Consider the following action for a free scalar field $\phi$ in a curved background $$S=\int dx\Big( \frac12g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi+\gamma \phi R\Big).$$

Here $g_{\mu\nu}$ is a metric and $R$ is the corresponding scalar curvature. The first term is the standard free field Lagrangian while the last term gives non-minimal coupling to gravity. I've encountered this Lagrangian in the context of 2d CFT, where one uses the last term to introduce a "background charge" in the system (Di Francesco CFT, chap.9). There the stress-energy tensor is computed in the flat space limit, which gives $$T_{\mu\nu}=\partial_\mu \phi\partial_\nu \phi-\frac12\eta_{\mu \nu}(\partial\phi)^2+\gamma\Big(\partial_\mu\partial_\nu\phi-\frac12\eta_{\mu\nu}\partial^2\phi\Big).$$ The additional $\gamma$-term is hence consequential for further computations.

I cat get this result formally, but struggle with a physical meaning. If I first set metric to be flat the $\gamma$-term will just vanish. Alternatively, I can formally get the same energy-momentum tensor from the action $$S'=\int dx\Big(\frac12 g^{\mu\nu}\partial_\mu\phi\partial_\nu\phi+\gamma g^{\mu\nu}\partial_\mu\partial_\nu \phi\Big),$$ but the last term is just a total derivative in flat space so again, it shouldn't be consequential for physics.

I haven't formulated the question clearly, but perhaps it is: why does the curvature term affect the flat-space physics? Can one get the same stress-energy tensor from a flat space action?


1 Answer 1


The stress tensor is not unique, whether it's defined using Noether's theorem or defined by varying the action with respect to the metric. The non-uniqueness of the Noether-derived version is familiar. This answer addresses the non-uniqueness of the other version.

The stress tensor $T^{ab}$ is defined by $$ \delta S\propto \int d^Dx\ \sqrt{g}\,T^{ab}(x)\delta g_{ab}(x), \tag{1} $$ where the metric is the only thing being varied. At first glance, this $T^{ab}$ seems to be unique, but it isn't. For simplicity, suppose that the prescribed background metric is flat. To use (1), we temporarily generalize the action to an arbitrary metric, then apply the definition (1), and then set the metric equal to the desired background metric — in this case the flat metric. The temporary generalization to an arbitrary metric is not unique. For example, in a scalar-field model, we can include a term $\sim \int \sqrt{g}\,R\phi^2$ in the temporary generalization (as shown in the OP), where $R$ is the Ricci scalar, with an arbitrary coefficient. The resulting stress tensor still depends on this coefficient after making the metric flat (even though $R=0$ for a flat metric), because $$ \left.\frac{\delta}{\delta g^{ab}}\sqrt{g}\,R\phi^2\right|_{g=\eta} = \partial_c\partial_d\big((\eta_{ab}\eta^{cd}-\delta^c_a\delta^d_b)\phi^2\big) \neq 0 \tag{2} $$ where $\eta$ is the flat metric. Therefore, the stress tensor defined by (1) is not unique. Equation (2) is a copy of an in-line equation from section 5.3.1 in [1].

For confirmation, here is an excerpt from [2]:

Thus conformal invariance is equivalent to the existence of a traceless stress tensor, although an arbitrarily chosen stress tensor will have a [non-zero] trace...

The specific example of a free massless scalar field in flat space-time is reviewed in section 2 of [3], which says:

this model is a conformal field theory. As it is well known, it admits a conserved, traceless stress-energy tensor... $$ T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi -\frac{1}{4(d-1)}\big((d-2)\partial_\mu\partial_\nu+g_{\mu\nu}\partial^2\big)\phi^2. $$

In contrast, the stress tensor that would be obtained by using the simplest generalization of the action in (1) is $$ T_{\mu\nu}=\partial_\mu\phi\partial_\nu\phi -\frac{1}{2} g_{\mu\nu}(\partial_\rho\phi)(\partial^\rho\phi), $$ which is conserved but not traceless (except when $d=2$).

Here's a related post, where the non-uniqueness of $T_{\mu\nu}$ is mentioned in the context of the relationship between conformal invariance and tracelessness:

Trace of stress tensor vanishes $\implies$ Weyl invariant


[1] Pons (2009), "Noether symmetries, energy-momentum tensors and conformal invariance in classical field theory," https://arxiv.org/abs/0902.4871

[2] Polchinski (1988), "Scale and conformal invariance in quantum field theory", Nuclear Physics B, 303: 226-236

[3] Giombi (2016), "TASI Lectures on the Higher Spin - CFT duality," https://arxiv.org/abs/1607.02967

  • $\begingroup$ This is interesting. I always assumed that the energy-momentum tensor is unique up to a tensor which is conserved off-shell. This seems not to be the case in your examples. Doesn't this mean that there actually are additional symmetries and the difference between different definitions of $T^{\mu\nu}$ comes from some other conserved currents? $\endgroup$ Commented Mar 25, 2019 at 13:46
  • $\begingroup$ @WeatherReport That's an interesting question, and I'm not sure. In flat spacetime, the quantity in my equation (2) is conserved by itself because of the equation of motion, but I don't know what extra symmetry would be associated with that quantity, unless it is related to conformal symmetry, or just scale invariance. Related: Noether's Theorem and scale invariance and Noether's theorem for scale invariance $\endgroup$ Commented Mar 25, 2019 at 23:54
  • $\begingroup$ I'm still confused because your answer only addresses a non-uniqueness of the energy-momentum tensor. However, in the context I've encountered it this modification is claimed to affect the physics. For example, in CFT the OPE with energy-momentum tensor determines scaling dimensions of the fields, so these are different whether we add the $\phi R$ term or not (this was actually a reason for its introduction). $\endgroup$ Commented Mar 26, 2019 at 9:13
  • $\begingroup$ @WeatherReport I think that statement about the energy-momentum tensor assumes that it's traceless. The $R$ term doesn't affect the flat-space physics, but using the OPE with $T_{ab}$ to get scaling dim's assumes a particular version of $T_{ab}$. The $R$ term is only needed if we want to get that particular version by varying the action with respect to the metric. Often, results that can be derived without referring to anything outside the particular case of interest (e.g. flat spacetime) can also be derived by treating that case as a special case of something more general (all spacetimes). $\endgroup$ Commented Mar 26, 2019 at 23:56
  • 2
    $\begingroup$ It is probably worth stressing that $\phi^2 R$ term is precisely the term that needs to be added to get a Weyl-invariant action. This is often referred to as the "conformally coupled scalar." Stress-tensor Ward identities (which imply the stuff about OPE) can be derived by starting with an action on a general metric background, so for the derivation to go through you have to use the conformally-coupled action. $\endgroup$ Commented Mar 28, 2019 at 20:31

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