Light-cone coordinate and Lorentz transformation

[Chenkb]

Assume that, in cartesian coordinate, we have a quark with momentum $k=(k_0,0,k_0sin\theta,k_0\cos\theta)$ and a fragmented hadron $p=(p_0,0,0,p_0)$.
Define, in light-cone coordinate, $k^+ = k_0 + k_3 = k_0(1+cos\theta)$, and $p^+ = p_0 + p_3 = 2p_0$.
And the longitudinal momentum fraction of the hadron is $z^+ = p^+/k^+$.

Obviously, we take the hadron direction as z-axis, and the quark has a transverse momentum $|\vec{k_\perp}| = k_0sin\theta$.
I wonder whether it is possible to make a suitable Lorentz transformation, so that the quark gets no transverse momentum, and maintain $z^+$ unchanged?

 

[LightHero]

No, it is not possible to make a suitable Lorentz transformation that will make the quark have no transverse momentum and keep the longitudinal momentum fraction of the hadron unchanged. The only way to do this would be to rotate the quark's momentum vector so that the transverse component was aligned with the longitudinal component, but this would change the longitudinal momentum fraction of the hadron since it would reduce the quark's momentum in the z-direction.