Monday 30 December 2019

Exercise 3.14 Killing vectors on two-sphere


Consider the three Killing vectors of the two-sphere, (3.188). Show that their commutators satisfy the following algebra:
\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


The three Killing vectors at 3.188 were
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

The metric for the 3-sphere in coordinates ##x^\mu=##  ##\left(\psi,\theta,\phi\right)## is $$
$$(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$$
$$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.


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
\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
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!
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


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 round 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.


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

·        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

·        A run (123456)

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


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

Thrown dice

Maximum score

One 5

One 1 and two 5s

Three 1s

Three 4s =400 + one 5

Three pairs

One 1

Nothing scores

One 1 and one 5

A run

Three 4s + three 3s

Three 2s

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.


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.


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


three 1s
three of a kind
three pairs


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).