26
$\begingroup$

In classical field theory, the stress-energy tensor can be defined in terms of the variation of the action with respect to the metric field, or with respect to a frame field if spinors are involved. Of course, this assumes that the field theory is expressed in terms of an arbitrary metric (or frame) field, even if it's only a background field, so that we can define the variation. This is nicely reviewed in "Currents and the Energy-Momentum Tensor in Classical Field Theory — A fresh look at an Old Problem" (https://arxiv.org/abs/hep-th/0307199).

A similar definition works in quantum field theory, too, ignoring issues of regularization.

We can also have a non-lagrangian field theory whose equations of motion are not necessarily derivable from any lagrangian. We can still express the equations in terms of an arbitrary background metric (or frame) field, but how do we define the stress-energy tensor in this case? Do non-lagrangian field theories still have a stress-energy tensor?

  • I understand that Noether's theorem assumes a lagrangian, but the stress-energy tensor seems more fundamental, because even non-lagrangian systems couple to gravity, right? Shouldn't that mean that they have a stress-energy tensor?

  • The existence of a stress-energy tensor seems to be a standard axiom in non-lagrangian formulations of conformal field theory, but I don't know if these theories could be formulated in terms of a lagrangian, or if they are truly non-lagrangianable. (Can we please make that a word?)


Other posts about non-lagrangian(able) theories:

$\endgroup$
1
  • 5
    $\begingroup$ I have an additional ill-defined question. Gravity (just considering standard GR here) does have a Lagrangian description, which is one way to get that the stress-energy is the variation of the matter Lagrangian w.r.t the metric. In some vague sense, that implies that a non-Lagrangian theory with a stress-energy tensor would destroy the Lagrangian formulation of GR. How do I understand that? $\endgroup$ – kaylimekay Dec 26 '20 at 16:12
15
$\begingroup$

Most theories do not have a conserved energy-momentum tensor, regardless of whether they are Lagrangian or not. For example you need locality and Lorentz invariance. When you have those, you can define the energy-momentum tensor via the partition function (which always exists, it basically defines the QFT): $$ \langle T_{\mu\nu}\rangle:=\frac{\delta}{\delta g^{\mu\nu}}Z[g] $$ You can define arbitrary correlation functions of $T$ by including insertions. And these functions define the operator $T$ itself. So $T$ is defined whenever $Z$ is a (differentiable) function of the metric, i.e., when you have a prescription to probe the dependence of the theory on the background metric. And this prescription is part of the definition of the theory: in order to define a QFT you must specify how the partition function is to be computed, for an arbitrary background. If you cannot or do not want to specify $Z$ for arbitrary $g$, then the derivative cannot be evaluated and $T$ is undefined. (This is not an unreasonable situation, e.g. I may be dealing with a theory that is anomalous and is only defined for a special set of metrics, e.g. Kähler. This theory does not make sense for arbitrary $g$, so I may not be able to evaluate $Z[g]$ for arbitrary $g$, and therefore $T$ may not exist).

If the theory admits an action, then the dependence of $Z$ on $g$ is straightforward: it is given by whatever the path-integral computes. If the theory does not admit an action, then you must give other prescriptions by which to compute $Z$. This prescription may or may not include a definition for arbitrary $g$; if it does not, then $T$ is in principle undefined.

But anyway, for fun consider the following very explicit example: $\mathcal N=3$ supersymmetry in $d=4$. This theory is known to be non-lagrangian. Indeed, if you write down the most general Lagrangian consistent with $\mathcal N=3$ SUSY, you can actually prove it preserves $\mathcal N=4$ SUSY as well. So any putative theory with strictly $\mathcal N=3$ symmetry cannot admit a Lagrangian. Such a theory was first constructed in arXiv:1512.06434, obtained almost simultaneously with arXiv:1512.03524. This latter paper analyses the consequences of the anomalous Ward identities for all symmetry currents, in particular the supercurrent and the energy-momentum tensor.

$\endgroup$
11
  • $\begingroup$ Quick question: in (classical) GR & almost all modified gravitaional theories, we have a covariantly conserved stress-energy tensor almost by definition (assuming on-shell and a diffeomorphism invariant action). How does this fit in with your answer here, as you said most theories don't have a conserved energy-momemtum tensor, even in the Lagrangian case. Was the point that most field theories (even with Lagrangian descriptions) aren't Lorentz invariant & local, or am I missing something? $\endgroup$ – Eletie Dec 26 '20 at 22:00
  • 1
    $\begingroup$ @Eletie Yes indeed, that is exactly what I meant. $\endgroup$ – AccidentalFourierTransform Dec 26 '20 at 22:05
  • $\begingroup$ In the classical case I'm used to us writing down some non-descript matter Lagrangian where we assume both those things, so that's interesting to hear. Thanks! $\endgroup$ – Eletie Dec 26 '20 at 22:08
  • 1
    $\begingroup$ @ChiralAnomaly In general $T:=\delta Z/\delta g$ is not conserved. You need, at the very least, translation invariance. If the theory is non-lagrangian, what does it mean for it to be invariant? Well, a possible definition is that $T$ is conserved. So according to that definition, $\nabla T=0$ is trivially satisfied. Of course, if you have other definitions of invariance, then the conservation will no longer be a tautology, but I dont know of any other useful definition of translation invariance for non-lagrangian theories. $\endgroup$ – AccidentalFourierTransform Dec 28 '20 at 14:44
  • 1
    $\begingroup$ (Not to mention the fact that, as you know well enough, Ward identities may be violated by anomalies, so even if you do have translation invariance at some formal level, the conservation law may break down. I don't think we can make any general statements other than: $T$ may or may not be conserved, depending on the theory. Which is a rather boring statement...) $\endgroup$ – AccidentalFourierTransform Dec 28 '20 at 14:47
11
$\begingroup$

Not all non-Lagrangian theories have a stress-energy tensor, an example of this is the critical point of the long-range Ising model, which can be expressed as a "defect" field theory where the action consists of two pieces integrated over spaces of different dimensionality and hence has no single Lagrangian that would describe it.

See chapter 6 of "Conformal Invariance in the Long-Range Ising Model" by Paulos, Rychkov, van Rees, Zan for a discussion of this formulation and what the "missing" stress-energy tensor means for the Ward identities. That paper also refers us in a footnote to "Conformal symmetry in non-local field theories" by Rajabpour, where a "non-local stress tensor" is constructed for a class of theories where the usual kinetic term with a local Laplacian is replaced by the non-local fractional Laplacian, but this object does not behave like one would usually like a stress-energy tensor in a CFT to behave, in particular its operator product expansions are "wrong".

$\endgroup$
1
  • 1
    $\begingroup$ Interesting article -- Thanks! For the benefit of others who might be interested, here's a little more info about one of the ideas in the article: We can start with a local lagrangian theory in a higher-dimensional AdS spacetime (which does have a stress-energy tensor) and integrate out the bulk part of the fields to get a non-local lower-dimensional theory that doesn't have a stress-energy tensor but that still has conformal invariance inherited from the higher-dimensional theory. So having conformal invariance does not imply having a stress-energy tensor. $\endgroup$ – Chiral Anomaly Dec 27 '20 at 3:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.