The energy of a closed universe is zero

[etotheipi]

In a globally hyperbolic spacetime you have a global time $t: M \longrightarrow \mathbf{R}$ which foliates it into Cauchy surfaces $\Sigma_t$ and the metric can be written $\mathrm{d}s^2 = -N^2 \mathrm{d}t^2 + h_{ij}(\mathrm{d}x^i + N^i \mathrm{d}t)(\mathrm{d}x^j + N^j \mathrm{d}t)$. By Gauss-Codacci the Ricci scalar $R'$ of the induced metric $h_{ij}$ satisfies $R' = R + 2R_{ij} n^i n^j - K^2 + K^{ij}K_{ij}$ and so the Einstein-Hilbert action is\begin{align*}S_{EH}[g] = \int_M \mathrm{d}^4 x \sqrt{-g} \left( R' - 2R_{ij} n^i n^j + K^2 - K^{ij}K_{ij}\right)\end{align*}This is supposed to simplify to\begin{align*}S_{EH}[g] = \int_M \mathrm{d}^4 x \sqrt{h} N \left( R' - K^2 + K^{ij}K_{ij}\right)\end{align*}how does that follow? We can write $R_{ij}n^i n^j = {R_{ikj}}^k n^i n^j = (\nabla_i \nabla_k - \nabla_k \nabla_i) n^i n^k$ and also $K_{ab} = h^c_a \nabla_c n_b$ but even when I wrote it all out it still didn't simplify nicely to the required action. The idea is to deduce the lagrangian $\mathcal{L}_{EH}$ and then evaluate the legendre transform $\int_{\Sigma} \mathrm{d}^3 x \left( \pi^{ij} \dot{h}_{ij} - \mathcal{L}_{EH} \right)$ but as it stands I can't simplify $\mathcal{L}_{EH}$. Is there a trick I am missing? Thank you

 

[PeterDonis]

This is supposed to simplify to

Where are you getting this problem from?

 

[etotheipi]

Where are you getting this problem from?

This one isn't actually an exercise, I'm trying to work through section 8.3 (starting page 111) of the Reall notes https://www.damtp.cam.ac.uk/user/hsr1000/black_holes_lectures_2020.pdf

 

[etotheipi]

It's just odd because I'd expected to end up with something into which you could substitute $K_{ab} = \nabla_a n_b$ and $K = \nabla_a n^a$ but instead...\begin{align*}

R_{ij}n^i n^j = {R_{ikj}}^k n^i n^j &= (\nabla_i \nabla_k - \nabla_k \nabla_i) n^i n^k \\

&= \nabla_i (n^i \nabla_k n^k + n^k \nabla_k n^i) - \nabla_k (n^i \nabla_i n^k + n^k \nabla_i n^i) \\ \\

&= (\nabla_i n^i )(\nabla_k n^k) + n^i \nabla_i \nabla_k n^k + (\nabla_i n^k)( \nabla_k n^i) + n^k \nabla_i \nabla_k n^i \\

&\quad \quad \quad - (\nabla_k n^i)(\nabla_i n^k) - n^i \nabla_k \nabla_i n^k - (\nabla_k n^k)(\nabla_i n^i) - n^k \nabla_k \nabla_i n^i \\ \\

&= n^i (\nabla_i \nabla_k n^k - \nabla_k \nabla_i n^k) + n^k (\nabla_i \nabla_k n^i - \nabla_k \nabla_i n^i) \\

&= n^i (\nabla_i \nabla_k n^k - \nabla_k \nabla_i n^k) + n^i (\nabla_k \nabla_i n^k - \nabla_i \nabla_k n^k) \\

&= 0 \quad \dots

\end{align*}which just cannot be true! So there must be a mistake

 

[PeterDonis]

which just cannot be true!

No, it's exactly what needs to be true in order for the simplification to work.

First, your signs are wrong for the last two terms on the RHS of your second equation in the OP; those terms should have the same signs as your first equation. (You swapped the order of the terms in the second equation, which may be where this error came from.) So for the terms inside the parentheses to simplify as needed, you need to have $R_{ij} n^i n^j = 0$.

Second, given the metric, you should be able to show that $\sqrt{-g} = \sqrt{h} N$.

 

[etotheipi]

No, it's exactly what needs to be true in order for the simplification to work.

First, your signs are wrong for the last two terms on the RHS of your second equation in the OP; those terms should have the same signs as your first equation. (You swapped the order of the terms in the second equation, which may be where this error came from.) So for the terms inside the parentheses to simplify as needed, you need to have $R_{ij} n^i n^j = 0$.

I might have overlooked something but I think the signs are correct as written; equation (3.23) of page 39 of the aforementioned notes give $R' = R + 2R_{ij} n^i n^j - K^2 + K^{ij}K_{ij}$ or rearranged as $R = R' - 2R_{ij} n^i n^j + K^2 - K^{ij} K_{ij}$. So when substituted into the action $\int d^4 x \sqrt{-g} R$ we should have the signs as written in the OP.

 

[etotheipi]

For what it's worth I found these notes: http://www.math.toronto.edu/mccann/assignments/426/Tong.pdf

which say that $R_{ab} n^a n^b = K^2 - K_{ac}K^{ac}$ and not zero. Funnily enough that's precisely what I need in order for it to work, but I don't follow their working...

 

[PeterDonis]

I might have overlooked something but I think the signs are correct as written; equation (3.23) of page 39 of the aforementioned notes give $R' = R + 2R_{ij} n^i n^j - K^2 + K^{ij}K_{ij}$ or rearranged as $R = R' - 2R_{ij} n^i n^j + K^2 - K^{ij} K_{ij}$. So when substituted into the action $\int d^4 x \sqrt{-g} R$ we should have the signs as written in the OP.

Ah, yes, I see, the $R'$ and $R$ terms are switched around in (3.23) so the signs of the $K$ terms have to flip in the simplification in (8.27) from the ones in (3.23).

I don't follow their working...

Is it the "the last two terms are divergences and can be neglected" part that is bothering you? I think that is because the whole thing is going to be plugged into an integral to get the action and the divergence terms will vanish in the integral.

 

[etotheipi]

It's just this part:

I don't know why it's valid to move the $n^a$ to the left side of the operator, and I also can't see how the third line follows from the second. Also, I'm not really sure what's wrong with my working in #4!

 

[cmb]

Could someone translate this into 'Joules' for me? Thanks. ;)

 

[stevendaryl]

It's just this part:

View attachment 282439

I don't know why it's valid to move the $n^a$ to the left side of the operator, and I also can't see how the third line follows from the second. Also, I'm not really sure what's wrong with my working in #4!

Are the indices $a$ and $c$ summed over?

Ignoring line 2, I can see how line 3 follows from line 1, although I don't know why they are writing it that way.

 

[etotheipi]

Are the indices $a$ and $c$ summed over?

Ignoring line 2, I can see how line 3 follows from line 1, although I don't know why they are writing it that way.

Yeah, the summation convention is understood. I think the second line is an unfortunate typo

 

[stevendaryl]

Yeah, the summation convention is understood. I think the second line is an unfortunate typo

Well, I don't understand if $a$ and $c$ are summed over. If you have a quantity $T_{ij}$ that is anti-symmetric, then summing over all $i$ and $j$ has to produce 0, right? Isn't the expression anti-symmetric in $a$ and $c$?

 

[etotheipi]

Well, I don't understand if $a$ and $c$ are summed over. If you have a quantity $T_{ij}$ that is anti-symmetric, then summing over all $i$ and $j$ has to produce 0, right? Isn't the expression anti-symmetric in $a$ and $c$?

I think that expression should evaluate to zero as in post #4 (and now that you mention it, it's indeed a contraction of a symmetric and antisymmetric part!), so I reckon they went wrong in the second line. But the mysterious thing is that their result is what's necessary to transform the integral $\int d^4 x \sqrt{-g} R$ into the one stated in Reall's notes. So I'm a bit flummoxed!