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RelativisticKinematics

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This is a personal summary from the Review of Particle Physics. In this summary {$c=1$}.

For one particle the 4-vector is:

{$ (1) \qquad\qquad {\cal P} =(E,\mbox{\it\bf p}) $}

and its modulus is:

{$ (2) \qquad\qquad{\cal P}\cdot{\cal P} = E^2-\mbox{\it\bf p}\cdot\mbox{\it\bf p}=m^2+p^2$}

whence {$E=\sqrt{m^2+p^2}$}

We now write the Lorentz transformation to a frame defined by the velocity {$\beta$} and by {$\gamma=1/\sqrt{1-\beta^2}$}. The variables {${\cal P}=(E,\mbox{\it\bf p})$} are in the original frame, whereas {${\cal P}^\prime=(E^\prime,\mbox{\it\bf p}^\prime)$} are in the new frame, and:

{$ (3) \qquad\qquad \begin{eqnarray} p^\prime_\perp&=& p_\perp \\ E^\prime &= &\gamma E - \gamma\beta p^\prime_\parallel \\p^\prime_\parallel& =& -\gamma\beta E + \gamma p_\parallel \end{eqnarray}$}

where the subscripts {$\parallel,\perp$} refer to the direction of the relative velocity {$\beta$}

It is straightforward to show that the combination {$ {\cal P}_1^\prime\cdot {\cal P}_2^\prime $} is equal to {$ {\cal P}_1\cdot {\cal P}_2 $}, i.e. the scalar product of 4- vectors is invariant. A special and evident case is that of Eq. (2), the modulus of a 4-vector.

Let us consider a proton hitting head-on a second proton at rest in the laboratory frame (e.g. for pion production). Let us find the center of mass (cm) frame by imposing that the two transformed momenta are equal and opposite. Here we use the primed variables for the lab frame:

{$ (4) \qquad\qquad \begin{eqnarray} E^\prime_1 & =& \sqrt{m^2+p^\prime^2} & \qquad & p^\prime_1=p^\prime \\ E^\prime_2& =& m & \qquad & p^\prime_2=0 \end{eqnarray} $}

For head-on hits there is only one direction, that of {$\mbox{\it\bf p}^\prime$}, and the products come off along the same direction, hence {$p_{1,2\,\perp}=0$} and we may write:

{$ \begin{eqnarray} p_1&=&-\gamma\beta \sqrt{m^2+p^\prime^2} + \gamma p^\prime\\ p_2&=&-\gamma\beta m \end{eqnarray}$}

Imposing that {$p_1+p_2=0$} one can calculate the cm velocity to be:

{$ (5) \qquad\qquad \beta=\frac {p^\prime} {\sqrt{m^2+p^\prime^2}+m} $}

This implies that:

{$ (6) \qquad\qquad \gamma= \frac {\sqrt{m^2+p^\prime^2} + m}{\sqrt{2m^2+2m\sqrt{m^2+p^\prime^2}}} $}

such that {$\gamma\beta=p^\prime/2m$} (like the classical cm velocity) and the momentum of the first proton in the cm is:

{$ (7a) \qquad\qquad p=p^\prime\, \frac {m }{\sqrt{2m^2+2m\sqrt{m^2+p^\prime^2}}} $}

the other proton having opposite momentum. Also the inverse of Eq. (7) is useful:

{$ (7b) \qquad\qquad p^\prime=\frac {2p} m \sqrt{m^2+p^2} $}

The energy of both particles in the cm is:

{$ (8) \qquad\qquad E=\frac {m\sqrt{m^2+p^\prime^2} +m^2} {\sqrt{2m^2+2m\sqrt{m^2+p^\prime^2}}} = \frac 1 2 \sqrt{2m^2+2m\sqrt{m^2+2p^\prime^2}} =\sqrt{\frac{m(E^\prime_1+E^\prime_2)} 2}$}

Note that the total energy in the cm, {$2E$}, corresponds to the Lorents invariant form {$({\cal P_1+P_2})\cdot({\cal P_1+P_2})$} since the total momentum in the cm is zero. In terms of {$p$}, of course, one has simply {$ E=\sqrt{m^2+p^2}$}.

The velocity of the lab frame seen from the cm is {$\beta^\prime=-\frac p {\sqrt{m^2+p^2}} $}, which corresponds to {$\gamma^\prime=\frac {\sqrt{m^2+p^2}} m$}. It is easy to check that transforming {$p$} and {$-p$} by Eq. (1) to this frame one finds respectively {$2p\sqrt{m^2+p^2}/m$} (cfr. Eq. (7b)) and {$0$}.

Two particles decay into three particles, e.g. two protons give rise to a proton, a neutron and a pion. Let's call {$M,M,m$} the mass of the proton, neutron and pion respectively, disregarding the small difference between the first two, Let's consider only the head-on case again. Working in the cm and equating the invariant form before and after the decay we get:

{$ \sqrt{2M^2+2M\sqrt{M^2+2q^\prime^2}} = 2\sqrt{M^2+q^2} = \sqrt{M^2+p^2}+\sqrt{M^2+(p-x)^2}+\sqrt{m^2+x^2} $}

where all the unprimed momenta are in the cm and add up to zero.

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