The k term in the Friedmann equation

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
Apashanka
Messages
427
Reaction score
15
One simple question whether the k in the friedmann equation
H(t)2=∑8πGε(t)/3c2-k/a2
is something related to curvature or is simply constt.??
If related to curvature whether it is 1/R where R is the radius of the 3-sphere.??
 
Last edited:
Astronomy news on Phys.org
haushofer said:
It's the normalized spatial curvature, which due to symmetries (space is maximal symmetric) can be described by one single constant.
This is accurate.

It can also be viewed as a constant which relates how fast the universe is expanding to how much matter/energy there is. If the initial conditions start the universe with not much matter and a lot of expansion, there's negative curvature. If the initial conditions start the universe with a lot of matter and not much expansion, then there's positive curvature. It's the equivalent in classical gravity of throwing a ball: if you throw it normally, it will fall back to the Earth. But if you're superhuman and throw it really fast, it will escape the Earth's gravity. Ignoring air friction (e.g. throwing it on the Moon instead), and you throw it at just the right speed, it will go all the way around the planet and hit you in the back of the head.

Another way to look at it is that the total amount of space-time curvature depends upon how much matter/energy there is. That space-time curvature is either going to show up as expansion or it's going to show up as spatial curvature. Slow expansion compared to density = positive spatial curvature. Fast expansion compared to density = negative spatial curvature.