Modulus of four acceleration

Question

The four acceleration is defined as $$\alpha^\mu = \gamma_V ^4 \left(\frac{\vec{v} \cdot \vec{a}}{c},\frac{\vec{v} \cdot \vec{a}}{c^2} \vec{v} + \frac{1}{\gamma_V ^2} \vec{a} \right)$$ where $\vec{v}$ is ordinary velocity and $\vec{a}$ ordinary acceleration in a certain reference frame.

Our professor told us that it comes straight forward from the definition that $$\alpha_\mu \alpha^\mu = \eta_{\mu \nu} \alpha^\mu \alpha^\nu = -\alpha^0 \alpha^0 + \vec{\alpha} \cdot \vec{\alpha} = |\vec{a}|^2$$ but I can't understand the last passage.

Answer 1

The equation your professor gave you ($\alpha^\mu \alpha_\mu = |\vec{a}|^2$) is only true in a frame where the object is instantaneously at rest ($\vec{v} = 0$). In this instance, we have $\alpha^\mu = (0, \vec{a})$ and the result follows fairly obviously.

A more general result can be obtained by plugging in the components of $\alpha^\mu$ into the definition of the four-vector norm: \begin{align*} \alpha^\mu \alpha_\mu &= -(\alpha^0)^2 + \vec{\alpha} \cdot \vec{\alpha} \\ &= \gamma_V^8 \left[ - \frac{(\vec{v} \cdot \vec{a})^2}{c^2} + \left( \frac{\vec{v} \cdot \vec{a}}{c^2} \vec{v} + \frac{1}{\gamma_V ^2} \vec{a} \right) \cdot \left( \frac{\vec{v} \cdot \vec{a}}{c^2} \vec{v} + \frac{1}{\gamma_V ^2} \vec{a} \right) \right] \end{align*} You can multiply this out, and it does simplify a little upon the application of the identity $(\gamma_V)^{-2} = 1 - |\vec{v}|^2/c^2$. But it does not reduce to $\alpha^\mu \alpha_\mu = |\vec{a}|^2$ unless $\vec{v} = 0$.