\frac{1}{vg} = \int_{∣ k ∣ < Λ} \frac{d ^{2} k}{( 2 π ) ^{2}} \frac{1}{2 E _{k}} coth (\frac{E _{k}}{2 T}) .

With

E = v^{2} k^{2} + m^{2}

, the radial measure obeys

k d k = E d E / v^{2}

, so

\frac{1}{vg} = \frac{T}{2 π v ^{2}} [lo g sinh (\frac{E}{2 T})]_{m}^{v^{2} Λ^{2} + m^{2}} .

Λ \to \infty

g_{c}

leaves

\frac{1}{vg} - \frac{1}{v g _{c}} = - \frac{T}{2 π v ^{2}} lo g [2 sinh (\frac{m}{2 T})] .

The zero-temperature relation with

m = Δ

\frac{1}{vg} - \frac{1}{v g _{c}} = - \frac{Δ}{4 π v ^{2}} .

Equating the finite parts gives

2 sinh (\frac{m}{2 T}) = e^{Δ/ (2 T)},

and therefore

m (T) = 2 T arsinh (\frac{1}{2} e^{Δ/ (2 T)}) .

The subtraction is a renormalization condition: it trades the cutoff-dependent bare coupling for the physical zero-temperature gap.

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