# Gauss-Codacci equation space-like foliation

1. Jun 5, 2013

### WannabeNewton

Hello! This is regarding problem 10.4 in Wald. Let $(M,g_{ab})$ be a space-time, $\Sigma_{t}$ be a one parameter family of space-like hypersurfaces foliating the space-time, and $n^{a}$ the unit normal vector field to the space-like hypersurfaces. We have the induced metric $h_{ab} = g_{ab} + n_a n_b$ on each $\Sigma$, which acts as a projection operator, taking space-time tensor fields to tensor fields over $\Sigma$. We also have an induced derivative operator $D_a$ associated with $h_{ab}$ which is defined by $D_{e}T^{a_1...a_n}{}{}{}_{b_{1}...b_m} = h_{e}{}{}^{f}h^{a_1}{}{}_{c_1}...h^{a_n}{}{}_{c_n}h_{b_1}{}{}^{d_1}...h_{b_m}{}{}^{d_m}\nabla_{f}T^{c_1...c_n}{}{}{}_{d_{1}...d_m}$ where $\nabla_{a}$ is the regular space-time derivative operator. Finally, define the extrinsic curvature $K_{ab}$ of $\Sigma$ with respect to $n^{a}$ by $K_{ab} = h_{a}{}{}^{c}\nabla_{c}n_{b}$.

So problem 10.4 is to show that the Gauss-Codacci equation $D_{a}K^{a}{}{}_{b} - D_{b}K^{a}{}{}_{a} = R_{cd}n^{d}h^{c}{}{}_{b}$ holds. I keep getting stuck near the end of the calculation and it's starting to get quite frustrating

$D_{a}K^{a}{}{}_{b} = h_{a}{}{}^{c}h^{a}{}{}_{d}h_{b}{}{}^{e}\nabla_{c}K^{d}{}{}_{e} = h^{cf}h_{b}{}{}^{e}\nabla_{c}\nabla_{f}n_{e} + h^{c}{}{}_{d}h_{b}{}{}^{e}\nabla_{c}h^{df}\nabla_{f}n_{e}$ and similarly $D_{b}K^{a}{}{}_{a} = h_{b}{}{}^{c}h^{a}{}{}_{d}h_{a}{}{}^{e}\nabla_{c}K^{d}{}{}_{e} = h_{b}{}{}^{c}h^{ef}\nabla_{c}\nabla_{f}n_{e} + h_{b}{}{}^{c}h^{e}{}{}_{d}\nabla_{c}h^{df}\nabla_{f}n_{e}$. Now, $h^{e}{}{}_{d}\nabla_{c}h^{df} = h^{e}{}{}_{d}n^{f}\nabla_{c}n^{d}$ and $h^{c}{}{}_{d}\nabla_{c}h^{df} = h^{c}{}{}_{d}n^{f}\nabla_{c}n^{d}$ because $\nabla_{a}g^{bc} = h_{ab}n^{b} = 0$.

Therefore, $D_{a}K^{a}{}{}_{b} - D_{b}K^{a}{}{}_{a} = h^{cf}h_{b}{}{}^{e}(\nabla_{c}\nabla_{f}n_{e} - \nabla_{e}\nabla_{f}n_{c}) + h^{c}{}{}_{d}h_{b}{}{}^{e}n^{f}(\nabla_{f}n_{e}\nabla_{c}n^{d} - \nabla_{f}n_{c}\nabla_{e}n^d)$.

Using the Leibniz rule for $\nabla_{a}$ you can also put the equation in the form $D_{a}K^{a}{}{}_{b} - D_{b}K^{a}{}{}_{a} = h^{cf}h_{b}{}{}^{e}(\nabla_{c}\nabla_{f}n_{e} - \nabla_{e}\nabla_{f}n_{c}) + h^{c}{}{}_{d}h_{b}{}{}^{e}n^{f}\nabla_{f}(n_{e}\nabla_{c}n^{d} - n_{c}\nabla_{e}n^{d})$.

As you can see, the final equation is antisymmetrized over $e$ and $c$ which, at face value, seems to be an issue because we want antisymmetrization over the indices on the space-time derivatives in order to get the Ricci tensor. I've tried everything but I can't seem to find a way to simplify the above equation any further. I tried using the fact that $n^{a}$ is hypersurface orthogonal, i.e. $n_{[a}\nabla_{b}n_{c]} = 0$, in as many different ways as possible but nothing seemed to work. Plugging in $h^{cf} = g^{cf} + n^{c}n^{f}$ only seemed to make things worse so there must be a way to simplify it before having to plug in for $h^{cf}$. All of the aforementioned issues apply to both forms of the equation written above. Does anyone have any hints / suggestions as to how I can show the desired result? Thank you very much in advance.

2. Jun 5, 2013

### Bill_K

[Hm, I've always seen it spelled Codazzi...]

You've started with the derivative expressions on the LHS and tried to come up with the Ricci tensor. My suggestion is to go the other way. Easier, I think, to start with the Riemann identity

ν] nσ = nλRλσμν

Contract both sides with something, hκν gσμ I guess, to get the RHS you want, and then try to manipulate the derivatives on the LHS into the desired expression.

3. Jun 5, 2013

### WannabeNewton

You're most likely right; I just wrote it the way Wald spelled it but I have no knowledge of the actual spelling. Sorry about that.

Ok cool, I didn't even think of doing it the way you suggested so let me try that and see how it goes. Funny enough, Wald's "hint" for the problem is "(Hint: Evaluate the left-hand side...)" lol (page 268 problem 4 if you're interested). Thank you very much Bill, I'll try what you said and hopefully it works out better!

4. Jun 5, 2013

### WannabeNewton

Well Bill, your method worked. It ended up giving the solution in just a few lines. I had thought that only math textbooks contained hints which took you in the longest and most difficult path to a solution but it looks like Wald has written a physics textbook that does the same :rofl: .

Here's how it worked out: using $\nabla_b n_{c} = K_{bc} - n_{b}n^{d}\nabla_{d}n_{c}$ we find that $$h^{e}{}{}_{a}h^{f}{}{}_{b}h^{g}{}{}_{c}n^{d}R_{efgd} = 2h^{e}{}{}_{a}h^{f}{}{}_{b}h^{g}{}{}_{c}\nabla_{[e}\nabla_{f]}n_{g} \\ = D_{a}K_{bc} - D_{b}K_{ac} - h^{e}{}{}_{a}h^{f}{}{}_{b}h^{g}{}{}_{c}n^{d}\nabla_{d}n_{g}\nabla_{e}n_{f} + h^{e}{}{}_{a}h^{f}{}{}_{b}h^{g}{}{}_{c}n^{d}\nabla_{d}n_{g}\nabla_{f}n_{e} \\ = D_{a}K_{bc} - D_{b}K_{ac} + h^{e}{}{}_{a}h^{g}{}{}_{c}n^{d}\nabla_{d}n_{g}K_{be} - h^{f}{}{}_{b}h^{g}{}{}_{c}n^{d}\nabla_{d}n_{g}K_{af} \\ = D_{a}K_{bc} - D_{b}K_{ac} +h^{g}{}{}_{c}n^{d}\nabla_{d}n_{g}K_{ba} - h^{g}{}{}_{c}n^{d}\nabla_{d}n_{g}K_{ab} \\ =D_{a}K_{bc} - D_{b}K_{ac}$$
where, in the last step, I used the fact that $K_{ab}n^{b} = 0$ and that $K_{[ab]} = 0$.
Thus, $D_{a}K^{a}{}{} _{b}- D_{b}K^{a}{}{}_{a} = h^{f}{}{}_{b}h^{ge}n^{d}R_{efgd} = h^{f}{}{}_{b}n^{d}R_{fd}$ after using the fact that $n^{e}n^{g}n^{d}R_{efgd} = n^{e}n^{(g}n^{d)}R_{ef[gd]} = 0$.

Thank you again Bill!

Last edited: Jun 5, 2013
5. Feb 6, 2014

### tommyj

Hey WNB sorry to bring up an old thread but above you said

$$2h^{e}{}{}_{a}h^{f}{}{}_{b}h^{g}{}{}_{c}\nabla_{[e}\nabla_{f]}n_{g} \\ = D_{a}K_{bc} - D_{b}K_{ac} - h^{e}{}{}_{a}h^{f}{}{}_{b}h^{g}{}{}_{c}n^{d}\nabla_{d}n_{g}\nabla_{e}n_{f} + h^{e}{}{}_{a}h^{f}{}{}_{b}h^{g}{}{}_{c}n^{d}\nabla_{d}n_{g}\nabla_{f}n_{e} \\$$

aren't you missing the terms $$h^{e}{}{}_{a}h^{f}{}{}_{b}h^{g}{}{}_{c}n^{d}[n_f\nabla _e\nabla _dn_g-n_e\nabla _f\nabla _dn_g]$$. one of the terms comes from eg. $$\nabla _e\nabla _fn_g=\nabla _e(K_{fg}-n_fn^d\nabla _dn_g)$$ Then ignoring the curvature term and the derivative of $n^d$ we have $$\nabla _e(n_f\nabla _dn_g)=\nabla _en_f\nabla _dn_g+n_f\nabla _e\nabla_dn_g$$. Or does this dissapear for a reason I haven't seen?

Last edited: Feb 6, 2014
6. Feb 6, 2014

### WannabeNewton

But anyways, these terms get combined with the $\nabla_{a}K_{bc}$ terms to form the $D_a K_{bc}$ terms. Try it out.

P.S. I'm still working on that other problem with Maxwell's equations.

7. Feb 7, 2014

### tommyj

Holy ish. I was just writing the thing out on here as I was still stuck, and I was just expanding things to make sure i copied it down correctyl when i realised I mixed up some letters. I see now that when you expand out all the others terms with the h's they all go to zero. Friday morning wasted but got there in the end, thanks!

8. Feb 7, 2014

### WannabeNewton

Haha I know that feeling. That's what happened to me with problem 7.5.