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Confusion from Weinberg's QFT vol1

  1. Dec 9, 2012 #1
    Hello! I came across with something that confused me while studying the book and I need some help. In section 7.3 , equation (7.3.4) should have the drivative of ε(x) in order to "vanish when ε(x) is constant" . But in the lines before this expression he says that the variation of action vanishes, not when ε(x) is constant but when the fields satisfy the dynamical equations. So why should (7.3.4) have the derivative of ε(x), since its constant value is not a criterion for vanishing variation?

    Also, some lines below, he deals with the case that ε(x) is not constant, but in equations (7.3.9) and (7.3.13) he treats ε(x) like a constant!

    What am I missing here???
     
  2. jcsd
  3. Dec 9, 2012 #2

    Bill_K

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    As I've said before, I think that when Weinberg wrote this book, he got paid by the subscript! :smile: He does have a way of making simple things sound complex, leaving the reader to wonder if it's really as complex as he makes it sound. Usually it's not.

    When he points out that (7.3.2) holds if the fields satisfy the dynamical equations, it's just a parenthetical remark, and he dismisses that case and goes on to say that the situation he's going to consider is more general than that.

    So he lets ε(x) be an arbitrary function, and considers the variation of the action. There are terms of two types: (a) the terms in which ε is not differentiated. These are just the ones that appear in (7.3.2), and we're assuming they vanish. (b) The terms in which ε(x) is differentiated once. They make up (7.3.4). That's all that's left. No higher derivatives of ε(x) occur, since the Lagrangian only depended on ψ and ψ·.

    He pulls the same kind of trick on the next page. (7.3.9) are the terms where ε(t) is undifferentiated, and they are assumed to vanish. So all that can be left are the terms containing dε/dt, and they make up (7.3.10).
     
  4. Dec 10, 2012 #3
    I sorry, but I can't find such terms using integration by parts. Either I will find terms like those in (7.3.2) but with no derivatives of ε(x), either terms with derivatives of ε(x) but with no terms like those in (7.3.2). Could someone demonstrate the process please?
     
  5. Dec 10, 2012 #4

    Bill_K

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    Integration by parts is not required. Look back at the unnumbered equation between (7.2.6) and (7.2.7), where he writes out the definition of the variational derivative. There are three terms, involving δψ, ∇δψ and δψ· respectively. The variation in ψ is given by (7.3.2): δψ = iεF. From this, ∇δψ = ∇(iεF) and δψ· = (iεF)·. Expand these out, and you will get ∇ε terms and ε· terms. In (7.3.4), Weinberg has combined them into a single 4-dimensional expression, using ∂ε/∂xμ ≡ (∇ε, ε·)
     
  6. Dec 10, 2012 #5
    That’s exactly what I do and I don’t get the correct result. Here are my computations:
    -using the un-numbered equation you mentioned I get:
    [tex]\delta I\left[ \Psi \right]=\sum\limits_{\ell }{\int{{{d}^{4}}x\left\{ \frac{\partial L}{\partial {{\Psi }^{\ell }}}i\epsilon \left( x \right){{F}^{\ell }}\left( x \right)+\frac{\partial L}{\partial \left( {{\partial }_{\mu }}{{\Psi }^{\ell }} \right)}{{\partial }_{\mu }}\left[ i\epsilon \left( x \right){{F}^{\ell }}\left( x \right) \right] \right\}}}[/tex]
    -expanding the derivative I get:
    [tex]\delta I\left[ \Psi \right]=\sum\limits_{\ell }{\int{{{d}^{4}}x\left\{ \left[ \frac{\partial L}{\partial {{\Psi }^{\ell }}}{{F}^{\ell }}\left( x \right)+\frac{\partial L}{\partial \left( {{\partial }_{\mu }}{{\Psi }^{\ell }} \right)}{{\partial }_{\mu }}{{F}^{\ell }}\left( x \right) \right]i\epsilon \left( x \right)+\frac{\partial L}{\partial \left( {{\partial }_{\mu }}{{\Psi }^{\ell }} \right)}{{F}^{\ell }}\left( x \right)i{{\partial }_{\mu }}\epsilon \left( x \right) \right\}}}[/tex]
    The first term, which is multiplied by ε(x), is not of the (7.32) form, since:
    [tex]\frac{\delta I\left[ \Psi \right]}{\delta {{\Psi }^{\ell }}\left( x \right)}i\epsilon \left( x \right){{F}^{\ell }}\left( x \right)=\left[ \frac{\partial L}{\partial {{\Psi }^{\ell }}}-{{\partial }_{\mu }}\frac{\partial L}{\partial \left( {{\partial }_{\mu }}{{\Psi }^{\ell }} \right)} \right]i\epsilon \left( x \right){{F}^{\ell }}\left( x \right) [/tex]
    and not:
    [tex]\frac{\delta I\left[ \Psi \right]}{\delta {{\Psi }^{\ell }}\left( x \right)}i\epsilon \left( x \right){{F}^{\ell }}\left( x \right)\ne \left[ \frac{\partial L}{\partial {{\Psi }^{\ell }}}{{F}^{\ell }}\left( x \right)+\frac{\partial L}{\partial \left( {{\partial }_{\mu }}{{\Psi }^{\ell }} \right)}{{\partial }_{\mu }}{{F}^{\ell }}\left( x \right) \right]i\epsilon \left( x \right) [/tex]
    So I use integration by parts to turn this the term term to:
    [tex]\left[ \frac{\partial L}{\partial {{\Psi }^{\ell }}}{{F}^{\ell }}\left( x \right)+\frac{\partial L}{\partial \left( {{\partial }_{\mu }}{{\Psi }^{\ell }} \right)}{{\partial }_{\mu }}{{F}^{\ell }}\left( x \right) \right]i\epsilon \left( x \right)\to \frac{\partial L}{\partial {{\Psi }^{\ell }}}{{F}^{\ell }}\left( x \right)-{{\partial }_{\mu }}\left[ \frac{\partial L}{\partial \left( {{\partial }_{\mu }}{{\Psi }^{\ell }} \right)}i\epsilon \left( x \right) \right]{{F}^{\ell }}\left( x \right) [/tex]
    But when I expand the derivative, the terms that contain derivatives of ε(x) cancel out. What is my mistake?
     
  7. Dec 10, 2012 #6

    Bill_K

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    Note carefully the arguments. The action is I[ψ], a functional of one variable. So (7.3.2) is little more than a definition, δI ≡ ∫d4x δI/δψ δψ. In that sense it's a completely general form and doesn't need to be derived! The only assertions in it are that δI = 0 and that ε may be pulled out of the integral.

    By contrast, the Lagrangian is L[ψ, ψ·], a functional of two variables. When you integrate by parts (d/dt) you get (7.2.4). The Lagrangian density ℒ[ψ, ∇ψ, ψ·] is a functional of all three. It's only when you're talking about δℒ that an integration by parts on ∇ ever comes in.
     
  8. Dec 10, 2012 #7
    In my previous post L stands for the lagrangian density (I don't know why it appeared like this)...
     
  9. Dec 10, 2012 #8

    Bill_K

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    But in what you're trying to prove, (7.3.2), only I[ψ] appears. Calculations involving L or ℒ aren't relevant.
     
  10. Dec 11, 2012 #9
    All these look quite ambiguous to me and I cannot approach rigorously these results. The only thing I can do in order to get some reasonable results, is to use the equations of motion but this is not the case. Anyaway, I will struggle it a little more and see what happens. Thaks a lot for your time!
     
  11. Dec 12, 2012 #10

    samalkhaiat

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    Weinberg is not the best person to teach you Noether Theorem. He usually uses the so called Levy and Gell-Mann procedure of the 1960. The procedure is useful for giving a quick way of finding the Noether currents and their divergences even in cases where the symmetry is not exact. So, the procedure is about finding the currents not proving the Noether theorem. It works fine for internal symmetries but it get very messy in the more complicated space-time symmetries such as conformal and super-Poincare symmetries. Below, I will explain the Levy & Gell-Mann method. But before I do that, I need to say some thing about the First Noether Theorem. When reading textbooks or listening to your professor, you often here the sentence “continuous symmetry implies conservation law”. This is, however, not what Noether proved in her first theorem. The correct statement of the theorem is “Global continuous symmetry implies, and is implied by, an identity, we now call The Noether Identity”. Mathematically, the first theorem says: giving a global group of transformations:
    [tex]
    \varphi_{r}(x) \rightarrow\varphi_{r}(x) + \Gamma_{r}{}^{a}( \varphi , \partial \varphi) \ \epsilon_{a},
    [/tex]
    where [itex]\epsilon_{a}[/itex] are the infinitesimal constant parameters of the group and [itex]\Gamma[/itex]'s are Lie algebra-valued functions, then
    [tex]
    \{ \frac{\delta S[ \varphi ]}{\delta \epsilon_{a}} = 0 \} \Longleftrightarrow \{ \frac{ \delta \mathcal{L}}{\delta \varphi_{s}} \ \Gamma_{s}{}^{a} (\varphi) + \partial^{\mu}\mathcal{J}_{\mu}^{a}=0 \}.
    [/tex]
    where
    [tex]
    \mathcal{J}^{a}{}_{\mu} = \frac{ \partial \mathcal{L}}{ \partial ( \partial^{\mu} \varphi_{r} )} \ \Gamma^{a}{}_{r}( \varphi ) + X^{a}{}_{\mu}(x) \mathcal{L},
    [/tex]
    is the Noether current,[itex]X^{a}{}_{\mu}[/itex] is the transformation matrix: [itex]\delta x^{\mu} = X^{\mu a}(x) \ \epsilon_{a}[/itex],
    [tex]
    \frac{\delta \mathcal{L}}{\delta \varphi_{r}} \equiv \frac{\partial \mathcal{L}}{\partial \varphi_{r}} - \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial \left( \partial_{\mu}\varphi_{r}\right)}\right),
    [/tex]
    is the Euler derivative, and
    [tex]S[ \varphi ] = \int_{\Omega}d^{n}x \ \mathcal{L}( x, \varphi, \partial \varphi ), [/tex]
    is the action integral over simply contractible domain [itex]\Omega \subseteq \mathbb{R}^{n}[/itex]. As you can see the statement “symmetry implies conserved current” is just the on-shell statement of the Noether identity, i.e. when the fields satisfy the Euler-Lagrange equations of motion
    [tex]\frac{\delta \mathcal{L}}{\delta \varphi_{r}}=0.[/tex]
    The proof is easily done by calculating the variational derivative of the action integral. This is made up of the sum of the variation of the Lagrangian and of the variation in the region of integration;
    [tex]\frac{ \delta S[\varphi]}{\delta \epsilon_{a}} = \int d^{n}x \ \frac{\delta \mathcal{L}}{\delta \epsilon_{a}} + \int \ \mathcal{L} \ \frac{ \delta }{ \delta \epsilon_{a}}( d^{n}x ),[/tex]
    where, from the Jacobian of the transformation,
    [tex]
    \frac{\delta}{\delta \epsilon_{a}}\ ( d^{n}x ) \approx d^{n}x \ \partial_{\mu}( \frac{ \delta x^{\mu}}{\delta \epsilon_{a}}) = d^{n}x \ \partial_{\mu}X^{\mu a}, \ \ (J)
    [/tex]
    and
    [tex]
    \frac{\delta\mathcal{L}}{\delta \epsilon_{a}}= \frac{ \partial \mathcal{L}}{ \partial x^{\mu}}\ \frac{\delta x^{\mu}}{\delta \epsilon_{a}} + \frac{ \partial \mathcal{L}}{\partial \varphi_{r}}\ \frac{\delta \varphi_{r}}{\delta \epsilon_{a}}+ \frac{\partial \mathcal{L}}{ \partial ( \partial_{\mu} \varphi_{r})} \ \partial_{\mu}( \frac{\delta \varphi_{r}}{ \delta \epsilon_{a}} ),
    [/tex]
    or
    [tex]
    \frac{ \delta \mathcal{L}}{ \delta \epsilon_{a}} = X^{\mu a}\ \partial_{\mu}\mathcal{L} + \{ \frac{ \partial \mathcal{L}}{ \partial \varphi_{r}} - \partial_{\mu}( \frac{\partial \mathcal{L}}{ \partial ( \partial_{\mu} \varphi_{r} ) } )\} \Gamma^{a}{}_{r} + \partial_{\mu} \left( \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi_{r} ) } \ \Gamma^{a}{}_{r} \right)
    [/tex]
    Putting this and eq(J) in the variation derivative of the action, we find
    [tex]
    \frac{\delta S[ \varphi ]}{\delta \epsilon_{a}} = \int_{\Omega} d^{n}x \ \left[ \frac{\delta\mathcal{L}}{\delta\varphi_{r}} \ \Gamma^{a}{}_{r} + \partial_{\mu} \left( X^{\mu a} \ \mathcal{L} + \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi_{r} ) }\ \Gamma^{a}{}_{r}\right) \right] .
    [/tex]
    The statement of Noether first theorem follows from the above equation, because [itex]\Omega[/itex] is simply contractible.

    Ok, let us try to explain the Levy and Gell-Mann method of generating Noether currents of the global symmetries of the Lagrangian [itex]\mathcal{L}(\varphi, \partial \varphi )[/itex]. It is assumed that for constant [itex]\epsilon[/itex] the Lagrangian is invariant, that is
    [tex]\frac{\delta \mathcal{L}}{\delta \epsilon} = 0.[/tex]
    Gell-Mann and Levy then considered the transformation
    [tex]
    \varphi \rightarrow \varphi + \epsilon (x) \ F( \varphi ),
    [/tex]
    and studied the change in [itex]\mathcal{L}[/itex]. From the fact that [itex]\epsilon (x)[/itex] is infinitesimal, so that [itex]\epsilon^{2}(x) \approx 0[/itex], it follows that both [itex]( \delta \mathcal{L}/ \delta \epsilon )[/itex] and [itex]( \delta \mathcal{L} / \delta \partial_{\mu}\epsilon)[/itex] are independent of [itex]\epsilon[/itex]. So, they could define the object
    [tex]
    J^{\mu}(x) \equiv \frac{\delta \mathcal{L}}{ \delta ( \partial_{\mu}\epsilon )},
    [/tex]
    and then prove that
    [tex]\frac{\delta\mathcal{L}}{\delta \epsilon} = \partial_{\mu} \left( \frac{ \delta \mathcal{L}}{ \delta ( \partial_{\mu}\epsilon )} \right) \equiv \partial_{\mu}J^{\mu},[/tex]
    when the field satisfies the Euler-Lagrange equation. In general, one can show that
    [tex]
    \frac{\delta \mathcal{L}}{\delta \epsilon(x)} - \partial_{\mu}\left( \frac{\delta \mathcal{L}}{\delta ( \partial_{\mu}\epsilon )}\right) = \left[ \frac{\partial \mathcal{L}}{\partial \varphi} - \partial_{\mu} \left( \frac{ \partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi )}\right) \right] \ F( \varphi )
    [/tex]
    Since for constant [itex]\epsilon[/itex] the transformation above is a global symmetry of the Lagrangian, [itex]\delta \mathcal{L}/ \delta \epsilon = 0[/itex], thus the “current” constructed for non-constant [itex]\epsilon[/itex] is the conserved current of the global symmetry. If the global transformation does not leave [itex]\mathcal{L}[/itex] invariant, then the divergence of the associated current is given directly by
    [tex]
    \partial_{\mu}J^{\mu} = \frac{\delta \mathcal{L}}{ \delta \epsilon (x)}.
    [/tex]
    So, when you make [itex]\epsilon \rightarrow \epsilon (x)[/itex] and calculate the change in [itex]\mathcal{L}[/itex], the Noether current of the global transformation will show up as the coefficient of [itex]\partial_{\mu}\epsilon[/itex], while [itex]\partial_{\mu}J^{\mu}[/itex] appears with [itex]\epsilon (x)[/itex],
    [tex]
    \delta \mathcal{L} = J^{\mu}\ \partial_{\mu}\epsilon + \epsilon \ \partial_{\mu}J^{\mu}.
    [/tex]

    Sam
     
  12. Dec 14, 2012 #11
    Thank you very much for your complete analysis. It will certainly clear the things up...
     
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