Stress tensors and improvement terms

Does a flat-space action determine a unique stress tensor? It determines conserved translation charges, subject to boundary conditions, but we can change the local density without changing those charges. Once we specify how the theory couples to a metric, the metric response chooses a particular stress tensor.

We will calculate this choice for a scalar field. The only required preparation is Sources, currents, and generating functionals. Most of the argument applies in either three or four dimensions and needs no supersymmetry. The final application uses the scalar R-charge introduced in R-symmetry and flavor symmetry.

A conserved tensor with a nonzero trace

Begin in Lorentzian dimension $d>2$, with a real scalar $\varphi$ and

$$\eta_{\mu\nu}=\operatorname{diag}(-1,+1,\ldots,+1),\qquad \Box=\eta^{\mu\nu}\partial_\mu\partial_\nu,\qquad S_0=-\frac12\int d^d x\, \left[(\partial\varphi)^2+m^2\varphi^2\right].$$

Here $(\partial\varphi)^2=\eta^{\mu\nu}\partial_\mu\varphi\,\partial_\nu\varphi$ and $m^2\ge0$. This scalar calculation uses the mostly-plus signature; the four-dimensional Weyl-spinor entrance uses mostly minus. No spinor identity is being imported between these conventions.

The equation of motion is $(\Box-m^2)\varphi=0$. Minimal metric coupling gives the symmetric tensor

$$T^{(0)}_{\mu\nu} =\partial_\mu\varphi\,\partial_\nu\varphi -\frac12\eta_{\mu\nu} \left[(\partial\varphi)^2+m^2\varphi^2\right], \qquad \partial^\mu T^{(0)}_{\mu\nu} =\left[(\Box-m^2)\varphi\right]\partial_\nu\varphi.$$

Thus it is conserved on solutions. Yet even at $m=0$ its trace is

$$T^{(0)\mu}{}_\mu =\left(1-\frac d2\right)(\partial\varphi)^2,$$

which need not vanish when $\Box\varphi=0$. The wave equation constrains second derivatives of the field, not the square of its gradient.

An improvement is a local change that preserves conservation without using another equation of motion. For any constant $\xi$,

$$\Delta T_{\mu\nu} =\xi\left(\eta_{\mu\nu}\Box-\partial_\mu\partial_\nu\right)\varphi^2, \qquad \partial^\mu\Delta T_{\mu\nu} =\xi(\partial_\nu\Box-\Box\partial_\nu)\varphi^2=0.$$

The two derivatives commute in flat space. Both terms are needed: keeping only $\eta_{\mu\nu}\Box\varphi^2$ would spoil conservation.

In units where a derivative has mass dimension one, $[\varphi]=(d-2)/2$ and $[T_{\mu\nu}]=d$. Consequently $\xi$ is dimensionless. For a massless solution,

$$\Box\varphi^2 =2(\partial\varphi)^2+2\varphi\Box\varphi =2(\partial\varphi)^2.$$

The improved trace therefore vanishes precisely when

$$1-\frac d2+2\xi(d-1)=0, \qquad \boxed{\xi_{\mathrm c}=\frac{d-2}{4(d-1)}}.$$

This gives $\xi_{\mathrm c}=1/8$ in $d=3$ and $1/6$ in $d=4$. These are statements about the classical massless scalar. Quantum trace anomalies and interacting fixed points require additional analysis.

What the improvement changes

The local energy density changes by a spatial divergence:

$$\Delta T_{00} =-\xi\,\boldsymbol{\nabla}^{\,2}\varphi^2,\qquad \Delta E=-\xi\int_{\partial\Sigma} dS\,n^i\partial_i\varphi^2.$$

Here $\Sigma$ is a constant-time spatial region and $n^i$ is its outward unit normal. Periodic fields, or adequate decay at spatial infinity, make this integral vanish. The momentum changes likewise reduce to boundary terms. Thus the translation charges agree under these assumptions. On a bounded region they need not agree without a corresponding treatment of boundary terms and boundary conditions.

The tensors remain different local operators. Their local values, and generally their correlation functions at separated points, can differ. Equality of total charges does not erase the distinction between ways of coupling the theory to geometry.

For an explicit example, take a standing wave, independent of the other spatial coordinates:

$$\varphi(t,x)=A\cos(kt)\cos(kx),\qquad k>0.$$

It satisfies $\Box\varphi=0$, but at $t=0$ its gradient square is $A^2k^2\sin^2(kx)$, so the unimproved trace is generally nonzero. At that instant,

$$T_{00} =A^2k^2\left[\frac12\sin^2(kx)+2\xi\cos(2kx)\right].$$

The second term follows by differentiating $\varphi^2=A^2\cos^2(kx)$ twice. At the conformal coefficient the energy density becomes $A^2k^2/4$ in $d=3$, and $A^2k^2[1+\cos^2(kx)]/6$ in $d=4$. The improvement redistributes energy. For either dimension its integral over one period, per unit transverse volume, is

$$E=\int_0^{2\pi/k}T_{00}\,dx=\frac{\pi A^2 k}{2},$$

independent of $\xi$. The cosine contribution integrates to zero, exactly as the boundary argument predicts.

Which curved action selects this tensor?

We now derive the improvement from metric variation. Use the curvature convention

$$[\nabla_\mu,\nabla_\nu]V^\rho =R^\rho{}_{\sigma\mu\nu}V^\sigma,\qquad R_{\sigma\nu}=R^\rho{}_{\sigma\rho\nu}.$$

The connection is Levi-Civita, and its curvature starts with $R^\rho{}_{\sigma\mu\nu} =\partial_\mu\Gamma^\rho_{\nu\sigma} -\partial_\nu\Gamma^\rho_{\mu\sigma}+\cdots$. Denote its scalar by $R_+$: after Euclidean continuation, a round $d$-sphere of radius $\ell$ has $R_+=d(d-1)/\ell^2$. The plus label distinguishes it from the opposite scalar-curvature convention used later.

Consider the family of actions

$$S_L[g,\varphi] =-\frac12\int d^dx\,\sqrt{-g}\, \left[g^{\mu\nu}\partial_\mu\varphi\,\partial_\nu\varphi +(m^2+\xi R_+)\varphi^2\right].$$

Here $g=\det g_{\mu\nu}$. All members reduce to $S_0$ on a flat metric, but their variations away from flat space differ. This is a concrete instance of the curved-extension ambiguity discussed in Festuccia and Seiberg, introduction, printed page 2.

Hold the scalar fixed and vary the inverse metric, defining

$$T_{\mu\nu}=-\frac{2}{\sqrt{-g}} \frac{\delta S_L}{\delta g^{\mu\nu}}, \qquad \delta S_L=-\frac12\int\sqrt{-g}\,T_{\mu\nu}\delta g^{\mu\nu} =+\frac12\int\sqrt{-g}\,T^{\mu\nu}\delta g_{\mu\nu}.$$

The last equality uses $\delta g^{\mu\nu}=-g^{\mu\rho}g^{\nu\sigma}\delta g_{\rho\sigma}$. Assume compactly supported metric variations, or no boundary; the following bulk calculation discards integrated divergences.

Let $f=\varphi^2$. The curvature variation is

$$\delta R_+ =R_{\mu\nu}\delta g^{\mu\nu} +g_{\mu\nu}\Box\delta g^{\mu\nu} -\nabla_\mu\nabla_\nu\delta g^{\mu\nu}, \qquad \delta\sqrt{-g}=-\frac12\sqrt{-g}\,g_{\mu\nu}\delta g^{\mu\nu}.$$

Here $\Box=\nabla^\mu\nabla_\mu$ on a curved metric. Integrate each derivative term twice. In particular,

$$\int\sqrt{-g}\,f\,g_{\mu\nu}\Box\delta g^{\mu\nu} =\int\sqrt{-g}\,g_{\mu\nu}(\Box f)\delta g^{\mu\nu},$$

and the term with the two covariant derivatives becomes $-\int\sqrt{-g}\,(\nabla_\mu\nabla_\nu f)\delta g^{\mu\nu}$. Combining these with the volume variation gives

$$\delta\int\sqrt{-g}\,fR_+ =\int\sqrt{-g}\, \left[fG_{\mu\nu} +(g_{\mu\nu}\Box-\nabla_\mu\nabla_\nu)f\right]\delta g^{\mu\nu},$$

where $G_{\mu\nu}=R_{\mu\nu}-g_{\mu\nu}R_+/2$ is the Einstein tensor. Therefore

$$\boxed{ \begin{aligned} T_{\mu\nu} ={}&\nabla_\mu\varphi\,\nabla_\nu\varphi -\frac12g_{\mu\nu}\left[(\nabla\varphi)^2+m^2\varphi^2\right]\\ &+\xi\left[ G_{\mu\nu}\varphi^2 +g_{\mu\nu}\Box\varphi^2 -\nabla_\mu\nabla_\nu\varphi^2\right]. \end{aligned}}$$

Setting the metric flat after varying reproduces the improvement. Setting $R_+=0$ in the action before varying would remove the very information that determines it.

There is a useful curved-space check. The contracted Bianchi identity gives $\nabla^\mu G_{\mu\nu}=0$, while commuting derivatives on the gradient of $f$ gives

$$\nabla^\mu(g_{\mu\nu}\Box-\nabla_\mu\nabla_\nu)f =-R_{\nu\rho}\nabla^\rho f.$$

Thus the divergence of the improvement bracket is $-\tfrac12R_+\partial_\nu f=-R_+\varphi\partial_\nu\varphi$. The full result is

$$\nabla^\mu T_{\mu\nu} =\underbrace{(\Box-m^2-\xi R_+)\varphi}_{\mathcal E}\, \partial_\nu\varphi.$$

It vanishes on the curved scalar equation $\mathcal E=0$. Simply covariantizing the flat improvement while omitting $G_{\mu\nu}\varphi^2$ would fail this check.

A second way to find the conformal coefficient

A Weyl transformation rescales the metric locally while leaving coordinates fixed. Set $m=0$ and vary

$$\delta_\omega g_{\mu\nu}=2\omega g_{\mu\nu}, \qquad \delta_\omega\varphi=-\Delta\omega\varphi, \qquad \Delta=\frac{d-2}{2}.$$

The scalar’s weight $\Delta$ is its free-field scaling dimension. To first order, $\delta_\omega\sqrt{-g}=d\omega\sqrt{-g}$ and $\delta_\omega R_+=-2\omega R_+-2(d-1)\Box\omega$. The terms without derivatives of $\omega$ cancel by dimensional counting. With $f=\varphi^2$, the remaining density variations are

$$\delta_\omega\!\left[\sqrt{-g}(\nabla\varphi)^2\right] =-\Delta\sqrt{-g}\,\nabla_\mu f\,\nabla^\mu\omega, \qquad \delta_\omega(\sqrt{-g}R_+f) =-2(d-1)\sqrt{-g}\,f\Box\omega.$$

Including the $-1/2$ in $S_L$ and integrating the first expression once gives

$$\delta_\omega S_L =\left[\xi(d-1)-\frac{\Delta}{2}\right] \int\sqrt{-g}\,f\Box\omega.$$

For compactly supported $\omega$ this vanishes at exactly $\xi=\xi_{\mathrm c}$. The curvature coupling restores classical local Weyl invariance, and the trace calculation is its flat-space manifestation. Indeed, varying both the field and metric also gives the local identity $T^\mu{}_\mu=\Delta\varphi\mathcal E$ at $m=0$ and $\xi=\xi_{\mathrm c}$. Only after imposing the equation of motion is the trace zero.

Return to Euclidean sources

For the scalar, continuing $t=-i\tau$ gives

$$S_E=\frac12\int\sqrt g\, \left[(\nabla\varphi)^2+(m^2+\xi R_+)\varphi^2\right], \qquad T^E_{\mu\nu}=+\frac2{\sqrt g}\frac{\delta S_E}{\delta g^{\mu\nu}}.$$

This is the Euclidean source convention of the preceding lesson: $\delta S_E=-\tfrac12\int\sqrt g\,T_E^{\mu\nu}\delta g_{\mu\nu}$. For $g_{\mu\nu}=\delta_{\mu\nu}+2h_{\mu\nu}$, the linear coupling is $-T_E^{\mu\nu}h_{\mu\nu}$. The opposite signs in the Lorentzian and Euclidean definitions accompany the corresponding action signs; the tensor formulas above continue with the metric.

If curvature is instead denoted $R_P=-R_+$, the Euclidean interaction is $-\xi R_P\varphi^2/2$. In this convention a round three-sphere has $R_P=-6/\ell^2$, as specified in the three-manifold paper, appendix A.2. A negative printed curvature term can therefore supply a positive conformal scalar potential.

The three-dimensional R-charge improvement

This application assumes R-symmetry and flavor symmetry. The general scalar derivation above is independent of it.

For a complex scalar set $\phi=(\varphi_1+i\varphi_2)/\sqrt2$ and $\rho=\widetilde\phi\phi$. On the usual scalar conjugate integration cycle, $\rho=(\varphi_1^2+\varphi_2^2)/2$. Away from that cycle, the Euclidean fields $\phi,\widetilde\phi$ remain independent. Summing the two real-scalar actions gives

$$\mathcal L_E=\partial_\mu\widetilde\phi\,\partial^\mu\phi+\xi R_+\rho, \qquad \Delta T^E_{\mu\nu} =2\xi(\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu)\rho.$$

The factor two follows from $\varphi_1^2+\varphi_2^2=2\rho$. For a chiral scalar of R-charge $r$, the curvature coefficient $-rR_P\rho/4$ selects $\xi=r/4$, hence $\Delta T^E_{\mu\nu}=(r/2)(\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu)\rho$. This reproduces the relation between equations (6.32) and (7.13). Mixing the R-symmetry with flavor changes this improvement; the full current multiplet will supply its supersymmetric organization.

At $r=1/2$, $\xi=1/8$ and the sphere contribution is $3\rho/(4\ell^2)$. For general $r$, additional background couplings matter. With vanishing central and gauge charges, $A_\mu=V_\mu=C_\mu=0$ and auxiliary background $H=-i/\ell$, the scalar coefficient is

$$-\frac r4R_P+r\left(r-\frac12\right)H^2 =\frac{r(2-r)}{\ell^2}.$$

The round-sphere chiral action derives the full supersymmetric result, including the fermions and auxiliary fields. The scalar curvature comparison alone does not establish supersymmetry.

Exercises

1. Can an improvement remove a mass?

In flat dimension $d>2$, take $\xi=\xi_{\mathrm c}$ but $m\ne0$. Find the trace on solutions of $(\Box-m^2)\varphi=0$. Explain why the massless improvement does not make this theory conformal.

First hint

Retain the mass term in both the tensor and the identity for $\Box\varphi^2$.

Second hint

On shell, $\Box\varphi^2=2(\partial\varphi)^2+2m^2\varphi^2$. The same value of $\xi$ cancels the gradient coefficient as before.

Solution

The trace is

$$T^\mu{}_\mu =\left(1-\frac d2\right)(\partial\varphi)^2 -\frac d2m^2\varphi^2 +\xi_{\mathrm c}(d-1)\Box\varphi^2.$$

The gradient coefficient vanishes, whereas the coefficient of $m^2\varphi^2$ is $-d/2+2\xi_{\mathrm c}(d-1)=-1$. Therefore

$$\boxed{T^\mu{}_\mu=-m^2\varphi^2\quad\text{on shell}.}$$

For example, the spatially uniform solution $\varphi=A\cos(mt)$ gives the nonzero trace $-m^2A^2\cos^2(mt)$. The dimensionful mass breaks scale invariance. A different constant $\xi$ cannot simultaneously cancel the independent gradient and mass terms: their cancellation conditions differ.

More generally the curved result at $\xi_{\mathrm c}$ is $T^\mu{}_\mu=\Delta\varphi\mathcal E-m^2\varphi^2$. It reduces to the preceding answer when $\mathcal E=0$, and to the Weyl identity when $m=0$.

2. When does the energy change?

In $d=3$, restrict the static massless solution $\varphi=vx$ to the spatial strip $0\le x\le L$. Compute its unimproved and improved energies per unit length in the other spatial direction. Evaluate them at $\xi=1/8$. Does this contradict the equality of translation charges?

First hint

The solution satisfies $\Box\varphi=0$, but $\partial_x\varphi^2$ does not agree at the two ends.

Second hint

Compute $\Delta T_{00}=-\xi\partial_x^2(v^2x^2)$ directly, then compare its integral with the endpoint formula.

Solution

Here $\partial_t\varphi=0$ and $(\partial\varphi)^2=v^2$, so

$$T^{(0)}_{00}=\frac12v^2,\qquad \Delta T_{00}=-2\xi v^2,\qquad E_\xi=\left(\frac12-2\xi\right)v^2L.$$

At $\xi=1/8$, $E_\xi=v^2L/4$, whereas $E_0=v^2L/2$. The difference is exactly

$$-\xi[\partial_x\varphi^2]_{0}^{L}=-2\xi v^2L.$$

The field is neither periodic nor decaying on this region. The assumption that removed the surface term therefore fails. To define a physical system on the strip, one must specify boundary conditions and any boundary contribution to its energy; comparing only these two bulk integrals does not settle that boundary problem.

3. Equal sphere potentials, different improvements

Use the three-dimensional application above. Compare $r=1/2$ and $r=3/2$ on a sphere of radius $\ell$. Both have scalar coefficient $U=r(2-r)/\ell^2$. The full chiral action contains the fermion term $-M_\psi\widetilde\psi\psi$, where $M_\psi=(r-1/2)/\ell$. Are these the same conformal coupling of the full multiplet? Check also the flat, massless scalar trace.

First hint

The coefficient $U$ combines the curvature interaction and the $H^2$ interaction. They carry different information before selecting a background.

Second hint

For the complex scalar, $K=\partial_\mu\widetilde\phi\,\partial^\mu\phi$ and the unimproved $d=3$ trace is $-K$. On the flat massless equations, $\partial^2\rho=2K$.

Solution

Both charges give $U=3/(4\ell^2)$, but $M_\psi$ is respectively zero and $1/\ell$. Moreover, their curvature couplings select $\xi=1/8$ and $3/8$. Only the first is the three-dimensional conformal coefficient.

The flat improvement contributes $(r/2)(d-1)\partial^2\rho=2rK$ to the trace in $d=3$. Consequently the scalar trace on shell is

$$T^\mu{}_\mu=(-1+2r)K,$$

which is zero for $r=1/2$ and $2K$ for $r=3/2$. The same value of a scalar potential on one selected metric does not fix the metric response, the fermion coupling, or the conformal stress tensor.

The calculation to carry forward is the order of operations: choose a curved action, vary its sources, and then specialize the background. That procedure identifies which stress tensor a curved-space observable probes.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 2, currents and rigid backgrounds
• Paper guide: Section 7, the linearized check