Graph, Wall, Tome: Difference between revisions

879 bytes added ,  29 January 2020
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* If $$dX^{1}$$ and $$dX^{2}$$ are perpendicular, then $$g_{12}$$ and $$g_{21}$$ would be 0, and we would get $$Length^{squared} = g_{11}(dX^{1})^{2} + g_{22}(dX^{2})^{2}$$
* If $$dX^{1}$$ and $$dX^{2}$$ are perpendicular, then $$g_{12}$$ and $$g_{21}$$ would be 0, and we would get $$Length^{squared} = g_{11}(dX^{1})^{2} + g_{22}(dX^{2})^{2}$$
* See: [https://www.youtube.com/watch?v=UfThVvBWZxM&t=14m27s the video @ 14m27s]
* See: [https://www.youtube.com/watch?v=UfThVvBWZxM&t=14m27s the video @ 14m27s]
===== Computing vector rotation due to parallel transport =====
Then, they show parallel transport when following a parallelogram, but over a curved 3D manifold. To compute the vector rotation by components, they show:
$$dV^{1} = dX^{1}dX^{2} (V^{1}R^{1}_{112} + V^{2}R^{1}_{212} + V^{3}R^{1}_{312})$$
$$dV^{2} = dX^{1}dX^{2} (V^{1}R^{2}_{112} + V^{2}R^{2}_{212} + V^{3}R^{2}_{312})$$
$$dV^{3} = dX^{1}dX^{2} (V^{1}R^{3}_{112} + V^{2}R^{3}_{212} + V^{3}R^{3}_{312})$$
or, using $$i$$ to summarize across all 3 components (difference vectors):
$$dV^{i} = dX^{1}dX^{2} (V^{1}R^{i}_{112} + V^{2}R^{i}_{212} + V^{3}R^{i}_{312})$$
or , using $$j$$ to index over all 3 components (original vector):
$$dV^{i} = dX^{1}dX^{2} \Sigma_{j} [(V^{j}R^{i}_{j12}]$$
* See: [https://www.youtube.com/watch?v=UfThVvBWZxM&t=19m33s the video @ 19m33s]
===== Putting it all together =====


=== How do they relate? ===
=== How do they relate? ===
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