- #1

Ich

Science Advisor

- 1,931

- 1

## Main Question or Discussion Point

Ok, I've looked a bit into your Posts. The problem is, I suck at math (i.e. differential geometry).

So please tell me if I got the basics right (your post #6):

[tex]

\begin{array}{rcl}

\vec{e}_1 & = & \partial_t\\

\vec{e}_2 & = & \partial_r \\

\vec{e}_3 & = & \frac{1}{r} \, \partial_\theta \\

\vec{e}_4 & = & \frac{1}{r \, \sin(\theta)} \, \partial_\phi

\end{array}

[/tex]

(Q: This is no longer a coordinate basis? Is there some explanation for dummies as to what "nonholonomic" means in this context? If not, don't bother, I'll come back to it later)

Boosting with

Correct so far?

Two corrections for your subsequent paragraph:

The frame is defined for r<t, and the Hubble 'constant' scales as 1/t.

Now for something completely different:

Thanks

So please tell me if I got the basics right (your post #6):

The Minkowski basis vetors are [itex]\partial_t,\partial_r,\partial_\theta,\partial_\phi[/itex], with length [itex]1,1,r,r \, \sin(\theta)[/itex], respectively. (Q: is it ok to think of [itex]\partial_t[/itex] as implicitly acting on the interval s? i.e. change in interval per change in coordinate value?) So we construct an orthonormal basis at each event with scaled vectorsThe Milne frame is

[tex]

\begin{array}{rcl}

\vec{e}_1 & = & \frac{t}{\sqrt{t^2-r^2}} \, \partial_t + \frac{r}{\sqrt{t^2-r^2}} \, \partial_r \\

\vec{e}_2 & = & \frac{r}{\sqrt{t^2-r^2}} \, \partial_t + \frac{t}{\sqrt{t^2-r^2}} \, \partial_r \\

\vec{e}_3 & = & \frac{1}{r} \, \partial_\theta \\

\vec{e}_4 & = & \frac{1}{r \, \sin(\theta)} \, \partial_\phi

\end{array}

[/tex]

[tex]

\begin{array}{rcl}

\vec{e}_1 & = & \partial_t\\

\vec{e}_2 & = & \partial_r \\

\vec{e}_3 & = & \frac{1}{r} \, \partial_\theta \\

\vec{e}_4 & = & \frac{1}{r \, \sin(\theta)} \, \partial_\phi

\end{array}

[/tex]

(Q: This is no longer a coordinate basis? Is there some explanation for dummies as to what "nonholonomic" means in this context? If not, don't bother, I'll come back to it later)

Boosting with

**v**=**r**/t yields the Milne frame.Correct so far?

Two corrections for your subsequent paragraph:

The frame is defined for r<t, and the Hubble 'constant' scales as 1/t.

Now for something completely different:

You would disagree with de Sitter space looking like an "inside out black hole"?#5 said:debunk the notion that FRW dusts (or any reasonably accurate cosmological model) can be considered as "an inside out black hole" [sic],

Thanks

Last edited: