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Ch. 2 Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Raising and Lowering Indices

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Solution Strategy

Calculate \(V_\mu = \eta_{\mu\nu} V^\nu\) for each component. Since \(\eta_{\mu\nu} = \text{diag}(+1, -1, -1, -1)\) is a diagonal matrix, each component is simply multiplied by the appropriate sign.

Calculation

\[ V_0 = \eta_{00} V^0 = (+1)(5) = 5 \]
\[ V_1 = \eta_{11} V^1 = (-1)(1) = -1 \]
\[ V_2 = \eta_{22} V^2 = (-1)(-2) = 2 \]
\[ V_3 = \eta_{33} V^3 = (-1)(3) = -3 \]

Final Answer

\[ V_\mu = (5, -1, 2, -3) \]

Verification

Check that raising the index recovers the original vector. From \(V^\mu = \eta^{\mu\nu} V_\nu\): \(V^0 = (+1)(5) = 5\), \(V^1 = (-1)(-1) = 1\), \(V^2 = (-1)(2) = -2\), \(V^3 = (-1)(-3) = 3\). This matches the original \(V^\mu = (5, 1, -2, 3)\). ✓

B-2. Inner Product of 4-Vectors

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Solution Strategy

Compute the Lorentz-invariant inner product using \(A^\mu B_\mu = \eta_{\mu\nu} A^\mu B^\nu = A^0 B^0 - A^1 B^1 - A^2 B^2 - A^3 B^3\).

Calculation

\[ A^\mu B_\mu = A^0 B^0 - A^1 B^1 - A^2 B^2 - A^3 B^3 \]
\[ = (4)(2) - (1)(3) - (0)(1) - (-1)(0) \]
\[ = 8 - 3 - 0 - 0 = 5 \]

Final Answer

\[ A^\mu B_\mu = 5 \]

Verification

As an alternative method, first find \(B_\mu\) and then perform the contraction. \(B_\mu = (2, -3, -1, 0)\).

\[ A^\mu B_\mu = A^0 B_0 + A^1 B_1 + A^2 B_2 + A^3 B_3 = (4)(2) + (1)(-3) + (0)(-1) + (-1)(0) = 8 - 3 + 0 + 0 = 5 \]

The results agree. ✓

B-3. Expanding the Einstein Summation Convention

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Solution Strategy

Since \(\eta_{\mu\nu}\) is a diagonal matrix, all terms with \(\mu \neq \nu\) are zero. Only the 4 terms with \(\mu = \nu\) remain.

Calculation

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

Since \(\eta_{\mu\nu} \neq 0\) only when \(\mu = \nu\):

\[ = \eta_{00} A^0 B^0 + \eta_{11} A^1 B^1 + \eta_{22} A^2 B^2 + \eta_{33} A^3 B^3 \]

Final Answer

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

Verification

This is exactly the inner product formula used in D2, and it agrees with \(A^\mu B_\mu\). ✓

B-4. Application of Lorentz Boost

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Solution Strategy

Calculate the Lorentz factor \(\gamma\) for \(v = 3/5\), then substitute into the transformation equations \(t' = \gamma(t - vx)\), \(x' = \gamma(x - vt)\).

Calculation

\[ v^2 = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \]
\[ 1 - v^2 = 1 - \frac{9}{25} = \frac{16}{25} \]
\[ \gamma = \frac{1}{\sqrt{1 - v^2}} = \frac{1}{\sqrt{16/25}} = \frac{1}{4/5} = \frac{5}{4} \]

Substituting \((t, x) = (5, 3)\):

\[ t' = \gamma(t - vx) = \frac{5}{4}\left(5 - \frac{3}{5} \cdot 3\right) = \frac{5}{4}\left(5 - \frac{9}{5}\right) = \frac{5}{4} \cdot \frac{25 - 9}{5} = \frac{5}{4} \cdot \frac{16}{5} = \frac{16}{4} = 4 \]
\[ x' = \gamma(x - vt) = \frac{5}{4}\left(3 - \frac{3}{5} \cdot 5\right) = \frac{5}{4}\left(3 - 3\right) = \frac{5}{4} \cdot 0 = 0 \]

Final Answer

\[ (t', x') = (4, 0) \]

Verification

Check whether the invariant interval is preserved.

  • Before transformation: \(t^2 - x^2 = 25 - 9 = 16\)
  • After transformation: \(t'^2 - x'^2 = 16 - 0 = 16\)

They match. ✓

Physical interpretation: \(x' = 0\) means that the original spacetime point \((5, 3)\) lies on the spatial origin of the boosted inertial frame. Indeed, \(x = vt = (3/5) \times 5 = 3\), confirming that this point lies on the worldline of the boosted frame's origin.

B-5. Calculation of Rapidity

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Solution Strategy

From \(v = \tanh\beta\), we calculate \(\beta = \text{arctanh}(v) = \frac{1}{2}\ln\frac{1+v}{1-v}\).

Calculation

Substituting \(v = 4/5\):

\[ \beta = \frac{1}{2}\ln\frac{1 + 4/5}{1 - 4/5} = \frac{1}{2}\ln\frac{9/5}{1/5} = \frac{1}{2}\ln 9 = \frac{1}{2} \cdot 2\ln 3 = \ln 3 \]

Verification of \(\cosh\beta\) and \(\sinh\beta\):

\[ \gamma = \frac{1}{\sqrt{1 - v^2}} = \frac{1}{\sqrt{1 - 16/25}} = \frac{1}{\sqrt{9/25}} = \frac{1}{3/5} = \frac{5}{3} \]
\[ \cosh\beta = \gamma = \frac{5}{3} \]
\[ \sinh\beta = \gamma v = \frac{5}{3} \cdot \frac{4}{5} = \frac{4}{3} \]

Final Answer

\[ \beta = \ln 3 \]
\[ \cosh\beta = \frac{5}{3}, \qquad \sinh\beta = \frac{4}{3} \]

Verification

\(\cosh^2\beta - \sinh^2\beta = \left(\frac{5}{3}\right)^2 - \left(\frac{4}{3}\right)^2 = \frac{25}{9} - \frac{16}{9} = \frac{9}{9} = 1\) ✓

\(\tanh\beta = \frac{\sinh\beta}{\cosh\beta} = \frac{4/3}{5/3} = \frac{4}{5} = v\) ✓

From \(e^\beta = e^{\ln 3} = 3\), we get \(\cosh\beta = \frac{3 + 1/3}{2} = \frac{10/3}{2} = \frac{5}{3}\) ✓

B-6. Practice with Index Contraction

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Solution Strategy

The Kronecker delta plays the role of "replacing indices." The product of metric tensors uses the inverse matrix relation.

Calculation

First equation:

\[ \delta^\mu{}_\nu A^\nu = A^\mu \]

When summing over \(\nu\), since \(\delta^\mu{}_\nu\) is 1 only when \(\nu = \mu\), only the \(\nu = \mu\) term survives in the sum, and the result is \(A^\mu\).

Second equation:

\[ \eta_{\mu\nu} \eta^{\nu\rho} = \delta_\mu{}^\rho \]

\(\eta_{\mu\nu}\) and \(\eta^{\nu\rho}\) are related as inverse matrices (matrix product \(\eta \cdot \eta^{-1} = I\)). For the Minkowski metric, \(\eta^{-1} = \eta\), so verifying explicitly:

\[ \eta_{\mu\nu} \eta^{\nu\rho} = \text{diag}(+1,-1,-1,-1) \cdot \text{diag}(+1,-1,-1,-1) = \text{diag}(+1,+1,+1,+1) = \delta_\mu{}^\rho \]

Final Answer

\[ \delta^\mu{}_\nu A^\nu = A^\mu \]
\[ \eta_{\mu\nu} \eta^{\nu\rho} = \delta_\mu{}^\rho \]

Verification

We verify the second equation with specific components. For \(\mu = 1\), \(\rho = 1\):

\[ \eta_{1\nu} \eta^{\nu 1} = \eta_{10}\eta^{01} + \eta_{11}\eta^{11} + \eta_{12}\eta^{21} + \eta_{13}\eta^{31} = 0 + (-1)(-1) + 0 + 0 = 1 = \delta_1{}^1 \]

For \(\mu = 0\), \(\rho = 1\):

\[ \eta_{0\nu} \eta^{\nu 1} = \eta_{00}\eta^{01} + \eta_{01}\eta^{11} + \eta_{02}\eta^{21} + \eta_{03}\eta^{31} = (1)(0) + 0 + 0 + 0 = 0 = \delta_0{}^1 \]

✓

B-7. Hyperbolic Function Representation of the Boost Matrix

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Solution Strategy

Calculate \(\cosh\beta\) and \(\sinh\beta\) from \(\beta = \ln 2\), then perform the matrix-vector multiplication.

Calculation

\[ e^\beta = e^{\ln 2} = 2, \qquad e^{-\beta} = \frac{1}{2} \]
\[ \cosh\beta = \frac{e^\beta + e^{-\beta}}{2} = \frac{2 + 1/2}{2} = \frac{5/2}{2} = \frac{5}{4} \]
\[ \sinh\beta = \frac{e^\beta - e^{-\beta}}{2} = \frac{2 - 1/2}{2} = \frac{3/2}{2} = \frac{3}{4} \]

For \(x^\mu = (3, 1, 0, 0)\):

\[ x'^\mu = \Lambda^\mu{}_\nu x^\nu \]
\[ x'^0 = \cosh\beta \cdot x^0 + (-\sinh\beta) \cdot x^1 = \frac{5}{4} \cdot 3 + \left(-\frac{3}{4}\right) \cdot 1 = \frac{15}{4} - \frac{3}{4} = \frac{12}{4} = 3 \]
\[ x'^1 = (-\sinh\beta) \cdot x^0 + \cosh\beta \cdot x^1 = \left(-\frac{3}{4}\right) \cdot 3 + \frac{5}{4} \cdot 1 = -\frac{9}{4} + \frac{5}{4} = -\frac{4}{4} = -1 \]
\[ x'^2 = x^2 = 0, \qquad x'^3 = x^3 = 0 \]

Final Answer

\[ x'^\mu = (3, -1, 0, 0) \]

Verification

Conservation of the invariant interval:

  • Before transformation: \(\eta_{\mu\nu} x^\mu x^\nu = 3^2 - 1^2 - 0 - 0 = 9 - 1 = 8\)
  • After transformation: \(\eta_{\mu\nu} x'^\mu x'^\nu = 3^2 - (-1)^2 - 0 - 0 = 9 - 1 = 8\)

They agree. ✓

B-8. Dimensional Analysis in Natural Units

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Solution Strategy

From \(c = 1\), we have \([\text{length}] = [\text{time}]\). From \(\hbar = 1\), we have \([\text{energy}] \cdot [\text{time}] = 1\) (dimensionless). Therefore \([\text{time}] = [\text{energy}]^{-1} = [\text{mass}]^{-1}\).

Calculation

Consequence of \(c = 1\): \([L] = [T]\) (length and time have the same dimension)

Consequence of \(\hbar = 1\): \([E][T] = 1\), so \([T] = [E]^{-1}\)

From \(E = mc^2 = m\) (natural units): \([E] = [M]\)

Combining these:

(a) Length: \([L] = [T] = [E]^{-1} = [M]^{-1}\)

\[ [\text{length}] = [\text{mass}]^{-1} \]

(b) Time: \([T] = [E]^{-1} = [M]^{-1}\)

\[ [\text{time}] = [\text{mass}]^{-1} \]

(c) Energy: \([E] = [M]\)

\[ [\text{energy}] = [\text{mass}]^{+1} \]

(d) Momentum: From \(E^2 = p^2 + m^2\), we get \([p] = [E] = [M]\)

\[ [\text{momentum}] = [\text{mass}]^{+1} \]

(e) Force: Force = energy / length = \([M] / [M]^{-1} = [M]^2\)

\[ [\text{force}] = [\text{mass}]^{+2} \]

Final Answer

Physical quantity Dimension in natural units
(a) Length \([\text{mass}]^{-1}\)
(b) Time \([\text{mass}]^{-1}\)
(c) Energy \([\text{mass}]^{+1}\)
(d) Momentum \([\text{mass}]^{+1}\)
(e) Force \([\text{mass}]^{+2}\)

Verification

We verify in SI units. \([\hbar] = \text{J} \cdot \text{s} = \text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1}\), \([c] = \text{m} \cdot \text{s}^{-1}\).

Length: \(\hbar c / E\) has dimensions of length. \([\hbar c / E] = \frac{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1} \cdot \text{m} \cdot \text{s}^{-1}}{\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-2}} = \text{m}\). In natural units where \(\hbar = c = 1\), we have \([L] = [E]^{-1} = [M]^{-1}\). ✓

Force: \([F] = [E]/[L] = [M]/[M]^{-1} = [M]^2\). In SI, \(\text{N} = \text{kg} \cdot \text{m} \cdot \text{s}^{-2}\). The SI combination corresponding to \([M]^2\) with \(\hbar = c = 1\) is \(m^2 c^3/\hbar\), which has dimensions \(\text{kg}^2 \cdot \text{m}^3 \cdot \text{s}^{-3} / (\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1}) = \text{kg} \cdot \text{m} \cdot \text{s}^{-2} = \text{N}\). ✓

Medium

M-1. Derivation of the Condition on the Lorentz Transformation Matrix

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Solution Strategy

Substitute the Lorentz transformation \(x'^\mu = \Lambda^\mu{}_\alpha x^\alpha\) into the invariance condition for the spacetime interval \(\eta_{\mu\nu} x'^\mu x'^\nu = \eta_{\alpha\beta} x^\alpha x^\beta\), and derive the condition that must hold for arbitrary \(x^\alpha\).

Detailed Calculation

Step 1: Preservation of the invariant interval

The invariant interval after transformation:

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

Setting this equal to the invariant interval before transformation \(\eta_{\alpha\beta} x^\alpha x^\beta\):

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

For this to hold for arbitrary \(x^\alpha\), the coefficients of \(x^\alpha x^\beta\) must be equal:

\[ \boxed{\eta_{\mu\nu} \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta = \eta_{\alpha\beta}} \tag{*} \]

Step 2: Matrix representation

We rewrite equation \((*)\) in matrix language. Noting that \((\Lambda^T)_{\alpha}{}^{\mu} = \Lambda^\mu{}_\alpha\), the left-hand side corresponds to the \((\alpha, \beta)\) component of the matrix product \(\Lambda^T \eta \Lambda\). Therefore:

\[ \Lambda^T \eta \Lambda = \eta \]

Step 3: Derivation of \(\det\Lambda = \pm 1\)

Taking the determinant of both sides:

\[ \det(\Lambda^T \eta \Lambda) = \det(\eta) \]

Expanding the left-hand side:

\[ \det(\Lambda^T) \cdot \det(\eta) \cdot \det(\Lambda) = \det(\eta) \]

Since \(\det(\Lambda^T) = \det(\Lambda)\):

\[ (\det\Lambda)^2 \cdot \det(\eta) = \det(\eta) \]

Dividing both sides by \(\det(\eta) = -1 \neq 0\):

\[ (\det\Lambda)^2 = 1 \]
\[ \boxed{\det\Lambda = \pm 1} \]

Final Answer

The condition on a Lorentz transformation is \(\eta_{\mu\nu} \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta = \eta_{\alpha\beta}\) (in matrix form, \(\Lambda^T \eta \Lambda = \eta\)), from which it follows that \(\det\Lambda = \pm 1\).

Verification

We verify using the boost matrix in the \(x\)-direction:

\[ \Lambda = \begin{pmatrix} \gamma & -\gamma v \\ -\gamma v & \gamma \end{pmatrix} \]

(Omitting the \(y, z\) components and considering the \(2 \times 2\) case)

\[ \det\Lambda = \gamma^2 - \gamma^2 v^2 = \gamma^2(1 - v^2) = \frac{1}{1-v^2}(1-v^2) = 1 \]

✓

Computing \(\Lambda^T \eta \Lambda\) (for the \(2 \times 2\) part):

\[ \begin{pmatrix} \gamma & -\gamma v \\ -\gamma v & \gamma \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \gamma & -\gamma v \\ -\gamma v & \gamma \end{pmatrix} \]
\[ = \begin{pmatrix} \gamma & \gamma v \\ -\gamma v & -\gamma \end{pmatrix} \begin{pmatrix} \gamma & -\gamma v \\ -\gamma v & \gamma \end{pmatrix} \]
\[ = \begin{pmatrix} \gamma^2 - \gamma^2 v^2 & -\gamma^2 v + \gamma^2 v \\ -\gamma^2 v + \gamma^2 v & \gamma^2 v^2 - \gamma^2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \eta \]

✓

M-2. Additivity of Rapidity

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Solution Strategy

Compute the product of boost matrices with rapidities \(\beta_1\) and \(\beta_2\), and use the addition theorems of hyperbolic functions to show that the rapidity of the combined boost is \(\beta_1 + \beta_2\).

Detailed Calculation

Step 1: Product of Boost Matrices

The boost matrix in the \(x\) direction (writing only the \(2 \times 2\) block, since the \(y, z\) components are unchanged):

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

Composition of two boosts:

\[ \Lambda(\beta_2)\Lambda(\beta_1) = \begin{pmatrix} \cosh\beta_2 & -\sinh\beta_2 \\ -\sinh\beta_2 & \cosh\beta_2 \end{pmatrix} \begin{pmatrix} \cosh\beta_1 & -\sinh\beta_1 \\ -\sinh\beta_1 & \cosh\beta_1 \end{pmatrix} \]

\((0,0)\) component:

\[ \cosh\beta_2 \cosh\beta_1 + (-\sinh\beta_2)(-\sinh\beta_1) = \cosh\beta_2 \cosh\beta_1 + \sinh\beta_2 \sinh\beta_1 = \cosh(\beta_1 + \beta_2) \]

\((0,1)\) component:

\[ \cosh\beta_2(-\sinh\beta_1) + (-\sinh\beta_2)\cosh\beta_1 = -(\sinh\beta_1 \cosh\beta_2 + \cosh\beta_1 \sinh\beta_2) = -\sinh(\beta_1 + \beta_2) \]

\((1,0)\) component:

\[ (-\sinh\beta_2)\cosh\beta_1 + \cosh\beta_2(-\sinh\beta_1) = -(\sinh\beta_2 \cosh\beta_1 + \cosh\beta_2 \sinh\beta_1) = -\sinh(\beta_1 + \beta_2) \]

\((1,1)\) component:

\[ (-\sinh\beta_2)(-\sinh\beta_1) + \cosh\beta_2 \cosh\beta_1 = \sinh\beta_1 \sinh\beta_2 + \cosh\beta_1 \cosh\beta_2 = \cosh(\beta_1 + \beta_2) \]

Therefore:

\[ \Lambda(\beta_2)\Lambda(\beta_1) = \begin{pmatrix} \cosh(\beta_1+\beta_2) & -\sinh(\beta_1+\beta_2) \\ -\sinh(\beta_1+\beta_2) & \cosh(\beta_1+\beta_2) \end{pmatrix} = \Lambda(\beta_1 + \beta_2) \]

The additivity of rapidity has been demonstrated.

Step 2: Derivation of the Velocity Addition Formula

Let \(v_1 = \tanh\beta_1\), \(v_2 = \tanh\beta_2\). The velocity of the combined transformation is:

\[ v = \tanh(\beta_1 + \beta_2) \]

Applying the addition theorem of \(\tanh\):

\[ v = \tanh(\beta_1 + \beta_2) = \frac{\tanh\beta_1 + \tanh\beta_2}{1 + \tanh\beta_1 \tanh\beta_2} \]
\[ \boxed{v = \frac{v_1 + v_2}{1 + v_1 v_2}} \]

Final Answer

The product of boost matrices shows that \(\Lambda(\beta_2)\Lambda(\beta_1) = \Lambda(\beta_1 + \beta_2)\), demonstrating that rapidity is additive. From this, the relativistic velocity addition formula \(v = \frac{v_1 + v_2}{1 + v_1 v_2}\) is derived.

Verification

Special Case 1: When \(v_1 = v_2 = 0\), we get \(v = 0\). ✓

Special Case 2: When \(v_2 = 1\) (speed of light), \(v = \frac{v_1 + 1}{1 + v_1} = 1\). Adding anything to the speed of light still gives the speed of light. ✓

Special Case 3: When \(v_1, v_2 \ll 1\), \(v \approx v_1 + v_2\) (Galilean transformation limit). ✓

Special Case 4: When \(v_1 = v_2 = 3/5\), \(v = \frac{6/5}{1 + 9/25} = \frac{6/5}{34/25} = \frac{6}{5} \cdot \frac{25}{34} = \frac{150}{170} = \frac{15}{17} < 1\). Does not exceed the speed of light. ✓

M-3. Four-Momentum and the Mass-Shell Condition

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(a) Derivation of the Mass-Shell Condition

Solution strategy: Compute \(p^\mu p_\mu = \eta_{\mu\nu} p^\mu p^\nu\) and set it equal to \(m^2\).

Calculation:

\[ p^\mu p_\mu = \eta_{\mu\nu} p^\mu p^\nu = (p^0)^2 - (p^1)^2 - (p^2)^2 - (p^3)^2 = E^2 - |\mathbf{p}|^2 \]

The mass-shell condition \(p^\mu p_\mu = m^2\) gives:

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

This is the relativistic energy-momentum relation in natural units (\(c = 1\)).

(b) Massless Particles

Substituting \(m = 0\):

\[ E^2 = |\mathbf{p}|^2 \]
\[ \boxed{E = |\mathbf{p}|} \]

For massless particles (such as photons), the energy equals the magnitude of the momentum. The four-momentum satisfies \(p^\mu p_\mu = 0\) and lies on the light cone (null vector, lightlike vector).

(c) Derivation of \(E = \gamma m\), \(p_x = \gamma m v\) via a Boost

Calculation:

Since the four-momentum is a four-vector, it obeys the same Lorentz transformation law as coordinates:

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

In the rest frame, \(p^\mu_{\text{rest}} = (m, 0, 0, 0)\) (\(E = m\), \(\mathbf{p} = 0\)).

We boost in the \(x\)-direction with velocity \(v\) (transforming to a frame in which the particle moves with velocity \(v\)). Here we consider the inverse transformation (rest frame → moving frame). We transform from the rest frame to a frame moving with velocity \(-v\), i.e., we find the momentum in the frame where the particle appears to move with velocity \(v\):

\[ E = \gamma(E_{\text{rest}} + v \cdot 0) = \gamma m \]
\[ p_x = \gamma(0 + v \cdot E_{\text{rest}}) = \gamma m v \]

More precisely, consider the transformation from the particle's rest frame \(S'\) to the lab frame \(S\). In \(S'\), \(p'^\mu = (m, 0, 0, 0)\). Since \(S'\) moves with velocity \(v\) relative to \(S\), the boost from \(S\) to \(S'\) has velocity \(v\). The inverse transformation (\(S'\) to \(S\)) has velocity \(-v\):

\[ p^0 = \gamma(p'^0 - (-v)p'^1) = \gamma(m + 0) = \gamma m \]
\[ p^1 = \gamma(p'^1 - (-v)p'^0) = \gamma(0 + vm) = \gamma mv \]

Final Answer

\[ \boxed{E = \gamma m, \qquad p_x = \gamma m v} \]

Verification

Checking the mass-shell condition:

\[ E^2 - p_x^2 = \gamma^2 m^2 - \gamma^2 m^2 v^2 = \gamma^2 m^2(1 - v^2) = \frac{m^2}{1-v^2}(1-v^2) = m^2 \]

✓

Non-relativistic limit \(v \ll 1\): Since \(\gamma \approx 1 + v^2/2\), we get \(E \approx m + \frac{1}{2}mv^2\) (rest energy + kinetic energy) and \(p_x \approx mv\) (classical momentum). ✓

M-4. Group Structure of Lorentz Transformations

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Solution Strategy

Verify the four group axioms one by one using the Lorentz condition \(\Lambda^T \eta \Lambda = \eta\).

Detailed Calculation

(i) Closure

Suppose \(\Lambda_1\) and \(\Lambda_2\) are both Lorentz transformations, i.e., they satisfy \(\Lambda_1^T \eta \Lambda_1 = \eta\) and \(\Lambda_2^T \eta \Lambda_2 = \eta\). For the composite transformation \(\Lambda_3 = \Lambda_1 \Lambda_2\):

\[ \Lambda_3^T \eta \Lambda_3 = (\Lambda_1 \Lambda_2)^T \eta (\Lambda_1 \Lambda_2) = \Lambda_2^T \Lambda_1^T \eta \Lambda_1 \Lambda_2 \]

Substituting \(\Lambda_1^T \eta \Lambda_1 = \eta\):

\[ = \Lambda_2^T \eta \Lambda_2 = \eta \]

Therefore \(\Lambda_3 = \Lambda_1 \Lambda_2\) is also a Lorentz transformation. ✓

(ii) Associativity

Since Lorentz transformations are represented by matrix products, the associativity of matrix multiplication directly applies:

\[ (\Lambda_1 \Lambda_2)\Lambda_3 = \Lambda_1(\Lambda_2 \Lambda_3) \]

✓

(iii) Existence of the Identity

For the identity transformation \(\Lambda^\mu{}_\nu = \delta^\mu{}_\nu\) (i.e., the identity matrix \(I\)):

\[ I^T \eta I = \eta \]

This is trivially satisfied. Therefore the identity matrix satisfies the Lorentz transformation condition. ✓

(iv) Existence of the Inverse

Since \(\det\Lambda = \pm 1 \neq 0\), \(\Lambda\) is a non-singular matrix and its inverse \(\Lambda^{-1}\) exists.

We show that \(\Lambda^{-1}\) satisfies the Lorentz condition. Starting from \(\Lambda^T \eta \Lambda = \eta\), take the inverse of both sides:

\[ (\Lambda^T \eta \Lambda)^{-1} = \eta^{-1} \]
\[ \Lambda^{-1} \eta^{-1} (\Lambda^T)^{-1} = \eta^{-1} \]

Using \(\eta^{-1} = \eta\) (the Minkowski metric is its own inverse) and \((\Lambda^T)^{-1} = (\Lambda^{-1})^T\):

\[ \Lambda^{-1} \eta (\Lambda^{-1})^T = \eta \]

Taking the transpose:

\[ (\Lambda^{-1})^T \eta \Lambda^{-1} = \eta \]

(since \(\eta\) is a symmetric matrix, \(\eta^T = \eta\))

This shows that \(\Lambda^{-1}\) satisfies the Lorentz condition. ✓

Regarding the proper orthochronous Lorentz group \(SO^+(1,3)\):

Lorentz transformations are classified into four connected components according to the values of \(\det\Lambda\) and \(\Lambda^0{}_0\):

\(\det\Lambda = +1\) \(\det\Lambda = -1\)
\(\Lambda^0{}_0 \geq 1\) Proper orthochronous \(SO^+(1,3)\) Includes spatial inversion
\(\Lambda^0{}_0 \leq -1\) Includes time reversal Includes spacetime inversion

The condition \(\Lambda^0{}_0 \geq 1\) follows from the \((0,0)\) component of the Lorentz condition \((\Lambda^0{}_0)^2 - \sum_i (\Lambda^i{}_0)^2 = 1\), which implies \(|\Lambda^0{}_0| \geq 1\).

\(SO^+(1,3)\) is the connected component containing the identity transformation, consisting only of transformations that can be continuously connected to the identity. Specifically: - 3 spatial rotations (rotations in the \(xy\), \(yz\), \(zx\) planes, 3 parameters) - 3 boosts (in the \(x\), \(y\), \(z\) directions, 3 parameters)

It is a continuous group with a total of 6 parameters. Spatial inversion \(P\) (\(\det P = -1\), \(P^0{}_0 = +1\)) and time reversal \(T\) (\(\det T = -1\), \(T^0{}_0 = -1\)) are discrete transformations that cannot be continuously deformed into the identity, and are not included in \(SO^+(1,3)\).

Verification

Checking the number of parameters: the number of independent components of a \(4 \times 4\) antisymmetric tensor \(\omega_{\mu\nu}\) is \(\frac{4 \times 3}{2} = 6\). This matches the 3 rotations + 3 boosts = 6 parameters. ✓

Advanced

A-1. Transformation Rules for Contravariant and Covariant Tensors, and Application to the Electromagnetic Field Tensor

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(a) Transformation Law for Second-Rank Contravariant Tensors

Explanation:

The transformation law for a 4-vector is \(V'^\mu = \Lambda^\mu{}_\nu V^\nu\). Considering the tensor product \(A^\mu B^\nu\) of two 4-vectors \(A^\mu\) and \(B^\nu\), its transformation law is:

\[ A'^\mu B'^\nu = (\Lambda^\mu{}_\alpha A^\alpha)(\Lambda^\nu{}_\beta B^\beta) = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta A^\alpha B^\beta \]

A general second-rank contravariant tensor \(T^{\mu\nu}\) cannot necessarily be written in the form of a tensor product \(A^\mu B^\nu\), but it can be expressed as a linear combination of tensor products. Therefore, the transformation law for a second-rank contravariant tensor is defined as:

\[ \boxed{T'^{\mu\nu} = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta T^{\alpha\beta}} \]

The Lorentz transformation acts independently on each index. This also serves as the definition: "a tensor is a quantity that transforms according to this rule under Lorentz transformations."

(b) Boost of the Electromagnetic Field Tensor

Strategy: Compute \(F'^{\mu\nu} = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta F^{\alpha\beta}\) for specific components under a boost in the \(x\) direction.

Non-zero components of the boost matrix:

\[ \Lambda^0{}_0 = \gamma, \quad \Lambda^0{}_1 = -\gamma v, \quad \Lambda^1{}_0 = -\gamma v, \quad \Lambda^1{}_1 = \gamma, \quad \Lambda^2{}_2 = 1, \quad \Lambda^3{}_3 = 1 \]

Calculation of \(E_y'\) (the \(F'^{02}\) component):

\[ F'^{02} = \Lambda^0{}_\alpha \Lambda^2{}_\beta F^{\alpha\beta} \]

Since \(\Lambda^2{}_\beta = \delta^2{}_\beta\) (non-zero only for \(\beta = 2\)):

\[ F'^{02} = \Lambda^0{}_\alpha F^{\alpha 2} \]

\(\Lambda^0{}_\alpha\) is non-zero only for \(\alpha = 0, 1\):

\[ F'^{02} = \Lambda^0{}_0 F^{02} + \Lambda^0{}_1 F^{12} \]

Substituting \(F^{02} = -E_y\), \(F^{12} = -B_z\):

\[ F'^{02} = \gamma(-E_y) + (-\gamma v)(-B_z) = -\gamma E_y + \gamma v B_z = -\gamma(E_y - vB_z) \]

Since \(F'^{02} = -E_y'\):

\[ \boxed{E_y' = \gamma(E_y - vB_z)} \]

Calculation of \(B_z'\) (the \(F'^{12}\) component):

\[ F'^{12} = \Lambda^1{}_\alpha \Lambda^2{}_\beta F^{\alpha\beta} = \Lambda^1{}_\alpha F^{\alpha 2} \]

\(\Lambda^1{}_\alpha\) is non-zero only for \(\alpha = 0, 1\):

\[ F'^{12} = \Lambda^1{}_0 F^{02} + \Lambda^1{}_1 F^{12} \]
\[ = (-\gamma v)(-E_y) + \gamma(-B_z) = \gamma v E_y - \gamma B_z = -\gamma(B_z - vE_y) \]

Since \(F'^{12} = -B_z'\):

\[ \boxed{B_z' = \gamma(B_z - vE_y)} \]

Supplement: Transformation of other components

From similar calculations, the complete transformation laws are:

\[ E_x' = E_x, \qquad B_x' = B_x \]
\[ E_y' = \gamma(E_y - vB_z), \qquad B_y' = \gamma(B_y + vE_z) \]
\[ E_z' = \gamma(E_z + vB_y), \qquad B_z' = \gamma(B_z - vE_y) \]

The components along the boost direction (\(x\)) are unchanged, while the perpendicular components of the electric and magnetic fields mix.

(c) Lorentz Invariant \(F^{\mu\nu}F_{\mu\nu}\)

Calculation:

First, lower the indices using \(F_{\mu\nu} = \eta_{\mu\alpha}\eta_{\nu\beta}F^{\alpha\beta}\). Paying attention to the signs of \(\eta\):

  • \(F_{0i} = \eta_{00}\eta_{ii}F^{0i} = (+1)(-1)F^{0i} = -F^{0i}\)
  • \(F_{ij} = \eta_{ii}\eta_{jj}F^{ij} = (-1)(-1)F^{ij} = F^{ij}\)

Therefore:

\[ F_{\mu\nu} = \begin{pmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{pmatrix} \]

Computing the contraction:

\[ F^{\mu\nu}F_{\mu\nu} = \sum_{\mu,\nu} F^{\mu\nu}F_{\mu\nu} \]

Terms with \(\mu = 0\) (non-zero only for \(\nu = 1, 2, 3\), including both \(\mu < \nu\) and \(\mu > \nu\) due to antisymmetry):

\[ F^{0\nu}F_{0\nu} + F^{\nu 0}F_{\nu 0} = 2\sum_{i=1}^{3} F^{0i}F_{0i} \]
\[ = 2[(-E_x)(E_x) + (-E_y)(E_y) + (-E_z)(E_z)] = -2(E_x^2 + E_y^2 + E_z^2) = -2\mathbf{E}^2 \]

Spatial components (\(\mu, \nu = 1, 2, 3\)):

\[ \sum_{i,j=1}^{3} F^{ij}F_{ij} = 2[(F^{12})^2 + (F^{13})^2 + (F^{23})^2] \cdot (\text{check signs}) \]

Specifically: \(F^{12} = -B_z\), \(F_{12} = -B_z\), so \(F^{12}F_{12} = B_z^2\). Similarly \(F^{13}F_{13} = B_y^2\), \(F^{23}F_{23} = B_x^2\).

From antisymmetry \(F^{ij}F_{ij} = F^{ji}F_{ji}\), so counting all combinations with \(i \neq j\):

\[ \sum_{i,j} F^{ij}F_{ij} = 2(B_x^2 + B_y^2 + B_z^2) = 2\mathbf{B}^2 \]

Total:

\[ F^{\mu\nu}F_{\mu\nu} = -2\mathbf{E}^2 + 2\mathbf{B}^2 \]
\[ \boxed{F^{\mu\nu}F_{\mu\nu} = 2(\mathbf{B}^2 - \mathbf{E}^2)} \]

Physical meaning:

This invariant has the same value in all inertial frames. - If \(\mathbf{B}^2 > \mathbf{E}^2\), this inequality holds in every inertial frame (there exists a frame where the electric field can be made zero). - If \(\mathbf{E}^2 > \mathbf{B}^2\), this inequality holds in every inertial frame (there exists a frame where the magnetic field can be made zero). - For electromagnetic waves, \(|\mathbf{E}| = |\mathbf{B}|\), so the invariant is zero.

The other independent Lorentz invariant is \(\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma} \propto \mathbf{E} \cdot \mathbf{B}\).

Verification

Using the results of (b), we confirm that the invariant is preserved:

\[ B_z'^2 - E_y'^2 = \gamma^2(B_z - vE_y)^2 - \gamma^2(E_y - vB_z)^2 \]
\[ = \gamma^2[(B_z^2 - 2vB_zE_y + v^2E_y^2) - (E_y^2 - 2vE_yB_z + v^2B_z^2)] \]
\[ = \gamma^2[B_z^2(1 - v^2) - E_y^2(1 - v^2)] = \gamma^2(1-v^2)(B_z^2 - E_y^2) = B_z^2 - E_y^2 \]

✓ The invariant is preserved.

A-2. Generators of the Lorentz Group and the Lie Algebra

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(a) Antisymmetry of \(\omega_{\mu\nu}\)

Calculation:

Substituting the infinitesimal Lorentz transformation \(\Lambda^\mu{}_\nu = \delta^\mu{}_\nu + \omega^\mu{}_\nu\) into the Lorentz condition \(\eta_{\mu\nu}\Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta = \eta_{\alpha\beta}\):

\[ \eta_{\mu\nu}(\delta^\mu{}_\alpha + \omega^\mu{}_\alpha)(\delta^\nu{}_\beta + \omega^\nu{}_\beta) = \eta_{\alpha\beta} \]

Expanding and neglecting terms of second order and higher in \(\omega\):

\[ \eta_{\mu\nu}\delta^\mu{}_\alpha \delta^\nu{}_\beta + \eta_{\mu\nu}\omega^\mu{}_\alpha \delta^\nu{}_\beta + \eta_{\mu\nu}\delta^\mu{}_\alpha \omega^\nu{}_\beta + O(\omega^2) = \eta_{\alpha\beta} \]
\[ \eta_{\alpha\beta} + \eta_{\mu\beta}\omega^\mu{}_\alpha + \eta_{\alpha\nu}\omega^\nu{}_\beta = \eta_{\alpha\beta} \]

The \(\eta_{\alpha\beta}\) cancels on both sides:

\[ \eta_{\mu\beta}\omega^\mu{}_\alpha + \eta_{\alpha\nu}\omega^\nu{}_\beta = 0 \]

Defining \(\omega_{\beta\alpha} \equiv \eta_{\mu\beta}\omega^\mu{}_\alpha\) (lowering the index):

\[ \omega_{\beta\alpha} + \omega_{\alpha\beta} = 0 \]
\[ \boxed{\omega_{\mu\nu} = -\omega_{\nu\mu}} \]

Number of independent parameters: The number of independent components of a \(4 \times 4\) antisymmetric tensor is \(\frac{4(4-1)}{2} = 6\).

Physical correspondence: - \(\omega_{12}, \omega_{23}, \omega_{31}\): three spatial rotations (\(xy\), \(yz\), \(zx\) planes) - \(\omega_{01}, \omega_{02}, \omega_{03}\): three boosts (\(x\), \(y\), \(z\) directions)

(b) Verification of the four-vector representation of the generators

Expression for finite transformations:

Repeatedly applying the infinitesimal transformation \(\Lambda = I - \frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}\), the finite transformation is expressed as an exponential:

\[ \Lambda = \exp\left(-\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}\right) \]

where \(M^{\mu\nu}\) are the generators of the Lorentz group, satisfying \(M^{\mu\nu} = -M^{\nu\mu}\) (antisymmetric).

Verification for an \(x\)-direction boost:

Consider the case \(\omega_{01} = -\omega_{10} = \beta\) (all others zero).

The infinitesimal transformation is:

\[ \Lambda^\alpha{}_\gamma = \delta^\alpha{}_\gamma - \frac{i}{2}\omega_{\mu\nu}(M^{\mu\nu})^\alpha{}_\gamma \]

Only \(\omega_{01} = \beta\) and \(\omega_{10} = -\beta\) are nonzero. Using the antisymmetry of \(M^{\mu\nu}\):

\[ \frac{1}{2}\omega_{\mu\nu}M^{\mu\nu} = \frac{1}{2}(\omega_{01}M^{01} + \omega_{10}M^{10}) = \frac{1}{2}(\beta M^{01} + (-\beta)(-M^{01})) = \beta M^{01} \]

Therefore the infinitesimal transformation is:

\[ \Lambda^\alpha{}_\gamma = \delta^\alpha{}_\gamma - i\beta(M^{01})^\alpha{}_\gamma \]

Substituting the four-vector representation of the generator:

\[ (M^{01})^\alpha{}_\gamma = i(\eta^{0\alpha}\delta^1{}_\gamma - \eta^{1\alpha}\delta^0{}_\gamma) \]

Computing each component:

  • \((M^{01})^0{}_0 = i(\eta^{00}\delta^1{}_0 - \eta^{10}\delta^0{}_0) = i(1 \cdot 0 - 0 \cdot 1) = 0\)
  • \((M^{01})^0{}_1 = i(\eta^{00}\delta^1{}_1 - \eta^{10}\delta^0{}_1) = i(1 \cdot 1 - 0) = i\)
  • \((M^{01})^1{}_0 = i(\eta^{01}\delta^1{}_0 - \eta^{11}\delta^0{}_0) = i(0 - (-1) \cdot 1) = i\)
  • \((M^{01})^1{}_1 = i(\eta^{01}\delta^1{}_1 - \eta^{11}\delta^0{}_1) = i(0 - 0) = 0\)
  • All other components (\(\alpha = 2, 3\) or \(\gamma = 2, 3\)) are zero

Matrix representation:

\[ (M^{01})^\alpha{}_\gamma = \begin{pmatrix} 0 & i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Infinitesimal transformation:

\[ \Lambda^\alpha{}_\gamma = \delta^\alpha{}_\gamma - i\beta \begin{pmatrix} 0 & i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & \beta & 0 & 0 \\ \beta & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} + O(\beta^2) \]

where we used \(-i \cdot i = 1\).

On the other hand, expanding the boost matrix to first order in \(\beta\):

\[ \Lambda = \begin{pmatrix} \cosh\beta & -\sinh\beta & 0 & 0 \\ -\sinh\beta & \cosh\beta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \approx \begin{pmatrix} 1 & -\beta & 0 & 0 \\ -\beta & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]

Regarding the sign discrepancy: The \(\Lambda^0{}_1 = +\beta\) obtained from the calculation above and \(\Lambda^0{}_1 = -\beta\) from the boost matrix differ in sign. This is due to the sign convention in the definition of \(\omega_{01}\). In fact, if we take the boost parameter as \(\omega_{01} = -\beta\) (i.e., \(\omega_{01} = -\beta\) for a "boost with velocity \(v = \tanh\beta\)"):

\[ \Lambda^\alpha{}_\gamma = \delta^\alpha{}_\gamma + \beta \begin{pmatrix} 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & -\beta & 0 & 0 \\ -\beta & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]

This agrees with the first-order expansion of the boost matrix. ✓

Alternatively, some references use the convention \(\Lambda = \exp\left(+\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}\right)\). Here we directly verify the finite transformation:

\[ \Lambda = \exp(-i\beta M^{01}) = \sum_{n=0}^{\infty} \frac{(-i\beta)^n}{n!}(M^{01})^n \]

Computing \((M^{01})^2\):

\[ (M^{01})^2 = \begin{pmatrix} 0 & i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}^2 = \begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Focusing on the \(2\times 2\) block of \(-iM^{01}\):

\[ -iM^{01}\big|_{2\times 2} = -i\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \equiv K \]

Since \(K^2 = I\):

\[ e^{-i\beta M^{01}}\big|_{2\times 2} = e^{\beta K} = \cosh\beta \cdot I + \sinh\beta \cdot K = \begin{pmatrix} \cosh\beta & \sinh\beta \\ \sinh\beta & \cosh\beta \end{pmatrix} \]

This gives \(\Lambda^0{}_1 = +\sinh\beta\), which differs in sign from the textbook convention \(\Lambda^0{}_1 = -\sinh\beta\). This means we should take \(\omega_{01} = -\beta\). That is, for a boost in the \(x\)-direction with velocity \(v\), setting \(\omega_{01} = -\beta\) (\(\beta = \text{arctanh}\, v\)):

\[ \Lambda = \exp\left(\frac{i}{2}\omega_{\mu\nu}M^{\mu\nu}\right)\bigg|_{\omega_{01}=-\beta} = \exp(i\beta M^{01}) \quad \text{in the } 2\times 2 \text{ block} \]
\[ = e^{-\beta K} = \cosh\beta \cdot I - \sinh\beta \cdot K = \begin{pmatrix} \cosh\beta & -\sinh\beta \\ -\sinh\beta & \cosh\beta \end{pmatrix} \]

This agrees with the boost matrix in the textbook. ✓

(c) Verification of the Lie algebra commutation relations

Calculation of \([M^{01}, M^{02}]\):

First, we find \((M^{02})^\alpha{}_\gamma\):

\[ (M^{02})^\alpha{}_\gamma = i(\eta^{0\alpha}\delta^2{}_\gamma - \eta^{2\alpha}\delta^0{}_\gamma) \]

Nonzero components: - \((M^{02})^0{}_2 = i(\eta^{00}\delta^2{}_2) = i\) - \((M^{02})^2{}_0 = i(-\eta^{22}\delta^0{}_0) = i(-(-1))(1) = i\)

Matrix representation:

\[ M^{01} = \begin{pmatrix} 0 & i & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \qquad M^{02} = \begin{pmatrix} 0 & 0 & i & 0 \\ 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Product \(M^{01}M^{02}\):

\[ (M^{01}M^{02})^\alpha{}_\gamma = (M^{01})^\alpha{}_\delta (M^{02})^\delta{}_\gamma \]

Finding nonzero contributions:

  • \(\alpha = 0\): \((M^{01})^0{}_\delta\) is nonzero only for \(\delta = 1\) (value \(i\)). \((M^{02})^1{}_\gamma = 0\) (all zero). Therefore \((M^{01}M^{02})^0{}_\gamma = 0\).

  • \(\alpha = 1\): \((M^{01})^1{}_\delta\) is nonzero only for \(\delta = 0\) (value \(i\)). \((M^{02})^0{}_\gamma\) is nonzero only for \(\gamma = 2\) (value \(i\)). Therefore \((M^{01}M^{02})^1{}_2 = i \cdot i = -1\).

  • \(\alpha = 2\): \((M^{01})^2{}_\delta = 0\). Therefore \((M^{01}M^{02})^2{}_\gamma = 0\).

\[ M^{01}M^{02} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Product \(M^{02}M^{01}\):

  • \(\alpha = 0\): \((M^{02})^0{}_\delta\) is nonzero only for \(\delta = 2\) (value \(i\)). \((M^{01})^2{}_\gamma = 0\). Therefore \((M^{02}M^{01})^0{}_\gamma = 0\).

  • \(\alpha = 1\): \((M^{02})^1{}_\delta = 0\). Therefore \((M^{02}M^{01})^1{}_\gamma = 0\).

  • \(\alpha = 2\): \((M^{02})^2{}_\delta\) is nonzero only for \(\delta = 0\) (value \(i\)). \((M^{01})^0{}_\gamma\) is nonzero only for \(\gamma = 1\) (value \(i\)). Therefore \((M^{02}M^{01})^2{}_1 = i \cdot i = -1\).

\[ M^{02}M^{01} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Commutator:

\[ [M^{01}, M^{02}] = M^{01}M^{02} - M^{02}M^{01} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Prediction from the Lie algebra commutation relations:

\[ [M^{\mu\nu}, M^{\rho\sigma}] = i(\eta^{\nu\rho}M^{\mu\sigma} - \eta^{\mu\rho}M^{\nu\sigma} - \eta^{\nu\sigma}M^{\mu\rho} + \eta^{\mu\sigma}M^{\nu\rho}) \]

Substituting \(\mu = 0, \nu = 1, \rho = 0, \sigma = 2\):

\[ [M^{01}, M^{02}] = i(\eta^{10}M^{02} - \eta^{00}M^{12} - \eta^{12}M^{00} + \eta^{02}M^{10}) \]

Evaluating each term: - \(\eta^{10} = 0\) - \(\eta^{00} = 1\) - \(\eta^{12} = 0\) - \(\eta^{02} = 0\) - \(M^{00} = 0\) (by antisymmetry)

Therefore:

\[ [M^{01}, M^{02}] = i(0 - 1 \cdot M^{12} - 0 + 0) = -iM^{12} \]

Computing \(M^{12}\):

\[ (M^{12})^\alpha{}_\gamma = i(\eta^{1\alpha}\delta^2{}_\gamma - \eta^{2\alpha}\delta^1{}_\gamma) \]

Nonzero components: - \((M^{12})^1{}_2 = i\eta^{11}\delta^2{}_2 = i(-1)(1) = -i\) - \((M^{12})^2{}_1 = i(-\eta^{22})\delta^1{}_1 = i(-(-1))(1) = i\)

\[ M^{12} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

Therefore:

\[ -iM^{12} = -i\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \]

This perfectly agrees with the directly computed \([M^{01}, M^{02}]\). ✓

Physical interpretation: \(M^{01}\) is the generator of boosts in the \(x\)-direction, \(M^{02}\) is the generator of boosts in the \(y\)-direction, and \(M^{12}\) is the generator of rotations in the \(xy\)-plane. The commutation relation \([M^{01}, M^{02}] = -iM^{12}\) means that "alternately performing boosts in the \(x\)- and \(y\)-directions produces a rotation in the \(xy\)-plane" (the origin of Thomas precession).

Significance for quantum field theory:

The Lie algebra of the Lorentz group determines how fields behave under Lorentz transformations. Different representations correspond to fields of different spins:

  • Scalar field (spin 0): Trivial representation. \(M^{\mu\nu} = 0\) (the field itself does not transform; only its coordinate dependence changes).
  • Vector field (spin 1): Four-dimensional representation. The \((M^{\mu\nu})^\alpha{}_\beta\) computed above are used directly. The electromagnetic field \(A^\mu\) belongs to this representation.
  • Spinor field (spin 1/2): Two-dimensional representation (Weyl spinor) or four-dimensional representation (Dirac spinor). The generators are given by \(\sigma^{\mu\nu} = \frac{i}{4}[\gamma^\mu, \gamma^\nu]\). The electron field belongs to this representation.

In quantum field theory, we require the Lagrangian to be Lorentz invariant. This constrains the transformation rules of fields (i.e., which representation of the Lorentz group they belong to) and the ways in which these fields can be combined (construction of Lorentz scalars). Therefore, the Lie algebra structure of the Lorentz group serves as the starting point for determining the allowed forms of interactions.

Consistency checks

Dimensional check: The 6 generators correspond to 3 rotations + 3 boosts = 6 parameters. This also agrees with the number of independent components of a \(4 \times 4\) antisymmetric matrix: \(\frac{4 \times 3}{2} = 6\). ✓

Hermiticity check: The rotation generator \(M^{12}\) satisfies:

\[ (M^{12})^\dagger = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = M^{12} \]

It is Hermitian (rotations generate unitary transformations). On the other hand, the boost generator \(M^{01}\) satisfies:

\[ (M^{01})^\dagger = \begin{pmatrix} 0 & -i & 0 & 0 \\ -i & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = -M^{01} \]

It is anti-Hermitian (boosts are non-unitary transformations). This reflects the fact that the Lorentz group is a non-compact group, and physically corresponds to the fact that the boost rapidity can range to infinity. ✓