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Voulkos
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If \begin{align} \mathbf{v} & =\textrm{velocity 3-vector} \tag{01a}\\ \mathbf{a} & =\textrm{acceleration 3-vector} \tag{01b}\\ \mathbf{U} &=\textrm{velocity 4-vector} \tag{01c}\\ \mathbf{A} &=\textrm{acceleration 4-vector} \tag{01d} \end{align} then \begin{align} \mathbf{A} & = \gamma_{v}\left(\gamma_{v}\,\mathbf{a}+\dfrac{\mathrm{d}\gamma_{v}}{\mathrm{d}t}\mathbf{v},\dfrac{\mathrm{d}\gamma_{v}}{\mathrm{d}t}c \right) \tag{02a}\\ \Vert\mathbf{A}\Vert^{2} & = -\gamma_{v}^{4}\left[\Vert\mathbf{a}\Vert^{2}+\left(\gamma_{v}^{2}-1\right) \left(\dfrac{\mathrm{d}v}{\mathrm{d}t}\right) ^{2} \right] \tag{02b} \end{align} You have all answers from equations (02).

Note that in the rest frame of the particle $\:(\mathbf{v}_{0}\equiv \boldsymbol{0}, \gamma_{v}=1,\mathbf{a}=\mathbf{a}_{0})\:$

\begin{align} \mathbf{A}_{0} & = \left(\mathbf{a}_{0},0 \right) \tag{03a}\\ \Vert\mathbf{A}_{0}\Vert^{2} & = -\Vert\mathbf{a}_{0}\Vert^{2} \tag{03b} \end{align}

Voulkos
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