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Using this question:

How can we recover the Newtonian gravitational potential from the metric of general relativity?

I want to formulate my question:

Can the first term of the energy-stress tensor, the energy density, be put in some cases as a function of a space? Just for an approximation? I mean can we for example schematize the energy density as if it is a quadratic funtion from a point in order to have an acceleration in a point that is linear in space? (in this case I put an example when there is an acceleration that is similar to that we see in the strong force but not asking for an unification theory answer, is just an example of schematization that I want to achieve)

Qmechanic
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Anto
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1 Answers1

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The acceleration mentioned if the linked question:$$\frac{d^2\mathbf X}{dt^2} = -\nabla \phi$$ can be modified introducing a variable $\tau = at + b$ and expressed by components of the vectors to: $$\frac{d^2 X^j}{d\tau^2} + \frac{\partial \phi}{\partial x^j} \left(\frac{dt}{d\tau}\right)\left(\frac{dt}{d\tau}\right) = 0$$

It has the form of the geodesic equation, where $$\Gamma^j_{00} = \frac{\partial \phi}{\partial x^j}$$ and all other connections vanishes. What means that the universal classic gravitational law can be view as a geodesic equation.

In order to visualize a correspondence with the Einstein field equation we have to take the second derivative of the scalar potential, (or more exactly, its Laplacian): $\nabla^2 \phi = -4\pi G\rho$. Here the left side is the $R_{00}$, one of the terms of the Ricci tensor. The density $\rho$ is the energy density $T_{00}$.

This is the Cartan approach to Newton gravitation, without any reference to relativity. It is of course not the same as the GR equations, but shows how the differential geometry can also be used to classical gravitation and its correspondence with the known equations.

Claudio Saspinski
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