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Decoding the Graph-Wall-Tome Connection (view source)
Revision as of 16:59, 1 November 2020
, 16:59, 1 November 2020no edit summary
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<div style="font-weight:bold;line-height:1.6;">What are $$R_{\mu v}$$ and $$R$$ geometrically?</div> | <div style="font-weight:bold;line-height:1.6;">What are $$R_{\mu v}$$ and $$R$$ geometrically?</div> | ||
<div class="mw-collapsible-content"> | <div class="mw-collapsible-content"> | ||
[https://www.youtube.com/watch?v=UfThVvBWZxM&t=12m6s Explanation of $$R$$] | [https://www.youtube.com/watch?v=UfThVvBWZxM&t=12m6s Explanation of $$R$$] | ||
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[[Further thoughts on the meaning of R]] | [[Further thoughts on the meaning of R]] | ||
</div></div> | |||
<div class="toccolours mw-collapsible mw-collapsed" style="width:400px; overflow:auto;"> | |||
<div style="font-weight:bold;line-height:1.6;">Further thoughts on the meaning of R</div> | |||
<div class="mw-collapsible-content"> | |||
===== Computing length in non-orthogonal bases ===== | |||
First, just describing the length of a vector on a curved space is hard. It is given by: | |||
$$Length^{squared} = g_{11}dX^{1}dX^{1} + g_{12}dX^{1}dX^{2} + g_{21}dX^{2}dX^{1} + g_{22}dX^{2}dX^{2}$$ | |||
Some notes: | |||
* This is not Pythagorean theorem, because $$dX^{1}$$ and $$dX^{2}$$ are not perpendicular. | |||
* Instead, looks like a formula to get the diagonal from two opposite vertices in a parallelogram. | |||
* 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] | |||
===== 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] | |||
Open questions: | |||
* Why a parallelogram? | |||
* How to properly overlay the parallelogram onto the 3d manifold, in order to do the parallel transport? | |||
* How does this relate to the length computation above? | |||
===== Putting it all together ===== | |||
Now, moving to 4D, we can compute $$R_{\mu v}$$ as: | |||
$$R_{00} = R^{0}_{000} + R^{1}_{010} + R^{2}_{020} + R^{3}_{030}$$ | |||
$$R_{10} = R^{0}_{100} + R^{1}_{110} + R^{2}_{120} + R^{3}_{130}$$ | |||
$$R_{01} = R^{0}_{001} + R^{1}_{011} + R^{2}_{021} + R^{3}_{030}$$ | |||
etc. | |||
Indexing i over all 4 component vectors / dimensions, we get: | |||
$$R_{00} = \Sigma_{i} R^{i}_{0i0}$$ | |||
$$R_{10} = \Sigma_{i} R^{i}_{1i0}$$ | |||
$$R_{01} = \Sigma_{i} R^{i}_{0i1}$$ | |||
etc. | |||
Summarizing on $$\mu$$, we get: | |||
$$R_{\mu 0} = \Sigma_{i} R^{i}_{\mu i0}$$ | |||
$$R_{\mu 1} = \Sigma_{i} R^{i}_{\mu i1}$$ | |||
etc | |||
Summarizing on $$v$$, we get: | |||
$$R_{\mu v} = \Sigma_{i} R^{i}_{\mu iv}$$ | |||
Open questions: | |||
* If we hadn't moved from 3D to 4D, what would this all have looked like? | |||
* What does this have to do with the parallelogram? | |||
* Why are there two indices? | |||
</div></div> | </div></div> | ||