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Ch. 4 Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Conversion to Natural Units and Restoring \(c\)

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Solution Strategy: Use dimensional analysis to insert or remove factors of \(c\) so that dimensions match on both sides.

(a) \(E^2 = p^2 c^2 + m^2 c^4\) in natural units

In natural units we set \(c = 1\), so we replace all factors of \(c^2\) and \(c^4\) with 1:

\[ \boxed{E^2 = p^2 + m^2} \]

In this expression, \(E\), \(p\), and \(m\) are all treated as having the same dimension (e.g., kg or eV).

(b) \(\gamma = 1/\sqrt{1 - v^2}\) in SI units

In SI units, velocity \(v\) has dimensions of m/s. Since \(v^2\) has dimensions of \(\text{m}^2/\text{s}^2\), it must be made dimensionless before subtracting from \(1\). We replace \(v\) with \(v/c\):

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

(c) Value of \(E = 5\) in natural units converted to SI units

In natural units, energy and mass are measured in the same units. To convert back to SI units, we use the relation \(E_{\text{SI}} = E_{\text{natural}} \cdot c^2\). However, the result depends on what unit "5" is expressed in (kg or eV).

If the energy in natural units is \(E = 5\) kg:

\[ E_{\text{SI}} = 5 \times c^2 = 5 \times (3 \times 10^8)^2 \approx 4.5 \times 10^{17}\;\text{J} \]

If the energy in natural units is \(E = 5\) eV:

Since eV is already a unit of energy, we simply convert to joules:

\[ E_{\text{SI}} = 5\;\text{eV} \times 1.602 \times 10^{-19}\;\text{J/eV} \approx 8.0 \times 10^{-19}\;\text{J} \]

(d) Rest energy of the electron

SI units (joules):

\[ E = m_e c^2 = 9.11 \times 10^{-31} \times (3 \times 10^8)^2 \approx 8.2 \times 10^{-14}\;\text{J} \]

Converting to electron volts: \(\approx 511\;\text{keV} = 0.511\;\text{MeV}\).

Natural units (kg):

In natural units \(E = m\), so

\[ \boxed{E = m_e = 9.11 \times 10^{-31}\;\text{kg}} \]

Verification

(a) When converting \(E^2 = p^2 + m^2\) from natural units back to SI, matching dimensions of \(p^2\) to \(E^2\) requires multiplying by \(c^2\) (\([p^2] = (\text{kg}\cdot\text{m/s})^2\), \([E^2/c^2] = (\text{kg}\cdot\text{m/s})^2\)). Matching \(m^2\) to \(E^2\) requires multiplying by \(c^4\). This yields \(E^2 = p^2 c^2 + m^2 c^4\), which is consistent. ✓

(c) For the electron, in natural units \(E = m_e = 9.11 \times 10^{-31}\) kg. Converting to SI gives \(E = m_e c^2 \approx 8.2 \times 10^{-14}\) J. This matches the well-known electron rest energy of \(0.511\) MeV. ✓

B-2. Time and Length in Natural Units

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Solution strategy: In natural units, \(c = 1\), so time and length are measured in the same units. We use \(c = 3 \times 10^8\) m/s as the conversion factor.

(a) 1 second in terms of length

The distance light travels in 1 second is \(c \times 1\;\text{s} = 3 \times 10^8\) m. In natural units, this is the length expression of "1 second":

\[ \boxed{1\;\text{s} = 3 \times 10^8\;\text{m}} \]

(b) 1 meter in terms of time

The time it takes light to travel 1 meter is \(1\;\text{m}/c = 1/(3 \times 10^8)\;\text{s}\):

\[ \boxed{1\;\text{m} = \frac{1}{3 \times 10^8}\;\text{s} \approx 3.33\;\text{ns}} \]

(c) Earth-Sun distance in terms of time

\[ \frac{1.5 \times 10^{11}\;\text{m}}{3 \times 10^8\;\text{m/s}} = 500\;\text{s} \approx 8\;\text{min}\;20\;\text{s} \]

Physical meaning: This is the time it takes light to travel from the Sun to Earth. In natural units, distance is expressed as "the time for light to arrive," so we can say "the Sun is 500 seconds away from Earth." The Sun we see is its appearance from 500 seconds ago.

(d) Human walking speed in natural units

\[ v = \frac{1\;\text{m/s}}{c} = \frac{1}{3 \times 10^8} \approx 3.33 \times 10^{-9} \]

In natural units, velocity is dimensionless (a ratio with the speed of light set to 1). Human walking speed is approximately one three-billionth of the speed of light.

Verification

(a)(b) \(c \times 1\;\text{s} = 3 \times 10^8\) m, \(1\;\text{m}/c = 1/(3\times 10^8)\) s, and in natural units both represent the same quantity. This is precisely the unit system in which \(c = 1\). ✓

(c) This matches the actual time for light to travel from the Sun to Earth (about 8 minutes). The everyday expression "the Sun is 8 light-minutes away" is precisely this natural-units way of expressing distance. ✓

(d) The fact that everyday human speeds are on the order of \(10^{-9}\) of the speed of light shows why special relativistic effects (\(\gamma \approx 1 + v^2/2\)) are difficult to observe in daily life. If \(v = 3.3 \times 10^{-9}\), then \(\gamma - 1 \sim 10^{-17}\), which is difficult to detect even with atomic clocks. ✓

B-3. Calculating the Minkowski Inner Product

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Solution strategy: Since \(\eta_{\mu\nu}\) is the diagonal matrix \(\mathrm{diag}(-1, +1, +1, +1)\), only the \(\mu = \nu\) terms contribute.

Calculation:

By Einstein's summation convention,

\[ \eta_{\mu\nu}A^\mu B^\nu = \sum_{\mu=0}^{3}\sum_{\nu=0}^{3}\eta_{\mu\nu}A^\mu B^\nu \]

Since \(\eta_{\mu\nu} = 0\) (for \(\mu \neq \nu\)),

\[ = \eta_{00}A^0 B^0 + \eta_{11}A^1 B^1 + \eta_{22}A^2 B^2 + \eta_{33}A^3 B^3 \]
\[ = (-1)(5)(2) + (+1)(3)(1) + (+1)(0)(0) + (+1)(0)(0) \]
\[ = -10 + 3 + 0 + 0 = -7 \]

Final answer:

\[ \boxed{\eta_{\mu\nu}\,A^\mu\,B^\nu = -7} \]

Verification: Writing it another way, \(-A^0 B^0 + A^1 B^1 + A^2 B^2 + A^3 B^3 = -10 + 3 = -7\). ✓

B-4. Components of a Covariant Vector

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Solution strategy: Compute \(A_\mu = \eta_{\mu\nu}A^\nu\) for each \(\mu\).

Calculation:

\[ A_0 = \eta_{0\nu}A^\nu = \eta_{00}A^0 = (-1)\cdot E = -E \]
\[ A_1 = \eta_{1\nu}A^\nu = \eta_{11}A^1 = (+1)\cdot p_x = p_x \]
\[ A_2 = \eta_{2\nu}A^\nu = \eta_{22}A^2 = (+1)\cdot p_y = p_y \]
\[ A_3 = \eta_{3\nu}A^\nu = \eta_{33}A^3 = (+1)\cdot p_z = p_z \]

Final answer:

\[ \boxed{A_0 = -E, \quad A_1 = p_x, \quad A_2 = p_y, \quad A_3 = p_z} \]

That is, \(A_\mu = (-E,\, p_x,\, p_y,\, p_z)\).

Verification: \(A_\mu A^\mu = -E^2 + p_x^2 + p_y^2 + p_z^2\). This agrees with \(\eta_{\mu\nu}A^\mu A^\nu\). ✓

B-5. Expansion of the Spacetime Interval into 16 Terms

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Solution strategy: Write out \(ds^2 = \eta_{\mu\nu}\,dx^\mu\,dx^\nu\) explicitly for all 16 terms with \(\mu, \nu = 0, 1, 2, 3\).

Calculation:

\[ ds^2 = \sum_{\mu=0}^{3}\sum_{\nu=0}^{3}\eta_{\mu\nu}\,dx^\mu\,dx^\nu \]

Writing out the 16 terms explicitly (with \(x^0 = t,\, x^1 = x,\, x^2 = y,\, x^3 = z\)):

\(\mu \backslash \nu\) 0 1 2 3
0 \(\eta_{00}\,dt\,dt\) \(\eta_{01}\,dt\,dx\) \(\eta_{02}\,dt\,dy\) \(\eta_{03}\,dt\,dz\)
1 \(\eta_{10}\,dx\,dt\) \(\eta_{11}\,dx\,dx\) \(\eta_{12}\,dx\,dy\) \(\eta_{13}\,dx\,dz\)
2 \(\eta_{20}\,dy\,dt\) \(\eta_{21}\,dy\,dx\) \(\eta_{22}\,dy\,dy\) \(\eta_{23}\,dy\,dz\)
3 \(\eta_{30}\,dz\,dt\) \(\eta_{31}\,dz\,dx\) \(\eta_{32}\,dz\,dy\) \(\eta_{33}\,dz\,dz\)

Since \(\eta_{\mu\nu}\) is a diagonal matrix, all 12 terms with \(\mu \neq \nu\) are zero. The remaining 4 terms are:

\[ ds^2 = \eta_{00}(dx^0)^2 + \eta_{11}(dx^1)^2 + \eta_{22}(dx^2)^2 + \eta_{33}(dx^3)^2 \]

Substituting \(\eta_{00} = -1\), \(\eta_{11} = \eta_{22} = \eta_{33} = +1\):

\[ \boxed{ds^2 = -dt^2 + dx^2 + dy^2 + dz^2} \]

Verification: For light, \(dx^2 + dy^2 + dz^2 = dt^2\), so \(ds^2 = 0\). This is consistent with the invariance of the speed of light. ✓

B-6. Relabeling Dummy Indices

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Solution strategy: Write out both sides as explicit sums and confirm they are the same expression.

Calculation:

Writing out the left-hand side explicitly:

\[ \eta_{\mu\nu}A^\mu B^\nu = \sum_{\mu=0}^{3}\sum_{\nu=0}^{3}\eta_{\mu\nu}A^\mu B^\nu \]
\[ = \eta_{00}A^0 B^0 + \eta_{01}A^0 B^1 + \cdots + \eta_{33}A^3 B^3 \]

Writing out the right-hand side explicitly:

\[ \eta_{\alpha\beta}A^\alpha B^\beta = \sum_{\alpha=0}^{3}\sum_{\beta=0}^{3}\eta_{\alpha\beta}A^\alpha B^\beta \]
\[ = \eta_{00}A^0 B^0 + \eta_{01}A^0 B^1 + \cdots + \eta_{33}A^3 B^3 \]

Both sides sum over all combinations from \(0\) to \(3\). The only difference is that the summation variables (dummy indices) have been renamed from \((\mu, \nu)\) to \((\alpha, \beta)\), and the value of each term is identical. Therefore,

\[ \boxed{\eta_{\mu\nu}\,A^\mu\,B^\nu = \eta_{\alpha\beta}\,A^\alpha\,B^\beta} \]

Verification: This has the same structure as \(\sum_{i=1}^{N} a_i = \sum_{j=1}^{N} a_j\). Since the summation indices simply "run" over their range, their names carry no meaning. ✓

B-7. Verification of the Four-Velocity Normalization Condition

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Solution strategy: Directly compute \(\eta_{\mu\nu}U^\mu U^\nu\).

Calculation:

\[ U^\mu = \gamma(1,\, v,\, 0,\, 0) \]
\[ \eta_{\mu\nu}U^\mu U^\nu = -(U^0)^2 + (U^1)^2 + (U^2)^2 + (U^3)^2 \]
\[ = -\gamma^2 \cdot 1^2 + \gamma^2 \cdot v^2 + 0 + 0 \]
\[ = \gamma^2(-1 + v^2) \]

Substituting \(\gamma^2 = 1/(1 - v^2)\) (with \(c = 1\)),

\[ = \frac{-1 + v^2}{1 - v^2} = \frac{-(1 - v^2)}{1 - v^2} = -1 \]

Final answer:

\[ \boxed{\eta_{\mu\nu}\,U^\mu\,U^\nu = -1} \]

Verification: When \(v = 0\), \(U^\mu = (1, 0, 0, 0)\) and \(\eta_{\mu\nu}U^\mu U^\nu = -1\). ✓

B-8. Low-Speed Limit of Relativistic Energy

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(a) Recovery of Newtonian Kinetic Energy in the Low-Speed Limit

Solution strategy: Taylor expand \(\gamma = (1 - v^2/c^2)^{-1/2}\). Apply the approximation \((1 + x)^n \approx 1 + nx\) with \(x = -v^2/c^2\), \(n = -1/2\).

Calculation:

\[ \gamma = \left(1 - \frac{v^2}{c^2}\right)^{-1/2} \approx 1 + \left(-\frac{1}{2}\right)\left(-\frac{v^2}{c^2}\right) = 1 + \frac{1}{2}\frac{v^2}{c^2} \]

Substituting into \(E = \gamma mc^2\):

\[ E \approx \left(1 + \frac{1}{2}\frac{v^2}{c^2}\right)mc^2 = mc^2 + \frac{1}{2}mv^2 \]
\[ \boxed{E \approx mc^2 + \frac{1}{2}mv^2} \]

The first term is the rest energy, and the second term is the Newtonian kinetic energy.

(b) Evaluation at Bullet-Train Speeds

Solution strategy: Substitute \(v = 100\) m/s, \(c \approx 3 \times 10^8\) m/s and compute the ratio.

Calculation:

\[ \frac{v^2}{c^2} = \frac{(100)^2}{(3 \times 10^8)^2} = \frac{10^4}{9 \times 10^{16}} \approx 1.11 \times 10^{-13} \]

From (a), the ratio of kinetic energy to rest energy is:

\[ \frac{\frac{1}{2}mv^2}{mc^2} = \frac{1}{2}\frac{v^2}{c^2} \approx 5.56 \times 10^{-14} \]
\[ \boxed{\frac{v^2}{c^2} \approx 1.1 \times 10^{-13}, \qquad \frac{(\text{kinetic } E)}{(\text{rest } E)} \approx 5.6 \times 10^{-14}} \]

Physical significance: At everyday speeds, the kinetic energy is only on the order of \(10^{-13}\) of the rest energy. In Newton's era, it was perfectly sufficient to describe motion in terms of changes in \(\frac{1}{2}mv^2\), while the underlying \(mc^2\) (approximately \(9 \times 10^{16}\) J, over a thousand times the Hiroshima atomic bomb) remained completely hidden. It was only in nuclear reactions that a tiny fraction (\(\sim 10^{-3}\)) of \(mc^2\) is released, making the existence of relativistic energy directly observable.

(c) Massless Particles

Solution strategy: Evaluate both \(E^2 = |\vec{p}|^2 c^2 + m^2 c^4\) and \(E = \gamma mc^2\) at \(m = 0\).

Calculation:

Substituting \(m = 0\) into \(E^2 = |\vec{p}|^2 c^2 + m^2 c^4\):

\[ E^2 = |\vec{p}|^2 c^2 \quad \Longrightarrow \quad \boxed{E = |\vec{p}|\,c} \]

Even with zero mass, a particle can carry finite energy as long as it has momentum \(|\vec{p}|\).

On the other hand, if we try to use \(E = \gamma mc^2\) with \(m = 0\) and \(E \neq 0\):

\[ E = \gamma \cdot 0 \cdot c^2 = 0 \quad (\text{for finite } \gamma) \]

which leads to a contradiction. To avoid this, we need \(\gamma \to \infty\), and \(\gamma = 1/\sqrt{1 - v^2/c^2} \to \infty\) implies \(v \to c\). Therefore, a massless particle with finite energy must necessarily travel at the speed of light.

The photon satisfies \(E = |\vec{p}|c\), and the graviton, which appears in Ch. 25, satisfies the same relation.

Verification:

  • (a) As \(v \to 0\), \(E \to mc^2\). This correctly reduces to the rest energy. ✓
  • (a) As \(v \to c\), the approximation breaks down (\(\gamma \to \infty\)). The expansion to second order is valid only for \(v \ll c\). ✓
  • (c) The photon energy-momentum relation \(E = pc\) can also be derived independently from electromagnetism, and the results agree. ✓

Medium

M-1. Tensor Contraction and Classification of Indices

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Solution Strategy: Identify free indices and dummy indices based on the definition of contraction.

Calculation

Writing \(T^{\mu\nu}A_\nu\) explicitly,

\[ T^{\mu\nu}A_\nu = \sum_{\nu=0}^{3} T^{\mu\nu}A_\nu = T^{\mu 0}A_0 + T^{\mu 1}A_1 + T^{\mu 2}A_2 + T^{\mu 3}A_3 \]

In this expression:

  • Dummy index (contracted index): \(\nu\) — appears in both upper and lower positions, and is summed from \(0\) to \(3\). Since it is a summation variable, renaming it does not change the value (for example, writing \(T^{\mu\alpha}A_\alpha\) gives the same result).

  • Free index: \(\mu\) — appears only once in the upper position and is not summed over. It yields a different value for each value of \(\mu\) (\(0, 1, 2, 3\)).

Rank of the Tensor

\(T^{\mu\nu}\) is a rank-2 contravariant tensor (2 free indices), and \(A_\nu\) is a rank-1 covariant tensor (1 free index). In the contraction \(T^{\mu\nu}A_\nu\), one index \(\nu\) is eliminated, so the only remaining free index is \(\mu\).

\[ \boxed{T^{\mu\nu}A_\nu \text{ is a rank-1 (contravariant) tensor (= 4-vector). The free index is } \mu\text{, and the dummy index is } \nu\text{.}} \]

Verification

General rule for tensor rank: when contracting an \(n\)th-rank tensor with an \(m\)th-rank tensor over \(k\) index pairs, the result is a tensor of rank \((n + m - 2k)\). In this problem, \(n = 2\), \(m = 1\), \(k = 1\), so \(2 + 1 - 2 = 1\)st rank. ✓

Confirming with a concrete example: the contraction \(T^{\mu\nu}U_\nu\) of the energy-momentum tensor \(T^{\mu\nu}\) with the 4-velocity \(U_\nu\) represents the flow of energy-momentum and is a 4-vector, which is physically meaningful as a rank-1 tensor. ✓

Advanced

A-1. Four-Velocity and Four-Acceleration

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(a) Orthogonality of 4-Velocity and 4-Acceleration

Solution strategy: Differentiate \(\eta_{\mu\nu}U^\mu U^\nu = -1\) with respect to \(\tau\).

Calculation:

\[ \frac{d}{d\tau}\left(\eta_{\mu\nu}U^\mu U^\nu\right) = \frac{d}{d\tau}(-1) = 0 \]

Differentiating the left-hand side. Since \(\eta_{\mu\nu}\) is constant,

\[ \eta_{\mu\nu}\frac{dU^\mu}{d\tau}U^\nu + \eta_{\mu\nu}U^\mu\frac{dU^\nu}{d\tau} = 0 \]

Substituting \(a^\mu \equiv dU^\mu/d\tau\):

\[ \eta_{\mu\nu}a^\mu U^\nu + \eta_{\mu\nu}U^\mu a^\nu = 0 \]

Relabeling the dummy indices \(\mu \leftrightarrow \nu\) in the first term and using \(\eta_{\mu\nu} = \eta_{\nu\mu}\) (symmetry),

\[ \eta_{\nu\mu}a^\nu U^\mu + \eta_{\mu\nu}U^\mu a^\nu = 2\,\eta_{\mu\nu}U^\mu a^\nu = 0 \]

Therefore,

\[ \boxed{\eta_{\mu\nu}\,U^\mu\,a^\nu = 0} \]

The 4-velocity and 4-acceleration are always orthogonal in the sense of the Minkowski inner product.

(b) 4-Acceleration in the Instantaneous Rest Frame

Solution strategy: Substitute \(U^\mu = (1, 0, 0, 0)\) in the instantaneous rest frame into the result of (a).

Calculation:

In the instantaneous rest frame, \(U^\mu = (1, 0, 0, 0)\). From the result of (a),

\[ \eta_{\mu\nu}U^\mu a^\nu = 0 \]
\[ \eta_{0\nu}U^0 a^\nu = 0 \quad (\text{only the $\mu = 0$ term survives since $U^i = 0$}) \]
\[ \eta_{00} \cdot 1 \cdot a^0 = 0 \quad \Longrightarrow \quad -a^0 = 0 \]
\[ \boxed{a^0 = 0} \]

Therefore, in the instantaneous rest frame, \(a^\mu = (0, a^1, a^2, a^3)\). Its Minkowski norm is

\[ \eta_{\mu\nu}a^\mu a^\nu = -(a^0)^2 + (a^1)^2 + (a^2)^2 + (a^3)^2 = (a^1)^2 + (a^2)^2 + (a^3)^2 \geq 0 \]

Equality holds only when \(a^\mu = 0\) (no acceleration). For an accelerating particle,

\[ \eta_{\mu\nu}a^\mu a^\nu > 0 \]

That is, the 4-acceleration is a purely spacelike vector.

(c) Worldline of Uniform Acceleration

Solution strategy: Determine \(U^\mu\) as a function of \(\tau\) and integrate to obtain the worldline.

Calculation:

For one-dimensional motion (along the \(x\)-direction only), \(U^\mu = (U^0, U^1, 0, 0)\).

Normalization condition:

\[ \eta_{\mu\nu}U^\mu U^\nu = -(U^0)^2 + (U^1)^2 = -1 \tag{I} \]

This can be parametrized as \(U^0 = \cosh f(\tau)\), \(U^1 = \sinh f(\tau)\) (automatically satisfied by the identity \(\cosh^2 - \sinh^2 = 1\)).

4-acceleration:

\[ a^\mu = \frac{dU^\mu}{d\tau} = \left(\dot{f}\sinh f,\, \dot{f}\cosh f,\, 0,\, 0\right) \]

where \(\dot{f} = df/d\tau\).

Condition of constant proper acceleration:

\[ \eta_{\mu\nu}a^\mu a^\nu = -\dot{f}^2\sinh^2 f + \dot{f}^2\cosh^2 f = \dot{f}^2(\cosh^2 f - \sinh^2 f) = \dot{f}^2 = g^2 \]

Therefore \(\dot{f} = g\) (choosing \(g > 0\)). Integrating,

\[ f(\tau) = g\tau + \text{const.} \]

Initial condition at \(\tau = 0\), \(U^\mu = (1, 0, 0, 0)\):

\[ U^0(0) = \cosh f(0) = 1, \quad U^1(0) = \sinh f(0) = 0 \]

This gives \(f(0) = 0\), so the constant \(= 0\).

\[ \boxed{U^0 = \cosh(g\tau), \qquad U^1 = \sinh(g\tau)} \]

Integration for the worldline:

\[ t(\tau) = \int_0^\tau U^0\,d\tau' = \int_0^\tau \cosh(g\tau')\,d\tau' = \frac{1}{g}\sinh(g\tau) \]
\[ x(\tau) = x(0) + \int_0^\tau U^1\,d\tau' = \frac{1}{g} + \int_0^\tau \sinh(g\tau')\,d\tau' = \frac{1}{g} + \frac{1}{g}[\cosh(g\tau) - 1] = \frac{\cosh(g\tau)}{g} \]

Here we used the initial condition \(x(0) = 1/g\).

Verification of the hyperbola:

\[ x^2 - t^2 = \frac{\cosh^2(g\tau)}{g^2} - \frac{\sinh^2(g\tau)}{g^2} = \frac{1}{g^2} \]
\[ \boxed{x^2 - t^2 = \frac{1}{g^2}} \]

This is a hyperbola in the \(x\)-\(t\) plane, known as hyperbolic motion.

Verification:

  • At \(\tau = 0\): \(t = 0\), \(x = 1/g\). Consistent with the initial conditions. ✓
  • Direct verification of \(\eta_{\mu\nu}a^\mu a^\nu = g^2\): Since \(a^\mu = (g\sinh(g\tau),\, g\cosh(g\tau),\, 0,\, 0)\), we have \(\eta_{\mu\nu}a^\mu a^\nu = -g^2\sinh^2(g\tau) + g^2\cosh^2(g\tau) = g^2(\cosh^2(g\tau) - \sinh^2(g\tau)) = g^2\). ✓
  • In the non-relativistic limit \(g\tau \ll 1\): \(t \approx \tau\), \(x \approx 1/g + \frac{1}{2}g\tau^2\). This gives \(x - 1/g \approx \frac{1}{2}gt^2\), recovering Newton's uniformly accelerated motion. ✓

A-2. General Direction Lorentz Boost

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(a) Verification of the Invariance of the Spacetime Interval

Solution strategy: Compute \(ds'^2 = -dt'^2 + d\mathbf{x}' \cdot d\mathbf{x}'\) and show that it equals \(ds^2 = -dt^2 + d\mathbf{x} \cdot d\mathbf{x}\).

Calculation:

The given transformations (\(c = 1\)):

\[ t' = \gamma(t - \mathbf{v} \cdot \mathbf{x}) \]
\[ \mathbf{x}' = \mathbf{x} + (\gamma - 1)\frac{(\mathbf{v} \cdot \mathbf{x})}{v^2}\mathbf{v} - \gamma\mathbf{v}\,t \]

Differential forms:

\[ dt' = \gamma(dt - \mathbf{v} \cdot d\mathbf{x}) \tag{i} \]
\[ d\mathbf{x}' = d\mathbf{x} + (\gamma - 1)\frac{(\mathbf{v} \cdot d\mathbf{x})}{v^2}\mathbf{v} - \gamma\mathbf{v}\,dt \tag{ii} \]

Computation of \(dt'^2\):

\[ dt'^2 = \gamma^2(dt - \mathbf{v} \cdot d\mathbf{x})^2 = \gamma^2\left[dt^2 - 2(\mathbf{v} \cdot d\mathbf{x})dt + (\mathbf{v} \cdot d\mathbf{x})^2\right] \]

Computation of \(d\mathbf{x}' \cdot d\mathbf{x}'\):

Decompose \(d\mathbf{x}\) into components parallel and perpendicular to \(\mathbf{v}\). We use the shorthand \(\mathbf{v} \cdot d\mathbf{x} \equiv \alpha\) and \(\hat{\mathbf{v}} = \mathbf{v}/v\).

\[ d\mathbf{x}_\parallel = \frac{\alpha}{v^2}\mathbf{v}, \qquad d\mathbf{x}_\perp = d\mathbf{x} - \frac{\alpha}{v^2}\mathbf{v} \]

Rewriting equation (ii):

\[ d\mathbf{x}' = d\mathbf{x}_\perp + \frac{\alpha}{v^2}\mathbf{v} + (\gamma - 1)\frac{\alpha}{v^2}\mathbf{v} - \gamma\mathbf{v}\,dt = d\mathbf{x}_\perp + \gamma\frac{\alpha}{v^2}\mathbf{v} - \gamma\mathbf{v}\,dt \]
\[ = d\mathbf{x}_\perp + \frac{\gamma\mathbf{v}}{v^2}(\alpha - v^2\,dt) \]

Since \(d\mathbf{x}_\perp\) and \(\mathbf{v}\) are orthogonal:

\[ |d\mathbf{x}'|^2 = |d\mathbf{x}_\perp|^2 + \frac{\gamma^2}{v^2}(\alpha - v^2\,dt)^2 \]

where \(|d\mathbf{x}_\perp|^2 = |d\mathbf{x}|^2 - \alpha^2/v^2\).

\[ |d\mathbf{x}'|^2 = |d\mathbf{x}|^2 - \frac{\alpha^2}{v^2} + \frac{\gamma^2}{v^2}(\alpha - v^2\,dt)^2 \]

Expanding the third term:

\[ \frac{\gamma^2}{v^2}(\alpha^2 - 2\alpha v^2\,dt + v^4\,dt^2) \]

Therefore:

\[ |d\mathbf{x}'|^2 = |d\mathbf{x}|^2 - \frac{\alpha^2}{v^2} + \frac{\gamma^2\alpha^2}{v^2} - 2\gamma^2\alpha\,dt + \gamma^2 v^2\,dt^2 \]
\[ = |d\mathbf{x}|^2 + \frac{\alpha^2}{v^2}(\gamma^2 - 1) - 2\gamma^2\alpha\,dt + \gamma^2 v^2\,dt^2 \]

Computation of \(ds'^2\):

\[ ds'^2 = -dt'^2 + |d\mathbf{x}'|^2 \]
\[ = -\gamma^2 dt^2 + 2\gamma^2\alpha\,dt - \gamma^2\alpha^2 + |d\mathbf{x}|^2 + \frac{\alpha^2(\gamma^2 - 1)}{v^2} - 2\gamma^2\alpha\,dt + \gamma^2 v^2\,dt^2 \]

The \(2\gamma^2\alpha\,dt\) terms cancel. Collecting terms:

\[ = dt^2(-\gamma^2 + \gamma^2 v^2) + |d\mathbf{x}|^2 + \alpha^2\left(-\gamma^2 + \frac{\gamma^2 - 1}{v^2}\right) \]

Coefficient of \(dt^2\): \(\gamma^2(v^2 - 1) = \gamma^2 \cdot (-1/\gamma^2) = -1\).

Coefficient of \(\alpha^2\):

\[ -\gamma^2 + \frac{\gamma^2 - 1}{v^2} = \frac{-\gamma^2 v^2 + \gamma^2 - 1}{v^2} = \frac{\gamma^2(1 - v^2) - 1}{v^2} = \frac{1 - 1}{v^2} = 0 \]

Therefore:

\[ \boxed{ds'^2 = -dt^2 + |d\mathbf{x}|^2 = ds^2} \]

The spacetime interval is preserved. ✓

(b) The Case \(\mathbf{v} = (v, 0, 0)\)

When \(\mathbf{v} = (v, 0, 0)\), we have \(\mathbf{v} \cdot \mathbf{x} = vx\), \(v^2 = v^2\).

Time component:

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

Spatial components:

\[ x' = x + (\gamma - 1)\frac{vx}{v^2}v - \gamma v\,t = x + (\gamma - 1)x - \gamma vt = \gamma x - \gamma vt = \gamma(x - vt) \quad \checkmark \]
\[ y' = y + (\gamma - 1)\frac{v \cdot 0}{v^2}v_y - \gamma v_y\,t = y + 0 - 0 = y \quad \checkmark \]
\[ z' = z \quad \checkmark \]

This reduces to the standard boost in the \(x\)-direction.

(c) Thomas Rotation

Solution strategy: Show that the matrix of a pure boost is symmetric, and argue that since the composition of two boosts in different directions is generally not symmetric, a rotation component must be present.

Symmetry of a pure boost:

We write down the matrix components of a general boost \(\Lambda(\mathbf{v})\). With \(\beta^i = v^i\), \(\beta = |\mathbf{v}|\) (setting \(c = 1\)):

\[ \Lambda^{0}{}_{0} = \gamma \]
\[ \Lambda^{0}{}_{i} = \Lambda^{i}{}_{0} = -\gamma\beta^i \]
\[ \Lambda^{i}{}_{j} = \delta^i{}_j + (\gamma - 1)\frac{\beta^i\beta^j}{\beta^2} \]

Here \(\Lambda^{0}{}_{i} = \Lambda^{i}{}_{0}\), and \(\Lambda^{i}{}_{j}\) is symmetric in \(i, j\) (since \(\beta^i\beta^j = \beta^j\beta^i\)). Therefore, the \(4 \times 4\) matrix of a pure boost is a symmetric matrix:

\[ \Lambda^{\mu}{}_{\nu} = \Lambda^{\nu}{}_{\mu} \]

Non-symmetry of the composition:

Let \(\mathbf{v}_1\) and \(\mathbf{v}_2\) both lie in the \(x\)-\(y\) plane with \(\mathbf{v}_1 \times \mathbf{v}_2 \neq \mathbf{0}\). Then \(\Lambda_1 = \Lambda(\mathbf{v}_1)\) and \(\Lambda_2 = \Lambda(\mathbf{v}_2)\) are both symmetric matrices.

Consider the transpose of the composite transformation \(\Lambda_{21} = \Lambda_2 \Lambda_1\):

\[ \Lambda_{21}^T = (\Lambda_2 \Lambda_1)^T = \Lambda_1^T \Lambda_2^T = \Lambda_1 \Lambda_2 \]

If \(\Lambda_{21}\) were symmetric, then \(\Lambda_{21} = \Lambda_{21}^T\), i.e., \(\Lambda_2\Lambda_1 = \Lambda_1\Lambda_2\). However, boosts in different directions are generally non-commutative (\(\Lambda_2\Lambda_1 \neq \Lambda_1\Lambda_2\)), so:

\[ \Lambda_{21}^T \neq \Lambda_{21} \]

That is, the composite transformation \(\Lambda_{21}\) is not a symmetric matrix.

Identification of the rotation component:

Any proper Lorentz transformation can be uniquely decomposed into the product of a pure boost \(B\) (symmetric matrix) and a spatial rotation \(R\) (orthogonal matrix) via the polar decomposition:

\[ \Lambda_{21} = B \cdot R \]

The fact that \(\Lambda_{21}\) is not symmetric means \(R \neq I\) (not the identity), i.e., a non-trivial spatial rotation is present.

This rotation is called the Thomas rotation (or Wigner rotation). When \(\mathbf{v}_1\) and \(\mathbf{v}_2\) lie in the \(x\)-\(y\) plane, the rotation axis is the \(z\)-axis, and the rotation angle \(\Omega\) depends on \(v_1\), \(v_2\), and the angle between them.

Physical significance: The Thomas rotation is the origin of the Thomas factor of \(1/2\) in the spin-orbit interaction in atomic physics, and is essential for the accurate calculation of the fine structure of the hydrogen atom. When an electron undergoes curved motion around the nucleus, the direction of the instantaneous boost changes continuously, and the composition of these boosts generates a rotation.

Consistency checks:

  • For boosts in the same direction (\(\mathbf{v}_1 \parallel \mathbf{v}_2\)): \(\Lambda_1\) and \(\Lambda_2\) commute, and the composition is a symmetric matrix (a pure boost). The Thomas rotation is zero. ✓
  • In the limit \(v_1, v_2 \ll 1\), the Thomas rotation angle is of order \(\Omega \approx \frac{1}{2}(\mathbf{v}_1 \times \mathbf{v}_2)\), a second-order small quantity. It vanishes in the non-relativistic limit. ✓