Doubt about the constant $\kappa$ of Einstein's field equations I'm trying to understand each of the terms in this equation intuitively
but I'm having a little trouble. I know that we can represent the equation in the following way:
$G_{\mu\nu}= \kappa T_{\mu\nu}$
where $\kappa=\cfrac{8\pi G}{c^4}$
But my question is: How can I remove the value of $\kappa$ from the equation to verify that its value is $\cfrac{8\pi G}{c^4}$? or is it a given thing?
 A: In general relativity this theorem is called the correspondence principle. It describes that, under particular conditions, using the  slow-motion approximation as well as the weak-field approximation, the einstein field equations will reduce to Newtons general theory of gravity. Which would mean that the metric and its derivatives are approximately static. For this derivation, one must first start with the geodesic equation. Applying these simplifying assumptions to the spatial components of the geodesic equation therefore concludes:
\begin{align*}
\frac{d^{2}q^{\omega}}{d\lambda^{2}}+\Gamma^{\omega}_{{\mu}{\nu}}\frac{dq^{\mu}}{d\lambda}\frac{dq^{\nu}}{d\lambda}=0{\,}\Longrightarrow{\,}c^{2}\Gamma^{\omega}_{{0}{0}}=-\frac{d^{2}q^{\omega}}{d\lambda^{2}}
\end{align*}
Our simplifying assumptions make the Christoffel symbols disappear together with the time derivatives, leaving the Ricci tensor as:
\begin{align*}
R_{{0}{0}}{\,}{\equiv}{\,}\partial_{\omega}\Gamma^{\omega}_{{0}{0}}=-\frac{1}{c^{2}}\partial_{\omega}\frac{d^{2}q^{\omega}}{d\lambda^{2}}=-\frac{\mathrm{div}\mathbf{g}}{c^{2}}=\frac{\Delta{\Phi}}{c^{2}}
\end{align*}
Since Newtonian gravitation can be described as the theory of gravitational vector fields, which again are just conservative force fields of a gravitational potential scalar field that, independently to any particular gravitational force, in itself yields the fundamental Poisson equation $\Delta{\Phi}=4\pi{G}\rho$ as stated in Gauß's law for gravity.
Supplementary to previous calculations, one can write the field equation in the traced reversed form by just substituting in it's contraction in the initial formula, therefore inducing the following result:
\begin{align*}
R_{{\mu}{\nu}}-\frac{1}{2}Rg_{{\mu}{\nu}}={\kappa}T_{{\mu}{\nu}}{\,}{\Longrightarrow}{\,}R_{{\mu}{\nu}}=\kappa[T_{{\mu}{\nu}}-\frac{1}{2}Tg_{{\mu}{\nu}}]
\end{align*}
Turning to the Einstein equations, we only need the time components. The low speed and static field assumptions furthermore imply that:
\begin{align*}
T_{{\mu}{\nu}}\approx\mathrm{diag}[{\rho}{c^{2}},0,0,0]{\,}\Longrightarrow{\,}{T}\approx{{\rho}{c^{2}}}
\end{align*}
Combining the above equations together reduces to:
\begin{align*}
\frac{\Delta{\Phi}}{{c}^{2}}=\frac{{\kappa}{\rho}{c^{2}}}{2}
\end{align*}
Which only will occur if our prior mentioned constant maintains the propensity of being able to be expressed as the following exact term:
\begin{align*}
\kappa=\frac{8{\pi}G}{c^{4}}
\end{align*}
$\mathfrak{Q.E.D.}$
