# Zero longitudinal momentum in Special Relativity

A particle is moving with an arbitrary momentum $p$ in K-reference frame. According to Jackson, "there is a unique Lorentz transformation in the $z$-direction to a frame K' where the particle has no $z$ component of momentum.", p.539, Classical Electrodynamics. He uses the proper vector transform:

$$r'_0 =\gamma (r_0 - \pmb \beta \cdot \pmb r)\\ r'_{||} = \gamma (r_{||} - \beta r_0)\\ \pmb r'_{\perp} = \pmb r_{\perp}$$

We apply it to the momentum vector. To get zero $p'_{||}$ one needs $p_{||} = \beta p_0$, hence $p_{||} = \frac{|v|}{c} p_0$. With this we finally get:

$$p_{||} = \frac{E|v|}{c^2} = \gamma m |v|$$

where $\pmb v$ is the velocity of K' with respect to K. This is the momentum of this particle as if it is moving with respect to the observer with the velocity equal to the velocity of K' frame. Then all Jackson's conclusions follow.

However, this equation doesn't exactly make perfect sense. I can understand this result thinking of the same sort of compensation as happens in Galilean mechanics: if I run as fast as the escalator goes but in the opposite direction, then I have zero velocity with respect to the observer on the ground. However, the statement of this equation doesn't say anything about the direction of $p_{||}$, only the magnitude. It could be that $\pmb p_{||}|| \pmb v$, which doesn't make a lot of sense to me.

In the Lorentz transform, the quantity that you've labeled $p_{\parallel}$ is the component of momentum parallel to the boost - that is, in this case, it's the $z$ component of the momentum $\mathbf{p}$. That's inherent in the definition of the transform. Because it's merely a component of a vector, and not a vector in its own right, there's no direction involved. Once you calculate the "magnitude"1 $p_{\parallel}$, you know everything there is to know about that particular component.
Of course, to find the momentum vector in the primed frame, you will need to combine the $z$ component $p_{\parallel}'$ with the $x$ and $y$ components $\mathbf{p}_{\perp}$. Or to find the four-momentum, you would also need to include the time component $p_0$.
• Thanks! How do I extract the direction? And by "not a vector", do you mean that $g=\{\pm1,\mp1,\mp1,\mp1\}$? The meaning of spatial orientation is still preserved, and vectorial momentum conservation should somehow make sense, no? – MsTais Aug 27 '18 at 14:15