I promised to say something about twistors. This will take a few posts and be spread out over some time – sorry ! Also I want to make it clear that my comments relate to the twistor theory I vaguely understood 25 years ago. I’m not familiar with what may have been done on the subject in the intervening time, including twistor string theory.
Post # 1: Spinors
Twistor theory was developed in the hope that it would provide a new way to formulate physics and hopefully lead to some new predictions. I have to say right up front that, as far as I’m aware, the latter hope was never realized. However, the former was realized and there were advantages to be gained from this. My own view of the twistor approach is that, rightly or wrongly, I think of it a bit like people think of Fourier transformations. In communications theory, people transform between the time domain and the frequency domain. Some operations which are complicated in one domain are simple in the other domain and vice versa.
The important transformation in twistor theory is the
Penrose transform which maps Spacetime < - - > Twistor Space
Twistor theory uses quite a few fairly abstract mathematical constructs, much of which are taken from the field of algebraic geometry, so unavoidably, I’m going to have to write down a teensy bit of mathematics.
The definition of twistors relies heavily on much better-known objects called spinors, and it’s virtually impossible to talk twistor without talking spinor language, so I’ll briefly describe spinors in this post. I make no pretence that any of this is rigorous or complete – I merely intend to give a flavour of the subject. OK caveats done, let’s start with spin and spinors.
We’ll restrict ourselves for now to flat spacetime, Minkowski space. This is the set of points with Cartesian coordinates (x
0, x
1, x
2, x
3). x
0 is the time coordinate and x
1, x
2, x
3 are the three space coordinates (x, y, z). These coordinates are considered to define events in an inertial frame, and the whole big deal about special relativity is that you can transform to a new inertial frame with coordinates (x’
0, x’
1, x’
2, x’
3) and the transformation will preserve the Minkowski metric i.e.
(dx’
0)
2-(dx’
1)
2- (dx’
2)
2- (dx’
3)
2 = (dx
0)
2-(dx
1)
2- (dx
2)
2- (dx
3)
2........(1)
The mapping (x
0, x
1, x
2, x
3) -> (x’
0, x’
1, x’
2, x’
3), known as a Lorentz transformation, is linear and hence can be represented by a 4x4 matrix.
For example for relative motion in a straight line in the x direction (people call this type of transformation a “boost”) would be represented by the matrix
(ɣ ...-βɣ...0...0)
(-βɣ...ɣ....0...0)...............(2)
(0 ....0.....1...0)
(0.....0 ....0...1)
where ɣ = 1/sqrt(1-v
2/c
2), and β = v/c. I apologize for the shitty looking depiction of matrices, this is the best I could manage with Unicode. Also, I had to put the dots in to separate the entries – I was having trouble with the spaces (but maybe that’s because I was drafting in MS Word, which has a mind of its own).
In addition to transformations like this, which describe relative motion, by mixing up the time and space coordinates (x
0 and x
1 in the example) the matrix can also describe rotations, which define new space coordinates which are tilted with respect to the old ones. Not any old 4x4 matrix is an allowable Lorentz transformation – they have certain restrictions (such as ensuring (1) is respected, for example). The set of allowed Lorentz transformations is a group – you can do stuff like compose two transformations to get a new one, and define inverse transformations etc. Mathematicians and physicists like groups. They like them very much indeed.
The Lorentz group is an abstract mathematical structure, and a large part of particle physics/field theory is concerned with
representations of this group (strictly the Poincare group, but I’ll ignore the difference for now). What’s a representation you may ask.
Spin 1Well you’ve already met the first example of a representation of the Lorentz group, namely 4x4 matrices of which (2) is an example. We can think of a representation as a mapping of the abstract group to a bunch of matrices. These 4x4 matrices, together with the vectors they act upon constitute the vector representation, which is descibed as “having spin equal to 1”. What this means intuitively is that, restricting attention to the spatial (3x3) part of the matrix and forgetting about the boosts for the moment, the matrix performs rotations on the vectors on which it acts. For example a rotation through angle θ anticlockwise about the z axis would be the matrix
(cosθ...sinθ....0)
(-sinθ...cosθ...0) ...................... (3)
(0.........0......1)
It’s easy to see that as θ goes from 0 to 2π, the vector on which the matrix acts rotates and goes back to where it started from. We’ll say that this sort of behaviour, i.e. rotation by 2π bringing you back to where you started from is “spin 1” behaviour. We can think of the rotation as acting on an arrow, anchored at its tail, the arrow representing the vector.
- Spin1.jpg (12.65 KiB) Viewed 5045 times
Spin 2All very easy to imagine. Now there is another type of behaviour, called spin 2. This sort of behaviour is exhibited, not by a vector, but by an object which rejoices in the name of “symmetric traceless second rank tensor”. Instead of the single index possessed by a vector (which labels the vector’s components), take an object with a pair of indices T
ij. The pair gives it the “second rank” designation. “Symmetric” means T
ij = T
ji. “Traceless” means = T
ii = 0. Note here we adopt the Einstein summation convention, whereby if an index is repeated in a term, that index is assumed to be summed over. Thus
T
ii = Sum over i ( T
ii )
With this convention, the rule for transforming a vector by multiplying it by a matrix is written (where L
ij are the components of the matrix, and the primed values V' are the new components of V after the rotation)
V’
i = L
ijV
jAnd the rule for transforming a second rank tensor is just
T’
ij = L
ikL
jlT
kl...........(4)
If you perform the rotation (3) about the z azis, and just look at the x and y directions, the relevant part of the matrix (3) is just
(cosθ... sinθ)
(-sinθ...cosθ)........... (5)
Using this in (4), we find that, for θ = π, T’
ij = T
ij, i.e. T has rotated back to itself. In other words, you only need to rotate through π to get back where you started. So, you can visualise a spin 2 "object" as a double headed arrow - a 180deg rotation gets you back to where you started.
- Spin2.jpg (11.11 KiB) Viewed 5045 times
Gravity is modelled as a spin 2 field.
Spin 1/2So far we have spin 1 and spin 2. The key spin for our twistorial purposes, however, will be spin ½. Given that you have to rotate through 2π to restore originality for spin 1 and through π for spin 2, you might guess that you’d have to rotate through 4π to restore originality for spin ½. Indeed you do, and the object in question is called a spinor.
To define spinors, again it’s the rotation part of the Lorentz group we’re concentrating on. A generic rotation in the x y z space is given by a matrix parameterised by 3 angles (the Euler angles ϕ
1, θ, ϕ
2)
(cosϕ
2cosϕ
1-cosθsinϕ
1sinϕ
2....-cosϕ
2sinϕ
1-cosθcosϕ
1sinϕ
2......sinϕ
2sinθ)
(sinϕ
2cosϕ
1+cosθsinϕ
1cosϕ
2....-sinϕ
2sinϕ
1+cosθcosϕ
1cosϕ
2....-cosϕ
2sinθ)
(sinθsinϕ
1...........................sinθcosϕ
1...............................-cosθ)
Rotation matrices can be mapped to complex 2x2 matrices, by pasting the Euler angles into the matrix
(cos(θ/2)e
i(ϕ2+ϕ1)/2... isin(θ/2)e
i(ϕ2-ϕ1)/2)
(isin(θ/2)e
i(ϕ2-ϕ1)/2... cos(θ/2)e
i(ϕ2+ϕ1)/2).............(6)
As a 2x2 matrix, this one operates on column vectors which have 2 complex components. The group of such matrices is called SU(2) and the complex “vectors” on which it operates are called spinors. The mapping from the 3 real-dimensional rotation matrices to the two complex-dimensional SU(2) matrices is double valued i.e. there are two SU(2) matrices for each rotation matrix.
Now the thing to note about the SU(2) matrix is that the angles appearing in there are divided by 2. This is why rotating a spinor (by multiplying by the SU(2) matrix (6) )requires a 4π rotation angle to restore it to its original state.
We had the pictures of the arrow and the double headed arrow to visualise spin 1 and spin 2 “objects”, but we’re going to have to work a bit harder to visualise spin ½. Given any physical object, 2π rotation will put it back where it started, so we need something more than just an object ! In fact what we use is an object together with its relationship with the environment. This is quite nicely illustrated by
this video. The object is the glass, and the “relationship with the environment” is dude’s arm. Half way through, the glass has rotated through 2π, but its relationship with the environment is different (dude’s arm is now twisted). A further rotation of 2π again puts it back where it started and dude’s arm is untwisted. This is sometimes referred to as the “orientation-entanglement” view of the spinor. There are other illustrations of it – for example there’s a nice picture in
The Road to Reality, but it would be naughty to copy it here.
Now a little aside from the main story – we all know the spin-statistics theorem – in particular that spin ½ particles are fermions. We now know that spin ½ particles are described by spinors which have this orientation-entanglement interpretation. We can use this to heuristically motivate the fact that these must be fermions. (Note I said "motivate", not "prove". This is just a bit of handwaving !)
- SpinStats.jpg (44.31 KiB) Viewed 5045 times
Being a fermion means that the system's wavefunction changes sign when you swap places with another fermion. Look at the diagram to see what happens when we interchange two spin ½ particles. For each particle, part of its “environment” is the other particle and we show its relation to the other particle by attaching a pair of strings. Label the particles black and white and exchange them. We end up with the situation in step 3. The particles are exchanged, but the strings are twisted. To untwist I can flip the black one twice, i.e. rotate it through 2π.
So, for spinor particles, one exchange is equivalent to a 2π rotation. Now a 2π rotation using the matrix (6) simply multiplies the spinor by -1, so there's something distinctly "fermionic" going on !
Note that I've talked about spatial rotations to simplify things, so I end up with the spinors being representations of SU(2). However, there are also boosts in the Lorentz group, and if you include those in the picture, the 2x2 matrices that act on the spinors are now elements of the group SL(2,ℂ) rather than SU(2), but as far as the heuristics above are concerned, this is just a technicality.
Next time - a bit more about spinors and Minkowski space (we'll get to twistors eventually).
Edit: Got the signature of the metric wrong !