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Ch. 3 Solutions

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Table of Contents

Basic

Medium

Basic

B-1. Derivation of the Hyperbolic Function Identity

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Problem:

Derive the hyperbolic function identity \(\cosh^2\varphi - \sinh^2\varphi = 1\) directly from the definitions

\[ \cosh\varphi = \frac{e^\varphi + e^{-\varphi}}{2}, \qquad \sinh\varphi = \frac{e^\varphi - e^{-\varphi}}{2} \]

Strategy: Square the definitions of \(\cosh\varphi\) and \(\sinh\varphi\) respectively, then take the difference.

Calculation:

\[ \cosh^2\varphi = \left(\frac{e^\varphi + e^{-\varphi}}{2}\right)^2 = \frac{e^{2\varphi} + 2e^0 + e^{-2\varphi}}{4} = \frac{e^{2\varphi} + 2 + e^{-2\varphi}}{4} \]
\[ \sinh^2\varphi = \left(\frac{e^\varphi - e^{-\varphi}}{2}\right)^2 = \frac{e^{2\varphi} - 2e^0 + e^{-2\varphi}}{4} = \frac{e^{2\varphi} - 2 + e^{-2\varphi}}{4} \]

Taking the difference,

\[ \cosh^2\varphi - \sinh^2\varphi = \frac{(e^{2\varphi} + 2 + e^{-2\varphi}) - (e^{2\varphi} - 2 + e^{-2\varphi})}{4} = \frac{4}{4} = 1 \]

Final answer:

\[ \boxed{\cosh^2\varphi - \sinh^2\varphi = 1} \]

Verification: Substituting \(\varphi = 0\) gives \(\cosh 0 = 1\), \(\sinh 0 = 0\), so \(1 - 0 = 1\). ✓

B-2. Relationship between Rapidity and the Lorentz Factor

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Problem:

Using the relationship between rapidity \(\varphi\) and velocity \(v\), given by \(\tanh\varphi = v/c\), derive \(\cosh\varphi = \gamma\) and \(\sinh\varphi = \gamma v/c\). Here, \(\gamma = 1/\sqrt{1 - v^2/c^2}\).

Solution strategy: Combine \(\tanh\varphi = v/c\) with the identity \(\cosh^2\varphi - \sinh^2\varphi = 1\).

Calculation:

From \(\tanh\varphi = \sinh\varphi / \cosh\varphi = v/c\), we have:

\[ \sinh\varphi = \frac{v}{c}\cosh\varphi \tag{1} \]

Substituting this into the identity:

\[ \cosh^2\varphi - \frac{v^2}{c^2}\cosh^2\varphi = 1 \]
\[ \cosh^2\varphi\left(1 - \frac{v^2}{c^2}\right) = 1 \]
\[ \cosh^2\varphi = \frac{1}{1 - v^2/c^2} \]

Since \(\cosh\varphi > 0\) (obvious from the definition),

\[ \boxed{\cosh\varphi = \frac{1}{\sqrt{1 - v^2/c^2}} = \gamma} \]

Substituting into equation (1),

\[ \boxed{\sinh\varphi = \frac{v/c}{\sqrt{1 - v^2/c^2}} = \gamma\frac{v}{c}} \]

Verification: \(\cosh^2\varphi - \sinh^2\varphi = \gamma^2 - \gamma^2 v^2/c^2 = \gamma^2(1 - v^2/c^2) = 1\). ✓

B-3. Contraction Calculation of the Lorentz Transformation Matrix

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Problem:

Using the Lorentz transformation matrix to an inertial frame moving with velocity \(v\) in the \(x\) direction (in units where \(c = 1\)):

\[ \Lambda^{\mu'}{}_{\nu} = \begin{pmatrix} \gamma & -\gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]

compute \(dx^{1'} = \Lambda^{1'}{}_{\nu}\,dx^\nu\) for \(dx^\mu = (dt,\, dx,\, 0,\, 0)\) following the contraction convention, and verify that \(dx' = \gamma(dx - v\,dt)\).

Solution strategy: Expand \(dx^{1'} = \Lambda^{1'}{}_{\nu}\,dx^\nu\) over \(\nu = 0, 1, 2, 3\).

Calculation:

\[ dx^{1'} = \Lambda^{1'}{}_{0}\,dx^0 + \Lambda^{1'}{}_{1}\,dx^1 + \Lambda^{1'}{}_{2}\,dx^2 + \Lambda^{1'}{}_{3}\,dx^3 \]

Reading each component from the second row (the \(\mu' = 1\) row) of the matrix,

\[ \Lambda^{1'}{}_{0} = -\gamma v, \quad \Lambda^{1'}{}_{1} = \gamma, \quad \Lambda^{1'}{}_{2} = 0, \quad \Lambda^{1'}{}_{3} = 0 \]

Substituting \(dx^\mu = (dt,\, dx,\, 0,\, 0)\),

\[ dx^{1'} = (-\gamma v)\,dt + \gamma\,dx + 0 + 0 \]
\[ \boxed{dx' = \gamma(dx - v\,dt)} \]

Verification: Setting \(dx' = 0\) (the origin of \(S'\)) gives \(dx = v\,dt\), meaning the origin of \(S'\) moves at velocity \(v\) in the \(S\) frame. ✓

Medium

M-1. Detailed Calculation for Determining the Lorentz Transformation Coefficients

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Problem:

In Section 3.5 of the main text, we obtained the system of simultaneous equations that determine the coefficients of the Lorentz transformation:

\[ \begin{aligned} -c^2\,a_1^2 + a_6^2\,v^2 &= -c^2 \quad &\text{(coefficient of }dt^2\text{)} \\ -c^2\,a_2^2 + a_6^2 &= 1 \quad &\text{(coefficient of }dx^2\text{)} \\ -2\,c^2\,a_1\,a_2 - 2\,a_6^2\,v &= 0 \quad &\text{(coefficient of the cross term }dt\,dx\text{)} \end{aligned} \]

Solve this system to derive

\[ a_1 = a_6 = \frac{1}{\sqrt{1 - v^2/c^2}}, \qquad a_2 = -\frac{v/c^2}{\sqrt{1 - v^2/c^2}} \]

Furthermore, show that the continuity condition requiring the transformation to reduce to the identity transformation (\(a_1 = a_6 = 1\), \(a_2 = 0\)) in the limit \(v \to 0\) fixes the sign of \(a_6\) to be \(+\).

Solution strategy: Express \(a_2\) in terms of \(a_1, a_6\) from the cross-term equation, then simultaneously solve the \(dx^2\) and \(dt^2\) equations to determine \(a_1, a_6\).

(a) Expressing \(a_2\) from the cross-term equation:

From the equation for the coefficient of the cross-term \(dt\,dx\):

\[ -2\,c^2\,a_1\,a_2 - 2\,a_6^2\,v = 0 \]

we obtain:

\[ a_2 = -\frac{a_6^2\,v}{c^2\,a_1} \tag{1} \]

(b) Substituting into the \(dx^2\) equation:

Substituting equation (1) into the equation for the coefficient of \(dx^2\):

\[ -c^2\,a_2^2 + a_6^2 = 1 \]

gives:

\[ -c^2 \cdot \frac{a_6^4\,v^2}{c^4\,a_1^2} + a_6^2 = 1 \]
\[ -\frac{a_6^4\,v^2}{c^2\,a_1^2} + a_6^2 = 1 \]

Multiplying both sides by \(c^2\,a_1^2\):

\[ -a_6^4\,v^2 + a_6^2\,c^2\,a_1^2 = c^2\,a_1^2 \tag{2} \]

(c) Expressing \(a_1^2\) from the \(dt^2\) equation:

From the equation for the coefficient of \(dt^2\):

\[ -c^2\,a_1^2 + a_6^2\,v^2 = -c^2 \]

we obtain:

\[ c^2\,a_1^2 = c^2 + a_6^2\,v^2 \tag{3} \]

(d) Solving for \(a_6^2\):

Substituting equation (3) into equation (2):

\[ -a_6^4\,v^2 + a_6^2\,(c^2 + a_6^2\,v^2) = c^2 + a_6^2\,v^2 \]

Expanding the left-hand side:

\[ -a_6^4\,v^2 + a_6^2\,c^2 + a_6^4\,v^2 = c^2 + a_6^2\,v^2 \]

The \(a_6^4\,v^2\) terms cancel, leaving:

\[ a_6^2\,c^2 = c^2 + a_6^2\,v^2 \]

Solving for \(a_6^2\):

\[ a_6^2\,(c^2 - v^2) = c^2 \]
\[ a_6^2 = \frac{c^2}{c^2 - v^2} = \frac{1}{1 - v^2/c^2} \]

(e) Determining the sign (continuity argument):

Since the transformation must reduce to the identity transformation \(a_6 \to +1\) as \(v \to 0\), the sign of \(a_6\) must be positive:

\[ a_6 = +\frac{1}{\sqrt{1 - v^2/c^2}} = \gamma \]

(Choosing the minus sign would introduce a spatial inversion, and the transformation would not continuously reduce to the identity in the \(v = 0\) limit.)

(f) Determining \(a_1\) and \(a_2\):

Substituting \(a_6^2 = \gamma^2\) into equation (3):

\[ c^2\,a_1^2 = c^2 + \gamma^2\,v^2 = c^2 + \frac{v^2}{1 - v^2/c^2} = \frac{c^2(1 - v^2/c^2) + v^2}{1 - v^2/c^2} = \frac{c^2}{1 - v^2/c^2} \]

Therefore \(a_1^2 = 1/(1 - v^2/c^2) = \gamma^2\). By continuity, \(a_1 = +\gamma\).

Substituting \(a_1 = \gamma\) and \(a_6 = \gamma\) into equation (1):

\[ a_2 = -\frac{\gamma^2\,v}{c^2\,\gamma} = -\frac{\gamma\,v}{c^2} = -\frac{v/c^2}{\sqrt{1 - v^2/c^2}} \]

Final Answer:

\[ \boxed{a_1 = a_6 = \frac{1}{\sqrt{1 - v^2/c^2}} = \gamma, \qquad a_2 = -\frac{v/c^2}{\sqrt{1 - v^2/c^2}} = -\frac{\gamma\,v}{c^2}} \]

Verification: Substituting these into the transformation equations from Section 3.3:

\[ t' = a_1\,t + a_2\,x, \qquad x' = a_6(x - v\,t) \]

gives:

\[ t' = \gamma\,t - \frac{\gamma\,v}{c^2}\,x = \gamma\!\left(t - \frac{v\,x}{c^2}\right), \qquad x' = \gamma(x - v\,t) \]

which yields \(ct' = \gamma(ct - \beta\,x)\), \(x' = \gamma(x - \beta\,c\,t)\) (where \(\beta = v/c\)). This agrees with the boxed equations in Section 3.6. ✓

M-2. Metric Preservation Condition under Lorentz Transformation

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Problem:

Verify the condition for the Lorentz transformation matrix \(\Lambda^{\mu'}{}_{\nu}\) to preserve the Minkowski metric

\[ \eta_{\mu'\nu'} = \Lambda^{\alpha}{}_{\mu'}\,\Lambda^{\beta}{}_{\nu'}\,\eta_{\alpha\beta} \]

by substituting the explicit \(\Lambda\) for a boost in the \(x\)-direction and checking the \((\mu', \nu') = (0, 0)\) and \((\mu', \nu') = (0, 1)\) components.

Note: The \(\Lambda^{\alpha}{}_{\mu'}\) in the problem statement represents the "transformation matrix from primed indices (\(S'\) frame components) to unprimed indices (\(S\) frame components)." In other words, it corresponds to the inverse of the forward transformation \(x^{\mu'} = \Lambda^{\mu'}{}_{\nu}x^\nu\). However, since the matrix for an \(x\)-direction boost is symmetric, in this problem the calculation works by "reading the 0th column of the forward transformation matrix \((\gamma, -\gamma v, 0, 0)\) as \(\Lambda^{\alpha}{}_{0'}\)."

Solution strategy: Substitute the specific \(\Lambda\) into the condition \(\eta_{\mu'\nu'} = \Lambda^{\alpha}{}_{\mu'}\Lambda^{\beta}{}_{\nu'}\eta_{\alpha\beta}\) and verify.

On the interpretation of indices: The \(\Lambda^{\alpha}{}_{\mu'}\) in the problem statement represents the matrix components of the transformation from primed to unprimed coordinates (the inverse transformation). For a boost in the \(x\)-direction, the inverse transformation is obtained by \(v \to -v\), so

\[(\Lambda^{-1})^{\alpha}{}_{\mu'} = \begin{pmatrix} \gamma & \gamma v & 0 & 0 \\ \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\]

However, since the matrix for an \(x\)-direction boost is symmetric, reading the "\(\mu'\)-th row, \(\nu\)-th column" of \(\Lambda^{\mu'}{}_{\nu}\) and the "\(\alpha\)-th row, \(\mu'\)-th column" of \((\Lambda^{-1})^{\alpha}{}_{\mu'}\) are unaffected by the transpose operation.

In fact, the metric preservation condition can be written in matrix form as \((\Lambda^{-1})^T \eta\, \Lambda^{-1} = \eta\), which is equivalent to \(\Lambda^T \eta\, \Lambda = \eta\) (multiply both sides by \(\Lambda^T\) from the left and \(\Lambda\) from the right). Below, following the hint in the problem, we compute using \(\Lambda^{\alpha}{}_{0'} = (\gamma, -\gamma v, 0, 0)\) (the 0th column of the forward transformation matrix). For an \(x\)-direction boost, since the matrix is symmetric, reading columns of the forward transformation gives the same result as reading rows of the inverse transformation.

Verification for \((\mu', \nu') = (0, 0)\):

\[ \eta_{0'0'} = \Lambda^{\alpha}{}_{0'}\Lambda^{\beta}{}_{0'}\eta_{\alpha\beta} = \sum_{\alpha}\sum_{\beta}\Lambda^{\alpha}{}_{0'}\Lambda^{\beta}{}_{0'}\eta_{\alpha\beta} \]

Since \(\eta_{\alpha\beta}\) is diagonal, only the \(\alpha = \beta\) terms survive:

\[ = (\Lambda^{0}{}_{0'})^2 \eta_{00} + (\Lambda^{1}{}_{0'})^2 \eta_{11} + (\Lambda^{2}{}_{0'})^2 \eta_{22} + (\Lambda^{3}{}_{0'})^2 \eta_{33} \]
\[ = \gamma^2 \cdot (-1) + (-\gamma v)^2 \cdot (+1) + 0 + 0 \]
\[ = -\gamma^2 + \gamma^2 v^2 = \gamma^2(v^2 - 1) = \frac{v^2 - 1}{1 - v^2} = -1 \]

This agrees with \(\eta_{0'0'} = -1\). ✓

Verification for \((\mu', \nu') = (0, 1)\):

\[ \eta_{0'1'} = \Lambda^{\alpha}{}_{0'}\Lambda^{\beta}{}_{1'}\eta_{\alpha\beta} = \sum_{\alpha}\Lambda^{\alpha}{}_{0'}\Lambda^{\alpha}{}_{1'}\eta_{\alpha\alpha} \]
\[ = \Lambda^{0}{}_{0'}\Lambda^{0}{}_{1'}\eta_{00} + \Lambda^{1}{}_{0'}\Lambda^{1}{}_{1'}\eta_{11} \]
\[ = (\gamma)(-\gamma v)(-1) + (-\gamma v)(\gamma)(+1) \]
\[ = \gamma^2 v - \gamma^2 v = 0 \]

This agrees with \(\eta_{0'1'} = 0\). ✓

Final answer: Both components are confirmed to satisfy the Minkowski metric preservation condition.

Cross-check: One can also verify \(\Lambda^T \eta \Lambda = \eta\) in matrix form. Since \(\det\Lambda = \gamma^2 - \gamma^2 v^2 = \gamma^2(1-v^2) = 1\), \(\Lambda\) is a proper Lorentz transformation. ✓

M-3. Quantitative Consequences of the Relativity of Simultaneity

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Problem:

Using the Lorentz transformation equations, find the time difference \(\Delta t'\) in the \(S'\) frame for two events that are simultaneous in the \(S\) frame (\(\Delta t = 0\)) and separated by a distance \(\Delta x = L\). Explain in words what consequence of the "relativity of simultaneity" this result demonstrates.

Solution strategy: Substitute \(\Delta t = 0\) and \(\Delta x = L\) into the time component of the Lorentz transformation.

Calculation:

The time component of the Lorentz transformation (writing \(c\) explicitly) is:

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

For two events that are simultaneous in the \(S\) frame (\(\Delta t = 0\)) and separated by a distance \(\Delta x = L\):

\[ \Delta t' = \gamma\left(0 - \frac{v}{c^2} \cdot L\right) = -\frac{\gamma v L}{c^2} \]
\[ \boxed{\Delta t' = -\frac{\gamma v L}{c^2}} \]

Consequences of the Relativity of Simultaneity:

This result demonstrates the following:

  1. Simultaneity is not absolute: Two events that are simultaneous in the \(S\) frame (\(\Delta t = 0\)) are generally not simultaneous in the \(S'\) frame (\(\Delta t' \neq 0\)). The condition \(\Delta t' = 0\) holds only when \(v = 0\) (both frames are identical) or \(L = 0\) (events at the same location).

  2. The time difference is proportional to the separation: \(|\Delta t'| = \gamma v L / c^2\), so the greater the spatial separation \(L\) between the two events, the larger the time difference in the \(S'\) frame.

  3. The temporal ordering depends on the sign of \(v\): When \(v > 0\), we have \(\Delta t' < 0\) (the event with the larger \(x\)-coordinate occurs first); when \(v < 0\), we have \(\Delta t' > 0\) (the event with the smaller \(x\)-coordinate occurs first). The temporal ordering of spatially separated simultaneous events can be reversed depending on the observer's state of motion.

  4. Consistency with causality: The spacetime interval between two events with \(\Delta t = 0\) and \(\Delta x = L \neq 0\) is \(\Delta s^2 = -c^2(\Delta t)^2 + (\Delta x)^2 = L^2 > 0\) (spacelike). Since there is no causal relationship between spacelike-separated events, causality is not violated even though the temporal ordering depends on the observer.

Verification:

Dimensional check: \([\gamma v L / c^2] = (\text{m/s})(\text{m})/(\text{m/s})^2 = \text{s}\). Correct dimensions of time. ✓

Limit \(v \ll c\): \(\gamma \approx 1\) gives \(\Delta t' \approx -vL/c^2\). At everyday scales (\(v \sim 10\;\text{m/s}\), \(L \sim 1\;\text{m}\)), we get \(|\Delta t'| \sim 10^{-16}\;\text{s}\), which is extremely small, making the breakdown of simultaneity undetectable. ✓

M-4. Relationship Between Proper Time and Coordinate Time

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Problem:

Using the proper time \(d\tau^2 = -ds^2 = dt^2 - dx^2 - dy^2 - dz^2\) (with \(c = 1\)), derive the relationship between coordinate time \(t\) and proper time \(\tau\):

\[ \frac{d\tau}{dt} = \frac{1}{\gamma} \]

Furthermore, explain from this result that "a moving clock runs slow."

Solution strategy: Factor out \(dt^2\) from \(d\tau^2 = -ds^2\) and substitute the three-velocity.

Calculation:

Definition of proper time (\(c = 1\)):

\[ d\tau^2 = -ds^2 = dt^2 - dx^2 - dy^2 - dz^2 \]

Factor out \(dt^2\):

\[ d\tau^2 = dt^2\left(1 - \frac{dx^2}{dt^2} - \frac{dy^2}{dt^2} - \frac{dz^2}{dt^2}\right) \]

Using the three-velocity \(v^i = dx^i/dt\), we have \(v^2 = (dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2\), so:

\[ d\tau^2 = dt^2(1 - v^2) \]
\[ \frac{d\tau}{dt} = \sqrt{1 - v^2} = \frac{1}{\gamma} \]

Final answer:

\[ \boxed{\frac{d\tau}{dt} = \frac{1}{\gamma}} \]

Physical meaning: Since \(\gamma \geq 1\) (with equality only when \(v = 0\)), we have \(d\tau \leq dt\). That is, a clock co-moving with the particle (ticking proper time \(\tau\)) always runs slower than coordinate time \(t\). This is time dilation.

For finite time intervals,

\[ \Delta\tau = \int \frac{dt}{\gamma} < \Delta t \]

"Moving clocks run slow" — compared to the coordinate time \(\Delta t\) of a stationary observer, the proper time \(\Delta\tau\) of a moving particle is shorter.

Verification: When \(v = 0\), \(\gamma = 1\) and \(d\tau = dt\) (a clock at rest ticks at the same rate as coordinate time). ✓
Dimensional analysis: restoring \(c\) gives \(d\tau/dt = \sqrt{1 - v^2/c^2}\), and for \(v \ll c\) we get \(d\tau \approx dt\). ✓

M-5. Derivation of the Velocity Addition Formula

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Problem:

A problem involving two successive applications of the Lorentz transformation. Consider a frame \(S'\) moving with velocity \(v_1\) in the \(x\) direction relative to frame \(S\), and a frame \(S''\) moving with velocity \(v_2\) in the same \(x\) direction relative to frame \(S'\). Using the additivity of rapidity \(\varphi_{12} = \varphi_1 + \varphi_2\), derive the relativistic velocity addition formula

\[ v_{12} = \frac{v_1 + v_2}{1 + v_1 v_2} \]

(with \(c = 1\)).

Solution strategy: Use the additivity of rapidity and the addition formula for \(\tanh\).

Calculation:

Let \(\varphi_1\) be the rapidity of the boost \(S \to S'\), and \(\varphi_2\) be the rapidity of the boost \(S' \to S''\). Since the Lorentz boost matrix is written in terms of hyperbolic functions, the composition of two boosts is

\[ \Lambda(\varphi_2)\Lambda(\varphi_1) = \Lambda(\varphi_1 + \varphi_2) \]

That is, the rapidity of the composition is

\[ \varphi_{12} = \varphi_1 + \varphi_2 \]

The combined velocity \(v_{12}\) is given by \(\tanh\varphi_{12} = v_{12}\) (with \(c = 1\)). Using the addition formula for \(\tanh\),

\[ v_{12} = \tanh(\varphi_1 + \varphi_2) = \frac{\tanh\varphi_1 + \tanh\varphi_2}{1 + \tanh\varphi_1\,\tanh\varphi_2} \]

Substituting \(\tanh\varphi_1 = v_1\) and \(\tanh\varphi_2 = v_2\),

\[ \boxed{v_{12} = \frac{v_1 + v_2}{1 + v_1 v_2}} \]

Verification:

  • When \(v_1, v_2 \ll 1\), we get \(v_{12} \approx v_1 + v_2\) (reduces to the Galilean velocity addition). ✓
  • When \(v_1 = 1\) (speed of light), \(v_{12} = (1 + v_2)/(1 + v_2) = 1\) (the speed of light remains the speed of light). ✓
  • When \(v_1 = v_2 = 0.9\), \(v_{12} = 1.8/1.81 \approx 0.9945 < 1\) (does not exceed the speed of light). ✓

Supplement: Proof of the additivity of rapidity

Writing the boost matrix in the \(x\)-direction in terms of rapidity,

\[ \Lambda(\varphi) = \begin{pmatrix} \cosh\varphi & -\sinh\varphi \\ -\sinh\varphi & \cosh\varphi \end{pmatrix} \]

(showing only the \(t\)-\(x\) block, with \(c = 1\)). Computing the product of two matrices,

\[ \Lambda(\varphi_2)\Lambda(\varphi_1) = \begin{pmatrix} \cosh\varphi_2\cosh\varphi_1 + \sinh\varphi_2\sinh\varphi_1 & -(\cosh\varphi_2\sinh\varphi_1 + \sinh\varphi_2\cosh\varphi_1) \\ -(\sinh\varphi_2\cosh\varphi_1 + \cosh\varphi_2\sinh\varphi_1) & \sinh\varphi_2\sinh\varphi_1 + \cosh\varphi_2\cosh\varphi_1 \end{pmatrix} \]

By the addition formulas for hyperbolic functions, \(\cosh(\varphi_1+\varphi_2) = \cosh\varphi_1\cosh\varphi_2 + \sinh\varphi_1\sinh\varphi_2\) and \(\sinh(\varphi_1+\varphi_2) = \sinh\varphi_1\cosh\varphi_2 + \cosh\varphi_1\sinh\varphi_2\), this becomes

\[ = \Lambda(\varphi_1 + \varphi_2) \quad \checkmark \]

M-6. Relativity of Simultaneity (Concrete Example)

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Problem:

Consider two events \(A\) and \(B\) that occur simultaneously at \(t = 0\) in the \(S\) frame, at positions \(x_A = 0\) and \(x_B = L\). In the \(S'\) frame moving with velocity \(v\) in the \(x\) direction, (a) find the time difference \(\Delta t' = t'_B - t'_A\) between the two events, and (b) determine which event occurs first, distinguishing cases by the sign of \(v\).

Solution Strategy: Apply the Lorentz transformation to events \(A = (0, 0)\) and \(B = (0, L)\) in frame \(S\).

Calculation:

(a) Calculating the time difference:

Time component of the Lorentz transformation (with \(c = 1\)):

\[ t' = \gamma(t - vx) \]

Event \(A\): \(t_A = 0\), \(x_A = 0\)

\[ t'_A = \gamma(0 - v \cdot 0) = 0 \]

Event \(B\): \(t_B = 0\), \(x_B = L\)

\[ t'_B = \gamma(0 - vL) = -\gamma vL \]

Time difference:

\[ \boxed{\Delta t' = t'_B - t'_A = -\gamma vL} \]

(b) Which event occurs first:

  • When \(v > 0\) (\(S'\) moves in the \(+x\) direction): \(\Delta t' = -\gamma vL < 0\), i.e., \(t'_B < t'_A\). Event \(B\) occurs first.

  • When \(v < 0\) (\(S'\) moves in the \(-x\) direction): \(\Delta t' = -\gamma vL > 0\), i.e., \(t'_B > t'_A\). Event \(A\) occurs first.

  • When \(v = 0\): \(\Delta t' = 0\). The events remain simultaneous.

Final Answer: Two events that are simultaneous in frame \(S\) are, in general, not simultaneous in frame \(S'\). Which event occurs first depends on the sign of \(v\).

Verification: The spacetime interval between the two events is \(\Delta s^2 = -0 + L^2 = L^2 > 0\) (spacelike). This is consistent with the general result that the temporal ordering of spacelike-separated events depends on the inertial frame. ✓
Restoring \(c\): \(\Delta t' = -\gamma vL/c^2\). For \(v \ll c\), \(\Delta t' \approx -vL/c^2 \approx 0\), so the breakdown of simultaneity is negligible at everyday scales. ✓