I've seen this same question before Why is there an extra term in definition of Noether current for spacetime translations? but I didn't understand the answer that was given so I would like to ask again:

If the fields/coordinates are changed in a way s.t. the E.L. equations are still obeyed after the transformation there is a conserved Noether current:

$$ J^\mu = \frac{\partial L}{\partial (\partial_{\mu}\phi)}\partial_{\mu}(\delta\phi)+L\delta x^\mu $$

This is because if the E.L. equations are obeyed, we have $\delta S=0$. But:

$$\delta S=\int \delta L dx^{4}$$ $$\delta L=\frac{\partial L}{\partial x^{\mu}}\delta x^{\mu}+\frac{\partial L}{\partial \phi}\delta \phi+\frac{\partial L}{\partial (\partial_{\mu}\phi)}\delta (\partial_{\mu}\phi)$$

The first term on the right exist only when the Lagrangian has explicit coordinate dependence. We assume the system obeys the E.L. equations and so the middle term can be replaced and the last term adjusted:

$$\delta L=\frac{\partial L}{\partial x^{\mu}}\delta x^{\mu}+\frac{\partial}{\partial x^{\mu}}\left(\frac{\partial L}{\partial (\partial_{\mu}\phi)}\right)\delta \phi+\frac{\partial L}{\partial (\partial_{\mu}\phi)}\partial_{\mu}(\delta \phi)$$

if we add the term $L\delta(\partial_{\mu}x)=0$ on the right side and we impose $\delta L=0$, which we can do as long as $\delta\phi$ and $\delta x^{\mu}$ are not zero at the boundary, we get:

$$\partial_{\mu}\left(\frac{\partial L}{\partial (\partial_{\mu}\phi)}\partial_{\mu}(\delta\phi)+L\delta x^\mu\right)=0$$

Clearly for a Lagrangian that does not depend explicitly on the coordinates the $L\delta x^{\mu}$ should be absent, yet I still see this formula used for Lagrangians that do not have explicit coordinate dependence. Can somebody please explain this to me?

  • 1
    $\begingroup$ What is it that you don't understand about the answers given in the linked post? $\endgroup$
    – my2cts
    Commented Feb 1, 2021 at 16:09

1 Answer 1


Consider the total derivative of the Langrange density with respect to position,

$$ \frac{dL}{dx}=\frac{\partial\Phi}{\partial x}\frac{\partial L}{\partial \Phi} +\frac{\partial d\Phi}{\partial x}\frac{\partial L}{\partial d\Phi}+\frac{\partial L}{\partial x} $$

With $$ \delta\Phi = \delta a \frac{\partial\Phi}{\partial x}, d\delta\Phi = d(\delta a \frac{\partial\Phi}{\partial x})=\delta a \frac{\partial d\Phi}{\partial x} $$ We find the variation of Langrange density with respect to our field: $$ \delta_{\Phi}L=\delta\Phi \frac{\partial L}{\partial \Phi} + d\delta\Phi\frac{\partial L}{\partial d\Phi}=\delta a (\frac{\partial\Phi}{\partial x}\frac{\partial L}{\partial \Phi} +\frac{\partial d\Phi}{\partial x}\frac{\partial L}{\partial d\Phi}) $$

One can see, that if $\frac{\partial L}{\partial x}=0$, which means when we do not have explicit coordinate dependence,following holds: $$ \delta_a L = \delta_\Phi L $$ And this so called invariance condition gives you the term you ask. Invariance with respect to field: $$ \delta_\Phi L=\partial_\mu(\frac{\partial L}{\partial(\partial_\mu \Phi_k)})(\delta\Phi_k) $$ Invariance with respect to position: $$ \delta_a L= \partial_\mu L\delta x^{\mu} $$ Rearranging gives the desired solution. As you can see, the invariance condition does not makes the last term absent, this term is neccecary. You can see this as follows. When you make a variation to your Langrange density, there are two effects: First, the field changes; second , the basis changes. The last term gives you the effect of "change of basis" at a given point.

A second way to "expect for" a coordinate dependence can be explained from the interpretation of the invariance condition. Invariance means, that the equations of motion in a system under a translation do not change. Then we must consider the transformations of the following kind: $$ \mathcal{L}\rightarrow \mathcal{L}+\partial_{\mu}J^{\mu} $$ In the variation of action one can simplify the second term to a surface integral with Stokes Theorem, and the surface terms are considered to be zero in the variation calculation. Therefore we can see the term $J^{\mu}$ is conserved as well, and the Noether current must be dependent on this term too. This term can be calculated as the derivation above.


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