Check of derivation from EFE to Newtonian allowed?

[JMessenger]

Is it allowed for me to ask for maths help concerning the Einstein field equations to the Newtonian gradient? I wouldn't be asking for any validation or opinion on the physical theory, only whether I have made a mathematical error. Having trouble finding anyone who can give it a review.

[CaptainPanic]

Sure!

If you show us what you already did, and that you made some effort, then I am sure someone will help.

[JMessenger]

Hi CaptainPanic, thanks!

 

I am treating the cosmological constant as just a constant of integration. Ignoring all physical meanings, from the fundamental theorem of calculus I should be able to derive down to two equivalent Newtonian gradients whether I have [math]\Lambda[/math]=0 or not equal to zero.

[math]R_{\mu \nu}-\frac{1}{2}g_{\mu \nu}R=G_{\mu \nu}=g_{\mu \nu}\Lambda-\Pi_{\mu \nu}[/math]

 

My full derivation is here (just looking for any help on the mathematical part, don't want to get into any arguments on the physical part)

http://www.vixra.org...1203.0025v1.pdf

 

My biggest problem is that in the book General relativity:an introduction for physicists they derive a Newtonian gradient with the cosmological constant as

[math]\vec{g}=-\nabla\Phi=-\frac{GM}{r^{2}}\hat{\vec{r}}+\frac{\Lambda c^{2}r}{3}\hat{\vec{r}}[/math]

which treats the cosmological constant as just a term with the correct units and opposing sign.

 

But for my derivation I get

[math]=-\frac{\Lambda_{\mathrm{vac}} c^{2}r}{6}\hat{\vec{r}}+\frac{G\rho_{\mathrm{res}}V}{r^{2}}\hat{\vec{r}}[/math].

 

My thoughts are that I should be getting the same denominator under the cosmological constant, but having looked through their derivation I am not sure how they end up with a 3.

[JMessenger]

Would this be better asked within a specific subforum?