Compare classical and relativistic momentum, derive p = gamma m v, and analyse the role of relativistic momentum in particle accelerators

Speed: $v = 0.95 c = 2.85 \times 10^8$ m/s.

Lorentz factor:

$\gamma = 1 / \sqrt{1 - 0.9025} = 1 / \sqrt{0.0975} = 1 / 0.3122 = 3.203$.

Relativistic momentum:

$p = \gamma m v = 3.203 \times 9.11 \times 10^{-31} \times 2.85 \times 10^8 = 8.31 \times 10^{-22}$ kg m/s.

Classical momentum at the same speed:

$p_c = m v = 9.11 \times 10^{-31} \times 2.85 \times 10^8 = 2.60 \times 10^{-22}$ kg m/s.

Ratio: $p / p_c = \gamma = 3.20$.

At $0.95c$ the relativistic momentum is more than three times the classical value, so classical mechanics underestimates the momentum substantially. Markers reward correct $\gamma$, both momenta, and an explicit comparison showing that the discrepancy grows rapidly as $v \to c$.

A charged particle of momentum $p$ moving perpendicular to a magnetic field $B$ follows a circular path of radius:

$r = p / (q B)$

with $p = \gamma m v$ in the relativistic case. As the particle is accelerated, its speed quickly saturates close to $c$, but $\gamma$ continues to grow without bound (it tends to infinity as $v \to c$). The momentum, and the kinetic energy $KE = (\gamma - 1) m c^2$, continue to grow with $\gamma$ even though $v$ barely changes.

To keep a particle of growing $\gamma$ on the same circular path, either $B$ must be increased (synchrotrons ramp $B$ in lockstep with $\gamma$ during acceleration) or $r$ must be made very large (LHC has $r \approx 4$ km). Otherwise the particle's radius would exceed the beam pipe.

The non-relativistic formula $r = m v / (q B)$ would predict the radius levels off as $v \to c$, but in reality the radius keeps growing as $\gamma m v$ grows. This is direct evidence in operating accelerators that relativistic momentum is the right expression.

Markers reward the $r = p/(qB)$ relationship, the role of growing $\gamma$, and connection to the design choices in real machines.

$\gamma = \dfrac{1}{\sqrt{1 - v^2/c^2}} = \dfrac{1}{\sqrt{1 - 0.80^2}} = \dfrac{1}{\sqrt{1 - 0.64}} = \dfrac{1}{\sqrt{0.36}} = \dfrac{1}{0.60}$

$\gamma = 1.7$ (2 s.f.).

Marks: one for correctly substituting $v/c = 0.80$ into the Lorentz-factor formula, one for the correct value $\gamma = 1.7$.

(a) Classical momentum. $v = 0.60 \times 3.00 \times 10^8 = 1.80 \times 10^8\ \text{m s}^{-1}$.

$p_{cl} = m_e v = 9.109 \times 10^{-31} \times 1.80 \times 10^8 = 1.64 \times 10^{-22}\ \text{kg m s}^{-1}$.

(b) Relativistic momentum. $\gamma = \dfrac{1}{\sqrt{1 - 0.60^2}} = \dfrac{1}{\sqrt{0.64}} = 1.25$.

$p = \gamma m_e v = 1.25 \times 1.64 \times 10^{-22} = 2.05 \times 10^{-22}\ \text{kg m s}^{-1}$.

Marks: one for $p_{cl} = 1.64 \times 10^{-22}\ \text{kg m s}^{-1}$, one for a correct $\gamma = 1.25$, one for $p = 2.05 \times 10^{-22}\ \text{kg m s}^{-1}$ with unit.

(a) Speed. Rearrange $\gamma = 1/\sqrt{1 - v^2/c^2}$ to $v = c\sqrt{1 - 1/\gamma^2}$.

$v = c\sqrt{1 - 1/1.25^2} = c\sqrt{1 - 0.64} = c\sqrt{0.36} = 0.60c = 1.80 \times 10^8\ \text{m s}^{-1}$.

(b) Momentum. $p = \gamma m_p v = 1.25 \times 1.673 \times 10^{-27} \times 1.80 \times 10^8 = 3.76 \times 10^{-19}\ \text{kg m s}^{-1}$.

Marks: one for correctly rearranging for $v$, one for $v = 0.60c$, one for $p = 3.76 \times 10^{-19}\ \text{kg m s}^{-1}$ with unit.

(a) Lorentz factor. The kinetic energy gained is $KE = eV = 1.602 \times 10^{-19} \times 2.0 \times 10^6 = 3.204 \times 10^{-13}\ \text{J}$.

Rest energy: $m_e c^2 = 9.109 \times 10^{-31} \times (3.00 \times 10^8)^2 = 8.198 \times 10^{-14}\ \text{J}$.

$\gamma = 1 + \dfrac{KE}{m_e c^2} = 1 + \dfrac{3.204 \times 10^{-13}}{8.198 \times 10^{-14}} = 1 + 3.91 = 4.91$.

(b) Speed. $v = c\sqrt{1 - 1/\gamma^2} = c\sqrt{1 - 1/4.91^2} = c\sqrt{1 - 0.0415} = 0.979c$.

(c) Momentum. $p = \gamma m_e v = 4.91 \times 9.109 \times 10^{-31} \times (0.979 \times 3.00 \times 10^8) = 1.31 \times 10^{-21}\ \text{kg m s}^{-1}$.

Marks: one for $KE = eV$ and rest energy $m_e c^2$ correctly found, one for $\gamma = 4.91$, one for $v = 0.979c$, one for $p = 1.31 \times 10^{-21}\ \text{kg m s}^{-1}$ with unit.

(a) For small $v/c$ the two curves nearly coincide (both close to a straight line), but as $v/c$ increases the relativistic curve bends upward and rises ever more steeply above the straight classical line, diverging sharply as $v/c \to 1$.

(b) From the plotted values, at $v/c = 0.6$: $p_{cl} \approx 0.301 \times 10^{-18}\ \text{kg m s}^{-1}$ and $p \approx 0.376 \times 10^{-18}\ \text{kg m s}^{-1}$, so $p/p_{cl} \approx 1.25$. At $v/c = 0.9$: $p_{cl} \approx 0.452 \times 10^{-18}\ \text{kg m s}^{-1}$ and $p \approx 1.036 \times 10^{-18}\ \text{kg m s}^{-1}$, so $p/p_{cl} \approx 2.29$. In both cases this ratio equals $\gamma$ at that speed (since $p/p_{cl} = \gamma m v / (m v) = \gamma$).

(c) $\gamma = 1/\sqrt{1 - v^2/c^2}$ has a denominator that tends to zero as $v \to c$, so $\gamma$, and hence $p = \gamma m v$, grows without bound; the classical line $p_{cl} = m v$ has no such factor and keeps rising at a constant, finite gradient. This is why the relativistic curve visually shoots upward near $v/c = 1$ while the straight line does not.

Marks: one for describing the near-coincidence at low speed and the divergence at high speed, one for correctly reading both momentum pairs from the graph, one for identifying the ratio as $\gamma$, one for the asymptote explanation linking it to $\gamma \to \infty$ as $v \to c$.

As $v \to c$, the factor $\sqrt{1 - v^2/c^2} \to 0$, so $\gamma = 1/\sqrt{1 - v^2/c^2} \to \infty$. Since $p = \gamma m v$ and $m$ is fixed and non-zero, the momentum (and correspondingly the kinetic energy $KE = (\gamma - 1) m c^2$) would have to become infinite for $v$ to actually reach $c$.

Because only a finite amount of work can be supplied by any real accelerator, $\gamma$ can be made arbitrarily large but never infinite, so $v$ can be pushed arbitrarily close to $c$ but never equal to it. This is a dynamical limit (an energy requirement), not merely a mathematical one.

Marks: one for stating $\gamma \to \infty$ as $v \to c$, one for linking this to momentum/energy needing to become infinite, one for the conclusion that only finite work is available so $v = c$ is unreachable for a massive particle.

Band-6 plan. (1) State the classical prediction for radius $r = mv/(qB)$ and note it saturates as $v \to c$. (2) State the relativistic radius $r = p/(qB) = \gamma m v/(qB)$ and explain why $\gamma$ keeps growing. (3) Explain the practical design consequence (ramped $B$, or very large fixed radius). (4) Conclude that observed accelerator behaviour is direct evidence for relativistic momentum.

Model answer. A charged particle moving in a circle in a magnetic field has the magnetic force supplying the centripetal force, $qvB = mv^2/r$, giving the classical radius $r = mv/(qB)$. As $v$ approaches $c$, this classical formula predicts the radius should level off, because $v$ itself is bounded and $m$ is constant, so a fixed-frequency cyclotron using this assumption should keep working at all energies.

In reality the correct expression for momentum is $p = \gamma m v$, so the radius of the circular path is $r = \dfrac{p}{qB} = \dfrac{\gamma m v}{qB}$. Although $v$ does saturate close to $c$, the Lorentz factor $\gamma = 1/\sqrt{1 - v^2/c^2}$ grows without bound as $v \to c$, so the momentum, and hence the radius (at fixed $B$), keeps increasing well past the point where the classical formula would predict it should stop growing.

This has two direct design consequences. First, a simple fixed-field cyclotron falls out of phase with its accelerating voltage at high energy, because the cyclotron (angular) frequency $\omega = qB/(\gamma m)$ decreases as $\gamma$ grows, so cyclotrons are limited to modest, non-relativistic energies. Second, modern accelerators are built as synchrotrons, which keep the orbit radius fixed and instead ramp the magnetic field $B$ in step with the rising $\gamma$ (the LHC ramps $B$ from about $0.5\ \text{T}$ to $8.3\ \text{T}$ as protons are pushed from $450\ \text{GeV}$ to $7\ \text{TeV}$), or else use a very large fixed radius (the LHC's $r \approx 4.3\ \text{km}$) to keep the required field manageable.

The fact that real accelerators need ever-increasing field strength or ever-larger radii to keep confining particles whose speed has already nearly saturated is direct experimental evidence that momentum keeps growing via $\gamma$, confirming $p = \gamma m v$ over the classical $p = m v$.

Marker's note: the top band states both radius formulas explicitly, explains why $\gamma$ (not $v$) is the quantity that keeps growing, and names a concrete design consequence (ramped $B$ in a synchrotron, or a large fixed radius) rather than a vague "accelerators need more energy" statement. A response that only asserts "particles cannot reach $c$" without connecting this to the radius/momentum relationship caps in the middle band.

Band-6 plan. Take a clear position (the statement is false) and justify with three strands: (1) derive/quote $p = \gamma m v$ and show $\gamma \to \infty$ as $v \to c$ despite $v$ being bounded, (2) explain physically why momentum must be unbounded (particle accelerator behaviour, need for consistent momentum conservation across frames), (3) connect to the energy side ($KE \to \infty$) to reinforce the dynamical speed limit. Finish with an explicit judgement.

Model answer. The statement is false. While it is true that a massive particle's speed is strictly bounded, $v < c$, its momentum is not bounded by the same limit, because momentum in special relativity is $p = \gamma m v$ with $\gamma = 1/\sqrt{1 - v^2/c^2}$, not the classical $p = m v$. As $v \to c$, the term $1 - v^2/c^2 \to 0$, so $\gamma \to \infty$. Even though $v$ itself only approaches the finite value $c$, the product $\gamma v$ grows without any upper limit, so $p$ increases without bound.

This unbounded momentum is not just a mathematical curiosity - it is required for momentum to remain a conserved quantity across all inertial reference frames, which was the original motivation for redefining momentum in relativity (Newtonian $p = mv$ fails to be conserved in every frame once particles move at a significant fraction of $c$). The relativistic form was constructed precisely so that conservation of momentum continues to hold when analysed by observers moving at different velocities, consistent with Einstein's postulates.

The unbounded growth of momentum is also observed directly: real particle accelerators must supply ever-increasing magnetic fields or ever-larger radii to keep confining particles whose speed has already saturated close to $c$, because the orbit radius $r = p/(qB)$ keeps growing with $p$ even though $v$ barely changes. This is only explicable if $p$, not $v$, is the quantity that grows without bound.

Finally, this ties directly to energy: the kinetic energy $KE = (\gamma - 1)mc^2$ also diverges as $v \to c$, which is exactly why $v = c$ is unreachable for a massive particle in finite time - not because momentum is capped, but because reaching $c$ would require infinite momentum and infinite energy, both of which are physically impossible to supply.

Overall, the bounded nature of speed and the unbounded nature of momentum are two separate and entirely consistent features of special relativity; conflating them, as the statement does, misunderstands the role of the Lorentz factor.

Marker's note: the top band explicitly refutes the statement (not just describes the equation), shows the mathematical mechanism ($\gamma \to \infty$ while $v \to c$ stays finite), connects momentum's unboundedness to the requirement that it remain conserved across frames, and uses accelerator evidence and/or the energy relation to reinforce the argument. A response that only restates $p = \gamma m v$ without addressing the "must momentum be bounded" claim caps in the middle band.