Relativistic Mechanics

With the 4-vector machinery in hand, we can now build a complete relativistic mechanics. The results are stunning: momentum and energy are unified into a single 4-vector, mass becomes a form of energy, and the simple Newtonian formula $p = mv$ is revealed as an approximation valid only at low speeds. This is not just an academic exercise, relativistic mechanics is the working framework of every particle physics experiment in the world.

Relativistic Momentum

In Newtonian mechanics, momentum is $\mathbf{p} = m\mathbf{v}$, and it is conserved in collisions. But this definition fails at relativistic speeds: if we use $p = mv$, momentum is not conserved in all frames. The fix is to use the spatial part of the 4-momentum:

As $v \to c$, $\gamma \to \infty$, and the momentum diverges. This means no finite force can accelerate a massive particle to the speed of light, it would require infinite momentum. The speed of light is a true speed limit for massive particles.

Relativistic Energy

The time component of the 4-momentum $p^\mu = (E/c, \mathbf{p})$ gives the relativistic energy:

The relativistic kinetic energy is the total energy minus the rest energy:

$$K = E - mc^2 = (\gamma - 1)mc^2$$

Let's verify this reduces to the Newtonian result at low speeds. For $v \ll c$:

$$\gamma \approx 1 + \frac{1}{2}\frac{v^2}{c^2} + \cdots$$ $$K \approx \frac{1}{2}\frac{v^2}{c^2} \cdot mc^2 = \frac{1}{2}mv^2$$

The familiar Newtonian kinetic energy emerges as the leading correction to the rest energy.

The Energy-Momentum Relation

The invariant magnitude of the 4-momentum gives the most important equation in relativistic mechanics:

$$p^\mu p_\mu = -\frac{E^2}{c^2} + |\mathbf{p}|^2 = -m^2c^2$$

This relation has two important special cases:

• At rest ($p = 0$): $E = mc^2$: the famous mass-energy equivalence
• Massless particles ($m = 0$): $E = pc$: photons, gravitons, and (approximately) neutrinos

The energy-momentum relation is a Lorentz invariant: the right-hand side $m^2c^4$ is the same in all frames, even though $E$ and $p$ individually change under boosts. This is precisely the invariant "length" of the 4-momentum vector.

Massless Particles: Photons

Photons are massless ($m = 0$), so their energy-momentum relation is simply $E = pc$. From quantum mechanics, $E = h\nu$ and $p = h/\lambda = h\nu/c$, which is consistent.

What is the 4-momentum of a photon? Since $m = 0$, we cannot write $p^\mu = m\gamma(c, \mathbf{v})$ (which would give $0 \cdot \infty$). Instead, the photon 4-momentum is:

$$p^\mu = \frac{E}{c}(1, \hat{\mathbf{n}})$$

where $\hat{\mathbf{n}}$ is the unit vector in the direction of propagation. The invariant magnitude is $p^\mu p_\mu = 0$: the photon 4-momentum is a null vector, consistent with $m = 0$.

Photons always travel at $c$ in every frame, but their energy and momentum (and hence frequency and wavelength) do change under Lorentz transformations. This is the relativistic Doppler effect.

Conservation of 4-Momentum

In any collision or decay, the total 4-momentum is conserved:

$$\sum_{\text{initial}} p^\mu_i = \sum_{\text{final}} p^\mu_f$$

This is a single 4-vector equation that encodes four conservation laws simultaneously: conservation of energy ($\mu = 0$) and conservation of 3-momentum ($\mu = 1, 2, 3$). Because it is a 4-vector equation, it holds in all inertial frames.

The Center-of-Momentum Frame

For any system of particles, we can always find a frame where the total 3-momentum is zero: $\mathbf{p}_{\text{tot}} = 0$. This is the center-of-momentum (CM) frame.

In the CM frame, the total energy equals the invariant mass times $c^2$:

$$E_{\text{CM}} = Mc^2 = \sqrt{-P^\mu P_\mu} \cdot c$$

The CM frame is the natural frame for analyzing collisions because the physics is simplest when the total momentum vanishes. The available energy for creating new particles is $E_{\text{CM}}$, not the total energy in the lab frame.

Relativistic Collisions

Example: Particle Creation Threshold

One of the most important applications in particle physics is finding the threshold energy for particle creation. Consider a proton beam hitting a stationary proton target. What is the minimum beam energy needed to produce a proton-antiproton pair?

$$p + p \to p + p + p + \bar{p}$$

At threshold, all four final-state particles are created at rest in the CM frame. The invariant mass of the final state is $M_f = 4m_p$. Since invariant mass is conserved:

$$M_i^2 c^4 = M_f^2 c^4 = (4m_p c^2)^2$$

In the lab frame (target at rest), the initial 4-momenta are:

$$p_1^\mu = (E/c, p, 0, 0), \quad p_2^\mu = (m_p c, 0, 0, 0)$$

Computing the invariant mass:

$$M_i^2 c^4 = (E + m_p c^2)^2 - (pc)^2 = E^2 + 2Em_p c^2 + m_p^2 c^4 - (E^2 - m_p^2 c^4)$$ $$= 2Em_p c^2 + 2m_p^2 c^4$$

Setting this equal to $(4m_p c^2)^2 = 16m_p^2 c^4$:

$$2Em_p c^2 + 2m_p^2 c^4 = 16m_p^2 c^4$$ $$E = 7m_p c^2 \approx 6.6 \text{ GeV}$$

Compton Scattering

As another classic example, consider a photon scattering off an electron at rest (Compton scattering). Using 4-momentum conservation:

$$p^\mu_\gamma + p^\mu_e = p'^\mu_\gamma + p'^\mu_e$$

Rearranging and squaring (using $p^\mu p_\mu$ invariants):

$$(p^\mu_\gamma + p^\mu_e - p'^\mu_\gamma)^2 = (p'^\mu_e)^2 = -m_e^2 c^2$$

This yields the Compton formula for the change in photon wavelength:

$$\lambda' - \lambda = \frac{h}{m_e c}(1 - \cos\theta)$$

where $\theta$ is the scattering angle. The quantity $h/(m_e c) = 2.43 \times 10^{-12}$ m is the Compton wavelength of the electron. This result, which has no classical explanation, was one of the key pieces of evidence for the particle nature of light.

Invariant Mass and Particle Discovery

The invariant mass is the primary tool for discovering new particles in accelerator experiments. When a particle decays, the invariant mass of its decay products equals the mass of the parent particle:

$$M^2 c^4 = \left(\sum E_i\right)^2 - \left|\sum \mathbf{p}_i c\right|^2$$

By measuring the energies and momenta of the decay products, experimentalists can reconstruct the invariant mass and look for peaks in the distribution. The Higgs boson was discovered in 2012 by finding a peak at $125$ GeV in the invariant mass spectrum of photon pairs and lepton pairs.

Key Insights

• Relativistic momentum $\mathbf{p} = \gamma m\mathbf{v}$ diverges as $v \to c$, making the speed of light an absolute speed limit
• Total relativistic energy $E = \gamma mc^2$ includes the rest energy $mc^2$
• The energy-momentum relation $E^2 = (pc)^2 + (mc^2)^2$ is a Lorentz invariant
• Massless particles have $E = pc$ and always travel at $c$
• 4-momentum conservation unifies energy conservation and momentum conservation into a single equation
• The center-of-momentum frame is the natural frame for analyzing collisions
• Threshold energies for particle creation grow as $M_f^2$ in fixed-target experiments but only as $M_f$ in colliders
• Invariant mass reconstruction is how new particles are discovered at colliders

Looking Ahead

We have arrived at the equation $E = mc^2$ several times in this lesson. In the next lesson, we will explore what this equation truly means, the equivalence of mass and energy, and its profound consequences: nuclear binding energy, pair creation and annihilation, and the deepest connection between the principle of least action and the structure of spacetime.