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With $x^0=ct$, so that ${\partial }_0=c^{-1}{\partial }_t$, the $\nu =0$ component of
${\partial }_{\mu }{T}^{\mu \nu }=0$ isand the $\nu =j$ component isThese are exactly the energy and momentum conservation laws from parts (a) and (b), now combined into conservation of the electromagnetic stress-energy tensor.

The first term of the action is the free boson conformal field theory. For the curvature coupling, insert the supplied first variation of $\sqrt{g}{R}^{(2)}$, integrate by parts twice and discard the boundary term. Its metric variation is the stress-tensor improvementin the conventions of the question. In a locally flat complex coordinate, the holomorphic component of the complete stress-energy tensor is thereforeas required. Thus the coupling $q\int \sqrt{g}X{R}^{(2)}$ makes $X$ a background-charge scalar field, equivalently a worldsheet coordinate in a linear dilaton conformal field theory.

By the state–operator correspondence, a CFT operator of scaling dimension $\Delta$ creates a cylinder state of energy $E=\Delta /R$. The identity gives the vacuum; the first nontrivial low-energy single-trace operator is the conserved stress-energy tensor, with $\Delta =3$ in three dimensions; and one translation raises the dimension by one. Taking the scheme-dependent vacuum energy to vanish,All other single-trace primaries are heavy by assumption, while the first two-graviton state begins at $6/R$.

The symmetric stress-energy tensor obtained by metric variation isIt is conserved on shell. With signature $(+---)$ its energy density isThe canonical stress-energy tensor differs from this symmetric tensor by the divergence of $F^{\mu \rho }{A}^{\nu }$ plus terms proportional to the field equation. Thus its integrated energy agrees after discarding the corresponding total spatial derivative, and positivity holds on shell.

In the Page–Geilker experiment, radioactive-decay counts supplied a quantum random decision that determined which of two macroscopically distinct configurations of lead masses the experimenters placed around a Cavendish torsion balance. In each observed branch, the balance deflected in the direction predicted by the mass configuration recorded in that branch, with a strong correlation between the decision and the measured gravitational torque.

The unobserved alternative was the prediction of the simplest semiclassical Einstein equation, in which one classical metric is sourced by the expectation value $⟨T_{\mu \nu }⟩$ of the matter stress-energy tensor. If the universal wavefunction does not collapse, the two nearly equally weighted mass configurations both contribute to that expectation value. Their opposite torques should then largely average away, and the balance should show little branch-correlated deflection. That behavior was not seen.

The experiment was designed to go beyond earlier observations of ordinary gravity and quantum matter. It deliberately created macroscopically different mass distributions in different quantum branches and tested whether the gravitational field followed the branch actually observed or the expectation-value average over all branches. It ruled out that simplest no-collapse semiclassical coupling under the experiment's assumptions. It supplied indirect evidence for quantum gravity, while leaving more elaborate classical-quantum couplings logically possible.

Under the finite dilation, the volume element transforms as $d^4{x}^{\prime }=e^{4\alpha }{d}^{4}x$, whereas ${\partial }_{\mu }^{\prime }{\varphi }^{\prime }(x^{\prime })=e^{-2\alpha }{\partial }_{\mu }\varphi (x)$. Thus the kinetic term and quartic interaction both acquire the factor $e^{-4\alpha }$. Their factors cancel the Jacobian determinant, provingA mass term would instead acquire a net factor $e^{2\alpha }$ in the action and break this scale invariance.

For the Noether current, use the variation at the same coordinate, rather than comparing values at different coordinates:At $m=0$, the corresponding Lagrangian density variation is ${\delta }_0\mathcal{L}=-\alpha (4+x\cdot \partial )\mathcal{L}=-\alpha {\partial }_{\mu }(x^{\mu }\mathcal{L})$. Thus $J_0^{\mu }=-x^{\mu }\mathcal{L}$, and the stated Noether current convention givesThe frequently used dilatation current $D^{\mu }=x_{\nu }{T}^{\mu \nu }+\varphi {\partial }^{\mu }\varphi$ is $-j^{\mu }$; this is solely an overall parameter convention. In the convention $D^{\mu }=x_{\nu }{T}^{\mu \nu }-V^{\mu }$, the virial current is $V^{\mu }=-\varphi {\partial }^{\mu }\varphi$.

Using solution and the trace computed in solution gives the useful identityIt follows that $\boxed{{\partial }_{\mu }{j}^{\mu }=0}$ in the massless theory on shell. Equivalently, $x_{\nu }{\Omega }^{\mu \nu }$ is a conserved current, since its divergence is the trace of the improved stress-energy tensor on shell. These are classical scale invariance statements; the nonzero beta function in solution explains why the interacting quantum theory need not retain them.

Use signature $(+---)$, coordinates $x^0=ct$ and four-velocity $u^{\mu }=d{x}^{\mu }/d\tau$ normalized by $u_{\mu }{u}^{\mu }=c^2$. The derivative is the metric-compatible Levi-Civita connection. Fix the curvature sign by $[{\nabla }_{\alpha },{\nabla }_{\beta }]V^{\mu }=R^{\mu }{}_{\nu \alpha \beta }{V}^{\nu }$. Then the required contractions areWith $G^{\mu \nu }=R^{\mu \nu }-g^{\mu \nu }R/2$, the contracted Bianchi identity gives ${\nabla }_{\mu }{G}^{\mu \nu }=0$. The Einstein field equations $G^{\mu \nu }+\Lambda g^{\mu \nu }=(8\pi G/c^4)T^{\mu \nu }$, with constant $\Lambda$, therefore require covariant conservation of stress-energy. This is a consistency relation between geometry and the total matter/field source. It does not mean ${\partial }_{\mu }{T}^{\mu \nu }=0$ in arbitrary coordinates, nor guarantee a global conserved energy in an arbitrary curved spacetime. The geometric identity alone, without the field equations or matter dynamics, is not an equation of motion for a freely specified material source.

For relativistic dust, $\rho$ is proper rest-mass density, measured in the fluid rest frame, and $u^{\mu }$ is its timelike four-velocity with the $c^2$ normalization above. In a local Cartesian inertial frame $u^{\mu }=\gamma (c,\mathbf{v})$, soAccordingly $T^{0i}/c$ is momentum density and $c{T}^{0i}$ is energy flux in direction $i$. The equality of momentum-density and energy-flux components after the appropriate factors of $c$ reflects the symmetry of the stress-energy tensor. In the comoving frame the spatial stress is zero, as required for pressureless matter.

Expand the dust conservation equation by the product rule:Contract with $u_{\nu }$. metric compatibility and constant normalization give $u_{\nu }{\nabla }_{\mu }{u}^{\nu }={\nabla }_{\mu }(u_{\nu }{u}^{\nu })/2=0$, leaving $c^2{\nabla }_{\mu }(\rho u^{\mu })=0$. Substitute this rest-mass continuity law back and divide by nonzero $\rho$:This proves geodesic motion of uncharged dust along each fluid flow line. pressure forces or another exchange of momentum would change the argument; at zero density there is no dust worldline to constrain.

The electromagnetic field tensor is antisymmetric and has six independent components. In the stated local inertial coordinates choose $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, givingThe fields are those measured by that local observer; changing observer mixes them through the Lorentz transformation of $F$. With ${ϵ}^{0123}=+1$ and ${\tilde{F}}^{\mu \nu }={ϵ}^{\mu \nu \alpha \beta }{F}_{\alpha \beta }/2$, two independent quadratic contractions areThe second is a pseudoscalar: it is invariant under proper Lorentz transformations and changes sign under an orientation reversal. Opposite choices for the orientation tensor change its written sign. A nonzero null electromagnetic field can have both contractions zero, so the two numbers do not determine whether the tensor itself vanishes.

For charged relativistic dust, insert $j^{\alpha }=(q/m)\rho u^{\alpha }$ into the force-density equation:Contraction with $u_{\nu }$ again gives the continuity law, because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$ by antisymmetry. After removing the continuity term,In a local inertial frame the spatial components reduce to $d(\gamma m\mathbf{v})/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and the temporal component gives $d(\gamma m{c}^2)/dt=q\mathbf{E}\cdot \mathbf{v}$. This is the relativistic Lorentz force, with gravity represented by the connection. The force is orthogonal to $u$ and preserves the rest-mass normalization. Matter alone now exchanges energy and momentum with the electromagnetic field; adding the field tensor's stress-energy restores zero divergence of the combined source.

For a scalar field, ${\nabla }_{\mu }\varphi ={\partial }_{\mu }\varphi$. Apply the product rule to the stress-energy tensor, using metric compatibility to pass the covariant derivative through $g^{\mu \nu }$:The middle terms cancel because the Levi-Civita connection gives a symmetric Hessian of a scalar field. Thus, with the sign conventions of this question, the scalar stress-energy divergence identity becomesThe last equality is the Klein-Gordon equation. This stress-energy conservation follows from the field equation and metric compatibility; it does not require imposing the Einstein field equations.

Use signature $(+---)$ and local inertial coordinates $x^0=ct$. Write $u^{\mu }=\gamma (c,\mathbf{v})$ and $w=\rho c^2+p$, where $\rho$ is rest mass density. The perfect fluid in general relativity has componentsHere $T^{00}$ is energy density, $c{T}^{i0}$ is energy flux, $T^{0i}/c$ is momentum density, and $T^{ij}$ is the flux of the $j$th momentum component through a surface normal to $i$. In the local rest frame these reduce to energy density $\rho c^2$, zero energy and momentum flow, and isotropic pressure $p$. The equation ${\nabla }_{\mu }{T}^{\mu \nu }=0$ expresses local stress-energy conservation. Its temporal and spatial components give energy and momentum balance; in flat coordinates, for example, ${\partial }_t{T}^{00}+{\partial }_i(c{T}^{i0})=0$.

For relativistic dust, $T^{\mu \nu }=\rho u^{\mu }{u}^{\nu }$. Contracting its divergence with $u_{\nu }$, and using metric compatibility together with $u_{\nu }{u}^{\nu }=c^2$, givesConsequently ${\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting this continuity equation back into the divergence leaves $\rho u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$. Wherever matter is present,so the individual particles follow geodesics.

The electromagnetic four-force law $f^{\mu }=q{F}^{\mu \nu }{u}_{\nu }$ must preserve the particle's rest mass. Thus $u_{\mu }{f}^{\mu }=0$, and requiring this for arbitrary timelike four-velocities makes the electromagnetic field tensor antisymmetric. Its six independent components comprise the electric field and magnetic field. With the stated signature and $x^0=ct$, one consistent convention isIt gives the spatial force $d\mathbf{p}/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and energy transfer $dE/dt=q\mathbf{E}\cdot \mathbf{v}$. Under a Lorentz transformation, $F^{\mu \nu }$ transforms as a rank-two contravariant tensor; its electric and magnetic components consequently mix. A four-potential supplies $F_{\mu \nu }={\nabla }_{\mu }{A}_{\nu }-{\nabla }_{\nu }{A}_{\mu }$, unchanged by $A_{\mu }\mapsto A_{\mu }+{\nabla }_{\mu }\chi$. The Maxwell equations areThese combine the source equations and the homogeneous equations into covariant tensor statements; $j^{\mu }$ is the four-current.

For charged relativistic dust, the matter tensor exchanges four-momentum with the electromagnetic field: ${\nabla }_{\mu }{T}_{\mathrm{m}}^{\mu \nu }=F^{\nu \alpha }{j}_{\alpha }$, while their total stress-energy tensor is conserved. Identical particles of rest mass $m$ and electric charge $q$ have $j^{\alpha }=(q/m)\rho u^{\alpha }$. Contraction with $u_{\nu }$ again gives mass continuity because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$. Removing that term givesUsing zero divergence for the matter tensor alone in this last step would omit the electromagnetic Lorentz force.

In the same Minkowski metric and electromagnetic units, $T^{\mu \nu }=-F^{\mu }{}_{\lambda }{F}^{\nu \lambda }+\frac{1}{4}{g}^{\mu \nu }{F}_{\alpha \beta }{F}^{\alpha \beta }$. Differentiating givesTo simplify the last two terms, contract ${\partial }_{\mu }{F}_{\nu \lambda }+{\partial }_{\nu }{F}_{\lambda \mu }+{\partial }_{\lambda }{F}_{\mu \nu }=0$ with $F^{\mu \lambda }$. Its first and third terms are equal, and its middle term is $-\frac{1}{2}{\partial }_{\nu }(F_{\alpha \beta }{F}^{\alpha \beta })$. Raising $\nu$ therefore gives $F^{\mu }{}_{\lambda }{\partial }_{\mu }{F}^{\nu \lambda }=\frac{1}{4}{\partial }^{\nu }(F_{\alpha \beta }{F}^{\alpha \beta })$. The cancellation leavesThis is the electromagnetic stress-energy divergence identity in rationalized units. As a sign check, $T^{00}=(E^2+B^2)/2$ is positive and $T^{0i}=(\mathbf{E}\times \mathbf{B}{)}_i$ is the Poynting vector. The integrated four-momentum $P^{\nu }=\int T^{0\nu }{d}^{3}x$ is conserved if the boundary flux vanishes. With matter sources, the electromagnetic field alone instead obeys ${\partial }_{\mu }{T}^{\mu \nu }=-F^{\nu \lambda }{J}_{\lambda }$: energy and momentum transfer to matter restore conservation of the total stress-energy tensor.

For metric signature $(+---)$, a perfect fluid has $T_{ab}=(\rho +p)u_a{u}_b-p{g}_{ab}$. The Einstein field equations and static Killing field is a Ricci eigenvector imply $(\rho +p)(k\cdot u)u_{[a}{k}_{b]}=0$. Two nonzero timelike vectors cannot be orthogonal, so $k\cdot u\ne 0$. If $\rho +p\ne 0$, the four-velocity is parallel to the static Killing vector. Choosing both future-directed gives the displayed normalization. For $\rho +p=0$, the stress-energy tensor is proportional to the metric tensor and does not determine a preferred fluid velocity; the strict enthalpy hypothesis is therefore material.

The temporal-temporal component of the photon stress-energy tensor in a local orthonormal tetrad. For an isotropic one-particle distribution function expressed using comoving energy $ϵ=aE$, it is ${\bar{\rho }}_{\gamma }=4\pi a^{-4}\int dϵ{ϵ}^3\bar{f}(ϵ)$. A thermal photon temperature perturbation gives a first-order directional energy perturbation four times its fractional temperature change. Boundary conditions on the distribution justify this factor through integration by parts.

Expand the Einstein tensor about the Minkowski metric. The quadratic term in the metric perturbation defines $t_{\mu \nu }=-G_{\mu \nu }^{(2)}[h,h]/(8\pi )$, a symmetric tensor relative to the background. For a first-order vacuum solution, the expanded contracted Bianchi identity gives ${\partial }^{\mu }{t}_{\mu \nu }=0$. Equivalently, a second-order correction $j$ obeys $G^{(1)}[j]=8\pi t$, whose flat divergence vanishes identically. This conservation is an on-shell statement. The local $t$ depends on perturbative gauge and the background split; conservation alone does not make it a gauge-independent local gravitational stress-energy tensor.

Relativistic dust is a continuum of noninteracting massive particles with one local four-velocity $u^{\mu }$ and negligible pressure. With signature $(+---)$, $u^{\mu }{u}_{\mu }=c^2$ and rest mass density $\rho$, its stress-energy tensor has the displayed form. Contracting stress-energy conservation with $u_{\nu }$ and using metric compatibility gives $c^2{\nabla }_{\mu }(\rho u^{\mu })=0$. Substituting back yields $u^{\mu }{\nabla }_{\mu }{u}^{\nu }=0$ wherever $\rho >0$: each particle follows a geodesic. The cosmological pressureless matter model is a homogeneous application.

For the real scalar field with metric $(+,-)$ and potential $U={\kappa }^2({\varphi }^2-1{)}^2/2$, the static phi-four kink is ${\varphi }_0(x)=\tanh (\kappa x)$ and satisfies ${\varphi }_0^{\prime 2}=2U({\varphi }_0)$. Its mass is $M=\int {\varphi }_0^{\prime 2}dx=4\kappa /3$. A Lorentz boost gives $\varphi (t,x)={\varphi }_0(\gamma (x-vt))$. The stress-energy tensor gives $P=-\int {\varphi }_t{\varphi }_{x}dx=\gamma v\int {\varphi }_0^{\prime 2}dx=\gamma Mv$. Similarly, $E=\frac{1}{2}[\gamma (1+v^2)+{\gamma }^{-1}]M=\gamma M$, proving the energy–momentum relation $E^2-P^2=M^2$.

For a minimally coupled scalar field, put $q^2=h^{ij}{D}_i\varphi D_j\varphi$ using the spatial covariant derivative. Spatial and normal projections of its stress-energy tensor give $\rho ={\Pi }^2/2+q^2/2+V$, ${\mathcal{J}}_i=-\Pi D_i\varphi$, and $S_{ij}=D_i\varphi D_j\varphi +h_{ij}({\Pi }^2/2-q^2/2-V)$. The trace-free stress is $D_i\varphi D_j\varphi -h_{ij}{q}^2/3$, with coefficient one. This makes the scalar anisotropic stress second order in a spatial gradient expansion.

Semiclassical gravity couples a classical spacetime metric to quantum matter. In its simplest expectation-value form, the classical gravitational field is sourced by the quantum expectation of the matter stress-energy tensor.

For a symmetric stress-energy tensor with compact support and ${\partial }_b{T}^{ab}=0$ in Minkowski spacetime, let $I^{ij}=\int T^{00}{x}^i{x}^j{d}^{3}x$. Two integrations by parts give ${\dot{I}}^{ij}=\int (T^{0i}{x}^j+T^{0j}{x}^i)d^{3}x$ and the displayed identity. Compact support removes the boundary fluxes. The support should stay bounded over the time interval used when differentiating the integrals.

A stress-energy superpotential is an antisymmetric tensor $B^{\rho \mu \nu }=-B^{\mu \rho \nu }$ whose divergence supplies a stress-energy tensor improvement. In the Maxwell field, the potential-dependent expression $B^{\rho \mu }{}_{\nu }=F^{\mu \rho }{A}_{\nu }$ produces the difference between canonical and gauge-invariant stress-energy tensors on shell. Antisymmetry ensures identically conserved improvement terms.

Adding ${\partial }_{\rho }{B}^{\rho \mu \nu }$ with $B^{\rho \mu \nu }=-B^{\mu \rho \nu }$ to a stress-energy tensor preserves its divergence identically, since commuting partial derivatives annihilates the antisymmetric pair. The total four-momentum changes only by a spatial boundary term. Improvements can make local densities symmetric or gauge-invariant; equality of integrated charges requires the chosen boundary conditions. Equation-of-motion terms can also relate representatives on shell.