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Ch. 4 Solutions

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Basic

Medium

Advanced

Basic

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

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Problem:

Rewrite the following expressions given in SI units into natural units (\(c = 1\)). Conversely, convert expressions in natural units back to SI units.

(a) Write the SI expression \(E^2 = p^2 c^2 + m^2 c^4\) in natural units.

(b) Write the natural units expression \(\gamma = 1/\sqrt{1 - v^2}\) in SI units.

(c) In natural units, a statement reads "energy \(E = 5\)." Assuming the rest mass \(m\) (in kg) is nonzero, express the corresponding energy in SI units (joules) in terms of \(m\).

(d) Express the rest energy of the electron with rest mass \(m_e \approx 9.11 \times 10^{-31}\) kg in both SI units (joules) and natural units (kg).

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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Problem:

In natural units (\(c = 1\)), time and length are measured in the same units. Answer the following questions.

(a) Express 1 second in units of length (meters). How many meters is it?

(b) Express 1 meter in units of time (seconds). How many seconds is it?

(c) The distance from the Earth to the Sun is approximately \(1.5 \times 10^{11}\) m. Express this in units of time (seconds). Physically, what does this value mean?

(d) Take the walking speed of a person to be \(v \approx 1\) m/s. What is the numerical value of this speed in natural units (where the speed of light equals 1)?

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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Problem:

For the 4-vectors \(A^\mu = (5,\, 3,\, 0,\, 0)\) and \(B^\mu = (2,\, 1,\, 0,\, 0)\), calculate the Minkowski inner product

\[ \eta_{\mu\nu}\,A^\mu\,B^\nu \]

using Einstein's summation convention (in units where \(c = 1\)).

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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Problem:

Given the 4-vector \(A^\mu = (E,\, p_x,\, p_y,\, p_z)\), express each component \(A_0,\, A_1,\, A_2,\, A_3\) of the covariant vector \(A_\mu = \eta_{\mu\nu}\,A^\nu\) in terms of \(E,\, p_x,\, p_y,\, p_z\).

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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Problem:

Write out all 16 terms of the spacetime interval \(ds^2 = \eta_{\mu\nu}\,dx^\mu\,dx^\nu\) with respect to \(\mu\) and \(\nu\), and show that since the off-diagonal components of \(\eta_{\mu\nu}\) are zero, one obtains \(ds^2 = -dt^2 + dx^2 + dy^2 + dz^2\) (with \(c = 1\)).

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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Problem:

Practice relabeling dummy indices. Verify that the following equality is correct under the summation convention by writing out the sums explicitly.

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

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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Problem:

For a particle with three-velocity \(\mathbf{v} = (v,\, 0,\, 0)\), verify by direct calculation that the four-velocity

\[ U^\mu = \gamma(1,\, v,\, 0,\, 0) \]

satisfies \(\eta_{\mu\nu}\,U^\mu\,U^\nu = -1\) (with \(c = 1\)).

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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Problem:

Regarding the low-speed limit of relativistic energy \(E = \gamma mc^2\), show the following.

(a) Using the approximation \((1 + x)^n \approx 1 + nx\) (for \(|x| \ll 1\)), show that when \(v \ll c\):

\[ E \approx mc^2 + \frac{1}{2}mv^2 \]

That is, in the low-speed limit, the total energy becomes the sum of the rest energy and the Newtonian kinetic energy.

(b) For an object with mass \(m = 1\) kg and speed \(v = 100\) m/s (approximately the speed of a bullet train), find the value of \(v^2/c^2\) and the ratio of the kinetic energy \(\frac{1}{2}mv^2\) to the rest energy \(mc^2\). Using this ratio, explain "why Newton's era could not have noticed rest energy."

(c) For a massless particle, substitute \(m = 0\) into \(E^2 = |\vec{p}|^2 c^2 + m^2 c^4\) to obtain \(E = |\vec{p}|c\). Furthermore, combining this with the relation \(E = \gamma mc^2\), argue that a massless particle with finite energy must necessarily travel at the speed of light.

(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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Problem:

The contraction \(T^{\mu\nu}A_\nu\) of a rank-2 tensor \(T^{\mu\nu}\) and a 4-vector \(A_\nu\) is a tensor of what rank? Identify the free index and the dummy index.

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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Problem:

Regarding the four-velocity \(U^\mu\) and the four-acceleration \(a^\mu \equiv dU^\mu/d\tau\), show the following.

(a) By differentiating both sides of \(\eta_{\mu\nu}\,U^\mu\,U^\nu = -1\) with respect to proper time \(\tau\), derive \(\eta_{\mu\nu}\,U^\mu\,a^\nu = 0\). That is, the four-velocity and four-acceleration are always orthogonal in the sense of the Minkowski inner product.

(b) In the instantaneous rest frame of the particle (\(U^\mu = (1, 0, 0, 0)\)), show from the result of (a) that \(a^0 = 0\), and explain why the four-acceleration is a purely spacelike vector (\(\eta_{\mu\nu}\,a^\mu\,a^\nu > 0\)).

(c) For one-dimensional motion (in the \(x\)-direction only) with constant proper acceleration \(g\) (\(\eta_{\mu\nu}\,a^\mu\,a^\nu = g^2 = \text{const.}\)), show that the worldline of the particle is described by the hyperbola

\[ x^2 - t^2 = \frac{1}{g^2} \]

(with initial conditions \(x = 1/g\) and \(U^\mu = (1, 0, 0, 0)\) at \(t = 0\)).

(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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Problem:

Consider a general Lorentz boost between two inertial frames \(S\) and \(S'\). When \(S'\) is moving with velocity \(\mathbf{v} = (v_x, v_y, v_z)\) (magnitude \(v = |\mathbf{v}|\)) relative to \(S\), the transformation is given by

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

(with \(c = 1\)). Show the following.

(a) Verify that this transformation leaves the spacetime interval \(ds^2 = -dt^2 + d\mathbf{x} \cdot d\mathbf{x}\) invariant.

(b) Show that in the case \(\mathbf{v} = (v, 0, 0)\), this reduces to the standard \(x\)-direction boost derived in this chapter.

(c) Discuss why the composition \(\Lambda(\mathbf{v}_2)\Lambda(\mathbf{v}_1)\) of two boosts \(\Lambda(\mathbf{v}_1)\) and \(\Lambda(\mathbf{v}_2)\) in different directions (\(\mathbf{v}_1 \times \mathbf{v}_2 \neq \mathbf{0}\)) is not, in general, a pure boost, but rather a boost plus a spatial rotation. This spatial rotation is called the Thomas rotation (Wigner rotation). For the case where both \(\mathbf{v}_1\) and \(\mathbf{v}_2\) lie in the \(x\)-\(y\) plane, show that the composite transformation contains a spatial rotation by using a symmetry argument for the transformation matrices.

(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. ✓