# How to get space component of weak field (linearized) metric?

For Minkowski space with a weak gravitational field the metric takes the form $$ds^2 = (1+2\phi)dt^2 -(1-2\phi)(dx^2+dy^2+dz^2),$$ where $$\phi$$ is the Newtonian gravitational potential.

You can get the $$(1+2\phi)$$ factor in front of the $$dt^2$$ by starting with the geodesic equation and going to the "Newtonian limit" of slow speeds and a slowly changing field.

But is there a way to get the $$(1-2\phi)$$ factor for the spatial part of the metric by a similar procedure? What about some clever thought experiment?

1. The spatial components of the metric $$g_{\mu\nu}~=~\eta_{\mu\nu}+h_{\mu\nu} ~=~\begin{bmatrix} \mp (1+2\phi/c^2) & 0 \cr 0 & \pm(1-2\phi/c^2){\bf 1}_{3\times 3}\end{bmatrix}_{4\times 4} \tag{1}$$ can be deduced from the fact that only the temporal component $$\bar{h}_{00}~=~\mp4\phi/c^2\tag{2}$$ of the trace-reversed metric perturbation $$\bar{h}_{\mu\nu}~:=~h_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}h \qquad\Leftrightarrow\qquad h_{\mu\nu}~=~\bar{h}_{\mu\nu}-\frac{1}{2}\eta_{\mu\nu}\bar{h}. \tag{3}$$ should survive in the Newtonian limit in order for the linearized EFE to decouple, cf. e.g. this & this related Phys.SE posts. [Here the $$\mp$$ sign refers to the Minkowski signature convention $$(\mp,\pm,\pm,\pm)$$.]

2. As for OP's suggestion to use the geodesic equation, note that the spatial components of the metric cannot be reliable extracted in the Newtonian limit as it would be higher order in speed.

Start with Einstein's equation $$$$R_{ab}-\frac{1}{2} g_{ab} R = \kappa\,T_{ab}. \tag{1}$$$$ Contract both sides of (1) with $$g^{ab}$$ to get $$R-\frac{N}{2} R = \kappa\,T \tag{2}$$ where $$N$$ is the number of spacetime dimensions and $$T$$ denotes the trace of $$T_{ab}$$, just like $$R$$ denotes the trace of $$R_{ab}$$. Use (2) in (1) to get $$R_{ab} = \kappa\,\left(T_{ab}+ \frac{g_{ab}}{2-{N}}T\right). \tag{3}$$ In the weak-field approximation and with a convenient choice of gauge (as described in many textbooks), we have $$R_{ab}\approx -\frac{1}{2}\partial^2 h_{ab} \tag{4}$$ with $$h_{ab}\equiv g_{ab}-\eta_{ab}$$. Using this in (3) gives $$-\frac{1}{2}\partial^2 h_{ab}\approx \kappa\,\left(T_{ab}+ \frac{g_{ab}}{2-{N}}T\right). \tag{5}$$ In the usual approximation with $$T_{00}$$ being the only significant component of $$T_{ab}$$, we have $$T\approx T_{00}. \tag{6}$$ Use this in (5) to get $$-\frac{1}{2}\partial^2 h_{ab}\approx \kappa\,\left(T_{ab}+ \frac{g_{ab}}{2-{N}}T_{00}\right). \tag{7}$$ In particular, equation (7) implies $$-\frac{1}{2}\partial^2 h_{00}\approx \kappa\,\frac{3-{N}}{2-{N}}T_{00} \tag{8}$$ and $$-\frac{1}{2}\partial^2 h_{jk}\approx -\kappa\,\frac{\delta_{jk}}{2-{N}}T_{00} \tag{9}$$ for $$j,k\neq 0$$. For the physically-relevant case $$N=4$$, equations (8)-(9) are consistent with $$h_{jk}\approx \delta_{jk}h_{00}. \tag{10}$$ This is consistent with the equation written in the OP, with $$2\phi\equiv h_{00}$$.

• Is the argument going from (8) and (9) to (10) that $h_{ii}$ and $h_{00}$ satisfy the same partial differential equation and both are zero at infinity and therefore they're equal? Anyway, I am still hoping there's a more direct method without going through Einstein's eqn (like the argument that gives you $h_{00}$). – Alex Feb 13 at 22:08
• @Alex Right, that's what I had in mind in going from (8)-(9) to (10). I don't know how deduce the result from a concise thought experiment, but that would be interesting. – Chiral Anomaly Feb 14 at 1:30

The spatial part of the metric is important when test particles have relativistic velocities, like photons. So you need this form of the metric when studying deflection of light by the sun, which was one of the first and significant experimental verifications of GR.

• Yes! Light deflection is how I came to this question in the first place. But getting the $g_{tt}$ component using the geodesic eqn requires already knowing what the geodesic should be for a non-relativistic particle. How can we get $g_{xx}$ without knowing how a photon ought to be deflected by a very weak gravitational field? Or is there a way to know this deflection without going all the way to Einstein's eqns? – Alex Feb 13 at 13:28
• How can one write the geodesic equation without knowing what the metric is? The Christoffel symbols in the geodesic equation depend on the metric. – Avantgarde Feb 13 at 20:57
• To get the $g_{tt}$ component you first write out the Christoffel symbols in terms of the perturbation $h$ to the metric ($g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}$). Then you plug the $\Gamma$'s into the geodesic equation and see that the geodesic equation looks like the eqn of motion ($-\nabla \phi=a$) in a Newtonian potential. That lets you figure out what $h_{tt}$ is in terms of the Newtonian potential. But this is only one equation for one unknown and I'm not sure how to get $h_{xx}$ along these same lines. – Alex Feb 13 at 22:04