- #1

davidbenari

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## Homework Statement

(Working with geometrised units)

Consider the EFE

##G^{\alpha \beta }+\Lambda g^{\alpha \beta} = 8 \pi T^{\alpha \beta} ##

work out (using weak-field considerations) an upper bound for the cosmological constant knowing that the radius of Pluto's orbit is 5.9 x 10^12 m.

## Homework Equations

##G^{\alpha \beta} = -\frac{1}{2}\Box \bar{h}^{\alpha \beta}##

## The Attempt at a Solution

The only important component will be ##\bar{h}^{00}##. Also, we will have to approximate the metric tensor as just ##g^{\alpha \beta} = \eta^{\alpha \beta}##. Also, we neglect the temporal derivatives of ##\bar{h}## since we are assuming non relativistic speeds.

With this in mind our EFE is written as

##\nabla^2\bar{h}^{00}=-16\pi \rho - 2\Lambda ##

Writing ##\bar{h}^{00}=-4\phi ## where ##\phi## is our Newtonian potential we get

##\nabla^2 \phi = 4 \pi \rho + \Lambda / 2 ##

The trivial part of ##\phi## is ##-M/r## with ##M## the mass of the sun. We propose the other solution as ##Cr^2##

Writing out the Laplacian we should arrive at

##6C=\Lambda/2##

Where we get that ##C=\Lambda / 12 ## and we conclude that

##\phi=\frac{-M}{r} + \frac{\Lambda}{12}r^2##

Our upper-bound (so that effects of the cosmological constant will become negligible) will be written as

##\frac{\Lambda}{12}r^2 < M/r##

where we arrive at ##\Lambda = 8.6 \times 10^{-35}m^-2 ## as an upper bound.

**Now my question is: Chapter 8 18(a) http://www.aei.mpg.de/~schutz/download/FirstCourseGR2.Solutions.1_0.pdf**

cites exactly half of what I've got.

http://www.physik.uzh.ch/lectures/agr/GR_Exercises/ex11.pdf

says the solution to the EFE equation is ##\phi=-M/r+\frac{\Lambda}{6} r^2 ##

which is slightly different from what I've got, and gives the "correct" upper limit cited in the first link.

I've been checking for hours now my solution and can't find where that factor on my solution to the diff eq went to. Any help will be much appreciated.

Thanks.

cites exactly half of what I've got.

http://www.physik.uzh.ch/lectures/agr/GR_Exercises/ex11.pdf

says the solution to the EFE equation is ##\phi=-M/r+\frac{\Lambda}{6} r^2 ##

which is slightly different from what I've got, and gives the "correct" upper limit cited in the first link.

I've been checking for hours now my solution and can't find where that factor on my solution to the diff eq went to. Any help will be much appreciated.

Thanks.

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