Recent content by Messenger

Thanks, then that wouldn't work. I was under the impression that the Ricci scalar can't vanish if there is a cosmological constant present, even if the Ricci tensor does. I need to make sure why I misunderstood that part. I do know that the small curvature from WMAP implies a very small...

True, I should probably write: R_{ab}-\frac{1}{2}Rg_{ab}=\Lambda g_{ab} R_{ab}=0 when g_{ab}=diag(1,-1,-1,-1) -\frac{1}{2}Rg_{ab}=\Lambda g_{ab}=diag(\Lambda,-\Lambda,-\Lambda,-\Lambda) -\frac{1}{2}Rg_{00}=\Lambda g_{00} -\frac{1}{2}R\frac{1}{\Lambda}g_{00}=g_{00}=1 I do understand...

It would be something like R=\Lambda g_{ab} \frac{R}{\Lambda}=g_{ab}=diag(1,-1,-1,-1) g_{ab}-\frac{R}{\Lambda}=diag(1-1,-1+1,-1+1,-1+1)=\delta_{ab}-h_{ab} \lim_{r\rightarrow\infty} h_{00} \rightarrow \frac{R_{00}}{\Lambda_{00}}=\frac{f_3}{C}=1 \lim_{r\rightarrow 0} h_{00} \rightarrow...

Thanks! That would be something I would be interested in looking at later on down the road It wouldn't be the first time I had remembered something incorrectly :) The Kaehler manifold isn't something I had considered yet and will look more into that. The basics of it are fairly simple...

I am researching a hypothesis and looking for anyone who is familiar with differential topology (specifically Einstein manifolds). I have access to the Besse book Einstein Manifolds but am also looking for any open questions in differential topology that I am not aware of. I am attempting to...

Hi Peter, Looking into the "zeroth derivative". I take it from the Ricci scalar being composed of second derivatives, it is never possible for the action principle to = 0 with 2\Lambda=-R since this would mean that either the Ricci scalar or the cosmological constant would have to be the...

Gives me quite a bit to look into and think about. I really do appreciate you taking your time on this board Peter. Thanks.

I had not seen this, looks interesting. This paper made me think quite a bit, oldie but a goodie:http://ajp.aapt.org/resource/1/ajpias/v39/i8/p901_s1

I need to read up more on the derivation of the LaGrangian that you showed. I would think there would be some correlation between R and \Lambda for them to show up together in it, similar to potential and kinetic energy.

Based upon: http://en.wikipedia.org/wiki/Scalar_curvature so to let it also "vanish" with a constant present in Euclidean space (Minkowski), wouldn't R have to become equal to -2\Lambda?

Peter, If for \frac{1}{16 \pi} R \sqrt{-g} R \rightarrow 0 does that mean for \frac{1}{16 \pi} \left( R + 2 \Lambda \right) \sqrt{-g} R \rightarrow -2\Lambda?

From http://eagle.phys.utk.edu/guidry/astro490/lectures/lecture490_ch9.pdf it states that g22 and g33 are the same as flat spacetime, and that g00 and g11 are subject to the boundary conditions that spacetime becomes flat at infinity. Isn't the 1-\frac{2M}{r} the same derivation for...

Peter, Thanks for some thought provoking comments, I will be back later with more questions after researching more in depth what you have stated.

I see where MTW:Gravitation show that equation as a "redefintion" of the Einstein tensor, but they state that the theory is no longer geometric when written that way. If there is no proof for that, doesn't that bother you? They write it as: "G_{\alpha\beta}"=R_{ab} - \frac{1}{2} g_{ab} R +...

My thoughts on this were that R=0 implies Minkowski space [-1,1,1,1] (such that g^{\mu\nu}g_{\mu\nu}=4, so that \Lambda g_{\mu\nu} is a rank 0 tensor implying zero curvature but that would also imply the Einstein tensor has vanished. If the Einstein tensor also became a four dimensional rank 0...