I am rather stumped by David Tong's derivation of the energy-momentum tensor for a translationally invariant theory because it appears it doesn't assume any type of Lagrangian at all.

A Lagrangian $\mathcal{L}(\phi,\partial_\mu \phi)$ has a symmetry $\phi \rightarrow \phi + \delta \phi $ if the off-shell variation is given to first order by a total derivative

$$ \delta \mathcal{L} = \partial_\mu F^\mu(\phi). $$

Tong argues that if we substitute in a particular field configuration into the Lagrangian then we can define a function $\mathcal{L}(x)\equiv \mathcal{L}(\phi(x),\partial_\mu \phi(x))$. Under a translation of the fields $\phi(x) \rightarrow \phi'(x) = \phi(x-\epsilon) $, we have

$$ \mathcal{L}(x) \rightarrow \mathcal{L}(x-\epsilon) = \mathcal{L}(x) - \epsilon^\mu \partial_\mu\mathcal{L}(x)$$

which is a total derivative, as shown in (1.40). This did not assume any form of the Lagrangian so is this telling me that all Lagrangians are translationally invariant? This seems to apply to Lorentz transformations too, despite not assuming a Lorentz invariant Lagrangian, as seen in equation (1.53).

If I take this idea further and suppose I performed a conformal transformation described by a transformation $ x^\mu \rightarrow x^\mu + \epsilon^\mu(x)$, where

$$ \epsilon^\mu(x) = a_\mu + b_{\mu \nu}x^\nu + c_{\mu \nu \rho} x^\nu x^\rho $$

as given by equation (2.7) of "Intro to CFT by Blumenhagen and Plauschinn", then I would say the off-shell variation is, from the arguments above, given by

$$ \delta \mathcal{L} = - \epsilon^\mu(x) \partial_\mu \mathcal{L}(x) $$

which can't be massaged into a total derivative. Is this telling me that no theories are conformally invariant? I know this is not true but I do not know how one could write this as a total derivative to fulfil the definition of a symmetry.

  • $\begingroup$ Lagrangians of what? The Lagrangian of a simple harmonic oscillator has explicit dependence on position. $\endgroup$
    – G. Smith
    Apr 28, 2020 at 0:45
  • $\begingroup$ @G.Smith I am talking about Lagrangian densities of the general form $\mathcal{L}(\phi,\partial_\mu \phi)$ as stated in my question. I am talking about quantum field theory as my tags show, not point particle mechanics $\endgroup$ Apr 28, 2020 at 0:59
  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Apr 29, 2020 at 4:12

3 Answers 3


When you write $\mathcal{L}(x)=\mathcal{L}(\phi(x),\partial_\mu\phi(x))$, you're assuming translational invariance. A more general Lagrangian is written $\mathcal{L}(x)=\mathcal{L}(\phi(x),\partial_\mu\phi(x),x)$.

  1. Yes they are all translationally invariant as long as they don't depend explicitly on $x^\mu$. As Tong assumes by writing $\mathcal{L}(\phi,\partial_\mu\phi)$.

  2. No, the argument wouldn't go through for Lorentz transformations as they also affect the indices $\mu$ in $\partial_\mu$ and in whatever other field that is not a scalar. That is, $\phi'_{\mu_1\cdots \mu_\ell}(x)$ is not $\phi_{\mu_1\cdots \mu_\ell}(x-\epsilon)$.

  3. The argument wouldn't go through for conformal transformations neither. Both for the reason above and for the fact that if $\epsilon$ depends on $x$ then it does not go through derivatives and so $\partial_\mu\phi$ transforms differently than $\phi$.

  4. That's not necessarily the form of the Lagrangian after a conformal transformation.

  • $\begingroup$ Thank you for your answer. In equation (1.53) of the notes, it is stated that under a Lorentz transformation the field varies as $ \delta \phi = - \omega^\mu_\nu x^\nu \partial_\mu \phi $. Tong then states that the Lagrangian transforms the same way too as $ \delta \mathcal{L} = - \omega^\mu_\nu x^\nu \partial_\mu \mathcal{L}$ which can be written as a total derivative. Surely no assumption of Lorentz invariance has been made? $\endgroup$ Apr 28, 2020 at 0:32
  • $\begingroup$ Surely if I transform my fields via any infinitesimal space-dependent transformation, I would have, by Taylor’s theorem, that $\mathcal{L}(x-\epsilon^\mu(x)) = \mathcal{L} -\epsilon^\mu \partial_\mu \mathcal{L}$. This doesn’t care about the exact functional form of the Lagrangian, just that it is some function of x? That is a function of x when a field has been substituted in $\endgroup$ Apr 28, 2020 at 0:39
  • $\begingroup$ @Matt0410 Re your last comment: yes, what you say is true, but this by itself does not imply anything. You still need to use Noether's theorem, and in the proof of that theorem equation (1.36) would fail if there were an explicit $x$-dependence in the Lagrangian. $\endgroup$ Apr 28, 2020 at 4:15

One approach to this is to invoke Nother's theorem. For every symmetry of the system, there is a corresponding conserved quantity. For spatial transnational symmetry, that conserved quantity is momentum (linear momentum). So all systems which conserve momentum will have this symmetry.

One way you can avoid conservation of momentum is to design a system which couples to an external force, permitting momentum to be transferred out of the system.


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