Why is the non-zero value of spatial curvature +/- 1?

[sunrah]

Going from the Newtonian to relativistic version of Friedmann's equation we use the substitution
[itex]kc^{2} = -\frac{2U}{x^{2}}[/itex]

The derivation considers the equation of motion of a particle with classic Newtonian dynamics. I can sought of see that if space is flat the radius of curvature will be infinite, so spatial curvature will vanish, but why exactly is |k| = 1 when space is not flat ?

Also I'm guessing that k is actually a ratio of something over that things absolute value, e.g.
[itex]k = \frac{thing}{\|thing\|}[/itex], because why else would it be +/- 1 or 0?

 

[Chalnoth]

but why exactly is |k| = 1 when space is not flat ?

It's purely a convention. By this convention, when [itex]k = \pm 1[/itex], the scale factor is the radius of curvature.

An alternative convention is to set the scale factor equal to 1 today. With this convention, the curvature parameter can take on any value, and is proportional to the inverse of the radius of curvature squared.

 

[sunrah]

Thanks, do you know where I can learn more about his? The introductory cosmology books just state this without justifying it and that bugs me.

 

[Chalnoth]

It's explained here:
https://en.wikipedia.org/wiki/Friedmann_equations

See the points following, "There are two commonly used choices for a and k which describe the same physics".

Hope that helps.

 

[sunrah]

So the radius of curvature is a function of the scale factor, but why? That's what I'd like to know. It isn't obvious to me. Thanks

 

[Chalnoth]

So the radius of curvature is a function of the scale factor, but why? That's what I'd like to know. It isn't obvious to me. Thanks

If the universe is described by a sphere, then the radius of curvature is the radius of the sphere. If you increase the distances between every two objects by a factor of two, you increase the radius of the sphere by a factor of two, and therefore increase the radius of curvature by a factor of two.

 

[Chronos]

That only works for a finite universe. But, then again, Einstein strongly favored that option having realized his field equations are not well behaved in the presence of infinities.

 

[Jorrie]

So the radius of curvature is a function of the scale factor, but why?

Actually, it is a function of the curvature constant [itex]\Omega_k= 1-\Omega_{total}[/itex]. The scale factor is just what it says, something to scale the effect of all the Omegas to earlier times. For a small positive curvature (i.e. a small negative curvature constant, unfortunately, due to legacy) the radius of spatial curvature is [itex]R_k \approx R_{Hubble}/\sqrt{\Omega_{total} - 1}[/itex]. For spatially flat it blows up to infinity and for a spatially open universe it comes out as an imaginary number.