1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
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.
|
