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Derivative of Mean Curvature and Scalar field

  1. Mar 20, 2016 #1
    1. The problem statement, all variables and given/known data
    Page 16 (attached file)
    [tex]\frac{dH}{dt}|_{t=0} = Δ_{Σ}φ + Ric (ν,ν)φ+|A|^{2}φ[/tex]
    [tex]\frac{d}{dt}(dσ_{t})|_{t=0} = - φHdσ[/tex]
    H = mean curvature of surface Σ
    A = the second fundamental of Σ
    ν = the unit normal vector field along Σ
    φ = the scalar field on three manifold M
    [tex]φ∈C^{∞}(Σ)[/tex]

    2. Relevant equations
    Now I want to find [tex]\frac{dφ}{dt} = ...? [/tex]
    with [tex]φ≠\frac{1}{H} [/tex]

    3. The attempt at a solution
    [tex]\frac{dH}{dt} = Δ_{Σ}φ + Ric (ν,ν)φ+|A|^{2}φ[/tex]
    [tex]\frac{1}{Δ_{Σ}+ Ric (ν,ν)+|A|^{2}} \frac{dH}{dt} = φ[/tex]
    [tex]\frac{d}{dt}\left ( \frac{1}{Δ_{Σ}+ Ric (ν,ν)+|A|^{2}} \frac{dH}{dt} \right )= \frac{dφ}{dt}[/tex]
    But I am not sure about this.
     
    Last edited: Mar 20, 2016
  2. jcsd
  3. Mar 25, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Apr 11, 2016 #3
    Further information (file attached, Appendix A, page 99):
    [tex]∂_{t} = φ\vec{ν} [/tex]
    So the derivation of [itex]φ[/itex] with respect to [itex]t[/itex] would be:
    [tex]\frac{dφ}{dt} = \frac{d}{dt} \left (\frac{1}{ν} \frac{∂}{∂t} \right )[/tex]
    [tex]\frac{dφ}{dt} = \frac{1}{ν} \frac{∂}{∂t} \left ( \frac{∂}{∂t} \right ) + \frac{∂}{∂t} \frac{d}{dt} \frac{1}{ν}[/tex]
    And now after this I don't know what to do
     

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