Fields and Stress-Energy Bridge

Einstein gravity couples geometry to matter through stress–energy, but that tensor is not guessed from a list of densities. It follows from a matter model, its action, and metric variation. This bridge checks the chain from field action to equation of motion, symmetry or gauge freedom, stress–energy, observer measurements, and local balance. Full field theory remains with Physical Foundations and General Relativity.

Required background. Calculus and Linear Algebra for Relativity supplies index manipulation, differentiation, and integration by parts. Helpful background. Variational Mechanics and Symmetries supplies stationary action and Noether reasoning.

You should leave able to relate actions, symmetries, stress–energy, and conservation, then perform a changed metric-variation or balance check.

Fields and stress-energy diagnostic

Use flat spacetime with the inherited $(-,+,+,+)$ signature unless a prompt says otherwise.

1. For a real scalar field, $$S[\phi]=\int d^4x\left[-\frac12\partial_\mu\phi\,\partial^\mu\phi-V(\phi)\right],$$ derive the Euler–Lagrange equation, retaining the boundary term until its condition is stated.
2. Electromagnetism uses $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$. Show that $F_{\mu\nu}$ is unchanged by $A_\mu\mapsto A_\mu+\partial_\mu\chi$. Explain why this is a redundancy rather than a new physical solution with a different field strength.
3. The scalar stress–energy tensor is $$T_{\mu\nu}=\partial_\mu\phi\,\partial_\nu\phi -g_{\mu\nu}\left(\frac12\partial_\alpha\phi\,\partial^\alpha\phi+V\right).$$ For a spatially homogeneous field, compute the energy density measured by the inertial observer $u^\mu=(1,0,0,0)$.
4. Show in flat spacetime that the divergence of this $T_{\mu\nu}$ vanishes when the scalar equation of motion holds.

Hints and solutions

Hints for 1

Vary $\phi$, integrate the derivative on $\delta\phi$ by parts, and state either compact support for $\delta\phi$ or a boundary condition that kills the surface term.

Solution to 1

The first variation is

$$\delta S=\int d^4x\left[-\partial^\mu\phi\,\partial_\mu\delta\phi-V'(\phi)\delta\phi\right].$$

Integration by parts gives

$$\delta S=\int d^4x\,[\Box\phi-V'(\phi)]\delta\phi -\int_{\partial\mathcal M}d\Sigma_\mu\,\partial^\mu\phi\,\delta\phi.$$

For compactly supported variations, fixed boundary value $\delta\phi|_{\partial\mathcal M}=0$, or sufficient falloff, the boundary term vanishes. Arbitrary interior variations then imply

$$\Box\phi-V'(\phi)=0.$$

For $V=m^2\phi^2/2$, this is $(\Box-m^2)\phi=0$, consistent with $\Box=-\partial_t^2+\nabla^2$.

Hints for 2

Substitute the shifted potential and use commutativity of ordinary partial derivatives on a smooth scalar.

Solution to 2

The transformed field strength is

$$F'_{\mu\nu}=F_{\mu\nu} +\partial_\mu\partial_\nu\chi -\partial_\nu\partial_\mu\chi =F_{\mu\nu}.$$

Thus the electric and magnetic fields and gauge-invariant observables are unchanged. The two potentials are different representatives of the same physical configuration, subject to global and boundary qualifications. Gauge freedom cannot be counted as an extra propagating degree of freedom.

Hints for 3

For $\partial_i\phi=0$, first compute $\partial_\alpha\phi\,\partial^\alpha\phi$. The measured density is $T_{\mu\nu}u^\mu u^\nu=T_{00}$.

Solution to 3

Homogeneity gives $\partial_\alpha\phi\,\partial^\alpha\phi=-\dot\phi^2$. Therefore

$$T_{00}=\dot\phi^2-g_{00}\left(-\frac12\dot\phi^2+V\right) =\frac12\dot\phi^2+V(\phi).$$

The kinetic term is positive. This is a useful sign check on both the metric and stress–energy convention.

Hints for 4

Differentiate both terms, use symmetry of second derivatives, and look for a factor multiplying $\partial_\nu\phi$.

Solution to 4

With constant metric,

$$\partial^\mu T_{\mu\nu} =(\Box\phi)\partial_\nu\phi +\partial_\mu\phi\,\partial^\mu\partial_\nu\phi -\partial_\nu\left(\frac12\partial_\alpha\phi\,\partial^\alpha\phi+V\right).$$

The two terms containing second derivatives cancel because

$$\partial_\nu\left(\frac12\partial_\alpha\phi\,\partial^\alpha\phi\right) =\partial_\alpha\phi\,\partial_\nu\partial^\alpha\phi.$$

Hence

$$\partial^\mu T_{\mu\nu} =[\Box\phi-V'(\phi)]\partial_\nu\phi=0$$

on solutions. Conservation is therefore linked to the field equation; it is not an independent algebraic property of an arbitrary $\phi$.

Interpret the result

• Secure: you can pass from an action to the equation and boundary term, distinguish gauge representatives from observables, compute an observer’s density, and prove on-shell local balance.
• Uncertain: one link in that chain needs a hint. Review only that link and repeat it for a different potential or field configuration.
• Not yet: the boundary term is silently discarded, gauge freedom is counted as physical, or $T_{00}$ is read without the metric and observer. Complete the focused review before studying matter-coupled Einstein equations.

Questions 1–2 sample classical-field reasoning; questions 3–4 sample stress–energy and conservation.

Focused review

Action, boundary data, and equations

The variation of a local action has a bulk Euler–Lagrange term and a boundary term. A well-posed variational principle specifies what is fixed or which boundary term is added. Classical Fields and Their Actions develops this structure and distinguishes ordinary global symmetries from gauge redundancy.

Stress–energy from metric variation

The site’s convention is

$$T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_{\rm m}}{\delta g^{\mu\nu}}.$$

This Hilbert tensor is symmetric for a diffeomorphism-covariant matter action and is the object that couples to the Einstein tensor. The derivation, including variation with respect to the inverse metric, belongs to Flat-Background Hilbert Stress-Energy from Metric Variation. Canonical, improved, effective, and gravitational pseudotensors must not be conflated with it.

Measurements and balance

An observer $u^\mu$ measures energy density $\rho=T_{\mu\nu}u^\mu u^\nu$, spatial energy flux by one orthogonal projection, and stress by two. In curved spacetime the local balance law is $\nabla_\mu T^{\mu\nu}=0$ on the coupled matter equations. This is local; a global conserved energy additionally needs suitable symmetry and boundary conditions.

Transfer task: a scalar wave carries energy

For $V=m^2\phi^2/2$, consider the plane wave

$$\phi=A\cos(kx-\omega t).$$

Find the dispersion relation required by the field equation. Compute $T_{00}$ and $T_{0x}$, average them over one period, and verify the relation between average energy flux and average energy density.

Hint 1

Apply $(\Box-m^2)$ to the wave. For the stress tensor, use $\dot\phi=A\omega\sin(kx-\omega t)$ and $\partial_x\phi=-Ak\sin(kx-\omega t)$.

Hint 2

$T_{0x}=\partial_0\phi\,\partial_x\phi$ because $g_{0x}=0$. Use $\langle\sin^2\rangle=\langle\cos^2\rangle=1/2$ and raise the first index carefully when interpreting flux.

Solution

The field equation gives

$$\omega^2=k^2+m^2.$$

The energy density is

$$T_{00}=\frac12\left(\dot\phi^2+(\partial_x\phi)^2+m^2\phi^2\right).$$

Averaging and using the dispersion relation,

$$\langle T_{00}\rangle =\frac{A^2}{4}(\omega^2+k^2+m^2) =\frac12A^2\omega^2.$$

Meanwhile

$$T_{0x}=-A^2\omega k\sin^2(kx-\omega t), \qquad \langle T^{0x}\rangle=-\langle T_{0x}\rangle =\frac12A^2\omega k.$$

Thus

$$\langle T^{0x}\rangle =\frac{k}{\omega}\langle T^{00}\rangle.$$

The ratio $k/\omega$ is the group velocity for this dispersion relation. It approaches one in the massless limit and is below one for $m\gt0$, providing both a physical and dimensional check.

Common errors

• Varying a derivative term without retaining its surface contribution.
• Treating gauge fixing as if it changed gauge-invariant field strength.
• Confusing canonical and Hilbert stress–energy without stating which one is used.
• Inferring a globally conserved energy from $\nabla_\mu T^{\mu\nu}=0$ without a symmetry and boundary conditions.

Return to your route

After the transfer task, return to Relativistic Matter and Compact Objects if stress–energy was the gap, to Gravity Beyond General Relativity if field actions were the gap, or to the First Graduate General Relativity course if this was course preparation.