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Appendix C Solutions

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Basic

Medium

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Basic

B-1. Klein-Gordon \(\partial \mathcal{L}/\partial \phi\)

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Problem:

\[ \mathcal{L} = -\frac{1}{2}\,\eta^{\mu\nu}(\partial_\mu \phi)(\partial_\nu \phi) - \frac{m^2}{2}\,\phi^2 \]

is given. Find the partial derivative with respect to \(\phi\): \(\dfrac{\partial \mathcal{L}}{\partial \phi}\).

Solution:

The only term containing \(\phi\) itself is the mass term \(-\frac{m^2}{2}\phi^2\). The derivative term depends on \(\partial_\mu\phi\), not on \(\phi\) itself.

\[ \frac{\partial\mathcal{L}}{\partial\phi} = -\frac{m^2}{2}\cdot 2\phi = \boxed{-m^2\phi} \]

Verification: This agrees with Step 1 in Section C.5 of the main text.

B-2. Klein-Gordon \(\partial \mathcal{L}/\partial(\partial\phi)\)

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Problem:

For the Klein-Gordon field Lagrangian density

\[ \mathcal{L} = -\frac{1}{2}\eta^{\alpha\beta}(\partial_\alpha\phi)(\partial_\beta\phi) - \frac{1}{2}m^2\phi^2 \]

find the partial derivative with respect to \(\partial_\mu \phi\): \(\dfrac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\).

Solution:

The term containing \(\partial_\mu\phi\) is \(-\frac{1}{2}\eta^{\alpha\beta}(\partial_\alpha\phi)(\partial_\beta\phi)\). Differentiating with respect to \(\partial_\mu\phi\), there are two contributions: one from the case \(\alpha = \mu\) and one from the case \(\beta = \mu\):

\[ \frac{\partial}{\partial(\partial_\mu\phi)}\left[\eta^{\alpha\beta}(\partial_\alpha\phi)(\partial_\beta\phi)\right] = \eta^{\mu\beta}\partial_\beta\phi + \eta^{\alpha\mu}\partial_\alpha\phi = 2\eta^{\mu\nu}\partial_\nu\phi \]

(We used the symmetry of \(\eta^{\alpha\beta}\): \(\eta^{\alpha\beta} = \eta^{\beta\alpha}\).)

Combining with the prefactor \(-\frac{1}{2}\):

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = -\frac{1}{2}\cdot 2\eta^{\mu\nu}\partial_\nu\phi = \boxed{-\eta^{\mu\nu}\partial_\nu\phi} \]

Verification: This agrees with Step 2 in Section C.5 of the main text.

B-3. Partial Derivatives of the String Lagrangian

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Problem:

For

\[ \mathcal{L} = \frac{\rho}{2}\left(\frac{\partial \psi}{\partial t}\right)^2 - \frac{\mathcal{T}}{2}\left(\frac{\partial \psi}{\partial x}\right)^2 \]

find \(\dfrac{\partial \mathcal{L}}{\partial \psi}\), \(\dfrac{\partial \mathcal{L}}{\partial(\partial_t \psi)}\), and \(\dfrac{\partial \mathcal{L}}{\partial(\partial_x \psi)}\) respectively.

Solution:

Since \(\psi\) itself does not appear explicitly in \(\mathcal{L}\):

\[ \boxed{\frac{\partial\mathcal{L}}{\partial\psi} = 0} \]
\[ \boxed{\frac{\partial\mathcal{L}}{\partial(\partial_t\psi)} = \rho\,\frac{\partial\psi}{\partial t}} \]
\[ \boxed{\frac{\partial\mathcal{L}}{\partial(\partial_x\psi)} = -\mathcal{T}\,\frac{\partial\psi}{\partial x}} \]

Verification: Dimensional check. \([\rho\,\partial_t\psi] = (\text{kg/m})\cdot(\text{m/s}) = \text{kg/s}\) (dimensions of momentum density), which is consistent.

B-4. \(\partial \mathcal{L}/\partial \phi\) in \(\phi^4\) Theory

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Problem:

For the Lagrangian density

\[ \mathcal{L} = -\frac{1}{2}\,\eta^{\mu\nu}(\partial_\mu \phi)(\partial_\nu \phi) - \frac{\lambda}{4!}\,\phi^4 \]

find \(\dfrac{\partial \mathcal{L}}{\partial \phi}\).

Solution:

The only term containing \(\phi\) is \(-\frac{\lambda}{4!}\phi^4\).

\[ \frac{\partial\mathcal{L}}{\partial\phi} = -\frac{\lambda}{4!}\cdot 4\phi^3 = -\frac{\lambda}{3!}\phi^3 = \boxed{-\frac{\lambda}{6}\phi^3} \]

Verification: \(4! = 24\), \(4/24 = 1/6 = 1/3!\). ✓

B-5. Explicit Expression of the d'Alembert Operator

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Problem:

Write out the d'Alembert operator \(\Box \equiv \eta^{\mu\nu}\partial_\mu\partial_\nu\) explicitly in terms of partial derivatives with respect to \((t, x, y, z)\), for the case \(\eta^{\mu\nu} = \mathrm{diag}(-1,+1,+1,+1)\).

Solution:

Since \(\eta^{\mu\nu} = \mathrm{diag}(-1,+1,+1,+1)\), all terms with \(\mu \neq \nu\) vanish. Only the diagonal components contribute:

\[ \Box = \eta^{00}\partial_0\partial_0 + \eta^{11}\partial_1\partial_1 + \eta^{22}\partial_2\partial_2 + \eta^{33}\partial_3\partial_3 \]
\[ \boxed{\Box = -\frac{\partial^2}{\partial t^2} + \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} = -\frac{\partial^2}{\partial t^2} + \nabla^2} \]

Verification: \(\Box\phi = 0\) agrees with the wave equation \(\frac{\partial^2\phi}{\partial t^2} = \nabla^2\phi\) (with speed of light \(c=1\)). ✓

B-6. Euler–Lagrange Equation for a 2-Dimensional Scalar Field

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Problem:

Write down the field Euler–Lagrange equation for

\[ \mathcal{L} = \frac{1}{2}(\partial_t \phi)^2 - \frac{1}{2}(\partial_x \phi)^2 - V(\phi) \]

where \(V(\phi)\) is an arbitrary function of \(\phi\).

Solution:

The Euler–Lagrange equation for fields is:

\[ \frac{\partial\mathcal{L}}{\partial\phi} - \partial_t\left(\frac{\partial\mathcal{L}}{\partial(\partial_t\phi)}\right) - \partial_x\left(\frac{\partial\mathcal{L}}{\partial(\partial_x\phi)}\right) = 0 \]

Computing each term:

\[ \frac{\partial\mathcal{L}}{\partial\phi} = -V'(\phi) \]
\[ \frac{\partial\mathcal{L}}{\partial(\partial_t\phi)} = \partial_t\phi \quad \Longrightarrow \quad \partial_t\left(\frac{\partial\mathcal{L}}{\partial(\partial_t\phi)}\right) = \partial_t^2\phi \]
\[ \frac{\partial\mathcal{L}}{\partial(\partial_x\phi)} = -\partial_x\phi \quad \Longrightarrow \quad \partial_x\left(\frac{\partial\mathcal{L}}{\partial(\partial_x\phi)}\right) = -\partial_x^2\phi \]

Substituting:

\[ -V'(\phi) - \partial_t^2\phi - (-\partial_x^2\phi) = 0 \]
\[ \boxed{\frac{\partial^2\phi}{\partial t^2} - \frac{\partial^2\phi}{\partial x^2} + V'(\phi) = 0} \]

Verification: Setting \(V(\phi) = \frac{m^2}{2}\phi^2\) gives \(V'(\phi) = m^2\phi\), which reduces to the Klein–Gordon equation \(\partial_t^2\phi - \partial_x^2\phi + m^2\phi = 0\). ✓

B-7. \(\sqrt{-g}\) for the Minkowski Metric

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Problem:

For the determinant \(g = \det(g_{\mu\nu})\) of the metric tensor \(g_{\mu\nu}\), find the values of \(g\) and \(\sqrt{-g}\) when the metric is the Minkowski metric \(\eta_{\mu\nu} = \mathrm{diag}(-1, 1, 1, 1)\).

Solution:

The determinant of a diagonal matrix is the product of its diagonal elements:

\[ g = \det(\eta_{\mu\nu}) = (-1)\cdot 1\cdot 1\cdot 1 = -1 \]
\[ \boxed{g = -1, \qquad \sqrt{-g} = \sqrt{1} = 1} \]

Verification: In flat spacetime, the volume element correction factor should be 1. ✓

B-8. \(\sqrt{-g}\) for the Schwarzschild Metric

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Problem:

\[ ds^2 = -\!\left(1 - \frac{2M}{r}\right)dt^2 + \left(1 - \frac{2M}{r}\right)^{-1}dr^2 + r^2 d\theta^2 + r^2\sin^2\!\theta\, d\varphi^2 \]

For the metric above, compute \(g = \det(g_{\mu\nu})\) and find \(\sqrt{-g}\).

Solution:

Since the metric is diagonal:

\[ g = g_{tt}\cdot g_{rr}\cdot g_{\theta\theta}\cdot g_{\varphi\varphi} \]

Reading off each component:

\[ g_{tt} = -\left(1 - \frac{2M}{r}\right), \quad g_{rr} = \left(1 - \frac{2M}{r}\right)^{-1}, \quad g_{\theta\theta} = r^2, \quad g_{\varphi\varphi} = r^2\sin^2\theta \]

Computing the product:

\[ g_{tt}\cdot g_{rr} = -\left(1 - \frac{2M}{r}\right)\cdot\left(1 - \frac{2M}{r}\right)^{-1} = -1 \]
\[ g_{\theta\theta}\cdot g_{\varphi\varphi} = r^2\cdot r^2\sin^2\theta = r^4\sin^2\theta \]

Therefore:

\[ g = (-1)\cdot r^4\sin^2\theta = -r^4\sin^2\theta \]
\[ \boxed{g = -r^4\sin^2\theta, \qquad \sqrt{-g} = r^2\sin\theta} \]

Check: In the limit \(M \to 0\), the Schwarzschild metric reduces to the Minkowski metric in spherical coordinates \(ds^2 = -dt^2 + dr^2 + r^2 d\Omega^2\). In that case we also have \(\sqrt{-g} = r^2\sin\theta\), which is consistent with the spherical coordinate volume element \(r^2\sin\theta\,dr\,d\theta\,d\varphi\). ✓

Medium

M-1. Euler–Lagrange Derivation of the Wave Equation for a String

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Problem:

Apply the field Euler–Lagrange equation to

\[ \mathcal{L} = \frac{\rho}{2}\left(\frac{\partial \psi}{\partial t}\right)^2 - \frac{\mathcal{T}}{2}\left(\frac{\partial \psi}{\partial x}\right)^2 \]

and derive the wave equation

\[ \rho\,\frac{\partial^2 \psi}{\partial t^2} = \mathcal{T}\,\frac{\partial^2 \psi}{\partial x^2} \]

Also, express the wave propagation speed \(v\) in terms of \(\mathcal{T}\) and \(\rho\).

Solution Strategy:

Using the results from D3, substitute into the 2-dimensional field Euler–Lagrange equation.

Detailed Calculation:

The field Euler–Lagrange equation (2-dimensional version):

\[ \frac{\partial\mathcal{L}}{\partial\psi} - \partial_t\left(\frac{\partial\mathcal{L}}{\partial(\partial_t\psi)}\right) - \partial_x\left(\frac{\partial\mathcal{L}}{\partial(\partial_x\psi)}\right) = 0 \]

Substituting the results from D3:

\[ 0 - \partial_t\left(\rho\,\partial_t\psi\right) - \partial_x\left(-\mathcal{T}\,\partial_x\psi\right) = 0 \]

Since \(\rho\) and \(\mathcal{T}\) are constants:

\[ -\rho\,\frac{\partial^2\psi}{\partial t^2} + \mathcal{T}\,\frac{\partial^2\psi}{\partial x^2} = 0 \]

Final Answer:

\[ \boxed{\rho\,\frac{\partial^2\psi}{\partial t^2} = \mathcal{T}\,\frac{\partial^2\psi}{\partial x^2}} \]

Rewriting this in the form \(\frac{\partial^2\psi}{\partial t^2} = v^2\frac{\partial^2\psi}{\partial x^2}\):

\[ \boxed{v = \sqrt{\frac{\mathcal{T}}{\rho}}} \]

Verification:

  • Dimensional analysis: \([\mathcal{T}] = \text{N} = \text{kg}\cdot\text{m/s}^2\), \([\rho] = \text{kg/m}\). Therefore \([\mathcal{T}/\rho] = \text{m}^2/\text{s}^2\), so \(v\) has dimensions of velocity. ✓
  • Physical intuition: Greater tension leads to faster propagation; greater density leads to slower propagation. ✓
  • Special case: If \(\mathcal{T} = 0\) then \(v = 0\) and no wave propagates (without tension there is no restoring force). ✓

M-2. Equation of Motion for \(\phi^4\) Theory

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Problem:

Apply the field Euler–Lagrange equation to the Lagrangian density

\[ \mathcal{L} = -\frac{1}{2}\,\eta^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi) - \frac{\lambda}{4!}\,\phi^4 \]

and derive the equation of motion for \(\phi\). Explain how the resulting equation differs from the massless Klein–Gordon equation \(\Box\phi = 0\), including the physical meaning of the difference.

Solution Strategy:

Substitute the results from D2 and D4 into the Euler–Lagrange equation.

Detailed Calculation:

Step 1: From D4,

\[ \frac{\partial\mathcal{L}}{\partial\phi} = -\frac{\lambda}{6}\phi^3 \]

Step 2: The same calculation as D2 (just without the mass term) gives

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} = -\eta^{\mu\nu}\partial_\nu\phi \]
\[ \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right) = -\eta^{\mu\nu}\partial_\mu\partial_\nu\phi = -\Box\phi \]

Step 3: Substituting into the Euler–Lagrange equation:

\[ -\frac{\lambda}{6}\phi^3 - (-\Box\phi) = 0 \]

Final Answer:

\[ \boxed{\Box\phi = \frac{\lambda}{6}\phi^3} \]

Differences from the massless Klein–Gordon equation \(\Box\phi = 0\):

  • In the limit \(\lambda = 0\), it reduces to \(\Box\phi = 0\). ✓
  • The term \(\frac{\lambda}{6}\phi^3\) on the right-hand side is a nonlinear self-interaction term. The Klein–Gordon equation is linear, and the superposition of solutions holds, but the equation of motion for \(\phi^4\) theory is nonlinear, and the superposition principle does not hold.
  • Physically, the \(\phi^4\) term represents the scalar field interacting with itself. In quantum field theory, this term corresponds to a vertex where four \(\phi\) particles interact at a single point (4-point interaction). The coupling constant \(\lambda\) controls the strength of the interaction.

Verification:

  • Dimensional analysis: \([\Box] = [\text{length}^{-2}]\). If \(\phi\) is a scalar field with \([\phi] = [\text{length}^{-1}]\) (in natural units), then the left-hand side gives \([\Box\phi] = [\text{length}^{-3}]\), and the right-hand side gives \([\lambda\phi^3] = [\lambda][\text{length}^{-3}]\), so \(\lambda\) is dimensionless. ✓ (The coupling constant of \(\phi^4\) theory in 4-dimensional spacetime is dimensionless.)
  • Limit \(\lambda \to 0\): Reduces to \(\Box\phi = 0\). ✓

M-3. Massless Scalar Field in Curved Spacetime

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Problem:

For the action

\[ S = \int d^4x\,\sqrt{-g}\left[-\frac{1}{2}\,g^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi)\right] \]

derive the equation of motion following the steps below:

(a) Take the variation \(\phi \to \phi + \delta\phi\) and compute \(\delta S\). Note that \(\sqrt{-g}\) does not depend on \(\phi\).

(b) Perform integration by parts and derive the equation of motion for \(\phi\) from \(\delta S = 0\). Verify that in flat spacetime, \(\partial_\mu(\sqrt{-g}\,g^{\mu\nu}\partial_\nu\phi) \to \eta^{\mu\nu}\partial_\mu\partial_\nu\phi = \Box\phi\).

(a) Calculation of the variation \(\delta S\):

Solution strategy: Substitute \(\phi \to \phi + \delta\phi\) and compute the first-order variation of the action. Since \(\sqrt{-g}\) and \(g^{\mu\nu}\) do not depend on \(\phi\), they are not varied.

\[ S = \int d^4x\,\sqrt{-g}\left[-\frac{1}{2}g^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi)\right] \]

Under \(\phi \to \phi + \delta\phi\), we have \(\partial_\mu\phi \to \partial_\mu\phi + \partial_\mu(\delta\phi)\), so:

\[ (\partial_\mu\phi + \partial_\mu\delta\phi)(\partial_\nu\phi + \partial_\nu\delta\phi) = (\partial_\mu\phi)(\partial_\nu\phi) + (\partial_\mu\phi)(\partial_\nu\delta\phi) + (\partial_\mu\delta\phi)(\partial_\nu\phi) + O(\delta\phi^2) \]

Keeping only the first-order variation:

\[ \delta S = \int d^4x\,\sqrt{-g}\left[-\frac{1}{2}g^{\mu\nu}\left((\partial_\mu\phi)(\partial_\nu\delta\phi) + (\partial_\mu\delta\phi)(\partial_\nu\phi)\right)\right] \]

Using the symmetry of the metric \(g^{\mu\nu} = g^{\nu\mu}\), the two terms are equal:

\[ g^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\delta\phi) = g^{\nu\mu}(\partial_\nu\delta\phi)(\partial_\mu\phi) = g^{\mu\nu}(\partial_\mu\delta\phi)(\partial_\nu\phi) \]

(In the last equality, the indices \(\mu \leftrightarrow \nu\) were relabeled.)

Therefore:

\[ \boxed{\delta S = -\int d^4x\,\sqrt{-g}\,g^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\delta\phi)} \]

(b) Integration by parts and derivation of the equation of motion:

Solution strategy: Use the product rule to transfer the derivative from \(\partial_\nu\delta\phi\) onto \(\delta\phi\).

By the product rule:

\[ \partial_\nu\left[\sqrt{-g}\,g^{\mu\nu}(\partial_\mu\phi)\,\delta\phi\right] = \partial_\nu\left[\sqrt{-g}\,g^{\mu\nu}(\partial_\mu\phi)\right]\delta\phi + \sqrt{-g}\,g^{\mu\nu}(\partial_\mu\phi)\,\partial_\nu(\delta\phi) \]

Therefore:

\[ \sqrt{-g}\,g^{\mu\nu}(\partial_\mu\phi)\,\partial_\nu(\delta\phi) = \partial_\nu\left[\sqrt{-g}\,g^{\mu\nu}(\partial_\mu\phi)\,\delta\phi\right] - \partial_\nu\left[\sqrt{-g}\,g^{\mu\nu}(\partial_\mu\phi)\right]\delta\phi \]

Substituting into \(\delta S\):

\[ \delta S = -\int d^4x\left\{\partial_\nu\left[\sqrt{-g}\,g^{\mu\nu}(\partial_\mu\phi)\,\delta\phi\right] - \partial_\nu\left[\sqrt{-g}\,g^{\mu\nu}\partial_\mu\phi\right]\delta\phi\right\} \]

The first term is a total divergence, which by Gauss's theorem converts to a boundary integral. Since \(\delta\phi = 0\) on the boundary, this term vanishes.

The remainder gives:

\[ \delta S = \int d^4x\,\partial_\mu\left(\sqrt{-g}\,g^{\mu\nu}\partial_\nu\phi\right)\delta\phi = 0 \]

(The indices \(\mu \leftrightarrow \nu\) have been relabeled.)

For this to vanish for arbitrary \(\delta\phi\), we require:

\[ \boxed{\frac{1}{\sqrt{-g}}\,\partial_\mu\!\left(\sqrt{-g}\,g^{\mu\nu}\,\partial_\nu\phi\right) = 0} \]

Flat spacetime check: In Minkowski spacetime, \(g^{\mu\nu} = \eta^{\mu\nu}\) and \(\sqrt{-g} = 1\), so:

\[ \frac{1}{1}\,\partial_\mu\!\left(1\cdot\eta^{\mu\nu}\partial_\nu\phi\right) = \eta^{\mu\nu}\partial_\mu\partial_\nu\phi = \Box\phi = 0 \]

This reduces to the massless Klein–Gordon equation. ✓

Verification:

  • Consistency with the covariant form: The covariant d'Alembert operator acting on a scalar field in curved spacetime is defined as \(\Box_g\phi = \frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}\,g^{\mu\nu}\partial_\nu\phi)\). The equation we obtained is \(\Box_g\phi = 0\), which agrees with the well-known covariant wave equation. ✓
  • Case \(m \neq 0\): Adding a mass term \(-\frac{m^2}{2}\phi^2\) introduces the contribution \(\frac{\partial\mathcal{L}}{\partial\phi} = -m^2\phi\), yielding \(\Box_g\phi - m^2\phi = 0\). ✓

M-4. Derivation of the Energy-Momentum Tensor

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Problem:

\[ T_{\mu\nu} = -\frac{2}{\sqrt{-g}}\,\frac{\delta S_m}{\delta g^{\mu\nu}} \]

Using this definition, derive \(T_{\mu\nu}\) from the Lagrangian density of a free scalar field:

\[ \mathcal{L}_m = -\frac{1}{2}\,g^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi) - \frac{m^2}{2}\,\phi^2 \]

When varying \(S_m = \int d^4x\,\sqrt{-g}\,\mathcal{L}_m\) with respect to \(g^{\mu\nu}\), you may use the identity \(\dfrac{\delta(\sqrt{-g})}{\delta g^{\mu\nu}} = -\dfrac{1}{2}\sqrt{-g}\,g_{\mu\nu}\).

Solution Strategy:

Vary \(S_m = \int d^4x\,\sqrt{-g}\,\mathcal{L}_m\) with respect to \(g^{\mu\nu}\). The variation acts on two places: (i) the direct contribution to \(g^{\mu\nu}\) in \(\mathcal{L}_m\), and (ii) the contribution through \(\sqrt{-g}\).

Detailed Calculation:

\[ S_m = \int d^4x\,\sqrt{-g}\left[-\frac{1}{2}g^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi) - \frac{m^2}{2}\phi^2\right] \]

Let \(g^{\mu\nu} \to g^{\mu\nu} + \delta g^{\mu\nu}\) and take the first-order variation.

(i) Variation of \(g^{\mu\nu}\) in \(\mathcal{L}_m\):

\(g^{\mu\nu}\) appears explicitly only in the first term:

\[ \delta\left[-\frac{1}{2}g^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi)\right] = -\frac{1}{2}(\partial_\mu\phi)(\partial_\nu\phi)\,\delta g^{\mu\nu} \]

(ii) Variation of \(\sqrt{-g}\):

From the given formula \(\frac{\delta(\sqrt{-g})}{\delta g^{\mu\nu}} = -\frac{1}{2}\sqrt{-g}\,g_{\mu\nu}\):

\[ \delta\sqrt{-g} = -\frac{1}{2}\sqrt{-g}\,g_{\mu\nu}\,\delta g^{\mu\nu} \]

(iii) Total variation:

\[ \delta S_m = \int d^4x\left[\sqrt{-g}\left(-\frac{1}{2}(\partial_\mu\phi)(\partial_\nu\phi)\right)\delta g^{\mu\nu} + \left(-\frac{1}{2}\sqrt{-g}\,g_{\mu\nu}\right)\mathcal{L}_m\,\delta g^{\mu\nu}\right] \]
\[ = \int d^4x\,\sqrt{-g}\left[-\frac{1}{2}(\partial_\mu\phi)(\partial_\nu\phi) - \frac{1}{2}g_{\mu\nu}\mathcal{L}_m\right]\delta g^{\mu\nu} \]

(iv) Extracting \(T_{\mu\nu}\):

From the definition \(T_{\mu\nu} = -\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g^{\mu\nu}}\):

\[ \frac{\delta S_m}{\delta g^{\mu\nu}} = \sqrt{-g}\left[-\frac{1}{2}(\partial_\mu\phi)(\partial_\nu\phi) - \frac{1}{2}g_{\mu\nu}\mathcal{L}_m\right] \]
\[ T_{\mu\nu} = -\frac{2}{\sqrt{-g}}\cdot\sqrt{-g}\left[-\frac{1}{2}(\partial_\mu\phi)(\partial_\nu\phi) - \frac{1}{2}g_{\mu\nu}\mathcal{L}_m\right] \]
\[ T_{\mu\nu} = (\partial_\mu\phi)(\partial_\nu\phi) + g_{\mu\nu}\mathcal{L}_m \]

Final Answer:

Substituting \(\mathcal{L}_m = -\frac{1}{2}g^{\alpha\beta}(\partial_\alpha\phi)(\partial_\beta\phi) - \frac{m^2}{2}\phi^2\):

\[ \boxed{T_{\mu\nu} = (\partial_\mu\phi)(\partial_\nu\phi) - g_{\mu\nu}\left[\frac{1}{2}g^{\alpha\beta}(\partial_\alpha\phi)(\partial_\beta\phi) + \frac{m^2}{2}\phi^2\right]} \]

Verification:

  • Symmetry: \(T_{\mu\nu} = T_{\nu\mu}\) is evident (the first term satisfies \((\partial_\mu\phi)(\partial_\nu\phi) = (\partial_\nu\phi)(\partial_\mu\phi)\), and \(g_{\mu\nu}\) in the second term is symmetric). ✓
  • Trace: \(g^{\mu\nu}T_{\mu\nu} = g^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi) + 4\mathcal{L}_m = g^{\mu\nu}(\partial_\mu\phi)(\partial_\nu\phi) + 4\left[-\frac{1}{2}g^{\alpha\beta}(\partial_\alpha\phi)(\partial_\beta\phi) - \frac{m^2}{2}\phi^2\right] = -g^{\alpha\beta}(\partial_\alpha\phi)(\partial_\beta\phi) - 2m^2\phi^2\). For \(m = 0\), \(T^\mu{}_\mu = -g^{\alpha\beta}(\partial_\alpha\phi)(\partial_\beta\phi)\), which is not conformally invariant (the scalar field in 4 dimensions is not traceless unless conformally coupled). This is consistent with known results. ✓
  • \(T_{00}\) component in flat spacetime with \(m = 0\): \(T_{00} = (\partial_t\phi)^2 - \eta_{00}\left[\frac{1}{2}\eta^{\alpha\beta}(\partial_\alpha\phi)(\partial_\beta\phi)\right] = (\partial_t\phi)^2 + \frac{1}{2}\left[-(\partial_t\phi)^2 + (\nabla\phi)^2\right] = \frac{1}{2}(\partial_t\phi)^2 + \frac{1}{2}(\nabla\phi)^2\). This is the correct energy density for a scalar field. ✓

Advanced

A-1. Maxwell Equations from the Electromagnetic Field Lagrangian

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Problem:

The Lagrangian density of the electromagnetic field in 4-dimensional Minkowski spacetime is given by

\[ \mathcal{L}_{\text{EM}} = -\frac{1}{4}\,\eta^{\mu\alpha}\,\eta^{\nu\beta}\,F_{\mu\nu}\,F_{\alpha\beta} \]

where \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\) is the electromagnetic field tensor (Faraday tensor), and \(A_\mu\) is the electromagnetic four-potential.

(a) Verify that \(\mathcal{L}_{\text{EM}}\) can be written as \(\mathcal{L}_{\text{EM}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\) using \(F^{\mu\nu} \equiv \eta^{\mu\alpha}\eta^{\nu\beta}F_{\alpha\beta}\).

(b) Apply the field Euler–Lagrange equation with respect to \(A_\nu\) to derive the vacuum Maxwell equations (source-free):

\[ \partial_\mu F^{\mu\nu} = 0 \]

(c) Show that the \(\nu = 0\) component and the \(\nu = i\) (\(i = 1,2,3\)) components correspond to Gauss's law \(\nabla \cdot \mathbf{E} = 0\) and the Ampère–Maxwell law \(\nabla \times \mathbf{B} = \frac{\partial \mathbf{E}}{\partial t}\) (source-free), respectively.

(a) Verification of \(\mathcal{L}_{\text{EM}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}\):

The definition of \(F^{\mu\nu}\) is \(F^{\mu\nu} \equiv \eta^{\mu\alpha}\eta^{\nu\beta}F_{\alpha\beta}\).

The given Lagrangian density is:

\[ \mathcal{L}_{\text{EM}} = -\frac{1}{4}\eta^{\mu\alpha}\eta^{\nu\beta}F_{\mu\nu}F_{\alpha\beta} \]

Since \(\eta^{\mu\alpha}\eta^{\nu\beta}F_{\alpha\beta} = F^{\mu\nu}\):

\[ \mathcal{L}_{\text{EM}} = -\frac{1}{4}F_{\mu\nu}\cdot\eta^{\mu\alpha}\eta^{\nu\beta}F_{\alpha\beta} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} \]
\[ \boxed{\mathcal{L}_{\text{EM}} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}} \quad \checkmark \]

(b) Derivation of \(\partial_\mu F^{\mu\nu} = 0\) from the Euler–Lagrange equation:

Strategy: The field variable is \(A_\nu\). The Euler–Lagrange equation is:

\[ \frac{\partial\mathcal{L}}{\partial A_\nu} - \partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\right) = 0 \]

Step 1: \(A_\nu\) does not appear explicitly in \(\mathcal{L}\) (since \(\mathcal{L}\) depends only on \(\partial_\mu A_\nu\)):

\[ \frac{\partial\mathcal{L}}{\partial A_\nu} = 0 \]

Step 2: Compute \(\frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)}\).

Since \(\mathcal{L} = -\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}\) and \(F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha\), we examine the dependence of \(F_{\alpha\beta}\) on \(\partial_\mu A_\nu\):

\[ \frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)} = \delta^\mu_\alpha\,\delta^\nu_\beta - \delta^\mu_\beta\,\delta^\nu_\alpha \]

Using the chain rule:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)} = -\frac{1}{4}\left[\frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)}F^{\alpha\beta} + F_{\alpha\beta}\frac{\partial F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)}\right] \]

\(F_{\alpha\beta}F^{\alpha\beta}\) is the contraction of \(F_{\alpha\beta}\) and \(F^{\alpha\beta}\), and since \(F^{\alpha\beta} = \eta^{\alpha\gamma}\eta^{\beta\delta}F_{\gamma\delta}\), the derivative with respect to \(\partial_\mu A_\nu\) has the same structure as the derivative of \(F_{\alpha\beta}\). Computing explicitly:

First term:

\[ \frac{\partial F_{\alpha\beta}}{\partial(\partial_\mu A_\nu)}F^{\alpha\beta} = (\delta^\mu_\alpha\delta^\nu_\beta - \delta^\mu_\beta\delta^\nu_\alpha)F^{\alpha\beta} = F^{\mu\nu} - F^{\nu\mu} = 2F^{\mu\nu} \]

(using the antisymmetry of \(F^{\mu\nu}\): \(F^{\nu\mu} = -F^{\mu\nu}\).)

Second term: Similarly

\[ F_{\alpha\beta}\frac{\partial F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)} \]

Since \(F^{\alpha\beta} = \eta^{\alpha\gamma}\eta^{\beta\delta}F_{\gamma\delta}\):

\[ \frac{\partial F^{\alpha\beta}}{\partial(\partial_\mu A_\nu)} = \eta^{\alpha\gamma}\eta^{\beta\delta}(\delta^\mu_\gamma\delta^\nu_\delta - \delta^\mu_\delta\delta^\nu_\gamma) = \eta^{\alpha\mu}\eta^{\beta\nu} - \eta^{\alpha\nu}\eta^{\beta\mu} \]
\[ F_{\alpha\beta}(\eta^{\alpha\mu}\eta^{\beta\nu} - \eta^{\alpha\nu}\eta^{\beta\mu}) = F^{\mu\nu} - F^{\nu\mu} = 2F^{\mu\nu} \]

Therefore:

\[ \frac{\partial\mathcal{L}}{\partial(\partial_\mu A_\nu)} = -\frac{1}{4}(2F^{\mu\nu} + 2F^{\mu\nu}) = -F^{\mu\nu} \]

Step 3: Substituting into the Euler–Lagrange equation:

\[ 0 - \partial_\mu(-F^{\mu\nu}) = 0 \]
\[ \boxed{\partial_\mu F^{\mu\nu} = 0} \]

(c) Verification of Gauss's law and the Ampère–Maxwell law:

Using the correspondence between the components of the electromagnetic field tensor and the electric and magnetic fields:

\[ F^{0i} = -E^i, \qquad F^{ij} = -\epsilon^{ijk}B_k \]

(where \(i, j, k = 1, 2, 3\) are spatial components.)

Case \(\nu = 0\) (Gauss's law):

\[ \partial_\mu F^{\mu 0} = \partial_0 F^{00} + \partial_i F^{i0} = 0 \]

Since \(F^{00} = 0\) (by antisymmetry) and \(F^{i0} = -F^{0i} = E^i\):

\[ \partial_i E^i = 0 \quad \Longrightarrow \quad \boxed{\nabla\cdot\mathbf{E} = 0} \]

This is the source-free Gauss's law. ✓

Case \(\nu = j\) (spatial component) (Ampère–Maxwell law):

\[ \partial_\mu F^{\mu j} = \partial_0 F^{0j} + \partial_i F^{ij} = 0 \]

Since \(F^{0j} = -E^j\), we have \(\partial_0 F^{0j} = -\frac{\partial E^j}{\partial t}\).

Since \(F^{ij} = -\epsilon^{ijk}B_k\):

\[ \partial_i F^{ij} = -\epsilon^{ijk}\partial_i B_k \]

We verify the relation to the \(j\)-component of \((\nabla\times\mathbf{B})\). Since \((\nabla\times\mathbf{B})^j = \epsilon^{jab}\partial_a B_b\) and \(\epsilon^{ijk} = \epsilon^{jki}\):

\[ -\epsilon^{ijk}\partial_i B_k = -\epsilon^{jki}\partial_i B_k \]

Relabeling indices \(i \to a\), \(k \to b\) gives \(-\epsilon^{jba}\partial_a B_b = +\epsilon^{jab}\partial_a B_b = (\nabla\times\mathbf{B})^j\).

(using \(\epsilon^{jba} = -\epsilon^{jab}\).)

Therefore \(\partial_i F^{ij} = (\nabla\times\mathbf{B})^j\).

The equation becomes:

\[ -\frac{\partial E^j}{\partial t} + (\nabla\times\mathbf{B})^j = 0 \]
\[ \boxed{(\nabla\times\mathbf{B})^j = \frac{\partial E^j}{\partial t} \quad \Longleftrightarrow \quad \nabla\times\mathbf{B} = \frac{\partial\mathbf{E}}{\partial t}} \]

This is the source-free Ampère–Maxwell law. ✓

Verification:

  • Regarding the remaining Maxwell equations: \(\partial_\mu F^{\mu\nu} = 0\) corresponds to half of Maxwell's equations (the source-free version of the sourced equations). The other half (\(\nabla\cdot\mathbf{B} = 0\) and Faraday's law \(\nabla\times\mathbf{E} = -\frac{\partial\mathbf{B}}{\partial t}\)) are automatically satisfied by the definition \(F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu\) (the Bianchi identity \(\partial_{[\lambda}F_{\mu\nu]} = 0\)). ✓
  • When adding source terms: If \(\mathcal{L} = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu} - J^\nu A_\nu\), then \(\frac{\partial\mathcal{L}}{\partial A_\nu} = -J^\nu\) is added, yielding \(\partial_\mu F^{\mu\nu} = J^\nu\). ✓

A-2. Einstein Equation with Cosmological Constant

→ Back to problem

Problem:

Consider the action obtained by adding a cosmological constant \(\Lambda\) to the Einstein–Hilbert action:

\[ S = \frac{1}{16\pi G}\int d^4x\,\sqrt{-g}\,(R - 2\Lambda) + \int d^4x\,\sqrt{-g}\,\mathcal{L}_m \]

(a) Vary the \(\sqrt{-g}\,\Lambda\) term with respect to \(g^{\mu\nu}\). You may use \(\dfrac{\delta(\sqrt{-g})}{\delta g^{\mu\nu}} = -\dfrac{1}{2}\sqrt{-g}\,g_{\mu\nu}\).

(b) Vary the full action above with respect to \(g^{\mu\nu}\) and set \(\delta S = 0\) to show that the Einstein equation with cosmological constant

\[ G_{\mu\nu} + \Lambda\, g_{\mu\nu} = 8\pi G\, T_{\mu\nu} \]

is obtained. You may use as known that the variation of \(\sqrt{-g}\,R\) with respect to \(g^{\mu\nu}\) gives \(\sqrt{-g}\left(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R\right) = \sqrt{-g}\,G_{\mu\nu}\).

(c) In the vacuum case (\(T_{\mu\nu} = 0\)), discuss the physical effect that \(\Lambda > 0\) has on spacetime, based on the structure of the Einstein equation.

(a) Variation of \(\sqrt{-g}\,\Lambda\) with respect to \(g^{\mu\nu}\):

Since \(\Lambda\) is a constant, the variation acts only on \(\sqrt{-g}\):

\[ \delta\!\left(\sqrt{-g}\,\Lambda\right) = \Lambda\,\delta(\sqrt{-g}) \]

Using the given formula \(\frac{\delta(\sqrt{-g})}{\delta g^{\mu\nu}} = -\frac{1}{2}\sqrt{-g}\,g_{\mu\nu}\):

\[ \delta(\sqrt{-g}) = -\frac{1}{2}\sqrt{-g}\,g_{\mu\nu}\,\delta g^{\mu\nu} \]

Therefore:

\[ \boxed{\delta\!\left(\sqrt{-g}\,\Lambda\right) = -\frac{1}{2}\sqrt{-g}\,\Lambda\,g_{\mu\nu}\,\delta g^{\mu\nu}} \]

(b) Derivation of the Einstein equation with cosmological constant:

The total action is:

\[ S = \frac{1}{16\pi G}\int d^4x\,\sqrt{-g}\,(R - 2\Lambda) + \int d^4x\,\sqrt{-g}\,\mathcal{L}_m \]

We compute the variation of each term.

First term: Variation of \(\sqrt{-g}\,R\) (used as a known result)

\[ \delta(\sqrt{-g}\,R) = \sqrt{-g}\left(R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R\right)\delta g^{\mu\nu} = \sqrt{-g}\,G_{\mu\nu}\,\delta g^{\mu\nu} \]

Second term: Variation of \(-2\sqrt{-g}\,\Lambda\)

Using the result from (a):

\[ \delta(-2\sqrt{-g}\,\Lambda) = -2\left(-\frac{1}{2}\sqrt{-g}\,\Lambda\,g_{\mu\nu}\right)\delta g^{\mu\nu} = \sqrt{-g}\,\Lambda\,g_{\mu\nu}\,\delta g^{\mu\nu} \]

Third term: Variation of the matter action

From the definition of the energy-momentum tensor \(T_{\mu\nu} = -\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g^{\mu\nu}}\):

\[ \delta S_m = \int d^4x\,\sqrt{-g}\left(-\frac{1}{2}T_{\mu\nu}\right)\delta g^{\mu\nu} \]

Setting the total variation to zero:

\[ \delta S = \int d^4x\,\sqrt{-g}\left[\frac{1}{16\pi G}\left(G_{\mu\nu} + \Lambda\,g_{\mu\nu}\right) - \frac{1}{2}T_{\mu\nu}\right]\delta g^{\mu\nu} = 0 \]

For this to vanish for arbitrary \(\delta g^{\mu\nu}\):

\[ \frac{1}{16\pi G}\left(G_{\mu\nu} + \Lambda\,g_{\mu\nu}\right) = \frac{1}{2}T_{\mu\nu} \]

Multiplying both sides by \(16\pi G\):

\[ G_{\mu\nu} + \Lambda\,g_{\mu\nu} = 8\pi G\,T_{\mu\nu} \]
\[ \boxed{G_{\mu\nu} + \Lambda\,g_{\mu\nu} = 8\pi G\,T_{\mu\nu}} \]

(c) Physical effects of \(\Lambda > 0\) in vacuum (\(T_{\mu\nu} = 0\)):

When \(T_{\mu\nu} = 0\), the Einstein equation becomes:

\[ G_{\mu\nu} = -\Lambda\,g_{\mu\nu} \]

This can be rewritten as:

\[ G_{\mu\nu} = 8\pi G\left(-\frac{\Lambda}{8\pi G}\,g_{\mu\nu}\right) \]

The right-hand side is equivalent to an energy-momentum tensor

\[ T_{\mu\nu}^{(\text{eff})} = -\frac{\Lambda}{8\pi G}\,g_{\mu\nu} \]

This corresponds to the perfect fluid energy-momentum tensor \(T_{\mu\nu} = (\rho + p)u_\mu u_\nu + p\,g_{\mu\nu}\) with

\[ \rho_\Lambda = \frac{\Lambda}{8\pi G}, \qquad p_\Lambda = -\frac{\Lambda}{8\pi G} = -\rho_\Lambda \]

That is, \(\Lambda > 0\) behaves as a uniform energy density with equation of state \(p = -\rho\).

Physical effects:

  1. Repulsive gravitational effect: The negative pressure \(p = -\rho < 0\) acts as a gravitational repulsion in general relativity. From the Friedmann acceleration equation \(\ddot{a}/a = -\frac{4\pi G}{3}(\rho + 3p) = -\frac{4\pi G}{3}(\rho_\Lambda - 3\rho_\Lambda) = \frac{8\pi G}{3}\rho_\Lambda > 0\), it accelerates the expansion of the universe.

  2. de Sitter spacetime: In the vacuum case with \(\Lambda > 0\), spacetime becomes de Sitter spacetime, describing an exponentially expanding universe. The scale factor is \(a(t) \propto e^{Ht}\) (where \(H = \sqrt{\Lambda/3}\)).

  3. Vacuum energy: The \(\Lambda\) term implies that energy exists in spacetime even in the absence of matter, corresponding to the simplest model of "vacuum energy" or "dark energy."

Verification:

  • Limit \(\Lambda = 0\): Reduces to the ordinary Einstein equation \(G_{\mu\nu} = 8\pi G\,T_{\mu\nu}\). ✓
  • Trace check: Taking the trace of \(G_{\mu\nu} + \Lambda g_{\mu\nu} = 8\pi G T_{\mu\nu}\), with \(g^{\mu\nu}G_{\mu\nu} = -R\) (in 4 dimensions) and \(g^{\mu\nu}g_{\mu\nu} = 4\), gives \(-R + 4\Lambda = 8\pi G T\). In vacuum \(T = 0\), we get \(R = 4\Lambda\). This is the well-known result for the Ricci scalar of de Sitter spacetime. ✓
  • Consistency with the Bianchi identity: Since \(\nabla^\mu G_{\mu\nu} = 0\) and \(\nabla^\mu(\Lambda g_{\mu\nu}) = 0\) (\(\Lambda\) is a constant and \(\nabla^\mu g_{\mu\nu} = 0\)), the energy-momentum conservation law \(\nabla^\mu T_{\mu\nu} = 0\) is automatically satisfied. ✓