04. Covariant and Contravariant

If the bases of a coordinate system are NOT orthogonal to each other, it's not so easy to get a metric tensor.

If we have three linearly independent vectors {e⃗_1, e⃗_2, e⃗_3} that form a new base, so that not neccessarily e⃗_i∙e⃗_j=δ_{i,j}, then we can represent a (point in space) x like this:

x⃗=x^1⋅e⃗_1+x^2⋅e⃗_2+x^3⋅e⃗_3 contravariant coordinates

(x^1,x^2,x^3) are the contravariant coordinates of x. Note that the index is at the top (and does not mean exponentiation).

There is also another basis with different vectors \{e⃗^1, e⃗^2, e⃗^3\} where:

x⃗=x_1⋅e⃗^1+x_2⋅e⃗^2+x_3⋅e⃗^3 covariant

(x_1,x_2,x_3) are the covariant coordinates of x

x_1=x⃗∙e⃗_1 covariant coordinate

Scalar Product

If we multiply the two different representations of the point x with each other, we get the scalar product:

x⃗⋅x⃗=(x^1⋅e⃗_1+x^2⋅e⃗_2+x^3⋅e⃗_3)⋅(x_1⋅e⃗^1+x_2⋅e⃗^2+x_3⋅e⃗^3)
x⃗⋅x⃗=x_i⋅x^i
x^1=x⃗∙e⃗^1
g_{i,j}:=g_{j,i}=e⃗_i∙e⃗_j
x_i=g_{i,j}⋅x^j
g^{i,j}:=g^{j,i}=e⃗^i∙e⃗^j
x^i=g^{i,j}⋅x_j
g_{i,j}⋅g^{j,k}=δ_i^k

Inner Product

If we multiply two identical representations of the point x with each other, we get the inner product.

|x⃗|²=g_{i,j}⋅x^i⋅x^j=g^{i,j}⋅x_i⋅x_j

Author: Danny (remove the ".nospam" to send)

Last modification on: Sat, 04 May 2024 .