Relativistic addition question

[SiennaTheGr8]

And now I'm thinking I was overthinking this, and that it should just be $\phi$ every time:

$\vec f = \dfrac{m}{\cosh{\phi}} \, \dfrac{d}{d \tau} \left( \sinh{\phi} \, \hat{v} \right) = m \left( \dfrac{d \phi}{d \tau} \hat{v} + v \dfrac{d \hat{v}}{d \tau} \right).$

$\vec f_{\parallel \vec v} = m \left[ \left( \dfrac{d \phi}{d \tau} \hat{v} \right)_{\parallel \vec v} + \left( v \dfrac{d \hat{v}}{d \tau} \right)_{\parallel \vec v} \right]$

$\vec f_{\parallel \vec v} = m \left( \dfrac{d \phi}{d \tau} \hat{v} + \vec 0 \right),$

so:

$f_{\parallel \vec v} = m \dfrac{d \phi}{d \tau},$

invariant under a boost in the $\pm \hat{v}$ direction (though in general $\hat{v}$ may rotate as the force is applied).

If that's right, then I guess I was getting myself confused earlier by the fact that $\Delta \phi$ isn't in the $\pm \hat{v}$ direction. That felt off (still kind of does).

 

[SiennaTheGr8]

If that's right, then I guess I was getting myself confused earlier by the fact that $\Delta \phi$ isn't in the $\pm \hat{v}$ direction. That felt off (still kind of does).

And now I'm very confident that it's right. The key point is that $\vec v$ is (at least) momentarily parallel to the boost axis, and so the infinitesimal $d \phi$ is indeed invariant under a collinear boost. Nothing feels off anymore, and I was wrong to bring the component $\phi_{\parallel \vec v}$ into this at all.

(Sorry if I've derailed the thread. That's my last word on this unless someone responds.)