In Taylor, read sections 15.13 - 15.17 for this week.

u = γ[(d**x**/dt), c(dt/dt)] = γ(**v**, c)

p = mu = (γm**v**, γmc).

Massless particles don't make sense in classical mechanics; a massless particle has zero kinetic energy and zero momentum, and may just as well be ignored. At first glance, relativity also seems to have no room for massless particles since p = γβmc and E = γmc².

But on closer inspection, it is seen that they can exist, with finite momentum and energy.

Collisions are one of the most important applications of special relativity. The study of subatomic objects (electrons, nuclei, protons, neutrons, quarks, ...) is only possible by probing one object with another through collisions, and the energies involved mean that the colliding particles are normally relativisitic.

Collisions are commonly discussed in one of two frames, the *lab frame* and the *CM frame*.
The lab frame is whatever frame the experiment is carried out in, physically.
The CM frame, as we found in earlier chapters, is useful since the total momentum of the system is zero.
This still holds in relativistic problems, only now we use the relativistic four-momentum and find the frame where the sum of the spatial components (p1, p2, and p3) are zero

P_{CM} = ∑p = (0, 0, 0, P_{4})

The fourth component is just the total relativisitic energy of the system divided by c, POne particular advantage to this frame is in two particle scattering, the final 3-momenta of the two emerging particles are equal in magnitude and opposite in sign. This is a nice simplification to the overall problem.

The lab frame normally represents the frame of reference in which a measurement apparatus is at rest, and is therefore the frame in which the scattered particles are measured. An example of this is the Rutherford scattering experiment, where an incoming beam of alpha particles strikes a target foil, at rest in the lab frame, and the outgoing alpha particles are observed in an apparatus, also at rest in the lab frame. The scattering angle is measured in the lab frame. In the text, the target nucleus is assumed to be much, much heavier than the alpha particle, so that the lab and CM frames are essentially equal, but for light nuclei (aluminum for instance) this is not exactly true and the scattering angle should be transformed to the CM frame for comparison with the theoretical expression.

High energy collisions of two photons can yield electron-positrons pairs. Due to the cosmic microwave background radiation (black body spectrum with T = 2.725K), this process sets an upper limit to the energy of photons that can freely propagate across galactic distances. Find the threshold energy (in eV) for photons to scatter (inelastically) into an electron-positron pair in the CM frame. Determine the threshold energy for a photon to scatter with a micrwave background photon (peak energy of the spectrum) and produce an electron-positron pair (this is the lab frame).

A positron is the anti-particle of an electron; it has the same mass (m_{e}c² = 511keV) but opposite charge.
The energy of the scattered particles (electron and positron) is the sum of their rest energies plus kinetic energies, E_{f} = 2m_{e}c² + T_{e} + T_{p}.
The threshold energy is the smallest possible energy for the process to proceed, and this occurs when the kinetic energies are zero, E_{f} = 2m_{e}c².
The photon is a massless particle, so it's four-vector is __p__ = (**p**_{3}, p_{3}), where **p**_{3} is the 3-momentum.
In the CM frame, the two photons have equal and opposite 3-momenta, so their total energy (fourth component) is 2p_{3}c.
Setting this equal to the threshold energy yields the photon energy of E_{γ} = p_{3}c = m_{e}c².

The energy of a photon at the peak of a blackbody spectrum is E_{μ} = hc/λ_{max} = 2π(hbar)cT/b = 2π(197.3×10^{-9} eV m)(2.725 K)/(2.898×10^{-3} m K) = 1.165×10^{-3}eV.
To determine the threshold energy for a colliding photon to produce an electron-positron pair, note that in the CM, the sum of the photon four-momenta is __P__ = (0, 0, 0, 2m_{e}c;), and its square is an invariant, P² = -4(m_{e}c)².
For scattering off microwave photons, __P__ = __p___{γ} + __p___{μ} = (p_{γ} - p_{μ}, 0, 0, p_{γ} + p_{μ}), and equating the squares, we find (p_{γ} - p_{μ})² - (p_{γ} + p_{μ})² = -4p_{γ}p_{μ} = -4(m_{e}c)².
Therefore the threshold energy of the photon is Eth = p_{γ}c = m_{e}²c^{4} / E_{μ} = (5.11×10_{5 eV)² / (1.165×10-3eV) = 2.24×1014eV.
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