Thursday, 13 August 2020

Big Bang!

 Now we want to do a conformal diagram for an expanding universe. The metric equation is$$
{ds}^2=-{dt}^2+t^{2q}\left({dr}^2+r^2{d\Omega}^2\right)
$$and ##0<q<1\ ,0<t<\infty\ ,\ 0\le r<\infty##. It should be pretty easy because we did most of the heavy lifting when we did the conformal diagram for flat spacetime. However I think Carroll made another mistake!

We introduce the coordinate ##\eta## with ##{dt}^2=t^{2q}{d\eta}^2## and we get a metric$$
{ds}^2=\left[\left(1-q\right)\eta\right]^{2q/\left(1-q\right)}\left(-{d\eta}^2+{dr}^2+r^2{d\Omega}^2\right)
$$The part on the right is the same as the flat metric with ##t\rightarrow\eta## so we can use all the work we did before to transform that into$$
{ds}^2=\omega^{-2}\left[-{dT}^2+{dR}^2+\sin^2{R}{d\Omega}^2\right]
$$with$$
\omega^{-2}=\left(\frac{\left[\left(1-q\right)\eta\right]^{q/\left(1-q\right)}}{\left(\cos{T}+\cos{R}\right)}\right)^2
$$and a bit of work on that gives $$
\omega=\left[\left(1-q\right)\sin{T}\right]^{q/\left(q-1\right)}\left(\cos{T}+\cos{R}\right)^{1/\left(1-q\right)}
$$But Carroll says that$$
\omega=\left(\frac{\cos{T}+\cos{R}}{2\sin{T}}\right)^{2q}\left(\cos{T}+\cos{R}\right)
$$I'm pretty sure that Carroll is wrong, even though his formula is more attractive. I also worked out how he went wrong. Carroll writes "The precise form of the conformal factor is actually not of primary importance" (because you throw it away for the diagram). Perhaps that's why he did not check it very carefully.

And here's the diagram

At the singularity very near ##t=0## space can apparently be as big as you like. Never fear: ##r## might be big but ##t^{2q}## will be very small, so distances are very small too.

Read all the details at
Commentary App H Conformal Diagram Expanding Universe.pdf (6 pages including a diversion on values of ##q##)


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