Chapter 7 • Theory & Derivations

Special Theory of Relativity: Kinematics & Transformations

Historical failure of Galilean relativity, the Michelson-Morley null experiment, Einstein's two postulates of Special Relativity, rigorous derivation of the Lorentz transformation equations, relativity of simultaneity, kinematic time dilation, Lorentz-Fitzgerald length contraction, relativistic velocity addition theorem, and relativistic longitudinal & transverse Doppler shifts.

§7.1 Breakdown of Galilean Relativity & The Michelson-Morley Experiment

1. The Galilean Transformation & The Classical Velocity Addition

In Newtonian mechanics, space and time are absolute entities independent of the observer. For two inertial reference frames $S$ and $S'$ with parallel axes where $S'$ moves along the positive $x$-axis with constant speed $v$ relative to $S$, the Galilean transformation is:

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

Differentiating with respect to the absolute time $t = t'$ yields the Galilean velocity addition law:

\vec{u}' = \vec{u} - \vec{v}

While Newton's laws of mechanics are invariant under Galilean transformations, Maxwell's equations of electrodynamics are not. Maxwell's equations predict that electromagnetic waves propagate in vacuum at a universal constant speed:

c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \approx 2.9979 \times 10^8 \text{ m/s}

Under Galilean transformations, an observer moving through the putative luminiferous ether at speed $v$ should measure a variable light speed $c' = c \pm v$.

2. The Michelson-Morley Interferometer Experiment (1887)

Albert A. Michelson and Edward W. Morley devised a high-precision optical interferometer floating on a pool of mercury to detect the Earth's orbital velocity through the ether ($v_{\text{Earth}} \approx 30 \text{ km/s}$, $v/c \approx 10^{-4}$). A monochromatic beam of light from source $S$ is split by a half-silvered mirror into two mutually perpendicular paths of length $L_1$ and $L_2$:

Longitudinal Arm (Parallel to Ether Wind): The light travels downstream at speed $c - v$ and upstream at $c + v$. The round-trip transit time is: $t_\parallel = \frac{L_1}{c - v} + \frac{L_1}{c + v} = \frac{2 L_1 c}{c^2 - v^2} = \frac{2 L_1}{c} \left( 1 - \frac{v^2}{c^2} \right)^{-1} \approx \frac{2 L_1}{c} \left( 1 + \frac{v^2}{c^2} \right)$
Transverse Arm (Perpendicular to Ether Wind): By the Pythagorean theorem, the effective light speed is $\sqrt{c^2 - v^2}$. The round-trip time is: $t_\perp = \frac{2 L_2}{\sqrt{c^2 - v^2}} = \frac{2 L_2}{c} \left( 1 - \frac{v^2}{c^2} \right)^{-1/2} \approx \frac{2 L_2}{c} \left( 1 + \frac{1}{2}\frac{v^2}{c^2} \right)$

Setting $L_1 = L_2 = L$, the round-trip time difference is:

\Delta t = t_\parallel - t_\perp \approx \frac{L}{c} \frac{v^2}{c^2}

Rotating the apparatus by $90^\circ$ exchanges the roles of the arms, doubling the time difference to $2\Delta t$. The expected fringe shift on the detector was:

\Delta N = \frac{c (2\Delta t)}{\lambda} = \frac{2 L v^2}{\lambda c^2} \approx 0.4 \text{ fringes}

The experiment was sensitive to $0.01$ fringes, yet no fringe shift was observed. The ether drift was conclusively zero.

§7.2 Einstein's Postulates & Derivation of Lorentz Transformations

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

Albert Einstein resolved the conflict between Newtonian mechanics and Maxwellian electrodynamics by abandoning absolute Newtonian space and time, founding the Special Theory of Relativity on two postulates:

The Principle of Relativity: The laws of physics take identical mathematical forms in all inertial reference frames. No physical experiment can distinguish between absolute rest and uniform rectilinear motion.
The Constancy of the Speed of Light: The speed of light in vacuum is an absolute universal constant $c$ in all inertial frames, independent of the motion of the emitting source or the observer.

2. Mathematical Derivation of the Lorentz Transformations

Let frame $S'$ move at speed $v$ along the $x$-axis of frame $S$. Due to homogeneity of space and time, the transformation must be linear:

x' = \gamma(x - v t), \quad y' = y, \quad z' = z

By the relativity postulate, the inverse transformation must be identical with $v \to -v$:

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

Consider a spherical light wave emitted from the coincident origins at $t = t' = 0$. The wavefront is described in both frames by:

x^2 + y^2 + z^2 - c^2 t^2 = 0, \quad x'^2 + y'^2 + z'^2 - c^2 t'^2 = 0

Since $y' = y$ and $z' = z$, along the $x$-axis $x = c t$ and $x' = c t'$. Substituting into the transformation equations:

c t' = \gamma(c - v) t, \quad c t = \gamma(c + v) t'

Multiplying the two equations:

c^2 t t' = \gamma^2 (c^2 - v^2) t t' \implies \gamma^2 = \frac{c^2}{c^2 - v^2} = \frac{1}{1 - v^2/c^2}

Taking the positive root (preserving direction of time):

\gamma = \frac{1}{\sqrt{1 - \beta^2}}, \quad \beta = \frac{v}{c}

Substituting $x' = \gamma(x - vt)$ into $x = \gamma(x' + vt')$ to solve for $t'$:

x = \gamma[\gamma(x - vt) + vt'] \implies \frac{x}{\gamma} - \gamma(x - vt) = \gamma vt'

t' = \gamma \left( t - \frac{v x}{c^2} \right)

Collecting the Lorentz Transformation Equations:

x' = \gamma (x - v t), \quad y' = y, \quad z' = z, \quad t' = \gamma \left( t - \frac{v x}{c^2} \right)

§7.3 Relativity of Simultaneity, Time Dilation & Length Contraction

1. The Relativity of Simultaneity

Consider two spatial events $A$ and $B$ that occur simultaneously at different spatial locations in frame $S$: $\Delta t = t_B - t_A = 0$ with $\Delta x = x_B - x_A \neq 0$. In the moving frame $S'$, the time difference is:

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

Crucial Consequence: Two spatially separated events that are simultaneous in one inertial frame are never simultaneous in any other inertial frame moving relative to it. Absolute simultaneity is physically meaningless.

2. Kinematic Time Dilation

Let a clock be located at a fixed position in frame $S'$ ($x_1' = x_2'$, so $\Delta x' = 0$). The time interval between two ticks measured by this clock is the proper time $\Delta t_0 = \Delta t' = \tau$. In frame $S$, using the inverse Lorentz transformation $t = \gamma(t' + v x'/c^2)$:

\Delta t = \gamma \Delta t' = \gamma \Delta t_0 = \frac{\Delta t_0}{\sqrt{1 - v^2/c^2}} > \Delta t_0

Moving Clocks Run Slow: A clock moving relative to an observer ticks slower by a factor of $\gamma$.

Experimental Proof: High-energy cosmic-ray muons created in the upper atmosphere ($\sim 15\text{ km}$) travel at $v \approx 0.999c$ ($\gamma \approx 22.4$). Their laboratory lifetime expands from $\tau_0 = 2.2 \;\mu\text{s}$ to $\Delta t = 49.3 \;\mu\text{s}$, allowing them to survive and reach ground-level detectors.

3. Lorentz-Fitzgerald Length Contraction

Let a rod lie at rest along the $x'$-axis in frame $S'$. Its proper length is $L_0 = x_2' - x_1'$. To measure the length $L = x_2 - x_1$ in frame $S$, the coordinates of both ends must be measured simultaneously at $t_1 = t_2$ ($\Delta t = 0$). Applying the Lorentz transformation $x' = \gamma(x - vt)$:

L_0 = x_2' - x_1' = \gamma(x_2 - x_1) - \gamma v(t_2 - t_1) = \gamma L

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

Moving Objects Contract: The physical length of an object is contracted along its direction of motion by factor $1/\gamma$. Dimensions perpendicular to the velocity ($y, z$) remain completely unaffected.

§7.4 Relativistic Velocity Addition & The Relativistic Doppler Effect

1. Relativistic Velocity Transformation Formula

Let a particle move with velocity $\vec{u} = (u_x, u_y, u_z) = (dx/dt, dy/dt, dz/dt)$ in frame $S$. In frame $S'$:

dx' = \gamma(dx - v dt), \quad dy' = dy, \quad dz' = dz, \quad dt' = \gamma\left(dt - \frac{v dx}{c^2}\right)

Dividing $dx'$ by $dt'$:

u_x' = \frac{dx'}{dt'} = \frac{\gamma(dx - v dt)}{\gamma\left(dt - \frac{v dx}{c^2}\right)} = \frac{\frac{dx}{dt} - v}{1 - \frac{v}{c^2}\frac{dx}{dt}} = \frac{u_x - v}{1 - \frac{u_x v}{c^2}}

For transverse components:

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

Universal Speed Limit: If a photon travels at $u_x = c$, then in any frame moving at speed $v$:

u_x' = \frac{c - v}{1 - \frac{c v}{c^2}} = \frac{c - v}{1 - v/c} = c

The speed of light $c$ is invariant under velocity addition. No compounding of sub-luminal velocities can ever exceed $c$.

2. Relativistic Doppler Effect

For an electromagnetic source emitting frequency $\nu_0$ moving at speed $v$ relative to an observer:

Longitudinal Doppler Effect (Source Moving Away): $\nu = \nu_0 \sqrt{\frac{1 - v/c}{1 + v/c}} = \nu_0 \frac{\sqrt{1 - \beta^2}}{1 + \beta}$ The observed wavelength is redshifted: $\lambda = \lambda_0 \sqrt{\frac{1 + \beta}{1 - \beta}}$.
Transverse Doppler Effect ($\theta = 90^\circ$): When the source moves perpendicular to the line of sight: $\nu_{\perp} = \frac{\nu_0}{\gamma} = \nu_0 \sqrt{1 - \frac{v^2}{c^2}}$ A purely relativistic phenomenon resulting directly from time dilation (predicted to be exactly zero in classical physics).

EXAMINATION PROBLEM SET

Standard University Exam Solved Problems

Solved Problem Example 7.1: Muon Atmospheric Decay: Time Dilation vs. Length Contraction

Muons are created at an altitude of $h = 10.0\text{ km}$ above sea level and travel vertically downward toward Earth at $v = 0.998 c$. The proper rest lifetime of a muon is $\tau_0 = 2.20 \;\mu\text{s}$. (a) Calculate the fraction of muons that survive to reach sea level in the Earth observer frame. (b) Re-evaluate the problem in the muon's rest frame using length contraction.

Step 1: Compute the Lorentz Factor γ

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

The relativistic time dilation and length contraction factor is $\gamma \approx 15.82$.

Step 2: Earth Frame Calculation (Time Dilation)

$$\Delta t_{\text{lab}} = \frac{h}{v} = \frac{10^4 \text{ m}}{0.998 \times 3 \times 10^8 \text{ m/s}} = 33.40 \;\mu\text{s}, \quad \tau_{\text{lab}} = \gamma \tau_0 = 15.82 \times 2.20 = 34.80 \;\mu\text{s}$$

The survival fraction according to radioactive exponential decay $N/N_0 = e^{-\Delta t / \tau}$ is $e^{-33.40 / 34.80} = e^{-0.9598} \approx 0.383$ (38.3% survive). Without relativity, $N/N_0 = e^{-33.40/2.20} = e^{-15.18} \approx 2.5 \times 10^{-7}$.

Step 3: Muon Frame Calculation (Length Contraction)

$$h' = \frac{h}{\gamma} = \frac{10.0 \text{ km}}{15.82} = 632.1 \text{ m}, \quad \Delta t' = \frac{h'}{v} = \frac{632.1}{0.998 c} = 2.112 \;\mu\text{s}$$

In the muon's frame, the clock ticks at normal proper rate $\tau_0 = 2.20 \;\mu\text{s}$, but the atmosphere is contracted to $632.1\text{ m}$. Survival fraction: $N/N_0 = e^{-\Delta t' / \tau_0} = e^{-2.112 / 2.20} = e^{-0.960} \approx 0.383$, yielding identical physical predictions.

Final Answer & Physical Insight

Survival fraction is 38.3% in both frames, demonstrating the physical equivalence of time dilation (Earth frame) and length contraction (muon frame).

Solved Problem Example 7.2: Relativistic Head-On Spacecraft Velocity Composition

Two starships $A$ and $B$ approach each other head-on as observed from a space station. Starship $A$ travels at $u_A = 0.80 c$ to the right, and starship $B$ travels at $u_B = 0.90 c$ to the left. (a) Calculate the speed of starship $B$ relative to starship $A$. (b) If starship $A$ fires a laser pulse toward $B$, calculate the speed of the laser pulse as measured by $B$.

Step 1: Set Up the Relativistic Velocity Transformation

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

Let frame $S$ be the space station. Frame $S'$ is attached to starship $A$, so $v = +0.80 c$. The velocity of starship $B$ in $S$ is $u_x = -0.90 c$.

Step 2: Calculate Relative Velocity of Starship B

$$u_B' = \frac{-0.90 c - 0.80 c}{1 - \frac{(-0.90 c)(0.80 c)}{c^2}} = \frac{-1.70 c}{1 - (-0.72)} = \frac{-1.70 c}{1.72} \approx -0.9884 c$$

The relative approach speed is $0.9884 c$, strictly less than $c$. Classical Galilean addition would have given an unphysical $1.70 c$.

Step 3: Laser Pulse Speed Measured by B

$$u_{\text{laser}}' = \frac{c - (-0.9884 c)}{1 - \frac{c(-0.9884 c)}{c^2}} = \frac{c(1 + 0.9884)}{1 + 0.9884} = c$$

By Einstein's second postulate, light travels at speed $c$ in all inertial reference frames regardless of relative velocities.

Final Answer & Physical Insight

(a) Relative speed of B as seen by A is 0.9884 c. (b) Laser pulse speed measured by B is exactly c.

Solved Problem Example 7.3: Cosmological Redshift & Relativistic Doppler Shift of a Quasar

A distant quasar emits the Lyman-alpha spectral line of atomic hydrogen at rest wavelength $\lambda_0 = 121.6\text{ nm}$. The spectral line is observed on Earth at a redshifted wavelength $\lambda_{\text{obs}} = 486.4\text{ nm}$. Determine the recession velocity $v/c$ of the quasar.

Step 1: Relate Observed and Rest Wavelengths to Relativistic Doppler Formula

$$\frac{\lambda_{\text{obs}}}{\lambda_0} = \sqrt{\frac{1 + \beta}{1 - \beta}} = 1 + z$$

The measured redshift parameter is $z = \frac{\lambda_{\text{obs}} - \lambda_0}{\lambda_0} = \frac{486.4 - 121.6}{121.6} = \frac{364.8}{121.6} = 3.0$.

Step 2: Square Both Sides and Solve for β = v/c

$$\frac{1 + \beta}{1 - \beta} = (1 + z)^2 = (1 + 3.0)^2 = 4^2 = 16$$

Cross-multiplying: $1 + \beta = 16(1 - \beta) = 16 - 16\beta$.

Step 3: Solve for Dimensionless Velocity β

$$17 \beta = 15 \implies \beta = \frac{v}{c} = \frac{15}{17} \approx 0.8824$$

The quasar is receding away from Earth at $88.24\%$ of the speed of light ($v \approx 2.65 \times 10^8\text{ m/s}$).

Final Answer & Physical Insight

Recession velocity: v/c = 15/17 ≈ 0.8824 (88.24% of light speed).