Similarly $$ \overline{u}(p_{2}')\gamma_{5}u(p_{2})=\sqrt{\frac{(E_{2}'+m)(E_{2}+m)} {4m^{2}}}\left(\frac{\sigma_{2}\cdot \mathbf{p}_{2}}{E_{2}+m}- \frac{\sigma_{2}\cdot\mathbf{p'}_{2}}{E_{2}'+m}\right). $$ In the CM system we have \( \mathbf{p}_{2}=-\mathbf{p}_{1} \), \( \mathbf{p'}_{2}= -\mathbf{p'}_{1} \) and so \( E_{2}=E_{1} \), \( E_{2}'=E_{1}' \). We can then write down the relativistic contribution to the NN potential in the CM system: $$ \begin{eqnarray} V^{pv}&=&-\frac{f_{\pi}^{2}}{m_{\pi}^{2}}4m^{2}\frac{1}{(p_{1}-p_{1}')^{2}- m_{\pi}^{2}}\frac{(E_{1}+m)(E_{1}'+m)}{4m^{2}} \nonumber \\ &\times&\left(\frac{\sigma_{1}\cdot\mathbf{p}_{1}}{E_{1}+m}-\frac{\sigma_{1} \cdot\mathbf{p'}_{1}}{E_{1}'+m}\right)\left(\frac{\sigma_{2}\cdot\mathbf{p}_{1}} {E_{1}+m}-\frac{\sigma_{2}\cdot\mathbf{p'}_{1}}{E_{1}'+m}\right). \nonumber \end{eqnarray} $$