Derivation of Einstein Field Equations w/o Poisson or Least Action

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SUMMARY

The discussion centers on the derivation of the Einstein Field Equations without relying on the Poisson equation or the least action principle. It references David Meldgin's 2011 paper, which argues that through appropriate coordinate transformations, one can set the metric to Minkowskian form and manipulate the derivatives of the metric. The conclusion drawn is that the field equations can be linear in the second order partial derivatives of the metric, leading to unique solutions represented by the Minkowski metric components. This approach challenges traditional methods of deriving the field equations.

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  • Understanding of Riemann curvature tensor components
  • Familiarity with coordinate transformations in differential geometry
  • Knowledge of partial derivatives and their applications in physics
  • Basic grasp of Einstein's Field Equations
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EagleH
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I would be grateful if some one would consider my following thought and indicate to me the likely mistakes, which I cannot do.
Following the paper “Why the Riemann Curvature Tensor needs twenty independent components” by David Meldgin UC Davis 2011, I understand that with a coordinates transformation, using the first and the second partial derivatives of the coordinates transformation to set the metric Minkowskian and the first partial derivatives of the metric equal to zero, while leaving the second order partial derivatives of the metric unchanged, one may using the 80 third order and the 140 fourth order partial derivatives of the coordinates transformation set the 200 third order partial derivatives of the metric equal to zero.

Therefore, by making an error of the fourth infinitesimal order, the highest derivatives of the metric to appear in the field equations would be the second order partial derivatives of it. The components of the spacetime tensor in the field equations would be functions of the second order partial derivatives of the metric only.

Suppose the exact form of the spacetime tensor in the field tensor equation is not known.

If the components of the stress-energy tensor and boundary conditions are given, it is expected that the components of the metric that will be the solutions derived from the “unknown” field equations have to be just the components of the Minkowski metric. Hence, these equations have unique solutions. This would not be possible if spacetime was represented in the field equations by nonlinear functions of the second order partial derivatives of the metric.

If the above is correct, it is derived that the field equations are linear in the second order partial derivatives of the metric.

Therefore, spacetime is represented in the field equations by a tensor of the form
If the above is correct, I think there would be no need to make reference to the Poisson equation, or to the least action principle to derive the field equations.
 
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EagleH said:
the paper “Why the Riemann Curvature Tensor needs twenty independent components” by David Meldgin UC Davis 2011

Can you provide a link?
 
EagleH said:
Therefore, by making an error of the fourth infinitesimal order, the highest derivatives of the metric to appear in the field equations would be the second order partial derivatives of it.

You're mixing up two different things. The reference you gave is talking about determining how many independent components are needed to determine the Riemann curvature tensor, by looking at how to fix the values of the metric coefficients and their derivatives by an appropriate coordinate transformation. That is a separate question from the question of how to derive the field equation and what kind of tensor appears in it.
 
EagleH said:
I understand that with a coordinates transformation, using the first and the second partial derivatives of the coordinates transformation to set the metric Minkowskian and the first partial derivatives of the metric equal to zero, while leaving the second order partial derivatives of the metric unchanged

Not quite. You can set the metric coefficients to be Minkowski and all of the first derivatives to zero, and you can set all but twenty of the second derivatives to zero. The twenty remaining second derivatives that you can't set to zero represent the twenty independent components of the Riemann tensor in the particular coordinates you have chosen. Note, btw, that in order to do all this you have to use up the first, second, and third derivatives of the coordinate transformation.

EagleH said:
one may using the 80 third order and the 140 fourth order partial derivatives of the coordinates transformation set the 200 third order partial derivatives of the metric equal to zero

As above, the 80 third derivatives of the coordinate transformation get used up in setting 80 of the 100 second derivatives of the metric to zero. That leaves twenty second derivatives that are the twenty independent components of the Riemann tensor in your chosen coordinates.

The reference you give does not talk at all about using the fourth derivatives of the coordinate transformation or setting the third derivatives of the metric. Where are you getting that from?
 

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