Technicality with Noether's theorem

[CAF123]

If we consider a transformation of a field $\Phi \rightarrow \Phi + \alpha \frac{\partial \Phi}{\partial \alpha}$ which is not a symmetry of a lagrangian then one can show that the Noether current is not conserved but that instead $\partial_{\mu}J^{\mu} = \frac{\partial L}{\partial \alpha}$.

I think the way this is derived is as follows $$\delta S = \int d^4 x \left( \left( \frac{\partial L}{\partial \Phi} \delta \Phi - \partial_{\mu} \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \right) \delta \Phi + \partial_{\mu} \left(\frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) \right)$$ Then the first term is zero due to the equations of motion and so we are left with the second term.

Writing out $\delta L = \frac{\partial L}{\partial \Phi}\delta \Phi + \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta (\partial_{\mu} \Phi)$, inserting what $\delta \Phi$ is we see that $\delta L = \alpha \frac{\partial L}{\partial \alpha}$ Then we can compare with the above and deduce the result more or less.

My questions are:
What permits the use of the equations of motion here? If the equations of motion hold then $\delta S = 0$ identically. Using the equations of motion gives me $\int \partial_{\mu} J^{\mu} d^4 x = \delta S$ but since I used the equations of motion isn't this just equal to zero?

 

[CAF123]

Basically I am asking: for a transformation which is not a symmetry of the action then $\delta S \neq 0$. However to derive $\delta S = \int d^4 x\, \alpha \partial_{\mu} J^{\mu}$ we had to use the EL equations which impose $\delta S = 0$ and the vanishing of the BC's. So what is the reconcilation of these two statements? I think the resolution is intricate and just wondered if anyone could provide any clarification.

 

[vanhees71]

While a symmetry refers to the action functional (i.e., it leaves the variation of the action unchanged as a functional of the fields), the conservation of the Noether charge (i.e., the validity of the continuity equation for the Noether current) is (in general) for the solutions of the field equations of motion, which are given by the Euler-Lagrange equations of Hamilton's principle.

I don't, however, understand your statements about a transformation which is not a symmetry in this sense. How do you conclude that the variation is a total four-divergence as written in your OP? Have read this in some textbook and if so in which one? (I hope, it's not from a peer-reviewed paper, because then I'd be troubled about a maybe failed review process ;-)).

 

[CAF123]

Hi vanhees71, what I write in my OP is the solution to the posed question in one of my tutorial questions for a QFT course that I did last year :)

 

[vanhees71]

Hm, but how do you prove that you get this four divergence if the transformation is not a symmetry transformation?

 

[CAF123]

For general variation of the action we get $$\delta S = \int d^4x \, \left[ \left( \frac{\partial L}{\partial \phi} - \partial_{\mu} \frac{\partial L}{\partial( \partial_{\mu} \phi)} \right) \delta \phi + \partial_{\mu} \left (\frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta \phi \right)\right]$$ Now, if we make this non symmetry transformation upon a trajectory that solves the EL equations then the first term vanishes and we are left with $$\delta S = \int d^4 x\, \partial_{\mu} \left (\frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta \phi \right)$$

This is what my solutions say to the exercise but
1) for a non symmetry transformation $\delta S \neq 0$ but (on shell) the above shows it is equal to integral over four divergence which yields a boundary term but isn't BC's always assumed to be zero? If not, why not?
2) EL equations were derived by imposing $\delta S = 0$ so when we used the EL equations should that not enforce $\delta S = 0$ regardless of the transformation made?

 

[strangerep]

For general variation of the action we get $$\delta S = \int d^4x \, \left[ \left( \frac{\partial L}{\partial \phi} - \partial_{\mu} \frac{\partial L}{\partial( \partial_{\mu} \phi)} \right) \delta \phi + \partial_{\mu} \left (\frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta \phi \right)\right]$$

In your post #1 you said "not a symmetry of the Lagrangian", but in post #7 you seem to be trying to talk about a non-symmetry of the action. These are not the same thing in general, so this possible confusion needs to be straightened out. I'm guessing that what you're really working with is a group of transformations which does not leave $L$ invariant, but does leave the action $S$ invariant. (If it doesn't leave $S$ invariant, then what's the point??)

(I'll defer any further attempted explanation until we straighten this out, since it's crucial.)

Btw, do you still have a copy of Greiner & Reinhardt "Field Quantization" (which I used earlier when we discussed Noether's thm a while back)? It might be easier to resolve the current difficulty if we could refer to a common textbook.

 

[CAF123]

Hi strangerep,

Yes, sorry I seem to have been a bit sloppy there - although actually perhaps it is ambiguous too as to whether the action is invariant or not because the question doesn't state that the resulting transformation changes the lagrangian by a total derivative (needed for S to be invariant). If the action is indeed invariant then wouldn't that mean that we would still have had $\partial_{\mu} J^{\mu} = 0$? ( the solution to the question posed was $\partial_{\mu} J^{\mu} = \partial L/ \partial \alpha$). Perhaps this solution indicates that the transformation really is just arbitrary leaving the lagrangian and action non invariant?

I will check if the library has a copy of Greiner.
Thanks!

 

[strangerep]

[...] perhaps it is ambiguous too as to whether the action is invariant or not because the question doesn't state that the resulting transformation changes the lagrangian by a total derivative (needed for S to be invariant).

In general, one must also consider a possible change of integration variable. But that's probably not what's going on here.

If the action is indeed invariant then wouldn't that mean that we would still have had $\partial_{\mu} J^{\mu} = 0$? ( the solution to the question posed was $\partial_{\mu} J^{\mu} = \partial L/ \partial \alpha$). Perhaps this solution indicates that the transformation really is just arbitrary leaving the lagrangian and action non invariant?

You still haven't shown the full wording of the original question, which makes it hard for me to straighten this out.

I will check if the library has a copy of Greiner.

I just checked. The calculations in Greiner don't explicitly cover this specific case (iiuc). But it doesn't matter, since (I think) we can still use Greiner as a template to perform a slightly different variational calculation properly...

 

[CAF123]

The full question as posed on the problem sheet is

'Show that if a transformation $\Phi \rightarrow \Phi + \alpha \partial \Phi/\partial \alpha$ is not a symmetry of the lagrangian then the Noether current is no longer conserved, but rather $\partial_{\mu}J^{\mu} = \partial L/ \partial \alpha$. Use this result to show that for a massive Dirac fermion the conservation of the chiral current is softly broken by the mass term, $\partial_{\mu} j^{\mu, 5} = -2m \bar \psi \gamma^5 \psi$. '

The bit about the dirac fermion I can do, it's just the first part I have questions about - their solution is given in the OP with my questions highlighted thereafter.
Thanks!

 

[haushofer]

Hi,

it's not strange that you're confused. I find it confusing also, and a lot of clarity is muddled by pedantic notation :P The point is that we regard two different things:

(1) To derive the EOM, we consider arbitrary variations of the fields which vanish at a boundary. The field configurations themselves, as a result, satisfy the EOM.
(2) To derive the symmetries, we consider certain specified field variations, while a priori the field configurations themselves do not satisfy any equations at all!

Important: the field configurations $\Phi(x)$ are independent of the field variations $\delta\Phi(x)$!

In (1), the field variations $\delta\Phi(x)$ do not obey certain equations, but the fields $\Phi(x)$ themselves do, while in case (2) the opposite is true: symmetries demand a very specific form for $\delta\Phi(x)$. In (1), the boundary term vanishes because the (otherwise arbitrary!) variations of the fields are zero at the boundary. In (2), we don't have such a restriction, because the field variations are not arbitrary. So, be very aware of what is restricted: the field configurations versus the field variations.

Now, what you do, is to regard a completely general variation of the action induced by a change of field configuration (not by a change of coordinates, but I think that's clear, otherwise you should also vary the volume element and you cannot commute variations of the fields with the derivatives wrt coordinates):

$\delta S = \int d^4 x \left( \left( \frac{\partial L}{\partial \Phi} - \partial_{\mu} \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \right) \delta \Phi + \partial_{\mu} \left(\frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) \right)$

This $\delta\Phi(x)$ is completely general. No funny stuff here. Next, you impose a condition on the field configuration (not on the field variation!): it obeys the EOM. So the first term, which are the Euler-Lagrange euqations, vanishes. And only after that, on top of the restriction that the field configurations obey the EOM, you also impose a condition on the field variation $\delta\Phi(x)$: it is not just any variation, but a symmetry of the action, hence it obeys a certain equation. And hence in that case you get a conserved current.

So now you have considered the variation of the action, in which both the field configuration $\Phi(x)$ and the field variation $\delta\Phi(x)$ is constrained. But you do that one step at a time.

Hope this helps :)

 

[Orodruin]

And only after that, on top of the restriction that the field configurations obey the EOM, you also impose a condition on the field variation δΦ(x)δΦ(x)\delta\Phi(x): it is not just any variation, but a symmetry of the action, hence it obeys a certain equation. And hence in that case you get a conserved current.

So to boil it down to what happens when the transformation is not a symmetry, i.e., OP's concern, $\delta S$ is not zero in this case and that is what gives you the source term in the continuity equation for the corresponding current.

 

[CAF123]

Thanks for all the replies,

So let me write down two cases: 1) If we consider a field variation that is constrained to be a symmetry of the action then $\delta S = 0$. If also the field configs are constrained to satisfy EL equations then we get $$\delta S = 0 = \int \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) d^4 x = \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) |_{\text{boundary}}$$ So this means the boundary term has to vanish in this case too right so that it agrees with the $\delta S=0$ on the lhs?

I just wondered about this because in the above posts, it was said that arbitrary variations of the fields vanish and in this case the field variation is not arbitrary but constrained so that $\delta S=0$.

2) The second case is similar to the above. Let's suppose the field variation is not constrained so that $\delta S \neq 0$, but again suppose field configs satisfy EL. Then $$\delta S \neq 0 = \int \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)}\delta \Phi \right) d^4 x $$ and so here the field variation should not vanish on the boundary in order to agree with the lhs?

I see that locally when we remove the integral signs 1) gives $\partial_{\mu}J^{\mu} = 0$ while the second case would give a source term (in accordance with symmetry of actions giving rise to conserved curents) but it's just the boundary terms I'm left to understand it seems with the above questions.

Thanks to all!

 

[strangerep]

1) If we consider a field variation that is constrained to be a symmetry of the action then $\delta S = 0$. If also the field configs are constrained to satisfy EL equations then we get $$\delta S = 0 = \int \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) d^4 x = \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) |_{\text{boundary}}$$ So this means the boundary term has to vanish in this case too right so that it agrees with the $\delta S=0$ on the lhs?

Are you sure you're applying the Divergence theorem correctly there? I.e., are you sure you don't need a surface integral on the rhs? (You've written "evaluate at boundary", instead of "integrate over boundary surface".)

A similar comment applies to your case (2).

(Aside: I've been trying re-write a derivation of this stuff in a more rigorous way. E.g., I presume the expression in the question: $$\phi ~\to~ \phi + \alpha \partial \phi/\partial \alpha $$is really just a sloppy way to (try and) say the following:

Let a 1-parameter Lie group of transformations act on the system, with real parameter denoted by $\alpha$, with $\alpha=0$ corresponding to the identity transformation. Since we work with a Lie group, the transformed field can be expanded as a Taylor series in $\alpha$ around the identity $\alpha=0$, as follows:$$\phi(\alpha) ~\approx~ \phi(0) ~+~ \alpha \left[ \frac{\partial \phi(\alpha)}{\partial\alpha} \right]_{\alpha=0} ~+~ \dots $$By assumption, these transformations simply change how the field is denoted, but do not map the physical system into some other (distinct, different) physical system. In other words, $\phi(\alpha)$ satisfies the equations of motion for the system for all $\alpha$ (else it could become a physically different system for some values of $\alpha$).

We abbreviate the transformed Lagrangian as $L(\alpha)$, i.e., $$L(\alpha) ~\equiv~ L\Big(\phi(\alpha)\,,\, \partial_\mu \phi(\alpha) \Big) ~.$$If(?) we agree that this correctly captures the intent of these symbols in the original question, I'll continue later in a subsequent post. [Edit: this is no longer necessary in view of @samalkhaiat's extensive explanation later in this thread.]

 

[CAF123]

Are you sure you're applying the Divergence theorem correctly there? I.e., are you sure you don't need a surface integral on the rhs? (You've written "evaluate at boundary", instead of "integrate over boundary surface".)

A similar comment applies to your case (2).

I don't think I had in mind the divergence theorem when writing what I wrote - just that I was trying to say that the integral of a four divergence of $X$ over all space-time is just $X$ evaluated at the boundary.

(c.f in 1D $$\int_{-\infty}^{\infty} dx \frac{d}{dx} X = X|_{-\infty}^{\infty}$$)

(Aside: I've been trying re-write a derivation of this stuff in a more rigorous way. E.g., I presume the expression in the question: $$\phi ~\to~ \phi + \alpha \partial \phi/\partial \alpha $$is really just a sloppy way to (try and) say the following:

Let a 1-parameter Lie group of transformations act on the system, with real parameter denoted by $\alpha$, with $\alpha=0$ corresponding to the identity transformation. Since we work with a Lie group, the transformed field can be expanded as a Taylor series in $\alpha$ around the identity $\alpha=0$, as follows:$$\phi(\alpha) ~\approx~ \phi(0) ~+~ \alpha \left[ \frac{\partial \phi(\alpha)}{\partial\alpha} \right]_{\alpha=0} ~+~ \dots $$By assumption, these transformations simply change how the field is denoted, but do not map the physical system into some other (distinct, different) physical system. In other words, $\phi(\alpha)$ satisfies the equations of motion for the system for all $\alpha$ (else it could become a physically different system for some values of $\alpha$).

We abbreviate the transformed Lagrangian as $L(\alpha)$, i.e., $$L(\alpha) ~\equiv~ L\Big(\phi(\alpha)\,,\, \partial_\mu \phi(\alpha) \Big) ~.$$If(?) we agree that this correctly captures the intent of these symbols in the original question, I'll continue later in a subsequent post.

Yup, I think that is a correct interpretation and I think it is correct to say that (in general) such a transformation causes the solution trajectories of the EL equations to change after the transformation but in the special case that these solutions do not change, we speak of the transformation as being a symmetry. (which on the practical level, amounts to a transformation that leaves the lagrangian invariant up to a total four divergence, thereby leaving the action invariant).

 

[Orodruin]

I don't think I had in mind the divergence theorem when writing what I wrote - just that I was trying to say that the integral of a four divergence of $X$ over all space-time is just $X$ evaluated at the boundary.

(c.f in 1D $$\int_{-\infty}^{\infty} dx \frac{d}{dx} X = X|_{-\infty}^{\infty}$$)

That is just the divergence theorem in one dimension ... If you add dimensions, the divergence theorem tells you that
$$
\int_V \partial_\mu J^\mu dV = \oint_S J^\mu dS_\mu,
$$
where $S$ is the boundary of $V$.

 

[CAF123]

I see, do you mean the $dV$ there to be $d^4x$? If taken to be a volume element then dV is dimension [V] and $d^4x$ of dimension [VT].

 

[Orodruin]

I see, do you mean the $dV$ there to be $d^4x$? If taken to be a volume element then dV is dimension [V] and $d^4x$ of dimension [VT].

In the special case of Minkowski space, yes. It is generally true whenever $dV$ is a product of the coordinate differentials.

 

[CAF123]

Ok thanks, then the rhs term in my post #14 in case 1) becomes $$\delta S = 0 = \oint_S \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) dS_{\mu}$$ _{(I see now there had to be additional structure in the rhs otherwise I have a stray index)}

For this to be zero, then the net variation of the field has to be zero over the boundary? (or in current language, I guess no leakage of current is allowed outside the boundary, hence the notion of 'conserved')

 

[samalkhaiat]

Thanks for all the replies,

So let me write down two cases: 1) If we consider a field variation that is constrained to be a symmetry of the action then $\delta S = 0$. If also the field configs are constrained to satisfy EL equations then we get $$\delta S = 0 = \int \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) d^4 x = \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) |_{\text{boundary}}$$ So this means the boundary term has to vanish in this case too right so that it agrees with the $\delta S=0$ on the lhs?

I just wondered about this because in the above posts, it was said that arbitrary variations of the fields vanish and in this case the field variation is not arbitrary but constrained so that $\delta S=0$.

2) The second case is similar to the above. Let's suppose the field variation is not constrained so that $\delta S \neq 0$, but again suppose field configs satisfy EL. Then $$\delta S \neq 0 = \int \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)}\delta \Phi \right) d^4 x $$ and so here the field variation should not vanish on the boundary in order to agree with the lhs?

I see that locally when we remove the integral signs 1) gives $\partial_{\mu}J^{\mu} = 0$ while the second case would give a source term (in accordance with symmetry of actions giving rise to conserved curents) but it's just the boundary terms I'm left to understand it seems with the above questions.

Thanks to all!

You need to understand the meaning of symmetry transformations, variation principle and Noether theorem. So, pay attention to what I am going to say.

1. Symmetry Transformations

Definition: given a particular (infinitesimal) field transformation $\delta_{\epsilon}\Phi$, and a differentiable action $$S[\Phi] = \int_{D} d^{4}x \ \mathcal{L}(\Phi , \partial\Phi),$$ the integral of a local Lagrangian $\mathcal{L}$ over bounded and arbitrarily contractible region $D$ of space-time. We say that $\delta_{\epsilon}\Phi$ is a symmetry transformation if (and only if), without using the EOM, the action changes according to $$\delta S[\Phi , \delta_{\epsilon}\Phi] \equiv S[\Phi + \delta_{\epsilon}\Phi] - S[\Phi] = \int_{D} d^{4}x \ \partial_{\mu}\Lambda^{\mu}(\Phi , \delta_{\epsilon}\Phi) , \ \ \ (1)$$ for some $\Lambda^{\nu}$. Eq(1) should be understood as a restriction on the given transformation $\delta_{\epsilon}\Phi$. Also, since the equations of motion have not been used in deriving Eq(1), it must hold for all $\Phi$.

So, if under some transformation $\delta_{\eta}\Phi$, you obtain $$\delta S[\Phi , \delta_{\eta}\Phi] = \int_{D} d^{4}x \ K(\Phi , \delta_{\eta}\Phi), \ \ \ \mbox{forall} \ \Phi(x) , \ \ \ \ (2)$$ for some function $K (\phi , \delta_{\eta}\Phi) \neq \partial_{\mu} F^{\mu}$, then $\delta_{\eta}\Phi$ is not a symmetry transformation.

Advise: always pay attention and understand the assumptions used in deriving mathematical expressions.

So, I will repeat: The function $\delta_{\epsilon}\Phi$ is a symmetry transformation if, and only if, it satisfies Eq(1) for all $\Phi(x)$.

Before we consider examples of (1) and (2), let us understand why Eq(1) is a good definition for symmetries: if $\Lambda^{\mu} = 0$, then $S[\Phi + \delta_{\epsilon}\Phi] = S[\Phi]$. Thus, if $\varphi (x)$ solves the EOM, so will the transformed field $\varphi (x;\epsilon) = \varphi (x) + \delta_{\epsilon}\varphi(x)$. This is also the case when $\Lambda^{\mu} \neq 0$, because $\mathcal{L}$ and $\mathcal{L} + \partial_{\mu}\Lambda^{\mu}$ lead to the same Euler-Lagrange equation.

1.1 Example of symmetry transformation:

Under space-time translation $x \to x + \epsilon$ by the infinitesimal constant $\epsilon$, an arbitrary field $\Phi$ changes according to $$\delta_{\epsilon}\Phi (x) = - \epsilon^{\mu}\partial_{\mu}\Phi . \ \ \ \ \ \ \ (3)$$ This induces the following change in the action

$$\delta S[\Phi , \delta_{\epsilon}\Phi] = \int_{D} d^{4}x \left( \frac{\partial \mathcal{L}}{\partial \Phi} \delta_{\epsilon}\Phi + \frac{\partial\mathcal{L}}{\partial(\partial_{\nu}\Phi)} \partial_{\nu}(\delta_{\epsilon}\Phi) \right) .$$ Assuming that $\mathcal{L}$ has no explicit dependence on $x^{\mu}$, and substituting Eq(3), we find that the change in the action is given by the following non-vanishing integral of a total divergence $$\delta S[\Phi , \delta_{\epsilon}\Phi] = \int_{D} d^{4}x \ \partial_{\mu}(- \epsilon^{\mu}\mathcal{L}) \ \ \ \forall \Phi(x) .$$ Thus, according to Eq(1), spacetime translation is a symmetry with $\Lambda^{\mu} (\Phi , \delta_{\epsilon}\Phi) = - \epsilon^{\mu} \mathcal{L}$.2. Arbitrary (algebraic) variation & The Noether Identity

An arbitrary variation $\delta\Phi$ of an arbitrary field $\Phi$, induces the following variation on the action integral

$$\delta S[\Phi , \delta\Phi] = \int_{D} d^{4}x \left( E(\Phi) \delta\Phi + \partial_{\mu}\left( \Pi^{\mu} \delta\Phi \right) \right) , \ \ \ \ \ \ (4)$$ where $$\Pi^{\mu} \equiv \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} ,$$ is the symplectic momentum, and $$E(\Phi) \equiv \frac{\partial \mathcal{L}}{\partial \Phi} - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\Phi)} \right) ,$$ is the Euler derivative. If $E(\psi) = 0$, we say that $\psi(x)$ is a solution to the EOM. So, keep in mind that $E(\Phi)$ is an operator acting on the arbitrary field $\Phi(x)$. This means that equation (4) is a valid expression for all $\Phi(x)$ and for all the variation $\delta\Phi$. Thus, it must also be valid for all $\Phi$ and for the symmetry variation $\delta_{\epsilon}\Phi$ as defined in equation (1). So, if we set $\delta \Phi = \delta_{\epsilon}\Phi$ in Eq(4) and subtract Eq(1) from it, we get $$\int_{D} d^{4}x \left( E(\Phi) \delta_{\epsilon}\Phi + \partial_{\mu}J^{\mu} \right) = 0 , \ \ \ \ \ \ \ (5)$$ where $$J^{\mu} \equiv \Pi^{\mu}\delta_{\epsilon}\Phi - \Lambda^{\mu} . \ \ \ \ \ \ \ \ \ \ (6)$$ Since $D \subset \mathbb{R}^{4}$ is an arbitrary contractible region, the integrand in Eq(5) must vanish identically. Thus, for the symmetry transformation $\delta_{\epsilon}\Phi$, the following (Noether) identity holds for arbitrary field $\Phi(x)$

$$E(\Phi) \delta_{\epsilon}\Phi + \partial_{\mu}J^{\mu} = 0 , \ \ \ \forall \Phi(x) . \ \ \ \ \ (7)$$

Thus, in presence of a continuous symmetry, a continuity equation $\partial_{\mu}J^{\mu} = 0$ is satisfied for the on-shell configuration $E(\psi) = 0$. And that is the statement of the (first) Noether theorem.

2.1 Example of a transformation that is not symmetry (explicit symmetry breaking):

Consider n fermion fields $\psi_{i}$ with different masses $m_{i}$. The Dirac lagrangian for such system can be written as $$\mathcal{L} = \bar{\Psi} \left( {\not} \partial + M \right) \Psi ,$$ where $\Psi = (\psi_{1}, \psi_{2}, \cdots , \psi_{n})^{T}$ and $M$ is $n \times n$ mass matrix not proportional to the identity matrix $I_{n}$, i.e., there is no mass degeneracy. Now, consider the finite $SU(n)$ transformations $$\Psi \to U(\eta) \Psi , \ \ \ \bar{\Psi} \to \bar{\Psi}U^{\dagger}(\eta) , \ \ U(\eta) = \exp (\eta^{a}T_{a}) .$$ Infinitesimally, we have $$\delta_{\eta}\Psi = \eta^{a}T_{a}\Psi , \ \ \ \delta_{\eta}\bar{\Psi} = - \eta^{a} \bar{\Psi}T_{a} .$$ Transforming the fields in the Lagrangian, we get $$\mathcal{L} \to \mathcal{L} (\eta) = \bar{\Psi} \left( {\not} \partial + U^{\dagger}(\eta) M U(\eta) \right) \Psi .$$ Expanding both sides to first order in $\eta$, we obtain

$$\mathcal{L} (\eta) = \mathcal{L} (0) + \eta^{a} \frac{d}{d \eta^{a}} \mathcal{L}(0) = \mathcal{L} (0) + \eta^{a} \bar{\Psi} \left[ M , T_{a} \right] \Psi ,$$ where $\mathcal{L}(0) = \mathcal{L}$ is the un-transformed Lagrangian. From that we get $$\delta_{\eta} \mathcal{L} = \eta^{a} \frac{d}{d \eta^{a}} \mathcal{L}(0) = \eta^{a} \bar{\Psi} \left[ M , T_{a} \right] \Psi . \ \ \ \ (8)$$

Clearly, the RHS of (8) cannot be written as total divergence, also it does not vanish since for our model $M \neq m I_{n}$. Thus, the $SU(n)$ transformation $U(\eta)$ is not symmetry transformation.

Now, let us subject $\mathcal{L}$ to an arbitrary variation and find

$$\delta \mathcal{L} = \delta \bar{\Psi} \left( {\not} \partial + M \right) \Psi + \bar{\Psi} \left( {\not} \partial + M \right) \delta\Psi .$$ In terms of the Euler derivatives, $E(\Psi)$ and $E(\bar{\Psi})$, we can rewrite the above as $$\delta \mathcal{L} = \delta \bar{\Psi} \ E(\Psi) + E(\bar{\Psi}) \ \delta \Psi + \partial_{\mu} \left( \bar{\Psi} \gamma^{\mu} \delta \Psi \right) .$$ To compare this equation with Eq(8), we set $\delta = \delta_{\eta}$

$$\delta_{\eta} \mathcal{L} = \delta_{\eta} \bar{\Psi} \ E(\Psi) + E(\bar{\Psi}) \ \delta_{\eta} \Psi + \eta^{a} \partial_{\mu} \left( \bar{\Psi} \gamma^{\mu} T_{a} \Psi \right) . \ \ \ (9)$$ Thus, from (8) and (9), we obtain the would-be Noether identity

$$\delta_{\eta} \bar{\Psi} \ E(\Psi) + E(\bar{\Psi}) \ \delta_{\eta} \Psi + \eta^{a} \partial_{\mu} \left( \bar{\Psi} \gamma^{\mu} T_{a} \Psi \right) = \eta^{a} \bar{\Psi} \left[ M , T_{a} \right] \Psi .$$ Finally, if we go for the on-shell configuration $E(\psi) = E(\bar{\psi}) = 0$, we find

$$\partial_{\mu} j^{\mu}_{a} = \bar{\psi} [ M , T_{a} ] \psi , \ \mbox{where} \ j^{\mu}_{a} \equiv \bar{\psi} \gamma^{\mu}T_{a} \psi .$$

This example shows that Noether’s theorem is more intelligent than us. It has detected that our model is not invariant under $SU(n)$, and calculated for us the divergence of the would-be-conserved symmetry current.

3. Boundary integrals: in the action principle and in Noether’s theorem

This is a tough business, and still is an active research area in gauge theory and GR. So, we will not discuss long rage forces (massless gauge fields) , but deal only with massive fields having compact support, i.e., $\Phi(t , r) \to 0$ as $r \to \infty$.

The principle of least action states that the action must be stationary, $\delta S[\Phi] = 0$, under arbitrary variation of the fields $\delta\Phi$, and fixed initial and final field configurations. That is to say that the variation satisfy $\delta \Phi|_{t_{1}} = \delta \Phi|_{t_{2}} = 0$ but arbitrary inside the integration domain (the bulk) $D$. Thus, in order for the “least action principle” to make sense, the action must be differentiable, i.e., it must possesses well-defined functional derivatives. So, we must have $$\delta S[\Phi] = \int_{D} d^{4}x \ ( \mbox{Stuff} ) \ \delta \Phi , \ \ \ \ \ \ (3.1)$$ with no extra non-vanishing terms. But we saw, Eq(4), that

$$\delta S[\Phi] = \int_{D} d^{4}x \ E(\Phi) \delta\Phi + \int_{D} d^{4}x \ \partial_{\mu} \left( \frac{ \partial \mathcal{L}}{ \partial ( \partial_{\mu}\Phi ) } \delta \Phi \right) . \ \ \ (3.2)$$

Thus, the second integral in Eq(3.2) must vanish in order to conclude that $\delta S[\Phi] = 0 \ \Leftrightarrow \ E(\Phi) = 0$ in $D$ and that, by fixing initial and final data, there exists a unique solution to the evolution equation $E(\Phi) = 0$. So, given $\delta \Phi|_{t_{1}} = \delta \Phi|_{t_{2}} = 0$, and assuming compact support, that is $\Phi(t,r)$ tends to zero sufficiently fast as $r \to \infty$, we need to show that $$\int_{D} d^{4}x \ \partial_{\mu}\left( \frac{ \partial \mathcal{L}}{ \partial ( \partial_{\mu}\Phi )} \delta \Phi \right) = 0 .$$ We take $D$ to be a “fat” world-tube containing the fields. Let $\partial D$ be the continuous, orientable, and piecewise smooth boundary of $D$. Let $t = t_{1}$ and $t = t_{2}$ be 3-dimensional cross-sections of the tube denoted by $\partial D(t_{1})$ and $\partial D(t_{2})$ respectively and $\partial D(r = \infty)$ the “cylindrical surface” that connects the two (constant time) cross-sections at $r = \infty$. Thus, it is clear that $$\partial D = \partial D(t_{1}) \cup \partial D(t_{2}) \cup \partial D(r = \infty ) . \ \ \ \ \ \ \ \ \ \ (3.3)$$ Applying Gauss’s theorem to the second integral in Eq(3.2), we get

$$\int_{D} d^{4}x \ \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} \delta\Phi \right) = \int_{\partial D} d^{3}x \ n_{\mu} \ \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} \delta \Phi , \ \ \ \ \ (3.4)$$ where $n_{\mu}$ is the outer pointing unit normal on $\partial D$: $$d^{3}x \ n_{\mu}|_{\partial D(t_{2})} = - d^{3}x \ n_{\mu}|_{\partial D(t_{1})} = (d^{3} \vec{x} , \vec{0}), \ \ d^{3}x \ n_{\mu}|_{\partial D(r = \infty)} = r^{2} d\Omega dt \hat{r} . \ \ \ \ (3.5)$$ Substituting (3.3) and (3.5) in RHS of (3.4), we find$$\int_{D} d^{4}x \ \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} \delta \Phi \right) = \int_{t_{2}} d^{3}\vec{x} \frac{\partial \mathcal{L}}{ \partial ( \partial_{0} \Phi)} \delta \Phi - \int_{t_{1}} d^{3}\vec{x} \frac{\partial \mathcal{L}}{ \partial ( \partial_{0} \Phi ) } \delta \Phi + \int_{r = \infty} \ d\Omega \ dt \ r^{2} \frac{\partial \mathcal{L}}{\partial (\partial_{r}\Phi)} \delta \Phi .$$ Now, the first two integrals on the RHS both vanish because of the fixed initial and final data $\delta \Phi (t_{1}) = \delta \Phi (t_{2}) = 0$. And in the absence of massless gauge fields and other constraints, the third integral vanishes because of the compact support assumption: at large $r$, $\frac{ \partial \mathcal{L} }{ \partial ( \partial_{r} \Phi ) }$ decays faster than the growth of $r^{2}$. Thus, in non-gauge field theories, the second integral in Eq(3.2) does indeed vanish leaving us with

$$\delta S[ \Phi ] = \int_{D} d^{4}x \ E(\Phi) \ \delta \Phi .$$ Since $\delta \Phi$ is an arbitrary function in the bulk $D$, then $\delta S[ \Phi ] = 0$ implies (and is implied by) $E(\Phi) = 0$ in $D$.

In gauge field theories and other constraint systems, boundary integrals require more careful analysis. Basically, the question of whether the boundary integral vanishes or not, is expressed in terms of asymptotic symmetries, i.e., the set of all symmetry transformations (having a non-zero conserved Noether charge) that preserve the asymptotic boundary conditions on the fields. For those who are interested I recommend the following paper and text[1] T. Regge and C. Teitelboim, “Role of Surface Integrals in the Hamiltonian Formulation of General Relativity" Annals Phys., 88, 286, 1974.

[2] M. Henneaux and C. Teitelboim, Quantization of Gauge Systems. Princeton University Press, 1992.

The last boundary integral, I leave for you as exercise. For the conserved symmetry current $J^{\mu}$, show that $$\int_{D} d^{4}x \ \partial_{\mu}J^{\mu}(t, \vec{x}) = 0 \ \Rightarrow \ Q(t_{1}) = Q(t_{2}) ,$$ where $$Q(t) \equiv \int d^{3}\vec{x} \ J^{0}(t , \vec{x})$$ is the Noether charge associated with the symmetry group.

 

[strangerep]

[...] So, pay attention to what I am going to say. [...]

Your post #21 deserves to be turned into an Insight.

Btw, is your book still "just over the horizon"?

 

[haushofer]

A reference which could also help:

https://arxiv.org/abs/1601.03616

 

[samalkhaiat]

Your post #21 deserves to be turned into an Insight.

I don't know if the original poster agree with you

Btw, is your book still "just over the horizon"?

So much work to do in so little time, my friend. It would be available “just over the horizon” if I could spare just two extra hours a day. With my current status, that means that I have to sleep two hour less for at least one month! The problem is this, my brain does not function properly if I sleep less than 7 hours a day.

Plus, there is this conversation between myself my daughter the other day:

My daughter: We do not see you, you spend your day and night working. Do you love physics more than you love us?
Myself: No, I love you more than anything else. Actually, it would the greatest achievement in theoretical physics, if I could understand why I love you more than physics.

 

[strangerep]

I don't know if the original poster agree with you

It's probably taking him a while to digest it. Also, you do realize that your opening admonition was a bit scary, right?

The problem is this, my brain does not function properly if I sleep less than 7 hours a day.

I know the feeling -- I now need at least 10.

Plus, there is this conversation between myself my daughter the other day:

My daughter: We do not see you, you spend your day and night working. Do you love physics more than you love us?

Hmm -- that's not first conversation of that kind that I've heard of.

A wise person once told me to cherish the (very few) years when a child actually wants you to be around. That will stop all too soon.

Myself: No, I love you more than anything else. Actually, it would the greatest achievement in theoretical physics, if I could understand why I love you more than physics.

Well, I can solve your last puzzle: although theoretical physics can be as damned ill-behaved as a young child, there is zero possibility that physics could ever love you back.

 

[CAF123]

Thanks for the nice post, yes I was taking my time to digest it before reply - I also agree it should go in the insights section.

Let me ask something about the last section with regards to boundary conditions. If I understand this part correctly, you are demonstrating a proof why one can usually ignore the boundary term in field theory calculations? In my last post, I said that for a symmetry transformation that is made on an already on shell configuration of fields then $$\delta S = 0 = \int d^4 x \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)}\delta \Phi\right)$$ while for a non symmetry transformation made on an on shell configuration of fields we get $$0 \neq \delta S = \int d^4 x \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)}\delta \Phi\right)$$

So at what point in your derivation of the vanishing of this boundary term do we differentiate the two cases? It seems we imposed the assumption about compact support but why would this only be true for the symmetry transformation case and not for the non symmetry one?

Thanks to everyone!:)

 

[MathematicalPhysicist]

You need to understand the meaning of symmetry transformations, variation principle and Noether theorem. So, pay attention to what I am going to say.

1. Symmetry Transformations

Definition: given a particular (infinitesimal) field transformation $\delta_{\epsilon}\Phi$, and a differentiable action $$S[\Phi] = \int_{D} d^{4}x \ \mathcal{L}(\Phi , \partial\Phi),$$ the integral of a local Lagrangian $\mathcal{L}$ over bounded and arbitrarily contractible region $D$ of space-time. We say that $\delta_{\epsilon}\Phi$ is a symmetry transformation if (and only if), without using the EOM, the action changes according to $$\delta S[\Phi , \delta_{\epsilon}\Phi] \equiv S[\Phi + \delta_{\epsilon}\Phi] - S[\Phi] = \int_{D} d^{4}x \ \partial_{\mu}\Lambda^{\mu}(\Phi , \delta_{\epsilon}\Phi) , \ \ \ (1)$$ for some $\Lambda^{\nu}$. Eq(1) should be understood as a restriction on the given transformation $\delta_{\epsilon}\Phi$. Also, since the equations of motion have not been used in deriving Eq(1), it must hold for all $\Phi$.

So, if under some transformation $\delta_{\eta}\Phi$, you obtain $$\delta S[\Phi , \delta_{\eta}\Phi] = \int_{D} d^{4}x \ K(\Phi , \delta_{\eta}\Phi), \ \ \ \mbox{forall} \ \Phi(x) , \ \ \ \ (2)$$ for some function $K (\phi , \delta_{\eta}\Phi) \neq \partial_{\mu} F^{\mu}$, then $\delta_{\eta}\Phi$ is not a symmetry transformation.

Advise: always pay attention and understand the assumptions used in deriving mathematical expressions.

So, I will repeat: The function $\delta_{\epsilon}\Phi$ is a symmetry transformation if, and only if, it satisfies Eq(1) for all $\Phi(x)$.

Before we consider examples of (1) and (2), let us understand why Eq(1) is a good definition for symmetries: if $\Lambda^{\mu} = 0$, then $S[\Phi + \delta_{\epsilon}\Phi] = S[\Phi]$. Thus, if $\varphi (x)$ solves the EOM, so will the transformed field $\varphi (x;\epsilon) = \varphi (x) + \delta_{\epsilon}\varphi(x)$. This is also the case when $\Lambda^{\mu} \neq 0$, because $\mathcal{L}$ and $\mathcal{L} + \partial_{\mu}\Lambda^{\mu}$ lead to the same Euler-Lagrange equation.

1.1 Example of symmetry transformation:

Under space-time translation $x \to x + \epsilon$ by the infinitesimal constant $\epsilon$, an arbitrary field $\Phi$ changes according to $$\delta_{\epsilon}\Phi (x) = - \epsilon^{\mu}\partial_{\mu}\Phi . \ \ \ \ \ \ \ (3)$$ This induces the following change in the action

$$\delta S[\Phi , \delta_{\epsilon}\Phi] = \int_{D} d^{4}x \left( \frac{\partial \mathcal{L}}{\partial \Phi} \delta_{\epsilon}\Phi + \frac{\partial\mathcal{L}}{\partial(\partial_{\nu}\Phi)} \partial_{\nu}(\delta_{\epsilon}\Phi) \right) .$$ Assuming that $\mathcal{L}$ has no explicit dependence on $x^{\mu}$, and substituting Eq(3), we find that the change in the action is given by the following non-vanishing integral of a total divergence $$\delta S[\Phi , \delta_{\epsilon}\Phi] = \int_{D} d^{4}x \ \partial_{\mu}(- \epsilon^{\mu}\mathcal{L}) \ \ \ \forall \Phi(x) .$$ Thus, according to Eq(1), spacetime translation is a symmetry with $\Lambda^{\mu} (\Phi , \delta_{\epsilon}\Phi) = - \epsilon^{\mu} \mathcal{L}$.2. Arbitrary (algebraic) variation & The Noether Identity

An arbitrary variation $\delta\Phi$ of an arbitrary field $\Phi$, induces the following variation on the action integral

$$\delta S[\Phi , \delta\Phi] = \int_{D} d^{4}x \left( E(\Phi) \delta\Phi + \partial_{\mu}\left( \Pi^{\mu} \delta\Phi \right) \right) , \ \ \ \ \ \ (4)$$ where $$\Pi^{\mu} \equiv \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} ,$$ is the symplectic momentum, and $$E(\Phi) \equiv \frac{\partial \mathcal{L}}{\partial \Phi} - \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\Phi)} \right) ,$$ is the Euler derivative. If $E(\psi) = 0$, we say that $\psi(x)$ is a solution to the EOM. So, keep in mind that $E(\Phi)$ is an operator acting on the arbitrary field $\Phi(x)$. This means that equation (4) is a valid expression for all $\Phi(x)$ and for all the variation $\delta\Phi$. Thus, it must also be valid for all $\Phi$ and for the symmetry variation $\delta_{\epsilon}\Phi$ as defined in equation (1). So, if we set $\delta \Phi = \delta_{\epsilon}\Phi$ in Eq(4) and subtract Eq(1) from it, we get $$\int_{D} d^{4}x \left( E(\Phi) \delta_{\epsilon}\Phi + \partial_{\mu}J^{\mu} \right) = 0 , \ \ \ \ \ \ \ (5)$$ where $$J^{\mu} \equiv \Pi^{\mu}\delta_{\epsilon}\Phi - \Lambda^{\mu} . \ \ \ \ \ \ \ \ \ \ (6)$$ Since $D \subset \mathbb{R}^{4}$ is an arbitrary contractible region, the integrand in Eq(5) must vanish identically. Thus, for the symmetry transformation $\delta_{\epsilon}\Phi$, the following (Noether) identity holds for arbitrary field $\Phi(x)$

$$E(\Phi) \delta_{\epsilon}\Phi + \partial_{\mu}J^{\mu} = 0 , \ \ \ \forall \Phi(x) . \ \ \ \ \ (7)$$

Thus, in presence of a continuous symmetry, a continuity equation $\partial_{\mu}J^{\mu} = 0$ is satisfied for the on-shell configuration $E(\psi) = 0$. And that is the statement of the (first) Noether theorem.

2.1 Example of a transformation that is not symmetry (explicit symmetry breaking):

Consider n fermion fields $\psi_{i}$ with different masses $m_{i}$. The Dirac lagrangian for such system can be written as $$\mathcal{L} = \bar{\Psi} \left( {\not} \partial + M \right) \Psi ,$$ where $\Psi = (\psi_{1}, \psi_{2}, \cdots , \psi_{n})^{T}$ and $M$ is $n \times n$ mass matrix not proportional to the identity matrix $I_{n}$, i.e., there is no mass degeneracy. Now, consider the finite $SU(n)$ transformations $$\Psi \to U(\eta) \Psi , \ \ \ \bar{\Psi} \to \bar{\Psi}U^{\dagger}(\eta) , \ \ U(\eta) = \exp (\eta^{a}T_{a}) .$$ Infinitesimally, we have $$\delta_{\eta}\Psi = \eta^{a}T_{a}\Psi , \ \ \ \delta_{\eta}\bar{\Psi} = - \eta^{a} \bar{\Psi}T_{a} .$$ Transforming the fields in the Lagrangian, we get $$\mathcal{L} \to \mathcal{L} (\eta) = \bar{\Psi} \left( {\not} \partial + U^{\dagger}(\eta) M U(\eta) \right) \Psi .$$ Expanding both sides to first order in $\eta$, we obtain

$$\mathcal{L} (\eta) = \mathcal{L} (0) + \eta^{a} \frac{d}{d \eta^{a}} \mathcal{L}(0) = \mathcal{L} (0) + \eta^{a} \bar{\Psi} \left[ M , T_{a} \right] \Psi ,$$ where $\mathcal{L}(0) = \mathcal{L}$ is the un-transformed Lagrangian. From that we get $$\delta_{\eta} \mathcal{L} = \eta^{a} \frac{d}{d \eta^{a}} \mathcal{L}(0) = \eta^{a} \bar{\Psi} \left[ M , T_{a} \right] \Psi . \ \ \ \ (8)$$

Clearly, the RHS of (8) cannot be written as total divergence, also it does not vanish since for our model $M \neq m I_{n}$. Thus, the $SU(n)$ transformation $U(\eta)$ is not symmetry transformation.

Now, let us subject $\mathcal{L}$ to an arbitrary variation and find

$$\delta \mathcal{L} = \delta \bar{\Psi} \left( {\not} \partial + M \right) \Psi + \bar{\Psi} \left( {\not} \partial + M \right) \delta\Psi .$$ In terms of the Euler derivatives, $E(\Psi)$ and $E(\bar{\Psi})$, we can rewrite the above as $$\delta \mathcal{L} = \delta \bar{\Psi} \ E(\Psi) + E(\bar{\Psi}) \ \delta \Psi + \partial_{\mu} \left( \bar{\Psi} \gamma^{\mu} \delta \Psi \right) .$$ To compare this equation with Eq(8), we set $\delta = \delta_{\eta}$

$$\delta_{\eta} \mathcal{L} = \delta_{\eta} \bar{\Psi} \ E(\Psi) + E(\bar{\Psi}) \ \delta_{\eta} \Psi + \eta^{a} \partial_{\mu} \left( \bar{\Psi} \gamma^{\mu} T_{a} \Psi \right) . \ \ \ (9)$$ Thus, from (8) and (9), we obtain the would-be Noether identity

$$\delta_{\eta} \bar{\Psi} \ E(\Psi) + E(\bar{\Psi}) \ \delta_{\eta} \Psi + \eta^{a} \partial_{\mu} \left( \bar{\Psi} \gamma^{\mu} T_{a} \Psi \right) = \eta^{a} \bar{\Psi} \left[ M , T_{a} \right] \Psi .$$ Finally, if we go for the on-shell configuration $E(\psi) = E(\bar{\psi}) = 0$, we find

$$\partial_{\mu} j^{\mu}_{a} = \bar{\psi} [ M , T_{a} ] \psi , \ \mbox{where} \ j^{\mu}_{a} \equiv \bar{\psi} \gamma^{\mu}T_{a} \psi .$$

This example shows that Noether’s theorem is more intelligent than us. It has detected that our model is not invariant under $SU(n)$, and calculated for us the divergence of the would-be-conserved symmetry current.

3. Boundary integrals: in the action principle and in Noether’s theorem

This is a tough business, and still is an active research area in gauge theory and GR. So, we will not discuss long rage forces (massless gauge fields) , but deal only with massive fields having compact support, i.e., $\Phi(t , r) \to 0$ as $r \to \infty$.

The principle of least action states that the action must be stationary, $\delta S[\Phi] = 0$, under arbitrary variation of the fields $\delta\Phi$, and fixed initial and final field configurations. That is to say that the variation satisfy $\delta \Phi|_{t_{1}} = \delta \Phi|_{t_{2}} = 0$ but arbitrary inside the integration domain (the bulk) $D$. Thus, in order for the “least action principle” to make sense, the action must be differentiable, i.e., it must possesses well-defined functional derivatives. So, we must have $$\delta S[\Phi] = \int_{D} d^{4}x \ ( \mbox{Stuff} ) \ \delta \Phi , \ \ \ \ \ \ (3.1)$$ with no extra non-vanishing terms. But we saw, Eq(4), that

$$\delta S[\Phi] = \int_{D} d^{4}x \ E(\Phi) \delta\Phi + \int_{D} d^{4}x \ \partial_{\mu} \left( \frac{ \partial \mathcal{L}}{ \partial ( \partial_{\mu}\Phi ) } \delta \Phi \right) . \ \ \ (3.2)$$

Thus, the second integral in Eq(3.2) must vanish in order to conclude that $\delta S[\Phi] = 0 \ \Leftrightarrow \ E(\Phi) = 0$ in $D$ and that, by fixing initial and final data, there exists a unique solution to the evolution equation $E(\Phi) = 0$. So, given $\delta \Phi|_{t_{1}} = \delta \Phi|_{t_{2}} = 0$, and assuming compact support, that is $\Phi(t,r)$ tends to zero sufficiently fast as $r \to \infty$, we need to show that $$\int_{D} d^{4}x \ \partial_{\mu}\left( \frac{ \partial \mathcal{L}}{ \partial ( \partial_{\mu}\Phi )} \delta \Phi \right) = 0 .$$ We take $D$ to be a “fat” world-tube containing the fields. Let $\partial D$ be the continuous, orientable, and piecewise smooth boundary of $D$. Let $t = t_{1}$ and $t = t_{2}$ be 3-dimensional cross-sections of the tube denoted by $\partial D(t_{1})$ and $\partial D(t_{2})$ respectively and $\partial D(r = \infty)$ the “cylindrical surface” that connects the two (constant time) cross-sections at $r = \infty$. Thus, it is clear that $$\partial D = \partial D(t_{1}) \cup \partial D(t_{2}) \cup \partial D(r = \infty ) . \ \ \ \ \ \ \ \ \ \ (3.3)$$ Applying Gauss’s theorem to the second integral in Eq(3.2), we get

$$\int_{D} d^{4}x \ \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} \delta\Phi \right) = \int_{\partial D} d^{3}x \ n_{\mu} \ \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} \delta \Phi , \ \ \ \ \ (3.4)$$ where $n_{\mu}$ is the outer pointing unit normal on $\partial D$: $$d^{3}x \ n_{\mu}|_{\partial D(t_{2})} = - d^{3}x \ n_{\mu}|_{\partial D(t_{1})} = (d^{3} \vec{x} , \vec{0}), \ \ d^{3}x \ n_{\mu}|_{\partial D(r = \infty)} = r^{2} d\Omega dt \hat{r} . \ \ \ \ (3.5)$$ Substituting (3.3) and (3.5) in RHS of (3.4), we find$$\int_{D} d^{4}x \ \partial_{\mu}\left( \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\Phi)} \delta \Phi \right) = \int_{t_{2}} d^{3}\vec{x} \frac{\partial \mathcal{L}}{ \partial ( \partial_{0} \Phi)} \delta \Phi - \int_{t_{1}} d^{3}\vec{x} \frac{\partial \mathcal{L}}{ \partial ( \partial_{0} \Phi ) } \delta \Phi + \int_{r = \infty} \ d\Omega \ dt \ r^{2} \frac{\partial \mathcal{L}}{\partial (\partial_{r}\Phi)} \delta \Phi .$$ Now, the first two integrals on the RHS both vanish because of the fixed initial and final data $\delta \Phi (t_{1}) = \delta \Phi (t_{2}) = 0$. And in the absence of massless gauge fields and other constraints, the third integral vanishes because of the compact support assumption: at large $r$, $\frac{ \partial \mathcal{L} }{ \partial ( \partial_{r} \Phi ) }$ decays faster than the growth of $r^{2}$. Thus, in non-gauge field theories, the second integral in Eq(3.2) does indeed vanish leaving us with

$$\delta S[ \Phi ] = \int_{D} d^{4}x \ E(\Phi) \ \delta \Phi .$$ Since $\delta \Phi$ is an arbitrary function in the bulk $D$, then $\delta S[ \Phi ] = 0$ implies (and is implied by) $E(\Phi) = 0$ in $D$.

In gauge field theories and other constraint systems, boundary integrals require more careful analysis. Basically, the question of whether the boundary integral vanishes or not, is expressed in terms of asymptotic symmetries, i.e., the set of all symmetry transformations (having a non-zero conserved Noether charge) that preserve the asymptotic boundary conditions on the fields. For those who are interested I recommend the following paper and text[1] T. Regge and C. Teitelboim, “Role of Surface Integrals in the Hamiltonian Formulation of General Relativity" Annals Phys., 88, 286, 1974.

[2] M. Henneaux and C. Teitelboim, Quantization of Gauge Systems. Princeton University Press, 1992.

The last boundary integral, I leave for you as exercise. For the conserved symmetry current $J^{\mu}$, show that $$\int_{D} d^{4}x \ \partial_{\mu}J^{\mu}(t, \vec{x}) = 0 \ \Rightarrow \ Q(t_{1}) = Q(t_{2}) ,$$ where $$Q(t) \equiv \int d^{3}\vec{x} \ J^{0}(t , \vec{x})$$ is the Noether charge associated with the symmetry group.

Isn't all what you wrote already covered in existing QFT books, such as Ryder's and Brown's books?

 

[samalkhaiat]

If I understand this part correctly, you are demonstrating a proof why one can usually ignore the boundary term in field theory calculations?

In general, ignoring boundary terms is not a smart thing to do in field theory. However, when you want to derive the EOM from an action principle, you can ignore such term provided that the fields have compact support. So, in the action principle the boundary term vanishes because of a) the fact that initial and final data are fixed, i.e., $\delta \phi (t_{1}) = \delta \phi (t_{2}) = 0$, and b) the assumption that, as $r \to \infty$, $\phi (t,r) \to 0$ faster than any polynomial in $r$. This is what I did in the last part of my post which was about the action principle.

On the other hand, in Noether theorem, you are given a particular (symmetry or non-symmetry) transformation $\delta_{\epsilon}\phi$. Therefore, it makes no sense to write $\delta_{\epsilon} \phi (t_{1}) = \delta_{\epsilon} \phi (t_{2}) = 0$ for transformation other than the identity transformation. Unless you provide further analysis, you have no reason to consider the initial and the final configurations to be invariant under the non-trivial transformation $\delta_{\epsilon}\phi$. It is this fact that makes boundary integrals in Noether theorem ideologically different from those in the action principle. Do the exercise at the end of #21, or see below.

I said that for a symmetry transformation that is made on an already on shell configuration of fields then $$\delta S = 0 = \int d^4 x \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)}\delta \Phi\right)$$

In general, this is not true. I have already given you an example. We have seen that spacetime translation is a symmetry with
$$\delta S[\phi , \delta_{\epsilon} \phi ] = \int_{D} d^{4}x \ \partial_{\sigma} \left( - \epsilon^{\sigma} \mathcal{L} \right) , \ \ \ \forall \phi (x) .$$
IF you subtract this from the following on-shell symmetry variation $$\delta S[\phi , \delta_{\epsilon} \phi ] = \int_{D} d^{4}x \ \partial_{\sigma} \left( \pi^{\sigma} (-\epsilon^{\mu}\partial_{\mu}\phi ) \right) , \ \ \ E(\phi) = 0 ,$$ you get $$0 = \int_{D} d^{4}x \ \partial_{\sigma} \left( \epsilon^{\mu} T^{\sigma}{}_{\mu}\right) , \ \ \ \ \ \ \ (1)$$ where $$T^{\sigma}{}_{\mu} \equiv \pi^{\sigma} \partial_{\mu}\phi - \delta^{\sigma}_{\mu} \ \mathcal{L} ,$$ is the energy-momentum tensor. Since $D$ is an arbitrary contractible region, Eq(1) gives you the continuity equation $\partial_{\sigma}T^{\sigma \rho} = 0$ on $E(\phi) = 0$.

The equation that you wrote is correct only for compact internal symmetry groups. These symmetries leave the Lagrangian (hence the action) invariant, i.e., without using the EOM, you obtain $$\delta S[\phi , \delta_{\epsilon}\phi ] = 0 , \ \ \ \ \forall \phi (x) .$$ But for an on-shell symmetry variation, you have $$\delta S[\phi , \delta_{\epsilon}\phi ] = \int_{D} d^{4}x \ \partial_{\mu} \left( \pi^{\mu} \delta_{\epsilon}\phi \right) .$$ Again, by comparing the two expressions, you get $$0 = \int_{D} d^{4}x \ \partial_{\mu} \left( \pi^{\mu} \delta_{\epsilon}\phi \right) , \ \ \ E(\phi) = 0 . \ \ \ \ (2)$$ Again, since $D$ is an arbitrary contractible region, Eq(2) gives you the continuity equation $\partial_{\mu} (\pi^{\mu} \delta_{\epsilon}\phi) = 0$ on $E(\phi) = 0$. However, since $D$ has continuous, orientable and piecewise smooth boundary $\partial D$, we can also use the divergence theorem on the RHS of Eq(2) and obtain $$0 = \int_{\partial D} d^{3}x \ n_{\mu} \ (\pi^{\mu} \delta_{\epsilon}\phi) , \ \ \ \ \ \ \ (3)$$ where (as before) $n_{\mu}$ is the outward-pointing unit normal on the boundary $\partial D$. Again, as in post #21, if we take
$$\partial D = \partial D(t_{1}) \cup \partial D(t_{2}) \cup \partial D(r = \infty ) ,$$ then Eq(3) becomes
$$0 = \int_{t_{2}} d^{3}\vec{x} ( \pi^{0}\delta_{\epsilon}\phi ) - \int_{t_{1}} d^{3}\vec{x} ( \pi^{0}\delta_{\epsilon}\phi ) + \int_{r = \infty} d\Omega dt r^{2} (\pi^{r}\delta_{\epsilon}\phi) .$$
Again by compact support, the third integral vanishes because at $r = \infty$, $\pi^{r}$ goes to zero faster than the growth of $r^{2}$. Therefore
$$\int_{x^{0} = t_{1}} d^{3}\vec{x} \ ( \pi^{0}\delta_{\epsilon}\phi ) = \int_{x^{0} = t_{2}} d^{3}\vec{x} \ ( \pi^{0}\delta_{\epsilon}\phi ) .$$ This means that the charge $$Q(t) \equiv \int_{x^{0} = t} d^{3}\vec{x} \ ( \pi^{0} \ \delta_{\epsilon}\phi ),$$ is independent of the space-like hyper-plane $x^{0} = t$ used to evaluate it. In other words, $Q$ is time-independent, i.e., $dQ / dt = 0$. This is the solution to the exercise I gave you in #21.

while for a non symmetry transformation made on an on shell configuration of fields we get $$0 \neq \delta S = \int d^4 x \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \Phi)}\delta \Phi\right)$$

In this case (when the transformation, $\delta_{\eta}\phi$, is not symmetry), applying the divergence theorem, together with the compact support assumption gives you
$$0 \neq \int_{t_{2}} d^{3}\vec{x} \ ( \pi^{0}\delta_{\eta}\phi ) - \int_{t_{1}} d^{3}\vec{x} \ ( \pi^{0}\delta_{\eta}\phi ) + 0 .$$
Thus, if the transformation is not symmetry transformation, you get $Q(t_{1}) \neq Q(t_{2})$. In other words, the non-symmetry transformation, $\delta_{\eta}\phi$, does not imply a time-independent charge, i.e., the quantity $Q(t)$ defined by the integral $\int d^{3}\vec{x} \ (\pi^{0} \delta_{\eta}\phi )$ is not conserved.

So at what point in your derivation of the vanishing of this boundary term do we differentiate the two cases?

Your question makes no sense. In post #21, I only did the boundary integral in the context of the action principle, i.e., when the arbitrary variation satisfies $\delta \phi (t_{1}) = \delta \phi (t_{2}) = 0$. I did not do any boundary integral with $\delta \phi$ being a particular symmetry or non-symmetry transformations (In this post, I did both cases). However, at the end of #21, I asked you to do the boundary integral $$0 = \int_{D} d^{4}x \ \partial_{\mu}J^{\mu} \equiv \int_{D} d^{4}x \ \partial_{\mu}(\pi^{\mu} \delta_{\epsilon}\phi) ,$$ when $\delta_{\epsilon}\phi$ is a symmetry transformation.

It seems we imposed the assumption about compact support but why would this only be true for the symmetry transformation case and not for the non symmetry one?

If the fields have compact support, they have compact support full stop. The meaning of the variation $\delta \phi$ (symmetry, non-symmetry or arbitrary variation) does not change that fact. I’ve just used the compact support assumption in the non-symmetry case $\delta_{\eta}\phi$.