Vacuum solution with nonzero cosmological constant

  • Thread starter La Guinee
  • Start date
  • #1
24
0
Consider the vacuum solution to Einstein's equations with non-zero cosmological constant. Following Carroll, we can find the equation for the radial geodesic with the aid of killing vectors. It takes the standard form: E = T + V. But, with non-zero cosmological constant V(r) now has a term proportional to r^2. This obviously doesn't reduce to ordinary Newtonian gravity. If there were actually a term proportional to r^2 we would have experimentally detected it. What's going on?
 

Answers and Replies

  • #2
1,997
5
Consider the vacuum solution to Einstein's equations with non-zero cosmological constant. Following Carroll, we can find the equation for the radial geodesic with the aid of killing vectors. It takes the standard form: E = T + V. But, with non-zero cosmological constant V(r) now has a term proportional to r^2. This obviously doesn't reduce to ordinary Newtonian gravity. If there were actually a term proportional to r^2 we would have experimentally detected it. What's going on?
General Relativity with a non zero cosmological constant is not compatible with special relativity in flat spacetimes or Newtonian gravity in the limit.
 
  • #3
24
0
General Relativity with a non zero cosmological constant is not compatible with special relativity in flat spacetimes or Newtonian gravity in the limit.
Ok. But if a nonzero cosmological constant has an r^2 potential it can't be compatible with our universe either. How does one reconcile this with the fact that people think the cosmological constant isn't zero?
 
  • #4
1,997
5
But if a nonzero cosmological constant has an r^2 potential it can't be compatible with our universe either.
How do you conclude as such?
 
  • #5
24
0
How do you conclude as such?
Wouldn't we have experimentally detected a potential that grows as r^2?
 
  • #6
DrGreg
Science Advisor
Gold Member
2,303
962
Wouldn't we have experimentally detected a potential that grows as r^2?
Not if the constant of proportionality is exceedingly tiny!

The effect of a tiny cosmological constant is only noticeable on huge (cosmological!) scales, and it's only in the last decade or so that we've found astronomical evidence to suggest it isn't zero.
 
  • #7
24
0
Not if the constant of proportionality is exceedingly tiny!

The effect of a tiny cosmological constant is only noticeable on huge (cosmological!) scales, and it's only in the last decade or so that we've found astronomical evidence to suggest it isn't zero.
I don't see how that makes sense. No matter how small the cosmological constant is, I can find an r much much greater than the cosmological constant. Unless you're saying the cosmological constant is small even compared to the size of the universe.
 
  • #8
24
0
I don't see how that makes sense. No matter how small the cosmological constant is, I can find an r much much greater than the cosmological constant. Unless you're saying the cosmological constant is small even compared to the size of the universe.
I take back what I just said. The condition would be that Lambda R^2 would have to be much less than 1, where R is the maximum distance over which Newtonian gravity has been observed to hold. Is this consistent with current estimates of the cosmological constant? In particular, the current upper bound would have to be less than or equal to the upper bound implied by this criterion. Anyone know if this is true?
 
  • #9
24
0
I found a paper that gives the estimate Lambda ~ 10^-52. That is indeed pretty small.
 

Related Threads on Vacuum solution with nonzero cosmological constant

  • Last Post
Replies
9
Views
2K
Replies
3
Views
4K
Replies
11
Views
2K
Replies
4
Views
3K
Replies
3
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
4
Views
681
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
5
Views
2K
Top