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.

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    $\begingroup$ Hello and welcome to Physics SE! When you refer to the "passages that lead to the results", are there any steps/calculations available that lead to this result (in which case it would be helpful to edit them into your post and show if there is some specific step that is troubling you etc.) or did the prof immediately "jump" from the second equation to the first? Thanks! $\endgroup$
    – jng224
    Apr 24, 2021 at 11:04
  • $\begingroup$ Thank you! I edited with all the steps I have found, but I can't understand the last step $\endgroup$
    – Jones
    Apr 24, 2021 at 13:15
  • $\begingroup$ The four acceleration is defined as... That is a hideous and unmotivated way to define it. It should be defined as $a^\mu=du^\mu/d\tau$ so that it is manifestly a four-vector. $\endgroup$
    – G. Smith
    Apr 24, 2021 at 19:31

1 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 $\vec{v} = 0$.

  • $\begingroup$ Just to say this is correct and a good way to write the result is $\alpha^\mu \alpha_\mu = a_0^2$ where $a_0$ is the proper acceleration, i.e. the three-acceleration in the instantaneous rest frame. $\endgroup$ Apr 24, 2021 at 17:57

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