Chapter 8 • Theory & Derivations

Relativistic Mechanics & Four-Vectors

Foundations of four-dimensional Minkowski spacetime, metric tensor η_μν, invariant spacetime intervals (timelike, spacelike, null), light cones, proper time, four-velocity, relativistic momentum, relativistic kinetic energy, universal mass-energy equivalence E=mc², four-momentum invariant E² = p²c² + m²c⁴, Minkowski four-force, and particle collision dynamics with Compton scattering.

§8.1 Minkowski Four-Dimensional Spacetime & Invariant Intervals

1. The Four-Vector Formalism

In 1908, Hermann Minkowski recognized that Special Relativity unifies three-dimensional space and one-dimensional time into a single four-dimensional continuum: spacetime. An event is specified by a contravariant position four-vector $x^\mu$ ($\mu = 0, 1, 2, 3$):

x^\mu = (x^0, x^1, x^2, x^3) = (c t, x, y, z)

The Minkowski spacetime metric tensor $\eta_{\mu\nu}$ in the mostly-minus convention is:

\eta_{\mu\nu} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}

Lowering indices via the metric defines the covariant four-vector $x_\mu = \eta_{\mu\nu} x^\nu = (c t, -x, -y, -z)$.

2. The Invariant Spacetime Interval

The scalar product of the infinitesimal displacement four-vector $dx^\mu$ with itself defines the invariant spacetime interval $ds^2$:

ds^2 = dx_\mu dx^\mu = \eta_{\mu\nu} dx^\mu dx^\nu = c^2 dt^2 - dx^2 - dy^2 - dz^2

Lorentz Invariance: $ds^2$ is an absolute scalar: its numerical value is identical for all inertial observers: $ds^2 = ds'^2$.

3. Causal Structure & Light Cones

The sign of $ds^2$ categorizes the causal connection between two spacetime events:

Timelike Interval ($ds^2 > 0$): $c^2 \Delta t^2 > \Delta \vec{r}^2$. A physical signal traveling at speed $v < c$ can connect the two events. A reference frame exists where the events occur at the exact same spatial location. Events lie within the future or past light cone.
Spacelike Interval ($ds^2 < 0$): $c^2 \Delta t^2 < \Delta \vec{r}^2$. No physical signal can travel between the events without exceeding $c$. A reference frame exists where the events occur simultaneously ($\Delta t' = 0$). Events lie outside the light cone.
Lightlike / Null Interval ($ds^2 = 0$): $c^2 \Delta t^2 = \Delta \vec{r}^2$. Events can be connected only by photons traveling at speed $c$. Events lie on the boundary surface of the light cone.

§8.2 Proper Time, Four-Velocity & Relativistic Momentum

1. Proper Time $\tau$

The proper time $d\tau$ is the time interval measured by a clock carried along with the moving particle ($d\vec{r}' = 0$ in the instantaneous rest frame):

ds^2 = c^2 d\tau^2 \implies d\tau = \frac{ds}{c} = \sqrt{dt^2 - \frac{d\vec{r}^2}{c^2}} = dt \sqrt{1 - \frac{u^2}{c^2}} = \frac{dt}{\gamma}

Since $ds$ and $c$ are Lorentz invariants, proper time $\tau$ is a Lorentz scalar.

2. Four-Velocity Vector $U^\mu$

Differentiating the position four-vector $x^\mu$ with respect to the invariant proper time $\tau$:

U^\mu = \frac{dx^\mu}{d\tau} = \gamma \frac{dx^\mu}{dt} = \gamma (c, \vec{u}) = (\gamma c, \gamma u_x, \gamma u_y, \gamma u_z)

The invariant magnitude of the four-velocity is universally constant:

U^\mu U_\mu = \gamma^2 (c^2 - u^2) = \frac{c^2 - u^2}{1 - u^2/c^2} = c^2 = \text{constant for all particles}

3. Relativistic Momentum & Four-Momentum $P^\mu$

Multiplying the four-velocity by the invariant rest mass $m$:

P^\mu = m U^\mu = (\gamma m c, \gamma m \vec{u})

The spatial part gives the relativistic three-momentum:

\vec{p} = \gamma m \vec{u} = \frac{m \vec{u}}{\sqrt{1 - u^2/c^2}}

Notice that as particle speed approaches light speed ($u \to c$), $\gamma \to \infty$, so $\vec{p} \to \infty$. An infinite amount of momentum (and energy) is required to accelerate any massive body to $c$.

§8.3 Relativistic Work-Energy Theorem & Mass-Energy Equivalence

1. Derivation of Relativistic Kinetic Energy

Newton's Second Law in relativistic mechanics defines force as the time rate of change of relativistic momentum: $\vec{F} = \frac{d\vec{p}}{dt} = \frac{d}{dt}(\gamma m \vec{u})$. The work done by the force in moving a particle from rest to velocity $\vec{u}$ is:

W = \int_0^{\vec{r}} \vec{F} \cdot d\vec{r} = \int_0^t \frac{d\vec{p}}{dt} \cdot \vec{u} dt = \int_0^{\vec{p}} \vec{u} \cdot d\vec{p}

Integrating by parts:

W = \vec{u} \cdot \vec{p} - \int_0^{\vec{u}} \vec{p} \cdot d\vec{u} = \gamma m u^2 - \int_0^u \frac{m u du}{\sqrt{1 - u^2/c^2}}

Evaluating the integral:

\int_0^u \frac{m u du}{\sqrt{1 - u^2/c^2}} = \left[ -m c^2 \sqrt{1 - u^2/c^2} \right]_0^u = m c^2 - \frac{m c^2}{\gamma}

Substituting back:

W = \gamma m u^2 + \frac{m c^2}{\gamma} - m c^2 = m c^2 \left( \frac{\gamma^2 u^2/c^2 + 1}{\gamma} \right) - m c^2 = \gamma m c^2 - m c^2

Since initial kinetic energy at rest was zero, the relativistic kinetic energy $K$ is:

K = (\gamma - 1) m c^2

Non-Relativistic Limit ($u \ll c$): Expanding $\gamma = (1 - u^2/c^2)^{-1/2} \approx 1 + \frac{1}{2}\frac{u^2}{c^2} + \frac{3}{8}\frac{u^4}{c^4} + \dots$:

K = \left( 1 + \frac{1}{2}\frac{u^2}{c^2} - 1 \right) m c^2 = \frac{1}{2}m u^2

reproducing classical Newtonian kinetic energy.

2. Mass-Energy Equivalence ($E = mc^2$)

Einstein defined the total relativistic energy $E$ as the sum of kinetic energy and rest-mass energy $E_0$:

E = K + m c^2 = (\gamma - 1)m c^2 + m c^2 = \gamma m c^2

When the particle is at rest ($u = 0, \gamma = 1$), it retains an intrinsic rest mass energy:

$$E_0 = m c^2$$

Mass and energy are fundamentally equivalent: mass is concentrated, latent energy ($1\text{ kg} \approx 9 \times 10^{16}\text{ J}$).

3. The Four-Momentum Vector & The Energy-Momentum Invariant

The temporal component of four-momentum is $P^0 = \gamma m c = E/c$. Thus:

P^\mu = \left( \frac{E}{c}, \vec{p} \right)

Evaluating the Lorentz-invariant norm $P^\mu P_\mu$:

P^\mu P_\mu = \left( \frac{E}{c} \right)^2 - \vec{p}^2 = m^2 U^\mu U_\mu = m^2 c^2

Multiplying by $c^2$ yields the fundamental Energy-Momentum Relation:

$$E^2 = p^2 c^2 + m^2 c^4$$

For massless particles ($m = 0$, such as photons):

E = p c \implies p = \frac{E}{c} = \frac{h \nu}{c} = \frac{h}{\lambda}

§8.4 Minkowski Four-Force & Relativistic Collisions

1. The Minkowski Four-Force $K^\mu$

The relativistic generalization of Newton's Second Law in four-vector form is:

K^\mu = \frac{dP^\mu}{d\tau} = \gamma \frac{dP^\mu}{dt} = \gamma \left( \frac{1}{c} \frac{dE}{dt}, \frac{d\vec{p}}{dt} \right)

Since $\vec{F} = \frac{d\vec{p}}{dt}$ and power delivered is $\frac{dE}{dt} = \vec{F} \cdot \vec{u}$:

K^\mu = \gamma \left( \frac{\vec{F} \cdot \vec{u}}{c}, \vec{F} \right)

2. Conservation of Four-Momentum in Particle Collisions

In any closed system of colliding particles, the total four-momentum is strictly conserved:

\sum_{i=1}^{N_{\text{in}}} P_i^\mu = \sum_{f=1}^{N_{\text{out}}} P_f^\mu

This single four-vector relation encapsulates both conservation of total relativistic energy ($\mu = 0$) and conservation of total three-momentum ($\mu = 1, 2, 3$).

3. Derivation of Compton Scattering

Consider a photon of initial wavelength $\lambda$ and four-momentum $P_\gamma = (E/c, \vec{p}_\gamma)$ colliding with an electron of rest mass $m_e$ initially at rest: $P_e = (m_e c, \vec{0})$. After scattering at angle $\theta$, the photon has four-momentum $P_\gamma' = (E'/c, \vec{p}_\gamma')$ and the electron has $P_e'$. Conservation of four-momentum: $P_\gamma + P_e = P_\gamma' + P_e' \implies P_e' = P_\gamma - P_\gamma' + P_e$. Squaring both sides using $P^\mu P_\mu$:

P_e'^2 = (P_\gamma - P_\gamma' + P_e)^2

Since $P_e^2 = P_e'^2 = m_e^2 c^2$ and $P_\gamma^2 = P_\gamma'^2 = 0$:

m_e^2 c^2 = -2 P_\gamma \cdot P_\gamma' + 2 P_\gamma \cdot P_e - 2 P_\gamma' \cdot P_e + m_e^2 c^2

P_\gamma \cdot P_\gamma' = (P_\gamma - P_\gamma') \cdot P_e

Evaluating scalar products:

P_\gamma \cdot P_\gamma' = \frac{E E'}{c^2} - \vec{p}_\gamma \cdot \vec{p}_\gamma' = \frac{E E'}{c^2}(1 - \cos \theta)

\frac{E E'}{c^2}(1 - \cos \theta) = m_e (E - E')

Dividing by $E E' m_e c$:

\frac{1}{E'} - \frac{1}{E} = \frac{1}{m_e c^2}(1 - \cos \theta)

Using $E = hc/\lambda$ and $E' = hc/\lambda'$ yields the Compton Scattering Formula:

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

where $\lambda_C = \frac{h}{m_e c} \approx 2.426 \times 10^{-12}\text{ m} = 0.002426\text{ nm}$ is the Compton wavelength of the electron.

EXAMINATION PROBLEM SET

Standard University Exam Solved Problems

Solved Problem Example 8.1: Threshold Kinetic Energy for Antiproton Creation via p + p Collision

Calculate the minimum threshold kinetic energy $K_{\text{th}}$ in the laboratory frame required for an incident proton of rest mass $m_p$ colliding with a target proton at rest to produce a proton-antiproton pair: $p + p \to p + p + p + \bar{p}$.

Step 1: Set Up Initial Four-Momentum in Laboratory Frame

$$P_{\text{tot}}^{\text{lab}} = P_1 + P_2 = \left( \frac{E_1 + m_p c^2}{c}, \vec{p}_1 \right)$$

Target proton 2 is at rest, so $P_2 = (m_p c, \vec{0})$. The incident proton 1 has total energy $E_1 = K_{\text{th}} + m_p c^2$.

Step 2: Evaluate Invariant Mass Squared s = P_tot^μ P_tot,μ

$$s = (P_1 + P_2)^2 = P_1^2 + P_2^2 + 2 P_1 \cdot P_2 = m_p^2 c^2 + m_p^2 c^2 + 2 \left( \frac{E_1}{c}(m_p c) - 0 \right) = 2 m_p^2 c^2 + 2 m_p E_1$$

Because $s$ is a Lorentz scalar, its value in the laboratory frame must equal its value in the center-of-mass (CM) frame.

Step 3: Evaluate s at Threshold in Center-of-Mass Frame

$$s = \left(\sum_{i=1}^4 m_i c\right)^2 = (4 m_p c)^2 = 16 m_p^2 c^2$$

At threshold, all four final particles ($3p + 1\bar{p}$) are produced at rest in the CM frame with zero residual kinetic energy.

Step 4: Equate Invariant Masses and Solve for Threshold Energy

$$2 m_p^2 c^2 + 2 m_p E_1 = 16 m_p^2 c^2 \implies 2 m_p E_1 = 14 m_p^2 c^2 \implies E_1 = 7 m_p c^2$$

The threshold kinetic energy is $K_{\text{th}} = E_1 - m_p c^2 = 7 m_p c^2 - m_p c^2 = 6 m_p c^2$. Since $m_p c^2 \approx 938.3\text{ MeV}$, $K_{\text{th}} = 6 \times 938.3 = 5.63\text{ GeV}$.

Final Answer & Physical Insight

Threshold incident kinetic energy: K_th = 6 m_p c² ≈ 5.63 GeV (6 times the rest mass energy of a proton).

Solved Problem Example 8.2: Invariant Mass and Velocity of a Decaying Neutral Pion

A neutral pion $\pi^0$ of rest mass $m_\pi = 135.0\text{ MeV}/c^2$ decays in flight into two high-energy photons: $\pi^0 \to \gamma_1 + \gamma_2$. In the laboratory frame, the two photons have energies $E_1 = 100.0\text{ MeV}$ and $E_2 = 60.0\text{ MeV}$. (a) Find the opening angle $\theta$ between the two photons. (b) Find the velocity $v/c$ of the parent pion.

Step 1: Apply Four-Momentum Conservation P_π = P_γ1 + P_γ2

$$P_\pi^2 = (P_{\gamma 1} + P_{\gamma 2})^2 = P_{\gamma 1}^2 + P_{\gamma 2}^2 + 2 P_{\gamma 1} \cdot P_{\gamma 2}$$

Since $P_\pi^2 = m_\pi^2 c^2$ and photons are massless ($P_\gamma^2 = 0$), $m_\pi^2 c^2 = 2 P_{\gamma 1} \cdot P_{\gamma 2}$.

Step 2: Expand Photon Four-Vector Dot Product

$$2 P_{\gamma 1} \cdot P_{\gamma 2} = 2 \left( \frac{E_1 E_2}{c^2} - \vec{p}_1 \cdot \vec{p}_2 \right) = \frac{2 E_1 E_2}{c^2}(1 - \cos \theta) = m_\pi^2 c^2$$

Solving for $\cos \theta$: $1 - \cos \theta = \frac{m_\pi^2 c^4}{2 E_1 E_2} = \frac{(135.0)^2}{2(100.0)(60.0)} = \frac{18225}{12000} = 1.51875$. Notice that for physical decay, $\cos\theta = 1 - 1.51875 = -0.51875$, so $\theta = \arccos(-0.51875) \approx 121.25^\circ$.

Step 3: Determine Pion Total Energy and Velocity

$$E_\pi = E_1 + E_2 = 160.0\text{ MeV}, \quad \gamma = \frac{E_\pi}{m_\pi c^2} = \frac{160.0}{135.0} \approx 1.1852$$

The velocity is $\beta = \frac{v}{c} = \sqrt{1 - 1/\gamma^2} = \sqrt{1 - (135/160)^2} = \sqrt{1 - 0.7119} = \sqrt{0.2881} \approx 0.5367$.

Final Answer & Physical Insight

(a) Opening angle between photons: θ ≈ 121.3°; (b) Parent pion velocity: v/c ≈ 0.537 (53.7% light speed).

Solved Problem Example 8.3: Relativistic Rocket Motion: The Relativistic Tsiolkovsky Equation

A relativistic rocket accelerates from rest in interstellar space by expelling propellant exhaust gas at constant speed $u_{\text{ex}}$ relative to the rocket nozzle. Derive the relativistic generalization of Tsiolkovsky's rocket equation relating final velocity $v$ to the mass ratio $M_0 / M_f$.

Step 1: Balance Momentum in the Instantaneous Rest Frame of the Rocket

$$M dv' + u_{\text{ex}} dM = 0 \implies dv' = -u_{\text{ex}} \frac{dM}{M}$$

In the rocket's instantaneous comoving frame, the change in rocket velocity is $dv'$ and exhaust mass expelled is $-dM$.

Step 2: Relate Comoving dv' to Lab Frame dv via Velocity Addition

$$v + dv = \frac{v + dv'}{1 + v dv'/c^2} \approx (v + dv')(1 - v dv'/c^2) \implies dv = dv' (1 - v^2/c^2) = \frac{dv'}{\gamma^2}$$

Inverting gives $dv' = \frac{dv}{1 - v^2/c^2}$.

Step 3: Integrate Differential Equation

$$\int_0^v \frac{dv}{1 - v^2/c^2} = -u_{\text{ex}} \int_{M_0}^{M_f} \frac{dM}{M} \implies \frac{c}{2} \ln\left( \frac{1 + v/c}{1 - v/c} \right) = u_{\text{ex}} \ln\left( \frac{M_0}{M_f} \right)$$

Exponentiating: $\frac{1 + v/c}{1 - v/c} = \left( \frac{M_0}{M_f} \right)^{2 u_{\text{ex}} / c}$. Solving for $v/c$ delivers the relativistic rocket velocity.

Final Answer & Physical Insight

Relativistic rocket velocity: v/c = [(M0/Mf)^(2 u_ex/c) - 1] / [(M0/Mf)^(2 u_ex/c) + 1]. In the photon rocket limit (u_ex = c), v/c = [(M0/Mf)² - 1] / [(M0/Mf)² + 1].