I am confused about a statement from do Caromo's proof of the Rauch Comparison Theorem. This is on page 215 of his textbook.
He says the following:
Since $\langle J, \gamma' \rangle \gamma'=\langle J'(0),\gamma'(0)\rangle t\gamma' + \langle J(0),\gamma'(0)\rangle \gamma'$
the tangential components of $J$ and $\tilde{J}$ have, by hypothesis, the same length.
Therefore we can suppose that $\langle J, \gamma'\rangle = 0 = \langle \tilde{J},\tilde{\gamma}' \rangle$.
I don't understand why this simplification. In particular, why does the first statement imply they have the same length and why can we (and why are we) assuming the second statement?
Let $A =(J',\gamma ')(0),\ B=(J,\gamma')(0)$,
Let $$J= f\gamma '+W$$ where $f=A t + B,\ \gamma'\perp W$.
(1) Here $f\gamma'$ is Jacobi so that so is $W$. Hence now we will check that $W,\ \widetilde{W}$ satisfies the assumptions.
(2) $J(0)=B \gamma' + W(0)$
(3) $(J',\gamma ')(0)= (A \gamma ' +W' ,\gamma') =A+ (W',\gamma')(0)$
(4) $|J'(0)|^2 =| A\gamma' +W'|^2 = A^2+|W'|^2$ since
$$(\gamma',W')(0)=(\gamma',J'-A\gamma')=0$$