4

In coordinates given by $x^\mu = (ct,x,y,z)$ the line element is given

$$ (ds)^2 = g_{00} (cdt)^2 + 2g_{0i}(cdt\;dx^i) + g_{ij}dx^idx^j, $$

where the $g_{\mu\nu}$ are the components of the metric tensor and latin indices run from $1$-$3$. In the first post-Newtonian approximation the space time metric is completely determined by two potentials $w$ and $w^i$. The Newtonian potential is contained within $w$ and the relativistic potential is contained with $w^i$. What I don't understand is:

Often in the literature concerning the first Post-Newtonian (PN) approximation, it is just quoted that the components of the metric tensor are given by:

$$\begin{split}g_{00} &= -\exp(-2w/c^2), \\ g_{oi} &= -4w^i/c^3, \\ g_{ij}&= - \delta_{ij}\left( 1 +2w/c^2 \right).\end{split}$$

Due to the fact that they are often just given without derivation, I'm assuming that I'm missing something very trivial here.

How are these metric components derived?

Qmechanic
  • 201,751
Rumplestillskin
  • 1,968

2 Answers2

3

Section IV.5.3 Post-Newtonian approximation of Introduction to General Relativity, Black Holes, and Cosmology, Y. Choquet-Bruhat discusses a derivation of that metric. It seems to originate from L. Blanchet and T. Damour, 1989, Post-newtonian generation of gravitational waves.

The starting point are Einsteins field equations (EFE) in harmonic coordinates and their 1PN expansion and the ansatz \begin{align} g_{00}\equiv&-\exp\left(-\frac{2}{c^2} V\right)\\ g_{0i}\equiv&-\frac{4}{c^3}V_i\\ g_{ij}\equiv&\exp\left(+\frac{2}{c^2} V\right)\gamma_{ij} \end{align} Using the EFE in harmonic coordinates one can show in 1PN

$$\gamma_{ij}=\delta_{ij}+O(c^{-4}).$$

Further one can find linear equations relating $V$ and $V_i$ to the source terms: \begin{align} \Delta V -\frac{1}{c^2}\partial_t^2V&=-\frac{4\pi G}{c^2} \left(T^{00}+T^{ss}\right)\\ \Delta V_i&=-\frac{4\pi G}{c}T^{0i}. \end{align}

For details of the involved computation I recommend the book and paper referenced previously and references therein.

N0va
  • 3,001
3

Personally I wouldn't say you're missing something 'trivial' - maybe the result is 'standard' in some way that is taken for granted by people who know the field.

The earliest reference containing the result appears to be Blanchet, Luc, and Thibault Damour. "Post-Newtonian generation of gravitational waves." Annales de l'IHP Physique théorique. Vol. 50. No. 4. 1989 (see equations $2.10a.$ - $c.$ therein).

The entire derivation is quite involved, but the level of complexity depends on what your starting point is.

In the linked paper the result is derived for the near-field, where the post-Newtonian expansion is assumed valid (i.e. $v/c << 1$, and a weak gravitational field). This allows them to make a number of assumptions, including that $\partial_0 g_{\mu\nu}^{in}/{\partial_i g_{\mu\nu}^{in}} = O(\frac{1}{c})$, where $g_{\mu\nu}^{in}$ is the metric in the inner domain where this near-field expansion is used.

If we start with the Einstein equation in harmonic coordinates, the following set of linearised equations can then be derived:

$\Box\ln(-g^{in}_{00}) = \frac{8\pi G}{c^4}(T^{00} + T^{ss}) + O(\frac{1}{c^6})$

$\Box g_{0i}^{in} = \frac{16\pi G}{c^4}T^{0i} + O(\frac{1}{c^5})$

$\Box g_{ij}^{in} = -\frac{8\pi G}{c^4}\delta_{ij}(T^{0i} + T^{ss}) + O(\frac{1}{c^4})$,

where $\Box$ is the d'Alembertian (using the mostly $+$ metric), and the various components of the energy-momentum tensor $T$ are of known order in $c$.

For convenience the authors then define

$\sigma = c^{-2}(T^{00} + T^{ss})$ and

$\sigma_i = c^{-1}T^{0i}$.

This is a clever trick that allows them to simplify the form of the resultant equations

They then define certain potentials (eq. $2.8$ and $2.9$), from which the desired results follow (along with a number of intermediate intuitive leaps to justify the calculations). These potentials themselves satisfy certain linear equations.

I hope this helps, and apologies for any typos. I think the best thing would be to read the linked paper. It is a bit convoluted, and in a sense I think the result you posted is just one possible parametrisation.

Martin C.
  • 1,429