Decoding the Graph-Wall-Tome Connection: Difference between revisions

Decoding the Graph-Wall-Tome Connection (view source)

Revision as of 16:59, 1 November 2020

2,836 bytes added , 1 November 2020

no edit summary

Aardvark

Bureaucrats, Interface administrators, Suppressors, Administrators, Widget editors

6,480

edits

@@ Line 21: / Line 21: @@
 <div style="font-weight:bold;line-height:1.6;">What are $$R_{\mu v}$$ and $$R$$ geometrically?</div>
 <div class="mw-collapsible-content">
 [https://www.youtube.com/watch?v=UfThVvBWZxM&t=12m6s Explanation of $$R$$]
@@ Line 34: / Line 33: @@
 [[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>