Quick questions on Relativistic momentum and particle accelerators: HSC Physics Module 7

Newtonian mechanics gives momentum as $p = m v$. Two clues that this must break at high speed:

The correct expression for momentum of a particle of rest mass $m$ moving at velocity $\vec{v}$ is:

The total relativistic energy is $E = \gamma m c^2$. Combining with $p = \gamma m v$, eliminating $\gamma$ and $v$:

The kinetic energy is $KE = (\gamma - 1) m c^2$. As $v \to c$, $KE \to \infty$, which means no finite amount of work can accelerate a massive particle to the speed of light. Massless particles travel at $c$ and cannot be accelerated or decelerated (they exist only at $c$ in vacuum).

The whole job of an accelerator is to push charged particles to extremely high energies for collision experiments. Relativistic momentum dominates the design.

The reachable physics is set not by the lab-frame energy but by the centre-of-mass energy $\sqrt{s}$ available to make new particles. For a fixed-target collision of a particle with rest energy $m c^2$ on a target of the same kind:

At $E = 7$ TeV, $E_0 = m_p c^2 = 0.938$ GeV.

A particle of momentum $p$ in a perpendicular magnetic field $B$ has radius:

A linac uses successive RF cavities to add small kicks to the particle's energy along a straight line. Relativistic momentum determines the spacing of the drift tubes: as $\gamma$ grows, $v$ saturates near $c$ but $p$ keeps increasing, so cavity spacings only need to grow modestly along the line.

At $0.95 c$ the classical value is off by a factor of $\gamma \approx 3.2$. At $0.99 c$ it is off by a factor of $7$.

Photons have $p = E / c = h f / c = h / \lambda$ despite being massless. Important for radiation pressure and Compton scattering.

The correct form is $r = p/(qB) = \gamma m v / (q B)$. This is why cyclotrons need to be replaced by synchrotrons at high energies.