From f61026119df4700f69eb73e95620bc5928ca0fcb Mon Sep 17 00:00:00 2001
From: User <bnewbold@daemon.robocracy.org>
Date: Tue, 13 Oct 2009 02:52:09 +0000
Subject: Grand rename for gitit transfer

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+============================================
+Tensors, Differential Geometry, Manifolds
+============================================
+
+.. note:: Most of this content is based on a 2002 Caltech course taught by
+    Kip Thorn [PH237]_
+
+
+On a manifold, only "short" vectors exist. Longer vectors are in a space tangent to the manifold.
+
+There are points (:m:`$P$`), separation vectors (:m:`$\Delta \vector P$`),
+curves (:m:`$Q(\zeta)$`), tangent vectors (:m:`$\delta P / \delta \zeta \equiv
+\lim_{\Delta \zeta \rightarrow 0} \frac{ \vector{ Q(\zeta+\delta \zeta) -
+Q(\zeta) } }{\delta \zeta}$`)
+
+Coordinates: :m:`$\Chi^\alpha (P)$`, where :m:`$\alpha = 0,1,2,3$`;
+:m:`$Q(\Chi_0, \Chi_1, ...)$`
+    there is an isomorphism between points and coordinates
+
+Coordinate basis: :m:`$\vector{e_\alpha} \equiv \left( \frac{\partial
+Q}{\partial \Chi^\alpha} \right$`)
+    
+    for instance, on a sphere with angles :m:`$\omega, \phi$`: 
+    
+    :m:`$\vector{e_\phi} = \left( \frac{\partial Q(\phi, \theta)}{\partial \phi}\right)_\theta$`
+
+Components of a vector:
+
+    :m:`$\vector{A} = \frac{\partial P}{\partial \Chi^\alpha }$`
+
+Directional Derivatives: consider a scalar function defined on a manifold \Psi(P)
+    :m:`$\partial_\vector{A} \Psi = A^\alpha \frac{\partial \Psi}{\partial \Chi^\alpha}$`
+
+Mathematicians like to say that the coordinate bases are actually directional derivatives
+
+Tensors
+------------
+
+A **tensor** :m:`$\bold{T}$` has a number of slots (called it's **rank**), takes a vector in each slot, and returns a real number. It is linear in vectors; 
+as an example for a rank-3 tensor:
+
+:m:`$$\bold{T} ( \alpha \vector{A} + \beta \vector{B}, \vector{C}, \vector{D}) =
+\alpha \bold{T} (\vector{A}, \vector{C}, \vector{D}) + \beta \bold{T}
+(\vector{B}, \vector{C}, \vector{D}) $$`
+
+Even a regular vector is a tensor: pass it a second vector and take the 
+inner product (aka dot product) to get a real.
+
+Define the **metric tensor** 
+:m:`$\bold{g}(\vector{A}, \vector{B}) = \vector{A} \cdot \vector{B}$`. The 
+metric tensor is rank two and symetric (the vectors A and B could be swapped 
+without changing the scalar output value) and is the same as the inner product. 
+
+:m:`$$\Delta P \cdot \Delta P \equiv \Delta P^2 \equiv (length of \Delta P)^2 A \cdot B = 1/4[ (A+B)^2 - (A-B)^2 ]$$`
+
+Starting with individual vectors, we can construct tensors by taking the 
+product of their inner products with empty slots; for example
+
+:m:`$$\vector{A} \crossop \vector{B} \crossop \vector{C} (\_ ,\_ ,\_)$$` 
+:m:`$$\vector{A} \crossop \vector{B} \crossop \vector{C} (\vector{E}, \vector{F}, \vector{G}) = ( \vector{A} \cdot \vector{E})(\vector{B} \cdot \vector{F})(\vecotr{C} \cdot \vector{G}) $$` 
+
+Spacetime
+--------------
+
+Two types of vectors.
+
+Timelike: :m:`$\vector{\Delta P}$`
+    :m:`$(\vector{\Delta P})^2 = -(\Delta \Tau)^2$`
+
+Spacelike: :m:`$\vector{\Delta Q}$`
+    :m:`$(\vector{\Delta Q})^2 = +(\Delta S)^2$`
+
+Because product of "up" and "down" basis vectors must be a positive Kronecker 
+delta, and timelikes squared come out negative, the time "up" basis must be  negative of the time "down" basis vector.
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