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Problematic derivations of Noether's theorem

  1. Oct 15, 2008 #1
    I am confused by various derivations of the Noether current in various text books. However, they either contradict with each other or exist many flaws.
    For example, originally I thought the best derivation is at the end of the book of classical mechanics by Goldstein. But I found that in the result eq(13.147) (3rd edition), only global symmetry can be used.

    Even in Weinberg's section 7.2, the derivation of the energy-momentum tensor (spacetime translational symmetry), the derivation is quite unreasonable. Suddenly appearing of [tex]\epsilon^\mu\partial_\mu\mathcal{L}[/tex] (this term is usually obtained from the deviation of the measure [tex]d^4x[/tex] in other QFT texts).

    Moreover, in Ryder's QFT book, his derivation seems to neglect the difference of two variations (even Weinberg made this mistake I thought):
    [tex]\Delta\phi(x) \equiv \phi'(x') - \phi(x)[/tex]
    [tex]\delta\phi(x) \equiv \phi'(x) - \phi(x)
    where [tex]\left[\Delta,\partial_\mu\right]\neq 0[/tex] while [tex]\left[\delta,\partial_\mu\right] = 0[/tex].

    The best derivation I have ever seen is Maggiore's.
    Is there anybody agrees with me?
    Or is there good and reasonable derivations?
    Thanks in advance.
  2. jcsd
  3. Oct 15, 2008 #2


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    I really hate Goldstein's book, and the section about Noether's therorem is probably the part I hate the most. I tried asking about his proof in another thread. Link. Let me know if you find it useful.

    The replies I got didn't really help me, but I was able to make some progress on my own. The most important posts in the thread are #3 (because it explains the notation) and #11.
  4. Oct 15, 2008 #3


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    I like Srednicki's derivation in his QFT book (available free online).
  5. Oct 16, 2008 #4


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    Last edited: Oct 16, 2008
  6. Oct 16, 2008 #5


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    That's an awesome series of posts sam. Now I know where to look if I want to read up on the conformal group. :smile:

    Regarding the derivation of Noether's theorem, it would be great if someone (anyone who's interested in an easy derivation) could take a look at what I did in the other thread (link in my previous post). I didn't get any feedback there. Post #3 is the derivation of the Euler-Lagrange equations. It's included only as an explanation of the notation. Post #11 is a derivation of the conserved currents from an invariance condition. (I omitted steps in both derivations, but I'll explain them if someone asks for it).

    I really don't think there's an easier way to do these things than the way I'm doing it. It's almost too easy. I just hope that doesn't mean I've made a mistake.
  7. Oct 16, 2008 #6


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    Fredik, your derivation is essentially the same as the one in Srednicki's book.
  8. Oct 16, 2008 #7
  9. Oct 17, 2008 #8


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    We can think of the Lagrangian either as a multi-variable polynomial

    [tex](\phi(x),\partial_\mu\phi(x))\mapsto\mathcal L(\phi(x),\partial_\mu\phi(x))[/tex]

    or a set of functionals (one for each x)

    [tex]\phi\mapsto\mathcal L_x[\phi]=\mathcal L(\phi(x),\partial_\mu\phi(x))[/tex]

    I prefer to define a variation of the field as a function [itex]\epsilon\mapsto\phi_\epsilon[/itex] where [itex]\phi_0=\phi[/itex] minimizes the action, rather than as a "small" function that we add to the field.

    I'm only mentioning these things to make sure that my notation isn't going to be confusing. I'll get to the point in my next post. I'll end this one here, because I'm getting "database error" a lot.
    Last edited: Oct 17, 2008
  10. Oct 17, 2008 #9


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    OK, here's the point: Suppose that we're considering the variation [itex]b\mapsto\phi_b[/itex], where [itex]\phi_b[/itex] is defined by [itex]\phi_b(x)=\phi(x+b)[/itex].

    [tex]\mathcal L_x[\phi_b]=\mathcal L(\phi(x+b),\partial_\mu\phi(x+b))=\mathcal L_{x+b}[\phi] \approx\mathcal L_x[\phi]+b^\mu\partial_\mu\mathcal L_x[\phi][/tex]

    In this post "[itex]\approx[/itex]" means "equal to first order in the parameters".

    Wow..."database error" for that. I can't preview this. :grumpy:
    Last edited: Oct 17, 2008
  11. Oct 17, 2008 #10


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    I didn't find it useful to distinguish between the two. See #4 in the thread I linked to above.

    Definitely not if you define a variation the way I'm doing it. [itex]\delta S[/tex] would just be

    [tex]\epsilon^r\frac{d}{d\epsilon^r}\bigg|_0 S[\phi_\epsilon][/tex]

    I'm not sure how it works if we define it using functional derivatives instead, and the textbook "definition" never made sense to me. ("[itex]\delta S[/tex] is a small change in S caused by a small change in [itex]\phi[/itex]").

    It seems to me that you only have to worry about the [itex]d^4x[/itex] if you're trying to minimize the function

    [tex](\phi,U)\mapsto\int_U d^4x \mathcal L_x[\phi][/tex]

    instead of

    [tex]\phi\mapsto\int_U d^4x \mathcal L_x[\phi][/itex]

    with a fixed U.
    Last edited: Oct 17, 2008
  12. Oct 17, 2008 #11
    Thanks, Fredrik!
    But, sorry for that, sometimes whenever I was stuck on something, I have to take much time to get other people's point...

    Actually, I thought this is the key point that why and where every text book is not consistent with each other.

    Due to the functional dependence, as long as we include the space-time symmetry, the situation becomes complicated. If there are internal symmetries only, then the derivation is clear, e.g. Weinberg's derivation in section 7.2.

    If we consider the following symmetry transformation, which includes space-time symmetry,
    [tex] x^\mu \rightarrow x'^\mu = x^\mu + \delta x^\mu(x)\cdots(*-1)[/tex]
    [tex]\phi(x) \rightarrow \phi'(x') = \phi(x) + \delta\phi(x)(*-2)[/tex]
    The symmetry of the theory means the action is invariant, i.e. the space-time integration of the Lagrangian, which depends on [itex]\phi'(x')[/tex], over the new variables [itex]x'[/tex](hence the integration domain is changed, so we need Jacobian here, in general), specifically,
    [tex]\delta S = \int d^4x'\mathcal{L}(\phi'(x'),\partial'_\mu\phi'(x')) - \int d^4x\mathcal{L}(\phi(x),\partial_\mu\phi(x)) = 0[/tex]

    However, the problem I confused in Weinberg's book is, for the derivation of space-time translational symmetry, he just considered the variation of the action as this,
    [tex]\delta S = \int d^4x\mathcal{L}(\phi(x'),\partial_\mu\phi(x')) - \int d^4x\mathcal{L}(\phi(x),\partial_\mu\phi(x))[/tex]
    Then, using the trick of releasing the global symmetry to be local, one can obtain the current.

    He didn't consider the Jacobian of the measure. It seems that he considered the space-time translation equivalently as the transformation on the fields,
    [tex]\phi(x) \rightarrow \phi(x+\epsilon(x)) = \phi(x) + \epsilon^\mu\partial_\mu\phi(x) = \phi'(x)[/tex]

    hmm...after relaxed by having a cup of coffee, I seem to figure out where I got stuck. The key point is, since the space-time coordinates are integrated over, the transformations eq(*-1,2) are equivalently equal to the following internal transformation,
    [tex]x^\mu\rightarrow x^\mu[/tex]
    [tex]\phi(x)\rightarrow \phi(x) + \delta\phi - \delta x^\mu\partial_\mu\phi(x)[/tex]

    I think I have understood the derivation of energy-momentum tensor by Weinberg's method. However, I will reread the articles by you guys to know the points you made.

  13. Oct 20, 2008 #12


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  14. Oct 22, 2008 #13


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    The equation

    \delta \mathcal{L} = \int \mathcal{L} \delta \left(d^{4}x \right) + \int d^{4}x \delta \mathcal{L} = \int d^{4}x \left( \delta^{*} \mathcal{L} + \partial_{a}( \delta x^{a} \mathcal{L}) \right),

    is always correct. This is because the local variation symbol

    [tex]\delta (...) = \delta^{*} (...) + \delta x^{a}\partial_{a} (...)[/tex]

    is a derivation, i.e., it acts like a differential operator. So, a transformation is certainly an invariance (symmetry) transformation if

    1) the Lagrangian itself is (form) invariant;

    [tex]\delta^{*}\mathcal{L} = \bar{\mathcal{L}}(x) - \mathcal{L}(x) = 0[/tex]

    2) the functional form of the Lagrangian changes by a total derivative

    [tex]\delta^{*}\mathcal{L} = \partial_{a} \Omega^{a}[/tex]

    Notice that in both cases, [itex]\mathcal{L}[/itex] and [itex]\bar{\mathcal{L}}[/itex] lead to the same equation of motion, and this what “symmetry” is all about.
    Let me illustrate this. Consider the free particle Lagrangian

    [tex]L = \frac{m}{2}\dot{x}^{2}[/tex]

    which gives the equation of motion

    [tex]\ddot{x} = 0[/tex]

    Let us study the possible invariance of this system under a transformation of the form

    [tex]\bar{x} = x + f, \ \ \dot{\bar{x}}= \dot{x} + \dot{f}[/tex]

    where f may be some function of time. We construct the Lagrangian [itex]\bar{L}[/itex] for the new coordinate [itex]\bar{x}[/itex],

    [tex]\bar{L}(\bar{x}) = \frac{m}{2} ( \dot{\bar{x}} - \dot{f})^{2} = L(\bar{x}) - m \dot{f}\dot{\bar{x}} + \frac{m}{2}\dot{f}^{2}[/tex]

    This leads to the equation of motion

    [tex]\ddot{\bar{x}} = \ddot{f}[/tex]

    Therefore, the transformation above is a symmetry if

    [tex]\ddot{f} = 0[/tex]


    [tex]f(t) = vt + x_{0}[/tex]

    Indeed, the transformations

    [tex]\bar{x} = x + vt + x_{0}, \ \ \dot{\bar{x}} = \dot{x} + v[/tex]

    are nothing but the Galilean symmetry transformations of the free particle. Also notice that

    [tex]\delta^{*}L = \bar{L}(\bar{x}) - L(\bar{x}) = \frac{d}{dt}\left( \frac{m}{2}v^{2}t - mv\bar{x} \right)[/tex]

    i.e., under the Galilean symmetry transformation above, the variation in the form of the Lagrangian function is a total time derivative.


    Last edited: Oct 22, 2008
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