Cosmology Chapter Mistake: Reflections on Relativity

[m4r35n357]

http://mathpages.com/rr/s7-01/7-01.htm

I am completely unable to follow the following sequence of working between equations 2 and 3. AFAIK the final answer is correct, but the intermediate steps seem to be a "casserole of nonsense". I would appreciate feedback from anyone who can follow this, or who can correct it if it is wrong . . .
--------------------------------------------------------------------------------------------------------------------------------------------

Now suppose we embed a Euclidean three-dimensional space (x,y,z) in a four-dimensional space (w,x,y,z) whose metric ishttp://mathpages.com/rr/s7-01/7-01_files/image004.gif

where k is a fixed constant equal to either +1 or -1. If k = +1 the four-dimensional space is Euclidean, whereas if k = -1 it is pseudo-Euclidean (like the Minkowski metric). In either case the four-dimensional space is "flat", i.e., has zero Riemannian curvature. Now suppose we consider a three-dimensional subspace comprising a sphere (or pseudo-sphere), i.e., the locus of points satisfying the conditionhttp://mathpages.com/rr/s7-01/7-01_files/image005.gif From this we have w2 = (1 - r2)/k = k - kr2, and thereforehttp://mathpages.com/rr/s7-01/7-01_files/image006.gif Substituting this into the four-dimensional line element above gives the metric for the three-dimensional sphere (or pseudo-sphere)http://mathpages.com/rr/s7-01/7-01_files/image007.gif Taking this as the spatial part of our overall spacetime metric (2) that satisfies the Cosmological Principle, we arrive athttp://mathpages.com/rr/s7-01/7-01_files/image008.gif

 

[WannabeNewton]

What exactly in the derivation is confusing you? It looks perfectly fine and sensible to me.

 

[m4r35n357]

What exactly in the derivation is confusing you? It looks perfectly fine and sensible to me.

OK, firstly "From this we have w2 = (1 - r2)/k = k - kr2". I'm OK with the first equality but the second makes me want to gag. Is he invoking some power series/binomial approximation without using an approximation sign?

Secondly, "and therefore[PLAIN]http://mathpages.com/rr/s7-01/7-01_files/image006.gif " is OK from the first equality, but does not follow from the second.

Thirdly, I can't get the final answer by squaring [PLAIN]http://mathpages.com/rr/s7-01/7-01_files/image006.gif and substituting into the polar metric at the top. I get $$(dw)^2 = k^2r^2 / (1 - kr^2)(dr)^2$$ so $$k(dw)^2 + (dr)^2 = (k^3r^2/(1 - kr^2) + 1) (dr)^2$$ and not $$k(dw)^2 + (dr)^2 = (1 /(1 - kr^2)) (dr)^2$$

I must be missing some invisible steps I suppose, can you tell me where?

 

[Fredrik]

OK, firstly "From this we have w2 = (1 - r2)/k = k - kr2". I'm OK with the first equality but the second makes me want to gag. Is he invoking some power series/binomial approximation without using an approximation sign?

It's much simpler than that. :) Since $k=\pm1$, we have $1/k=k$.

 

[PAllen]

k is +1 or -1, by stipulation.

[edit: cross posted with Fredrik].

 

[m4r35n357]

Good catch guys, I need to revisit my calculations . . .

 

[Fredrik]

Maybe I'm making some mistake too, but I keep getting
$$dw=\frac{-kr}{\sqrt{k(1-r^2)}}dr$$ and therefore
$$dw^2=\frac{r^2}{k(1-r^2)}dr^2.$$ How are you guys getting a factor of $k$ in the $r^2$ term in the denominator (and not in the other term as well)?

 

[m4r35n357]

OK, so $dw^2 = k - kr^2$ because $k = 1/k$. Therefore $w = \sqrt(k - kr^2)$ and $$\frac{dw}{dr} = \frac{1}{2 \sqrt(k - kr^2)} . -2kr = \frac{-kr}{\sqrt(k -kr^2)}$$
So I still haven't quite got there. I'm not going to attempt substituting back into the metric until I get this right . . . any clues?

 

[PAllen]

Maybe I'm making some mistake too, but I keep getting
$$dw=\frac{-kr}{\sqrt{k(1-r^2)}}dr$$ and therefore
$$dw^2=\frac{r^2}{k(1-r^2)}dr^2.$$ How are you guys getting a factor of $k$ in the $r^2$ term in the denominator (and not in the other term as well)?

I get the same as you. Previously, I answered only the initial question. Further, if you plug this back into the metric, k drops out altogether.

 

[m4r35n357]

Hmm, I'm not entirely sure this can be saved, that would be a shame as it's nice & concise. I've attempted working backwards from the metric but not come up with a sensible intermediate step as yet . . .

I think we need $$\frac{dw}{dr} = \frac {\pm r} {\sqrt(1 - kr^2)}$$

 

[PAllen]

Well, everything works out fine if one changes the definition of [pseudo-]sphere to have the k in front of the r² term to start. Maybe that is the mistake?

 

[m4r35n357]

You mean as in $w^2 + kr^2 = 1$ ?

 

[PAllen]

You mean as in $w^2 + kr^2 = 1$ ?

Yes. If I work out everything from there, I get the right intermediate expressions, and the right final metric.

 

[m4r35n357]

Cool, thanks for that, I think I'll check it all out tomorrow, my brain hurts!