From 0b1a56db66d4dd7d8c19f1a1dc2aa5cd3f893227 Mon Sep 17 00:00:00 2001
From: bryan newbold <bnewbold@snark.mit.edu>
Date: Sun, 13 Jul 2008 02:00:24 -0400
Subject: spell checked some files; moved and added images; fixed some links;
 added warnings

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 math/tensors | 68 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
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 create mode 100644 math/tensors

(limited to 'math')

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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 (P), separation vectors (\Delta \vector P), curves ( Q(\zeta) ), tangent vectors ( \delta P / \delta \zeta \equiv \lim_{\Delta \zeta \rightarrow 0} \frac{ \vector{ Q(\zeta+\delta \zeta) - Q(\zeta) } }{\delta \zeta} )
+
+Coordinates: \Chi^\alpha (P), where \alpha = 0,1,2,3; Q(\Chi_0, \Chi_1, ...)
+    there is an isomorphism between points and coordinates
+
+Coordinate basis: \vector{e_\alpha} \equiv \left( \frac{\partial Q}{\partial \Chi^\alpha} \right)
+    for instance, on a sphere with angles \omega, \phi: 
+        \vector{e_\phi} = \left( \frac{\partial Q(\phi, \theta)}{\partial \phi}\right)_\theta
+
+Components of a vector:
+    \vector{A} = \frac{\partial P}{\partial \Chi^\alpha }
+
+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.
+
+\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}) 
+
+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 dot
+product to get a real.
+
+Define the metric tensor g(\vector{A}, \vector{B}) = \vector{A} \dot \vector{B}
+
+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 ]
+
+Tensor Product:
+    ????????????????
+
+Spacetime
+--------------
+
+Two types of vectors.
+
+Timelike: \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.
+
+
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