# The ADM energy of gravitational waves?

I have been looking for books about this question for several days. However, almost all books use Landau–Lifshitz pseudotensor to calculate the energy of gravitational waves. And they said the result of Gravitational Waves' energy doesn't depend on the kinds of pseudotensor. So, I want to try to use another way to calculate the energy of gravitational waves, such as the ADM Energy.

First of all, we let $$g_{ab}=\eta_{ab}+\gamma_{ab} \,.$$ Then use the linear Einstein's Field Equation,$$R_{ab}=0 ~~\Rightarrow~~ \Box^2\gamma_{ab}=0 \,.$$ For a plane wave propagating along the $x^3$-axis,we know that the only components of $\gamma_{\mu\nu}$ that are different from zero are $$\gamma_{11}=-\gamma_{22},\gamma_{12}=\gamma_{21}$$ So, $$\gamma_{jj}=\gamma_{11}+\gamma_{22}+\gamma_{33}=0$$ Consider the ADM Energy $$E=\frac{c^4}{16 \pi G} \lim_{r\to\infty} \iint_{S_{r}}\hspace{-34.5px}\subset\!\supset \left(\partial_{j}h_{ij}-\partial_{i}h_{jj}\right) \, \mathrm{d}S^i$$ Some calculation about $h_{ab}$,$$h_{ab}=g_{ab} \mp n_{a}n_{b} ~~\Rightarrow~~ h_{ij}=g_{ij}=\eta_{ij}+\gamma_{ij}$$ $$h_{jj}=\eta_{jj}+\gamma_{jj}=\eta_{jj}= \text{const}$$ $$\partial_{1}h_{ij}=\partial_{2}h_{ij}=0$$ Finally, we have $$E=0$$

The result is certainly wrong, but where is the mistake? I have been thinking for a long time, but I don't get anything.

• Why is $n_an_b$ zero? And is the space-time asymptoticly flat? – MBN Feb 27 '15 at 11:19
• @MBN : $(\Sigma,h_{ab})$ is a spacelike hypersurface of $(M,g_{ab})$,$x^1,x^2,x^3$ is the coordinate system for $\Sigma$,so the space components of $n_{a}n_{b}$ should be zero.(However,the time component isn't zero.) About asymptoticly flat,the answer seems to be no? Then why we can use Landau–Lifshitz pseudotensor? – goodqt Feb 27 '15 at 12:19
• Try Straumann.... – GRrocks Feb 27 '15 at 14:00
• Pseudotensors don't care about asymptotic flatness-they are quasi-local. The standard ADM energy-momentum integrals on the other hand only work for asymptotically flat space-times. – FenderLesPaul Feb 27 '15 at 18:35
• The ADM energy does not, I repeat, does not measure the radiated gravitational wave energy. It is defined a space like infinity and therefore captures only non-radiative data (no null ray can reach spatial infinity). The quantity that does capture the energy of radiation is the Bondi mass aspect. – Prahar Mar 16 '16 at 12:01

It's very important that $T^{(1)}_{ab}=G^{(1)}_{ab}=0$ but $T_{ab}\not=0$.
From this PDF, we can learn $$T^{03}=-\frac{1}{16\pi}\left[\left(\frac{\partial h_{11}}{\partial t}\right)^2+\left(\frac{\partial h_{12}}{\partial t}\right)^2 \right] \,,$$ noting that $c=1$ and $G=1.$ Because $$\sqrt{-g}(T^{\mu\nu}+t^{\mu\nu})=E=0$$ So $$t^{03}=\frac{1}{16\pi}\left[\left(\frac{\partial h_{11}}{\partial t}\right)^2+\left(\frac{\partial h_{12}}{\partial t}\right)^2\right] \,.$$ It is the same as the result of Landau–Lifshitz pseudotensor.