Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 33D / b

Table of contents
- Solution b

Solution

Substitute

U = \frac{1}{2} (u_{z} λ^{- 1} e^{u} λ^{- 1} e^{- u} - u_{z}), V = \frac{1}{2} (- u_{\overset{z}{ˉ}} λ e^{- u} λ e^{u} u_{\overset{z}{ˉ}}) .

The off-diagonal entries of

U_{\overset{z}{ˉ}} - V_{z} + [U, V]

cancel identically. Its upper-left entry is

u_{z \overset{z}{ˉ}} + \frac{1}{4} (e^{- 2 u} - e^{2 u}) = u_{z \overset{z}{ˉ}} - \frac{1}{2} sinh (2 u),

and its lower-right entry is the negative of this. Hence compatibility is equivalent to

u_{z \overset{z}{ˉ}} = \frac{1}{2} sinh (2 u) .

This is the Sinh-Gordon equation in complex light-cone coordinates.

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