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Physics
Special and General Relativity
Parallel Transport of a Tensor: Understand Equation
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[QUOTE="AndersF, post: 6532748, member: 677147"] [B]TL;DR Summary:[/B] Understanding the equation for a tensor to be parallel-transported According to my book, the equation that should meet a vector ##\mathbf{v}=v^i\mathbf{e}_i## in order to be parallel-transported in a manifold is: ##v_{, j}^{i}+v^{k} \Gamma_{k j}^{i}=0## Where ##v_{, j}^i## stands for ##\partial{v^i}{\partial y^j}##, that is, the partial derivative of the component ##v^i## of ##\mathbf{v}## with respect to the general coordinate ##y^j##. I see that there is a sum in ##k## form 1 to ##n##, and that this equation must be meet for all ##i,j=1,2,...,n##, being ##n## the dimenssion of the manifold. However, I find it difficult to understand how to read this formula describing the condition for parallel transport of a tensor: ##T_{j_{1} j_{2} \ldots j_{r}, k}^{i_{1} i_{2} \ldots i_{s}}+\sum_{m=1}^{s} T_{j_{1} j_{2} \ldots j_{r}}^{i_{1} i_{2} \ldots p_{m} \ldots i_{s}} \Gamma_{p_{m} k}^{i_{m}}-\sum_{n=1}^{r} T_{j_{1} j_{2} \ldots q_{n} \ldots j_{r}}^{i_{1} i_{2} \ldots i_{s}} \Gamma_{j_{n} k}^{q_{n}}=0## (My theory is that whoever wrote that formula probably did so to engage in a competition of convoluted mathematical notations... :confused: ) Could somebody please help me understand it how should be read? For example, how would it apply for a tensor of order three ##T^{a,b}_{\alpha,\beta}##? [/QUOTE]
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Parallel Transport of a Tensor: Understand Equation
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