Mathematics / Tensor Analysis Differential Invariants & Curvature Tensors 100% Free Open Access
Chapter 8 • Theory & Derivations

Unit 8: Physical Applications: Continuum Mechanics, Electromagnetism & Gravitation

Application of tensor analysis to the fundamental laws of classical physics and general relativity: Cauchy stress and the stress-energy-momentum tensor T^{\mu\nu}, relativistic conservation laws \nabla_\mu T^{\mu\nu} = 0, covariant formulation of Maxwell electrodynamics with electromagnetic field tensor F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu and Lorentz force, the geodesic principle and gravitational time dilation / redshift, derivation and physical interpretation of the Einstein Field Equations G_{\mu\nu} + \Lambda g_{\mu\nu} = (8\pi G / c^4) T_{\mu\nu}, the static spherically symmetric Schwarzschild solution, Mercury's anomalous perihelion precession, and gravitational deflection of light.

§8.1 Stress-Energy-Momentum Tensors & Conservation Laws in Curved Coordinates

1. Cauchy Stress Tensor in Continuum Mechanics

In classical 3D continuum mechanics, internal contact forces distributed across an infinitesimal area element $d\mathbf{S} = \mathbf{n} \, dS$ are characterized by the Cauchy stress tensor $\sigma^{ij}$:

$$dF^i = \sigma^{ij} n_j \, dS$$
  • Conservation of Linear Momentum: In the absence of body forces:
$$\nabla_j \sigma^{ij} = \rho \frac{D v^i}{Dt}$$

where $\nabla_j$ is the covariant derivative associated with the spatial Riemannian metric $g_{ij}$.

  • Conservation of Angular Momentum: By requiring the net torque on an infinitesimal volume element to vanish, the stress tensor must be symmetric:
$$\sigma^{ij} = \sigma^{ji}$$

2. The Relativistic Stress-Energy-Momentum Tensor $T^{\mu\nu}$

In 4-dimensional pseudo-Riemannian spacetime $(M, g_{\mu\nu})$, energy, momentum density, shear stress, and isotropic pressure are unified into a symmetric rank-2 contravariant tensor field $T^{\mu\nu}$:

$$\left( T^{\mu\nu} \right) = \begin{pmatrix} T^{00} & T^{01} & T^{02} & T^{03} \\ T^{10} & T^{11} & T^{12} & T^{13} \\ T^{20} & T^{21} & T^{22} & T^{23} \\ T^{30} & T^{31} & T^{32} & T^{33} \end{pmatrix} = \begin{pmatrix} \text{Energy Density } (\rho c^2) & \text{Energy Flux / } c \\ c \times \text{Momentum Density} & \text{Stress Tensor } (\sigma^{ij}) \end{pmatrix}$$
Perfect Fluid Stress-Energy Tensor:

For an ideal fluid characterized by proper energy density $\rho(x)$, isotropic pressure $p(x)$, and 4-velocity field $u^\mu = \frac{dx^\mu}{d\tau}$ (normalized such that $g_{\mu\nu} u^\mu u^\nu = -c^2$):

$$T^{\mu\nu} = \left( \rho + \frac{p}{c^2} \right) u^\mu u^\nu + p g^{\mu\nu}$$

3. Local Conservation Laws

In special relativity (flat Minkowski spacetime $\eta_{\mu\nu}$), conservation of energy and momentum is expressed as $\partial_\mu T^{\mu\nu} = 0$. According to the principle of minimal gravitational coupling (equivalence principle), the physical conservation law in curved spacetime is obtained by replacing ordinary derivatives with covariant derivatives:

Fundamental Conservation Law:

$$\nabla_\mu T^{\mu\nu} = 0 \iff \frac{1}{\sqrt{-g}} \frac{\partial}{\partial x^\mu} \left( \sqrt{-g} T^{\mu\nu} \right) + \Gamma^\nu_{\mu\alpha} T^{\mu\alpha} = 0$$

Projecting along and orthogonal to the 4-velocity $u^\nu$ yields the relativistic continuity equation and the relativistic Euler equation of fluid dynamics!

§8.2 Covariant Formulation of Maxwell's Electrodynamics ($F_{\mu\nu}$, $\nabla_\mu F^{\mu\nu} = \mu_0 J^\nu$)

1. The 4-Potential and Field Strength Tensor

Classical electrodynamics is governed by electric and magnetic vector fields $\mathbf{E}$ and $\mathbf{B}$, which mix under Lorentz coordinate transformations. Tensor analysis unifies them into a single antisymmetric rank-2 covariant tensor field $F_{\mu\nu}$, called the Faraday (electromagnetic field) tensor.

Let $A_\mu = (-\phi/c, \mathbf{A})$ be the covariant electromagnetic 4-potential.

Definition 8.1 (Electromagnetic Field Tensor):

$$F_{\mu\nu} \equiv \nabla_\mu A_\nu - \nabla_\nu A_\mu = \partial_\mu A_\nu - \partial_\nu A_\mu$$

Because the connection coefficients $\Gamma^\lambda_{\mu\nu} = \Gamma^\lambda_{\nu\mu}$ are symmetric, the Christoffel symbols cancel identically!

In matrix form with metric signature $(-, +, +, +)$:

$$(F_{\mu\nu}) = \begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & B_z & -B_y \\ E_y/c & -B_z & 0 & B_x \\ E_z/c & B_y & -B_x & 0 \end{pmatrix}$$

By raising indices with $g^{\mu\alpha} g^{\nu\beta}$, we obtain the contravariant tensor $F^{\mu\nu}$.


2. Covariant Maxwell Equations

Theorem 8.1 (Maxwell's Equations in Curved Spacetime): The four classical Maxwell equations reduce to two elegant, manifestly covariant tensor equations:

  1. Inhomogeneous Maxwell Equations (Sources):
$$\nabla_\mu F^{\mu\nu} = \mu_0 J^\nu \iff \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} F^{\mu\nu} \right) = \mu_0 J^\nu$$

where $J^\nu = (\rho c, \mathbf{J})$ is the 4-current density.

  1. Homogeneous Maxwell Equations (Bianchi Identity):
$$\nabla_\lambda F_{\mu\nu} + \nabla_\mu F_{\nu\lambda} + \nabla_\nu F_{\lambda\mu} = \partial_\lambda F_{\mu\nu} + \partial_\mu F_{\nu\lambda} + \partial_\nu F_{\lambda\mu} = 0$$
Automatic Charge Conservation:

Taking the covariant divergence of the inhomogeneous equation:

$$\mu_0 \nabla_\nu J^\nu = \nabla_\nu \nabla_\mu F^{\mu\nu} = \frac{1}{2} [\nabla_\nu, \nabla_\mu] F^{\mu\nu} = 0$$

because $F^{\mu\nu}$ is antisymmetric while the commutator contraction is symmetric. Thus, electric charge conservation $\nabla_\nu J^\nu = 0$ is an exact geometric identity!


3. The Lorentz Force Density

The electromagnetic 4-force density acting on charged matter is:

$$f^\mu = F^{\mu\nu} J_\nu$$

which unifies Coulomb electric force and magnetic Lorentz force $\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$.

§8.3 The Geodesic Principle & Gravitational Time Dilation / Redshift

1. The Geodesic Hypothesis

In Einstein's theory of General Relativity, gravity is not a physical Newtonian force, but the curvature of spacetime. Free particles experiencing no non-gravitational forces follow geodesics of the spacetime metric:

$$\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu_{\alpha\beta} \frac{dx^\alpha}{d\tau} \frac{dx^\beta}{d\tau} = 0$$

where $\tau$ is the proper time experienced by a clock moving along the worldline:

$$c^2 d\tau^2 = -g_{\mu\nu} dx^\mu dx^\nu$$

2. Gravitational Time Dilation

Consider a static gravitational field with metric:

$$ds^2 = g_{00}(x) (dx^0)^2 + g_{ij}(x) dx^i dx^j = -c^2 d\tau^2$$

For a stationary observer situated at fixed spatial coordinates ($dx^i = 0$):

$$-c^2 d\tau^2 = g_{00}(x) c^2 dt^2 \implies d\tau = \sqrt{-g_{00}(x)} \, dt$$

Let two clocks be stationed at positions $A$ and $B$ where the metric components are $g_{00}(A)$ and $g_{00}(B)$:

$$\frac{d\tau_A}{d\tau_B} = \frac{\sqrt{-g_{00}(A)}}{\sqrt{-g_{00}(B)}}$$

In a weak Newtonian gravitational potential $\Phi(x)$, $g_{00} \approx -\left( 1 + \frac{2\Phi}{c^2} \right)$:

$$d\tau \approx \left( 1 + \frac{\Phi}{c^2} \right) dt$$

Clocks situated deeper in a gravitational well ($\Phi < 0$) run measurably slower than clocks higher up!


3. Gravitational Redshift of Light

A light wave emitted at position $A$ with frequency $\nu_A$ and received at position $B$ undergoes a gravitational frequency shift:

$$\frac{\nu_B}{\nu_A} = \frac{\sqrt{-g_{00}(A)}}{\sqrt{-g_{00}(B)}}$$

For light escaping from the surface of a star of mass $M$ and radius $R$ ($g_{00} = -\left(1 - \frac{2GM}{c^2 R}\right)$) to a distant observer ($g_{00}(\infty) = -1$):

$$\nu_\infty = \nu_{\text{emit}} \sqrt{1 - \frac{2GM}{c^2 R}} < \nu_{\text{emit}}$$

The light is redshifted ($\Delta \lambda > 0$), directly confirming the curvature of the metric tensor!

§8.4 The Einstein Field Equations $G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$

1. The Search for the Gravitational Field Equations

In Newtonian gravity, the gravitational potential $\Phi$ satisfies Poisson's equation:

$$\nabla^2 \Phi = 4\pi G \rho$$

Einstein sought a tensorial equation of the form:

$$\mathcal{G}_{\mu\nu} = \kappa T_{\mu\nu}$$

where $\mathcal{G}_{\mu\nu}$ is a symmetric rank-2 geometric tensor constructed from the metric $g_{\mu\nu}$ and its first and second derivatives.

Essential Constraints:
  1. Since energy-momentum is conserved ($\nabla^\mu T_{\mu\nu} = 0$), the geometric tensor must satisfy the divergence-free identity:
$$\nabla^\mu \mathcal{G}_{\mu\nu} = 0$$
  1. In 1915, David Hilbert and Albert Einstein proved that the unique symmetric tensor containing at most second derivatives of the metric that satisfies this condition (Lovelock's theorem in 4D) is:
$$\mathcal{G}_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} + \Lambda g_{\mu\nu} = G_{\mu\nu} + \Lambda g_{\mu\nu}$$

2. The Einstein Field Equations

Definition 8.2 (The Einstein Field Equations):

$$G_{\mu\nu} + \Lambda g_{\mu\nu} = \frac{8\pi G}{c^4} T_{\mu\nu}$$

where:

  • $G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu}$ is the Einstein tensor.
  • $\Lambda$ is the cosmological constant.
  • $G = 6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}$ is Newton's gravitational constant.
  • $c$ is the speed of light.
  • $\kappa = \frac{8\pi G}{c^4} \approx 2.076 \times 10^{-43} \, \text{N}^{-1}$ is Einstein's gravitational coupling constant.

3. Trace-Reversed Form and Vacuum Equations

Taking the trace with $g^{\mu\nu}$:

$$g^{\mu\nu} G_{\mu\nu} + 4\Lambda = \kappa g^{\mu\nu} T_{\mu\nu} \implies -R + 4\Lambda = \kappa T \implies R = -\kappa T + 4\Lambda$$

Substituting back into $G_{\mu\nu}$:

$$R_{\mu\nu} = \frac{8\pi G}{c^4} \left( T_{\mu\nu} - \frac{1}{2} T g_{\mu\nu} \right) + \Lambda g_{\mu\nu}$$
Vacuum Field Equations ($\Lambda = 0, T_{\mu\nu} = 0$):

In empty spacetime outside mass distributions:

$$R_{\mu\nu} = 0$$

Spacetimes satisfying $R_{\mu\nu} = 0$ are called Ricci-flat. Crucially, $R_{\mu\nu} = 0$ does NOT imply flat spacetime! The full Riemann tensor $R^\rho_{\mu\sigma\nu}$ can still be non-zero through its Weyl tensor component, propagating gravitational waves and mediating gravitational attraction!

§8.5 The Schwarzschild Metric, Planetary Perihelion Precession & Gravitational Deflection

1. The Schwarzschild Solution (1916)

Karl Schwarzschild derived the exact vacuum solution ($R_{\mu\nu} = 0$) for a static, spherically symmetric mass $M$:

Definition 8.3 (Schwarzschild Metric):

$$ds^2 = -\left( 1 - \frac{r_s}{r} \right) c^2 dt^2 + \left( 1 - \frac{r_s}{r} \right)^{-1} dr^2 + r^2 \left( d\theta^2 + \sin^2\theta \, d\phi^2 \right)$$

where $r_s = \frac{2GM}{c^2}$ is the Schwarzschild radius (event horizon). For the Sun: $r_s \approx 2.95 \text{ km}$; for the Earth: $r_s \approx 8.87 \text{ mm}$.


2. Relativistic Planetary Orbits and Perihelion Precession

Consider motion confined to the equatorial plane $\theta = \frac{\pi}{2}$. From the geodesic equations, energy per unit mass $E$ and angular momentum per unit mass $L$ are conserved:

$$\left( 1 - \frac{r_s}{r} \right) c \frac{dt}{d\tau} = \frac{E}{c}, \qquad r^2 \frac{d\phi}{d\tau} = L$$

Substituting into the metric $g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu = -c^2$:

$$\frac{1}{2} \left(\frac{dr}{d\tau}\right)^2 + V_{\text{eff}}(r) = \frac{E^2 - c^4}{2c^2}$$

where the relativistic effective potential is:

$$V_{\text{eff}}(r) = -\frac{GM}{r} + \frac{L^2}{2r^2} - \frac{G M L^2}{c^2 r^3}$$

The third term $-\frac{GML^2}{c^2 r^3}$ is the general relativistic correction to Newtonian gravity.

Using $u = 1/r$, the orbital Binet equation becomes:

$$\frac{d^2 u}{d\phi^2} + u = \frac{GM}{L^2} + \frac{3GM}{c^2} u^2$$

By perturbation theory, an elliptical orbit precesses by an angle per revolution:

$$\Delta\phi = \frac{6\pi GM}{c^2 a (1 - e^2)}$$

For Mercury: $a = 5.79 \times 10^{10} \text{ m}, e = 0.2056$. This yields $\Delta\phi = 42.98'' \text{ per century}$, matching the observed anomalous 43 arcseconds per century unexplained by Newtonian planetary perturbations!


3. Gravitational Deflection of Starlight

For null geodesics (photons, $ds^2 = 0$), the orbital equation with impact parameter $b$ gives:

$$\frac{d^2 u}{d\phi^2} + u = \frac{3GM}{c^2} u^2$$

Integrating the perturbation from $\phi = -\pi/2$ to $\pi/2$ for light grazing the Sun at radius $R$:

Deflection Angle Formula:

$$\delta\theta = \frac{4GM}{c^2 R_{\odot}}$$

For the Sun ($M = M_\odot, R = R_\odot$):

$$\delta\theta \approx 1.751'' \text{ (arcseconds)}$$

Exactly twice the naive Newtonian corpuscular value ($0.875''$)! This famous prediction was confirmed by Arthur Eddington's 1919 solar eclipse expedition, providing historic observational confirmation of general relativity.

Rigorous Tiered Solved Examination Problems

Step-by-step unskipped derivations, complete proofs, and verification across Foundational, Advanced, and Honors tiers.

Fundamentals Example 8.1: Problem 8.1: Relativistic Energy-Momentum Tensor of Dust and Energy Density

Consider a pressureless cloud of non-interacting particles ('dust') in spacetime with proper mass density $\rho_0$ and 4-velocity field $u^\mu = \frac{dx^\mu}{d\tau}$. Its stress-energy tensor is given by:

$$T^{\mu\nu} = \rho_0 u^\mu u^\nu$$
  1. Verify that $T^{\mu\nu}$ is symmetric.
  2. In the rest frame of the dust, where $u^\mu = (c, 0, 0, 0)$ in Minkowski spacetime $\eta_{\mu\nu} = \text{diag}(-1, 1, 1, 1)$, write down the matrix components of $T^{\mu\nu}$ and $T^\mu_\nu$.
  3. Compute the scalar trace $T = g_{\mu\nu} T^{\mu\nu}$.
  4. Show that the conservation law $\nabla_\mu T^{\mu\nu} = 0$ decomposes into the continuity equation $\nabla_\mu (\rho_0 u^\mu) = 0$ and the geodesic equation $u^\mu \nabla_\mu u^\nu = 0$.

1. Symmetry

$$T^{\nu\mu} = \rho_0 u^\nu u^\mu = \rho_0 u^\mu u^\nu = T^{\mu\nu}$$

Symmetric by definition. $\blacksquare$


2. Matrix Components in Rest Frame

With $u^\mu = (c, 0, 0, 0)$:

$$T^{00} = \rho_0 c^2, \qquad T^{0i} = T^{i0} = T^{ij} = 0$$
$$(T^{\mu\nu}) = \begin{pmatrix} \rho_0 c^2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Lowering the second index with $\eta_{\nu\lambda} = \text{diag}(-1, 1, 1, 1)$:

$$T^0_0 = \eta_{00} T^{00} = -1(\rho_0 c^2) = -\rho_0 c^2$$
$$(T^\mu_\nu) = \begin{pmatrix} -\rho_0 c^2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \qquad \blacksquare$$

3. Trace $T$

$$T = T^\mu_\mu = g_{\mu\nu} T^{\mu\nu} = \rho_0 (g_{\mu\nu} u^\mu u^\nu) = \rho_0 (-c^2) = -\rho_0 c^2 \qquad \blacksquare$$

4. Decomposition of Conservation Law

Expand $\nabla_\mu T^{\mu\nu} = 0$:

$$\nabla_\mu (\rho_0 u^\mu u^\nu) = \left[ \nabla_\mu (\rho_0 u^\mu) \right] u^\nu + \rho_0 u^\mu \nabla_\mu u^\nu = 0$$

Contract this equation with the 4-velocity covector $u_\nu$:

$$\left[ \nabla_\mu (\rho_0 u^\mu) \right] (u^\nu u_\nu) + \rho_0 u^\mu (u_\nu \nabla_\mu u^\nu) = 0$$

Since $u^\nu u_\nu = -c^2 = \text{const}$:

$$u_\nu \nabla_\mu u^\nu = \frac{1}{2} \nabla_\mu (u_\nu u^\nu) = \frac{1}{2} \nabla_\mu (-c^2) = 0$$

Thus, the second term vanishes identically:

$$\left[ \nabla_\mu (\rho_0 u^\mu) \right] (-c^2) + 0 = 0 \implies \nabla_\mu (\rho_0 u^\mu) = 0$$

This is the relativistic mass-continuity equation (conservation of particle number).

Now substitute $\nabla_\mu (\rho_0 u^\mu) = 0$ back into the original conservation equation:

$$0 \cdot u^\nu + \rho_0 u^\mu \nabla_\mu u^\nu = 0 \implies u^\mu \nabla_\mu u^\nu = 0$$

This is the geodesic equation for the dust worldlines $\frac{D u^\nu}{d\tau} = 0$! Energy-momentum conservation automatically forces pressureless dust particles to follow spacetime geodesics! $\blacksquare$

Computational Triad Example 8.2: Problem 8.2: Electromagnetic Energy-Momentum Tensor and Trace-Free Property

The electromagnetic stress-energy tensor in 4D spacetime is defined as:

$$T^{\mu\nu}_{\text{EM}} = \frac{1}{\mu_0} \left( F^{\mu\alpha} F^\nu_{\ \alpha} - \frac{1}{4} g^{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} \right)$$
  1. Prove that $T^{\mu\nu}_{\text{EM}}$ is strictly symmetric.
  2. Prove that in $n = 4$ spacetime dimensions, the trace of $T^{\mu\nu}_{\text{EM}}$ vanishes identically: $T = g_{\mu\nu} T^{\mu\nu}_{\text{EM}} = 0$.
  3. Use Maxwell's equations $\nabla_\mu F^{\mu\nu} = \mu_0 J^\nu$ and $\nabla_{[\lambda} F_{\mu\nu]} = 0$ to calculate $\nabla_\mu T^{\mu\nu}_{\text{EM}}$ and show that $\nabla_\mu T^{\mu\nu}_{\text{EM}} = -F^{\nu\alpha} J_\alpha$.
  4. What physical principle does $\nabla_\mu (T^{\mu\nu}_{\text{matter}} + T^{\mu\nu}_{\text{EM}}) = 0$ express?

1. Proof of Symmetry

$$F^{\mu\alpha} F^\nu_{\ \alpha} = F^{\mu\alpha} g_{\alpha\beta} F^{\nu\beta} = g_{\alpha\beta} F^{\mu\alpha} F^{\nu\beta}$$

Using antisymmetry $F^{\mu\alpha} = -F^{\alpha\mu}$ and $F^{\nu\beta} = -F^{\beta\nu}$:

$$g_{\alpha\beta} (-F^{\alpha\mu})(-F^{\beta\nu}) = g_{\alpha\beta} F^{\alpha\mu} F^{\beta\nu} = F^\beta_{\ \mu} F^{\nu\beta} = F^{\nu\beta} F_{\mu\beta} = F^{\nu\alpha} F^\mu_{\ \alpha}$$

Thus $T^{\mu\nu}_{\text{EM}} = T^{\nu\mu}_{\text{EM}}$. $\blacksquare$


2. Trace-Free Property in 4 Dimensions

$$T = g_{\mu\nu} T^{\mu\nu}_{\text{EM}} = \frac{1}{\mu_0} \left( g_{\mu\nu} F^{\mu\alpha} F^\nu_{\ \alpha} - \frac{1}{4} g_{\mu\nu} g^{\mu\nu} F_{\alpha\beta} F^{\alpha\beta} \right)$$

Evaluate each term:

  • $g_{\mu\nu} F^{\mu\alpha} F^\nu_{\ \alpha} = F_\nu^{\ \alpha} F^\nu_{\ \alpha} = F^{\alpha\nu} F_{\alpha\nu} = F_{\alpha\beta} F^{\alpha\beta}$.
  • In 4D, $g_{\mu\nu} g^{\mu\nu} = \delta^\mu_\mu = 4$.

Substituting:

$$T = \frac{1}{\mu_0} \left( F_{\alpha\beta} F^{\alpha\beta} - \frac{1}{4}(4) F_{\alpha\beta} F^{\alpha\beta} \right) = \frac{1}{\mu_0} \left( F_{\alpha\beta} F^{\alpha\beta} - F_{\alpha\beta} F^{\alpha\beta} \right) = 0 \qquad \blacksquare$$

Because $T \equiv 0$, pure radiation has zero gravitational trace, meaning $R = 0$ in the presence of electromagnetic fields!


3. Covariant Divergence

$$\nabla_\mu T^{\mu\nu}_{\text{EM}} = \frac{1}{\mu_0} \left[ (\nabla_\mu F^{\mu\alpha}) F^\nu_{\ \alpha} + F^{\mu\alpha} \nabla_\mu F^\nu_{\ \alpha} - \frac{1}{4} g^{\mu\nu} \nabla_\mu (F_{\alpha\beta} F^{\alpha\beta}) \right]$$

Using $\nabla_\mu F^{\mu\alpha} = \mu_0 J^\alpha$:

$$\frac{1}{\mu_0} (\mu_0 J^\alpha) F^\nu_{\ \alpha} = F^\nu_{\ \alpha} J^\alpha = -F^{\nu\alpha} J_\alpha = -f^\nu_{\text{Lorentz}}$$

The remaining terms cancel by the homogeneous Maxwell equation $\nabla_{[\mu} F_{\alpha\beta]} = 0$:

$$F^{\mu\alpha} \nabla_\mu F^\nu_{\ \alpha} - \frac{1}{2} F^{\alpha\beta} \nabla^\nu F_{\alpha\beta} = 0$$

Therefore:

$$\nabla_\mu T^{\mu\nu}_{\text{EM}} = -F^{\nu\alpha} J_\alpha \qquad \blacksquare$$

4. Physical Conservation Principle

Since $\nabla_\mu T^{\mu\nu}_{\text{matter}} = +F^{\nu\alpha} J_\alpha$ (Lorentz force acting on matter):

$$\nabla_\mu \left( T^{\mu\nu}_{\text{matter}} + T^{\mu\nu}_{\text{EM}} \right) = F^{\nu\alpha} J_\alpha - F^{\nu\alpha} J_\alpha = 0$$

This expresses total energy-momentum conservation: whatever energy and momentum the electromagnetic field loses is transferred into kinetic energy and momentum of the charged matter! $\blacksquare$

Honors / Proof Challenge Example 8.3: Problem 8.3: Derivation of the Schwarzschild Metric from Vacuum Einstein Equations

Consider a general static, spherically symmetric spacetime metric in spherical coordinates $(t, r, \theta, \phi) = (x^0, x^1, x^2, x^3)$:

$$ds^2 = -e^{2\Phi(r)} c^2 dt^2 + e^{2\Lambda(r)} dr^2 + r^2 (d\theta^2 + \sin^2\theta \, d\phi^2)$$

where $\Phi(r)$ and $\Lambda(r)$ are unknown radial metric functions.

  1. Compute the non-zero Christoffel symbols $\Gamma^\lambda_{\mu\nu}$ for this metric.
  2. Compute the Ricci tensor components $R_{00}$, $R_{11}$, and $R_{22}$.
  3. Impose the vacuum Einstein field equations $R_{\mu\nu} = 0$:
  • Combine $R_{00}$ and $R_{11}$ to show that $\Phi'(r) + \Lambda'(r) = 0 \implies \Phi(r) = -\Lambda(r) + \text{const}$.
  • Use $R_{22} = 0$ to solve for $e^{-2\Lambda(r)}$.
  1. Determine the integration constant from the Newtonian weak-field limit and arrive at the exact Schwarzschild metric.

1. Christoffel Symbols

Using $\Gamma^\lambda_{\mu\nu} = \frac{1}{2} g^{\lambda\sigma} (\partial_\mu g_{\nu\sigma} + \partial_\nu g_{\mu\sigma} - \partial_\sigma g_{\mu\nu})$:

  • $\Gamma^0_{01} = \Phi'$, $\Gamma^1_{00} = c^2 \Phi' e^{2(\Phi - \Lambda)}$
  • $\Gamma^1_{11} = \Lambda'$, $\Gamma^1_{22} = -r e^{-2\Lambda}$, $\Gamma^1_{33} = -r \sin^2\theta e^{-2\Lambda}$
  • $\Gamma^2_{12} = \frac{1}{r}$, $\Gamma^2_{33} = -\sin\theta\cos\theta$
  • $\Gamma^3_{13} = \frac{1}{r}$, $\Gamma^3_{23} = \cot\theta$

2. Ricci Tensor Components

Evaluating $R_{\mu\nu} = \partial_\lambda \Gamma^\lambda_{\mu\nu} - \partial_\nu \Gamma^\lambda_{\mu\lambda} + \Gamma^\lambda_{\mu\nu} \Gamma^\sigma_{\lambda\sigma} - \Gamma^\lambda_{\mu\sigma} \Gamma^\sigma_{\nu\lambda}$:

$$\begin{aligned} R_{00} &= c^2 e^{2(\Phi - \Lambda)} \left[ \Phi'' + (\Phi')^2 - \Phi' \Lambda' + \frac{2}{r} \Phi' \right] \\ R_{11} &= -\left[ \Phi'' + (\Phi')^2 - \Phi' \Lambda' - \frac{2}{r} \Lambda' \right] \\ R_{22} &= e^{-2\Lambda} \left[ r(\Lambda' - \Phi') - 1 \right] + 1 \\ R_{33} &= R_{22} \sin^2\theta \end{aligned}$$

3. Vacuum Field Equations $R_{\mu\nu} = 0$

Step A: Sum of $R_{00}$ and $R_{11}$

Consider the linear combination:

$$\frac{e^{-2(\Phi - \Lambda)}}{c^2} R_{00} + R_{11} = \left[ \Phi'' + (\Phi')^2 - \Phi'\Lambda' + \frac{2}{r}\Phi' \right] - \left[ \Phi'' + (\Phi')^2 - \Phi'\Lambda' - \frac{2}{r}\Lambda' \right]$$
$$\frac{e^{-2(\Phi - \Lambda)}}{c^2} R_{00} + R_{11} = \frac{2}{r} (\Phi' + \Lambda') = 0$$

Since $r \ne 0$:

$$\Phi'(r) + \Lambda'(r) = 0 \implies \Phi(r) + \Lambda(r) = C_0$$

As $r \to \infty$, spacetime must be flat Minkowski space: $\Phi(\infty) = 0$ and $\Lambda(\infty) = 0 \implies C_0 = 0$. Thus:

$$\Phi(r) = -\Lambda(r) \implies e^{2\Phi} = e^{-2\Lambda}$$
Step B: Solve $R_{22} = 0$

Substituting $\Phi' = -\Lambda'$ into $R_{22}$:

$$R_{22} = e^{-2\Lambda} [ r(2\Lambda') - 1 ] + 1 = 0$$

Notice that:

$$\frac{d}{dr}\left( r e^{-2\Lambda} \right) = e^{-2\Lambda} - 2 r \Lambda' e^{-2\Lambda} = - \left( e^{-2\Lambda} [ 2 r \Lambda' - 1 ] \right)$$

Therefore, $R_{22} = 0$ is equivalent to:

$$1 - \frac{d}{dr}\left( r e^{-2\Lambda} \right) = 0 \implies \frac{d}{dr}\left( r e^{-2\Lambda} \right) = 1$$

Integrating with respect to $r$:

$$r e^{-2\Lambda} = r - r_s \implies e^{-2\Lambda(r)} = 1 - \frac{r_s}{r}$$

where $r_s$ is a constant of integration. Since $e^{2\Phi} = e^{-2\Lambda}$:

$$e^{2\Phi(r)} = 1 - \frac{r_s}{r}$$

4. Determination of Integration Constant $r_s$

At large distances $r \gg r_s$, the Newtonian limit requires:

$$g_{00} = -c^2 e^{2\Phi} \approx -c^2 \left( 1 + \frac{2\Phi_{\text{Newton}}}{c^2} \right) = -c^2 \left( 1 - \frac{2GM}{c^2 r} \right)$$

Comparing with $e^{2\Phi} = 1 - \frac{r_s}{r}$:

$$r_s = \frac{2GM}{c^2}$$

Substituting back into the line element:

$$ds^2 = -\left( 1 - \frac{2GM}{c^2 r} \right) c^2 dt^2 + \left( 1 - \frac{2GM}{c^2 r} \right)^{-1} dr^2 + r^2 \left( d\theta^2 + \sin^2\theta \, d\phi^2 \right)$$

This completes the exact, closed-form derivation of the Schwarzschild Metric! $\blacksquare$