Relativistic momentum and energy Knowledge-Based Study

Study notes

1. Relativistic Momentum

   In classical mechanics, the linear momentum of an object of mass m moving with velocity v is given by $p = mv$. However, when applying this definition to high-speed particles moving near the speed of light c, the principle of conservation of linear momentum fails to hold across different inertial reference frames under the Lorentz transformations.

   To ensure that the law of conservation of momentum remains universally valid in all inertial reference frames, Einstein redefined linear momentum in special relativity:

   $$p = \gamma m v = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}}$$

   Here, m represents the rest mass (invariant mass) of the object, and $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$ is the Lorentz factor.

   Key Characteristics:

   • Low-speed limit: When $v \ll c$, $v^2/c^2 \to 0$, which leads to $\gamma \approx 1$. Thus, relativistic momentum reduces smoothly to classical Newtonian momentum: $p \approx mv$.
   • High-speed limit: As $v \to c$, $\gamma \to \infty$, causing relativistic momentum $p$ to approach infinity. This shows that an infinite impulse (and thus infinite work) would be required to accelerate a massive object to the speed of light, making $c$ an unattainable speed limit for objects with non-zero rest mass.

   Relativistic Energy

   Applying the work-energy theorem with relativistic momentum leads to a profound revision of work and kinetic energy in special relativity. The total relativistic energy $E$ of a free particle is defined as:

   $$E = \gamma m c^2 = \frac{m c^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$

   Components of Relativistic Energy:

   • Rest Energy ($E_0$): When a particle is at rest ($v = 0$), $\gamma = 1$. Even with no motion, the particle possesses an intrinsic energy determined strictly by its mass: $$E_0 = m c^2$$ This demonstrates the fundamental equivalence of mass and energy.
   • Relativistic Kinetic Energy ($K$): Kinetic energy is the extra energy an object possesses due to its motion. It is the difference between total energy and rest energy: $$K = E - E_0 = (\gamma - 1)m c^2$$

   Classical Approximation of Kinetic Energy

   For velocities much smaller than the speed of light ($v \ll c$), we can expand $\gamma$ using the binomial theorem:

   $$\gamma = \left(1 - \frac{v^2}{c^2}\right)^{-1/2} \approx 1 + \frac{1}{2}\frac{v^2}{c^2} + \frac{3}{8}\frac{v^4}{c^4} + \dots$$

   Substituting this expansion into the kinetic energy equation gives:

   $$K \approx \left(1 + \frac{1}{2}\frac{v^2}{c^2} - 1\right) m c^2 = \frac{1}{2}mv^2$$

   Classical kinetic energy is therefore a first-order approximation valid only at speeds negligible compared to c.

   The Invariant Energy-Momentum Relation

   Total relativistic energy $E = \gamma m c^2$ and relativistic momentum $p = \gamma m v$ can be combined mathematically to eliminate velocity v. Squaring both terms reveals an essential invariant relation:

   $$E^2 = (pc)^2 + (m c^2)^2$$

   This equation, often called the relativistic energy-momentum dispersion relation, connects total energy, linear momentum, and rest mass directly.

   Crucial Implications:

   • Invariant Quantity: The difference $E^2 - (pc)^2 = (m c^2)^2$ is an invariant scalar. Its value is identical in every inertial reference frame, regardless of the observer's relative motion.
   • Massless Particles: For particles with zero rest mass ($m = 0$), such as photons: $$E = pc$$ Massless particles must always travel at the speed of light $c$ in a vacuum and carry momentum directly proportional to their energy: $p = \frac{E}{c}$.
   • Units in High-Energy Physics: In modern physics, momentum is frequently expressed in units of $\text{MeV}/c$ or $\text{GeV}/c$, and mass in $\text{MeV}/c^2$, allowing practical and direct calculations without converting through SI units.