Thursday, 17 September 2020

Curvature in two dimensions

Here we calculate formulas for curvature in two dimensions. Specifically we give coordinates, metrics, Christoffel symbols, Riemann tensors (only twice) and scalar curvature (or Ricci scalar) for ellipsoids, elliptic paraboloids and hyperbolic paraboloids.

Images from Wikipedia: Ellipsoid , Paraboloid

We also find a general formula for the scalar curvature in two dimensions which only requires one component of the Riemann tensor. On the way we find the formulas to calculate all the other non zero Riemann components from the one. With coordinates ##\left(\theta,\phi\right)## which are naturally used for the ellipsoid, the formula for the scalar curvature is $$
$$This is very similar to the formula given at the very end of the Wikipedia article on Gaussian curvature ##K## which is$$
K=-\frac{1}{E}\left(\frac{\partial}{\partial u}\Gamma_{12}^2-\frac{\partial}{\partial v}\Gamma_{11}^2+\Gamma_{12}^1\Gamma_{11}^2-\Gamma_{11}^1\Gamma_{12}^2+\Gamma_{12}^2\Gamma_{12}^2-\Gamma_{11}^2\Gamma_{22}^2\right)
$$The article also reveals that the scalar curvature is twice the Gaussian curvature. The Wikipedia formula might be better written with all the one indices replaced by ##u## and all the 2 indices replaced by ##v##. Then reverting to ##\theta,\phi## as indices, the thing in brackets in the first formula is ##R_{\ \ \ \phi\theta\phi}^\theta## and in the second is ##R_{\ \ \ \theta\theta\phi}^\phi##. The relationship mentioned above between Riemann components is
R_{\ \ \ \theta\theta\phi}^\theta&=\frac{g_{\theta\phi}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\
R_{\ \ \ \theta\phi\theta}^\theta&=-\frac{g_{\theta\phi}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\
R_{\ \ \ \phi\theta\phi}^\theta&=\partial_\theta\Gamma_{\phi\phi}^\theta-\partial_\phi\Gamma_{\theta\phi}^\theta+\Gamma_{\theta\theta}^\theta\Gamma_{\phi\phi}^\theta+\Gamma_{\theta\phi}^\theta\Gamma_{\phi\phi}^\phi-\Gamma_{\phi\theta}^\theta\Gamma_{\theta\phi}^\theta-\Gamma_{\phi\phi}^\theta\Gamma_{\theta\phi}^\phi&\phantom {10000}\nonumber\\
R_{\ \ \ \phi\phi\theta}^\theta&=-R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\
R_{\ \ \ \theta\theta\phi}^\phi&=-\frac{g_{\theta\theta}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\
R_{\ \ \ \theta\phi\theta}^\phi&=\frac{g_{\theta\theta}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\
R_{\ \ \ \phi\theta\phi}^\phi&=-\frac{g_{\theta\phi}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber\\
R_{\ \ \ \phi\phi\theta}^\phi&=\frac{g_{\theta\phi}}{g_{\phi\phi}}R_{\ \ \ \phi\theta\phi}^\theta&\phantom {10000}\nonumber
\end{align}So now it is easy to work out that the mysterious ##E## in the Wikipedia formula should be ##g_{\theta\theta}## which is the same as ##g_{11}##. 

For the record the scalar curvatures for the ellipsoids, elliptic paraboloids and hyperbolic paraboloids are respectively$$
$$These are a bit vague until you know what the coordinate systems are but you can see what the sign of the curvature is which is what I was interested in.

Why did I do all this?

This was quite a project. It was sparked off by the following:
On Physics Forums Ibix said:
I don't know where you [JoeyJoystick] are getting numbers for the mass of the universe from - our current understanding is that it's infinite in size and mass.
I said:
Universe infinite in size and mass? That's extraordinary. Where can I read more about it, please?
Ibix said:
I would think Carroll covers it - chapter 8 of his lecture notes certainly does. The flat and negative curvature FLRW metrics are infinite in extent and have finite density matter everywhere. Modern cosmological models are a bit more complicated, but retain those features.

Chapter 8 in the book is on Cosmology and about 50 pages long. Part 8 of the lecture notes is on Cosmology and contain 15 pages. I think those are the places to look, but after a quick look I did not find anything specifically about the size of the Universe.

If the universe has flat or negative curvature locally and then we presume that applies everywhere, that implies it is infinite in extent. Presumably shortly after the big bang the universe was finite in extent, so at some time it must have changed from finite to infinite! Maybe that's inflation?

So I decided to test the flatness idea in two dimensions....
Summary in Commentary 8 Curvatures 2D.pdf (6 pages), mostly pictures.
Full details in
Commentary 8 Curvatures 2D calculations.docx
or Commentary 8 Curvatures 2D calculations.pdf (24 pages)

Wednesday, 19 August 2020

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).

Thursday, 13 August 2020

Big Bang!

 Now we want to do a conformal diagram for an expanding universe. The metric equation is$$
$$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$$
$$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$$
$$and a bit of work on that gives $$
$$But Carroll says that$$
$$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##)

Saturday, 8 August 2020

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:$$
On the way to finding conformal coordinates we tried coordinates$$
\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) 

Tuesday, 4 August 2020

Exercise Appendix G1 Conformal Null Geodesics


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?


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).

Thursday, 23 July 2020

Conformal Transformations

I want to understand the conformal diagrams in section 5.7 so I must read appendix G then H. Most of this is just checking Carroll's formulas for the conformal 'dynamical variables' - things like the connection coefficients and the Riemann tensor. It is eye bogglingly dense.

Conformal transformations all start when you multiply each component of the metric by a scalar ##\omega## which may depend on the coordinates. So we have a conformal metric $$
$$Then we want find things like the Riemann tensor in the 'conformal frame'. It's quite easy to show that it is$$
{\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=R_{\ \ \ \sigma\mu\nu}^\rho+\nabla_\mu C_{\ \ \ \nu\sigma}^\rho-\nabla_\nu C_{\ \ \ \mu\sigma}^\rho+C_{\ \ \ \mu\lambda}^\rho C_{\ \ \ \nu\sigma}^\lambda-C_{\ \ \ \nu\lambda}^\rho C_{\ \ \ \mu\sigma}^\lambda
C_{\ \ \ \mu\nu}^\rho=\omega^{-1}\left(\delta_\nu^\rho\nabla_\mu\omega+\delta_\mu^\rho\nabla_\nu\omega-g^{\rho\lambda}g_{\mu\nu}\nabla_\lambda\omega\right)
Carrol then says "it is a matter of simply plugging in and grinding away to get"$$
{\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=R_{\ \ \ \sigma\mu\nu}^\rho-2\left(\delta_{[\mu}^\rho\delta_{\nu]}^\alpha\delta_\sigma^\beta-g_{\sigma[\mu}\delta_{\nu]}^\alpha g^{\rho\beta}\right)\omega^{-1}\nabla_\alpha\nabla_\beta\omega
+2\left(2\delta_{[\mu}^\rho\delta_{\nu]}^\alpha\delta_\sigma^\beta-2g_{\sigma[\mu}\delta_{\nu]}^\alpha g^{\rho\beta}+g_{\sigma[\mu}\delta_{\nu]}^\rho g^{\alpha\beta}\right)\ \omega^{-2}\left(\nabla_\alpha\omega\right)\left(\nabla_\beta\omega\right)
I'm glad it wasn't complicated because getting to that took two dense pages part of which is shown below. He's also used the antisymmetrisation operator [], which is very clever but hard work. It also screws up my latex generator which does not like ]'s in indices.
See that in searchable form at Commentary App G Conformal Transformations.pdf (10 gruelling pages)

Saturday, 18 July 2020

Coordinates and basis vectors

I was still not really sure what is meant by the statement that partials form a coordinate basis. Then if they do, is their character (null, timelike or spacelike) related to the metric? At last I asked on Physics forums and PeterDonis had the final word. So now I do know what is meant by partials forming a coordinate basis and there is some relationship to the metric.

Carroll writes$$
$$"Thus the partials ##\left\{\partial_\mu\right\}## do indeed represent a good basis for the vector space of the directional derivatives, which we can therefore safely identify with the tangent space."

We have a parabola parameterized by ##\lambda## given by$$t=\frac{\lambda^2}{10}\ ,\ \ x=\lambda$$so$$\frac{dt}{d\lambda}=\frac{\lambda}{5}\ \ ,\ \ \ \frac{dx}{d\lambda}=1$$and the tangent vector ##V=d/d\lambda## has components ##\left(dt / d\lambda , dx/ d \lambda\right)## at ##\left(t,x\right)##.

In plane polar coordinates ##\left(r,\theta\right)## the ##\partial_\theta## basis vector is along lines with parameters ##\left(k_r,\lambda\right)##. So the ##\theta## basis vectors lie on concentric circles around the origin. Similarly ##r## basis vectors are radial.
All you need for plane polar coordinates is a line segment to measure ##\theta## from and one end to serve as the origin. The line is normally drawn horizontally. That is not essential.

See how it all hangs together: Commentary 2.3 Coordinates and basis vectors.pdf 
And the thread on physics forums here.