Sect., we consider specifically the isovector magnetic form factor of the nucleon. We first show the limitations in extracting the magnetic moment from lattice data at the smallest available lattice momentum transfer. To show this is remedied by twisted boundary conditions, we calculate the isovector magnetic form factor in heavy-baryon  at finite volume with partially twisted boundary conditions.

Momentum Extrapolation {#s:momext}
----------------------

In principle the momentum dependence of the nucleon form factors predicted from heavy baryon  [@Jenkins:1990jv; @Jenkins:1991es; @Bernard:1992qa; @Bernard:1995dp] can be used to extrapolate lattice data down to zero momentum transfer in a model-independent way. In , the isovector magnetic form factor has the one-loop behavior [@Bernard:1992qa; @Jenkins:1993pi; @Bernard:1995dp; @Bernard:1998gv] $$\label{eq:ffinfvol}
F_2(q^2) = 
2 \mu_1 
- 
\frac{g_A^2 M}{2 \pi f^2} \int_0^1 dx \, m_\pi P_\pi(x,q^2)
-
\frac{g_{\D N}^2 M}{9 \pi^2 f^2}
\int_0^1 dx \, F[m_\pi P_\pi(x,q^2), \D]
,$$ where $P_\pi(x,q^2) = \sqrt{ 1 + x (1-x) q^2 / m_\pi^2}$, and the non-analytic function $F(m,\d)$ is given by $$F(m,\d) 
= 
- \d \log \frac{m^2}{4 \d^2} 
+ 
\sqrt{\d^2 - m^2} \log 
\left( 
\frac{\d - \sqrt{\d^2 - m^2} + i \varepsilon}{\d + \sqrt{\d^2 - m^2}+ i \varepsilon}
\right)
.$$

![Plot of the isovector magnetic form factor’s deviation from linear $q^2$ behavior. []{data-label="f:mom"}](mom.eps){width="50.00000%"}

