Does the Landau-Lifshitz pseudotensor contain all non-linear terms in $h_{\mu\nu}$? I was recently reading up some literature on gravitational waves in curved spaces and came across the following confusion. Basically starting from the Einstein Field Equation (EFE),
$$G_{\mu\nu} = 8\pi T_{\mu\nu}$$
we can expand it in powers of $h_{\mu\nu}$ (where $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$) as,
$$G^{(1)}_{\mu\nu} = 8\pi (T_{\mu\nu} + t_{\mu\nu})$$
where $t_{\mu\nu}$ is the LL pseudotensor and $G^{(1)}_{\mu\nu}$ contains only linear terms in $h$.
My question is starting from the definition of the Landau-Lifschitz pseudotensor, i.e.
$$t_{\mu\nu} = -\frac{1}{8\pi}G_{\mu\nu} + \frac{1}{16\pi}\{(-g)(g_{\mu\nu}g_{\alpha\beta} - g_{\mu\alpha}g_{\nu\beta})\},^{\alpha\beta}.$$
How is it evident that $t_{\mu \nu}$ contains all the higher powers of $h_{\mu\nu}$? Any hint whatsoever will be appreciated.
Note: Had the last expression looked like $G_{\mu\nu}$ minus the linear terms in $g_{\mu\nu}$ (or $h_{\mu\nu}$), it would have been easier to conclude possibly.
Edit 1: In case it is not clear, I want to understand mathematically how all terms in the LL pseudotensor are quadratic or higher in the metric and connections.
 A: The physics of the pseudo- tensor is quite independent of the weak field decomposition of the metric ($g_{\mu \nu}= \eta_{\mu \nu} +h_{\mu \nu}$). So, for the time being, lets forget about the decomposition and focus of the pseudo-tensor itself.
The key facts about the quantity
$$\tilde{t}^{\mu \nu}=\frac{1}{16\pi}  \partial_\alpha \partial_{\beta} \left[(-g) (g^{{\mu \nu}}g^{{\alpha \beta}}-g^{{\mu \alpha}}g^{{\nu \beta}} )\right]  $$
are:

*

*Its automatically (ordinarily, not covariantly) conserved as an identity owing to $\mu, \alpha$ antisymmetry.

*It reduces to $\frac{G^{\mu \nu}}{8\pi}$ in Riemann Normal Coordinates(RNC).

Remember, that in RNC, there are no gravitational waves, and no energy, momentum associated with gravity. Therefore in RNC, since $\frac{G^{\mu \nu}}{8\pi}$ is exactly equal to $T^{\mu \nu}_\text{matter}$ (by Einstein's equations) and there is there is no energy, momentum associated with gravity, this is also the full energy momentum tensor in RNC.
Now, LL define $\tilde{t}^{\mu \nu}$ to be the full energy momentum pseudo tensor (accounting for matter+gravity), in arbitrary frames. The motivation for this is:

*

*It is ordinarily conserved and therefore obeys ordinary conservation laws.

*Since it is a two derivative quantity, its integral over all space lives at spatial infinity (by successive application of integration theorems relating volume integration to surface integrals)

*Assuming asymptotically flat space, it is a tensor at spatial infinity, under Lorentz transformations.

Now, turning to your question. You have defined the quantity
$$t^{\mu \nu}= \tilde{t}^{\mu \nu} - \frac{G^{\mu \nu}}{8 \pi}$$
This equation, intuitively reads:
$$t^{\mu \nu} =T^{\mu \nu}_\text{full}- T^{\mu \nu}_\text{matter}$$.
$t^{\mu \nu}$ is therefore the energy momentum tensor associated with gravity. It needs to be fully non linear as you have pointed out.
With indices up, its clear how this quantity is non linear in $h_{\mu \nu}$:

*

*$G^{\mu \nu}$ is the full Einstein tensor and has an infinite expansion in $h$

*$(-g)$ has an infinite expansion in $h$.

*$g^{\mu \nu}$ is the inverse of $g_{\mu \nu}$ and therefore has an infinite expansion in $h$.

A: If you're making the assumption $$g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$$ then its very easy Landau, Lifshitz already gave us the ready to substitute formulae ($101.6$ in Vol2)  here
Because of the above assumption $$\Gamma^{\alpha}_{\mu\nu}=\frac{1}{2}\eta^{\alpha\rho}(\partial_{\mu}h_{\nu\rho}+\partial_{\nu}h_{\mu\rho}-\partial_{\rho}h_{\mu\nu})$$
where higher order terms are dropped.
Now $\Gamma=O(1)$ just looking at $101.6$ gives $$t^{\mu\nu}=(O(1)\times O(1))(O(0)+O(1))+l.c.$$
clearly $$t^{\mu\nu}=O(2)\times (O(0)+O(1))$$
so in no way it can have $O(1)$ terms in $h_{\mu\nu}$. That's it!
Just a heads up the original formulae given Landau book is for curved background $g_{\mu\nu}=g_{\mu\nu}^{B}+h_{\mu\nu}$ so if we have to show the above result we have to do a bit longer calculation, $\Gamma$ will have two contributing terms of zeroth and first order, the computation will require $30$ terms in total to be calculated $((6+2+2)\times3)$ which will be of order $1$ and they have to add up to zero which is really a non-trivial result. Let's hope someone by the time this bounty ends come up with a clever way to calculate those 30 terms or with some other solution.
A: You can check my blog post for all the mathematical details of the derivation of the LL pseudotensor. Basically, it's not a decomposition of the metric in higher or lower order terms. It is a separation of first from second derivatives. The local stress-energy tensor is so constructed from the Einstein tensor that it becomes a second derivative of an antisymmetric tensor which in the locality should turn to zero (see the LL book, chapter 96). In the process of its construction all first derivatives are ignored (see the detailed derivation in my post). However, in the general non-local case, this construct does not turn to zero and the difference contains all first derivatives but does not contain the second derivatives. This difference is namely the LL pseudotensor. It can be expressed in several different ways: as metric derivatives (16 terms), as Christiffels (16 terms), or as metric density derivatives (10 terms).
