In the December 12 1936 issue of Nature, Ludwik Silberstein publishes a letter entitled "Minimal lines and geodesics within matter: A fundamental difficulty of Einstein's theory." The issue Silberstein address was that Einstein equates geometry $G$ with a material tensor $T$ so that one should know what the other is talking about. Otherwise, you would be equating apples and oranges. Now according to Einstein's generalization of the Minkowski indefinite metric, $$ds^2=g_{\nu\mu}dx^{\nu}dx^{\mu}$$ its vanishing should still represent "light-lines". Its banal to show that in vacuo the light-lines $ds=0$ represent light trajectories that propagate at speed $c$, while in matter that is characterized by a material tensor $T_{\nu\mu}$, "the minimal lines manifestly cannot represent light propagation, even to a rough approximation." In such an isotropic medium, the light velocity is not $c$ but $c/n$, where $n$ is the index of refraction in the material medium. He uses the case of a Schwarzschild incompressible liquid sphere where the components of the mixed matter tensor are $$T_1^1=T_2^2=T_3^3=-p/c^2, \hspace{30pt} T_4^4=\varrho,$$ where $p$ is the hydrostatic pressure and $\varrho$ is the constant density of the liquid, Since the $T$'s determine the $g$'s via Einstein equation, if there is no trace of $n$, the index of refraction, in $T$, there can be no trace of it in the $g$'s. Wrong! The Schwarzschild metric is determined by Einstein's condition of emptiness, the vanishing of all the components of the Ricci tensor, $R_{\n\mu}=0.$ Due to the fact that the $g$'s depend on $r$ since the medium is isotropic and static, there will be a nonunity index of refraction. But, if we want to consider the Schwarzschild metric immersed in an incompressible liquid sphere who tells $T$ how to react since the index of refraction lies within the $g$'s and not in $T$ which cannot account for gravitational energy at all. Therefore, you are, in effect, equating apples and oranges. Moreover, the radius of curvature of the sphere is $c\surd(3/8\pi G\varrho)$, which is the product of $c$ and the Newtonian free-fall time. So says $T$, but what does $G$ say? It says that the radius of curvature is $c\surd(r^3/GM)$. So which is it? Are we working with a model of constant density, $\varrho$, or one with constant mass, $M$? According to Dirac, the "emptiness" of the system is unaffected by the gravitational field whereas other fields do. But, then there can be no mass, $M$ and the gravitational field lies in the $g$'s. For if there is mass, then there is a density of matter defined as $M/V$, where $V$ is the volume of the incompressible fluid sphere. It is therefore a logical inconsistency to consider the gravitational field to be buried in the metric and not reflected in the matter tensor which is equated to it via Einstein's equations. But if the gravitational field is in $G$ how can it simultaneously be in $T$? The linearized Einstein equations supposedly describe gravitational waves whose source is a pseudo-tensor density. Now the right-hand side of the equation accounts for the source of the gravitational field. So which is it: Is gravity geometry or gravity a force that can do work and should be included in the matter tensor? You can't have your cake and eat it too! Search