Scalar and Electromagnetic Field Sources

Canonical scalar and Maxwell fields are both obtained from local covariant actions, but they source Ricci curvature differently. A minimally coupled scalar generally has nonzero trace, including for a massless scalar in four dimensions, whereas the Maxwell tensor is traceless specifically in four spacetime dimensions. Their observer energy fluxes also have different structure: scalar flux follows the product of temporal and spatial field gradients, while electromagnetic flux is the Poynting vector. The discussion here is classical and uses the canonical two-derivative actions.

Required background. Classical Fields and Their Actions provides scalar and gauge-field action variation; Flat-Background Hilbert Stress-Energy from Metric Variation provides metric response and observer decomposition; Affine Connections and Covariant Derivatives provides covariant divergence and curvature commutators.

Helpful background. Events, Intervals, and Minkowski Geometry reviews local Lorentz frames; Tensor Fields and Natural Operations reviews forms and tensor traces; Continuum Kinematics and Balance Laws supplies energy-flux interpretation; Presymplectic Potentials and Currents prepares field-space and boundary variations.

Canonical scalar source

For a real scalar with potential $V(φ)$,

$$S_φ =-\int d^Dx\,\sqrt{-g} \left[ \frac12 g^{\mu\nu}\nabla_\muφ\nabla_\nuφ +V(φ) \right].$$

Varying φ, integrating by parts, and fixing the boundary variation gives

$$\delta_φ S_φ =\int d^Dx\,\sqrt{-g} \left(\Boxφ-V'(φ)\right)\deltaφ,$$

so the field equation is $\Boxφ-V'=0$. The boundary assumption and complete derivation are given in Reall 2013, §10.1, pp. 115–116, equations (10.1)–(10.6), PDF.

Metric variation uses $\delta\sqrt{-g}=-\tfrac12\sqrt{-g}g_{\mu\nu}\delta g^{\mu\nu}$ and yields

$$T^{(φ)}_{\mu\nu} =\nabla_\muφ\nabla_\nuφ -g_{\mu\nu} \left[ \frac12(\nablaφ)^2+V \right].$$

Its divergence is a direct equation-of-motion check:

$$\nabla_\mu T^{(φ)\mu}{}_{\nu} =\left(\Boxφ-V'\right)\nabla_\nuφ.$$

For an observer $u^\mu$, write $D_\muφ=h_\mu{}^\nu\nabla_\nuφ$ and $\dotφ=u^\mu\nabla_\muφ$. Then

$$\rho_φ =\frac12\dotφ^2 +\frac12D_\muφ D^\muφ+V, \qquad q^{(φ)}_\mu=-\dotφ\,D_\muφ.$$

Thus an inhomogeneous evolving scalar transports energy; a spatially homogeneous scalar has no flux for its comoving observer.

Maxwell source from metric variation

Let $F_{\mu\nu}=2\nabla_{[\mu}A_{\nu]}$ and

$$S_{\rm EM} =-\frac14\int d^Dx\,\sqrt{-g}\, F_{\rho\sigma}F^{\rho\sigma}.$$

Hold the one-form $A_\mu$ fixed in the metric variation. Antisymmetry makes $F_{\mu\nu}$ independent of the Levi-Civita connection, while

$$\delta(F_{\rho\sigma}F^{\rho\sigma}) =2F_{\mu\rho}F_\nu{}^\rho\delta g^{\mu\nu}.$$

Combining this with the determinant variation gives

$$\boxed{ T^{({\rm EM})}_{\mu\nu} =F_{\mu\rho}F_\nu{}^\rho -\frac14g_{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}.}$$

Varying $A_\mu$ gives $\nabla_\mu F^{\mu\nu}=0$, supplemented by $\nabla_{[\mu}F_{\nu\rho]}=0$. These equations imply $\nabla_\mu T^{({\rm EM})\mu\nu}=0$. Reall presents their minimally coupled form and the local energy, momentum-flux, and stress interpretation in Reall 2013, §§5.1–5.2, pp. 49–53, PDF.

For an observer, define the spatial electric and magnetic fields by $E_\mu=F_{\mu\nu}u^\nu$ and $B_\mu=(\star F)_{\mu\nu}u^\nu$. In four dimensions,

$$\rho_{\rm EM}=\frac12(E_\mu E^\mu+B_\mu B^\mu).$$

The invariant energy flux is $q_\mu=-h_\mu{}^\alpha T^{({\rm EM})}_{\alpha\beta}u^\beta$. To make its cross-product sign unambiguous, define the observer’s spatial volume form and product by

$$\varepsilon_{\mu\nu\rho}=\epsilon_{\mu\nu\rho\sigma}u^\sigma, \qquad (E\times B)_\mu=\varepsilon_{\mu\nu\rho}E^\nu B^\rho.$$

Then $q_\mu=(E\times B)_\mu$. With the stated dual convention, $\varepsilon_{123}=-1$ in a tetrad with $\epsilon_{0123}=+1$ and $u=e_{(0)}$; using instead a three-dimensional symbol with $+1$ requires reversing the magnetic-field convention as well. The tensor expression for $q_\mu$ is independent of that notation choice.

For the sourced convention used on Maxwell Fields in Curved Spacetime, $\nabla_\mu F^{\mu\nu}=-J^\nu$ and charged matter feels $f^\nu=F^\nu{}_{\rho}J^\rho$. Differentiating the electromagnetic tensor and using $\nabla_{[\mu}F_{\nu\rho]}=0$ gives

$$\nabla_\mu T^{({\rm EM})\mu\nu} =F^\nu{}_{\rho}\nabla_\mu F^{\mu\rho} =-F^\nu{}_{\rho}J^\rho.$$

Matter gains the four-momentum lost by the electromagnetic field. Only their closed combined system has vanishing total divergence.

Trace and Ricci comparison

Taking the traces in $D$ dimensions gives

$$T^{(φ)} =\left(1-\frac D2\right)(\nablaφ)^2-DV,$$

and

$$T^{({\rm EM})} =\left(1-\frac D4\right)F_{\rho\sigma}F^{\rho\sigma}.$$

Consequently Maxwell theory is traceless at the classical level when $D=4$. This dimension-specific cancellation is checked explicitly in Kontou–Sanders 2020, §2.3, pp. 14–15, equations (36)–(38), PDF. It is not true for a massive Proca field, generic nonlinear electrodynamics, or $D\ne4$.

Set $D=4$ and $\Lambda=0$. Trace reversal of Einstein’s equation gives

$$R_{\mu\nu} =8\pi G\left(T_{\mu\nu}-\frac12Tg_{\mu\nu}\right).$$

For the scalar this becomes

$$R_{\mu\nu} =8\pi G \left(\nabla_\muφ\nabla_\nuφ+Vg_{\mu\nu}\right),$$

whereas for Maxwell theory $R_{\mu\nu}=8\pi G T^{({\rm EM})}_{\mu\nu}$ and $R=0$. The last statement does not mean the spacetime is Ricci-flat: a traceless source can have nonzero Ricci tensor. Conversely, even a massless minimally coupled scalar generally has $T=-(\nablaφ)^2$ in four dimensions. The conformally improved scalar belongs to the nonminimal theory treated next.

Regimes and physical qualifications

The scalar energy density is nonnegative when $V\ge0$, but trace sign and pointwise energy conditions remain model- and configuration-dependent. A negative potential can make observer energy negative. Maxwell’s sum-of-squares rest-frame energy supports the usual classical energy conditions, as summarized in Kontou–Sanders 2020, §2.3, pp. 14–15, PDF, but quantum renormalized stress–energy is a different object and need not obey pointwise positivity.

Both tensors are local classical sources. Charged matter, boundary flux, nonminimal curvature terms, quantum anomalies, and effective higher-derivative operators alter either the separate conservation law or the trace. These qualifications must be fixed before using $T=0$ or an energy inequality as a geometric premise.

Their fluxes also distinguish standing from transported energy. A scalar configuration with $\dotφ\ne0$ and $D_\muφ\ne0$ carries a directed flux $-\dotφ D_\muφ$, but either a static profile or a spatially homogeneous profile can have nonzero energy density with zero flux for the chosen observer. Electromagnetic flux vanishes when the observer’s electric and magnetic fields have zero cross product, even though each field may contribute positive energy and anisotropic stress. These are observer-dependent decompositions of invariant tensors; changing $u^\mu$ mixes energy density, momentum density, and stress without changing the underlying $T_{\mu\nu}$.

The two models also respond differently to adding a mass scale. A scalar mass enters through $V=\tfrac12m^2φ^2$ and changes the trace directly. A vector mass requires a Proca potential term, removes Maxwell gauge invariance, and contributes additional stress. It cannot be represented by keeping the Maxwell tensor while merely changing the field equation.

Exercise: why four dimensions are special

Derive the Maxwell trace in arbitrary $D$ and, for $D>2$, determine the Ricci scalar sourced by a Maxwell field when $\Lambda=0$.

Solution

Contract the metric tensor:

$$\begin{aligned} T^{({\rm EM})} &=g^{\mu\nu}F_{\mu\rho}F_\nu{}^\rho -\frac D4F_{\rho\sigma}F^{\rho\sigma}\\ &=\left(1-\frac D4\right)F_{\rho\sigma}F^{\rho\sigma}. \end{aligned}$$

Tracing $G_{\mu\nu}=8\pi G T_{\mu\nu}$ in $D$ dimensions gives $(1-D/2)R=8\pi G T$, hence

$$R=-\frac{16\pi G}{D-2} \left(1-\frac D4\right)F^2.$$

For $D=4$, $T=0$ and $R=0$, while $R_{\mu\nu}$ can remain nonzero. The identity of tracelessness is special to four dimensions, although a particular field with $F^2=0$ can have zero trace in another dimension. The division by $D-2$ excludes two dimensions, where the Einstein tensor is identically zero.

References

• Kontou, Eleni-Alexandra, and Ko Sanders. “Energy Conditions in General Relativity and Quantum Field Theory.” Classical and Quantum Gravity 37 (2020): 193001. DOI. Open PDF.
• Reall, Harvey S. General Relativity. Cambridge Part III lecture notes, 2013. PDF.