curvature in $M_{T}$. Finally, when the width is $\geq2\Delta$ as in figure \[fig5\](d), the spin splitting becomes unobservable in $M_{T},$ and one is left with what appears as one unique oscillation around the center $\left(x_{1}+x_{2}\right)/2$.

Now, for a Fermi-Dirac distribution of the form $\left[1+e^{(y-y_{0})/a}\right]^{-1}$, the width $w$ around $y_{0}$ is determined by the condition $1/\left[1+e^{w/a}\right]=\sigma$, where $\sigma\ll1$ is the appreciation considered. From our experience, it is sufficient to take $\sigma\sim10^{-2}$, which implies a width $w\sim5a$. Therefore, following figure \[fig5\](d), the critical temperature for which the spin splitting becomes unobservable satisfies $2\Delta\simeq5k\tilde{T}$. Given that $k\tilde{T}=2kTx/\mu$, and $2\Delta=4\mu_{\mathrm{B}}/\mu$, we get the temperature $T_{s}=2\mu_{\mathrm{B}}/5kx$. This result basically means that, in order of magnitude, the thermal energy $kT$ equals the spin splitting energy $2\mu_{\mathrm{B}}/x$. One should replace in $x$ the value at which the broadening is computed. If the peak at $x_{1}$ is considered, the corresponding exponential factor is $\tilde{\beta}\left(x-x_{1}\right)$, and the condition $2\Delta\simeq5k\tilde{T}$ implies to take $x=x_{2}$. On the other hand, if we take the peak $x_{2}$, we should take $x=x_{1}$. This gives two different temperatures $T_{s,1}\simeq2\mu_{\mathrm{B}}/5 Kx_{2}$ and $T_{s,2}\simeq2\mu_{\mathrm{B}}/5 Kx_{1}$, but $\left|T_{s,1}-T_{s,2}\right|=T_{s,1}\left(2\Delta/x_{1}\right)=T_{s,2}\left(2\Delta/x_{2}\right)$ with $\Delta/x_{i}\ll1$ for both cases. Thus in general $T_{s,1}\simeq T_{s,2}$, and we can take the average temperature $T_{s}=\left(T_{s,1}+T_{s,2}\right)/2$, so

$$T_{s}\simeq\frac{2}{5}\frac{\mu_{\mathrm{B}}B}{k},\label{Ts}$$

