(You only have to answer two of the three questions below.)

**1.** [50 points total] Relativity

**A.** A particle of mass $m_1$ decays at rest into two particles with rest masses $m_2$ and $m_3$ respectively. The i-th particle has 4-momentum $p_i=m_i\gamma_i\{1,0,0,\beta_i\}$ where $\gamma_i=(1-\beta_i^2)^{-1/2},i=1,2,3,\beta_1=0$.

**i.** [5 points] Give an expression for the relation between the three 4-vectors. What conservation law(s) does it represent?

Greg’s Solution:

(1)This represents conservation of 4-momentum.

**ii.** [5 points] Find the invariant mass of the two-particle system following the decay.

Greg’s Solution:

(2)**iii.** [5 points] Find the kinetic energy of particle 3 in terms of $m_1,m_2,m_3,\beta$.

Greg’s Solution:

(3)**B.** [15 points] In system S, two events at coordinates $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ occur simultaneously at time $t_0$. System S’ moves at a velocity $\beta_c$ along the x-axis with respect to system S. Show that the two events are not simultaneous in S’, but instead have a time separation of $\Delta t=-\beta \gamma \Delta x/c$, where $\gamma=1/(1-\beta^2)^{1/2}$ and $\Delta x=x_1-x_2$.

Greg’s Solution:

The Lorentz transform is,

(4)Thus, the two points are transformed to,

(5)And so we see that the difference between the two time components is indeed $\gamma\beta (x_2-x_1)/c$.

**C.** [20 points] The CERN SPS accelerator produces a beam of $\,^{208}\text{Pb}$ nuclei with a laboratory total energy of 158 GeV/nucleon. This beam strikes a fixed target of $\,^{208}\text{Pb}$ atoms. (Note: the rest mass of a nucleon is 0.938 GeV. Ignore effects of binding energy.)

**i.** [6 points] Find the Lorentz $\gamma$ factor of the beam.

Greg’s Solution:

The total energy is $\gamma m$, so we have that

(6)**ii.** [7 points] For Pb-Pb collisions, find the invariant mass per nucleon of the system and the center of mass collision energy per nucleon.

Greg’s Solution:

(7)where $m$ is the rest mass of a nucleon of lead; note that this has the correct limit that as $\gamma\to 1$, $m_{\text{tot}}^2\to (2m_{\text{nucleon}})^2$.

The first component of the momentum tensor is is the total energy; to obtain the kinetic energy we just subtract the mass of the two nucleons from it,

(8)**iii.** [7 points] If a collider produced colliding beams of lead ions at the same energy, what would be the center of mass collision energy per nucleon?

Greg’s Solution:

In this case, the total 4-momentum of the system would be

(9)Subtract the mass of the two nucleons to obtain the collision energy,

(10)
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