From 0f704e6dfacb902fd0c81ae96a1fff2ad58f5a59 Mon Sep 17 00:00:00 2001 From: bnewbold Date: Sun, 24 Jan 2010 08:36:50 +0000 Subject: hmmmm, needs some work still --- math/tensors.page | 37 ++++++++++++++++++------------------- 1 file changed, 18 insertions(+), 19 deletions(-) (limited to 'math/tensors.page') diff --git a/math/tensors.page b/math/tensors.page index 7ea1848..feb9a01 100644 --- a/math/tensors.page +++ b/math/tensors.page @@ -1,33 +1,30 @@ Tensors, Differential Geometry, Manifolds ============================================ -Most of this content is based on a 2002 Caltech course taught by Kip Thorn (PH237). - +*References: 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 ($\delta P / \delta \zeta \equiv -\lim_{\Delta \zeta \rightarrow 0} \frac{ \vector{ Q(\zeta+\delta \zeta) - -Q(\zeta) } }{\delta \zeta}$) +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 +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: $\vector{e_\alpha} \equiv \left( \frac{\partial -Q}{\partial \Chi^\alpha} \right$) +Coordinate basis: +$$\vector{e_{\alpha}} \equiv \left( \frac{\partial Q}{\partial \chi^\alpha} \right$$ - for instance, on a sphere with angles $\omega, \phi$: +for instance, on a sphere with angles $\omega, \phi$: - $\vector{e_\phi} = \left( \frac{\partial Q(\phi, \theta)}{\partial \phi}\right)_\theta$ +$\vector{e_{\phi}} = \left( \frac{\partial Q(\phi, \theta)}{\partial \phi}\right)_{\theta}$ Components of a vector: +$$\vector{A} = \frac{\partial P}{\partial \chi^\alpha }$$ - $\vector{A} = \frac{\partial P}{\partial \Chi^\alpha }$ - -Directional Derivatives: consider a scalar function defined on a manifold \Psi(P) - $\partial_\vector{A} \Psi = A^\alpha \frac{\partial \Psi}{\partial \Chi^\alpha}$ +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 @@ -46,7 +43,7 @@ inner product (aka dot product) to get a real. Define the **metric tensor** $\bold{g}(\vector{A}, \vector{B}) = \vector{A} \cdot \vector{B}$. The -metric tensor is rank two and symetric (the vectors A and B could be swapped +metric tensor is rank two and symmetric (the vectors A and B could be swapped without changing the scalar output value) and is the same as the inner product. $$\Delta P \cdot \Delta P \equiv \Delta P^2 \equiv (length of \Delta P)^2 A \cdot B = 1/4[ (A+B)^2 - (A-B)^2 ]$$ @@ -63,10 +60,12 @@ Spacetime Two types of vectors. Timelike: $\vector{\Delta P}$ - $(\vector{\Delta P})^2 = -(\Delta \Tau)^2$ +: $(\vector{\Delta P})^2 = -(\Delta \Tau)^2$ Spacelike: $\vector{\Delta Q}$ - $(\vector{\Delta Q})^2 = +(\Delta S)^2$ +: $(\vector{\Delta Q})^2 = +(\Delta S)^2$ 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. + +[PH237]: http://elmer.tapir.caltech.edu/ph237/ -- cgit v1.2.3