I Adapting Schwarzschild Metric for Nonzero Λ

Sciencemaster
Messages
129
Reaction score
20
TL;DR Summary
The Schwarzschild metric is designed with a cosmological constant in mind. Would it be feasible to make some modifications to it that would make it apply in a spacetime with a nonzero Λ?
So, there are a fair amount of metrics designed with a zero value for the cosmological constant in mind. I was wondering if there was some method to modify metrics to account for a nonzero cosmological constant. Say, for instance, the Schwarzschild metric due to its relative simplicity. A feature of the metric is that it has a stress-energy tensor of 0 due to both the ricci tensor and scalar curvature going to 0. However, this doesn't happen if the cosmological constant is nonzero. So, would there be a way to modify a metric to account for a nonzero cosmological constant while keeping fundamental features intact--in this case, the 0 stress energy tensor and general form of the Schwarzschild metric? I tried multiplying by a constant as well as a few other minor modifications, and none of them gave a SET of 0. So, would it be feasible to make some modification to a metric designed for Λ=0 in order for it to apply when Λ is nonzero while still keeping similar basic properties?
 
Physics news on Phys.org
That's Schwarzschild-deSitter spacetime, if I understand what you're trying to do.

I don't know about "keeping similar basic properties". There are some quite significant differences, such as a maximum mass for a black hole.
 
  • Like
Likes vanhees71 and PeterDonis
Sciencemaster said:
would there be a way to modify a metric to account for a nonzero cosmological constant while keeping fundamental features intact--in this case, the 0 stress energy tensor and general form of the Schwarzschild metric?
Of course you can't keep everything else exactly the same when you add a nonzero cosmological constant. But the Schwarzschild-de Sitter solution, which @Ibix mentioned, is what you get when you assume spherical symmetry, a nonzero cosmological constant, and no other stress-energy present. If you look it up, you will see that its metric is quite similar to the Schwarzschild metric; the only difference (at least in the coordinates that correspond to standard Schwarzschild coordinates) is an extra term in ##\Lambda## in ##g_{rr}## and ##g_{tt}##.
 
First off, I want to add that I made a mistake. I said something along the lines of, 'keeping the Schwarzschild metric's property of a 0 SET. However, I neglected the ambient energy density in the universe that contributes to the nonzero cosmological constant. So...yeah, I was thinking about a metric that gave $$T_{00}=\frac{\rho_{vacuum}}{c^2}$$. Ignoring that, thanks for suggesting the Schwarzschild-de Sitter, that definitely helps and might be close enough to what I'm looking for!
 
Sciencemaster said:
Schwarzschild-de Sitter, that definitely helps and might be close enough to what I'm looking for!
It is what you are looking for; it is the unique metric that satisfies the condition you give in post #4 and also contains a Schwarzschild black hole.
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
According to the General Theory of Relativity, time does not pass on a black hole, which means that processes they don't work either. As the object becomes heavier, the speed of matter falling on it for an observer on Earth will first increase, and then slow down, due to the effect of time dilation. And then it will stop altogether. As a result, we will not get a black hole, since the critical mass will not be reached. Although the object will continue to attract matter, it will not be a...

Similar threads

Replies
7
Views
7K
Replies
57
Views
3K
Replies
11
Views
2K
Replies
13
Views
3K
Replies
22
Views
3K
Replies
22
Views
3K
Replies
11
Views
2K
Back
Top