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As explained for example in this article by Prof. Strassler, modern twistor methods to calculate scattering amplitudes have already been proven immensely helpful to calculate the standard model background in searches for "new physics".

If I understand this correct, the "practical" power of these methods lies in their ability to greatly simplify the calculation of scattering processes, which are due to limited computer power for example, not feasable applying conventional Feynman diagrams.

Depending on the system considered, a renormalization group transformation involves the calculation or summation of complicated Feynman diagrams too, which usually has to be simplified to obtain renormalization group equations which are numerically solvable in a finite amount of time.

So my question is: Could the new twistor methods to calculate scattering amplitudes be applied to simplify investigations of the renormalization group flow, in particular investigations of the whole renormalization group flow field beyond a single fixed point, too? Are such things already going on at present?

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+1: Excellent question! –  Michael Brown Mar 1 '13 at 13:21
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That's a remarkably technically high-brow question. Dilaton. ;-) I will wait for someone's better answers but my guess would be No. In general, twistors (which are really identified with null lines or surfaces, via Penrose) are powerful for exactly conformal (and therefore scale-invariant) theories (with exactly massless particles) such as the finite N=4 gauge theory. For scale-invariant theories, the renormalization group operations are trivial (do nothing). When the scale invariance is broken, twistor calculation starts to be messy, unconstrained etc. –  Luboš Motl Mar 1 '13 at 13:55
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Well my twistor theory's 25 years out of date, but from the little I understand of the modern applications, Lubos is right - the limitations of the original twistor theory still apply: it is nicely tailored to handle conformally invariant theories, $\mathcal{N}=4$ SYM is the canonical modern example, but the methods don't lend themselves easily to models with mass. You need to bring in an "infinity twistor" to break conformal invariance by singling out the null cone at infinity and then things get difficult. –  twistor59 Mar 1 '13 at 14:17
    
Thanks @LubošMotl and twistor59, since some time I am slightly more seriously interested in the renormalization business :-). Scale invariant fixed points are boring, so it would be more fun if these twistor methods could be extended to work for not scale/conformal invariant cases... I'll wait and see what other people have to say for a while. –  Dilaton Mar 1 '13 at 17:07
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up vote 4 down vote accepted

These methods do not merely simplify known Feynman techniques. They uncover previously unknown structures in the final amplitudes by using entirely new (motivic) techniques.

The renormalization procedures of the Feynman method are quite hidden in the new formalism, because it does not begin by imposing locality on the underlying physics. It makes all internal lines (in the twistor diagrams) on shell. The Hopf algebraic structure of renormalization would appear before the traditional description. So yes, the twistor description should clarify renormalization, but probably not in the way you are expecting. For a start, the complex renormalization procedure is not even required in many computations.

No, these things are not being done now, because they don't really make sense with the current state of knowledge. To date, people have focused on solving N=4 SYM and related theories, usually supersymmetric, or else working on concrete gluon amplitudes for the LHC. Only now is it time to begin attacking QCD itself, and beyond.

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There's a chapter on applying these tools to normal (non-SUSY) theories in the latest paper by Nima Arkani-Hamed and friends: arxiv.org/abs/arXiv:1212.5605 –  Vibert Mar 7 '13 at 8:17
    
In the "Outlook" chapter, "A clear goal would be to understand the physics of the renormalization group along these lines". I suspect the short answer to Dilatons question is "yes" and he'd have to ask N A-H for the long answer. –  JollyJoker Jul 16 '13 at 9:34
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