- #1

Chenkb

- 41

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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?

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?

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