Tensors, Differential Geometry, Manifolds
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*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
($\delta P / \delta \zeta \equiv \lim_{\Delta \zeta \rightarrow 0} \frac{ vector{ Q(\zeta+\delta \zeta) - Q(\zeta) } }{\delta \zeta}$)
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
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A **tensor** $\bold{T}$ 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:
$$\bold{T} ( \alpha \vector{A} + \beta \vector{B}, \vector{C}, \vector{D}) =
\alpha \bold{T} (\vector{A}, \vector{C}, \vector{D}) + \beta \bold{T}
(\vector{B}, \vector{C}, \vector{D}) $$
Even a regular vector is a tensor: pass it a second vector and take the
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 symmetric (the vectors A and B could be swapped
without changing the scalar output value) and is the same as the inner product.
$$\Delta P \cdot \Delta P \equiv \Delta P^2 \equiv (length of \Delta P)^2 A \cdot B = 1/4[ (A+B)^2 - (A-B)^2 ]$$
Starting with individual vectors, we can construct tensors by taking the
product of their inner products with empty slots; for example
$$\vector{A} \crossop \vector{B} \crossop \vector{C} (\_ ,\_ ,\_)$$
$$\vector{A} \crossop \vector{B} \crossop \vector{C} (\vector{E}, \vector{F}, \vector{G}) = ( \vector{A} \cdot \vector{E})(\vector{B} \cdot \vector{F})(\vecotr{C} \cdot \vector{G}) $$
Spacetime
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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.
[PH237]: http://elmer.tapir.caltech.edu/ph237/