The transformation is a boundary Lorentz boost of rapidity $\beta$. It preserves $-d\tau^2+dX^2=-dT^2+dW^2$ and describes the charged thermal state in a frame moving with speed $|v|=\tanh\beta$. The gauge field becomes
$$
A=\mu L(1-z)(\cosh\beta\,dT+\sinh\beta\,dW).
$$
If the rest-frame contravariant current is $J^\mu=(\rho,0,0)$, then in the new coordinates
$$
\boxed{J^T=\rho\cosh\beta,
\qquad J^W=-\rho\sinh\beta,
\qquad J^Y=0},
$$
where the sign of the spatial component follows from the stated passive coordinate transformation. Reversing the boost convention reverses that sign. For small $\beta$, the induced spatial current is $J^W\simeq-\rho\beta$.

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Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 354 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-354/2/d/solution