Special Theory of Relativity

In classical physics, two events that occur at the same time are simultaneous for all observers. In Special Relativity, simultaneity is relative. Two events that are simultaneous in one inertial frame ($\Delta t = 0$) are generally not simultaneous in another inertial frame moving relative to the first. From the Lorentz transformation: $$ \Delta t' = \gamma\left(\Delta t - \frac{v\Delta x}{c^2}\right) $$ If $\Delta t = 0$ but $\Delta x \neq 0$, then $\Delta t' \neq 0$. Thus, the order and simultaneity of spatially separated events depend on the observer’s frame of reference.

• World Point: An event in spacetime is completely specified by four coordinates $(ct, x, y, z)$. This point in four-dimensional Minkowski spacetime is called a world point.
• World Line: The continuous path traced by a particle (or observer) through spacetime is called its world line. For a particle at rest, the world line is parallel to the time axis. For a particle moving with constant velocity, it is a straight line inclined to the time axis. Accelerated particles have curved world lines.

In Minkowski spacetime, the light cone of an event $O$ is the set of all events that can be connected to $O$ by a light signal.

• Future Light Cone: The region of spacetime that can be reached from $O$ by a light signal or slower signal ($ds^2 \ge 0$, future-directed). Events inside the future light cone are causally connected to $O$ and can be influenced by it.
• Past Light Cone: The region of spacetime from which a light signal or slower signal can reach $O$. Events inside the past light cone can influence the event at $O$.

The light cone divides spacetime into: - Timelike region (inside the cone) – causal influence possible - Spacelike region (outside the cone) – no causal influence possible - Lightlike (null) surface – the cone itself

Four-velocity The four-velocity $U^\mu$ of a particle is the derivative of its world-line coordinates with respect to proper time $\tau$: $$ U^\mu = \frac{dx^\mu}{d\tau} = \gamma(c, \mathbf{v}) $$ where $\gamma = \dfrac{1}{\sqrt{1-v^2/c^2}}$. It is a timelike four-vector with constant magnitude: $$ U^\mu U_\mu = c^2 $$ Four-momentum $$ P^\mu = m_0 U^\mu = \gamma m_0 (c, \mathbf{v}) = \left(\frac{E}{c}, \mathbf{p}\right) $$ where $E = \gamma m_0 c^2$ is the total energy and $\mathbf{p} = \gamma m_0 \mathbf{v}$ is the relativistic three-momentum. Four-force $$ F^\mu = \frac{dP^\mu}{d\tau} $$ It is the rate of change of four-momentum with respect to proper time. Proof of the energy-momentum relation Since $P^\mu P_\mu$ is a Lorentz invariant: $$ P^\mu P_\mu = m_0^2 c^2 $$ In components: $$ \left(\frac{E}{c}\right)^2 - p^2 = m_0^2 c^2 $$ Multiplying both sides by $c^2$: $$ E^2 - p^2 c^2 = m_0^2 c^4 $$ $$ \therefore \quad E^2 = p^2 c^2 + m_0^2 c^4 $$

Given: $\Delta x = 9 \times 10^9$ m, $\Delta t = 0$ (simultaneous), $v = 0.8c$ From Lorentz transformation: $$ \Delta t' = \gamma \left( \Delta t - \frac{v \Delta x}{c^2} \right) = -\gamma \frac{v \Delta x}{c^2} $$ $$ \gamma = \frac{1}{\sqrt{1 - (0.8)^2}} = \frac{1}{\sqrt{0.36}} = \frac{1}{0.6} = \frac{5}{3} $$ $$ \frac{v}{c^2} = \frac{0.8c}{c^2} = \frac{0.8}{c} = \frac{0.8}{3 \times 10^8} = 2.6667 \times 10^{-9}\ \text{s/m} $$ $$ \frac{v \Delta x}{c^2} = 2.6667 \times 10^{-9} \times 9 \times 10^9 = 24\ \text{s} $$ $$ |\Delta t'| = \frac{5}{3} \times 24 = 40\ \text{s} $$ Answer: The time interval is $40$ seconds.

Given: $m_0 c^2 = 105$ MeV, $E = 315$ MeV $$ \gamma = \frac{E}{m_0 c^2} = \frac{315}{105} = 3 $$ $$ \gamma = \frac{1}{\sqrt{1 - \beta^2}} = 3 \implies 1 - \beta^2 = \frac{1}{9} \implies \beta^2 = \frac{8}{9} $$ $$ \beta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \approx 0.9428 $$ $$ v = 0.9428\, c $$ Answer: Velocity of the muon is $0.9428\,c$ (or $\dfrac{2\sqrt{2}}{3}c$).

Lorentz Transformation (boost along $x$-axis with velocity $v$): $$ \begin{align*} x' &= \gamma (x - vt) \\ y' &= y \\ z' &= z \\ t' &= \gamma \left(t - \frac{vx}{c^2}\right) \end{align*} $$ where $\gamma = \dfrac{1}{\sqrt{1-v^2/c^2}}$. In four-vector notation (Minkowski coordinates $x^0 = ct$, $x^1 = x$, $x^2 = y$, $x^3 = z$): $$ \begin{align*} x'^0 &= \gamma (x^0 - \beta x^1) \\ x'^1 &= \gamma (x^1 - \beta x^0) \\ x'^2 &= x^2 \\ x'^3 &= x^3 \end{align*} $$ with $\beta = v/c$. Time Dilation via Minkowski Diagram In a Minkowski diagram, the time axis ($ct$) is vertical and the space axis ($x$) is horizontal. - The world-line of a clock at rest in frame $S'$ is a straight line inclined at an angle $\theta$ where $\tan\theta = v/c$. - Proper time $\Delta\tau$ is the time measured by a clock moving with the particle (interval along its own world-line). - Coordinate time $\Delta t$ in frame $S$ is the projection of this interval onto the $ct$-axis of $S$. Because of the hyperbolic geometry of Minkowski spacetime (the metric is $ds^2 = c^2dt^2 - dx^2$), the length of the world-line segment (proper time) is shorter than its projection on the time axis of another frame. Thus: $$ \Delta t = \gamma \Delta\tau > \Delta\tau $$ Moving clocks run slow — this is time dilation.

The quantity $x_\mu x^\mu$ is the squared interval (Minkowski norm): $$ x_\mu x^\mu = (ct)^2 - x^2 - y^2 - z^2 = c^2 t^2 - \mathbf{r}^2 $$ Under a Lorentz transformation (boost along $x$): $$ \begin{align*} ct' &= \gamma (ct - \beta x) \\ x' &= \gamma (x - \beta ct) \\ y' &= y \\ z' &= z \end{align*} $$ Compute $x'_\mu x'^\mu$: $$ \begin{align*} (ct')^2 - (x')^2 - (y')^2 - (z')^2 &= \gamma^2 (ct - \beta x)^2 - \gamma^2 (x - \beta ct)^2 - y^2 - z^2 \\ &= \gamma^2 \Big[ (ct)^2 - 2\beta ct\, x + \beta^2 x^2 - x^2 + 2\beta ct\, x - \beta^2 (ct)^2 \Big] - y^2 - z^2 \\ &= \gamma^2 \Big[ (1 - \beta^2)(ct)^2 - (1 - \beta^2)x^2 \Big] - y^2 - z^2 \\ &= \gamma^2 (1 - \beta^2) \big[(ct)^2 - x^2\big] - y^2 - z^2 \end{align*} $$ But $\gamma^2 (1 - \beta^2) = 1$, therefore: $$ (ct')^2 - (x')^2 - (y')^2 - (z')^2 = (ct)^2 - x^2 - y^2 - z^2 $$ $$ \therefore \quad x'_\mu x'^\mu = x_\mu x^\mu $$ Hence $x_\mu x^\mu$ is a Lorentz invariant.

Given: Kinetic energy $K = m_0 c^2$ Total energy $E = K + m_0 c^2 = 2 m_0 c^2$ $$ \gamma = \frac{E}{m_0 c^2} = 2 $$ $$ \frac{1}{\sqrt{1 - \beta^2}} = 2 \implies 1 - \beta^2 = \frac{1}{4} \implies \beta^2 = \frac{3}{4} $$ $$ \beta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \approx 0.866 $$ $$ v = \frac{\sqrt{3}}{2}\, c $$ Answer: Velocity of the particle is $\dfrac{\sqrt{3}}{2}\, c$.