Dale
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Let's stick to inertial frames. If you are going to be using non-inertial frames then the usual process that we already covered in great depth applies. If you use non-inertial frames then you are simply doing standard parallel transport and there is no conceptual difference between one non-inertial coordinate system and another.
Yes, exactly.Anamitra said:1)We take a chain of inertial frames from A to B along L1 wrt orthogonal bases whose axes are parallel to each other individually. This may be achieved by parallel transport of the axes.The basis at B is slided down to A by parallel transport to A along L2.
i) When a vector T[T^{0},T^{1}T^{2}T^{3} ] moves from A to B along L1 the components do not change.
ii) The two bases at A ate not identical. The same vector will have different components in the two bases at A.let A1 be the basis wrt L1 and A2 the basis wrt L2.
I may be misunderstanding what you are saying, but from how I read this statement that is already the case with the inertial coordinate system approach, by definition. Let's make this a little concrete. Suppose the vector at A is a timelike vector of magnitude 5, then we can choose an orthonormal basis at A such that T=(5,0,0,0). If we then parallel transport the bases and T from A to B along L1 and L2 we are guaranteed that at each point along the path (including B) the coordinates of T=(5,0,0,0) in the respective bases. But because the bases themselves are different, having the same coordinates does not imply that the vectors are the same.Anamitra said:4)Let the components of the tensor T at A be T^{0},T^{1}T^{2}T^{3} in the basis A1 and T^{'0},T^{'1}T^{'2}T^{'3} in the other[A2].WE choose a transformation in such a way that when we move from A to B along L2 the tensor at P has the components T^{0},T^{1}T^{2}T^{3} at P