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. $\endgroup$
    – Foshiba
    Dec 30, 2016 at 8:57

5 Answers 5

  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? $\endgroup$
    – Foshiba
    Dec 31, 2016 at 15:16
  • $\begingroup$ Well, that's a misunderstanding. I updated the answer with a hopefully clearer formulation. $\endgroup$
    – Qmechanic
    Dec 31, 2016 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? $\endgroup$
    – Foshiba
    Dec 31, 2016 at 17:02
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Dec 31, 2016 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? $\endgroup$
    – Foshiba
    Dec 31, 2016 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, 2016 at 19:20

The purpose of this answer is to elaborate on the answer by Knzhou. We have the following statement:

Theorem: Suppose that $\delta$ is a general (i.e. not necessarily vertical) quasisymmetry of the action $S$. Then there is an equivalent vertical quasisymmetry $\delta^\ast$, under which the action as also quasi-invariant with the same Noether current.

This statement has two caveats:

  • Although the full Noether currents are equivalent, the distribution of the "bare" Noether current and the "improvement term" are not the same in the two cases. In particular if $\delta$ is an exact symmetry, $\delta^\ast$ will be a quasisymmetry in general.
  • The corresponding vertical symmetry $\delta^\ast$ is a generalized symmetry in general, even if $\delta$ is an ordinary symmetry. The difference between the two will be explained in the body in the answer.

Suppose that the action is$$S[\phi]=\int_\Omega\mathcal L(x,\phi(x),\partial\phi(x))d^nx,$$ where $\Omega$ is a compact $n$ dimensional domain of integration, the field is $\phi^i(x)$ with $m$ components and $n$ independent variables $x^\mu$. For simplicity, a first order Lagrangian is assumed, but the result is qualitatively valid for the higher order case as well (i.e. the specific formulae are different, but the overall result is the same).

We first consider a variation $\delta$ given by $$ x^{\prime\mu}=x^\mu+\epsilon\delta x^\mu,\quad\phi^{\prime i}(x^\prime)=\phi^i(x)+\epsilon\delta\phi^i(x). $$ We assume this variation is an off-shell quasisymmetry of the action, i.e. $$ \delta S[\phi]=\int_\Omega d_\mu F^\mu\ d^nx$$ for some improvement current $F^\mu$. Let $L=\mathcal Ld^nx$ be the Lagrangian $n$-form and define the total variation $\delta_T\mathcal L$ of the Lagrangian density to be $$ \delta L=\delta_T\mathcal L\ d^nx. $$

As OP has correctly derived we have $$ \delta_T\mathcal L=E_i(L)(\delta\phi^i-\partial_\mu\phi^i\delta x^\mu)+d_\mu\left[\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta\phi^i-\left(\frac{\partial\mathcal L}{\partial\phi^i_\mu}\phi^i_\nu-\mathcal L\delta^\mu_\nu\right)\delta x^\nu\right] \\ = E_i(L)(\delta\phi^i-\partial_\mu\phi^i\delta x^\mu)+d_\mu\left[\frac{\partial\mathcal L}{\partial\phi^i_\mu}\left(\delta\phi^i-\partial_\nu\phi^i\delta x^\nu\right)+\mathcal L\delta x^\mu\right] \\ = E_i(L)\delta^\ast\phi^i+d_\mu\left[\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta^\ast\phi^i+\mathcal L\delta x^\mu\right],$$ where $E_i(L)$ is the Euler-Lagrange expression of the Lagrangian, and $$ \delta^\ast\phi^i=\delta\phi^i-\partial_\mu\phi^i\delta x^\mu $$is the vertical part of the variation.

The full Noether current is then $$ J^\mu=\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta^\ast\phi^i+\mathcal L\delta x^\mu-F^\mu . $$

Now consider making only $\delta^\ast\phi^i$ as a variation with no horizontal part. The variation of the action is $$ \delta^\ast S=\int_\Omega\left(E_i(L)\delta^\ast\phi^i+d_\mu\left(\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta^\ast\phi^i\right)\right)d^nx. $$ Comparing this with the integral of $\delta_T\mathcal L$, we find $$ \delta^\ast S=\delta S-\int_\Omega d_\mu(\mathcal L\delta x^\mu)d^nx. $$

It follows that if $\delta$ is a quasisymmetry of $S$ with improvement term $F^\mu$, then $\delta^\ast S$ is also a quasisymmetry of $S$ with improvement term $\bar F^\mu=F^\mu-\mathcal L\delta x^\mu$. The full Noether current from this symmetry is then $$ J^\mu=\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta^\ast\phi^i-\bar F^\mu=\frac{\partial\mathcal L}{\partial\phi^i_\mu}\delta^\ast\phi^i+\mathcal L\delta x^\mu-F^\mu, $$ which agrees with the previous current.


Clearly, if the improvement term vanishes for $\delta$, i.e. $F^\mu=0$, then the corresponding vertical variation still has an improvement term $-\mathcal L\delta x^\mu$. Thus - as stated in the intro - a non-vertical exact symmetry may only be replaced with a vertical quasisymmetry.

Secondly, the variations that appear in Noether's theorem are such that they are not variations about a specific field $\phi^i$, but rather, the variation of any field can be calculated. In other words, $\delta$ is a "vector field" rather than a single "tangent vector" in field space. Ordinarily, these variations have the functional form $$ \delta x^\mu=\xi^\mu(x),\quad\delta\phi^i(x)=\Xi^i(x,\phi(x)). $$ If the variations have this functional dependency, then they generate projectable flows in the $n+m$-dimensional total space of both dependent and independent variables $(x^\mu,y^i)$ (this point would be made much more understandable in a fibre bundle formulation, which I am not doing for the sake of accessibility).

However for vertical variations we may also consider variations of the form $$ \delta\phi^i(x)=\Xi^i(x,\phi(x),\partial\phi(x),...,\partial^r\phi(x)), $$ and this is called a generalized variation (and if a symmetry, then a generalized symmetry) (further remark: One could in principle consider any functional $\delta\phi^i(x)=\Xi^i[\phi](x)$, however in the spirit of locality, one usually considers finite order functionals only).

Generalized vertical variations do not generate flows in the $(x,y)$-space, but they generate flows in the field space, via the equation $$ \frac{\partial\phi^i_\epsilon}{\partial\epsilon}=\Xi^i(x,\phi_\epsilon(x),...,\partial^r\phi_\epsilon(x)). $$

With this in mind, if $\delta$ is an ordinary, non-vertical variation with $$ \delta x^\mu=\xi^\mu(x),\quad \delta\phi^i(x)=\Xi^i(x,\phi(x)), $$ then the corresponding vertical variation has the form $$ \delta^\ast\phi^i(x)=\mathrm Z^i(x,\phi(x),\partial\phi(x))=\Xi^i(x,\phi(x))-\partial_\mu\phi^i(x)\xi^\mu(x), $$ which shows that $\delta^\ast$ is actually a generalized vertical symmetry.

Thus - as stated in the intro - a non-vertical ordinary variation may only be replaced by a vertical generalized variation in general.


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.


You're missing the variational on the volume element $d^nx$, which is non-zero, if there are non-zero variationals on the coordinates; and are not distinguishing between total and internal variationals.

First: I'm going to write $φ^a_μ$ for the gradients $∂_μφ^a$ (and use $a$ as the field index, instead of $(i)$). Second: I'm going to write the derivatives with respect to the Lagrangian density $𝔏$ with inverted indices: $$φ^a_μ ≡ ∂_μ φ^a,\quad 𝔏_μ = \frac{∂𝔏}{∂x^μ},\quad 𝔏^μ_a = \frac{∂𝔏}{∂φ^a_μ}.$$ where $$∂_μ ≡ \frac{∂}{∂x^μ}.$$ The variational on the Lagrangian density $𝔏$ is: $$\begin{align} δ𝔏 &= 𝔏_μ δx^μ + 𝔏_a δφ^a + 𝔏^ν_a δφ^a_ν\\ &= 𝔏_μ δx^μ + 𝔏_a \left(\bar{δ}φ^a + δx^ν ∂_νφ^a\right) + 𝔏^ν_a \left(\bar{δ}φ^a_ν + δx^μ ∂_μ φ^a_ν\right)\\ &= \left(𝔏_μ + 𝔏_a ∂_μ φ^a + 𝔏^ν_a ∂_μ φ^a_ν\right) δx^μ + 𝔏_a \bar{δ}φ^a + 𝔏^μ_a \bar{δ}φ^a_μ. \end{align}$$ The total differential of the Lagrangian density $𝔏$, that chain-rules the $x$-dependence on the field components is: $$∂_μ𝔏 = 𝔏_μ + 𝔏_a ∂_μ φ^a + 𝔏^ν_a ∂_μ φ^a_ν$$ Thus $$δ𝔏 = ∂_μ𝔏 δx^μ + 𝔏_a \bar{δ}φ^a + 𝔏^μ_a \bar{δ}φ^a_μ.$$

There are two types of variationals to consider. The total variation $δF(x)$ is the infinitesimal form of $F'(x') - F(x)$, and the internal variational $\bar{δ}F(x)$ is the infinitesimal form of $F'(x) - F(x)$. Their chief difference is that $\bar{δ}(⋯) = 0$, for anything that involves $x$'s alone. For the field components and their gradients, they are related by $δ = \bar{δ} + δx^μ ∂_μ$.

For coordinate differentials: $$δdx^μ = dδx^μ = ∂_ν δx^μ dx^ν.$$ For the volume $n$-form: $$d^nx = dx^0 ∧ dx^1 ∧ ⋯ ∧ dx^{n-1}.$$ we have: $$δ\left(d^nx\right) = δdx^μ ∧ ι_μ d^nx = ∂_ν δx^μ dx^ν ∧ ι_μ d^nx = ∂_ν δx^μ δ^ν_μ d^nx = ∂_μ δx^μ d^nx.$$ Thus, for the action integral $S = \int 𝔏 d^n x$, we have: $$δS = \int δ\left(𝔏 d^n x\right) = \int \left(∂_μ𝔏 δx^μ + 𝔏_a \bar{δ}φ^a + 𝔏^μ_a \bar{δ}φ^a_μ + 𝔏 ∂_μ dx^μ\right) d^nx.$$

Internal variationals are subject to: $$\bar{δ}φ^a_μ = ∂_μ\bar{δ}φ^a\quad⇒\quad \bar{δ}φ^a_μ = \bar{δ}∂_μφ^a = ∂_μ\bar{δ}φ^a,$$ so we can integrate by parts to: $$δS = \left(𝔈_a \bar{δ}φ^a + ∂_μ\left(𝔏 δx^μ + 𝔏^μ_a \bar{δ}φ^a\right)\right) d^nx.$$ where $$𝔈_a ≡ 𝔏_a - ∂_μ𝔏^μ_a,$$ is the Euler-Lagrange term.

Under a quasi-symmetry: $$δS = \int ∂_μ Λ^μ d^nx$$ we then have the integral identity: $$ 0 = \int \left(𝔈_a \bar{δ}φ^a + ∂_μ Θ^μ\right) d^nx $$ where $$Θ^μ = 𝔏 δx^μ + 𝔏^μ_a \bar{δ}φ^a - Λ^μ$$ from which follows (using the usual "since the variational was general" argument) the vanishing of the integrand: $$𝔈_a \bar{δ}φ^a + ∂_μ Θ^μ = 0.\tag{1}\label{1}$$

Restoring total variationals, and using a little index-gynmastics (i.e. $𝔏 δx^μ = 𝔏 δ^μ_ν δx^ν$) one has: $$Θ^μ = \left(𝔏 δ^μ_ν - 𝔏^μ_a φ^a_ν\right) δx^ν + 𝔏^μ_a δφ^a - Λ^μ.$$

"On-shell" equality, denoted by $≅$ means subject to $𝔈_a ≅ 0$, while "off-shell" refers to the general case. So, the on-shell version of the identity ($\ref{1}$) is: $$∂_μ Θ^μ ≅ 0.$$


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