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Wikipedia states that

$$ \begin{align} \mathbf{A} =\frac{d\mathbf{U}}{d\tau} &= \left(\gamma_u\dot\gamma_u c,\gamma_u^2\mathbf a+\gamma_u\dot\gamma_u\mathbf u\right) \\ &= \left(\gamma_u^4\frac{\mathbf{a}\cdot\mathbf{u}}{c}, \gamma_u^2\mathbf{a}+\gamma_u^4\frac{\left(\mathbf{a}\cdot\mathbf{u}\right)}{c^2}\mathbf{u}\right) \\ &= \left(\gamma_u^4\frac{\mathbf{a}\cdot\mathbf{u}}{c}, \gamma_u^4\left(\mathbf{a}+\frac{\mathbf{u}\times \left(\mathbf{u}\times\mathbf{a}\right)}{c^2}\right)\right). \end{align} $$

I understand the first two lines perfectly. But I don't see how the gamma squared in the second line becomes gamma to the fourth in the third line. Am I missing something

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  • $\begingroup$ have you expanded the double rotational to see the result? $\endgroup$ Dec 20, 2020 at 15:22
  • $\begingroup$ I dont understand. Looking at the last term in the second equation "a dot u" is a scalar. The result should be gamma to the fourth times u (times a scalar) $\endgroup$
    – R. Emery
    Dec 20, 2020 at 15:34

1 Answer 1

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with: $$\gamma^2\,\boldsymbol a+\gamma^4\frac{\boldsymbol a\cdot \boldsymbol u}{c^2}\boldsymbol u=\gamma^4\left(\frac{1}{\gamma^2}\boldsymbol a+\frac{\boldsymbol a\cdot \boldsymbol u}{c^2}\boldsymbol u\right)$$

and : $$\gamma^2=\frac{1}{1-\frac {\boldsymbol u \cdot \boldsymbol u}{c^2}}$$ $\Rightarrow$ $$\left(\frac{1}{\gamma^2}\boldsymbol a+\frac{\boldsymbol a\cdot \boldsymbol u}{c^2}\boldsymbol u\right)=\boldsymbol a+\frac{\boldsymbol u\times (\boldsymbol u\times \boldsymbol a)}{c^2}$$

Edit:

use the identity

$$\boldsymbol u\times (\boldsymbol u\times \boldsymbol a)=(\boldsymbol u\cdot \boldsymbol a)\boldsymbol u-(\boldsymbol u\cdot \boldsymbol u)\,\boldsymbol a$$

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    $\begingroup$ Concise and effective answer. +1 $\endgroup$
    – Sebastiano
    Dec 20, 2020 at 17:21

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