Derivative of Mean Curvature and Scalar field

In summary, the given conversation discusses finding the derivative of the scalar field φ with respect to time, while also considering the mean curvature H, second fundamental form A, and unit normal vector field ν. The equations used involve the Laplace operator and the Ricci curvature, and it is stated that φ does not equal 1/H. Appendix A on page 99 provides additional information, including the use of the partial derivative ∂/∂t.
  • #1
darida
37
1

Homework Statement


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]

Homework Equations


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

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.
 

Attachments

  • D Maximo - Hawking mass and local rigidity of minimal two-spheres.pdf
    213.4 KB · Views: 251
  • D Maximo - Hawking mass and local rigidity of minimal two-spheres.pdf
    213.4 KB · Views: 226
Last edited:
  • #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
 

Attachments

  • HL Bray Thesis.pdf
    1.2 MB · Views: 262

1. What is the derivative of mean curvature?

The derivative of mean curvature is a measure of the rate of change of the mean curvature at a given point on a surface. It represents the change in the average curvature of the surface in a particular direction.

2. How is the derivative of mean curvature calculated?

The derivative of mean curvature is calculated using the Weingarten map or shape operator, which is a linear transformation that maps tangent vectors to their corresponding normal vectors at a given point on the surface. The derivative is then found by taking the derivative of this map.

3. What is the significance of the derivative of mean curvature in surface geometry?

The derivative of mean curvature provides important information about the shape and curvature of a surface. It can help determine critical points, such as umbilic points, and can be used to classify surfaces as convex, concave, or saddle-shaped.

4. What is the relationship between the derivative of mean curvature and scalar fields?

The derivative of mean curvature can be used to calculate the gradient of a scalar field defined on a surface. This allows for the analysis of how the scalar field changes along the surface and can aid in understanding the behavior of the surface itself.

5. Can the derivative of mean curvature be negative?

Yes, the derivative of mean curvature can be both positive and negative. A positive derivative indicates that the mean curvature is increasing in the given direction, while a negative derivative indicates that it is decreasing. The value of the derivative can also be zero, which would indicate a critical point on the surface.

Similar threads

  • Special and General Relativity
Replies
7
Views
591
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
4K
  • Advanced Physics Homework Help
Replies
1
Views
328
  • Advanced Physics Homework Help
Replies
2
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
7
Views
5K
  • Advanced Physics Homework Help
Replies
5
Views
1K
Back
Top