From ee2b54548a21ff05fe520b4b774a9b7ab7e13b39 Mon Sep 17 00:00:00 2001
From: bryan newbold <bnewbold@snark.mit.edu>
Date: Tue, 29 Jul 2008 03:33:50 -0400
Subject: progress? ...

---
 math/tensors | 39 ++++++++++++++++++++-------------------
 1 file changed, 20 insertions(+), 19 deletions(-)

(limited to 'math')

diff --git a/math/tensors b/math/tensors
index 8fda6a5..e95a96a 100644
--- a/math/tensors
+++ b/math/tensors
@@ -23,46 +23,47 @@ Components of a vector:
 Directional Derivatives: consider a scalar function defined on a manifold \Psi(P)
     \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 T has a number of slots (say 3) and takes a vector in each slot and returns a real number. It is linear in vectors.
+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:
 
-\epic{T} ( \alpha \vector{A} + \beta \vector{B}, \vector{C}, \vector{D}) =
-   \alpha \epic{T} (\vector{A}, \vector{C}, \vector{D}) +
-   \beta \epic{T} (\vector{B}, \vector{C}, \vector{D}) 
+:m:`$$\bold{T} ( \alpha \vector{A} + \beta \vector{B}, \vector{C}, \vector{D}) =
+   \alpha \bold{T} (\vector{A}, \vector{C}, \vector{D}) +
+   \beta \bold{T} (\vector{B}, \vector{C}, \vector{D}) $$`
 
-The number of "slots" is the rank of the tensor.
+Even a regular vector is a tensor: pass it a second vector and take the 
+inner product (aka dot product) to get a real.
 
-Even a regular vector is a tensor: pass it a second vector and take the dot
-product to get a real.
+Define the **metric tensor** 
+:m:`$\bold{g}(\vector{A}, \vector{B}) = \vector{A} \dot \vector{B}$`. 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. 
 
-Define the metric tensor g(\vector{A}, \vector{B}) = \vector{A} \dot \vector{B}
+:m:`$$\Delta P \dot \Delta P \equiv \Delta P^2 \equiv (length of \Delta P)^2 A \dot B = 1/4[ (A+B)^2 - (A-B)^2 ]$$`
 
-Inner Product:
-    \Delta P \dot \Delta P \equiv \Delta P^2 \equiv (length of \Delta P)^2
-    A \dot 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
 
-Tensor Product:
-    ????????????????
+:m:`$$\vector{A} \crossop \vector{B} \crossop \vector{C} (\_ ,\_ ,\_)$$` 
+:m:`$$\vector{A} \crossop \vector{B} \crossop \vector{C} (\vector{E}, \vector{F}, \vector{G}) = ( \vector{A} \dot \vector{E})(\vector{B} \dot \vector{F})(\vecotr{C} \dot \vector{G}) $$` 
 
 Spacetime
 --------------
 
 Two types of vectors.
 
-Timelike: \vector{\Delta P}
+Timelike: :m:`$\vector{\Delta P}$`
     (\vector{\Delta P})^2 = -(\Delta \Tau)^2
 
 Spacelike: \vector{\Delta Q}
     (\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.
-
+delta, and timelikes squared come out negative, the time "up" basis must be  negative of the time "down" basis vector.
 
+lides.pdf
+bnewbold@snark$ xzg
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