I have not seen any complete derivation for the metric in a limited newtonian case:

$$\begin{align} ds^{2} = -(1+2\Phi)dt^{2} +(1-2\Phi)\left(dr^{2} + r^{2}(d\theta^{2} + \sin^{2} \theta\ d\phi^{2})\right). \end{align}$$

I believe that we should begin from cartesian form of the metric and then apply the transformation for spherical coordinates:

$$\begin{align} ds^{2} = -(1+2\Phi)dt^{2} +(1-2\Phi)\delta_{ij}dx^{i}dx^{j} \end{align}$$

In the newtonian limit:

  1. Particles are moving slowly
  2. The gravity field is weak
  3. The field is static

With these conditions, we can perturb the metric linearly:

$$\begin{align} g_{\mu \nu} &= \eta_{\mu \nu} + h_{\mu \nu} \\ g^{\mu \nu} &= \eta^{\mu \nu} - h^{\mu \nu} \end{align}$$ where $\eta_{\nu \mu}$ is some canonical metric (Minkowski then in this case) and $|h_{\mu \nu}| \ll 1$ is a small perturbation.

If we then follow the time component our the geodesic equation:

$$\begin{align} \frac{d^{2}x^{u}}{d\tau^{2}} + \Gamma^{\mu}_{\nu \lambda}\frac{dx^{\nu}}{d\tau}\frac{dx^{\lambda}}{d\tau} = 0 \end{align}$$

and then solve for time and spatial components (while taking the time derivative of a static field):

$$\begin{align} \frac{d^{2}x^{\mu}}{d\tau^{2}} + \Gamma^{\mu}_{00} \left( \frac{dt}{d\tau} \right)^{2} &= 0 \\ \frac{d^{2}x^{\mu}}{d\tau^{2}} &= - \frac{1}{2} \eta^{\mu \lambda}\partial_{\lambda}h_{00} \left( \frac{dt}{d\tau} \right)^{2} \end{align}$$

we see that when $\mu=0$:

$$\begin{align} \frac{dt}{d\tau} = constant \end{align}$$

and when we see that when $\mu=i$:

$$\begin{align} \frac{d^{2}x^{i}}{dt^{2}} &= - \frac{1}{2} \partial_{i} h_{00} \end{align}$$

in which we have $h_{00} = -2 \Phi$, reminiscent of that of the acceleration $\vec{a} = -\nabla \Phi $ where $\Phi$ is the newtonian potential.

Thus $$\begin{align} g_{00} &= - (1+2\Phi). \end{align}$$

Now my problem is trying to solve for the spatial components $g_{ij}$ in a similar fashion.

When trying to work it out, my work starts to look convoluted and messy and I just get lost in translation:

$$\begin{align} \Gamma^{\mu}_{ij}&= \frac{1}{2} g^{\mu \nu} ( \partial_{i}g_{\nu j} + \partial_{j}g_{i \nu} - \partial_{\nu}g_{i j} ). \\ \end{align}$$

Taking $\mu=0$, the whole connection goes to zero. But for a spatial components while implementing the perturbed metric, I get stuck.


1 Answer 1


I might have found the solution.

In the spatial configuration: $$ \begin{align} \frac{d^{2}x^{\mu}}{d \tau^{2}} + \Gamma^{\mu}_{ij}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau} = 0 \end{align} $$ The affine connection expanded out is of the form: $$ \begin{align} \Gamma^{\mu}_{ij} = \frac{1}{2}g^{\mu \nu}( \partial_{i}g_{\nu j} + \partial_{j}g_{i \nu} - \partial_{\nu}g_{i j} ) \end{align} $$

If $\mu=0$ then the time derivatives of the static field will vanish and metrics non diagonal points are zero.

Taking the spatial component of the connections: $$ \begin{align} \Gamma^{k}_{ij} = \frac{1}{2}g^{k l}( \partial_{i}g_{l j} + \partial_{j}g_{i l} - \partial_{l}g_{i j} ) \end{align} $$ Equating the lower indices and taking the newtonian limit yields: $$ \begin{align} \Gamma^{k}_{ii} &= \frac{1}{2}\eta^{k l}( \partial_{i}h_{l i} + \partial_{i}h_{i l} - \partial_{l}h_{i i} ) \\ &= - \frac{1}{2}\eta^{k l} \partial_{l}h_{i i} \end{align} $$ by the symmetry of the metric.

With our geodesic in this form, and taking the newtonian limit: $$ \begin{align} \frac{d^{2}x^{k}}{d \tau^{2}} + \Gamma^{k}_{ij}\frac{dx^{i}}{d\tau}\frac{dx^{j}}{d\tau} &= 0 \\ \frac{d^{2}x^{k}}{d t^{2}} &= - \Gamma^{k}_{ii}\left( \frac{dx^{i}}{dt}\right)^{2} \\ \frac{d^{2}x^{k}}{d t^{2}} &= \frac{1}{2}\eta^{k l} \partial_{l}h_{i i} \left( \frac{dx^{i}}{dt}\right)^{2} \\ \frac{d^{2}x^{k}}{d t^{2}} &= \frac{1}{2} \partial_{k}h_{i i} \left( \frac{dx^{i}}{dt}\right)^{2} \end{align} $$

Comparing to the equation for a gravitational potential $\vec{a} = - \nabla \Phi$, I just deduce that $h_{ii} = -2 \Phi$, just like for the time component iff $i=j$.


$$ \begin{align} g_{ij} &= (1 -2 \Phi ) \delta_{ij} \end{align} $$

Completing the metric and hopefully by applying spherical coordinate transformation, produces the approximation metric for Earth.


Their is no need for a coordinate transformation. You can see that expanding out the line element we can tac on the metric for a two sphere and argue that the deviation of the radius is just $dr^{2} = dx^{2} + dy^{2} + dz^{2}$


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