Dropping the Baton

There seem to have been rather a lot of dropped batons in the relay races at the Olympics (e.g. see here).

It set me thinking about where I might have seen this sort of thing happening elsewhere, and I realised that dropping the baton is like annihilating the vacuum state.

How so?

The simplest possible algebra that one can use to model the process of baton-passing goes like this:

a^† increments (by 1) the number of hands holding the baton
a decrements (by 1) the number of hands holding the baton

|0> is the "vacuum" state where the baton has one hand holding it
a^†|0> is the state where the baton has two hands holding it

a|0> = 0 is the annihilation of the vacuum where the baton has zero hands holding it, i.e. a state from which there is no way to recover

Note that it is important to define the vacuum state as corresponding to one (rather than zero) hand holding the baton, otherwise the algebra (i.e. annihilation of the vacuum) doesn't correctly model the dropping of the baton. Thus the counting of hands holding the baton is really a measure of how many excess hands are holding the baton, because the case of one hand is actually the ground (or vacuum) state in a relay race.

Most of the time the state is |0>, and during a successful handover of the baton it passes through the transition state a^†|0>, after which it returns to the state |0>. However, during an unsuccessful handover of the baton it goes to the state a|0> which is 0, where the vacuum has been annihilated.

Successful handover: a a^†|0> = |0>
Unsuccessful handover: a^†a|0> = 0

The order in which the a and a^† operations are applied is important, and is neatly summarised by how their commutator a a^† - a^†a acts on |0> (take the difference of the above equations).

(a a^† - a^†a)|0> = |0>

A stronger form of this result is the operator relation

a a^† - a^†a = 1

This relation takes note of the fact that there are n ways of applying a to the state (a^†)ⁿ |0> (i.e. choose from 1 of n excess hands to decrement by 1 the number of excess hands holding the baton), but there is only 1 way of applying a^† to the state (a^†)ⁿ |0>. The case n=0 is when the vacuum gets annihilated by application of a.

The Olympic athletes who dropped the baton were the victims of a a^† - a^†a = 1 (rather than 0). I wonder whether they saw it that way.

Ethel the Aardvark Goes Quantity Surveying

Here is some Sunday afternoon entertainment.

There was a "cheese shop sketch" before the famous Cheese Shop Sketch that we all remember. I was reminded about it whilst browsing the Wikipedia entry for Marty Feldman.

Here it is (text copied from here). The customer is played by Marty Feldman and the shop assistant by John Cleese.

Assistant: Good morning, sir.
Customer: Good morning. Can you help me? Do you have a copy of 'Thirty Days In the Samarkand Desert with a Spoon' by A.E.J. Elliott?
Assistant: Um ... well, we haven't got it in stock, sir.
Customer: Never mind. How about 'A Hundred and One Ways to Start a Monsoon'?
Assistant: ... By ... ?
Customer: An Indian gentleman whose name eludes me for the moment.
Assistant: I'm sorry, I don't know the book, sir.
Customer: Not to worry, not to worry. Can you help me with 'David Copperfield'?
Assistant: Ah, yes. Dickens ...
Customer: No.
Assistant: ... I beg your pardon?
Customer: No, Edmund Wells.
Assistant: ... I think you'll find Charles Dickens wrote 'David Copperfield', sir.
Customer: No, Charles Dickens wrote 'David Copperfield' with two 'p's. This is 'David Coperfield' with one 'p' by Edmund Wells.
Assistant: (a little sharply) Well in that case we don't have it.
Customer: Funny, you've got a lot of books here.
Assistant: We do have quite a lot of books here, yes, but we don't have David Coperfield' with one 'p' by Edmund Wells. We only have 'David Copperfield' with two 'p's by Charles Dickens.
Customer: Pity - it's more thorough than the Dickens.
Assistant: More thorough?
Customer: Yes ... I wonder if it's worth having a look through all your 'David Copperfields'...
Assistant: I'm quite sure all our 'David Copperfields' have two 'p's.
Customer: Probably, but the first edition by Edmund Wells also had two 'p's. It was after that they ran into copyright difficulties.
Assistant: No, I can assure you that all our 'David Copperfields' with two 'p's are by Charles Dickens.
Customer: How about 'Grate Expectations?
Assistant: Ah yes, we have that ...
He goes to fetch it and returns to the counter.
Customer: ... That's 'G-r-a-t-e Expectations', also by Edmund Wells.
Assistant: I see. In that case, we don't have it. We don't have anything by Edmund Wells, actually - he's not very popular.
Customer: Not 'Knickerless Nickleby'? That's K-n-i-c-k-e-r
Assistant: No!
Customer: Or 'Quristmas Quarol 'with a Q?
Assistant: No, definitely ... not.
Customer: Sorry to trouble you.
Assistant: Not at all.
Customer: I wonder if you have a copy of 'Rarnaby Budge'?
Assistant: (rather loudly) No, as I say, we're right out of Edmund Wells.
Customer: No, not Edmund Wells - Charles Dikkens.
Assistant: Charles Dickens?
Customer: Yes.
Assistant: You mean 'Barnaby Rudge'.
Customer: No, 'Rarnaby Budge' by Charles Dikkens ... that's Dikkens with two 'k's, the well-known Dutch author.
Assistant: No, no - we don't have 'Rarnaby Budge' by Charles Dikkens with two 'k's the well-known Dutch author, and perhaps to save time I should add right away that we don't have 'Carnaby Fudge' by Daries Tikkens, nor 'Stickwick Stapers' by Miles Pikkens with four Ms and a silent Q, why don't you try the chemist?
Customer: I did. They sent me here.
Assistant: (making a mental note) ... Did they?
Customer: I wonder if you have ... 'The Amazing Adventures of Captain Gladys Stoat-Pamphlet and her Intrepid Spaniel Stig among the Giant Pygmies of Corsica', Volume Two.
Assistant: No, we don't have that one. Well, I mustn't keep you standing around all day ..
Customer: I wonder if ...
Assistant: No, no, we haven't got it. I'm closing for lunch now anyway.
The assistant moves rapidly away from the counter.
Customer: ... But I thought I saw it over there.
The assistant checks and turns slowly.
Assistant: ... What?
Customer: Over there.
He indicates a bookshelf.
Customer: 'Olsen's Standard Book of British Birds'.
Assistant: (very suspiciously) 'Olsen's Standard Book of British Birds'?
Customer: Yes.
Assistant: ... 0-l-s-e-n?
Customer: Yes!
Assistant: B-i-r-d-s?
Customer: Yes!
Assistant: Well, we do have that one, yes.
He goes and takes the book off a shelf.
Customer: ... The expurgated version, of course.
Assistant: ... I'm sorry, I didn't quite catch that.
Customer: The expurgated version.
Assistant: The expurgated version of 'Olsen's Standard Book of British Birds'?
Customer: Yes. The one without the gannet.
Assistant: The one without the gannet?! They've all got the gannet it's a standard bird, the gannet, it's in all the books.
Customer: Well I don't like them. They've got long nasty beaks! And they wet their nests.
Assistant: But ... but you can't expect them to produce a special edition for gannet-haters!
Customer: I'm sorry, I specially want the one without the gannet.
The assistant is speechless.
Assistant: All right!
He suddenly tears out the relevant page.
Assistant: Anything else?
Customer: Well, I'm not too keen on robins.
Assistant: Right! Robins, robins ...
He tears that one out too and slams the book on the counter.
Assistant: No gannets, no robins - there's your book!
Customer: I can't buy that. It's torn.
Assistant: ... So it is! He tosses it into the bin.
Customer: I wonder if you've got ...
Assistant: Go on! Ask me another.
Customer: How about 'Biggles Combs his Hair'?
Assistant: No, no, we haven't got that one, funny. Try me again.
Customer: 'The Gospel According to Charlie Drake'?
Assistant: No ...
Customer: Have you got 'Ethel the Aardvark Goes Quantity-Surveying'?
Assistant: No, no, we haven't ... which one?
Customer: 'Ethel the Aardvark Goes Quantity-Surveying'.
Assistant: 'Ethel the Aardvark'?! I've seen it! We've got it!!
He dashes to a bookshelf, finds it, and holds it up triumphantly.
Assistant: Here! Here!!! 'Ethel the Aardvark Goes Quantity Surveying'. Now - buy it!
He slams it on the desk. The customer stares in horror!
Customer: ... I haven't got enough money on me.
Assistant: (quickly) I'll take a deposit!
Customer: I haven't got any money on me.
Assistant: I'll take a cheque!
Customer: I haven't got a cheque book!
Assistant: It's all right, I've got a blank one!
Customer: I don't have a bank account!!
Assistant: ... All right!! I'll buy it for You!
He rings the purchase up and pays for it himself. He gives the change to the customer.
Assistant: There we are, there's your change - that's for the taxi home ...
Customer: Wait! Wait! Wait!
Assistant: What? What? What?!!!
Customer: ... I can't read ...
Assistant: Right! Sit!! ...
He sits the customer down on his knees and starts to read aloud.
Assistant: 'Ethel the Aardvark was trotting down the lane one lovely summer day, trottety-trottety-trot, when she saw a nice Quantity-Surveyor ...

Twerp Bollickagh - the website

Twerp Bollickagh

It could be the title of a journal!

I suspect that many of the experimentally accurate theoretical "predictions" given in a "Grand Unified Theory" that is available online (search for the string "The calculated relations between the lepton masses") were arrived at by exhaustive numerology, i.e. by searching through a large number of simple expressions to find the ones that gave the required results. Then a "proof" of each of these results was reverse-engineered using pseudo-physical explanations rather than using rigorous maths.

Let me show you an example of what I mean.

This "GUT" gives some simple expressions for various mass ratios. There is even an expression for the ratio of the neutron mass to the electron mass, which depends only on the electomagnetic coupling strength (i.e. fine structure constant) and not on the strong interaction strength. How do the quarks and gluons in the neutron know how to interact in order to give this amazing result?

The expression given by this "GUT" for the muon to electron mass ratio (i.e. mμ/me) is

(α^(-2) / 2π)^(2/3) (1 + 2π α^2 / 2) / (1 + α/2)

which produces a value 206.76828 that closely corresponds to the experimentally observed value 206.76827.

Let's see whether it is possible to "derive" this result by an exhaustive search of all simple expressions of this general type. The parameterisation that I will use is the most general form that is suggested by the mass ratio quoted above

f[{a1, a2, a3, a4}, {b1, b2, b3, b4}, {c1, c2, c3, c4}, {d1, d2, d3, d4}, α]] =
(a1/a2)^(a3/a4) (α)^(b1/b2) (2π α^2)^(b3/b4) (1 + (c1/c2)(α) + (c3/c4)(2π α^2)) / (1+ (d1/d2)(α) + (d3/d4)(2π α^2))

where all of the parameters are integers which are grouped in pairs to form rational fractions. To compute numerical results I inserted specific values for these parameters (avoiding singular cases), I use α=0.00729735, and a target mass ratio mμ/me=206.76827.

I then computed f[{a1, a2, a3, a4}, {b1, b2, b3, b4}, {c1, c2, c3, c4}, {d1, d2, d3,
d4}, α]] for all parameter values in the following small ranges (I have been rather cavalier and restricted the ranges to save time):
{a1, 1, 2}, {a2, 1, 2}, {a3, 1, 3}, {a4, 1, 3},
{b1, 0, -3, -1}, {b2, 1, 3}, {b3, 0, -3, -1}, {b4, 1, 3},
{c1, 0, 2}, {c2, 1, 2}, {c3, 0, 2}, {c4, 1, 2},
{d1, 0, 2}, {d2, 1, 2}, {d3, 0, 2}, {d4, 1, 2}
There is some repetition of trial solutions here, but this doesn't matter.

I then selected from this large set of trial solutions all of the cases that predicted a value for mμ/me that lay within 0.01 of the target value, and here they are (in decreasing order of goodness of fit) with the prediction errors shown in square brackets:

(1/(2π α^2))^(2/3) (1 + π α^2) / (1 + α/2) [0.0000110213]
(1/(2π α^2))^(2/3) (1 + 2π α^2) / (1 + α/2 + π α^2) [0.000130968]
(1/(2π α^2))^(2/3) (1 + α/2 + π α^2) / (1 + α) [0.00261802]
(1/(2π α^2))^(2/3) (1 + α/2 + 2π α^2) / (1 + α + π α^2) [0.0027371]

The best fit solution at the top of this list is the same as the one given by the "GUT".

What do we conclude from this little exercise?

It is really easy to do exhaustive searches to find best-fit solutions. The above fit works as well as it does because it starts with two different quantities (α) and (2π α^2) (where π is not a rational fraction), and combines them in various ways using lots of rational fractions to tailor the combination, which then leads to a dense set of candidate solutions from which the best-fit solution can then be picked.

Unless you happened to pick the physically correct parametric form to search over (Balmer got lucky with atomic spectra, but that is not to be used as a justification for this approach), then there is no physical significance to solutions that are obtained in this way. If you hedge your bets by searching over a large set of parametric forms, then you will almost certainly find many solutions that have a good fit to the target value, but this doesn't guarantee that any of them is physically significant. Interestingly, a related problem occurs in the context of the Landscape.

The approach used in this "GUT" is numerology, pure and simple. Of course, I only suspect that this is the way that the above expression for mμ/me was "derived"; I can't prove that this is the case.

Holiday in South West Cornwall

I have been neglecting this blog; my previous posting was more than half a year ago. Not only that, but the broken basin that I described in my previous posting has still not been fixed!

I have just returned from a few weeks camping in South West Cornwall (the area known as West Penwith), which is an area that I enjoy visiting because it is like leaving your "baggage" at home and going away to "the edge of the world". Here is a small sample of some of the photographs that I took whilst I was there.

The West Penwith area is rather exposed to the Atlantic weather systems, so you have to pitch your tent away from bushes that look like this:

Much of the local granite is beautifully weathered. I wonder how long it takes for granite exposed to the elements to begun to look like this:

As usual, I tried to do a bit of maths to exercise my mind but I kept losing factors of π, so I ended up just banging the rocks together:

To satisfy my curiosity, I visited the famous tin mine engine houses at the Botallack Crowns Mine, which are much more precariously located than they seem in this photograph (someone needs to invent a simple photographic system that gives the viewer a full 3D spatial awareness, like a feel for the yawning chasm just in front of them):

I watched a bit of Cornish cricket at the Lafrowda Festival in St Just, which seems to be much more fun than the activity called "cricket" that I have seen on TV:

I did quite a bit of moorland walking, but I was never quite sure whether I was inside or outside the areas of open moorland, as this gate in the middle of nowhere illustrates:

There appears to be only a loose correspondence between the moorland paths marked on the Ordnance Survey map and the actual paths on the ground (I double checked my position using my GPS locator), so I sometimes found myself wading through a sea of gorse and heather that was up to waist high in places. This scenery was very pretty to look at, but it tore my legs to shreds.

There is a move to enclose the moors and to graze cattle, which has caused uproar amongst some of the local inhabitants who have started a Save Penwith Moors campaign. My preference is for open moorland scenery unspoilt by fencing and cattle, and I hope yours is too.

Update (21 December 2008): I have just noticed that the 9 Maidens Common has had a reprieve from the cattle grazing plans (see here). Excellent news!