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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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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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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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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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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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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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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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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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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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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. ✓