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diff --git a/math/tensors b/math/tensors new file mode 100644 index 0000000..8fda6a5 --- /dev/null +++ b/math/tensors @@ -0,0 +1,68 @@ +============================================ +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. + + |