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PU-105 / L14 · lecture

Why Momentum Must Be Redefined

~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with the Lorentz factor \(\gamma\) and the concept of proper time from the previous lecture. Familiarity with Newton's second law and the definition of momentum as \(p = mv\) is assumed.

Why this matters

Classical momentum fails to conserve in relativistic collisions, necessitating a new definition that respects the speed of light limit.

In Newtonian mechanics, momentum conservation is a cornerstone of dynamics. However, when velocities approach the speed of light, the Galilean transformation of velocities no longer holds. If we insist that momentum is conserved in all inertial frames, the simple product of mass and velocity breaks down. This lecture explains why we must redefine momentum to include the Lorentz factor, ensuring that the laws of physics remain consistent across all reference frames.

The Failure of Classical Momentum

Imagine a collision between two particles. In Newtonian physics, if particle A hits particle B, the total momentum before the collision equals the total momentum after. This works perfectly at low speeds. However, special relativity tells us that velocities do not add linearly. If you observe a collision from a moving train, the velocities of the particles change according to the Lorentz velocity transformation, not the simple Galilean addition.

Here is the core problem: if you calculate the total momentum using the classical formula \(p = mv\) in one frame, and then transform those velocities to a second frame using the correct relativistic rules, the total momentum in the second frame will generally not equal the sum of the individual momenta calculated in that second frame. In other words, \(p_{\text{total}}\) is not invariant under Lorentz transformations if we use \(p = mv\). This violates the principle of relativity, which states that the laws of physics (including conservation laws) must hold in all inertial frames. Therefore, the definition of momentum itself must be modified to account for the relativistic nature of space and time.

Deriving the Relativistic Form

To fix this, we look for a new definition of momentum that reduces to \(mv\) at low speeds but behaves correctly at high speeds. The derivation (cited as D-098) shows that the only form that satisfies conservation in all frames is proportional to the velocity but scaled by the Lorentz factor \(\gamma\).

The Lorentz factor is defined as:

\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]

Thus, the relativistic momentum \(p\) is defined as:

\[ p = \gamma m v \]

Here, \(m\) is the rest mass of the particle, which is an invariant quantity (the same in all frames). The factor \(\gamma\) accounts for the relativistic effects. Notice that as \(v\) approaches \(c\), the term \(v^2/c^2\) approaches 1, making the denominator approach zero and \(\gamma\) approach infinity. This implies that momentum increases without bound as speed approaches the speed of light, preventing any massive object from reaching \(c\).

Low-Velocity Limit and Continuity

A crucial requirement for any new physical law is that it must reproduce the old, well-tested laws in the appropriate limit. In this case, when velocities are much smaller than the speed of light (\(v \ll c\)), the relativistic momentum must reduce to the Newtonian momentum.

Let us examine the behavior of \(\gamma\) when \(v \ll c\). We can use the binomial approximation for small \(x\), where \((1-x)^{-1/2} \approx 1 + \frac{1}{2}x\). Here, \(x = v^2/c^2\), which is very small.

\[ \gamma = \left(1 - \frac{v^2}{c^2}\right)^{-1/2} \approx 1 + \frac{1}{2}\frac{v^2}{c^2} \]

Substituting this into the momentum equation:

\[ p \approx \left(1 + \frac{1}{2}\frac{v^2}{c^2}\right) m v = mv + \frac{1}{2} m \frac{v^3}{c^2} \]

Since \(v \ll c\), the term \(\frac{v^3}{c^2}\) is negligible compared to \(v\). Therefore, \(p \approx mv\). This confirms that the relativistic definition is a smooth extension of the classical one, ensuring continuity with everyday experience.

Implications for Force and Acceleration

Newton's second law states that force is the time derivative of momentum: \(F = \frac{dp}{dt}\). In classical mechanics, since \(m\) is constant, this simplifies to \(F = ma\). However, in relativity, \(p = \gamma m v\), and \(\gamma\) depends on \(v\), which changes with time if the particle is accelerating.

Therefore, \(F = \frac{d}{dt}(\gamma m v)\). Because \(\gamma\) is not constant, we cannot simply pull it out of the derivative. This leads to a more complex relationship between force and acceleration. Specifically, the acceleration is no longer necessarily parallel to the force, and the "effective inertia" of the particle depends on the direction of the force relative to the velocity. This is a direct consequence of the redefined momentum and highlights that relativistic dynamics is fundamentally different from Newtonian dynamics, even though it reduces to it at low speeds.

The derivation

◆ DERIVATION EMBED · D-098 · not re-derived here

Relativistic Momentum from Conservation

If total momentum is to be conserved in every inertial frame under the Lorentz transformation of velocities, then the momentum of a single particle of rest mass m moving with velocity v cannot be mv; it must instead be p = γmv, where γ = 1/√(1 − v2/c2). This reduces to the Newtonian p = mv when v « c.

p = γ m v = m v / √(1 − v2/c2)
Open full derivation →
Tier 3 ยท watch out

A common misconception is that the mass of an object increases with velocity. This "relativistic mass" concept is outdated and can lead to confusion. In modern physics, mass \(m\) is strictly the invariant rest mass. It is the momentum and energy that increase with velocity, not the mass itself. Always use \(m\) for rest mass and \(\gamma m\) for the relativistic inertia factor. C

Worked example

Problem. An electron has a rest mass of \(m = 9.11 \times 10^{-31}\ \mathrm{kg}\). It is moving at a speed of \(v = 0.9c\), where \(c = 3.00 \times 10^8\ \mathrm{m/s}\). Calculate its relativistic momentum.

  1. Calculate the Lorentz factor \(\gamma\). \[ \gamma = \frac{1}{\sqrt{1 - \frac{(0.9c)^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.81}} = \frac{1}{\sqrt{0.19}} \] \[ \gamma \approx \frac{1}{0.4359} \approx 2.294 \]
  2. Calculate the velocity in \(\mathrm{m/s}\). \[ v = 0.9 \times 3.00 \times 10^8\ \mathrm{m/s} = 2.70 \times 10^8\ \mathrm{m/s} \]
  3. Calculate the relativistic momentum \(p = \gamma m v\). \[ p = 2.294 \times (9.11 \times 10^{-31}\ \mathrm{kg}) \times (2.70 \times 10^8\ \mathrm{m/s}) \] \[ p = 2.294 \times 9.11 \times 2.70 \times 10^{-31+8}\ \mathrm{kg \cdot m/s} \] \[ p \approx 56.4 \times 10^{-23}\ \mathrm{kg \cdot m/s} \] \[ p \approx 5.64 \times 10^{-22}\ \mathrm{kg \cdot m/s} \]
\[ \boxed{p \approx 5.64 \times 10^{-22}\ \mathrm{kg \cdot m/s}} \]

Units check. The units are \(\mathrm{kg} \times \mathrm{m/s}\), which is the standard unit for momentum. The calculation is consistent with SI units.

CHECK YOURSELF

Three quick tests

  1. What happens to the relativistic momentum as \(v\) approaches \(c\)?
    AnswerIt approaches infinity because \(\gamma\) approaches infinity.
  2. Why is the rest mass \(m\) preferred over "relativistic mass" in modern physics?
    AnswerRest mass is an invariant quantity, independent of the reference frame, whereas relativistic mass depends on velocity and can cause confusion.
  3. Does the relativistic momentum formula reduce to \(mv\) when \(v \ll c\)?
    AnswerYes, because \(\gamma \approx 1\) for small \(v/c\).
PROBLEMS

Problem set for PU-105 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-098 · Threads: force, energy, light