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Last modified: January 16, 2020
One Forms
---------
Basis One-forms
---------------
A one-form is like a machine that when fed with
a vector spits out a number which depends linearly
on the input. We can define dxμ to be the basis
one-form which gives 1 when you input the basis
tangent vector, ∂μ, and 0 when you input ∂ν for
μ ≠ ν. Thus,
[dxμ](∂ν) = (∂/∂xν)(xμ) = ∂xμ/∂xν = δμν
This should not be interpreted as ∂[∂ν]/∂xμ but
rather as (dxμ) 'acting' on (∂ν).
The basis one-forms are the gradients of the
scalar coordinate functions (i.e. xμ = xμ(p)
where (U,p) is a local chart. Equivalently
we can say φ: U -> Rn where φ consists of n
real valued functions, xμ). We can show this
as follows:
∇ := (∂/∂xν)eν
∇xμ = (∂xμ/∂xν)eν
Where ∂xμ/∂xν are the components of the gradient
(a dual vector) and eν are the dual basis vectors.
Therefore,
∇xμ = δμνeν
= eμ
≡ dxμ
Transformation Laws
-------------------
vectors one-forms
------- ---------
∂μ -> ∂'μ = (∂xμ/∂x'μ)∂μ dxμ -> dx'μ = (∂x'μ/∂xμ)dxμ
Vμ -> V'μ = (∂x'μ/∂xμ)Vμ ωμ -> ω'μ = (∂xμ/∂x'μ)ωμ
V'x = (∂x'/∂x)Vx + (∂x'/∂y)Vy
V'y = (∂y'/∂x)Vx + (∂y'/∂y)Vy
- - - - - -
| V'x | = | ∂x'/∂x ∂x'/∂y || Vx |
| V'y | | ∂y'/∂x ∂y'/∂y || Vy |
- - - - - -
Vector components have a contravariant transformation
law.
∂x' = (∂x/∂x')∂x + (∂y/∂x')∂y
∂y' = (∂x/∂y')∂x + (∂y/∂y')∂y
- - - - - -
| ∂x' ∂y' | = | ∂x ∂y || ∂x/∂x' ∂x/∂y' |
- - - - | ∂y/∂x' ∂y/∂y' |
- -
Basis vectors have a covariant transformation law.
∂φ/∂x' = (∂x/∂x')(∂φ/∂x) + (∂y/∂x')(∂φ/∂y)
∂φ/∂y' = (∂x/∂y')(∂φ/∂x) + (∂y/∂y')(∂φ/∂y)
- - - - - -
| ∂φ/∂x' ∂φ/∂y' | = | ∂φ/∂x ∂φ/∂y || ∂x/∂x' ∂x/∂y' |
- - - - | ∂y/∂x' ∂y/∂y' |
- -
One-form components have a covariant transformation
law.
dx' = (∂x'/∂x)dx + (∂x'/∂y)dy
dy' = (∂y'/∂x)dx + (∂y'/∂y)dy
- - - - - -
| dx' | = | ∂x'/∂x ∂x'/∂y || dx |
| dy' | | ∂y'/∂x ∂y'/∂y || dy |
- - - - - -
Basis one-forms have a contravariant transformation
law.
One-forms
---------
One-forms are also known as covariant/dual vectors.
They are often written as ω:
df ≡ ω = ωμdxμ
Where,
ωμ = {∂f/∂x1,∂f/∂x2, ...}
and,
dxμ = {dx1,dx2, ...}
Therefore,
df = ω = (∂f/∂x1)dx1 + (∂f/∂x2)dx2 ...
= (∂f/∂xμ)dxμ
Example: Conversion one-form in polar coorinates to
one-form in xy plane.
x = rcosθ
y = rsinθ
dx = (∂x/∂r)dr + (∂x/∂θ)dθ
dy = (∂y/∂r)dr + (∂y/∂θ)dθ
dx = cosθdr - rsinθdθ
dy = sinθdr + rcosθdθ
Likewise,
dr = (∂r/∂x)dx + (∂r/∂y)dy
dθ = (∂θ/∂x)dx + (∂θ/∂y)dy
∂r/∂x = x/√(x2 + y2) = x/r = cosθ
∂r/∂y = y/√(x2 + y2) = y/r = sinθ
∂θ/∂x = -y/(x2 + y2) = -y/r2 = -sinθ/r
∂θ/∂y = x/(x2 + y2) = x/r2 = cosθ/r
dr = (x/r)dx + (y/r)dy
dθ = (-y/r2)dx + (x/r2)dy
The product of a one-form with a vector is given by:
v = v1(∂/∂x'1) + v2(∂/∂x'2) + ...
= v1∂1 + v2∂1 + ...
= Σvμ∂μ
μ
ω = ω1dx1 + ω2dx2 + ...
= Σωμdxμ
ν
vω = Σvμ∂μ(Σωνdxν)
μ ν
= Σvμωνδνμ
μν
= Σvμωμ ∈ R
μ
Example: Product of vector and one-form in polar
coordinates.
dx = cosθdr - rsinθdθ
dy = sinθdr + rcosθdθ
∂u/∂x = (∂u/∂r)(∂r/∂x) + (∂u/∂θ)(∂θ/∂x)
= cosθ(∂u/∂r) - (1/r)sinθ(∂u/∂θ)
∂u/∂y = (∂u/∂r)(∂r/∂y) + (∂u/∂θ)(∂θ/∂y)
= sinθ(∂u/∂r) + (1/r)cosθ(∂u/∂θ)
(cosθdr - rsinθdθ)(cosθ(∂u/∂r) - (1/r)sinθ(∂u/∂θ))
= cos2θdr(∂u/∂r) - (1/r)cosθsinθdr(∂u/∂θ)
- rsinθcosθdθ(∂u/∂r) + sin2θdθ(∂u/∂θ)
= 1
(sinθdr + rcosθdθ)(sinθ(∂u/∂r) + (1/r)cosθ(∂u/∂θ))
= sin2θdr(∂u/∂r) + (1/r)sinθcosθdr(∂u/∂θ)
- rcosθsinθdθ(∂u/∂r) + cos2θdθ(∂u/∂θ)
= 1
Exterior Derivative
-------------------
df(V) := Vf
= Vμ∂μf
= (∂f/∂xμ)Vμ
The reasoning for this can be found by considering
the directional derivative d/dλ = (dxμ/dλ)∂μ.
df[(dxμ/dλ)(∂μ)] = df[(dxμ/dλ)(∂/∂xμ)]
----------
^ = df[(d/dλ)]
|
V = (d/dλ)[f]
= df/dλ
In other words, df when given a vector returns
the directional derivative in that direction.
It is worth noting that f is a scalar function
(0-form). df is therefore a 1-form (dual vector).
Therefore, df acting on the tangent vector dxμ/dλ)(∂μ)
produces a scalar, df/dλ.
Example:
f(x,y) = x2y at point (1,1) in the direction (-1,1).
= 2xy + x2
= (2,1) at (1,1)
(-1,1) => (-√2/2,√2/2) after normalization
(2,1).(-√2/2,√2/2) = -√2/2
^ ^ ^
| | |
∂f/∂xμ dxμ/dλ df/dλ
= ∂μf = Vμ
In terms of the one-form we get:
df(V) = Vμ(∂f/∂xμ)
= (∂f/∂xμ)Vμ
= 2xy(-√2/2) + x2(√2/2)
= -√2/2 at (1,1) as before.
Infinitesimal Version of the Chain Rule
---------------------------------------
From before we had:
df(V) = (∂f/∂xμ)Vμ
Now,
dxμ(V) = dxμ(Vν∂ν)
= Vνdxμ(∂ν)
= Vνδμν
= Vμ
Therefore,
df(V) = (∂f/∂xμ)dxμ(V)
Or,
df = (∂f/∂xμ)dxμ
In the context of one-forms, dx = (∂x/∂xμ)dxμ
is the infinitesimal version of the chain rule.
We can see the relationship as follows:
df/dxμ = (∂f/∂g(xμ))(dg(xμ)/dxμ) where f = f(g(xμ))
Therefore, rearranging:
df/dg(xμ) = (∂f/∂xμ)(dxμ/dg(xμ))
multiply by dg(xμ) to get:
df = (∂f/∂xμ)dxμ
Strictly speaking when we talk about infinitesimal
displacements we should be specifying differential
forms in the equations and not dxμ. This is because
displacement is a scalar that can only be obtained
by contracting a one-form with a tangent vector.
For example, the line segment, ds2 = ημνdxμdxν,
should really be ds2 = ημνdxμdxν. However, most
references do not distinguish between 'dx', the
informal notion of an infinitesimal displacement,
and 'dxμ', the rigorous notion of a basis one-form
given by the gradient of a coordinate function.
Example:
(ds)2 = (dx)2 + (dy)2
= ((∂x/∂r)dr + (∂x/∂θ)dθ)2 + ((∂y/∂r)dr + (∂y/∂θ)dθ)2
= (cosθdr - rsinθdθ)2 + (sinθdr + rcosθdθ)2
= cos2θ(dr)2 + r2sin2θ(dθ)2 + sin2θ(dr)2 + r2cos2θ(dθ)2
= (dr)2 + r2(dθ)2