3
$\begingroup$

I have a question concerning the derivation of Noether's theorem in Peskin & Schroeders introduction to QFT. Below is a picture of the first part of it.

The proof then proceeds by saying that one term vanishes because of the euler-lagrange equations. It seems to me like what they are saying is that the term only vanishes because we have a symmetry transformation. But to me this is the general behaviour of the lagrangian under a transformation and the term in the brackets will always vanish, regardless of the transformation. So my question is: Why does the term in the square brackets in eq. (2.11) only vanish for symmetry transformations?

Peskin

$\endgroup$
1
$\begingroup$

The writing is certainly less clear than it ought to be. Your instincts are right; they just didn't explain it well. In particular, they're not saying that the term in brackets only vanishes for symmetry transformations. It always vanishes for a field obeying the Euler-Lagrange equations.

What they're really saying is:

  1. A continuous transformation can be written as equation (2.9)
  2. The equation of motion will be invariant under that continuous transformation if the Lagrangian changes as equation (2.10)
  3. Let's see how the Lagrangian actually changes under the transformation of (2.9)
  4. It changes as a term that looks like $\alpha \partial_\mu \mathcal{J}^\mu$ plus a term that goes to zero because of the Euler-Lagrange equations for the original, un-transformed system
  5. The actual change we get is the change we demanded for the Lagrangian to be invariant up to a 4-divergence
  6. Therefore, the continuous transformation is a symmetry transformation

The text immediately following what the OP included says

The second term [in equation (2.11)] vanishes by the Euler-Lagrange equation. We set the remaining term equal to $\alpha \partial_\mu \mathcal{J}^\mu$ and find \begin{equation} \partial_\mu j^\mu(x) = 0\, \qquad \mathrm{for} \qquad j^\mu(x) = \frac{\partial \mathcal{L}} {\partial (\partial_\mu \phi)} \Delta \phi - \mathcal{J}^\mu. \end{equation}

This really makes it look like they're saying that $j^\mu$ is identically zero; in fact, I would argue that that's the most reasonable reading of the text. What they should have said is that we get the generic result \begin{equation} \Delta \mathcal{L} = \partial_\mu \left( \frac{\partial \mathcal{L}} {\partial (\partial_\mu \phi)} \Delta \phi \right) \tag{A} \end{equation} whenever $\phi$ satisfies the Euler-Lagrange equations, but to call $\Delta \phi$ a symmetry transformation we require the existence of some other field $\mathcal{J}^\mu$ such that \begin{equation} \Delta \mathcal{L} = \partial_\mu \mathcal{J}^\mu \tag{B} \end{equation} regardless of whether or not the Euler-Lagrange equations are satisfied. Again, (A) is generically true for any differentiable Lagrangian and field satisfying the E-L equations. On the other hand, you have to use the particular form of $\mathcal{L}$ and the particular form of $\Delta \phi$ to decide if (B) is true in any particular case. (Also note that $\mathcal{J}^\mu = 0$ is perfectly acceptable.)

It probably would have been clearer if P&S had begun by moving the explanation of what a symmetry transformation is [the four sentences up to equation (2.10)] to a previous paragraph and expanding on it. Then they could introduce a continuous transformation and show how the Lagrangian changes. But why bother editing a piece of text for clarity, when it's only explaining something as frivolous and useless as Noether's Theorem? ;)

These issues are made much clearer in @joshphysics's version of the proof — in particular with his introduction of separate $\mathcal{J}^\mu$ and $\mathcal{K}^\mu$ vector fields. They look like they perform the same function, and yet they're different, as he explains in the "Important Notes" section at the bottom of his answer, and in some of the comments.

I also highly recommend the excellent "Quantum Field Theory for the Gifted Amateur", which does a typically superb job on Noether's Theorem, and is really what I think should be the standard text. The title is deceptive because it's aimed at professional physicists, just not current professional quantum-field theorists. I encourage you to at least read the other answer, or even take up "Gifted Amateur" if you really want to understand QFT.

$\endgroup$
  • 1
    $\begingroup$ something as frivolous and useless as Noether's Theorem? LOL $\endgroup$ – yngabl Nov 8 '17 at 17:21
  • 1
    $\begingroup$ Thank you for your quick answer. It seems I understood the steps correctly, I just didn't understand what they were trying to prove. After your explanation I now believe that any transformation of the form (2.9) is a symmetry transformation that will leave the e.o.m. invariant. Is that correct? As I am typing it, I am already less convinved of this, as it looks like the most general form one could write a transformation of the field. So maybe as a follow up question: If I did understand your answer correctly, what would a transformation look like which is not a symmetry transformation? $\endgroup$ – Tim Lorenzo Fleischmann Nov 9 '17 at 0:38
  • $\begingroup$ Again, very good question; you're thinking clearly about this, and any confusion you're experiencing seems to be coming from poor explanations that you're getting (including mine). I've edited my answer to explain this a little more. $\endgroup$ – Mike Nov 9 '17 at 17:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.