math/tensors

Tensors, Differential Geometry, Manifolds

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(ζ)), tangent vectors ($\delta P / \delta \zeta \equiv \lim_{\Delta \zeta \rightarrow 0} \frac{ vector{ Q(\zeta+\delta \zeta) - Q(\zeta) } }{\delta \zeta}$)

Coordinates: χα(P), where α = 0, 1, 2, 3; Q(χ0,χ1,...) 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 ω, ϕ:

$\vector{e_{\phi}} = \left( \frac{\partial Q(\phi, \theta)}{\partial \phi}\right)_{\theta}$

Components of a vector: $$\vector{A} = \frac{\partial P}{\partial \chi^\alpha }$$

Directional Derivatives: consider a scalar function defined on a manifold Ψ(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 $\bold{T}$ 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.

ΔP ⋅ ΔP ≡ ΔP2 ≡ (lengthofΔP)2A ⋅ B = 1/4[(A+B)2−(AB)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

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.