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Topic 5 of 10

Relativistic Momentum & Dynamics

In special relativity, momentum is $\vec p=\gamma m\vec v$. The relativistic second law $\vec F=d\vec p/dt$ preserves causality — as $v\to c$, the momentum grows without bound, requiring infinite force and energy to reach $c$. This is why the speed of light is an absolute limit for massive objects.

Key Concepts

Relativistic Momentum

\vec p=\gamma m\vec v

. In the low-speed limit,

\gamma\to1

and

\vec p\to m\vec v

. As

v\to c

|\vec p|\to\infty

Relativistic Force

\vec F=d\vec p/dt

. This is Lorentz-covariant. Longitudinal force (

F_{||}

) relates to

\gamma^3 m a_{||}

; transverse force (

F_\perp

) to

\gamma m a_\perp

— the effective mass is direction-dependent.

Speed Limit

For a massive particle, reaching

v=c

requires infinite momentum (and energy). A constant force produces continuously decreasing acceleration as

v\to c

; the particle asymptotically approaches but never reaches

c

Four-Momentum

p^\mu=(E/c,\vec p)

. The invariant

p_\mu p^\mu=(mc)^2

gives the energy-momentum relation

E^2=(pc)^2+(mc^2)^2

Key Equations

Relativistic momentum

\vec{p} = \gamma m\vec{v}

Reduces to mv for v ≪ c; diverges as v → c.

Energy-momentum relation

E^2 = (pc)^2 + (mc^2)^2

Invariant relation; for massless particles (photons), E = pc.

Kinetic energy

K = (\gamma - 1)mc^2

Relativistic KE; reduces to ½mv² for v ≪ c (to first order in β²).

Worked Example

Proton Momentum and Kinetic Energy

Problem

A proton ( $m_p c^2=938.3$ MeV) moves at $v=0.80c$ . Find its (a) relativistic momentum (in MeV/c) and (b) kinetic energy (in MeV).

Solution

$\gamma=1/\sqrt{1-0.64}=5/3=1.667$ .

p = \gamma m_p v = \frac{5}{3}\times938.3\times0.80\text{ MeV}/c = \frac{5}{3}\times750.6 = 1251\text{ MeV}/c

K = (\gamma-1)m_pc^2 = (1.667-1)\times938.3 = 0.667\times938.3 = 625.8\text{ MeV}

Answer

p=1251

MeV/c,

K=625.8

MeV.

Practice

Exercises

7 problems

1 of 7

A proton ( $m_pc^2=938.3$ MeV) moves at $\beta=0.60$ ( $\gamma=1.25$ ). Find its momentum $p$ (in MeV/c).

Your answer

MeV/c

2 of 7

Same proton ( $\beta=0.60$ , $\gamma=1.25$ ). Find its kinetic energy $K$ (in MeV).

Your answer

MeV

3 of 7

An electron ( $m_ec^2=0.511$ MeV) moves at $\beta=0.90$ ( $\gamma=2.294$ ). Find its total energy $E=\gamma m_ec^2$ (in MeV).

Unlock Exercise 3