Mathematical Principles of Theoretical Physics

(Rick Simeone) #1

154 CHAPTER 3. MATHEMATICAL FOUNDATIONS


3.4.5 Scalar potential theorem


In Theorem3.26, if the vector fieldAinDAis zero, and the first Betti numberβ 1 (M) = 0
forM, then we have the following scale potential theorem. This result is also important for
the gravitational field equations and the theory of dark matter and dark energy introduced in
Chapter 7.


Theorem 3.28(Scalar Potential Theorem).Assume that the first Betti number ofMis zero,
i.e.β 1 (M) = 0. Let F be a functional of Riemannian metrics. Then there is a scalar field
φ∈H^2 (M)such that the extremum points{gij}of F with divergence-free constraint satisfy
the equation


(3.4.46) (δF(gij))kl=DkDlφ.


Proof.Let{gij}be an extremum point ofFunder the divergence-free constraint:


M

(δF(gij))klXkl


−gdx= 0 ∀X={Xkl}withDkXkl= 0.

By Theorem3.26andA=0 in (3.4.42),δF(gij)is in the form


(3.4.47) (δF(gij))kl=DkΦl,


for some{Φl} ∈H^1 (T∗M). By Theorem3.24,δF(gij)is symmetric. Hence we have


(3.4.48) DkΦl=DlΦk.


In addition, by


DkΦl=

∂Φl
∂xk

−ΓkljΦj,

andΓklj=Γlkj, it follows from (3.4.48) that


(3.4.49)


∂Φl
∂xk

=


∂Φk
∂xl

.


By assumption, the first Betti number ofMis zero, i.e. the first homology ofMis zero:
H 1 (M) =0. It follows from the de Rham theorem that if


d(Φkdxk) =

(


∂Φk
∂xl


∂Φl
∂xk

)


dxl∧dxk= 0 ,

then there exists a scalar functionφsuch that


dφ=
∂ φ
∂xk

dxk=Φkdxk.

Thus, we infer from (3.4.49) that


Φk=
∂ φ
∂xk

for someφ∈H^2 (M).

Therefore, we derive (3.4.46) from (3.4.47), and the proof is complete.

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