normal neighbourhood ${\mathcal{U}}$ orthogonal to $\dot\gamma(s)$ and for each $y\in{\mathcal{U}}$ parallel transport $\{ e^a_\alpha \}_{\alpha=0,1,2,3}$ along the unique geodesic connecting $y$ to $\gamma(s)$. In this manner we ‘sweep out’ a tube $\tau_\gamma\subset {\mathcal{M}}$ in spacetime. Importantly, the local frame (throughout $\tau_\gamma)$ is a local section of $F{\mathcal{M}}$ which may be identified with a smooth section $E$ of $S{\mathcal{M}}$. The details of this identification may be found in §3 of [@fewster; @verch]. For our purposes it is sufficient to know that, as a consequence of this construction, the form of $(T^\mathrm{split}_{00'})^{AB}$ simplifies when restricted to the diagonal. The precise statement, quoted from [@dawson; @fewster] (lem.4) is:

If $E$ is any local section of $S{\mathcal{M}}$ obtained in the above fashion from the curve $\gamma$, then $\Theta^{AB}|_\gamma=0$.

We shall assume that $E$ has been obtained in this way. Therefore, the finite contribution $\langle \rho^{\mathrm{fin}}\rangle_\omega$ to the energy density is given by: $$\langle \rho^{\mathrm{fin}}\rangle_\omega (\gamma(t)) = \frac{i}{2}\vartheta^*\bigg(\big( \mathbbm{1}\otimes e_0\cdot\nabla -e_0\cdot\nabla \otimes \mathbbm{1} \big)
\big({\mathrm{W}}_\omega - {\,}^\psi{\mathrm{H}}^{(+)}_1  \big)\bigg)(t,t)$$ where $\vartheta=\gamma\otimes\gamma$. Finally, it may be argued in analogy with [@fewster; @verch] that one may re-express $\langle \rho^{\mathrm{fin}}\rangle_\omega$ as $$\label{Had regularised energy}
\langle \rho^{\mathrm{fin}}\rangle_\omega (\gamma(t)) = \frac{1}{2}\bigg( \mathbbm{1} \otimes D - D\otimes\mathbbm{1}  \bigg)\vartheta^*\big( {\mathrm{W}}_\omega-{\,}^\psi{\mathrm{H}}^{(+)}_1 \big)(t,t)$$ where $D$ 