= Solution
Use coordinates $X^\mu=(ct,\mathbf x)$ and the mostly-plus <Minkowski metric> $\eta_{\mu\nu}=\operatorname{diag}(-1,1,1,1)$. Along a future-directed timelike <world line>, <proper time> satisfies
$$
c^2d\tau^2=-\eta_{\mu\nu}dX^\mu dX^\nu=c^2dt^2-d\mathbf x^2,
\qquad d\tau=dt\sqrt{1-v^2/c^2}.
$$
The <four-velocity> and <four-acceleration> are
$$
\boxed{U^\mu=\frac{dX^\mu}{d\tau}=\gamma(c,\mathbf v),\qquad
A^\mu=\frac{dU^\mu}{d\tau},\qquad\gamma=(1-v^2/c^2)^{-1/2}.}
$$
The <Minkowski norm> is fixed: $U^\mu U_\mu=-c^2$. Differentiating with respect to <proper time>, using the constant <Minkowski metric>, gives
$$
0=\frac{d}{d\tau}(U^\mu U_\mu)=2U_\mu A^\mu,
\qquad \boxed{\mathbf U\cdot\mathbf A=0.}
$$
In the instantaneous rest <reference frame>, $U^\mu=(c,0,0,0)$, so the orthogonality says $A^0=0$ there. This construction uses a timelike <world line>; a <massless particle> has no rest-frame <proper time> parameter.
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