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Vierbein (Frame) Fields
-----------------------
The action for a point particle is:
S = -mc∫ √(-g_{μν}(∂x^{μ}/∂τ)(∂x_{μ}/∂τ))dτ
We postulate that there is another action that is equivalent to this.
S' = (1/2)∫ (g_{μν}e^{-1}(∂x^{μ}/∂τ)(∂x_{μ}/∂τ) - em^{2}) dτ
This form of the action makes it look as if we have coupled the
worldline theory to 1D gravity, with the field e(τ). The equation
of motion for e is:
e^{2} = g_{μν}(∂x^{μ}/∂τ)(∂x_{μ}/∂τ)/m^{2}
If we plug this back into S' we get:
S' = (1/2)∫ (g_{μν}√(m^{2}/g_{μν}(∂x^{μ}/∂τ)(∂x_{μ}/∂τ))(∂x^{μ}/∂τ)(∂x_{μ}/∂τ)
- ((∂x^{μ}/∂τ)(∂x_{μ}/∂τ)/m^{2})m^{2}) dτ
= m∫ (√(g_{μν}(∂x_{μ}/∂τ))(∂x^{μ}/∂τ)) - g_{μν}(∂x^{μ}/∂τ)(∂x_{μ}/∂τ) dτ
= -m∫ (√(g_{μν}(∂x_{μ}/∂τ))(∂x^{μ}/∂τ)) dτ = S
e in this case is referrred to as an EINBEIN FIELD. This idea can
be extended to more dimensions. For 2 indeces we have a ZWEIBEIN
FIELD, for 3 indeces we have a DREIBEIN FIELD and for 4 indeces we
have a VIERBEIN FIELD. The latter case is also referred to as a
TETRAD.
The vierbein field can be interpreted in the following manner.
Consider a point, P, on a curved manifold. P is defined in the
coordinate frame, x, and there is a metric g_{μν}(x) associated with
it. However, it is also possible to construct a Lorentz frame
with coordinates, ξ^{a}(x), in tangent space that is locally flat at
P. A vector on the manifold at P, can now be mapped into this new
frame and expanded as a linear combination of orthonormal basis
vectors on a Lorentz manifold.. The vierbein field, therefore,
has two kinds of indices: those that label the spacetime
coordinates and those that label the frame coordinates.
The vierbein field theory is the most natural way to represent
a relativistic quantum field theory in curved space and can be
regarded as a gauge field for gravity. While it plays the role
of a gauge field it not behave in the same way as the vector
potential field.
We can view the vierbein e_{μ}^{a} as the transformation matrix
between arbitrary coordinates, x, and inertial coordinates, ξ.
dξ^{m} = (∂ξ^{m}/∂x^{μ})dx^{μ} ≡ e_{μ}^{m}(x)dx^{μ}
and the inverse,
dx^{μ} = (∂x^{μ}/∂ξ^{m})dξ^{m} ≡ e_{m}^{μ}(x)dξ^{m}
Therefore, since the inner product has to be the same in both
frames:
dξ^{m} = e_{μ}^{m}(x)dx^{μ} = e_{μ}^{m}(x)e_{m}^{μ}(x)dξ^{m}
ds^{2} = η_{mn}dξ^{m}dξ^{n} = η_{mn}e_{μ}^{m}e_{ν}^{n}dx^{μ}dx^{ν}
The vierbeins satisfy the following relationships:
e^{μ}_{a}e^{a}_{ν} = δ^{μ}_{ν} = I
e^{a}_{μ}e^{μ}_{b} = δ^{a}_{b} = I
g_{μν} = e^{μ}_{a}e^{ν}_{b}η^{ab} and g^{μν} = e_{μ}^{a}e_{ν}^{b}η_{ab}
η_{ab} = e^{μ}_{a}e^{ν}_{b}g_{μν} and η^{ab} = e^{a}_{μ}e^{b}_{ν}g^{μν}
det[g^{μν}] = det[e^{μ}_{a}e^{ν}_{b}η^{ab}]
det[g^{μν}] = det[e^{μ}_{a}]det[e^{ν}_{b}]det[η^{ab}]
det[g^{μν}] = (det[e^{μ}_{a}])^{2}(-1)
√(-det[g^{μν}]) = det[e^{μ}_{a}]