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Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 306 / 1 / b / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2023 iii Paper 306 1 b
2026-09-28  0 By others on same topic  0 Discussions Create my own version
An infinitesimal Lorentz transformation is δXμ=ωμν​Xν with ωμν​=−ωνμ​. Applying Noether theorem to this continuous symmetry gives the conserved Lorentz current
Jαμν​=2πα′1​(Xμ∂α​Xν−Xν∂α​Xμ),∂αJαμν​=0.
(1)
Its Noether charge is Mμν=∫0π​dσJτμν​. Substituting the open-string mode expansion and using orthogonality of the cosine modes yields
Mμν=xμpν−xνpμ−i∑n=1∞​n1​(α−nμ​αnν​−α−nν​αnμ​).
(2)
The first term is orbital angular momentum; the sum is the contribution of the string oscillators.

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