General Relativity from Newtonian Gravity

Posted on Posted in General Relativity, Physics

Start with Newtonian gravity field, the validity of which is locally verified for weak fields :

(1)   \begin{equation*} \displaystyle{\Delta \Phi (\mathbf{x})=4\pi G\rho (\mathbf{x})} \end{equation*}

which is, expressed in terms of the energy-momentum tensor

(2)   \begin{equation*} \displaystyle{\Delta g_{00}=\frac{8\pi G}{c^4}T_{00}} \end{equation*}

Now attempt a first ansatz to formulate this in a Lorentz-invariant manner :

(3)   \begin{equation*} \displaystyle{\square g_{\alpha \beta }=-\frac{8\pi G}{c^4}T_{\alpha \beta }} \end{equation*}

which further leads to a covariant formulation of the form

(4)   \begin{equation*} \displaystyle{G_{\mu \nu }=\frac{8\pi G}{c^4}T_{\mu \nu }} \end{equation*}

Our task will now be to determine the unknown tensor G. We impose the following conditions on that tensor :

  1. G is a Riemann tensor
  2. G is composed of the first and second derivatives of the metric tensor
  3. The energy-momentum tensor obeys the usual symmetry and conservation laws \displaystyle{T_{\mu \nu }=T_{\nu \mu }} and \displaystyle{T{^{\mu \nu }}_{||\nu }=0}; these properties then by default must also apply to our tensor G
  4. The theory must reduce to Newton’s gravity for weak fields

Using the above four points, the Bianchi identities, as well as the general ansatz

(5)   \begin{equation*} \displaystyle{G_{\mu \nu }=aR_{\mu \nu }+bRg_{\mu \nu }} \end{equation*}

plus a little tensor algebra, one find that the easiest tensor which satisfies all of the above conditions is

(6)   \begin{equation*} \displaystyle{R_{\mu \nu }-\frac{1}{2}g_{\mu \nu }R=\kappa T_{\mu \nu }} \end{equation*}

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