝜂𝑑𝑡2
+ 𝜂−1
𝑑𝑟2
+ 𝑟2
𝑑Ω2
where
(3.2)
𝑑Ω2
= 𝑑θ2
+ 𝑠𝑖𝑛2
θ 𝑑ϕ2
is the unit two-sphere that we are going to suppress so as to simplify the drills to our
convenience.
To simplify things let us have all motions confined on a plane where θ and ϕ are constants and
let us have the differentials of these coordinates vanish to reduce (3.1) into
(3.3)
𝑑𝑆2
= −𝜂𝑑𝑡2
+ 𝜂−1
𝑑𝑟2
𝜂 = 1 −
2𝐺𝑀 𝑞
𝑟
𝑥 𝜇
= (𝑥0
= 𝑡; 𝑥1
= 𝑟)
and we will only have two components of the parametrized coordinate paths 𝑥 𝜇
(𝜆) = (𝑥0
(𝜆) =
𝑡(𝜆); 𝑥1
(𝜆) = 𝑟(𝜆)). Such set up will give us the relevant components of the Christoffel
connections
(3.4)
Γ00
1
=
1
2
𝑑
𝑑𝑟
(
1
2
𝜂2
)
1
2
𝑑𝜂
𝑑𝑟
=
𝐺𝑀 𝑞
𝑟2
Γ01
0
= Γ10
0
= −Γ11
1
=
𝑑
𝑑𝑟
ln √ 𝜂 =
1
2
𝑑
𝑑𝑟
ln 𝜂
Thus, we write (1) explicitly into its two components
(4.1)
𝑑𝑉0
𝑑𝜆
+
1
2
1
𝜂
𝑑𝜂
𝑑𝑟
(
𝑑𝑥0
𝑑𝜆
𝑉1
+
𝑑𝑥1
𝑑𝜆
𝑉0
) = 0
and
(4.2)
𝑑𝑉1
𝑑𝜆
+
1
2
𝜂
𝑑𝜂
𝑑𝑟
𝑑𝑥0
𝑑𝜆
𝑉0
−
1
2
1
𝜂
𝑑𝜂
𝑑𝑟
𝑑𝑥1
𝑑𝜆
𝑉1
= 0
Now, our first set of drills is to consider reducing these parallel transport equations into
component equations of geodesic. We are able to do this particular set up by parallel transporting
the tangent vectors to the paths themselves. That is, our vectors to parallel transport are
(5.1)
𝑉 𝜇
=
𝑑𝑥 𝜇
𝑑𝜆
: 𝑉0
=
𝑑𝑡
𝑑𝜆
; 𝑉1
=
𝑑𝑟
𝑑𝜆
So now, (4.1) becomes
(5.2)
𝑑𝑉0
𝑑𝜆
+
1
𝜂
𝑑𝜂
𝑑𝜆
𝑉0
= 0
with a solution
(5.3)
𝑉0 ( 𝜆) =
𝜂(𝜆0)
𝜂(𝜆)
𝑉0( 𝜆0)
For (4.2) we have
(6.1)
𝑑𝑉1
𝑑𝜆
+
1
2
1
𝜂
𝑑𝜂
𝑑𝑟
((𝜂( 𝜆0) 𝑉0( 𝜆0))2
− (𝑉1
)2) = 0
The solution of which is given by
(6.2)
𝑉1( 𝜆) =
±𝑉1( 𝜆0)
√1 − 𝑐𝜇
2 𝜂( 𝜆0)
√1 − 𝑐𝜇
2 𝜂( 𝜆)
Returning to the geodesic equations (4.1) and (4.2) we will now take another set of drills wherein
we consider parallel transporting vectors along the path 𝑥1
(𝜆) = 𝑟(𝜆), while holding that 𝑥0
≠
𝑥0
(𝜆), for which
𝑑𝑥0
𝑑𝜆
= 0. Thus, from (4.1) we write
(7.1)
𝑑𝑉0
𝑑𝜆
+
1
2
1
𝜂
𝑑𝜂
𝑑𝜆
𝑉0
= 0
This has the solution
(7.2)
𝑉0( 𝜆) =
√𝜂( 𝜆0)
√𝜂( 𝜆)
𝑉0( 𝜆0)
The other parallel transport equation in this case, following from (4.2) is given by
(7.3)
𝑑𝑉1
𝑑𝜆
−
1
2
1
𝜂
𝑑𝜂
𝑑𝜆
𝑉1
= 0
In turn, this is satisfied by
(7.4)
𝑉1( 𝜆) =
√𝜂( 𝜆)
√𝜂( 𝜆0)
𝑉1( 𝜆0)
In the alternate set up we consider the parametric path 𝑥0
(𝜆) and this time it is 𝑥1
that is
independent of the parameter lambda so that
𝑑𝑥1
𝑑𝜆
= 0.
Thus, we write (4.1) and (4.2) simultaneously as
(8.1)
𝑑𝑉0
𝑑𝜆
+
1
2
1
𝜂
𝑑𝜂
𝑑𝑟
𝑑𝑥0
𝑑𝜆
𝑉1
= 0
and
(8.2)
𝑑𝑉1
𝑑𝜆
+
1
2
𝜂
𝑑𝜂
𝑑𝑟
𝑑𝑥0
𝑑𝜆
𝑉0
= 0
Based on these two equations, clearly each component vector is related to other component and
we write such relation as
(8.3)
𝑉0
= ±
1
𝜂
√(𝑉1)2 + (𝜇 𝑞 𝜂)2
Substituting this relation in (8.2) thus, we are led to
(8.4)
𝑑𝑉1
𝑑𝜆
±
1
2
𝑑𝜂
𝑑𝑟
𝑑𝑥0
𝑑𝜆
√(𝑉1)2 + (𝜇 𝑞 𝜂)2 = 0
Its vector solution takes the form
(8.5)
𝑉1( 𝜆) = ±𝜇 𝑞 𝜂𝑠𝑖𝑛ℎ (∓
1
2
𝑑𝜂
𝑑𝑟
𝑥0
(𝜆))
The other component vector can then be given as
(8.6)
𝑉0( 𝜆) = ±𝜇 𝑞 𝑐𝑜𝑠ℎ(∓
1
2
𝑑𝜂
𝑑𝑟
𝑥0
(𝜆))
which satisfies
(8.7)
𝑑𝑉0
𝑑𝜆
±
1
2
𝑑𝜂
𝑑𝑟
𝑑𝑥0
𝑑𝜆
√(𝑉0)2 − 𝜇 𝑞
2 = 0
Note:
(8.8)
√(𝑉0)2 − 𝜇 𝑞
2 = ±𝜇 𝑞 𝑠𝑖𝑛ℎ(∓
1
2
𝑑𝜂
𝑑𝑟
𝑥0
(𝜆))
We would like to elucidate more and do some cleaning up. Going back to (6.1), given (5.3) along
with (6.2) we write
(9.1)
𝑑𝑉1
𝑑𝜆
= −
1
2
𝑐𝜇
2
𝑑𝜂
𝑑𝑟
(𝑉1( 𝜆0))2
1 − 𝑐𝜇
2 𝜂( 𝜆0)
This together with (6.2) must satisfy (6.1) to yield
(9.2)
1
2
1
𝜂
𝑑𝜂
𝑑𝑟
[(𝜂( 𝜆0) 𝑉0( 𝜆0))2
−
(𝑉1( 𝜆0))2
1 − 𝑐𝜇
2 𝜂( 𝜆0)
] = 0
This enforces the initial condition
(9.3)
𝑉0( 𝜆0) = ±
1
𝜂( 𝜆0)
𝑉1( 𝜆0)
√1 − 𝑐𝜇
2 𝜂( 𝜆0)
To check for (8.4) we take note of (8.3) and from there we write
(10.1)
𝑑𝑉1
𝑑𝜆
+
1
2
𝑑𝜂
𝑑𝑟
𝜂
𝑑𝑥0
𝑑𝜆
𝑉0
= 0
with a note that 𝜂( 𝑟) and 𝑟 ≠ 𝑟(𝜆). We obtain the derivative of (8.5) with respect to the
parameter 𝜆 and then taking note of (8.3) we can arrive at (10.1) thereby, checking consistency in
(8.4).
In similar procedure we check for (8.7). This time we write (8.8) as
(10.2)
±√(𝑉0)2 − 𝜇 𝑞
2 = 𝜇 𝑞 𝑠𝑖𝑛ℎ(∓
1
2
𝑑𝜂
𝑑𝑟
𝑥0
(𝜆)) = 𝑉1
Thus, with this we can write (8.7) in the form already given by (8.1), where it is also noted that
𝜂( 𝑟) and 𝑟 ≠ 𝑟(𝜆). Then we take the derivative of (8.6) with respect to lambda in which in the
result we take note of (10.2) to arrive at (8.1) thus, checking consistency with this expression.
Ref’s
[1]Ohanian, H. C., GRAVITATION AND SPACETIME, New York: W. W. Norton and
Company Inc., copyright 1976
[2]Townsend, P. K., Blackholes – Lecture Notes, http://xxx.lanl.gov/abs/gr-qc/9707012
[3]Carroll, S. M., Lecture notes On General Relativity, http://www.arxiv.org/abs/gr-qc/9712019

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Parallel tansport sssqrd

  • 1. Refresher Drills Equations of Parallel Transport Roa, F.J.P. In this draft I will no longer go into the conceptual details concerning parallel transport of a vector along parametrized paths. This draft mainly presents few review drills related to parallel transport. Simply quoting the result, the equation for parallel transport of a vector is given by (1) 𝑑𝑉 𝜇 𝑑𝜆 + Γ𝜎𝜌 𝜇 𝑑𝑥 𝜎 𝑑𝜆 𝑉 𝜌 = 0 where the Christoffel connections Γ as specific to the given metric solutions to Einstein’s field equation. In this draft we will be dealing with Schwarzschild metric solutions. The vector that we parallel transport has the contravariant ((1,0) tensors) components 𝑉 𝜇 that we parallel transport along coordinate paths 𝑥 𝜇 (𝜆) that we parametrize in terms of an independent parameter 𝜆. These parametrized paths have their corresponding tangent vectors (2) 𝑊 𝜎 = 𝑑𝑥 𝜎 𝑑𝜆 As previously said we will be dealing with the Schwarzschild solutions whose fundamental spacetime line element has the form (3.1) 𝑑𝑆2 = −𝜂𝑑𝑡2 + 𝜂−1 𝑑𝑟2 + 𝑟2 𝑑Ω2 where (3.2) 𝑑Ω2 = 𝑑θ2 + 𝑠𝑖𝑛2 θ 𝑑ϕ2 is the unit two-sphere that we are going to suppress so as to simplify the drills to our convenience. To simplify things let us have all motions confined on a plane where θ and ϕ are constants and let us have the differentials of these coordinates vanish to reduce (3.1) into (3.3) 𝑑𝑆2 = −𝜂𝑑𝑡2 + 𝜂−1 𝑑𝑟2
  • 2. 𝜂 = 1 − 2𝐺𝑀 𝑞 𝑟 𝑥 𝜇 = (𝑥0 = 𝑡; 𝑥1 = 𝑟) and we will only have two components of the parametrized coordinate paths 𝑥 𝜇 (𝜆) = (𝑥0 (𝜆) = 𝑡(𝜆); 𝑥1 (𝜆) = 𝑟(𝜆)). Such set up will give us the relevant components of the Christoffel connections (3.4) Γ00 1 = 1 2 𝑑 𝑑𝑟 ( 1 2 𝜂2 ) 1 2 𝑑𝜂 𝑑𝑟 = 𝐺𝑀 𝑞 𝑟2 Γ01 0 = Γ10 0 = −Γ11 1 = 𝑑 𝑑𝑟 ln √ 𝜂 = 1 2 𝑑 𝑑𝑟 ln 𝜂 Thus, we write (1) explicitly into its two components (4.1) 𝑑𝑉0 𝑑𝜆 + 1 2 1 𝜂 𝑑𝜂 𝑑𝑟 ( 𝑑𝑥0 𝑑𝜆 𝑉1 + 𝑑𝑥1 𝑑𝜆 𝑉0 ) = 0 and (4.2) 𝑑𝑉1 𝑑𝜆 + 1 2 𝜂 𝑑𝜂 𝑑𝑟 𝑑𝑥0 𝑑𝜆 𝑉0 − 1 2 1 𝜂 𝑑𝜂 𝑑𝑟 𝑑𝑥1 𝑑𝜆 𝑉1 = 0 Now, our first set of drills is to consider reducing these parallel transport equations into component equations of geodesic. We are able to do this particular set up by parallel transporting the tangent vectors to the paths themselves. That is, our vectors to parallel transport are (5.1) 𝑉 𝜇 = 𝑑𝑥 𝜇 𝑑𝜆 : 𝑉0 = 𝑑𝑡 𝑑𝜆 ; 𝑉1 = 𝑑𝑟 𝑑𝜆 So now, (4.1) becomes (5.2) 𝑑𝑉0 𝑑𝜆 + 1 𝜂 𝑑𝜂 𝑑𝜆 𝑉0 = 0
  • 3. with a solution (5.3) 𝑉0 ( 𝜆) = 𝜂(𝜆0) 𝜂(𝜆) 𝑉0( 𝜆0) For (4.2) we have (6.1) 𝑑𝑉1 𝑑𝜆 + 1 2 1 𝜂 𝑑𝜂 𝑑𝑟 ((𝜂( 𝜆0) 𝑉0( 𝜆0))2 − (𝑉1 )2) = 0 The solution of which is given by (6.2) 𝑉1( 𝜆) = ±𝑉1( 𝜆0) √1 − 𝑐𝜇 2 𝜂( 𝜆0) √1 − 𝑐𝜇 2 𝜂( 𝜆) Returning to the geodesic equations (4.1) and (4.2) we will now take another set of drills wherein we consider parallel transporting vectors along the path 𝑥1 (𝜆) = 𝑟(𝜆), while holding that 𝑥0 ≠ 𝑥0 (𝜆), for which 𝑑𝑥0 𝑑𝜆 = 0. Thus, from (4.1) we write (7.1) 𝑑𝑉0 𝑑𝜆 + 1 2 1 𝜂 𝑑𝜂 𝑑𝜆 𝑉0 = 0 This has the solution (7.2) 𝑉0( 𝜆) = √𝜂( 𝜆0) √𝜂( 𝜆) 𝑉0( 𝜆0) The other parallel transport equation in this case, following from (4.2) is given by (7.3) 𝑑𝑉1 𝑑𝜆 − 1 2 1 𝜂 𝑑𝜂 𝑑𝜆 𝑉1 = 0 In turn, this is satisfied by
  • 4. (7.4) 𝑉1( 𝜆) = √𝜂( 𝜆) √𝜂( 𝜆0) 𝑉1( 𝜆0) In the alternate set up we consider the parametric path 𝑥0 (𝜆) and this time it is 𝑥1 that is independent of the parameter lambda so that 𝑑𝑥1 𝑑𝜆 = 0. Thus, we write (4.1) and (4.2) simultaneously as (8.1) 𝑑𝑉0 𝑑𝜆 + 1 2 1 𝜂 𝑑𝜂 𝑑𝑟 𝑑𝑥0 𝑑𝜆 𝑉1 = 0 and (8.2) 𝑑𝑉1 𝑑𝜆 + 1 2 𝜂 𝑑𝜂 𝑑𝑟 𝑑𝑥0 𝑑𝜆 𝑉0 = 0 Based on these two equations, clearly each component vector is related to other component and we write such relation as (8.3) 𝑉0 = ± 1 𝜂 √(𝑉1)2 + (𝜇 𝑞 𝜂)2 Substituting this relation in (8.2) thus, we are led to (8.4) 𝑑𝑉1 𝑑𝜆 ± 1 2 𝑑𝜂 𝑑𝑟 𝑑𝑥0 𝑑𝜆 √(𝑉1)2 + (𝜇 𝑞 𝜂)2 = 0 Its vector solution takes the form (8.5) 𝑉1( 𝜆) = ±𝜇 𝑞 𝜂𝑠𝑖𝑛ℎ (∓ 1 2 𝑑𝜂 𝑑𝑟 𝑥0 (𝜆)) The other component vector can then be given as
  • 5. (8.6) 𝑉0( 𝜆) = ±𝜇 𝑞 𝑐𝑜𝑠ℎ(∓ 1 2 𝑑𝜂 𝑑𝑟 𝑥0 (𝜆)) which satisfies (8.7) 𝑑𝑉0 𝑑𝜆 ± 1 2 𝑑𝜂 𝑑𝑟 𝑑𝑥0 𝑑𝜆 √(𝑉0)2 − 𝜇 𝑞 2 = 0 Note: (8.8) √(𝑉0)2 − 𝜇 𝑞 2 = ±𝜇 𝑞 𝑠𝑖𝑛ℎ(∓ 1 2 𝑑𝜂 𝑑𝑟 𝑥0 (𝜆)) We would like to elucidate more and do some cleaning up. Going back to (6.1), given (5.3) along with (6.2) we write (9.1) 𝑑𝑉1 𝑑𝜆 = − 1 2 𝑐𝜇 2 𝑑𝜂 𝑑𝑟 (𝑉1( 𝜆0))2 1 − 𝑐𝜇 2 𝜂( 𝜆0) This together with (6.2) must satisfy (6.1) to yield (9.2) 1 2 1 𝜂 𝑑𝜂 𝑑𝑟 [(𝜂( 𝜆0) 𝑉0( 𝜆0))2 − (𝑉1( 𝜆0))2 1 − 𝑐𝜇 2 𝜂( 𝜆0) ] = 0 This enforces the initial condition (9.3) 𝑉0( 𝜆0) = ± 1 𝜂( 𝜆0) 𝑉1( 𝜆0) √1 − 𝑐𝜇 2 𝜂( 𝜆0) To check for (8.4) we take note of (8.3) and from there we write (10.1)
  • 6. 𝑑𝑉1 𝑑𝜆 + 1 2 𝑑𝜂 𝑑𝑟 𝜂 𝑑𝑥0 𝑑𝜆 𝑉0 = 0 with a note that 𝜂( 𝑟) and 𝑟 ≠ 𝑟(𝜆). We obtain the derivative of (8.5) with respect to the parameter 𝜆 and then taking note of (8.3) we can arrive at (10.1) thereby, checking consistency in (8.4). In similar procedure we check for (8.7). This time we write (8.8) as (10.2) ±√(𝑉0)2 − 𝜇 𝑞 2 = 𝜇 𝑞 𝑠𝑖𝑛ℎ(∓ 1 2 𝑑𝜂 𝑑𝑟 𝑥0 (𝜆)) = 𝑉1 Thus, with this we can write (8.7) in the form already given by (8.1), where it is also noted that 𝜂( 𝑟) and 𝑟 ≠ 𝑟(𝜆). Then we take the derivative of (8.6) with respect to lambda in which in the result we take note of (10.2) to arrive at (8.1) thus, checking consistency with this expression. Ref’s [1]Ohanian, H. C., GRAVITATION AND SPACETIME, New York: W. W. Norton and Company Inc., copyright 1976 [2]Townsend, P. K., Blackholes – Lecture Notes, http://xxx.lanl.gov/abs/gr-qc/9707012 [3]Carroll, S. M., Lecture notes On General Relativity, http://www.arxiv.org/abs/gr-qc/9712019