Given a particle with mass $m$ and four momentum $p=\left(p_x,p_y,p_z,\sqrt{|\vec{p}|^2+m^2}\right)$. Is it possible to reach every point in phase space, i.e. every combination of values $p_x,p_y,p_z\in\mathbb{R}$, by a simple Lorentz-Boost?

I'm guessing no, but don't know why.

  • $\begingroup$ boosts are just translations in momentum space. so yes. you can reach any point with a certain translation. $\endgroup$
    – image357
    Mar 2 '17 at 13:11
  • $\begingroup$ Is this really a trivial statement? Could you explain this in more detail? $\endgroup$
    – FloodLuszt
    Mar 2 '17 at 15:14
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    $\begingroup$ Comment to the post (v2): Do you mean pure Lorentz-boosts, or are we also allowed to use other Lorentz-transformations, like e.g. spatial rotations? $\endgroup$
    – Qmechanic
    Mar 2 '17 at 15:53
  • $\begingroup$ I mean pure Lorentz-boosts as defined by one single 3-vector $\beta$ or $v$. $\endgroup$
    – FloodLuszt
    Mar 2 '17 at 15:57

Two boosts in orthogonal directions give you a boost and a rotation, so you are free to include rotations. But actually the answer is even simpler. Since the particle is massive, start in its rest frame where $p = (0,0,0,m)$ (note that we usually put time first, but I'm using your notation). Obviously you can boost this in any direction, so by rotational symmetry we can just consider the boosts along $x$. And it should be easy to convince yourself that boosts along $x$ can give you a momentum $p = (p_x,0,0,\sqrt{p_x^2 + m^2})$ for any $p_x \in \mathbb{R}$. So the answer to your question is yes.

Note also, be careful how you use words like "phase space". Since by definition, phase space is the space of states available, the answer to your question is trivially yes. The more interesting question is whether the phase space is all of $\mathbb{R}^3$, or just a subset.

  • $\begingroup$ Using both your ideas you prove that two four momenta with the same mass are connected by a sequence of two boosts or more. Instead OP's question refers to one boost if I understand well. I guess the answer is negative. $\endgroup$ Mar 2 '17 at 13:49
  • 1
    $\begingroup$ OK, but (as I hinted at slightly in the first sentence), boosts do not form a group. The physically-interesting question is whether two four-momenta are connected by a Lorentz transformation, which includes rotations as well as boosts. $\endgroup$ Mar 2 '17 at 14:11
  • $\begingroup$ Thanks for your answers. I was indeed wondering about one single boost, not a Lorentz transformation. So @Marcel's comment suggests that it is possible. I know that the generators of boosts are in fact momentum translation operators but I was hoping to get a more intuitive answer. $\endgroup$
    – FloodLuszt
    Mar 2 '17 at 14:49
  • $\begingroup$ Two pure boosts are always sufficient to connect two given momenta as you wrote in the second part: starting from the first momentum moves to the rest frame by means of a boost and then moves to the second momentum with another boost. No rotations are necessary (even if you may generate rotations by composing boosts as you correctly said). $\endgroup$ Mar 2 '17 at 15:25
  • $\begingroup$ But then you'd still need more than one boost. $\endgroup$
    – FloodLuszt
    Mar 2 '17 at 15:58

Let $\vec{p}$ be the original spatial part of the four-momentum, and let $\vec{p}'$ be the desired spatial part of the four-momentum. Let us assume that for any $\vec{p}'$, there exists a boost vector $\vec{\beta}$ under which $\vec{p}$ maps to $\vec{p}'$. This will lead to a contradiction.

We can choose our coordinate system such that $\vec{\beta}$ points in the $x$-direction. We can then rotate our coordinate system about this $x$-axis such that $\vec{p}$ points in the $xz$-plane (i.e., $p_y = 0$.) Under the action of the boost $\vec{\beta}$, we must have $p'_y = 0$ as well. (Remember, we defined our coordinates such that $\vec{\beta}$ was in the $x$-direction.) But we assumed that $\vec{p}'$ could be any vector in $\mathbb{R}^3$; we therefore have a contradiction.

We conclude that a single pure boost cannot map an arbitrary spatial momentum $\vec{p}$ to another arbitrary spatial momentum $\vec{p}'$.

  • $\begingroup$ It is not at all obvious to me how to describe the range of a given spatial vector under "pure boosts" (as a subset of $\mathbb{R}^3$). It would be interesting to find out. $\endgroup$ Mar 2 '17 at 16:16

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