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Could someone explain the correspondence between lines in twistor space and minkowski space-time points? a basic derivation would suffice

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3 Answers 3

up vote 9 down vote accepted

The ordinary twistor space is parameterized by $(\lambda^\alpha,\mu_{\dot\alpha})$. Here, the $\alpha$ is a 2-valued $SL(2,C)$ spinor index of one chirality and the dotted index is its complex conjugate, the index of the opposite chirality.

At the level of spinors, vectors are equivalent to "spintensors" with one undotted and one dotted index. $$ V_\mu = \sigma_\mu^{\alpha \dot\alpha} V_{\alpha \dot \alpha}.$$ This is a basic fact about the Lie algebras. $SO(3,1)$ is locally isomorphic to $SL(2,C)$ - they're the same 6-dimensional Lie groups - and the 4-vector is the tensor product of ${\bf 2}$ and ${\bf \overline{2}}$.

If you're unfamiliar with this equivalence of vectors of "spintensors" with two indices, notice that the components of a 4-vector may be organized as $$v^{\alpha\dot\alpha} = \left( \begin{array}{cc} v^0+v^3 & v^1-i v^2\\ v^1+i v^2 & v^0-v^3 \end{array} \right) $$ Note that the determinant of this matrix - a natural function of the matrix elements - is simply $v^\mu v_\mu$. The whole matrix may be understood as the appropriate combination of the three Pauli matrices - plus the "time-like Pauli" (identity) matrix multiplied by the time component of the vector. Sorry if my signs deviate from the prevailing convention.

So far, it has only been a story about the vectors or spinors. What about the twistors? Well, it's a simple one-line formula. If I have a point $x^\mu$ which is equivalent to the $x^{\alpha \dot\alpha}$ matrix - as any 4-vector - I may simply write an equation $$ \lambda^\alpha = x^{\alpha \dot\alpha} \mu_{\dot\alpha} $$ Note that it is a set of two complex linear equations - for $\alpha=0,1$ - so it defines a linear object. For a different value of $x^\mu$, I get different equations. Moreover, the equations link the $\lambda$ and $\mu$ objects which are coordinates on the twistor space.

Now, I have to explain why the equations above define a line. First, the $\lambda$ and $\mu$ objects are pairs of complex numbers, so in total, we have four complex coordinates. However, the twistor space is a projective space. Note that if the equations above are satisfied for some $\lambda$ and $\mu$ - four complex numbers - they will also be satisfied if you multiply both $\lambda$ and $\mu$ by an arbitrary complex number (the same one for both). This projectivity holds universally: $\mu$ is naturally scaled in the same way as $\lambda$ because it may be understood as a dimensionally inverse object to $\lambda^{\dot\alpha}$ (an object that appeared for the first time in this answer) that scales in the inverse way relatively to $\lambda^\alpha$ in order to keep e.g. the vector $\lambda^\alpha \lambda^{\dot\alpha}$ constant.

So the twistor space is really a complex projective space, $CP^3$, and the equations above are two complex conditions, so we're left with a one(-complex)-dimensional object, a complex line. Spacetime points are in one-to-one correspondence to complex lines in the twistor space.

Many more things may be translated. For example, if two spacetime points are separated by a null interval, the corresponding two lines in the twistor space intersect. The intersection - a point in the twistor space - may be identified with a null line in the Minkowski space, and I could derive many other things of this kind.

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great answer! thanks Lubos. Just one further clarification: you said that λαλα˙ must be kept constant while we change a scaling parameter applying both to λ and μ. I definitely agree that such scaling parameter will satisfy the equation for the same xμ, what i don't understand is why that scaling has the additional requirement of keeping λαλα˙ constant. –  lurscher Feb 2 '11 at 16:38
    
hi, thanks! It was just another example motivating why it's a projective space. $\lambda^\alpha\lambda^{\dot\alpha}$ is constant e.g. in the modern application of twistors on gauge theory because it is interpreted as the momentum $p^{\alpha\dot\alpha}$ of the external gluons whose scattering we calculate. Note that such a form of $p$ is automatically null i.e. light-like - the matrix is degenerate - which is OK because gluons are massless. So instead of $p$, the external gluons are labeled by $\lambda^\alpha$ and $\lambda^{\dot\alpha}$ which can be rescaled in opposite ways to keep $p$ fixed. –  Luboš Motl Feb 2 '11 at 17:17

I would like to add some further points to the answers above on the Twistor Space <--> Spacetime correspondence.

The Twistor space T is a four complex dimensional space with elements described by $(Z^0,Z^1,Z^2,Z^3)$ or $Z^{\alpha}=(\omega^A,\pi_{A'})$ in spinor terms. The incidence relation between Minkowski points and Twistors is given (in spinor form) by:

$\omega^A=ix^{AA'}\pi_{A'}$

(In addition to a different representation of $x^{AA'}$ to that of Lubos it should also be noted that there is a 4-fold ambiguity in the representation of a spacetime point by a Twistor $(Z^{\alpha},-Z^{\alpha},iZ^{\alpha},-iZ^{\alpha})$ are all the same point as there are two sets of double cover involved.)

This Twistor Space contains a norm

$||Z||=Z^AZ^c_{A}=\omega^A\pi^c_{A} + \omega^{cA'}\pi_{A'}$ where $^c$ denotes spinor/twistor complex conjugation.

There are two sets of reductions to Twistor space to obtain the Minkowski space correspondence:

  1. ||Z||=0 these are the null Twistors - the space sometimes denoted N.

  2. T modulo $\lambda Z^{\alpha}$ - the space denoted PT -projective Twistor Space.

Putting these two together we get PN - projective Null Twistor Space. It is PN which is mapped to Minkowski space via the incidence relation ie projective null twistors.

This raises the question as to what to more general cases of Twistor mappings correspond to.

In general PT maps to Complexified Minkowski space. This space has the $(x^0,x^1,x^2,x^3)$ become complex numbers and the metric becomes a complex analytic extension of the Minkowski metric. Non-null Twistors become complex planes in that space.

The mapping to Complexified Minkowski space had several attractions in the original Twistor theory including the idea that in a curved space context maybe the disappearance of a point as a Singularity could be seen as just its disappearance from real Minkowski space, but not the more basic Twistor space.

Finally in the Incidence relation we are assuming that $\pi_{A'}$ is non-zero. In that case the Incidence relation could be geometrically considered to be a mapping to a Compactified completion of Complex Minkowski Space which includes the null cone at Infinity and thereby the other constructs of Compactified Minkowski space used in Penrose diagrams.

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when you say T modulo $\lambda Z^{\alpha}$ i presume that includes lines NOT intersecting the origin –  lurscher Feb 7 '11 at 15:48
    
Correct - I had a slight formatting difficulty there too. It should be T-{0}. –  Roy Simpson Feb 7 '11 at 16:14
    
so, i think the question should arise - the fermion spin doesn't play any role in this mapping? –  lurscher Feb 9 '11 at 15:09
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Not in this basic construction. However this is not the end of the story. There are several interpretations of massless particles given the geometry above. The Twistor Norm can represent helicity of a massless particle moving in Minkowski space. –  Roy Simpson Feb 9 '11 at 15:39

Given Luboš answer, a good place to learn about this stuff is straight from the horse's mouth:Spinors and Space-Time: Volume 1, Two-Spinor Calculus and Relativistic Fields and Spinors and space-time: Spinor and twistor methods in space-time geometry.

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thanks for the references –  lurscher Feb 2 '11 at 16:58
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I was kind of obsessed with this book when I was at the high school... –  Luboš Motl Feb 2 '11 at 17:14

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