9
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Why is $n_an_b$ zero? And is the space-time asymptoticly flat? $\endgroup$
    – MBN
    Commented Feb 27, 2015 at 11:19
  • $\begingroup$ @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? $\endgroup$
    – lrh2000
    Commented Feb 27, 2015 at 12:19
  • $\begingroup$ Try Straumann.... $\endgroup$
    – GRrocks
    Commented Feb 27, 2015 at 14:00
  • $\begingroup$ 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. $\endgroup$ Commented Feb 27, 2015 at 18:35
  • $\begingroup$ 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. $\endgroup$
    – Prahar
    Commented Mar 16, 2016 at 12:01

2 Answers 2

1
$\begingroup$

OK, maybe I get where my mistake is.

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.

I hope my answer is correct.

$\endgroup$
-1
$\begingroup$

I am not sure whether it is true that the ADM energy depends on rest mass. But if it is true, an object with zero rest mass generate no gravity such as plane electromagnetic wave. The property of plane gravitational wave is analogous to plane electromagnetic wave, so it should has zero rest mass as well. You can apply it to gravitational solitons (a wave packet which travels at the speed of light) to see whether it is true.

If a gravitational wave packet has non zero rest mass, it should be regarded as "geon". I know there are some solution of them, but whether they are stable are not sure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.