Single-vortex energy-entropy balance ($\Delta F/(k_{B}T)=E_c/(k_{B}T)+(\pi K-2)\log (L/a)$)

Consider an original counting example: a square two-dimensional XY model of side $L$, a microscopic core radius $a$, and a fixed phase stiffness ${\rho }_s$. This deliberately neglects screening by other vortices. Away from a unit phase vortex, $\theta =\arg (x+iy)$ and $|\nabla \theta |=1/r$. Integrating the Goldstone-mode effective free energy over an annulus givesBoundary shape changes finite terms. There are order $(L/a{)}^2$ core positions, so the positional entropy is $S_{\mathrm{v}}=2k_B\log (L/a)+O(k_B)$, where $k_B$ is the Boltzmann constant. With $K={\rho }_s/(k_{B}T)$, subtracting $T{S}_{\mathrm{v}}$ gives the displayed free energy.

For $L/a=1000$ and $E_c/(k_{B}T)=2$, the dimensionless free energy cost is about $5.546$ at $K=0.8$ and $-0.965$ at $K=0.5$. The same logarithm appears in the cost and the positional reward: increasing the system size strengthens whichever has the larger coefficient. The corresponding one-defect Boltzmann factors, about $0.00390$ and $2.62$, are relative statistical weights, not normalized probabilities. The latter signals failure of a dilute single-defect estimate.

A single net phase vortex requires compatible boundaries: on a periodic sample use neutral pairs, as required by vortex charge neutrality. This argument identifies a possible threshold, but predicting the universal stiffness jump requires the renormalized phase stiffness, after smaller pairs have responded. It does not justify setting the transition temperature equal to a bare lattice coupling divided by a universal constant.