In chapter 13 of Goldstein's classical mechanics, on page 591 when talking about Noether's theorem, Goldstein says we need condition 3, which is

$$\tag{13.133} \int_{\Omega'}\mathcal{L}(\eta_\rho'(x^\mu),\eta'_{\rho,\nu}(x^\mu),x^\mu)\,dx^4- \int_{\Omega}\mathcal{L}(\eta_\rho(x^\mu),\eta_{\rho,\nu}(x^\mu),x^\mu)\,dx^4=0$$

How is this satisfied for the Klein Gordon equation under translation? Here $\eta'_\rho=\eta_\rho$ and $\eta'_{\rho,\nu}=\eta_{\rho,\nu}$.

Using $\phi$ to denote the scalar field, I think we want $$\int_{\Omega'}\frac{1}{2}(\partial_\mu\phi\partial^\mu\phi-m^2\phi^2)=\int_{\Omega}\frac{1}{2}(\partial_\mu\phi\partial^\mu\phi-m^2\phi^2)$$

Here $\Omega'$ is a translation of $\Omega$. Even taking into account the fact that $\phi$ satisfies the Klein Gordon equation, we need $$\int_{\Omega'}\partial_\mu(\phi\partial^\mu\phi)=\int_\Omega\partial_\mu(\phi\partial^\mu\phi).$$ There is no reason for this to hold even using the divergence theorem.

  • $\begingroup$ What is your goal/question exactly? $\endgroup$
    – schris38
    Commented Aug 9, 2022 at 18:18
  • $\begingroup$ @schris38 How is this satisfied for the Klein Gordon equation under translation? $\endgroup$ Commented Aug 9, 2022 at 18:19
  • $\begingroup$ What is the second half of your question? Why do you integrate the KG equation? I am trying to understand what you want to understand, as the question "how is something satisfied for the KG equation under translation" is something I haven't heard before. $\endgroup$
    – schris38
    Commented Aug 9, 2022 at 18:21
  • $\begingroup$ @schris38 I'm not integrating the KG equation. I'm translating 13.133 of Goldstein to the case of the KG equation. The KG Lagrangian is $\frac{1}{2}(\square-m^2)\phi$. There is no second half of my question. It's the same question, I'm just stating why I think 13.133 is not satisfied. $\endgroup$ Commented Aug 9, 2022 at 18:22
  • $\begingroup$ Are you certain you have seen somewhere $\frac{1}{2}(\Box-m^2)\phi$ being referred to as a Lagrangian? Usually kinetic terms in Lagrangians contain two powers of the field... $\endgroup$
    – schris38
    Commented Aug 9, 2022 at 18:25

3 Answers 3


Here $\eta'_\rho=\eta_\rho$ and $\eta'_{\rho,\nu}=\eta_{\rho,\nu}$.

No, this is wrong. The transformed fields have the same values at the same physical location, but the same physical location is described by different coordinates in each case.

So you have, for example: $\eta'(x') = \eta(x)$.

For example, if your function in one case is a gaussian peaked at zero like $e^{-x^2}$ and your transformation is an active translation by $a$ then your function in the second case is a gaussian peaked at $a$ like $e^{-(x-a)^2}$. The functions $e^{-x^2}$ and $e^{-(x-a)^2}$ are different functions.

To provide an even more explicit example, I could write $$ \eta(x) = e^{-x^2} $$ and $$ x' = x + a\;. $$ And the inverse: $$ x = x' - a\;. $$

And so we have: $$ \eta'(x') = \eta(x(x')) = e^{-(x(x'))^2} = e^{-(x' - a)^2} $$

Using $\phi$ to denote the scalar field, I think we want $$\int_{\Omega'}\frac{1}{2}(\partial_\mu\phi\partial^\mu\phi-m^2\phi^2)=\int_{\Omega}\frac{1}{2}(\partial_\mu\phi\partial^\mu\phi-m^2\phi^2)$$

Again, this is wrong, in general. You need the transformed fields in the transformed integral (on the LHS).


It looks like you want to understand the content of equation (13.133). This equation is just the statement that the action is invariant under some transformation $\phi\mapsto \phi'$. Being more precise, the action is a functional $S[\phi]$ of the fields in the theory. In general it is written as the integral of some Lagrangian density: $$S[\phi]=\int d^dx {\cal L}(\phi(x),\partial_\mu \phi(x),\partial_\mu \partial_\nu \phi(x))\tag{1}.$$

The Lagrangian density ${\cal L}$ is actually a function of finitely many variables which gets evaluated when these variables take on the values $\phi(x)$, $\partial_\mu \phi(x)$, $\partial_{\mu}\partial_\nu\phi(x)$ and so on for each $x$.

Now consider performing one transformation $\phi\mapsto \phi'$. Then presumably your action would change and this is quantified by $$\delta S=S[\phi']-S[\phi]\tag{2}.$$

Saying that $\delta S=0$ means that the field transformation leaves the action invariant and this is exactly (13.133). This is the definition of a symmetry: a symmetry of the action $S[\phi]$ is defined to be a transformation of the fields keeping the action invariant.

So to check a symmetry you need to know the transformation. A scalar field $\phi(x)$ transforms under translations as: $$\phi'(x)=\phi(x-a)\tag{3}.$$

This is a definition. Observe that $\phi'(x)$ and $\phi(x)$, when compared at the same point, are different. Since equality between functions is pointwise equality, this means that $\phi'(x)$ and $\phi(x)$ are two different functions on spacetime, as expected since you transformed your field configuration. How the field changes at each point is quantified by the variation $\delta \phi(x)=\phi'(x)-\phi(x)$.

Now, given (3) you can check the KG action is invariant. To do so, observe that using the chain rule, the derivatives transform just as $\phi$ itself, $\partial_\mu \phi'(x)=(\partial_\mu \phi)(x-a)$ and $\partial_\mu \partial_\nu\phi'(x)=(\partial_\mu \partial_\nu \phi)(x-a)$. The brackets around the derivative are just to remind that using the chain rule we will get the function $\partial_\mu\phi$ evaluated at $x-a$ and so on.

The point now is that we don't even need the specific KG action. All we need is that all terms in the KG Lagrangian are built from the field and its partial derivatives, this means that we have

$${\cal L}(\phi'(x),\partial_\mu \phi'(x),\partial_\mu \partial_\nu \phi'(x))={\cal L}(\phi(x-a),(\partial_\mu \phi)(x-a),(\partial_\mu \partial_\nu \phi)(x-a)).\tag{4}$$

Now substitute into the action. You will find that

$$S[\phi']=\int d^dx {\cal L}(\phi(x-a),(\partial_\mu \phi)(x-a),(\partial_\mu \partial_\nu \phi)(x-a))\tag{5}.$$

Now make a change of variables $x\to x+a$. This is just a translation so the measure is kept invariant, and comparing to (1) we see that



I am not sure I understand 100% what you are asking, but I will attempt to write down an answer. If that still does not make things clear, you can always comment.

So, from what I understand, you are trying to verify that, given a Klein-Gordon (free) field, performing a translation will leave the field invariant (correct me if I am wrong) and hence Eq. (13.133) in Goldstein's Classical Mechanics will hold $$\int_{\Omega'}d^4x\mathcal{L}\Big(\phi'(x^{\mu}),\phi'_{,\rho}(x^{\mu}),x^{\mu}\Big)- \int_{\Omega}d^4x\mathcal{L}\Big(\phi(x^{\mu}),\phi_{,\rho}(x^{\mu}),x^{\mu}\Big)=0$$ for the case of the scalar field! Recall that the Klein-Gordon Lagrangian is given by $$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi- \frac{1}{2}m^2\phi^2$$

So, you can substitute if you want with the result being $$\int_{\Omega'}d^4x\Big(\frac{1}{2}\partial_{\mu}\phi'\partial^{\mu}\phi'- \frac{1}{2}m^2\phi^{'2}\Big)- \int_{\Omega}d^4x\Big(\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi- \frac{1}{2}m^2\phi^2\Big)=0$$ and check that upon expanding according to $\phi\rightarrow\phi'=\phi+\delta \phi$, with $\delta\phi=a^{\mu}\partial_{\mu}\phi$ (under translations), one can obtain the Klein-Gordon equation.

If you prefer to think it in the reverse direction (despite that no one ususally thinks it that way), you can. Take the KG equation, multiply with another field and integrate (presumably you need to add and subtract some terms) and the result will be Eq. (13.133) of the book. If you need more details on how this can be done, please comment. However, I am not so keen in providing them, because usually we derive the equations of motion (K.G. Eq.) from the demand that the action be invariant under some set of transformations instead of taking the equations of motion (K.G. Eq.), assuming that they hold and reverse engineering to reach the conclusion that the action is invariant!

If I misunderstood your question or if anything is not clear, please comment. I hope this helps...


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