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#### Kiwi

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**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.