A How is eq. 1.5.3 written using three-vectors and how does it lead to eq. 1.5.4?

[SwetS]

I am going through https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf, here I don't understand how we wrote eq. 1.5.3 with the help of three-vectors to get eq. 1.5.4.

 

[Ibix]

The two expressions aren't the same, I think. 1.5.3 assumes that $\vec{\beta}_v=(\beta_v,0,0)^T$, while 1.5.4 makes no such assumption. You could just plug this assumption into 1.5.4 as a plausibility check. If you actually need to derive 1.5.4 then I'd start with four velocities and work from there, not from 1.5.3.

@vanhees71 might add more detail.

 

[Ibix]

A slightly better plausibility argument is to argue that $(\beta_v+\bar{\beta}_w^1,\bar\beta_w^2/\gamma_v,\bar\beta_w^3/\gamma_v)^T$ could be said to be $\vec\beta_v$ plus the component of $\vec{\bar\beta}_w$ parallel to $\vec\beta_v$ plus $1/\gamma_v$ times the component of $\vec{\bar\beta}_w$ perpendicular to $\vec\beta_v$.

The vector times the dot product in the last term in brackets in 1.5.4 pulls out the component of $\vec{\bar\beta}_w$ parallel to $\vec\beta_v$, which is then added/subtracted appropriately to get what I wrote in words above.

 

[vanhees71]

The idea is to calculate the three-velocity $\vec{w}$ first for the simplifying case that $\vec{v}=v \vec{e}_1$. Then one makes use of the fact that $\vec{w}=\vec{W}/W^0$ is a "three-vector", i.e., it transforms under rotations as a three-vector, and thus one can get the expression for an arbitrary $\vec{v}$ by writing (1.5.2) in a form that is kovariant under rotations; you can indeed check that when setting $\vec{v}=v \vec{e}_1$ in (1.5.3) you get back (1.5.2). Since (1.5.3) is written in a kovariant form under rotations, it must be correct for the general case, if it's correct for the special case.