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I Technicality with Noether's theorem

  1. Apr 22, 2017 #1

    CAF123

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    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?
     
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  3. Apr 27, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Apr 30, 2017 #3

    CAF123

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    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.
     
  5. Apr 30, 2017 #4

    vanhees71

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    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 ;-)).
     
  6. Apr 30, 2017 #5

    CAF123

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    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 :)
     
  7. May 1, 2017 #6

    vanhees71

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    Hm, but how do you prove that you get this four divergence if the transformation is not a symmetry transformation?
     
  8. May 1, 2017 #7

    CAF123

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    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?
     
  9. May 2, 2017 #8

    strangerep

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    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.
     
  10. May 2, 2017 #9

    CAF123

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    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!
     
  11. May 2, 2017 #10

    strangerep

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

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

    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...
     
  12. May 3, 2017 #11

    CAF123

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    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!
     
  13. May 3, 2017 #12

    haushofer

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    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 [itex]\Phi(x)[/itex] are independent of the field variations [itex]\delta\Phi(x)[/itex]!

    In (1), the field variations [itex]\delta\Phi(x)[/itex] do not obey certain equations, but the fields [itex]\Phi(x)[/itex] themselves do, while in case (2) the opposite is true: symmetries demand a very specific form for [itex]\delta\Phi(x)[/itex]. 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 [itex]\delta\Phi(x)[/itex] 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 [itex]\delta\Phi(x)[/itex]: 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 [itex]\Phi(x)[/itex] and the field variation [itex]\delta\Phi(x)[/itex] is constrained. But you do that one step at a time.

    Hope this helps :)
     
    Last edited: May 3, 2017
  14. May 3, 2017 #13

    Orodruin

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    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.
     
  15. May 4, 2017 #14

    CAF123

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    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!
     
  16. May 5, 2017 #15

    strangerep

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    Are you sure you're applying the Divergence theorem correctly there? :oldwink: 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.]
     
    Last edited: May 5, 2017
  17. May 5, 2017 #16

    CAF123

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    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 $$\int_{-\infty}^{\infty} dx \frac{d}{dx} X = X|_{-\infty}^{\infty}$$)

    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).
     
  18. May 5, 2017 #17

    Orodruin

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    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##.
     
  19. May 5, 2017 #18

    CAF123

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    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].
     
  20. May 5, 2017 #19

    Orodruin

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    In the special case of Minkowski space, yes. It is generally true whenever ##dV## is a product of the coordinate differentials.
     
  21. May 5, 2017 #20

    CAF123

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    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')
     
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