Let us consider $N$ independent scalar fields which satisfy the Euler-Lagrange equations of motion and are denoted by $\phi^{(i)}(x) \ ( i = 1,...,N)$, and are extended in a region $\Omega$ in a $D$-dimensional model spacetime $\mathcal{M}_D$. Now consider the classical Lagrangian density, $\mathcal{L}(\phi^{(i)}, \partial_\mu \phi^{(i)}, x^\mu)$. We apply the following infintesimal fixed-boundary transformation to $\mathcal{M}_D$. \begin{align*} x \to \widetilde{x}^\mu &\equiv x^\mu + \delta x^\mu (x), \tag{1} \\ \text{such that, }\ \delta x^\mu\Big{|}_{\partial\Omega}&=0, \tag{2} \\ \text{and the fields transform as: }\ \phi^{(i)}(x) &\to \widetilde{\phi}^{(i)}(\widetilde{x}) \equiv \phi^{(i)} (x) + \delta\phi^{(i)} (x). \tag{3} \\ \end{align*}

According to my calculations, up to first order in the variation, the Lagrangian density is given by: $$ \boxed{ \delta \mathcal{L} = \partial_\mu \Big( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\delta\phi^{(i)} - \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\partial_\nu \phi^{(i)} \delta x^\nu + \mathcal{L} \delta x^\mu \Big) - \mathcal{L} \partial_\mu (\delta x^\mu) }\tag{4} $$

Therefore, the conserved current is $$ \boxed{ J^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\delta\phi^{(i)} - \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\partial_\nu \phi^{(i)} \delta x^\nu + \mathcal{L} \delta x^\mu - F^\mu } \tag{5}$$ where $F^\mu$ is some arbitrary field that vanishes on $ \partial \Omega$.

However, most textbooks ignore the second and the third terms in the above expression. Compare, for example, with Peskin and Schroeder (p.18) which sets:

$$ J^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\delta\phi^{(i)} - F^\mu. \tag{6} $$

For another example, Schweber (p. 208) ignores all terms but the first in the variation of the Lagrangian density, and writes:

$$ \delta \mathcal{L} = \partial_\mu \Big( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\delta\phi^{(i)} \Big).\tag{7} $$

So what is going on here? Am I missing something? We seem to have set the same assumptions, but get different results. Am I wrong, or are they?

EDIT: Condition (2) is unnecessary, as it was never used in the derivation of the current. Please ignore its presence in the above text.

  • $\begingroup$ Here is a derivation of my result, if you would like to read. $\endgroup$ – Meghana Dec 30 '16 at 8:57
  1. Eq. (5) is (up to factors of the infinitesimal parameter $\varepsilon$) the standard expression for the full Noether current. Here:

    • $\delta x^{\mu}$ is the so-called horizontal component of the infinitesimal variation;
    • $\delta \phi -\frac{\partial \phi}{\partial x^{\mu}} \delta x^{\mu} $ is the so-called vertical component of the infinitesimal variation;
    • $F^{\mu}$ is an improvement term in case of quasisymmetry.
  2. The main point is that Schweber (7), Peskin & Schroeder (6) are only considering situations with purely vertical transformations, i.e. situations where $\delta x^{\mu}=0$.

  3. Let us mention that the last term in eq. (4) gets cancelled by the Jacobian contributions from the integration measure. Hence it is not present in eq. (5).

  4. Finally, it seems relevant to mention that OP's boundary condition (2) is often not fulfilled in important applications, such as the canonical stress-energy-momentum (SEM) tensor, which is the Noether current for spacetime translations. See e.g. this Phys.SE post. Therefore the boundary condition (2) should be relaxed appropriately. Similarly, the improvement term $F^{\mu}$ is not some arbitrary field that vanishes on the boundary, as OP claims (v3) under eq. (5). Instead the improvement term $F^{\mu}$ is dictated by the quasisymmetry, which fixes $F^{\mu}$ up to a divergence-free term.

  • $\begingroup$ The distinction between horizontal and vertical infinitesimal variations seems to suggest that they are linearly independent quantities, which they are clearly not. So what do you mean when you only consider vertical transformation? Do you set $ \delta x^\mu = 0 $? If $\delta \phi$ is induced by $ \delta x^\mu$, then that would mean the vanishing of the vertical variation as well. What am I missing? $\endgroup$ – Meghana Dec 31 '16 at 15:16
  • $\begingroup$ Well, that's a misunderstanding. I updated the answer with a hopefully clearer formulation. $\endgroup$ – Qmechanic Dec 31 '16 at 15:33
  • $\begingroup$ Thank you. I understand your point. I would like to ask one last question about your statement that $ F^\mu $ is necessarily dictated by quasi-symmetry. Since we know that $ \int_\Omega \partial_\mu F^\mu = F^\mu\Big|_{\partial \Omega}$, we can simply add an arbitrary divergence to the variation of the action, provided we satisfy $ F^\mu\Big|_{\partial \Omega} = 0 $. I do not see why we would need to restrict the structure of this field any further. Perhaps it is convenient to do so in certain situations, but it should not be necessary. Could you please explain your stand? $\endgroup$ – Meghana Dec 31 '16 at 17:02
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Dec 31 '16 at 17:20

The issue is that there are two ways to write an infinitesimal field transformation. As a simple example, let's consider a triplet of fields $\phi_i$ which transform as a vector in space, and suppose we're dealing with a rotational symmetry. We can write this symmetry in two ways:

  • Your method: the rotation changes the spatial coordinates (your $\delta x^\mu$) and changes the value of the field by rotation (your $\delta \phi^i$).
  • The more common method: the rotation only changes the value of the field while holding spatial coordinates constant, i.e. $\delta x^\mu = 0$.

While it looks like your method is more general, the second method works equally well, as any shift in the coordinates by a small $\delta x^\mu$ is equivalent to a shift in the field value by $\partial_\mu \phi^i \delta x^\mu$.

Setting $\delta x^\mu = 0$ in Peskin and Schroeder's answer gives yours, so they agree with you, except that their $\delta \phi$ will be more complicated. The Schweber book is a little more basic and probably dropped the total derivative just to simplify things.

  • $\begingroup$ You claim that in the more common method, $ \delta x^\mu = 0 $, and also maintain that $ \delta\phi^i \propto \partial_\mu \phi^i \delta x^\mu $. Wouldn't that mean $ \delta\phi^i = 0$ everywhere? How do you explain this? $\endgroup$ – Meghana Dec 31 '16 at 14:40
  • $\begingroup$ @Meghana Sorry, I used bad notation. In the first equation I mean that there's no spatial transformation, so the $\delta x^\mu$ terms in your version of the Noether current don't appear. In the second equation, I mean that an extra change should be added to $\delta \phi$ to compensate. In this context, $\delta x^\mu$ is simply a four vector and has no meaning beyond that. It's equal to whatever the spatial transformation would have been. $\endgroup$ – knzhou Dec 31 '16 at 19:20

A Noether current is always connected to some transformation. If you drop the second and third terms in the second box, you have the current for a pure field transformation with no coordinate transformation. Note that the field transformation has two parts: One originates from a given field shift, the other induced by a coordinate transformation. If, for example, you would set the pure field shift to zero and keep only the part induced by the coordinate shift, you would get the energy-momentum tensor of the theory.

Correction: You only get the energy-momentum tensor as Noether current if you set the coordinate transformation to be space-time translations.


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