Einstein-Rosen bridges: Wormholes in Schwarzschild spacetime

 

Nearing the end of section 5.7 Carroll discusses wormholes connecting regions IV and I of the Kruskal diagram. These wormholes are also called  Einstein-Rosen bridges.

It was supposed to be impossible to travel between regions I and IV of the Kruskal diagram and here Carroll shows us how it can almost be done. He is very brief, his diagram is wrong, but luckily I found a paper by Peter Collas and David Klein which goes into much more detail and helped me understand. I was even able to plot diagrams of the wormhole which takes you from the depths of region IV to the depths of region I. Sadly there is never enough time and my plot is, admittedly, a bodge. The real calculations would be too complicated.

There is another way for an intrepid explorer from region I (where we live) to get a glimpse of region IV. After they cross the event horizon (the dashed line ##r=R_s##) they could look 'down and to the left' and they could see light coming in from region IV. They could even meet another explorer from region IV. However they could never tell us back in region I what they learnt and would eventually perish in the singularity.

Read it here Commentary 5.7a Wormholes.pdf (3 pages).

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##)

Conformal Diagrams

Continuing my studies of conformal transformations and diagrams I move on to appendix H, follow Carroll's logic carefully and attempt to plot his conformal diagram of Minkowski space which he shows in Fig H.4 and I have copied above in the centre. My effort is on the right. The diagrams are similar except that the curves of constant ##t##, the Minkowski coordinate, have gradient 0 nowhere on his diagram and twice on mine. And for lines of constant ##r## the score is 1,3 (gradient ##\infty##). I was distressed. Carroll does not give explicit equations for the curves so there is quite a long chain of calculation to get them plotted. I triple checked it and could find no error so I ransacked the internet and found the short paper from from 2008 by Claude Semay, title "Penrose-Carter diagram for an uniformly accelerated observer". The first part is only about an inertial observer and Semay draws a conformal diagram for her with lines of constant ##t,r## just like mine. I have reproduced half his diagram on the left. So I think Carroll has made a mistake in his Figure H.4 - perhaps he just guessed at the curves!

Carroll lists the important parts of the diagram

##i^+=## future timelike infinity (##T=\pi,R=0##)

##i^0=## spatial infinity (##T=0,R=\pi##)

##i^-=## past timelike infinity (##T=-\pi,R=0##)

##J^+=## future null infinity (##T=\pi-R,0<R<\pi##)

##J^-=## past null infinity (##T=-\pi+R,0<R<\pi##)

Carroll use a symbol like ##\mathcal {J}## not ##J## which he calls "scri". It is hard to reproduce.

Conformal diagrams are spacetime diagrams with coordinates such that the whole of spacetime fits on a piece of paper and moreover light cones are at 45° everywhere. The latter makes it easy to visualize causality. Since Minkowski spacetime has 45° light cones, if coordinates can be found which have a metric which is a conformal transformation of the Minkowski metric, the job is done. The first part of appendix H is devoted to finding conformal coordinates for flat Minkowski spacetime, expressed in polar coordinates - presumably to ease our work later in spherically symmetrical manifolds such as Schwarzschild. So we start from that metric:$$
{ds}^2=-{dt}^2+{dr}^2+r^2\left({d\theta}^2+\sin^2{\theta}{d\phi}^2\right)
$$

On the way to finding conformal coordinates we tried coordinates$$
\bar{t}=\arctan{t}\ \ ,\ \ \bar{r}=\arctan{r}
$$which certainly pack spacetime into the range$$
-\frac{\pi}{2}<\bar{t}<\frac{\pi}{2}\ ,\ 0\le\bar{r}<\frac{\pi}{2}
$$as you will see below if you press the button. Carroll says it might be fun to draw the light cones on that, so I made a movie:

Light cone at various ##\bar{r}##

Carroll's Figure H.2 is also quite confusing. It does not show the ##u,v## axes and I naturally assumed that the ##u## axis pointed down and to the right. It does not. It does the opposite.

Read all the details at Commentary App H Conformal Diagrams.pdf (12 pages) 

Exercise Appendix G1 Conformal Null Geodesics

Question

Show that conformal transformations leave null geodesics invariant, that is, that the null geodesics of ##g_{\mu\nu}## are the same as those of ##\omega^2g_{\mu\nu}##. (We already know that they leave null curves invariant; you have to show that the transformed curves are still geodesics.) What is the relationship between the affine parameter in the original and conformal metrics?

Answer

The answer to this is a bit feeble I think - so feeble that I forgot to post it for ten days. It relies on Carroll's assertion in section 3.4 on the properties of geodesics that from some kind of equation like his 3.58 you can always find an equation that satisfies the geodesic equation and he gives us the relationship of the affine parameter to the magic equation. It would have been nice to prove the assertion but I think that was out of scope.

It's at: Ex G1 Conformal Null Geodesics.pdf (a mere two pages).