Special Relativity
Special relativity fundamentally restructures our understanding of space and time, replacing the Galilean concept of absolute time with a unified, four-dimensional spacetime continuum (Minkowski spacetime).
The Principia
Einstein’s formulation of Special Relativity rests on two invariant postulates:
- The Principle of Relativity (First Postulate): The laws of physics are invariant (identical) in all inertial frames of reference. There is no absolute or “preferred” inertial frame.
- The Invariance of the Speed of Light (Second Postulate): The speed of light in a vacuum, $c$, is constant and independent of the relative motion of the source and the observer in all inertial frames.
The Mathematical Framework: Minkowski Spacetime
To satisfy these postulates, space and time cannot be treated independently. We define a 4-dimensional spacetime manifold parameterized by the coordinates:
$$x^\mu = (x^0, x^1, x^2, x^3) = (ct, x, y, z)$$
The geometry of this spacetime is defined by the invariant spacetime interval, $ds^2$, which must be agreed upon by all inertial observers. Using the metric signature $(-, +, +, +)$, the interval is defined as:
$$ds^2 = \eta_{\mu\nu} dx^\mu dx^\nu = -c^2dt^2 + dx^2 + dy^2 + dz^2$$
Where $\eta_{\mu\nu}$ is the Minkowski metric tensor:
$$\eta_{\mu\nu} = \begin{pmatrix}
-1 & 0 & 0 & 0
0 & 1 & 0 & 0
0 & 0 & 1 & 0
0 & 0 & 0 & 1
\end{pmatrix}$$
Transformations between inertial frames that leave $ds^2$ invariant are called Lorentz Transformations. For a standard boost along the $x$-axis with relative velocity $v$, the transformation matrix $\Lambda^\mu_\nu$ is:
$$x’^\mu = \Lambda^\mu_\nu x^\nu$$
$$\Lambda^\mu_\nu = \begin{pmatrix}
\gamma & -\gamma \beta & 0 & 0
-\gamma \beta & \gamma & 0 & 0
0 & 0 & 1 & 0
0 & 0 & 0 & 1
\end{pmatrix}$$
Where $\beta = \frac{v}{c}$ and the Lorentz factor is $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$.
Kinematic Consequences
The invariance of $c$ directly leads to counterintuitive physical consequences for observers in relative motion.
-
Relativity of Simultaneity: Two events that are simultaneous in one reference frame are generally not simultaneous in a frame moving relative to the first. If $\Delta t = 0$ but $\Delta x \neq 0$ in frame $S$, then $\Delta t’ = -\gamma \frac{v}{c^2} \Delta x \neq 0$ in frame $S’$.
-
Time Dilation: Moving clocks tick slower. If proper time $d\tau$ is measured by a clock at rest in its own frame ($dx=dy=dz=0$), an observer moving at velocity $v$ relative to the clock measures the time interval $dt$:
$$dt = \gamma d\tau$$
- Length Contraction: Moving objects are measured to be shorter along the direction of motion. If $L_0$ is the proper length (measured in the object’s rest frame), the measured length $L$ in a moving frame is:
$$L = \frac{L_0}{\gamma}$$
Relativistic Energy-Momentum Formulation
To ensure that the laws of dynamics (conservation of energy and momentum) hold true in all inertial frames, we must upgrade classical 3-vectors to relativistic 4-vectors.
Four-Velocity Classical velocity $\mathbf{v} = \frac{d\mathbf{x}}{dt}$ is not a Lorentz covariant vector because $dt$ transforms between frames. We instead differentiate with respect to the invariant proper time $\tau$:
$$U^\mu = \frac{dx^\mu}{d\tau} = \gamma \frac{dx^\mu}{dt} = \gamma(c, \mathbf{v})$$
Note that the magnitude of the 4-velocity is always constant: $$U_\mu U^\mu = \eta_{\mu\nu} U^\mu U^\nu = -c^2$$
Four-Momentum Multiplying the 4-velocity by the invariant rest mass $m_0$ yields the four-momentum vector $P^\mu$:
$$P^\mu = m_0 U^\mu = (m_0 \gamma c, m_0 \gamma \mathbf{v})$$
We define the relativistic energy $E$ and relativistic 3-momentum $\mathbf{p}$ as:
- $E = \gamma m_0 c^2$
- $\mathbf{p} = \gamma m_0 \mathbf{v}$
Thus, the four-momentum is cleanly expressed as:
$$P^\mu = \left(\frac{E}{c}, \mathbf{p}\right)$$
Energy-Momentum Invariant (The Mass Shell Condition) Just as the square of the 4-velocity is an invariant, the square of the 4-momentum yields one of the most important invariants in physics. By taking the inner product of $P^\mu$ with itself:
$$P_\mu P^\mu = \eta_{\mu\nu} P^\mu P^\nu = -\left(\frac{E}{c}\right)^2 + |\mathbf{p}|^2$$
Since $P^\mu = m_0 U^\mu$, we also know that $P_\mu P^\mu = m_0^2 (U_\mu U^\mu) = -m_0^2 c^2$. Equating these gives the relativistic energy-momentum relation:
$$-\frac{E^2}{c^2} + p^2 = -m_0^2 c^2 \implies E^2 = (pc)^2 + (m_0 c^2)^2$$
For a particle at rest ($\mathbf{p} = 0$), this famously reduces to the rest-mass energy equivalence:
$$E = m_0 c^2$$
For a massless particle like a photon ($m_0 = 0$), the equation yields $E = pc$.