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Integral of normal curvature over a closed genus 0 surface

Kiwi

Active member
Feb 7, 2012
109
Melbourne, Australia
I'm studying "Introduction to tensor analysis and the calculus of moving surfaces" by Pavel Grinfeld. So far I have been unable to solve exercise 314 which reads:


Exercise 314
Show that: \(\frac{d}{dt} \int_S \bar{N} B^\alpha_\alpha dS=0\)


You will enjoy this exercise as it draws upon virtually all of the surface identities, including the Codazzi equation.


I have found the following partial solution online:




\(\begin{aligned}
\frac{d}{dt} \int_S \bar{N} B^\alpha_\alpha dS&=\frac{d}{dt} \int_S \nabla^\alpha \bar{S}_\alpha dS ; \; \text{by eqn 11.16}\\
&=\int_S \dot{\nabla} (\nabla^\alpha \bar{S}_\alpha) dS -\int_S C B^\alpha_\alpha \nabla^\beta \bar{S}_\beta dS ; \; \text{by eqn 15.56} \\
&=\int_S \frac{\partial(\nabla^\alpha \bar{S}_\alpha)}{\partial t} - V^\beta \nabla_\beta \nabla^\alpha \bar{S}_\alpha dS-\int_S C B^\alpha_\alpha \nabla^\beta \bar{S}_\beta dS ; \; \text{by eqn 15.37}\\
&=\int_S \frac{\partial(\nabla^\alpha \bar{S}_\alpha)}{\partial t}-V^i Z^\beta _i \nabla_\beta \nabla^\alpha \bar{S}_\alpha dS-\int_S C B^\alpha_\alpha \nabla^\beta \bar{S}_\beta dS ; \; \text{by exercise 299 and eqn 10.15}\\

\end{aligned}\)






And made my own attempts at continuing the solution:
Attempt 1:


\(\frac{d}{dt} \int_S \bar{N} B^\alpha_\alpha dS =\int_S T1 dS + \int_S T2 dS + \int_S T3 dS\)


where:


\(T1=\frac{\partial \bar{N} B^\alpha_\alpha}{\partial t}=\frac{\partial(\nabla^\alpha \bar{S}_\alpha)}{\partial t}=\frac{\partial(Z^\alpha_i\nabla^i (Z_\alpha^j\bar{Z}_j))}{\partial t}=\frac{\partial Z^\alpha_i \bar{Z}_j \nabla^i Z_\alpha^j}{\partial t}
=\bar{Z}_j Z^{im} \frac{\partial Z^\alpha_i \nabla_m Z_\alpha^j}{\partial t}
=\bar{Z}_j Z^{im} Z^\alpha_{it} \nabla_m Z_\alpha^j+\bar{Z}_j Z^{im} Z^\alpha_{i} \nabla_m Z_{\alpha t}^j
\)


\(T2=-V^i Z^\beta _i \nabla_\beta \nabla^\alpha \bar{S}_\alpha
=-V^j (N^iN_j+Z^i_\delta Z_j^\delta) Z^\beta _i \nabla_\beta \nabla^\alpha \bar{S}_\alpha
=-C N^i Z^\beta _i \nabla_\beta \nabla^\alpha \bar{S}_\alpha -V^j Z^i_\delta Z_j^\delta Z^\beta _i \nabla_\beta \nabla^\alpha \bar{S}_\alpha
=-C N^i Z^\beta _i \nabla_\beta \nabla^\alpha \bar{S}_\alpha -\bar{Z}_k V^j Z^\beta _j \nabla_\beta (N^k B^\alpha_\alpha) \)


\(T2=-V^i Z^\beta _i Z_\beta^j \nabla_j (N^k \bar{Z}_k B^\alpha_\alpha)\)


\(T3=- C B^\alpha_\alpha \nabla^\beta \bar{S}_\beta=- V^i N_i B^\alpha_\alpha Z^\beta_j \nabla^j (Z_\beta^k \bar{Z}_k)
=- \bar{Z}_k V^i Z^\beta_j N_i B^\alpha_\alpha \nabla^j Z_\beta^k \)


\(T2+T3=-V^i Z^\beta _i Z_\beta^j \nabla_j (N^k \bar{Z}_k B^\alpha_\alpha)- \bar{Z}_k V^i Z^\beta_j N_i B^\alpha_\alpha \nabla^j Z_\beta^k
=-V^i \bar{Z}_k (Z^\beta _i Z_\beta^j \nabla_j (N^k B^\alpha_\alpha)- Z^\beta_j N_i B^\alpha_\alpha \nabla^j Z_\beta^k)\)<p>
but this seems to be getting me no where.


Attempt 2
\(\begin{aligned}
\frac{d}{dt} \int_S \bar{N} B^\alpha_\alpha dS&=\int_S \dot{\nabla} (\bar{N} B^\alpha_\alpha) dS -\int_S C B^\beta_\beta (\bar{N} B^\alpha_\alpha) dS ; \; \text{} \\
&=\int_S \bar{N} \dot{\nabla} B^\alpha_\alpha + B^\alpha_\alpha \dot{\nabla} \bar{N} - B^\alpha_\alpha (C B^\beta_\beta \bar{N} ) dS ; \; \text{} \\
&=\int_S \bar{N} \dot{\nabla} B^\alpha_\alpha dS + \int_S B^\alpha_\alpha (\dot{\nabla} \bar{N} - C B^\beta_\beta \bar{N} ) dS ; \; \text{} \\
&=\int_S \bar{N} \dot{\nabla} B^\alpha_\alpha dS + \int_S u'v dS ; \; \text{where: } \\
\end{aligned}\)


\(u'=\dot{\nabla} \bar{N} - C B^\beta_\beta \bar{N} \text{ and } u = \int_S u' dS = \frac{d}{dt}\int_S\bar{N}ds = 0\)


\(v = B^\alpha_\alpha \text{ and } v' = \dot{\nabla} B^\alpha_\alpha\)


I can't do that with integration by parts on a surface can I?


The main problem to me seems to be to get rid of that time derivative. The only way I can see to do that is to make it only act on objects with ambient indicees? Or is there another way?


I can't see any reason/way to employ the Codazzi equation.
 

Kiwi

Active member
Feb 7, 2012
109
Melbourne, Australia
I'm still struggling with this one. Can someonw fault the following logic for me? It seems way to simple:

\(\int_S \bar{N} B^\alpha_\alpha dS=\int_S \nabla^\alpha \bar{S}_\alpha dS=0 \) by the divergence theorem on a closed surface.

So

\(\frac{d}{dt} \int_S \bar{N} B^\alpha_\alpha dS=0\)