Wednesday, 30 December 2020

Exercise 8.1 N+n+1-dimensional spacetime

My first exercise in almost five months!
Questions
Consider an ##\left(N+n+1\right)##-dimensional spacetime with coordinates ##\left\{t,x^I,y^i\right\}## where ##I## goes from 1 to ##N## and ##i## goes from 1 to ##n##. Let the metric be $$
{ds}^2=-{dt}^2+a^2\left(t\right)\delta_{IJ}{dx}^I{dx}^J+b^2\left(t\right)\gamma_{ij}\left(y\right){dy}^i{dy}^j
$$where ##\delta_{IJ}## is the usual Kronecker delta and ##\gamma_{ij}\left(y\right)## is the metric on an ##n##-dimensional maximally symmetric spatial manifold. Imagine we normalise the metric ##\gamma## such that the curvature parameter $$
k=\frac{R\left(\gamma\right)}{n\left(n-1\right)}
$$is either +1,0 or -1, where ##R\left(\gamma\right)## is the Ricci scalar for the metric ##\gamma_{ij}##.
(a) Calculate the Ricci tensor for this metric
(b) Define an energy-momentum tensor in terms of an energy density ##\rho## and pressure in the ##x^I## and ##y^i## directions, ##p^{\left(N\right)}## and ##p^{\left(n\right)}:##$$
T_{00}=\rho
$$$$
T_{IJ}=a^2p^{\left(N\right)}\delta_{IJ}
$$$$
T_{ij}=b^2p^{\left(n\right)}\gamma_{ij}
$$Plug the metric and ##T_{\mu\nu}## into Einstein's equation and to derive Friedmann like equations for ##a## and ##b## (three independent equations in all).
(c) Derive equations for the energy density and the two pressures at a static solution, where ##\dot{a}=\dot{b}=\ddot{a}=\ddot{b}=0##, in terms of ##k,n## and ##N##. Use these to derive expressions for the equation of state parameters ##w^{\left(N\right)}=\frac{p^{\left(N\right)}}{\rho}## and ##w^{\left(n\right)}=\frac{p^{\left(n\right)}}{\rho}##, valid at the static solution.
Answers
I got some answers and wonder if this is anywhere near string theory. The Ricci tensor was an exercise in tensor splitting and looked quite nice, the Friedmann like equations were less pretty and there was something very wrong with ##w^{\left(n\right)}## in part (c). 

My efforts are here. Seasons greetings: Ex 8.1 N+n+1 dimensions.pdf (7 pages) 

Thursday, 24 December 2020

The Friedmann equations

Expanding Universe
In section 3.8 we start with the Robertson-Walker metric, which expands at a rate ##a^2\left(t\right)## (##a## is the dimensionless scale factor), and the energy-momentum tensor of a perfect fluid which we use to model the universe, we then use the equation of state ##p=w\rho## and the conservation of the energy-momentum tensor ##\nabla_\mu T^{\mu\nu}=0## to find that the energy density is proportional to some power of the scale factor: ##\rho\propto a^{-3\left(1+w\right)}##. Energy conditions give us an idea of possible values of ##w##. 

But then we go on to find values of ##w## for a 
  • matter dominated universe (now) which gives ##\rho_M\propto a^{-3}## 
  • radiation dominated universe (early) which gives ##\rho_R\propto a^{-4}## 
  • vacuum dominated (late / de Sitter and anti-de Sitter) ##\rho_\Lambda\propto a^0## 
The first has an energy-momentum tensor for dust, the second for electromagnetism and the third from another sort of perfect fluid where ##p=-\rho## is the equation of state.

We then apply Einstein's equation and using the metric, the energy-momentum tensor and Ricci tensor for this metric that we found in the previous section we get the Friedmann equations which define Friedmann-Robertson-Walker (FRW) universes and can determine the evolution of the scale factor ##a##. 

Finally I tried evolving the ##a## in the matter dominated universe and rewrote the Friedmann equations using the conventional metric

Find out what the equations are and how to get them Commentary 8.3 Friedmann equation.pdf (7 pages)

Monday, 21 December 2020

Maxwell's equations have something missing

James C. Maxwell is peeved
I return to chapter one and Maxwell's equations.  Carroll's version of Maxwell's equations "in 19th century notation" at his equations 1.92 are$$
\nabla\times\mathbf{B}-\partial_t\mathbf{E}=\mathbf{J}$$
$$\nabla\bullet\mathbf{E}=\rho$$
$$\nabla\times\mathbf{E}+\partial_t\mathbf{B}=0 $$
$$\nabla\bullet\mathbf{B}=0$$
But that's not Maxwell's equations! In SI units they are

$$\nabla\times\mathbf{B}-\frac{1}{c^2}\partial_t\mathbf{E}=\mu_0\mathbf{J}$$
$$\nabla\bullet\mathbf{E}=\frac{\rho}{\epsilon_0}$$
$$\nabla\times\mathbf{E}+\partial_t\mathbf{B}=0$$
$$\nabla\bullet\mathbf{B}=0$$
Where ##\epsilon_0,\mu_0## are the electric and magnetic constants. Carroll has already announced that we were setting the speed of light ##c=1## but he does not say anything about ##\epsilon_0,\mu_0## here or later. They are just left out. I first noticed this when calculating the Energy-Momentum tensor for electro-magnetic radiation. I think that in Carrol's "natural units" (which might be a bit unorthodox) we also have ##\epsilon_0=1## and that means that, as ##c=1##, ##\mu_0=1## too.

See my reasoning in Commentary 1.8 Maxwells equations and units.pdf (two pages). 

Wednesday, 16 December 2020

Energy conditions

The equation of state of the prefect fluid that models the Friedmann universe is $$
p=w\rho
$$where ##p,\rho## are pressure and energy density and ##w## is a constant. In section 4.6 Carroll describes five 'popular' energy conditions on the energy-momentum tensor without saying why they are popular or even why they are interesting and shows that for any of them to be true we must have ##w\geq-1## in the equation of state. The energy conditions, which sound like characters in a robot movie, are WEC, NEC, DEC, NDEC and SEC. Possible values of ##\rho,p## are shown in the diagram below for these energy conditions along with possible values when ##w\geq-1## in the equation of state. Assuming that our Friedmann universe has ##\rho>0## (highly plausible) and satisfies one of the robots (haven't got the faintest idea) then we must have ##w\geq-1##.  

From the equation of state ##w\geq-1## gives$$
\frac{p}{\rho}\geq-1\Rightarrow p\geq-\rho\Rightarrow p+\rho\geq0
$$which is the same as the NEC. So why aren't the pictures (b) and (f) the same? I hope you've seen my error. Luckily I did and have avoided falsely accusing Carroll of a slip up.
Carroll's picture with my adornments.

































Here is my limited understanding: Commentary 4.6 Energy conditions.pdf (3 pages)

Saturday, 12 December 2020

What the flux? The energy momentum tensor for a perfect fluid.

I have a guilty secret: I have never really understood the energy momentum tensor (aka stress-energy tensor) ##T^{\mu\nu}##. Carrol talks about it often and gives some definition which I just accept. It is obviously important - it is one of the terms, the source of the gravitational field,  in Einstein's equation - and it's about time I understood it better. A good example is back in section 1.9 where we are given the energy momentum tensor for a perfect fluid in its rest frame as $$
T^{\mu\nu}=\left(\begin{matrix}\rho&0&0&0\\0&q&0&0\\0&0&q&0\\0&0&0&q\\\end{matrix}\right)
$$##\rho## is energy / mass density and ##q## is pressure. (I use ##q## not ##p## for pressure otherwise I get it confused with ##p^\mu## the four-momentum which we also use - immediately). Carroll writes "This symmetric (2,0) tensor tells us all we need to know about the energy-like aspects of a system: energy density, pressure, stress and so forth. A general definition of ##T^{\mu\nu}## is 'the flux of four-momentum ##p^\mu## across a surface of constant ##x^\nu##'."

I also realise that I have only a dim understanding of what 'flux' really means. Wikipedia has nine pages on it. It is not simple and all the examples are on three dimensions.

If you can get to that nice tensor above with all the zeros and one ##\rho## and three ##q## 's it is easy to move on to the general expression for the tensor in GR which is$$
T^{\mu\nu}=\left(\rho+q\right)U^\mu U^\nu+qg^{\mu\nu}
$$I now think that flux is the amount of stuff that crosses a surface. Obviously the bigger the area, the
more stuff crosses and you want the flux at a point, so we must have the amount of stuff per (unit) area that crosses the surface at the point. In four dimensions a surface will be flat against three other surfaces in three pairs of the other dimensions. Get it? So if you just add up the flux of stuff through each of the three areas then you get the total flux through the surface you started with. The diagram shows the case for flux of mass through three surfaces of constant time. Using an argument like this I was able to justify the ##\rho## and ##q##'s in the tensor. The zero's and the symmetry elude me but the latter becomes clear from the formula for ##T^{\mu\nu}##. I think that Wikipedia is more plausible than Carroll on the spatial components. The ones in the top row and left column are more mysterious.

For some bodgy calculations see Commentary 8.3 Energy-Momentum tensor.pdf (5 pages).

Friday, 4 December 2020

Robertson-Walker metrics

In section 8.2 we meet what Carroll calls the Robertson-Walker metrics:$$
{ds}^2=-{dt}^2+R^2\left(t\right)\left[\frac{{d\bar{r}}^2}{1-k{\bar{r}}^2}+{\bar{r}}^2{d\Omega}^2\right]=-{dt}^2+a^2\left(t\right)\left[\frac{{\rm dr}^2}{1-\kappa r^2}+r^2{d\Omega}^2\right]
$$The second version is Carroll's preferred form - 'flouting' conventional wisdom.


Four people found the equations in various teams. They are Alexander Friedmann (Russian), Georges Lemaรฎtre (Belgian), Howard Robertson (USian) and Arthur Walker (British) and the metrics are often named after one or some or all. Carroll favours the English speakers.

As usual I check Carroll's equations and (not so usual) have three minor complaints.
1) When he gives the Christoffel symbols at his equation 8.44 the third line is$$
\Gamma_{01}^1=\Gamma_{02}^2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Gamma_{03}^3=\frac{\dot{a}}{a}
$$there should be an ##=## sign between ##\Gamma_{02}^2## and ##\Gamma_{03}^3##!
2) Just after equation 8.35 he says that a flat 3-manifold could be described by 'a more complicated manifold such as the three-torus ##S^1\times S^1\times S^1##'. How can a torus, something generated by circles, be flat? I think he should have said a flat thee-torus. It can be generate from three circles but not as simply as the usual doughnut which is not flat.
3) Before that, just after equation 8.29, he says ##k## sets the curvature and therefore the size of the spatial surfaces. One might think that ##k=0## sets the size as infinite (flat curvature) but the aforementioned flat three-torus is a counter example with ##k=0## and finite size. So ##k=0## does not set the size as you might have expected. Similar arguments apply when ##k<0##.

I learned a few other things too: I suspected that if the metric is diagonal then the Ricci tensor must be diagonal. Not true! But on the plus side I found Win's supercharged formula for calculating Ricci tensor components in that case. It avoids the need to calculate Christoffel and Riemann components. I also learned a bit more about smooth isometric embedding.

I also noticed something slightly mysterious that Carroll does not draw to our attention: In the closed universe case (finite size?) the radial coordinate ##\bar{r}## is constrained by ##{\bar{r}}^2<1##. So the universe might be even more closed than we thought.

Read it all at Commentary 8.2 Robertson-Walker Metrics.pdf (7 pages + 5 of calculations)


 

Wednesday, 2 December 2020

Curvature of a torus

Torus from Wikipedia


Carroll seems to say that a three-torus can have curvature zero everywhere. A two-torus does not. 






It has a curvature ##S## $$S=\frac{2\cos{\theta}}{r\left(R+r\cos{\theta}\right)}$$where ##r,R,\theta## are shown on my wire diagram. 





So the curvature is not constant. So a torus is not maximally symmetric. The curvature vanishes at the top and bottom of the torus, is negative on the inside and positive on the outside which is exactly what you would expect. Note that it gets more negative than it gets positive - probably because the inside is closer to the centre and therefore has to curve harder. If ##R=20## the minima and maxima are almost equally far from zero. 

Wrong!

I asked for help on Physics Forums here and you can have a flat torus as shown in the diagram on the right. It is is like a rectangle size ##A\times B## with coordinates ##x,y## but if you go off an edge, you come back to the opposite edge, so adding multiples of ##A## to the ##x## coordinate of ##B## to the ##y## coordinate also keep you in the same place. Pretty similar to the regular torus where adding ##2\pi## to ##\theta## or ##\phi## does the same. The only trick is that you have to start with two circles in four dimensions not three! So Carroll was messing about in six dimensions.

Here's why Commentary 8.2 Tori.pdf. (4 pages).

Saturday, 28 November 2020

Diagonal metric and Ricci tensor

Gregorio Ricci-Curbastro
Just as I was wrapping up section 8.2 on the Robertson-Walker I wondered if the Ricci tensor was always diagonal if the metric was diagonal. The R-W metric and Ricci tensor are both diagonal and Carroll had not said anything about off diagonal components of the Ricci tensor except that they vanished. I was dissatisfied and about to ask on Physics Forums. Then I decided to try to check myself, fumbled for a bit and then found a 1996 paper by K.Z. Win. I suppose in this day and age of Mathematica and the like it has sunk into obscurity but nevertheless it was interesting and useful for me who lacks such tools. 

Win shows that for a diagonal metric the diagonal components of the Ricci tensor are$$
4R_{\mu\mu}=\left(\partial_\mu\ln{\left|g_{\mu\mu}\right|}-2\partial_\mu\right)\partial_\mu\ln{\left|\frac{g}{g_{\mu\mu}}\right|}-\sum_{\sigma\neq\mu}\left[\left(\partial_\mu\ln{\left|g_{\sigma\sigma}\right|}\right)^2+\left(\partial_\sigma\ln{\frac{\left|g\right|}{{g_{\mu\mu}}^2}}+2\partial_\sigma\right)g^{\sigma\sigma}\partial_\sigma g_{\mu\mu}\right]
$$and the off diagonal components are $$
4R_{\mu\nu}=\left(\partial_\mu\ln{\left|g_{\nu\nu}\right|}-\partial_\mu\right)\partial_\nu\ln{\left|\frac{g}{g_{\mu\mu}g_{\nu\nu}}\right|}+\left(\mu\leftrightarrow\nu\right)-\sum_{\sigma\neq\mu,\nu}{\partial_\mu\ln{\left|g_{\sigma\sigma}\right|}\partial_\nu\ln{\left|g_{\sigma\sigma}\right|}}
$$where ##\mu\neq\nu##, there is no summation over ##\mu,\nu## and ##\left(\mu\leftrightarrow\nu\right)##  stands for preceding terms with ##\mu,\nu## interchanged. The partial derivative operator ##\partial_\tau## usually only applies to what immediately follow. ##\partial_\tau\partial_\tau## is a second order derivative. The exception is the final ##2\partial_\sigma## in the first expression which applies to the whole of ##g^{\sigma\sigma}\partial_\sigma g_{\mu\mu}##.

It turns out that in two dimensions a diagonal metric does imply a diagonal Ricci tensor - but we knew that. In more dimensions it's not so simple. It is necessary to calculate the off diagonal components of the R-W Ricci tensor.

To calculate the Ricci tensor more directly one would use $$
R_{\mu\mu}=R_{\ \ \mu\sigma\mu}^\sigma=\partial_\sigma\Gamma_{\mu\mu}^\sigma-\partial_\mu\Gamma_{\sigma\mu}^\sigma+\Gamma_{\sigma\rho}^\sigma\Gamma_{\mu\mu}^\rho-\Gamma_{\mu\rho}^\sigma\Gamma_{\sigma\mu}^\rho
$$ and$$
R_{\mu\nu}=R_{\ \ \mu\sigma\nu}^\sigma=\partial_\sigma\Gamma_{\nu\mu}^\sigma-\partial_\nu\Gamma_{\sigma\mu}^\sigma+\Gamma_{\sigma\rho}^\sigma\Gamma_{\nu\mu}^\rho-\Gamma_{\nu\rho}^\sigma\Gamma_{\sigma\mu}^\rho
$$and once the Christoffel symbols have been calculated, which is quite easy in the R-W case,  the former takes about the same number of lines as Win's formula but the latter is probably much less efficient than Win. However Win's method avoids calculating Christoffel symbols at all.
 
Read the full details at Commentary 8.2 Diagonal metric and Ricci tensor.pdf (11 pages)

Tuesday, 17 November 2020

Constants and calculations

Babylonian equations
Carroll uses natural units in his book as do I on this web site. In natural units the speed of light, Planck's constant and Boltzmann's constants ##c=\hbar=h/2\pi=k=1##. In our equations these terms are left out. They need to be put in again to do real calculations and it's not always obvious how to do it. When Carroll does calculations he often uses centimeters and grams. I prefer to use SI units: meters and kilograms. 

It is also common to use geometric units which are units in which ##c=G=1## (so mass and energy have the same units as length and time). You cannot also have ##\hbar=1## in these units.

In natural units the Schwarzschild radius is ##R_S=2GM##. I suppose that in geometric units it is ##R_S=2M##. In SI units it is ##2GM/c^2##. The ##c^2## makes a big difference and is easy to forget!

The document contains a list in SI units of the values of universal constants and other handy constants like the mass of the sun. The table also gives the conversion factors from natural to SI units, which are derived from their dimensions. Two examples of their use are given at the end followed by a list of small and large prefixes (Terra, peta, pico etc) and their meanings. 

Tuesday, 10 November 2020

Plotting geodesics of Schwarzschild

I did some experiments with trying to plot solutions to the geodesics of Schwarzschild. They might give orbits of stars round the black hole Sagittarius A* at the centre of our galaxy. The geodesic equations become three simultaneous second order differential equations which give what I call the chugger equations below:
##t## is the coordinate time and ##r,\theta## are the usual plane polar coordinates. We are only considering curves in one plane. A prime indicates a derivative with respect to ##\lambda## which is the affine parameter from the geodesic equation and could be proper time. ##G## is old Newton's gravitational constant, ##M## is the mass of Sagittarius A* and ##c## is the speed of light.

We start with some initial values of ##t,t^\prime,r,r^\prime,\theta,\theta^\prime## and select some ##d\lambda## . Then we use those to get new values of  ##t,t^\prime,r,r^\prime,\theta,\theta^\prime## from the chugger equations again and again and again and put them in successive rows of a spreadsheet. The spreadsheet then uses the values of ##r,\theta## to plot the curve. One of the stars orbiting Sagittarius A* is S2 and we use that as our example. It completes an orbit about every 16 years. The first image was my first attempt. The shape is good (it follows the top part of the Newtonian ellipse) until it gets pretty close to the black hole. The second image shows a close up of the same curve near the black hole.

The dashed line is the ellipse that S2 would follow according to Newton's equations.



If I 'manually' decreased ##d\lambda## when near the black hole and then increased it again as S2 departed I could get the third image which shows the last leg of the approximation. To be able to do that conveniently I had to program the chugger equations in VBA. To do 1,000 iterations takes about three minutes.









Finally I did all the calculations in Excel adjusting ##d\lambda## as it went along and got the fourth image from a 20,000 row spreadsheet. The calculation time is about one second. It was the best so far but still not good enough. The furthest distance (apsis) of S2 from the black hole decreases by 4% - it should be the same. However the furthest distance did advance by 0.0012 radians. I calculated (with help from Carroll) that it should be 0.0035 radians.









For a bit of fun I also made an artist's impression of precession of the perihelion. It's simply done from the exact solution to Newton's equations which is an ellipse.

Further material including spreadsheets and VBA at

Tuesday, 3 November 2020

Solar orbit plotter

Here's how the plotter works
Newton's second law is$$
\vec{F}=m\vec{a}
$$Newton's law of gravity is$$
\vec{F}=\frac{GMm}{r^2}
$$They give you two differential equations of motion in polar coordinates$$
\frac{d^2r}{dt^2}-r\left(\frac{d\theta}{dt}\right)^2=-\frac{GM}{r^2}
$$$$
2\frac{dr}{dt}\frac{d\theta}{dt}+r\frac{d^2\theta}{dt^2}=0
$$Kepler's laws are
  1. The orbit of a planet is an ellipse with the Sun at one of the two foci.
  2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
  3. The square of a planet's orbital period is proportional to the cube of the length of the semi-major axis of its orbit.
The second differential equation quickly gives you Kepler's second law. It is more difficult to get$$
r=\frac{P}{1+e\cos{\left(\theta-\phi\right)}}
$$That is a circle when ##e=0##, an ellipse when ##\left|e\right|<1##, a parabola when ##\left|e\right|=1## and a hyperbola when ##\left|e\right|>1##. ##\phi## is the angle of the axis of symmetry of the curve. ##P## is a magic constant. The equation gives you Kepler's first law. 

Newton must have been very pleased when he did those. Part of the job was inventing calculus! 

The initial conditions determine ##P,e,\phi##. That requires more work if you only know the initial position and velocity.

Full details (except inventing calculus) in Commentary 5.4 Orbital toypdf. (9 pages)
The spreadsheets for plotting the curves are at 
Commentary 5.4 Orbital toy.xlsm (with animation macros)

Thursday, 22 October 2020

Maximally symmetric universes

In section 8.1 we meet maximally symmetric universes. That's universes where every point in spacetime is the same. I think the only ones are de Sitter, Minkowski and Anti de Sitter. We've done flat, boring Minkowski. De Sitter and Anti de Sitter are more interesting and we do conformal diagrams for both of them. At the end of the section Carroll does conformal anti de Sitter and casually draws some geodesics on it without saying how he plotted them. I was able to put my newly learnt skills to good use and show the same curves as he did๐Ÿ˜€. My version is on the left.

He also jumps back to section 3.1 and uses the equation there for the Riemann tensor. It is$$
R_{\rho\sigma\mu\nu}=\kappa\left(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu}\right)
$$where ##\kappa## is a constant. That equation easily becomes$$
R_{\ \ \ \sigma\mu\nu}^\rho=\kappa\left(\delta_\mu^\rho g_{\sigma\nu}-\delta_\nu^\rho g_{\sigma\mu}\right)
$$so it looks like there's a really way to calculate the fiendish Riemann tensor for these special cases. Unfortunately $$
\kappa=\frac{R}{n\left(n-1\right)}
$$where ##n## is the number of dimensions and ##R## is the curvature scalar (aka Ricci scalar) which is constant in a maximally symmetric universe. Nevertheless to calculate it you have to calculate the Riemann tensor first! B*gger.

I still wanted to check that it was all true. We already did it for a the surface of a unit sphere, S², and have found it was constant at 2 and its Riemann tensor does satisfy the formula above. So S² is maximally symmetric. All points on the surface of a sphere are equal. The metric for conformal anti de Sitter is$$
{ds}^2=\frac{\alpha^2}{\cos^2{\chi}}\left(-{dt^\prime}^2+{d\chi}^2+\sin^2{\chi}{d\Omega_2}^2\right)
$$where ##\alpha## is some constant. So its Riemann tensor is trickier!

As I had already calculated the Christoffel symbols for that to make the diagram above, I calculated its Riemann tensor and then its curvature scalar which is ##-12\alpha^{-2}##. So it is constant and satisfies the formula as I checked with a cunning spreadsheet.

Calculating that Riemann tensor, which is a relatively easy one, took 9 hours 45 minutes over three days and six sittings. I got faster as I went along, notwithstanding long lunches preceded by a refreshing dry martini.

Read all about it at
and

Friday, 9 October 2020

Plotting geodesics numerically

Geodesic equations are sets of second order differential equations and are usually somewhere between hard and impossible to solve analytically. The following are the geodesic equations for the surface of a sphere (S²):
$$\frac{d^2\theta}{d\lambda^2}-\sin{\theta}\cos{\theta}\left(\frac{d\phi}{d\lambda}\right)^2=0$$
$$\frac{d^2\phi}{d\lambda^2}+2\cot{\theta}\frac{d\phi}{d\lambda}\frac{d\theta}{d\lambda}=0$$They are about the simplest you can get and I still don't know if it's possible to solve them. When you have the solution you will be able to plot the geodesic curves as on the graph above.

This road block is very annoying and I have finally busted through it. It only took a couple of days! I started with some very simple examples to test that what I was doing was correct and tested the theory on said S² which I had explored in March 2019. It all works and the general procedure for constructing geodesics is quite straightforward really! What a nice surprise.๐Ÿ˜€

Read all about it here Plotting a differential equation.pdf (8 pages with lots of picture). It's a short instruction manual on how to do it yourself.

Wednesday, 30 September 2020

Very obscure bug in Google Blogger

I use Google Blogger for this blog. I also use MathJax for displaying equations which are in a code called Latex. Google recently introduced major changes to Blogger and it screwed up the equation display. I prepare text for a post in MS-Word so we might have something like this (but replace all £ signs by $ signs)
££\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0
££
and it should come out like this$$
\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0
$$but sadly it now comes out like this
$$
\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0
$$
which is not what is wanted!

To move the Latex code from MS-Word was simply a matter of copying it and pasting it as plane text (Ctrl+Shift+V).

In the old version of Blogger linefeeds produced HTML <br/>, in the new version they produce </div><div>, which screws up Mathjax. The old version and the new version of the HTML are shown below (once again, replace all £ signs by $ signs)

££ <br/>\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0<br/>££

££ 
</div><div>\chi_\nu\nabla_\mu\chi_\sigma+\chi_\sigma\nabla_\nu\chi_\mu+\chi_\mu\nabla_\sigma\chi_\nu=0 </div><div>££

Solution

  1. To begin with I just edited the latex removing linefeeds and reinstating with shift-enter. That replaces  </div><div> by <br/>. You also have to make sure that the text style is Normal not Paragraph. The editor sometimes seems to start in the latter mode.
  2. Then I discovered Search replace in HTML editing view. So now I just replace all </div><div> by <br/>.
It's still rather tedious. I wish the lovely people at Google would fix it.

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 $$
R=\frac{2}{g_{\phi\phi}}\left(\partial_\theta\Gamma_{\phi\phi}^\theta-\partial_\phi\Gamma_{\theta\phi}^\theta+\Gamma_{\theta\theta}^\theta\Gamma_{\phi\phi}^\theta+\Gamma_{\theta\phi}^\theta\Gamma_{\phi\phi}^\phi-\Gamma_{\phi\theta}^\theta\Gamma_{\theta\phi}^\theta-\Gamma_{\phi\phi}^\theta\Gamma_{\theta\phi}^\phi\right)
$$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
\begin{align}
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$$
R_{El}=\frac{2b^2}{\left(a^2\cos^2{\theta}+b^2\sin^2{\theta}\right)^2}
$$$$
R_{Ep}=\frac{2a^2}{\left(1+a^2r^2\right)^2}
$$$$
R_{Hp}=\frac{-8}{\left|g\right|^2a^2b^2}
$$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$$
{ds}^2=-{dt}^2+t^{2q}\left({dr}^2+r^2{d\Omega}^2\right)
$$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##)


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:$$
{ds}^2=-{dt}^2+{dr}^2+r^2\left({d\theta}^2+\sin^2{\theta}{d\phi}^2\right)
$$
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) 

Tuesday, 4 August 2020

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).

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 $$
{\widetilde{g}}_{\mu\nu}=\omega^2g_{\mu\nu}
$$Then we want find things like the Riemann tensor in the 'conformal frame'. It's quite easy to show that it is$$
{\widetilde{R}}_{\ \ \ \sigma\mu\nu}^\rho=R_{\ \ \ \sigma\mu\nu}^\rho+\nabla_\mu C_{\ \ \ \nu\sigma}^\rho-\nabla_\nu C_{\ \ \ \mu\sigma}^\rho+C_{\ \ \ \mu\lambda}^\rho C_{\ \ \ \nu\sigma}^\lambda-C_{\ \ \ \nu\lambda}^\rho C_{\ \ \ \mu\sigma}^\lambda
$$where$$
C_{\ \ \ \mu\nu}^\rho=\omega^{-1}\left(\delta_\nu^\rho\nabla_\mu\omega+\delta_\mu^\rho\nabla_\nu\omega-g^{\rho\lambda}g_{\mu\nu}\nabla_\lambda\omega\right)
$$
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$$
\frac{d}{d\lambda}=\frac{dx^\mu}{d\lambda}\partial_\mu
$$"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.

Tuesday, 14 July 2020

The Metric

Kruskal from From Stanford
I started to write this in May 2019 when I read section 2.5 and it has remained a work in progress until now. I have quite a collection of metrics: 3D Euclidean, Plane polar, Spherical polar, Surface of sphere (S2), Minkowski, Spherical Minkowski, Schwarzschild, Eddington-Finkelstein, Kruskal. I will add more as I find them.

We start with a bit of general information about a metric, then list each of the metrics. There's also something on null, spacelike and timelike coordinates; converting metrics to coordinate systems with an example on spherical metrics; four velocities and mysterious interchange of  ##\mathrm{d}x## and ##dx##.

On this page the coordinate infinitesimals (which are really one forms) should be written with an unitalicized ##\rm{d}## for example ##\mathrm{d}x## not ##dx##. That is very fiddly in Latex so I have not bothered except in the very first occurrence! (Most texts don't bother at all).

The glory of the metric

The metric ##g_{\mu\nu}## is a symmetric (0,2) tensor and Carroll lists the following to appreciate its "glory". He is following Sachs and Wu (1977).
  1. The metric supplies a notion of "past" and "future".
  2. The metric allows the computation of path length and proper time.
  3. The metric determines the "shortest distance" between two points, and therefore the motion of test particles.
  4. The metric replaces the Newtonian gravitational field ##\phi##.
  5. The metric provides a notion of locally inertial frames and therefore a sense of "no rotation".
  6. The metric determines causality, by defining the speed of light faster than which no signal can travel.
  7. The metric replaces the traditional Euclidean three-dimensional dot product of Newtonian mechanics.
  8. The (inverse) metric lowers (raises) indices: ##U_\mu=g_{\mu\nu}U^\nu\ ,\ U^\mu=g^{\mu\nu}U_\nu##
I added the last one. It is related to the second last.

Two and three dimensional geometry metrics

These are called Euclidean or Riemannian metrics.
3D Euclidean$${ds}^2=\mathrm{d}x^2+\mathrm{d}y^2+\mathrm{d}z^2$$$$g_{ij}=\left(\begin{matrix}1&0&0\\0&1&0\\0&0&1\\\end{matrix}\right)$$we explain why the equation and matrix formulations are equivalent. 
Plane polar coordinates
Coordinates ##(r,\theta)## radial, angle to ##x## axis$$g_{ij}=\left(\begin{matrix}1&0\\0&r^2\\\end{matrix}\right)$$Spherical polar 
Coordinates ##(r,\theta,\phi)## radial, polar, azimuthal (= longitude)$$g_{ij}=\left(\begin{matrix}1&0&0\\0&r^2&0\\0&0&r^2\sin^2{\theta}\\\end{matrix}\right)$$Surface of sphere (S2)
Coordinates ##(\theta,\phi)## polar, azimuthal$$g_{ij}=\left(\begin{matrix}1&0\\0&\sin^2{\theta}\\\end{matrix}\right)$$This metric is often written $${d\Omega}^2={d\theta}^2+\sin^2{\theta}{d\phi}^2$$

Relativistic metrics

These are called Lorentzian or pseudo-Riemannian metrics. 
We use a ##-+++## signature and the speed of light ##c## is conveniently set to 1.
Minkowski
Coordinates ##(t,x,y,z)## $$g_{\mu\nu}=\left(\begin{matrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\\end{matrix}\right)$$Spherical Minkowski
Coordinates ##(t,r,\theta,\phi)## ##r,\theta,\phi## as in spherical$$g_{\mu\nu}=\left(\begin{matrix}-1&0&0&0\\0&1&0&0\\0&0&r^2&0\\0&0&0&r^2\sin^2{\theta}\\\end{matrix}\right)$$We could also write$${ds}^2=-d\tau^2=-{dt}^2+{dr}^2+r^2{d\Omega}^2$$Schwarzschild metric (spherical polar)
Coordinates are ##(t,r,\theta,\phi)## time, radial, polar, azimuthal$${ds}^2=-d\tau^2=-\left(1-\frac{2GM}{r}\right){dt}^2+\left(1-\frac{2GM}{r}\right)^{-1}{dr}^2+r^2\left({d\theta}^2+\sin^2{\theta}{d\phi}^2\right)$$We often replace ##2GM## (twice Newton's gravitational constant times the central mass) by ##R_s## the Schwarzschild radius which is the radius of event horizon and the last part by ##r^2{d\Omega}^2##. 
Eddington-Finkelstein metric
The Eddington-Finkelstein metric is for the same spacetime as Schwarzschild but with coordinates ##\left(v,r,\theta,\phi\right)## $${ds}^2=-\left(1-\frac{R_s}{r}\right)dv^2+dvdr+drdv+r^2{d\Omega}^2$$$$v=t+r+R_s\ln{\left|\frac{r}{R_s}-1\right|}$$This is the first metric we have listed which is not diagonal.
Kruskal predecessor
On the way to the Kruskal metric we get coordinates ##\left(v^\prime,u^\prime,\theta,\phi\right)## with metric equation$${ds}^2=-\frac{2{R_s}^3}{r}e^{-\frac{r}{R_s}}\left(dv^\prime du^\prime+du^\prime dv^\prime\right)+r^2{d\Omega}^2$$where ##r## is implicitly defined in terms of ##u^\prime,v^\prime## as$$u^\prime v^\prime=-\left(\frac{r}{R_s}-1\right)e^{r/R_s}$$Kruskal
Kruskal coordinates are ##\left(T,R,\theta,\phi\right)## where in terms of Schwarzschild ##t,r## $$T=\left(\frac{r}{R_s}-1\right)^{1/2}e^\frac{r}{2R_s}\sinh{\left(\frac{t}{2R_s}\right)}$$$$R=\left(\frac{r}{R_s}-1\right)^{1/2}e^\frac{r}{2R_s}\cosh{\left(\frac{t}{2R_s}\right)}$$and they give a metric equation$${ds}^2=\frac{4{R_s}^3}{r}e^{-\frac{r}{R_s}}\left(-dT^2+dR^2\right)+r^2{d\Omega}^2$$with ##r## implicitly defined from$$T^2-R^2=\left(1-\frac{r}{R_s}\right)e^\frac{r}{R_s}$$

Read all that and more on  at Commentary 2.5 The Metric.pdf (11 pages)


Thursday, 9 July 2020

Kruskal coordinates and the maximally extended Schwarzschild solution

Kruskal coordinates ##\left(T,R,\theta,\phi\right)## are called 'maximally extended' because they cover the whole spacetime except the true singularity at ##r=0##. Indeed, they find some remarkable new regions of spacetime! The history of the discoveries spans 45 years from Einstein and Schwarzschild (1915) to Kruskal (1960).

The Kruskal coordinates are related to Schwarzschild coordinates ##\left(t,r,\theta,\phi\right)## by $$T=\left(\frac{r}{R_s}-1\right)^{1/2}e^\frac{r}{2R_s}\sinh{\left(\frac{t}{2R_s}\right)}$$
$$R=\left(\frac{r}{R_s}-1\right)^{1/2}e^\frac{r}{2R_s}\cosh{\left(\frac{t}{2R_s}\right)}$$and give a metric equation$${ds}^2=\frac{4{R_s}^3}{r}e^{-\frac{r}{R_s}}\left(-dT^2+dR^2\right)+r^2{d\Omega}^2$$with ##r## implicitly defined from$$T^2-R^2=\left(1-\frac{r}{R_s}\right)e^{r/R_s}$$From these we can draw a Kruskal diagram  showing lines of constant ##t## and ##r## and light cones which miraculously are always at 45° just like in flat spacetime.

The regions above the upper ##r=0## line and below the lower ##r=0## line are not part of spacetime. (##r<0## and ##t>+\infty## or ##t<-\infty## in them). The rest of the diagram is divided into four regions.

I    Right quarter. Normal space time outside the event horizon.๐Ÿ‘
II   Below the upper ##r=0## line and above the upper ##r=R_s## lines. Inside the event horizon.๐Ÿ‘Ž
III  Above the lower ##r=0## line and below the lower ##r=R_s## lines. The white hole.๐Ÿ’ฃ
IV  Left quarter. The unreachable mirror image of normal space time.๐Ÿ‘ป

The red light cones, which are always at 45°, are informative. In region I you can always maintain a fixed ##r## and you can always move up towards and into region II. You can never get into regions III or IV. Once in region II it is impossible to maintain constant ##r## because lines of constant ##r## are always flatter than 45° so you inevitably arrive at ##r=0##. Region IV is like region I and you can only get to region II from it. Carroll says region III is a the time reverse of II and can be thought of as a white hole. "There is a singularity in the past, out of which the universe appears to spring". Things can only come out of it. They can just get directly to II through the origin, more likely they will go into I or IV.

Regions II and III are allowed even though ##t>+\infty## in II and ##t<-\infty## in III.

Although region I and region IV are mutually unreachable, if an intrepid explorer went from region I into region II, they would be able to see some things that had happened in region IV. Likewise an explorer from region IV could have a brief look at region I before perishing.

Find out how to get to Kruskal and a few other things in Commentary 5.7 Kruskal coordinates.pdf (11 pages).


Saturday, 27 June 2020

Time travel inside a black hole

We can now use the geodesic that we have found and tested to plot a spacetime diagram all the way to the centre of a black hole. The graph shows the result for our famous beacon, or even for a foolish astronaut.

As we saw before they dawdle at the event horizon for ever (##t\rightarrow\infty## as ##r\rightarrow2GM##). But we know you can cross an event horizon and when that happens in the distant future they hurry backwards in time and soon get to a reasonable ##r,t##! 

Meanwhile the astronaut looking at a wristwatch sees proper time, ##\tau##, ticking by steadily and reaches the centre in finite time.

Luckily the astronauts time travel cannot be observed from the outside. Our red faces are saved.

Given everything we have done before the calculations are simple. It's a one pager in Commentary 5.6 Time Travel.pdf.

Schwarzschild Black Holes - The Geodesic


(1) and (2) are the ##t## and ##r## geodesic equations in the Schwarzschild metric which we found in section  5.4. (2) becomes a bit simpler on a radial path. That's (3). Geodesic equations are meant to give you trajectories of freely falling particles parametrised by ##\lambda##. (4) and (5) are the equation of a particle (or beacon) falling along a radius into a black hole from a distance ##r_*##. We used them to plot a beacon's path here. They come from (6) which we calculated in exercise 5.5 where we also calculated (7). 
##t## is coordinate time, ##r## is the distance from the centre. They are the coordinates.
##\lambda## is an affine parameter (it is proportionate to the length along the line).
##R_s## is the Schwarzschild radius (radius of event horizon).
##\theta,\phi## are the other spherical polar coordinates (polar and azimuth), which we can ignore.
##r_*## is the radial distance from which the test particle, or beacon, is dropped.
##\tau## is the proper time, which can be an affine parameter for a massive particle.

The question is: Is the path given by (4) and (5) a real geodesic? That is, does it satisfy (1) and (3)?

And the answer is YES.

I studied these geodesics three months ago and they have never yet been useful. This is the first time that they have even clicked with anything!

Here's how they click: Commentary 5.4#1 Geodesics of Schwarzschild.pdf (only 2 pages really)

Wednesday, 24 June 2020

Proper acceleration, Spaghettification and G2

After that Exercise 5.5 I thought I was an expert on falling into a black hole and and that I could calculate the 'proper acceleration' and spaghettification which is the term used for what happens when acceleration differs so much in different parts of your body that you get stretched out like the doomed lady on the right.

Then there is there is the matter of the gas cloud known as G2, which was discovered heading towards Sagittarius A*, the black hole at the centre of our galaxy, in 2011 by some folk at the Max Planck Institute. G2 was destined to come closest to Sagittarius A* in Spring 2014 "with a predicted closest approach of only 3000 times the radius of the event horizon". There was great excitement because spaghettification and great fireworks were expected. However nothing much happened and G2 continues on its way, orbiting Sagittarius A*.

I attempted to do some calculations and tested them on Physics Forums and got adverse comments from PeterDonis. Ibix was more positive "I think your maths is correct ...". PeterDonis showed me the 'correct' way of calculating proper acceleration and then dragged me back to the geodesic deviation equation which is the right way to calculate spaghettification. But I still don't fully understand said equation and how to use it๐Ÿ˜ญ. I also learnt a bit more about units: 'natural' and 'geometric'. PeterDonis is a hard task master.

The correct way to calculate proper acceleration gives infinite acceleration at the event horizon. One benefit of that is that it tells you that you cannot escape falling through it, once close enough. That is true. The drawback is that it makes the radial change in acceleration also infinite. So you will get ripped up at the event horizon. That is not true given a big enough black hole.

Here are my calculations (about four pages) and what I learnt (another three).
Commentary 5.6#4 Proper acceleration.pdf 

Monday, 15 June 2020

Beacon falling into a black hole, revisited

Having done Exercise 5.5 and discovered lots of things about a beacon being dropped into a black hole we can now revisit Commentary 5.6 where we struggled with Carroll's figure 5.8 and Carroll's claims about increasing intervals observed at a safe distance ##r_\ast## from the centre of a black hole. Back then I ran into several problems: 1) I could not calculate the equation for the radial geodesic, 2) When I used some invented curve, that was more or less the right shape, I had to estimate ##\Delta\tau_1##'s along the beacon path, 3) Having done that the intervals measured by the observer did often not increase.

Now we have an equation of motion for the beacon, that is the radial geodesic. I thought it would never be useful. We now find that it is useful it, and it is the intimidating$$
t=\frac{\sqrt{r_\ast-R_S}}{\sqrt{R_S}}\left[\sqrt{r_\ast-r}\sqrt r-\left(r_\ast+2R_S\right)\sin^{-1}{\left(\frac{\sqrt r}{\sqrt{r_\ast}}\right)}\right]
$$$$
+R_S\ln{\left|\frac{\left(r_\ast-R_S\right)\sqrt r+\sqrt{R_S}\sqrt{r_\ast-R_S}\sqrt{r_\ast-r}}{\left(r_\ast-R_S\right)\sqrt r-\sqrt{R_S}\sqrt{r_\ast-R_S}\sqrt{r_\ast-r}}\right|}
$$$$
+\frac{\pi\left(r_\ast+2R_S\right)\sqrt{r_\ast-R_S}}{2\sqrt{R_S}}
$$(##R_s=2GM## is the Schwarzschild radius) so we can plot that on a graph and the first problem is resolved. Moreover we found the proper speed of the beacon$$
\frac{dr}{d\tau}=-\sqrt{\frac{R_s\left(r_\ast-r\right)}{rr_\ast}}
$$By inverting and integrating that we have an expression for ##\tau## along the path and we can calculate ##\Delta\tau_1##'s which fixes the second problem. We already knew how to plot the return flight of the photon and when we do so, we find that the intervals measured by the observer do
increase for successive signals. So problem 3 was fixed! However, we are still not out of the woods. My second attempt is shown on the right with amounts in natural units. The observer is at ##r_\ast=15## and the event horizon at ##2GM=10##. We consider three photons emitted at ##a,b,c## separated by ##\Delta\tau_1=6##. We can measure the intervals seen by the observer and they do increase but the photon world line is disappointingly flat, unlike Carroll's. Spacetime is almost flat up to event ##b##. So we have to start very close to the event horizon with ##r_\ast=12## and zoom in to the area marked by the red rectangle. Eventually we achieve something like Carroll's (with a bonus photon from ##d##) as shown below. The diagram on the left is fairly bare like Carroll's the same one on the right has numbers added in natural units. ##\Delta\tau_1## was a very small ##0.5##  and the intervals observed at ##r_\ast## were an order of magnitude larger and increasing as Carroll predicted. The observer might have wanted to subtract out the, easily calculated, flight times of the returning photons. The intervals between emissions still increase as can be seen by the increasing vertical distances of ##a,b,c,d## on the diagram.
As an added bonus, from the equation for the proper speed of the beacon, we can calculate it's finite proper time to the event horizon and to the centre of the black hole. The increasing length along the geodesic for fixed proper time helps us intuit a resolution to the apparent paradox that the beacon 'never seems to get into the black hole'

Moreover if we get inside the event horizon, the proper time from there to ##r=0## is ##\pi GM## which is the maximum possible value that we calculated in exercise 5.3.

Get the details here: Commentary 5.6#3 Schwarzschild Black Holes.pdf (5 pages)

More intriguing puzzles remain:

  • To relate the monster formula for ##t## to the geodesic equation from section 5.4; 
  • To think about the beacon's (free falling) inertial coordinate system, its forward and backward light cones and its relationship to the Schwarzschild coordinate system 
  • Spaghettification and the mysterious G2 gas cloud

Saturday, 6 June 2020

Exercise 5.5 Observer and beacon outside black hole

Question

Consider a comoving observer sitting at constant spatial coordinates ##\left(r_\ast\ ,\theta_\ast\ ,\ \phi_\ast\right)## around a Schwarzschild black hole of mass ##M##. The observer drops a beacon onto the black hole (straight down along a radial trajectory). The beacon emits radiation at a constant wavelength ##\lambda_{em}## (in the beacon rest frame).

a) Calculate the coordinate speed of the beacon as a function of ##r##.
b) Calculate the proper speed of the beacon. That is, imagine there is a comoving observer at fixed ##r##, with a locally inertial coordinate system set up as the beacon passes by, and calculate the speed as measured by the comoving observer. What is it at ##r=2GM##?
c) Calculate the wavelength ##\lambda_{obs}##, measured by the observer at ##r_\ast##, as a function of the radius ##r_{em}## at which the radiation was emitted.
d) Calculate the time ##t_{obs}## at which a beam emitted by the beacon at radius ##r_{em}## will be observed at ##r_\ast##.
e) Show that at late times, the redshift grows exponentially: ##\lambda_{obs}/\lambda_{em}\propto e^{t_{obs}/T}##. Give an expression for the time constant ##T## in terms of the black hole mass ##M##.

Answers

This question is a fascinating can of worms. I needed the help of an unwitting mentor Jeriek Van den Abeele from the University of Oslo. The first crucial help was to use the timelike killing vector constant $$
E=\left(1-\frac{2GM}{r}\right)\frac{dt}{d\tau}
$$This came in useful not only for the beacon velocities but also for the  travel time of the photon back to the observer at ##\ r_\ast## in question d.

I also needed help from Physics Forums on comoving coordinates and I think I understand them now, although there was some dispute about how the (second)  comoving observer in question (b) would actually achieve that state. When people talk about comoving coordinates for the Universe, they are talking about something quite different from here: Coordinates comoving with the Universe are growing with the Universe.

Part of the worminess arises from confusion about what coordinates are referring to what. There was a profusion of subscripts: ##r_{em},r_\ast,\ R_s,t_\gamma,t_{em},t_{obs},t_b##. The last of those was introduced gratuitously by my mentor who I refer to as Oslo. In my opinion ##t_b\equiv t_{em},\ \ r_b\equiv r_{em}##. I try to avoid these subscripts as much as possible.

I got two answers to question (b). One was the proper speed calculated in the strange comoving inertial coordinate system, as Carroll asked, and the other was the proper speed in Schwarzschild coordinates. I called them ##dr^\prime/d\tau^\prime## and ##dr/d\tau##. The first was what Carroll asked for and can be used (but is not essential) in question (c). Apart from that I am not sure how useful it is. ##dr/d\tau## is what would be experienced by an astronaut falling with the beacon and it does not reach the speed of light at the event horizon.

Interestingly I calculated the answer to (c) using Doppler redshift + gravitational redshift and Oslo did it in one leap - which was more complex and contained a small error which had no effect. The formulas in the two answers looked quite different but when plotted gave the same lines. Eventually I proved that the formulas were in fact the same.

As usual the actual geodesic equations are not used to find out about all these geodesics. Nevertheless I will have a try to see if I can do better with that beacon in section 5.6.

Read it all at Ex 5.5 Observer and beacon.pdf (11 pages not including other documents)

Wednesday, 13 May 2020

Reciprocals of prime numbers

On a program about Gauss (BBC In Our Time), near 48:00, they said that the reciprocal of a prime number can be a recurring decimal and the maximum length of the recurring part is one less than the number. My curiosity was aroused.

Examples are
1/7=0.142857142857142857142857142857142857...
1/59=0.01694915254237288135593220338983050847457627118644067796610169491525423728813559322033898305084745762711864406779661...
the latter was found by a VBA program which could theoretically go to 2 billion digits because that is the maximum size of an array and length of a string. However it would run out of time or memory well before it got there. Calculation the reciprocals of the numbers from 80,000-89,999 and putting them into an Excel spreadsheet found 316 reciprocals which had recurring digits one less than the number you first thought of in 7 hours 50 minutes. 89,989 is prime 1/89,989 has 89,988 recurring digits and the VBA program  took 44 seconds just to calculate that. A different programming language would be much faster.

I now have a list (in several spreadsheets) of all the reciprocals from 2 to 999,999. It is fascinating.

A 'number whose reciprocal has recurring digits one less than the number' is a bit of a mouthful, so I abbreviate it to an ##R_-## ('R minus') number. It is also useful to define a function ##R\left(n\right)## which gives the number of Recurring digits in ##n##. The VBA program calculates ##R\left(n\right)##. For ##R_-## numbers, ##R\left(n\right)=n-1## . A related function is ##U\left(n\right)## which is the number of non-recurring digits, or Unique digits, in the reciprocal. So for example we have
##R\left(7\right)=6,\ U\left(7\right)=0## and
##\frac{1}{22}=0.0454545\ldots\ \ ,\ \ R\left(22\right)=2,\ U\left(22\right)=1##

The scatter chart contains a point at ##n,R\left(n\right)## for every number from 2 to 9,999.

Given positive integers ##m,n,p## greater than 0 or 1, I have proved that
1) ##R\left(n\right)<n## for all ##n##. This is shown by the graph. There are no data points above the main diagonal line which has gradient ##~1##.

2) ##R\left(n\right)=n-1## only for prime numbers. Or only prime numbers are ##R_-## numbers.

3) ##R_-## numbers occur about 2.6 times less often than prime numbers as shown by the density of the main diagonal line.

4) For multiples of ##R_-## numbers ##n## , ##R\left(m\times n\right)## is likely to be ##R\left(n\right)##. These are shown by the less dense diagonal straight lines which have gradients 1/2, 1/3, 1/5 ....

5) ##U\left(p\times2^m\times5^n\right)=m\ or\ n## whichever is larger of ##m,n##. (We can have ##m,n=0## in this case and ##p## must not have 2 or 5 as a factor.)

6) A consequence of 5 is that two fifths of consecutive numbers have ##Q\left(n\right)=0##.

7) Another consequence of 5 is that all prime numbers, except 2 and 5, and therefore all ##R_-## numbers have no non-recurring digits.

The first two have mathematical proofs, the rest are really conjectures based on experiments with lots of numbers. The mathematical proofs are algorithmic: One thinks of how to do the calculation and that proves the result.

Here's why Reciprocals of prime numbers.pdf (11 pages)

On Gauss and me

Carl Friedrich Gauss
On the program they mention the famous story about Gauss when he was age 10 at elementary school adding up all the numbers from 1 to 100. His class was asked to do this by the teacher who, no doubt, thought she could have some quiet time while the class was busy. Gauss added the numbers up in a minute, foiling the teacher's plan. Apparently Gauss used to love telling this story.

When I was about 10 or 11 and day dreaming at school assembly, I worked out a formula for the sum of the first ##n## numbers. It is of course ##\Sigma=n(n+1)/2##. I thought I had made an fantastic mathematical and nervously went to tell Mr Fillingham, head maths teacher. He gave me a strange look and showed me the general formulas for arithmetic and geometric progressions. I was crestfallen and never repeated the story.

Gauss was given a book of logarithm tables when he was 15 and it had a list of prime numbers in the back. He said that there was poetry in the tables.

He spent so much time day dreaming about mathematics that he memorised his times tables up to very large numbers. He would have loved computers. They give you the power to play with very large sets of numbers.