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author | bryan newbold <bnewbold@snark.mit.edu> | 2008-07-29 03:33:50 -0400 |
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committer | bryan newbold <bnewbold@snark.mit.edu> | 2008-07-29 03:33:50 -0400 |

commit | ee2b54548a21ff05fe520b4b774a9b7ab7e13b39 (patch) | |

tree | ff0b9bdba9d207c77908d963f1e3ae7670ec9f92 /math | |

parent | 8c62b232273c58cf9cd6a939ad61b05a9725ebce (diff) | |

download | knowledge-ee2b54548a21ff05fe520b4b774a9b7ab7e13b39.tar.gz knowledge-ee2b54548a21ff05fe520b4b774a9b7ab7e13b39.zip |

progress? ...

Diffstat (limited to 'math')

-rw-r--r-- | math/tensors | 39 |

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 |