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In general relativity the christoffel symbols 'pop' out of the metric $g^{\mu \nu} dx_\mu dx_\nu$. When $g^{\mu \nu} dx_\mu dx_\nu$ is integrated it is a measure of distance between the $2$ points. Using this as a starting point we have:

$$ \frac{8 \pi G}{c^4}T^{\mu \nu} = G^{\mu \nu}$$

where $T^{\mu \nu}$ is the stress energy tensor and $G^{\mu \nu}$ is the Einstien tensor. In the weak field limit one can derive Poisson's equation:

$$ \nabla^2 \phi = 4 \pi G \rho $$

However, when Newtonian gravity approximated and integrated from the Poisson equation, we get:

$$ U_g = - G \frac{m_1 m_2}{r}$$

where $r$ is the displacement. More precisely if I look at the action in general relativity:

$$ S_{GR} = \int (g^{\mu \nu} dx_\mu dx_\nu)^{1/2} $$

This is a measure of distance. However the action in Newtonian gravity is:

$$ S_{NG} = \int (\frac{1}{2}m_1 v^2 + G \frac{m_1 m_2}{r}) dt $$

where $r$ is displacement.

I feel I missed something subtle on how distance became displacement. Does anyone mind sharing a proof with this point in mind?

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    $\begingroup$ What is the difference between the distance between two points and the displacement between two points? $\endgroup$
    – probably_someone
    Commented May 7, 2020 at 3:30
  • $\begingroup$ also why do you need distance as starting point to Einstein field equation? And did you consider that density in case of two point masses is given by $ \rho=m_1\delta(r-r_1)+m_2\delta(r-r_2) $? Might be where your problem lies? $\endgroup$
    – Umaxo
    Commented May 7, 2020 at 4:38
  • $\begingroup$ @probably_someone I would makea difference when you integrate something like $\vec F.d \vec r$ $\endgroup$
    – More Anonymous
    Commented May 7, 2020 at 5:20
  • $\begingroup$ @Umaxo usually one does begin with distance though right? $\endgroup$
    – More Anonymous
    Commented May 7, 2020 at 5:23
  • $\begingroup$ @MoreAnonymous Can you explain what the difference is? I'm trying to understand precisely how you're defining these two terms. $\endgroup$
    – probably_someone
    Commented May 7, 2020 at 5:42

1 Answer 1

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The displacement is hidden in the metric.

Notation $(X^0,X^1,X^2,X^3)=(t,x_1,x_2,x_3)$. Newton's equations are given by $$m\ddot{x}^{\alpha}(t)=mf^{\alpha}(x(t)),$$ where $$f^\alpha(x(t))=-\left[\frac{G M}{r(t)^2}\right]_\alpha$$ i.e. $\alpha$ component of the field. Taking Newtonian gravity as a curvature in spacetime, the required equations of motion are

\begin{equation} \ddot{X}^0=0 \\ \ddot{X}^\alpha-f^{\alpha}(X(t)) \dot{X}^0 \dot{X}^0=0. \end{equation}

We know that the autoparallel (geodesic) equation in spacetime obtained by minimizing the action $$\int d\lambda \sqrt{g_{i j}(\gamma(\lambda))\dot{\gamma}^i(\lambda) \dot{\gamma}^j(\lambda)},$$ (where $\gamma$ is the curve parameterized by $\lambda$) is given by $\ddot{X}^a+\Gamma^{a}_{b c}\dot{X}^b \dot{X}^c=0.$

Comparing with the above equations of motion gives us Christoffel symbols as

$\Gamma^\alpha_{00}=-f^\alpha$ which is a function of displacement. All other Christoffel symbols are zeros.

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