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authorbryan newbold <bnewbold@snark.mit.edu>2008-07-29 03:33:50 -0400
committerbryan newbold <bnewbold@snark.mit.edu>2008-07-29 03:33:50 -0400
commitee2b54548a21ff05fe520b4b774a9b7ab7e13b39 (patch)
treeff0b9bdba9d207c77908d963f1e3ae7670ec9f92 /math
parent8c62b232273c58cf9cd6a939ad61b05a9725ebce (diff)
downloadknowledge-ee2b54548a21ff05fe520b4b774a9b7ab7e13b39.tar.gz
knowledge-ee2b54548a21ff05fe520b4b774a9b7ab7e13b39.zip
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Diffstat (limited to 'math')
-rw-r--r--math/tensors39
1 files changed, 20 insertions, 19 deletions
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