## Friday, 24 January 2020

### The energy-momentum tensor in SR Fluid flow by Thierry Dugnolle
Many moons ago I had skipped the end of section 9 and all of section 10 in chapter 1. It is time to return to them and flat space time starting with the end of section 9 it looks at the energy-momentum tensor for a perfect fluid. I had written at the top of the page "? don't follow much of this from 1.116. Now I do. It shows how to get from that tensor to two well known classical (non-relativistic) equations in the non-relativistic limit: a continuity equation and the Euler equation from fluid dynamics. This gives us faith that the tensor is correct - even though I had not heard of either of these equations.

Along the way we find some tricks with four-velocity $U^\mu={dx^\mu} / {d\tau}$, meet the projection tensor $P_{\ \ \ \ \nu}^\sigma$ which projects a vector orthogonal to another do some dimensional analysis and find out what energy density really is (I missed this in the book, or Carroll forgot to mention it). I had to consult the great Physics Forums on that and got liked for a facetious comment about measuring it in £sd 😀- I'm sure that oil companies must do it: kerosene is more valuable per kg than bunker fuel.

All the details here Commentary 1.9 Energy Momentum Tensor.pdf (6 pages)

## Friday, 17 January 2020

### Physics in curved spacetime

I've now started chapter 4 on Gravitation just in time for 2020. Very exciting! Fools straight line
Carroll first states two formulas of Newtonian Gravity his 4.1 and 4.2$$\mathbf{a}=-\nabla\Phi$$where $\mathbf{a}$ is the acceleration of a body in a gravitational potential $\Phi$. And Poisson's differential equation for the potential in terms of the matter density $\rho$ and Newton's gravitational constant $G$:$$\nabla^2\Phi=4\pi G\rho$$I had a long pause thinking about the various formulas for the Laplacian $\nabla^2$ here.
How to these tie up with the old-fashioned laws? Newton's law of gravity is normally stated as$$F=G\frac{m_1m_2}{r^2}$$which combined with Newton's second law $F=m\mathbf{a}$ gives us the acceleration of a mass in the presence of another as$$\mathbf{a}=G\frac{M}{r^2}$$In exercise 3.6 we were given 'the familiar Newtonian gravitational potential'$$\Phi=-\frac{GM}{r}$$A bit of rough reasoning shows these are equivalent.

At his 4.4 Carroll states that the next equation gives the path of a particle subject to no forces$$\frac{d^2x^i}{d\lambda^2}=0$$If we solve it in polar coordinates for $r,\theta$ instead of $x,y$ Carroll says we get a circle and he cheekily suggests that we might think free moving particles follow that path. But the solution is $$r=m\theta+k$$where $m,k$ are constants. We can plot that and, obviously if $m=0$ we get a circle of radius $k$ but if $m\neq0$ we get other more interesting lines which are equally wrong. See above.

Then we examine the equations in a near Newtonian environment and apart from a slight problem I had with equation 4.13 that $g^{\mu\nu}=\eta^{\mu\nu}-h^{\mu\nu}$ which I recon is $g^{\mu\nu}=\eta^{\mu\nu}+h^{\mu\nu}$ we arrive at the conclusion that in the near Newtonian environment we have the time, time component of the metric is $$g_{00}=-1-2\Phi$$which is also what we were given in Exercise 3.6.

## Thursday, 9 January 2020

### The Laplacian Laplace 1749-1827
The second equation in chapter 4 was Poisson's differential equation for the gravitational potential $\Phi$:
\begin{align}
\nabla^2\Phi=4\pi G\rho&\phantom {10000}(1)\nonumber
\end{align}What does $\nabla^2$ mean? A quick look on the internet reveals that it is the Laplace operator or Laplacian sometimes written $\Delta$ (capital delta) or possibly $∆$ (which Microsoft describes as increment). They come out different in latex \Delta and ∆ respectively. I should use the former.

I have found various formulations for the Laplacian and I want to check that they are all really the same. Two are from Wikipedia and the third is from Carroll. They are:
A Wikipedia formula in $n$ dimensions:
\begin{align}
\nabla^2=\frac{1}{\sqrt{\left|g\right|}}\frac{\partial}{\partial x^i}\left(\sqrt{\left|g\right|}g^{ij}\frac{\partial}{\partial x^j}\right)&\phantom {10000}(2)\nonumber
\end{align}A Wikipedia formula in "in 3 general curvilinear coordinates $(x^1,x^2,x^3)$":
\begin{align}
\nabla^2=g^{\mu\nu}\left(\frac{\partial^2}{\partial x^\mu\partial x^\nu}-\Gamma_{\mu\nu}^\lambda\frac{\partial}{\partial x^\lambda}\right)&\phantom {10000}(3)\nonumber
\end{align}And Carroll's formula (from exercise 3.4) which I did not understand at the time because I had not spotted the second step:
\begin{align}
\nabla^2=\nabla_\mu\nabla^\mu=g^{\mu\nu}\nabla_\mu\nabla_\nu&\phantom {10000}(4)\nonumber
\end{align}The Wikipedia also gives a formula for the Laplacian in spherical polar coordinates:
\begin{align}
\nabla^2f&=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial f}{\partial r}\right)+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left(\sin{\theta}\frac{\partial f}{\partial\theta}\right)+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2f}{\partial\phi^2}&\phantom {10000}(5)\nonumber\\

&=\frac{1}{r}\frac{\partial^2}{\partial r^2}\left(rf\right)+\frac{1}{r^2\sin{\theta}}\frac{\partial}{\partial\theta}\left(\sin{\theta}\frac{\partial f}{\partial\theta}\right)+\frac{1}{r^2\sin^2{\theta}}\frac{\partial^2f}{\partial\phi^2}&\phantom {10000}(6)\nonumber
\end{align}where $\phi$  represents the azimuthal angle and $\theta$ the zenith angle or co-latitude. So the metric will be
\begin{align}
g_{\mu\nu}=\left(\begin{matrix}1&0&0\\0&r^2&0\\0&0&r^2\sin^2{\theta}\\\end{matrix}\right)&\phantom {10000}(7)\nonumber
\end{align}I assumed that the coordinates are ordered $r,\theta,\phi$ although Wikipedia does not say that.

I want to prove that
A) (2), (3) and (4) both give (5) or (6) the Laplacian in spherical polar coordinates.
B) (4) is equivalent to (3) the general 3-dimensional expression.
C) (4) is equivalent to (2) the general 𝑛-dimensional expression.
A was quite easy. B follows immediately from the formula for the covariant derivative. I could not prove C. After some help from Physics forums I proved it for a diagonal metric apparently it is true for a non diagonal metric but I could not penetrate the hints.

View my struggles at Commentary 4.1 Laplacian.pdf. A and B are on the first two pages and the other seven are where it all starts to go wrong.

I spent far too long on this but
1) Revisited using the Levi-Civita symbol for expanding determinant (47)
2) Met the log, exponent, trace of a matrix (63) and for diagonal matrices (67)
3) Used diagonalizing matrices (65)
4) Used the product operator $\prod\$for the first time at (27) and again at (66)
5) Proved that $\ln{\left(\det{\left(A\right)}\right)}=tr{\left(\ln{\left(A\right)}\right)}$ (66)
6) Found a formula for the log of a derivative (73)
7)  Found a formula for the log of a matrix (76)

## Monday, 30 December 2019

### Question

Consider the three Killing vectors of the two-sphere, (3.188). Show that their commutators satisfy the following algebra:
\begin{align}
\left[R,S\right]=T&\phantom {10000}(1)\nonumber\\
\left[S,T\right]=R&\phantom {10000}(2)\nonumber\\
\left[T,R\right]=S&\phantom {10000}(3)\nonumber
\end{align}

The three Killing vectors at 3.188 were
\begin{align}
R&=\partial_\phi&\phantom {10000}(4)\nonumber\\
S&=\cos{\phi}\partial_\theta-\cot{\theta}\sin{\phi}\partial_\phi&\phantom {10000}(5)\nonumber\\
T&=-\sin{\phi}\partial_\theta-\cot{\theta}\cos{\phi}\partial_\phi&\phantom {10000}(6)\nonumber
\end{align}It is easy to prove the algebra as long as we remember that the $\partial_\theta,\partial_\phi$ are just the basis vectors and Carroll's 2.23 that $\left[X,Y\right]^\mu=X^\lambda\partial_\lambda Y^\mu-Y^\lambda\partial_\lambda X^\mu$ which we proved back in exercise 2.04. So first we could write (4),(5),(6) as $R=\left(0,1\right),\ S=\left(\cos{\phi},-\cot{\theta}\sin{\phi}\right),\ T=\left(-\sin{\phi},-\cot{\theta}\cos{\phi}\right)$.

What is more difficult to understand is what does this neat algebra mean? A common way to show a commutator (which Carroll uses when introducing the Riemann tensor) is thus

The commutator $\left[R,S\right]$ is the difference between doing $R$ then $S$ and $S$ then $R$. So the algebra that the Killing vectors satisfy corresponds to this geometry:

$R$ has always been shown as a unit vector in increasing $\phi$ direction, $T,S$ are more flexible. I'm not sure how this helps.

The proof is given at Ex 3.14 Killing vectors on two-sphere.pdf (2 pages)

## Tuesday, 24 December 2019

### Review chapter 3

Before we start I have again found a very important paragraph, it is the last in section 3.2 and describes how we have got from the beginning to here. The beginning was the concept of a set which became a manifold and we ended up with metrics and covariant derivatives!

After nearly getting to the end of chapter 3 I realised that my ideas about covariant derivatives needed refinement and that I did not really understand parallel transport. With the former it would seem that metric compatibility, $\nabla_\mu g_{\lambda\nu}=0$, arises out of the Leibnitz rule and the demand that the covariant derivative is a tensor and is not an 'additional property' - with the caveat that the manifold in question has a metric with the usual properties. On the latter I was still hazy until I explored further in this review.

I still don't think I have mastered every detail of this chapter to section 8 on Killing vectors but one thing is clear: If you have a tensor field $T\left(x^\mu\right)$ which you can express in terms of coordinates $x^\mu$ and we consider two points $x^\alpha,x^\beta$ then if you parallel transport the tensor from $x^\alpha$ along a geodesic to $x^\beta$ and call it $T\prime\left(x^\beta\right)$ there, then in all likelihood $T\prime\left(x^\beta\right)\neq T\left(x^\beta\right)$ and in some sense at least $T^\prime\left(x^\beta\right)=T\left(x^\alpha\right)$. I think. Equation (20) in the document is saying that for a vector.

The rest of chapter 3 was straightforward and finally we met the geodesic deviation equation $$A^\mu=\frac{D^2}{dt^2}S^\mu=R_{\ \ \nu\sigma\rho}^\mu T^\nu T^\rho S^\sigma$$$A^\mu$ is the "relative acceleration of (neighbouring) geodesics", $S^\mu$ is a vector orthogonal to a geodesic (pointing towards its neighbour) and $T^\mu$ is a vector tangent to the geodesic. The equation expresses the idea that the acceleration between two neighbouring geodesics is proportional to the curvature and, physically, it is the manifestation of gravitational tidal forces.

The six page document repeats the above and reviews covariant derivatives, parallel transport and geodesics as shown in the video, Riemann and Killing. It's at Commentary 3 Review chapter 3.pdf.

## Thursday, 5 December 2019

### Exercise 3.08 Vital statistics of a 3-sphere

Question
The metric for the 3-sphere in coordinates $x^\mu=$  $\left(\psi,\theta,\phi\right)$ is $${ds}^2={d\psi}^2+\sin^2{\psi}\left({d\theta}^2+\sin^2{\theta}{d\phi}^2\right)$$(a) Calculate the Christoffel connection coefficients. Use whatever method you like, but it is good practice to get the connection coefficients by varying the integral (3.49)

(b) Calculate the Riemann tensor $R_{\ \ \ \sigma\mu\nu}^\rho$ , Ricci tensor $R_{\mu\nu}$ and Ricci scalar $R$.

(c) Show that (3.191) is obeyed by this metric, confirming that the 3-sphere is a maximally symmetric space (as you might expect.) 3.191 was$$R_{\rho\sigma\mu\nu}=\frac{R}{n\left(n-1\right)}\left(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu}\right)$$Calculating the Christoffel coefficients was easy because I could use the code I had written before. I did not take the advice "get good practice by varying the integral"! I forgot to use the result of exercise 3.3! Grrr. The Riemann tensor was gruelling. It has 81 components but in this case there are really only three which are independent and each of them generate only three more by index symmetries (as shown in the picture). However, every one of the 81 needs checking. We also notice that each component is ± a metric component and wonder if there is some relationship like the relation between these components similar to exercise 3.3. Perhaps a future challenge! The Ricci bits are almost trivial. For (c) I was able to write a simple bit of code to compare all 81 equations. It did the job.

Note:
A 1-sphere is a circle - the set of points in $\mathbf{R}^2$ at an equal distance from the origin.
A 2-sphere is the surface of a sphere - the set of points in $\mathbf{R}^3$ at an equal distance from the origin.
A 3-sphere is the set of points in $\mathbf{R}^4$ at an equal distance from the origin.

(a) Christoffel connection coefficients
\begin{align}
\Gamma_{\theta\theta}^\psi&=-\sin{\psi}\cos{\psi}&\phantom {10000}(5)\nonumber\\
\Gamma_{\phi\phi}^\psi&=-\sin{\psi}\cos{\psi}\sin^2{\theta}&\phantom {10000}(6)\nonumber\\
\Gamma_{\psi\theta}^\theta=\Gamma_{\theta\psi}^\theta&=\cot{\psi}&\phantom {10000}(7)\nonumber\\
\Gamma_{\phi\phi}^\theta&=-\sin{\theta}\cos{\theta}&\phantom {10000}(8)\nonumber\\
\Gamma_{\psi\phi}^\phi=\Gamma_{\phi\psi}^\phi&=\cot{\psi}&\phantom {10000}(9)\nonumber\\
\Gamma_{\theta\phi}^\phi=\Gamma_{\phi\theta}^\phi&=\cot{\theta}&\phantom {10000}(10)\nonumber
\end{align}(b) Riemann tensor components are
\begin{align}
R_{\ \ \ \theta\psi\theta}^\psi&=\sin^2{\psi}&\phantom {10000}(13)\nonumber\\
R_{\ \ \ \theta\theta\psi}^\psi&=-\sin^2{\psi}&\phantom {10000}(14)\nonumber\\
{R}_{\ \ \ {\phi\psi\phi}}^{\psi}&=\sin^2{\psi}\sin^2{\theta}&\phantom {10000}(15)\nonumber\\
R_{\ \ \ \phi\phi\psi}^\psi&=-\sin^2{\psi}\sin^2{\theta}&\phantom {10000}(16)\nonumber\\
R_{\ \ \ \psi\psi\theta}^\theta&=-1&\phantom {10000}(17)\nonumber\\
R_{\ \ \ \psi\theta\psi}^\theta&=1&\phantom {10000}(18)\nonumber\\
R_{\ \ \ \phi\theta\phi}^\theta&=\sin^2{\psi}\sin^2{\theta}&\phantom {10000}(19)\nonumber\\
R_{\ \ \ \phi\phi\theta}^\theta&=-\sin^2{\psi}\sin^2{\theta}&\phantom {10000}(20)\nonumber\\
R_{\ \ \ \psi\psi\phi}^\phi&=-1&\phantom {10000}(21)\nonumber\\
R_{\ \ \ \psi\phi\psi}^\phi&=1&\phantom {10000}(22)\nonumber\\
R_{\ \ \ \theta\theta\phi}^\phi&=-\sin^2{\psi}&\phantom {10000}(23)\nonumber\\
R_{\ \ \ \theta\phi\theta}^\phi&=\sin^2{\psi}&\phantom {10000}(24)\nonumber
\end{align}The Ricci tensor is twice the metric!
\begin{align}
R_{\mu\nu}=2g_{\mu\nu}&\phantom {10000}(35)\nonumber
\end{align}The Ricci scalar is 6

(c) was solved by VBA
Complete answer with more links in Ex 3.08 Vital statistics of a 3-sphere.pdf

## Monday, 2 December 2019

### Zilch - How to play and probabilities

Zilch is a parlour game introduced to my family by Camilla who learnt it on her honeymoon with David from a man called Bali Bill. There are variations on the rules which we will not further discuss.

## How to play

### Start

The game is played with six ordinary dice by two to six people. Six is quite a lot and it usually gets out of hand with more. One of the players must be chosen as the scorer.

Each player takes it in turn to play and it is decided who starts by each player throwing just one dice and seeing who gets the highest number. The highest number starts. If two or more players get the same highest number, those players throw again until a starter is found. The game proceeds with the starter making a play. Once they have finished the next player on the left plays and so it goes found and round. It is important to note that every player gets the same number of plays.

In order to start scoring points a player must make 500 points in their first scoring play.

### Middle: Plays, scoring

→ When a player plays they start by throwing all six dice.

⇒ Whenever a player throws (any number of dice) there are two possible outcomes:
1) The thrown dice get no points. That is Zilch. Their play ends, Z for Zilch is written against their score. They lose any score they made that play. The next player plays.
2) The dice thrown score some points. The player may then stop and their accumulated total for that play is added to their score; the next player plays. Or the player keeps some of the scoring dice and throws the remainder again. Often there is no choice of how many dice to keep.  The total scored so far is accumulated for that play. Now go back to ⇒ (with less dice to throw).

On step 2 above their may be no dice left. This is excellent. The player may start again with all six dice and continue to accumulate points for that play. (Go back to →).

Since a player must get a score of 500 or more in one play to get into the game they must throw relentlessly until that happens. This can cause numerous zilches at the start of the game.

If a player gets four zilches in a row 500 is deducted off their score. Subsequent consecutive zilches incur the same penalty of -500. With bad luck, it is quite possible to go seriously negative at the start of the game. I have seen -5000 and the player involved ended the whole game on a record breaking zero.

#### Scoring

You might be wondering how you do score points in this game. Here's the answer:

 Thrown dice include Points scored ·        One 1 100 (and two 1s scores 200 points) ·        One 5 50 ·        Three of a kind 100×N where N is the number on the dice. So three 4s scores 400 points. But … ·        Three 1s 1000 (not 100 which would be very silly) ·        Three pairs 1000 ·        A run (123456) 1000

Clearly throws including the last three are very desirable as are three 6s, three 5s and two sets of three of a kind.

#### Examples

You may want to cover up the answers to test yourself! They are written quite faintly on the right.

 Thrown dice Maximum score Reason 235643 50 One 5 143575 200 One 1 and two 5s 321211 1000 Three 1s 542444 450 Three 4s =400 + one 5 255552 1000 Three pairs 231 100 One 1 364632 0=Z=zilch Nothing scores 51 150 One 1 and one 5 532641 1000 A run 434334 700 Three 4s + three 3s 23226 200 Three 2s 131113 1100 Three 1s +one 1

Some of the examples are deliberately tricky, but it is easy to make an error in the excitement of the game. For example a throw like 255552 might easily be scored as 550 (for three 5s and the singleton 5). It is also quite easy to not see three pairs or a run. The last example, 131113, is interesting because one could also take 1000 points for the three pairs and then throw all six dice again. This is normally the better strategy.

When less than six dice are thrown, a run or three pairs are impossible.  When only one or two dice are thrown, three of a kind or three 1s are also impossible.

The dice that were previously thrown, whose score is 'in the bank', have no effect on the score of the thrown dice.

The scores should be laid out as shown to the right. There are four players Alice, Bob, Doris and Chris. Alice was the starter, chosen as described above, so her score is in the first column. The players were not sitting in alphabetical order round the table. She is not doing well, she has scored zilch five times in a row. Bob was next. He scored 700 in his first play, zilch in his second then 200, 800. Doris 1000, 200, Z, 500. Scores for each play are not recorded. It is easy to tell who must play last because the scores are laid out so neatly.

 A B D C Z 700 1000 Z Z Z Z 900 1200 Z 1300 1900 -500 1700 1700 3000 -1000

Here is an example of Alice's first play: She threw the six dice and got 146523. She kept the 1 and the 5 (worth 150) and threw the other four dice again. With those she got four 2s. She kept three of those bringing her total on that play to 350. She has one dice left and needs 150 points to get into the game with 500. She throws it and it's a 1! She now has 450 and can throw all six again. But then she threw 364632 which is no points so she lost all her score that play and got Zilch. On Alice's fifth play she got zilch again so another 500 was deducted.

Alice must throw more than 500 in one play to get going on the right direction. If she got exactly 1000 in her sixth play 0 not Z would be put in her score. Z would be wrong and confusing.

Doris threw three 1's as her first throw and wisely stopped. 1000 went in her score and she was in the game. Next time round (after all the others got zilch) she threw 153562 and kept the score of 200. She probably would have been better to keep the 1 and throw five dice again.

### End

As stated above every player gets the same number of plays. The game ends when a player's score gets to 10,000 or more. We'll call that person Bob. When that happens any other players who have not had as many plays as Bob get one last chance to equal or overtake Bob. So that's Doris and Chris in our example. If Doris succeeded she would be declared the winner (unless Chris overtook her on the last play of the game.)

## Tactics and Etiquette

• If in a throw any dice fall off the table or are cocked (leaning at an angle due to other dice or some other obstacle) all the thrown dice must be thrown again.
• It's important not to get zilch in a play but also important to get a decent score. Therein lies the tension in each play and the judgement required.
• After any throw is made, nobody should touch the dice until a few players (particularly the scorer) have seen them all. Moving them around may be considered cheating. Peter ✠ used to wrap his arms around his thrown dice so only he could see them. This was banned.
• It may be best not to comment after a throw is made. The player may fail to spot a good score. There is no need to tell them (until it's too late). On the other hand you may innocently call a run a 100 and see if the player falls for your trap. Camilla says this is unsporting.
• Consider the scorer. They not only have to play but they have to add up everybody else's score. In particular do not start a new play until the scorer has written down the score from the last play.
• If you’re in Bali and Chris wins, for the next game someone will make him go and get the next round of drinks. While he’s away from the table they’ll move to sit in his place because they believe the seat is affecting his luck – and they want it!
• Cheating: It is surprisingly easy to cheat if all the players are discussing the latest gossip or otherwise entertained. A cheat may slyly turn over a dice as they are 'rearranging' a throw. If all the other players are really being so inattentive it serves them right. Be warned, be attentive and devise a punishment if a cheater persists.

## The Zilch Odds Table

We all know that the odds of getting a 1 with one dice is 1 in 6 = 1/6 = 17%. What are the odds (the probability) of getting a 1 with six dice? They aren't six times the odds of getting a 1 with one dice. That would be 100% - a dead cert. Here is a little table with some useful probabilities.

Zilch Odds Table
 Throwing dice Probability of 1 or 5 zilch three 1s three of a kind three pairs run 6 91% 2.5% 6% 36% 5% 1.5% 5 87% 16% 4% 21% 0 0 4 80% 10% 2% 10% 0 0 3 70% 27% .5% 3% 0 0 2 56% 44% 0 0 0 0 1 33% 67% 0 0 0 0

Notes:
1) Does not help much with calculating the odds of getting over 2000 in a play.
2) Not guaranteed correct.
3) These are all calculated in zilch (Click Read more) except for the first three zilch odds which were done with a simulator. All the others were checked by said simulator which gave the same answer.

By George, November 2019 with thanks to David and Camilla who checked this for me (and also told me that the rules I had been using were wrong).