Bianchi Identities [The Physics Travel Guide]

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--- advanced_tools:bianchi_identities [2018/05/03 11:53]
jakobadmin [Intuitive]
+++ advanced_tools:bianchi_identities [2019/01/16 14:35] (current)
jakobadmin [Abstract]
@@ Line 5: / Line 5: @@
 <tabbox Intuitive>
-Intuitively, Bianchi identities state that "the boundary of a boundary is zero".
+Intuitively, Bianchi identities state that "//the boundary of a boundary is zero//".
+For example, let's consider a disk: O
+The disk has a boundary, which is a circle. A circle has no boundary.
+Next, let's consider a line segment: |
+It is one-dimensional like the circle but has a boundary: the two endpoints. But then again, this boundary (=the two endpoints) don't have a boundary.
 <tabbox Concrete>
+For an extremely illuminating discussion see
+  * [[https://link.springer.com/article/10.1007%2FBF01882731|The Boundary of a Boundary Principle - a unified approach]] by Arkady Kheyfets.
+In addition, good discussion can be found in
   * See chapter 15 in "Gravitation" by Misner Thorne and Wheeler and also
   * page 253 in Gauge fields, knots, and gravity by J. Baez
+----
-<tabbox Abstract>
-<note tip>
+Bianchi identities express the fact that the boundary of a boundary is always zero. Mathematically this follows by applying Stoke's theorem twice. This is discussed explicitly in the book No-Nonsense Electrodynamics by Schwichtenberg
-The motto in this section is: //the higher the level of abstraction, the better//.
-</note>
+<tabbox Abstract>
+In general relativity, the Bianchi identity
+$$ \nabla R = \nabla \nabla \theta =0  $$
+roughly says "that the sum over a closed two-dimensional surface of rotations induced by Riemannian curvature is equal to zero. [...] Geometrically this means that the density of the moment of rotation induced by Riemannian curvature is equal to zero automatically." ([[https://link.springer.com/article/10.1007%2FBF01882731|Source]])
 <tabbox Why is it interesting?>
 <blockquote>

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