What is the alternative form of the extrinsic curvature in Carroll App D?

[LAHLH]

Hi,

I'm working through Carroll App D, and trying to show the alternative form of the extrinsic curvature that he says should take a few lines.

Starting with $$K_{\mu\nu}=P^{\alpha}_{\mu}P^{\beta}_{\nu}\nabla_{(\alpha}n_{\beta)}$$ where P is the projection tensor, and n is the normal to the hypersurface.

Then I expand out using the definition of the projection operator:

$$K_{\mu\nu}=\left(\delta^{\alpha}_{\mu}-\sigma n^{\alpha}n_{\mu}\right)\left(\delta^{\beta}_{\nu}-\sigma n^{\beta}n_{\nu}\right) \nabla_{(\alpha}n_{\beta)}$$

$$K_{\mu\nu}=\nabla_{(\mu}n_{\nu)}-\sigma n^{\beta}n_{\nu}\nabla_{(\mu}n_{\beta)} -\sigma n^{\alpha}n_{\mu}\nabla_{(\alpha}n_{\nu)}+n_{\mu}n_{\nu}n^{\alpha}n^{\beta}\nabla_{(\alpha}n_{\beta)}$$

focussing on the final term for a moment:

$$n_{\mu}n_{\nu}n^{\alpha}n^{\beta}\nabla_{\alpha}n_{\beta}+n_{\mu}n_{\nu}n^{\alpha}n^{\beta}\nabla_{\beta}n_{\alpha}$$
$$=2n_{\mu}n_{\nu}n^{\alpha}n^{\beta}\nabla_{\alpha}n_{\beta}$$
$$=n_{\mu}n_{\nu}n^{\alpha}\nabla_{\alpha}n^{\beta}n_{\beta}$$
$$=n_{\mu}n_{\nu}n^{\alpha}\nabla_{\alpha}\sigma$$
$$=0$$

(follows if n is parallel transported which I think it is). Getting rid of similar terms of the form $$n^{\beta}\nabla_{\alpha}n_{\beta}$$ by the same trick. Then I'm left with:

$$K_{\mu\nu}=\nabla_{(\mu}n_{\nu)}-\frac{1}{2}\sigma n_{\mu} n^{\alpha}\nabla_{\alpha} n_{\nu} -\frac{1}{2}\sigma n_{\nu}n^{\alpha}\nabla_{\alpha}n_{\mu}$$

Now the only way I can think of proceeding is using the fact that n is hypersuface orthogonal $$n_{[\mu}\nabla_{\nu}n_{\sigma]}=0$$. If you expand this and contact it with say $$n^{\mu}$$, multiply through by sigma, and use the above trick to eliminate a couple of terms, I get:

$$\nabla_{[\nu}n_{\mu]}-\sigma n_{\nu} n^{\alpha}\nabla_{\alpha} n_{\mu}+\sigma n_{\mu}n^{\alpha}\nabla_{\alpha}n_{\nu}=0$$

Subbing this into the above:

$$K_{\mu\nu}=\nabla_{(\mu}n_{\nu)}+ \nabla_{[\nu}n_{\mu]}-\frac{1}{2}\sigma n_{\nu} n^{\alpha}\nabla_{\alpha} n_{\mu}-\frac{1}{2}\sigma n_{\nu}n^{\alpha}\nabla_{\alpha}n_{\mu}$$

which is almost the desired relation, maybe I've made a few algebraic errors. But does anyone know if my general method is correct? seems more than 'a few lines...'

 

[Bill_K]

What is the form you're aiming for?

 

[LAHLH]

Hi, oops I forgot to post that: $$K_{\mu\nu}=\nabla_{\mu}n_{\nu}-\sigma n_{\mu}a_{\nu}$$ where the acceleration $$a_{\nu}=n^{\alpha}\nabla_{\alpha}n_{\nu}$$

With the method I was using I have managed to reproduce this now after correcting some factors of 2 I missed off and sign errors.

But I'm still kind of wondering about the assumptions I made: specifically $$n^{\alpha}\nabla_{\alpha}(n^{\beta}n_{\beta}) =0$$. n here is the normal vector, and has norm $$n^{\beta}n_{\beta}=\pm 1 =\sigma$$, it seems almost common sense that this should vanish, but the fact that Carroll said we're not assuming the integral curves of $$n^{\mu}$$ are geodesics is making me think, is the character of n really conserved as we move along the integral curve? why should it be necessarily if the curve is not a geodesic?

Carroll himself uses a similar argument to arrive at the answer zero in another equation of the appendix (D.19) , where he has $$Y^{\nu}_{(i)}\nabla_{\nu}(n_{\mu}n^{\mu})=0$$, but there you are taking the directional derivative along the $$Y^{\mu}_{(i)}$$ basis vector direction, which are the basis on the hypersurface itself, so I suppose it makes sense the norm of the normal vector wouldn't change as you moved over the surface.

I seem to get the right answer, I just don't know if my method is actually sound.