Lemma: The derivative of

In ${\mathcal{D}}^{\prime }({\mathbb{R}}^4)$, ${\partial }_{\alpha }\mathcal{B}=\frac{1}{2}{x}_{\alpha }R$. Consequently $\mathrm{supp}{\partial }_{\alpha }\mathcal{B}=\partial J^{+}$, and on $V_{+}$, $({\partial }_{\alpha }\mathcal{B})(d)=\frac{1}{4\pi }{d}_{\alpha }\delta (q)$.

Proof. Use the partial Fourier transform ${\mathcal{F}}_{\mathbf{x}}$ and the formulas of XVFpatch (Lemmas lem:R, lem:RR): ${\mathcal{F}}_{\mathbf{x}}R=\Theta (t)\sin (\rho t)/\rho$, ${\mathcal{F}}_{\mathbf{x}}\mathcal{B}=\Theta (t)u_2(t)$, $u_2=(\sin \rho t-\rho t\cos \rho t)/(2{\rho }^3)$, $u_2(0)=0$, $u_2^{\prime }=t\sin (\rho t)/(2\rho )$. For $\alpha =0$: ${\mathcal{F}}_{\mathbf{x}}({\partial }_t\mathcal{B})=\Theta (t)u_2^{\prime }(t)=\frac{t}{2}{\mathcal{F}}_{\mathbf{x}}R={\mathcal{F}}_{\mathbf{x}}(\frac{1}{2}tR)$, and $x_0=t$. For $\alpha =i\in \{1,2,3\}$: ${\mathcal{F}}_{\mathbf{x}}({\partial }_i\mathcal{B})=i{p}_i\Theta u_2$, while ${\mathcal{F}}_{\mathbf{x}}(x^{i}R)=i{\partial }_{p_i}{\mathcal{F}}_{\mathbf{x}}R=i\Theta (t)\frac{p_i}{\rho }\frac{d}{d\rho }\frac{\sin \rho t}{\rho }=-2i{p}_i\Theta u_2$, so ${\mathcal{F}}_{\mathbf{x}}(\frac{1}{2}{x}_{i}R)={\mathcal{F}}_{\mathbf{x}}(-\frac{1}{2}{x}^{i}R)=i{p}_i\Theta u_2$. Fourier injectivity on ${\mathcal{S}}^{\prime }$ gives the identity. The support statement follows from $\mathrm{supp}R=\partial J^{+}$ and $x_{\alpha }\ne 0$ somewhere on every open piece of the cone. The chart formula is (P4): $\frac{1}{2}{d}_{\alpha }\cdot \frac{1}{2\pi }\delta (q)$.