Or,
𝑑𝐴̅ 𝜇
𝑑𝜏
+ Γ̅ 𝜇
𝛼𝛽
𝑑𝑥̅ 𝛼
𝑑𝜏
𝐴̅ 𝛽 =
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
𝑑𝐴 𝑗
𝑑𝜏
+
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
Γ 𝑗
𝑝𝑞
𝑑𝑥 𝑝
𝑑𝜏
𝐴 𝑞 (3)
Using(2) with(3) we obtain
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
𝑑𝐴 𝑗
𝑑𝜏
+
𝑑
𝑑𝜏
(
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
)+ Γ̅ 𝜇
𝛼𝛽
𝑑𝑥̅ 𝛼
𝑑𝜏
𝐴̅ 𝛽 =
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
𝑑𝐴 𝑗
𝑑𝜏
+
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗 Γ 𝑗
𝑝𝑞
𝑑𝑥 𝑝
𝑑𝜏
𝐴 𝑞
Or,
𝑑
𝑑𝜏
(
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
) + Γ̅ 𝜇
𝛼𝛽
𝑑𝑥̅ 𝛼
𝑑𝜏
𝐴̅ 𝛽 = +
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
Γ 𝑗
𝑝𝑞
𝑑𝑥 𝑝
𝑑𝜏
𝐴 𝑞 (4)
Nowletrus consideranarbitrarychoice :
𝑑𝐴 𝑗
𝑑𝜏
+ Γ 𝑗
𝑝𝑞
𝑑𝑥 𝑝
𝑑𝜏
𝐴 𝑞 = 𝑋 𝑗 andconsequently,
𝑑𝐴̅ 𝜇
𝑑𝜏
+
Γ̅ 𝜇
𝛼𝛽
𝑑𝑥̅ 𝛼
𝑑𝜏
𝐴̅ 𝛽 = 𝑌 𝜇
Usingthis choice with(4) we obtain
𝑑
𝑑𝜏
(
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
) −
𝑑𝐴̅ 𝜇
𝑑𝜏
+ 𝑌 𝜇 =
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
[−
𝑑𝐴 𝑗
𝑑𝜏
+ 𝑋 𝑗]
𝑑
𝑑𝜏
(
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
) −
𝑑𝐴̅ 𝜇
𝑑𝜏
+ 𝑌 𝜇 = −
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
𝑑𝐴 𝑗
𝑑𝜏
+
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗 𝑋 𝑗](https://image.slidesharecdn.com/3264adfa-c6ff-4e38-9ba5-42b6a362af1d-160307105209/85/ParallelABX-1-320.jpg)
![But 𝑌 𝜇 =
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
𝑋 𝑗Or,
𝑑𝐴̅ 𝜇
𝑑𝜏
=
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
𝑑𝐴 𝑗
𝑑𝜏
+
𝑑
𝑑𝜏
(
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
) (5)
To note that
𝑑𝐴 𝑗
𝑑𝜏
+
𝑑
𝑑𝜏
(
𝜕𝑥̅ 𝜇
𝜕𝑥 𝑗
) =
𝑑𝑥 𝛽
𝑑𝜏
∇ 𝛽 𝐴𝑗 = 𝑈 𝛽∇ 𝛽 𝐴 𝑗
Equation(5) is the same as (2) or (1) withsuitable initial conditions..From(2) youcanarrive at (2.1) or
(4). 𝐼𝑡𝑖𝑠 𝑛𝑜𝑡 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦 𝑓𝑜𝑟 𝑋 𝑗[andhence for 𝑌 𝜇 to be the null tensor[rankone].Itmightappearthat
relation(2.1) will holdforanyarbitrarytransformationwithnon-singular determinantof the Jacobian.
But thisisnot true since the properspeedsontwosidesof (2.1) are relatedtothe transformation
elements.Properspeeds
𝑑𝑥 𝑝
𝑑𝜏
and
𝑑𝑥̅ 𝛼
𝑑𝜏
are connectedwiththe transformationitself ,itselementsbeing
𝜕𝑥̅ 𝛼
𝜕𝑥 𝑝
For parallel transportwe specificallyuse 𝑋 𝑗 = 0andconsequrently 𝑌 𝜇 = 0
𝑑𝐴 𝑗
𝑑𝜏
+ Γ 𝑗
𝑝𝑞
𝑑𝑥 𝑝
𝑑𝜏
𝐴 𝑞 = 𝑋 𝑗
We find 𝑓(𝑡, 𝑥, 𝑦, 𝑧) and 𝑔(𝑡, 𝑥, 𝑦, 𝑧) suchthat
𝑑( 𝑓𝐴 𝑗)
𝑑𝜏
+ Γ 𝑗
𝑝𝑞
𝑑𝑥 𝑝
𝑑𝜏
( 𝑓𝐴 𝑞) = 0
𝑑( 𝑔𝐴̅ 𝜇)
𝑑𝜏
+ Γ̅ 𝜇
𝛼𝛽
𝑑𝑥̅ 𝛼
𝑑𝜏
( 𝑔𝐴̅ 𝛽) = 0
𝑓
𝑑𝐴 𝑗
𝑑𝜏
+ Γ 𝑗
𝑝𝑞
𝑑𝑥 𝑝
𝑑𝜏
𝑓𝐴 𝑞 + 𝐴 𝑗
𝑑𝑓
𝑑𝜏
= 0
𝑓 (
𝑑𝐴 𝑗
𝑑𝜏
+ Γ 𝑗
𝑝𝑞
𝑑𝑥 𝑝
𝑑𝜏
𝐴 𝑞)+ 𝐴 𝑗
𝑑𝑓
𝑑𝜏
= 0
∇ 𝛽 𝐴 𝑗 is rank twotensorof mixedorderwhile 𝑈 𝛽∇ 𝛽 𝐴 𝑗 isarank one tensor: 𝛽 disappearson
contraction.’ 𝑗’isthe survivingindex[uncontractedindex]
An important property of parallel transport ispreservationof dot product as we move along the
concernedworld line : this is possible onlywhen 𝑿𝒋 = 𝟎 and 𝒀 𝝁 = 𝟎. Preservationof dot product
is also a salientattribute for coordinate transformations. This goesin favor of identifyingparallel
transport product as a transformation](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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- 1. On Parallel Transport Anamitra Palit.anamitra@gmail.com Part I Let 𝐴 𝜇be a tensordefinedwithrespecttosome transformation𝑥̅ 𝜇 ↔ 𝑥 𝑗: 𝑥̅ 𝜇 = 𝑥̅ 𝜇( 𝑥 𝑗);𝑖, 𝑗 = 0,1,2,3zerodenotingthe time component 𝐴̅ 𝜇 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝐴 𝑗 (1) Differentiatingwithrespecttopropertime we obtain: 𝑑𝐴̅ 𝜇 𝑑𝜏 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝑑𝐴 𝑗 𝑑𝜏 + 𝑑 𝑑𝜏 ( 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 ) (2) Againwe have the standardresult 𝑑𝐴̅ 𝜇 𝑑𝜏 + Γ̅ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 𝐴̅ 𝛽 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 [ 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞](2.1) Or, 𝑑𝐴̅ 𝜇 𝑑𝜏 + Γ̅ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 𝐴̅ 𝛽 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝑑𝐴 𝑗 𝑑𝜏 + 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 (3) Using(2) with(3) we obtain 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝑑𝐴 𝑗 𝑑𝜏 + 𝑑 𝑑𝜏 ( 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 )+ Γ̅ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 𝐴̅ 𝛽 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝑑𝐴 𝑗 𝑑𝜏 + 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 Or, 𝑑 𝑑𝜏 ( 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 ) + Γ̅ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 𝐴̅ 𝛽 = + 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 (4) Nowletrus consideranarbitrarychoice : 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 = 𝑋 𝑗 andconsequently, 𝑑𝐴̅ 𝜇 𝑑𝜏 + Γ̅ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 𝐴̅ 𝛽 = 𝑌 𝜇 Usingthis choice with(4) we obtain 𝑑 𝑑𝜏 ( 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 ) − 𝑑𝐴̅ 𝜇 𝑑𝜏 + 𝑌 𝜇 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 [− 𝑑𝐴 𝑗 𝑑𝜏 + 𝑋 𝑗] 𝑑 𝑑𝜏 ( 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 ) − 𝑑𝐴̅ 𝜇 𝑑𝜏 + 𝑌 𝜇 = − 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝑑𝐴 𝑗 𝑑𝜏 + 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝑋 𝑗
- 2. But 𝑌 𝜇 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝑋 𝑗Or, 𝑑𝐴̅ 𝜇 𝑑𝜏 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝑑𝐴 𝑗 𝑑𝜏 + 𝑑 𝑑𝜏 ( 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 ) (5) To note that 𝑑𝐴 𝑗 𝑑𝜏 + 𝑑 𝑑𝜏 ( 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 ) = 𝑑𝑥 𝛽 𝑑𝜏 ∇ 𝛽 𝐴𝑗 = 𝑈 𝛽∇ 𝛽 𝐴 𝑗 Equation(5) is the same as (2) or (1) withsuitable initial conditions..From(2) youcanarrive at (2.1) or (4). 𝐼𝑡𝑖𝑠 𝑛𝑜𝑡 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦 𝑓𝑜𝑟 𝑋 𝑗[andhence for 𝑌 𝜇 to be the null tensor[rankone].Itmightappearthat relation(2.1) will holdforanyarbitrarytransformationwithnon-singular determinantof the Jacobian. But thisisnot true since the properspeedsontwosidesof (2.1) are relatedtothe transformation elements.Properspeeds 𝑑𝑥 𝑝 𝑑𝜏 and 𝑑𝑥̅ 𝛼 𝑑𝜏 are connectedwiththe transformationitself ,itselementsbeing 𝜕𝑥̅ 𝛼 𝜕𝑥 𝑝 For parallel transportwe specificallyuse 𝑋 𝑗 = 0andconsequrently 𝑌 𝜇 = 0 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 = 𝑋 𝑗 We find 𝑓(𝑡, 𝑥, 𝑦, 𝑧) and 𝑔(𝑡, 𝑥, 𝑦, 𝑧) suchthat 𝑑( 𝑓𝐴 𝑗) 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 ( 𝑓𝐴 𝑞) = 0 𝑑( 𝑔𝐴̅ 𝜇) 𝑑𝜏 + Γ̅ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 ( 𝑔𝐴̅ 𝛽) = 0 𝑓 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝑓𝐴 𝑞 + 𝐴 𝑗 𝑑𝑓 𝑑𝜏 = 0 𝑓 ( 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞)+ 𝐴 𝑗 𝑑𝑓 𝑑𝜏 = 0 ∇ 𝛽 𝐴 𝑗 is rank twotensorof mixedorderwhile 𝑈 𝛽∇ 𝛽 𝐴 𝑗 isarank one tensor: 𝛽 disappearson contraction.’ 𝑗’isthe survivingindex[uncontractedindex] An important property of parallel transport ispreservationof dot product as we move along the concernedworld line : this is possible onlywhen 𝑿𝒋 = 𝟎 and 𝒀 𝝁 = 𝟎. Preservationof dot product is also a salientattribute for coordinate transformations. This goesin favor of identifyingparallel transport product as a transformation
- 3. 𝑓𝑋 𝑗 + 𝐴 𝑗 𝑑𝑓 𝑑𝜏 = 0 Similarly 𝑔𝑌 𝜇 + 𝐴̅ 𝑗 𝑑𝑔 𝑑𝜏 = 0 We are upgradingarbitraryvectorfieldstoonesthathave zero covariantderivativesalongcurvesunder investigationbythe use of scale factorslike f and g[foreachcomponentwe have a suitable scale factor] PointstoObserve 1) 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 = 0 𝑎𝑛𝑑 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑙𝑦 𝑑𝐴̅ 𝜇 𝑑𝜏 + Γ̅ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 𝐴̅ 𝛽 = 0 are parallel transportequations As we move alonga curve[worldline]fromone segmenttothe next 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝑔𝑒𝑡𝑠 𝑓𝑖𝑥𝑒𝑑 𝑢𝑝 .I can determine the tensorinthe nextsegment onlyif Ihave the knowledge of 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 .Informationof thistransformation rule inrelationto 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 isembeddedinthe properspeedsoneitherside of (2.1).[See Special note] 2) Covariantderivative=0impliesdirectionalderivative iszero. Duringdifferentiationwe have toevaluate quantitieslike 𝑑𝐴 𝑗 = 𝐴 𝑗( 𝜏 + Δ𝜏) − 𝐴 𝑗(𝜏).The vector 𝐴 𝑗( 𝜏 + Δ𝜏) has to be parallel translatedoveraninfinitesimal distancesothatthe initial pointsof 𝐴 𝑗( 𝜏 + Δ𝜏) and 𝐴 𝑗(𝜏)coincide.Duringparallel translationthe componentschange:thisistrue evenin flatspace time contextif the systemisnotrectangular.The Christoffel symbolstake care of the changes incomponentsdue toparallel translation[infinitesimal].Now 𝑑𝐴̅ 𝜇 𝑑𝜏 + Γ̅ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 𝐴̅ 𝛽 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 [ 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞] and 𝜕𝐴̅ 𝜇 𝜕𝑥̅ 𝛼 + Γ̅ 𝜇 𝛼𝛽 𝐴̅ 𝛽 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 [ 𝜕𝐴 𝑗 𝜕𝑥 𝑝 + Γ 𝑗 𝑝𝑞 𝐴 𝑞] are derivable fromeach otherbut 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 ≠ 𝜕𝐴 𝑗 𝜕𝑥 𝑝 + Γ 𝑗 𝑝𝑞 𝐴 𝑞 and 𝑑𝐴̅ 𝜇 𝑑𝜏 + Γ̅ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 𝐴̅ 𝛽 ≠ 𝜕𝐴̅ 𝜇 𝜕𝑥̅ 𝛼 + Γ̅ 𝜇 𝛼𝛽 𝐴̅ 𝛽 Properspeedof relatingtothe transformationitself isinvolvedin(2.1):a conspicuouspointtotake not of carefully. WhenI move throughdifferentpointsof acurve usingtransformingthe vectorthrougha chainof movingreference fames the validformulaisof course 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 = 0,our usual parallel transportformula 𝑑𝐴 𝑗 𝑑𝜏 = −Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 impliesthanlocally 𝐴 𝑗 changessimilarlyinrelationtocurvature elements[determinedbyChristoffelsymbols] andlocal properspeed If Γ 𝑗 𝑝𝑞 = 0
- 4. 𝑑𝐴 𝑗 𝑑𝜏 = 0 𝐴 𝑗 = 𝐶 𝑗[local constant] 𝐴 𝑗 𝐴 𝑘 = 𝐶 𝑗 𝐶 𝑗 Or, 𝐴 𝑗 = 𝐶𝐴 𝑘 is a scalingeffecttypical of lineartransformationsThe constantCis independentof proiper time If you lookat Lorentztransformationsthe original andtransformedsidesare notdependentonproper time considerations Importantto keepinmindthat 𝑑𝐴 𝑗 𝑑𝜏 = −Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 considersaparticulartype of covariantderivative whichiszero.In general itshouldnobe zero For flatspac e time youhave 𝑑𝐴 𝑗 𝑑𝜏 = 0insteadof 𝑑𝐴 𝑗 𝑑𝜏 = −Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞 Part II In transformationthe same objectisviewedindifferentframes.The normremainsinvariant.Since the objectremainsthe same, the transformation rule definesparallelismwithrespecttto differentframes. In parallel transportwe move a4-vectorA alonga curve parameterizedby 𝜏 accodingtothe equation: 𝑑𝐴 𝜇 𝑑𝜏 = −Γ 𝜇 𝛼𝛽 𝑑𝑥 𝛼 𝑑𝜏 𝐴 𝛽 (A) 𝑑𝐴 𝜇 𝑑𝜏 + Γ 𝜇 𝛼𝛽 𝑑𝑥 𝛼 𝑑𝜏 𝐴 𝛽 = 0 𝑑(𝑛𝐴 𝜇) 𝑑𝜏 + Γ 𝜇 𝛼𝛽 𝑑𝑥 𝛼 𝑑𝜏 𝑛𝐴 𝛽 = 0 (A’) N:constant We are movingthe vector 𝐴 𝜇 alonga curve 𝑥 𝛼 = 𝑥 𝛼(𝜏) We assume thatsome transformation 𝑥̅ 𝑗 = 𝑥̅ 𝑗( 𝑥 𝛼)aswe move alongthe curve from one segmentto another
- 5. Let forsome four vectorB we have 𝐵̅ 𝜇 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 𝐵 𝜇 as you move alongthe curve, 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 indicatingthe specifictransformationelements/rule [ 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 representthe appropriate transformationelementsforpassingfromone segmentof the curve to another] ]: Since B isfour vector We denote: 𝑑𝐵 𝜇 𝑑𝜏 + Γ 𝜇 𝛼𝛽 𝑑𝑥 𝛼 𝑑𝜏 𝐵 𝛽 = 𝑓 𝜇 (B) 𝑓 𝜇:fourvector component NowfromA’ and B we have, 𝑑( 𝐵 𝜇−𝑛𝐴 𝜇) 𝑑𝜏 + Γ 𝜇 𝛼𝛽 𝑑𝑥 𝛼 𝑑𝜏 ( 𝐵 𝛽 − 𝑛𝐴 𝜇) = 𝑓 𝜇 (C) Since n can be variedat random, 𝐵 𝛽 = 𝑛𝐴 𝛽 and 𝑓 𝜇 = 0 seemtobe readilyrecognizable solutionsof C It wouldbe appropriate tosolve A toobtainparallel transportresult [ 𝑑𝐴̅ 𝜇 𝑑𝜏 + Γ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 𝐴̅ 𝛽] = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 [ 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞](D) The above relationissatisfiedirrespective of the nature of transformationif A isa tensor. As I move the vectoralonghe curve I have to make sure thatI alwayshave a tensorad I proceedwith such movement Successof relation(D) will insome sense indicate that 𝐴 𝜇 isa vector If 𝑑𝐴 𝜇 𝑑𝜏 + Γ 𝜇 𝛼𝛽 𝑑𝑥 𝛼 𝑑𝜏 𝐴 𝛽 = 0 for everymovement: 𝐵 𝛽 = 𝑛𝐴 𝛽 and 𝑓 𝜇 = 0 ThenrelationDis automaticallyvalidforanyarbitrarytransferindicatingvisa 4-vector. Earlierwe have shownthat the appropriate transformationwasbeingfollowed: 𝑑𝐴 𝜇 𝑑𝜏 + Γ 𝜇 𝛼𝛽 𝑑𝑥 𝛼 𝑑𝜏 𝐴 𝛽 beingplausible .If youstart witha tensor,the transformationshave tobe specified.These transformationcanbe rbitrary. We choose suchtransformationthatwill transformitforframes alongthe curve.RelationDwill automaticallybe successfulevenwithnozero: 𝑑𝐴 𝜇 𝑑𝜏 + Γ 𝜇 𝛼𝛽 𝑑𝑥 𝛼 𝑑𝜏 𝐴 𝛽:But we will nothave 𝐵 𝛽 = 𝑛𝐴 𝛽 and 𝑓 𝜇 = 0 if 𝑑𝐴 𝜇 𝑑𝜏 + Γ 𝜇 𝛼𝛽 𝑑𝑥 𝛼 𝑑𝜏 𝐴 𝛽 ≠ 0 Special Note:
- 6. We recall: 𝑑𝐴̅ 𝜇 𝑑𝜏 + Γ̅ 𝜇 𝛼𝛽 𝑑𝑥̅ 𝛼 𝑑𝜏 𝐴̅ 𝛽 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 [ 𝑑𝐴 𝑗 𝑑𝜏 + Γ 𝑗 𝑝𝑞 𝑑𝑥 𝑝 𝑑𝜏 𝐴 𝑞](2.1) Again: Takingthe correspondence: 𝑥̅ 𝜇 = 𝑥̅ 𝜇( 𝑥𝑖)fromthe relations 𝑥 𝑗 = 𝑥 𝑗(𝜏)and 𝑥̅ 𝜇 = 𝑥̅ 𝜇(𝜏)inthe two segments we have , 𝑑𝑥̅ 𝜇 = 𝜕𝑥̅ 𝜇 𝜕 𝑥 𝑖 𝑑𝑥 𝑖 (A) 𝑑𝑥̅ 𝜇 𝑑𝑥 𝑘 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑖 𝑑𝑥 𝑖 𝑑𝑥 𝑘 (B) 𝑑𝑥̅ 𝜇 and 𝑑𝑥 𝑘 have beenconsideredalongthe twosegmentsof the curve andhence the d operator The quantities 𝑑𝑥 𝑖 𝑑𝑥 𝑘 and 𝑑𝑥̅ 𝜇 𝑑𝑥 𝑘 will be determinedbythe shape of the curves/worldlines: 𝑥 𝑗 = 𝑥 𝑗(𝜏)and 𝑥̅ 𝜇 = 𝑥̅ 𝜇(𝜏) 𝑑𝑥̅ 𝜇 𝑑𝑥 𝑘 = 𝑑𝑥̅ 𝜇 𝑑𝜏 𝑑𝑥 𝑘 𝑑𝜏 [We have to maintain 𝑑𝜏 same inthe numeratorand inthe denominatortopreserve the manifold picture] We have sixteenequationsof the type (B) andsixteenunknowns 𝜕𝑥̅ 𝜇 𝜕 𝑥 𝑖 tobe solvedoutif the proper speedsonthe twosectionsof the curve are knownto us The properspeedsleadusto the transformationelements 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑗 Transformationrule isembeddedinthe properspeedsoneitherside whichare knowntous Steps 1) From 𝑑𝑥̅ 𝜇 𝑑𝜏 and 𝑑𝑥 𝑘 𝑑𝜏 we find 𝑑𝑥̅ 𝜇 𝑑𝑥 𝑘 2) 𝑑𝑥 𝑖 𝑑𝑥 𝑘 is knownfromfromthe curve to be transformed 3) We solve 𝑑𝑥̅ 𝜇 𝑑𝑥 𝑘 = 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑖 𝑑𝑥 𝑖 𝑑𝑥 𝑘 tofind 𝜕𝑥̅ 𝜇 𝜕 𝑥 𝑖 Let usrecall the metricfor a moment
- 7. 𝑑𝜏2 = 𝑔𝑡𝑡 𝑑𝑡2 − 𝑔 𝑥𝑥 𝑑𝑡2 − 𝑔 𝑦𝑦 𝑑𝑡2 − 𝑔 𝑧𝑧 𝑑𝑡2 1 = 𝑔 𝑡𝑡 ( 𝑑𝑡 𝑑𝜏 ) 2 − 𝑔 𝑥𝑥( 𝑑𝑥 𝑑𝜏 ) 2 − 𝑔 𝑦𝑦 ( 𝑑𝑦 𝑑𝜏 ) 2 − 𝑔 𝑧𝑧 ( 𝑑𝑧 𝑑𝜏 ) 2 The Christoffel symbolsare functionsof the metriccoefficientsandtheirderivatives We mightconsiderthe metricfunctionsasfunctionsof the Christoffel symbolsanduse such substitutioninthe above metric. The quadruplet ( 𝑑𝑡 𝑑𝜏 , 𝑑𝑥 𝑑𝜏 , 𝑑𝑦 𝑑𝜏 , 𝑑𝑧 𝑑𝜏 )isconstrainedbythe metricandhence by the Christoffel symbols. Thisclearlyimpliesfromthe earlierpartof the special note thatthe transformationelements 𝜕𝑥̅ 𝜇 𝜕𝑥 𝑖 dependonthe Christofellsymbolsandconsequentlyonthe curvature tensor Again, We multiplybothsidesof the metricbythe constant n2 𝑛2 = 𝑔𝑡𝑡 ( 𝑛 𝑑𝑡 𝑑𝜏 ) 2 − 𝑔 𝑥𝑥 ( 𝑛 𝑑𝑥 𝑑𝜏 ) 2 − 𝑔 𝑦𝑦 ( 𝑛 𝑑𝑦 𝑑𝜏 ) 2 − 𝑔 𝑧𝑧 ( 𝑛 𝑑𝑧 𝑑𝜏 ) 2 Againwe obtain 1 = 𝑔 𝑡𝑡 ( 𝑑𝑡 𝑑𝜏 ) 2 − 𝑔 𝑥𝑥( 𝑑𝑥 𝑑𝜏 ) 2 − 𝑔 𝑦𝑦 ( 𝑑𝑦 𝑑𝜏 ) 2 − 𝑔 𝑧𝑧 ( 𝑑𝑧 𝑑𝜏 ) 2 That is , foreach fourdirectionwe have aunique quadruplet ( 𝑑𝑡 𝑑𝜏 , 𝑑𝑥 𝑑𝜏 , 𝑑𝑦 𝑑𝜏 , 𝑑𝑧 𝑑𝜏 ). [Will be continued]