## Wednesday, 30 December 2020

### Exercise 8.1 N+n+1-dimensional spacetime

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.
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 Lemaître (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

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).