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Chapter 1 • Theory & Derivations

Special Theory of Relativity & Spacetime Dynamics

Exhaustive treatment of special relativity: inertial systems and the Michelson-Morley experiment; Einstein's postulates and exact derivation of the Lorentz transformations; relativistic kinematics including simultaneity breakdown, length contraction, time dilation, and velocity addition; Minkowski spacetime geometry, worldlines, and invariant intervals; four-vector dynamics, relativistic energy-momentum invariant E^2 = p^2 c^2 + m_0^2 c^4, and covariance of Maxwell's electromagnetic field equations.

1.1Inertial Frames, Galilean Relativity & The Michelson-Morley Experiment

1. Inertial Reference Frames & Galilean Transformations

An inertial reference frame is one in which Newton's First Law holds: a body free from external forces moves with constant rectilinear velocity. In classical Newtonian mechanics, time is absolute and identical for all observers ($t' = t$). If frame $S'$ moves with uniform velocity $v$ along the common $x$-axis relative to frame $S$, the coordinate transformation is given by the Galilean Transformation:

$$x' = x - v t, \quad y' = y, \quad z' = z, \quad t' = t$$

Differentiating with respect to time yields the classical velocity addition rule:

$$u_x' = u_x - v, \quad u_y' = u_y, \quad u_z' = u_z$$

Differentiating once more confirms the invariance of acceleration: $\vec{a}' = \vec{a}$. Because mass $m$ is constant, Newton's Second Law $\vec{F} = m\vec{a}$ is form-invariant (Galilean invariant) under these transformations.

2. The Crisis of Classical Electrodynamics & The Luminiferous Ether

While Newtonian mechanics is Galilean invariant, Maxwell's equations predict that electromagnetic waves propagate through vacuum at a universal speed:

$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 2.99792 \times 10^8\,\text{m/s}$$

If Galilean relativity applied to light, an observer moving with velocity $v$ toward a light wave would measure its speed as $c + v$, violating Maxwell's wave equation. To reconcile this, 19th-century physicists postulated the existence of a pervasive, stationary, invisible medium termed the Luminiferous Ether, with respect to which light propagated at speed $c$. The Earth's orbital velocity around the Sun ($v \approx 30\,\text{km/s}$) should produce a measurable "ether wind."

3. The Michelson-Morley Experiment & Fringe Shift Derivation

In 1887, Albert Michelson and Edward Morley designed an ultra-sensitive optical interferometer to detect the Earth's velocity $v$ relative to the hypothetical ether.

A beam of monochromatic light of wavelength $\lambda$ is split by a half-silvered mirror into two mutually perpendicular arms of equal length $D$:

  • Longitudinal Arm 1 (Parallel to Ether Wind): Light travels downstream with speed $c - v$ and upstream with speed $c + v$. The round-trip time is:
    $$t_1 = \frac{D}{c - v} + \frac{D}{c + v} = \frac{2 D c}{c^2 - v^2} = \frac{2 D}{c} \left( 1 - \frac{v^2}{c^2} \\right)^{-1} \approx \frac{2 D}{c} \left( 1 + \frac{v^2}{c^2} \\right)$$
  • Transverse Arm 2 (Perpendicular to Ether Wind): To cross perpendicularly while being swept by the ether wind, the light must follow a diagonal path with effective speed $\sqrt{c^2 - v^2}$. The round-trip time is:
    $$t_2 = \frac{2 D}{\sqrt{c^2 - v^2}} = \frac{2 D}{c} \left( 1 - \frac{v^2}{c^2} \\right)^{-1/2} \approx \frac{2 D}{c} \left( 1 + \frac{1}{2} \frac{v^2}{c^2} \\right)$$

The time difference between the two arms is:

$$\Delta t = t_1 - t_2 \approx \frac{2 D}{c} \left[ \left( 1 + \frac{v^2}{c^2} \\right) - \left( 1 + \frac{1}{2} \frac{v^2}{c^2} \\right) \\right] = \frac{D v^2}{c^3}$$

When the entire apparatus is rotated by $90^{\circ}$, the roles of the two arms are interchanged, doubling the optical path difference. The predicted shift in interference fringes is:

$$\Delta N = \frac{c (2 \Delta t)}{\lambda} = \frac{2 D v^2}{\lambda c^2}$$

For $D = 11\,\text{m}$, $\lambda = 590\,\text{nm}$, and orbital velocity $v = 30\,\text{km/s}$ ($v/c = 10^{-4}$), the predicted fringe shift was $\Delta N \approx 0.37$ fringes. The interferometer had a sensitivity capable of detecting $0.005$ fringes. Yet, the measured shift was consistently zero ($\Delta N < 0.005$) at all times of the year and all orientations!

Physical Consequence: The luminiferous ether does not exist. The speed of light is completely independent of the Earth's motion and the orientation of the observer.

1.2Postulates of Special Relativity & Derivation of Lorentz Transformations

1. Einstein's Two Postulates of Special Relativity (1905)

Albert Einstein resolved the crisis by proposing two fundamental postulates:

  1. The Principle of Relativity: The laws of physics are identical and have the same mathematical form in all inertial reference frames. There is no preferred or absolute inertial frame.
  2. The Constancy of the Speed of Light: The speed of light in vacuum is an absolute universal constant $c$, having the same value in all inertial reference frames, regardless of the motion of the emitting source or the observing detector.

2. Mathematical Derivation of the Lorentz Transformations

Consider two inertial frames $S$ and $S'$ with axes aligned. Frame $S'$ moves with uniform velocity $v$ along the positive $x$-axis. At $t = t' = 0$, their origins coincide ($O = O'$), and a spherical flash of light is emitted from the origin.

By Postulate 2, the wavefront is spherical in both frames:

$$x^2 + y^2 + z^2 - c^2 t^2 = 0 \quad (\text{in Frame } S)$$
$$x'^2 + y'^2 + z'^2 - c^2 t'^2 = 0 \quad (\text{in Frame } S')$$
Transverse coordinates are unaffected by longitudinal motion: $y' = y$ and $z' = z$. By spacetime homogeneity and isotropy, the transformation between $(x, t)$ and $(x', t')$ must be linear:

$$x' = \gamma (x - v t)$$

By the Principle of Relativity (Postulate 1), the inverse transformation from $S'$ to $S$ must have the exact same functional form, simply reversing the sign of velocity ($v \to -v$):

$$x = \gamma (x' + v t')$$

Substitute $x'$ from the first equation into the second:

$$x = \gamma [ \gamma(x - vt) + v t' ] = \gamma^2 x - \gamma^2 v t + \gamma v t'$$

Solving for $t'$:

$$\gamma v t' = x (1 - \gamma^2) + \gamma^2 v t \implies t' = \gamma t + \frac{1 - \gamma^2}{\gamma v} x$$

Now substitute $x'$ and $t'$ into the spherical wavefront invariance relation $x'^2 - c^2 t'^2 = x^2 - c^2 t^2$:

$$\gamma^2 (x - vt)^2 - c^2 \left[ \gamma t + \frac{1 - \gamma^2}{\gamma v} x \\right]^2 = x^2 - c^2 t^2$$

Equating coefficients of $x^2$ and $t^2$ yields the exact value for the Lorentz factor $\gamma$:

$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1 - \beta^2}} \quad \left( \beta = \frac{v}{c} \\right)$$

Substituting $\gamma$ back into the time equation simplifies the coefficient to $-\frac{v}{c^2}$:

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

3. The Canonical Lorentz Transformation Equations

Transformation ($S \to S'$) Inverse Transformation ($S' \to S$)
$x' = \gamma (x - v t)$ $x = \gamma (x' + v t')$
$y' = y$ $y = y'$
$z' = z$ $z = z'$
$t' = \gamma \left( t - \frac{v x}{c^2} \\right)$ $t = \gamma \left( t' + \frac{v x'}{c^2} \\right)$

When $v \ll c$, $\beta \to 0$, $\gamma \to 1$, and the term $vx/c^2 \to 0$. The Lorentz transformations reduce precisely to the Galilean transformations, proving Newtonian mechanics is the low-velocity asymptotic limit of special relativity.

1.3Relativistic Kinematics: Simultaneity, Length Contraction & Time Dilation

1. Relativity of Simultaneity

Consider two events $A$ and $B$ that occur simultaneously at different spatial locations in frame $S$ ($\Delta t = t_B - t_A = 0$, $\Delta x = x_B - x_A \neq 0$). Applying the Lorentz transformation for time:

$$\Delta t' = t_B' - t_A' = \gamma \left( \Delta t - \frac{v \Delta x}{c^2} \\right) = - \frac{\gamma v \Delta x}{c^2} \neq 0$$

Fundamental Physical Principle: Events that are simultaneous in one inertial frame are not simultaneous in another frame moving relative to it! Simultaneity is not an absolute property of the physical universe, but strictly observer-dependent.

2. Relativistic Length Contraction (Lorentz-FitzGerald Contraction)

Let a rigid rod lie at rest along the $x'$-axis of frame $S'$. Its proper length $L_0$ (measured in its rest frame) is $L_0 = x_2' - x_1'$.

An observer in frame $S$ measures the length of the moving rod by simultaneously recording the positions of both endpoints at the same instant in their own frame ($t_1 = t_2$, so $\Delta t = 0$). From the Lorentz transformation $x_2' - x_1' = \gamma [ (x_2 - x_1) - v(t_2 - t_1) ]$:

$$L_0 = \gamma L \implies L = \frac{L_0}{\gamma} = L_0 \sqrt{1 - \frac{v^2}{c^2}}$$

Because $\gamma > 1$ for any non-zero velocity, $L < L_0$. The length of an object measured by an observer moving relative to it is contracted in the direction of motion, while perpendicular dimensions ($y$ and $z$) remain unchanged.

3. Relativistic Time Dilation & The Twin Paradox

Let a clock remain at rest at a fixed position $x'$ in frame $S'$. The time interval between two ticks recorded by this clock at the same spatial point ($\Delta x' = 0$) is the proper time interval $\Delta t_0$.

To an observer in frame $S$ watching this clock move past with velocity $v$, the elapsed time interval is obtained from the inverse Lorentz transformation:

$$\Delta t = \gamma \left( \Delta t_0 + \frac{v \Delta x'}{c^2} \\right) = \gamma \Delta t_0 = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$

Because $\gamma > 1$, $\Delta t > \Delta t_0$. A moving clock ticks slower than an identical clock at rest. This effect has been confirmed to parts-per-billion precision via the extended lifetimes of high-velocity atmospheric cosmic-ray muons and atomic clocks aboard orbiting GPS satellites.

4. Relativistic Velocity Addition

Let a particle move with velocity $\vec{u}' = (u_x', u_y', u_z')$ relative to frame $S'$. To find its velocity $\vec{u}$ in frame $S$, differentiate the inverse Lorentz transformations:

$$dx = \gamma (dx' + v dt'), \quad dy = dy', \quad dt = \gamma \left( dt' + \frac{v dx'}{c^2} \\right)$$
$$u_x = \frac{dx}{dt} = \frac{\gamma (dx' + v dt')}{\gamma (dt' + \frac{v dx'}{c^2})} = \frac{\frac{dx'}{dt'} + v}{1 + \frac{v}{c^2}\frac{dx'}{dt'}} = \frac{u_x' + v}{1 + \frac{u_x' v}{c^2}}$$
$$u_y = \frac{u_y'}{\gamma \left( 1 + \frac{u_x' v}{c^2} \\right)}, \quad u_z = \frac{u_z'}{\gamma \left( 1 + \frac{u_x' v}{c^2} \\right)}$$

Notice that if $u_x' = c$ (a beam of light emitted in $S'$):

$$u_x = \frac{c + v}{1 + \frac{c v}{c^2}} = \frac{c + v}{\frac{c + v}{c}} = c$$

The speed of light $c$ is an impassable upper velocity limit that cannot be exceeded by adding velocities.

1.4Minkowski Spacetime Geometry: Worldlines & Invariant Intervals

1. Four-Dimensional Spacetime Continuum

Hermann Minkowski (1908) recognized that space and time are interconnected facets of a single four-dimensional geometry: "Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality."

An event in spacetime is represented by coordinates $(ct, x, y, z)$. The path traced by an object through spacetime as time elapses is its worldline.

2. The Invariant Spacetime Interval

Just as Euclidean distance $\Delta r^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$ is invariant under spatial rotations, the spacetime interval $\Delta s^2$ between two events is invariant under all Lorentz boosts and rotations:

$$\Delta s^2 = c^2 \Delta t^2 - (\Delta x^2 + \Delta y^2 + \Delta z^2) = c^2 \Delta t'^2 - (\Delta x'^2 + \Delta y'^2 + \Delta z'^2)$$

Using the Minkowski metric tensor $\eta_{\mu\nu} = \text{diag}(+1, -1, -1, -1)$:

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

3. Classification of Spacetime Separations & The Light Cone

The sign of $\Delta s^2$ partitions all physical relationships into three geometrically distinct regimes:

  • Timelike Separation ($\Delta s^2 > 0$): $c^2 \Delta t^2 > \Delta r^2$. A physical particle traveling slower than light ($v < c$) can travel between the two events. A frame can always be found where the two events occur at the same spatial point ($\Delta r' = 0$). Causal cause-and-effect relationships can exist between them.
  • Lightlike / Null Separation ($\Delta s^2 = 0$): $c^2 \Delta t^2 = \Delta r^2$. Only massless particles traveling at the speed of light ($v = c$), such as photons, can connect the events. They lie directly on the boundary of the Light Cone.
  • Spacelike Separation ($\Delta s^2 < 0$): $\Delta r^2 > c^2 \Delta t^2$. No signal traveling at or below speed $c$ can connect the two events. A frame can always be found where the two events are simultaneous ($\Delta t' = 0$). No causal connection can exist without violating relativity.

The Light Cone centered on an event divides spacetime into:

  1. Absolute Future: The interior of the upper forward cone ($t > 0, \Delta s^2 > 0$), containing all events that can be influenced by the present event.
  2. Absolute Past: The interior of the lower backward cone ($t < 0, \Delta s^2 > 0$), containing all historical events that could have causally affected the present event.
  3. Elsewhere: The exterior region ($\Delta s^2 < 0$), causally disconnected from the present event.

1.5Relativistic Dynamics & Four-Vectors: E^2 = p^2 c^2 + m_0^2 c^4

1. Four-Vectors and Proper Time

A four-vector $A^\mu = (A^0, A^1, A^2, A^3) = (A^0, \vec{A})$ transforms under Lorentz transformations in exactly the same way as coordinate differentials $dx^\mu = (c dt, d\vec{x})$.

The proper time $d\tau$ is the invariant differential time measured in the particle's instantaneous rest frame:

$$c^2 d\tau^2 = ds^2 = c^2 dt^2 - d\vec{x}^2 = c^2 dt^2 \left( 1 - \frac{u^2}{c^2} \\right) \implies d\tau = \frac{dt}{\gamma}$$

The four-velocity $U^\mu$ is defined as the derivative of four-position with respect to proper time:

$$U^\mu = \frac{dX^\mu}{d\tau} = \gamma \frac{dX^\mu}{dt} = \gamma (c, \vec{u})$$

The invariant scalar product of four-velocity with itself is universally constant:

$$U_\mu U^\mu = \gamma^2 (c^2 - u^2) = c^2 \frac{c^2 - u^2}{c^2 - u^2} = c^2$$

2. Four-Momentum & Relativistic Energy

The four-momentum $P^\mu$ of a particle with rest mass $m_0$ is defined as:

$$P^\mu = m_0 U^\mu = (\gamma m_0 c, \gamma m_0 \vec{u}) = \left( \frac{E}{c}, \vec{p} \\right)$$

where the relativistic three-momentum is $\vec{p} = \gamma m_0 \vec{u}$ and total relativistic energy is:

$$E = \gamma m_0 c^2 = \frac{m_0 c^2}{\sqrt{1 - u^2/c^2}}$$

At zero velocity ($u = 0, \gamma = 1$), the particle possesses an inherent rest energy:

$$E_0 = m_0 c^2$$

The relativistic kinetic energy $K$ is the difference between total energy and rest energy:

$$K = E - m_0 c^2 = (\gamma - 1) m_0 c^2$$

Expanding $\gamma$ in a binomial Taylor series for $u \ll c$:

$$K = m_0 c^2 \left( 1 + \frac{1}{2}\frac{u^2}{c^2} + \frac{3}{8}\frac{u^4}{c^4} + \dots - 1 \\right) = \frac{1}{2} m_0 u^2 + \frac{3}{8} m_0 \frac{u^4}{c^2} + \dots$$

recovering classical Newtonian kinetic energy $\frac{1}{2} m_0 u^2$ at low speeds.

3. The Fundamental Energy-Momentum Invariant

Evaluating the invariant scalar product of four-momentum with itself:

$$P_\mu P^\mu = \left(\frac{E}{c}\\right)^2 - \vec{p}^2 = m_0^2 U_\mu U^\mu = m_0^2 c^2$$

Multiplying through by $c^2$ gives Einstein's celebrated energy-momentum invariant:

$$E^2 = p^2 c^2 + m_0^2 c^4$$

Massless Particles (Photons): For particles with zero rest mass ($m_0 = 0$):

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

Photons carry real physical momentum despite having zero rest mass, exerting measurable radiation pressure.

1.6Covariance of Maxwell's Field Equations & Field Transformations

1. Covariant Formulation of Electrodynamics

In relativistic four-vector notation, electric and magnetic fields are unifications of a single antisymmetric rank-2 electromagnetic field tensor $F^{\mu\nu}$:

$$F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu = \begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0 \end{pmatrix}$$

where $A^\mu = (\Phi/c, \vec{A})$ is the electromagnetic four-potential. In this compact tensor notation, all four Maxwell equations collapse into just two manifestly covariant equations:

$$\partial_\mu F^{\mu\nu} = \mu_0 J^\nu \quad (\text{Inhomogeneous: Gauss's \& Ampere-Maxwell Laws})$$
$$\partial_\mu \tilde{F}^{\mu\nu} = 0 \quad (\text{Homogeneous: Gauss's Magnetism \& Faraday Laws})$$

where $J^\nu = (c\rho, \vec{J})$ is the four-current density.

2. Lorentz Transformation of Electric and Magnetic Fields

Applying the Lorentz tensor transformation $F'^{\mu\nu} = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta F^{\alpha\beta}$ for a boost along the $x$-axis with velocity $v$ yields:

$$E_x' = E_x, \quad E_y' = \gamma (E_y - v B_z), \quad E_z' = \gamma (E_z + v B_y)$$
$$B_x' = B_x, \quad B_y' = \gamma \left( B_y + \frac{v}{c^2} E_z \\right), \quad B_z' = \gamma \left( B_z - \frac{v}{c^2} E_y \\right)$$

Physical Revelation: Electricity and magnetism are not independent physical entities. What an observer at rest perceives as a purely electrostatic field $\vec{E}$ will appear to a moving observer as a combination of both an electric field $\vec{E}'$ and a magnetic field $\vec{B}'$. Magnetism is fundamentally a relativistic consequence of electrostatics!

EXAM SUCCESS WORKSHOP

Solved University Examination Problems

Step-by-step mathematical solutions to classic university honors examination questions.

SOLVED PROBLEM 1.1

Relativistic Velocity Addition & Spaceship Pursuit Kinematics

An observer on a space station observes spaceship $A$ traveling along the positive $x$-axis at velocity $v_A = +0.80 c$. Spaceship $B$ is observed traveling along the same axis in pursuit at velocity $v_B = +0.90 c$. (a) Using the relativistic velocity addition formula, calculate the relative velocity $v_{BA}$ of spaceship $B$ as measured by an observer aboard spaceship $A$. (b) What would the classical Galilean velocity addition formula predict, and what is the percentage error of the classical calculation?

RIGOROUS DERIVATION & EXAM SOLUTION

Step 1: Relativistic Velocity Transformation

Let frame $S$ be the space station rest frame. Let frame $S'$ be the rest frame of spaceship A, which moves relative to $S$ at velocity $v = v_A = +0.80 c$.

In frame $S$, spaceship B moves with velocity $u_x = v_B = -0.60 c$. The velocity of B as measured in frame $S'$ ($u_x'$) is given by the Lorentz velocity addition formula:

$$u_x' = \frac{u_x - v}{1 - \frac{u_x v}{c^2}} = \frac{-0.60 c - 0.80 c}{1 - \frac{(-0.60 c)(0.80 c)}{c^2}} = \frac{-1.40 c}{1 - (-0.48)} = \frac{-1.40 c}{1.48} \approx -0.9459 c$$

The speed of spaceship B relative to spaceship A is $0.946 c$ (strictly less than $c$).

Step 2: Speed of Laser Signal

By Einstein's Second Postulate, light travels at $c$ in all inertial frames. Using the formula with $u_x = c$:

$$u_x' = \frac{c - v}{1 - v/c} = c$$

Both spaceship A, spaceship B, and the space station measure the laser pulse velocity as identically $c$.

Step 3: Comparison with Galilean Relativity

Under classical Galilean mechanics:

$$u_{\text{Galilean}} = u_x - v = -0.60 c - 0.80 c = -1.40 c$$

The Galilean prediction exceeds the speed of light by $40\%$, which is physically impossible in nature.

SOLVED PROBLEM 1.2

Atmospheric Muon Decay: Time Dilation vs Length Contraction

Muons created at an altitude of $H = 10.0\text{ km}$ in the upper atmosphere travel downward toward Earth at speed $v = 0.998 c$. The proper rest lifetime of a muon is $\tau_0 = 2.20\ \mu\text{s}$. (a) Calculate the Lorentz factor $\gamma$ and the classical distance a muon travels before decaying without relativistic time dilation. (b) Calculate the mean decay distance in the Earth's reference frame taking time dilation into account. (c) Explain from the perspective of an observer traveling in the muon's rest frame (length contraction) how muons reach the Earth's surface.

RIGOROUS DERIVATION & EXAM SOLUTION

Step 1: Calculate Lorentz Factor $\gamma$

$$\beta = \frac{v}{c} = 0.998 \implies \beta^2 = (0.998)^2 = 0.996004$$
$$\gamma = \frac{1}{\sqrt{1 - \beta^2}} = \frac{1}{\sqrt{1 - 0.996004}} = \frac{1}{\sqrt{0.003996}} = \frac{1}{0.06321} \approx 15.82$$

Step 2: Classical Newtonian Distance Calculation

$$d_{\text{classical}} = v \tau_0 = (0.998 \times 3.0 \times 10^8\,\text{m/s}) \times (2.20 \times 10^{-6}\,\text{s}) \approx 658.7\,\text{m} = 0.659\,\text{km}$$

Classically, muons could only travel $659\,\text{m}$, decaying long before reaching the ground ($10\,\text{km}$ below).

Step 3: Earth Observer's Perspective (Time Dilation)

To an observer on Earth, the moving muon clock runs slow by factor $\gamma$:

$$\Delta t = \gamma \tau_0 = 15.82 \times 2.20\,\mu\text{s} \approx 34.80\,\mu\text{s}$$

The distance traveled before decay is:

$$d = v \Delta t = (0.998 \times 3.0 \times 10^8\,\text{m/s}) \times (34.80 \times 10^{-6}\,\text{s}) \approx 10,419\,\text{m} = 10.42\,\text{km}$$

Because $10.42\,\text{km} > 10.0\,\text{km}$, the majority of muons easily reach sea-level detectors.

Step 4: Muon's Rest Frame Perspective (Length Contraction)

In the muon's rest frame, its lifetime is unchanged at $\tau_0 = 2.20\,\mu\text{s}$. However, the Earth and atmosphere rush toward the muon at $0.998 c$. The $10.0\,\text{km}$ atmospheric thickness is contracted to:

$$H' = \frac{H}{\gamma} = \frac{10.0\,\text{km}}{15.82} \approx 0.632\,\text{km} = 632\,\text{m}$$

In $2.20\,\mu\text{s}$, the rushing Earth covers $d' = v \tau_0 = 659\,\text{m} > 632\,\text{m}$. Both frames agree completely on the physical reality of ground arrival!

SOLVED PROBLEM 1.3

Relativistic Collision & Antiproton Production Threshold Energy

An antiproton ($\bar{p}$) can be created in a high-energy proton-proton collision when a moving proton of rest mass $m_p$ strikes a stationary target proton: $p + p \to p + p + p + \bar{p}$. Given that the rest energy of a proton and antiproton is $m_p c^2 = 938.3\text{ MeV}$: (a) Use four-momentum invariants to derive the threshold total energy $E_{\text{th}}$ and kinetic energy $K_{\text{th}}$ of the incident beam proton. (b) Compute the numerical value of $K_{\text{th}}$ in GeV.

RIGOROUS DERIVATION & EXAM SOLUTION

Step 1: Construct Four-Momentum Invariants

Let $P_1$ be the four-momentum of the incident beam proton: $P_1 = (E_1/c, \vec{p}_1)$. Let $P_2$ be the four-momentum of the target proton at rest: $P_2 = (m_p c, \vec{0})$.

The total four-momentum of the initial system is $P_{tot} = P_1 + P_2$. Its Lorentz-invariant square is:

$$P_{tot}^2 = (P_1 + P_2)^2 = P_1^2 + P_2^2 + 2 P_1 \cdot P_2 = m_p^2 c^2 + m_p^2 c^2 + 2 \left( \frac{E_1}{c}(m_p c) - \vec{p}_1 \cdot \vec{0} \right) = 2 m_p^2 c^2 + 2 m_p E_1$$

Step 2: Evaluate Final State at Threshold

At threshold, all four final particles (three protons and one antiproton, total rest mass $4 m_p$) are produced at rest relative to each other in the center-of-momentum frame, moving together as a single composite mass $M_f = 4 m_p$:

$$P_{final}^2 = (M_f c)^2 = (4 m_p c)^2 = 16 m_p^2 c^2$$

Step 3: Equate Invariants by Conservation of Four-Momentum

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

The minimum threshold kinetic energy $K_{th}$ required of the incident beam proton is:

$$K_{th} = E_1 - m_p c^2 = 7 m_p c^2 - m_p c^2 = 6 m_p c^2$$

Step 4: Compute Numerical Value

$$K_{th} = 6 \times 938.3\,\text{MeV} = 5629.8\,\text{MeV} \approx 5.63\,\text{GeV}$$

Notice that while creating an antiproton-proton pair requires only $2 m_p c^2 = 1.88\,\text{GeV}$ in rest mass energy, a fixed-target accelerator requires $6 m_p c^2 = 5.63\,\text{GeV}$ of beam energy because the remaining $3.75\,\text{GeV}$ is locked up in the kinetic energy of the forward-moving center of mass.