The finite-temperature gap equation is
$$
\frac1{vg}
=\int_{|\mathbf k|<\Lambda}
\frac{d^2k}{(2\pi)^2}
\frac1{2E_k}\coth\left(\frac{E_k}{2T}\right).
$$
With $E=\sqrt{v^2k^2+m^2}$, the radial measure obeys $k\,dk=E\,dE/v^2$, so
$$
\frac1{vg}
=\frac{T}{2\pi v^2}
\left[
\log\sinh\left(\frac{E}{2T}\right)
\right]_{m}^{\sqrt{v^2\Lambda^2+m^2}}.
$$
As $\Lambda\to\infty$, subtraction of the massless, zero-temperature equation at $g_c$ leaves
$$
\frac1{vg}-\frac1{vg_c}
=-\frac{T}{2\pi v^2}
\log\left[2\sinh\left(\frac{m}{2T}\right)\right].
$$
The zero-temperature relation with $m=\Delta$ is
$$
\frac1{vg}-\frac1{vg_c}
=-\frac{\Delta}{4\pi v^2}.
$$
Equating the finite parts gives
$$
2\sinh\left(\frac{m}{2T}\right)=e^{\Delta/(2T)},
$$
and therefore
$$
\boxed{
m(T)=2T\operatorname{arsinh}
\left(\frac12e^{\Delta/(2T)}\right).}
$$
The subtraction is a <renormalization condition>: it trades the cutoff-dependent bare coupling for the physical zero-temperature gap.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 337 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-337/1/e/solution