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In the book 'Conformal Field Theory' by Di Francesco et al, a derivation of Noether's theorem is demonstrated by imposing that, what I believe is said to be a more elegant approach, the parameter $\omega$ is explicitly $x$-dependent, so that $\omega = \omega(x)$ for a local transformation and then finally at the end considering a global transformation that consequently results in Noether's theorem. The derivation is on pages 39-41.

I understand all of the derivation, however, it has been brought to my attention that the whole derivation seems to rest on an inconsistent starting point that would therefore make the rest of the argument, while mathematically correct, completely useless.

On P.39, Di Francesco writes that the generic infinitesimal transformations of the coordinates and field are, respectively, $$x'^{\mu} = x^{\mu} + \omega_a \frac{\delta x^{\mu}}{\delta \omega_a}$$$$\Phi'(x') = \Phi(x) + \omega_a \frac{\delta F}{\delta \omega_a}(x).$$ This is written under the assumption that the $\omega_a$ are infinitesimal parameters ,not functions of $x$. So when he uses this result in his derivation on P.40, and says that he will make the supposition that the $\omega_a$ are dependent on $x$, how can he do this?

For clarity, in case I missed something, I was having this discussion here: http://www.physicsforums.com/showthread.php?t=760137 and in post 14 is where the subtlety arises. So is it really a flaw? I am just really looking for another opinion on this. I asked one of the professors at my university and he said provided the $x$ dependence is 'small' then it is valid, but I am not quite sure exactly what that means.

Qmechanic
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CAF
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1 Answers1

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I) OP's question (v2) seems to be essentially a question of mathematical precision vs. the way physicists express themselves concisely when speaking of "infinitesimals" without becoming too technical by introducing epsilons and deltas and what not. See also this related Phys.SE post.

Perhaps the easiest and most elementary way to make sense of the "infinitesimal function" $\omega(x)$ in Di Francesco et. al., CFT, is to think of it as a product$^1$

$$ \omega(x)~=~\delta~f(x), $$

where $f(x)$ is a bounded function$^2$, e.g. $|f(x)|\leq 1$; and $\delta$ is an "infinitesimal constant". An "infinitesimal constant" is just physics-jargon for a small number $\delta>0$ so small, that we can neglect higher order contributions of $\delta$ in the calculation, to the precision $\epsilon>0$ that we are working.

II) Concerning the derivation of Noether's theorem via the trick of an $x$-dependent $\omega(x)$, see this Phys.SE post.

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$^1$ We have dropped the index $a$ in $\omega_a(x)$ for simplicity.

$^2$ Technically, it may be convenient to further assume that the function $f(x)$ is differentiable with bounded derivative, perhaps even of compact support. The function $f(x)$ has a role not unlike a test function.

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Qmechanic
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  • Hi Qmechanic. So, just to clarify, Di Francesco's derivation seems ok to you? Basically, he is writing $\omega$ as a function of $x$ throughout, but this $x$ dependence expressed through your function $f(x)$ is suppressed by the constant $\delta$ which permits $\omega$ to always be viewed as an infinitesimal function of $x$? – CAF Jul 20 '14 at 13:54
  • As always when physicists write maths, it is up to the reader to dot the i's and cross the t's. In other words, if you can find a counterexample to one of Di Francesco et al's arguments, it is likely excluded by an implicitly implied assumption, such as, e.g., all functions are assumed to be sufficiently many times differentiable, etc. – Qmechanic Jul 20 '14 at 14:38
  • Ok, thanks. So the way Di Francesco originally wrote the infinitesimal transformations (as written in the OP) are for a constant $\omega$. Those equations are also true for a local transformation, $\omega = \omega(x)$ provided $\omega(x)$ is of the form that you wrote, $\omega(x) = \delta~ f(x)$. Would that be right? – CAF Jul 20 '14 at 15:15
  • Up to higher-order contributions, and modulo above caveats: Yes. – Qmechanic Jul 20 '14 at 15:27
  • I have a final question about the statement that a symmetry transformation induces a conserved quantity by virtue of Noether's theorem. When we say a 'symmetry transformation' I understand that this means one in which the action functional remains invariant, but is this invariancy caused by a trivial transformation of the field? E.g lorentzian symmetry transformations only correspond to those where the field is a lorentz scalar (which is to say it belongs to the trivial rep of the lorentz group)? – CAF Jul 21 '14 at 08:55
  • Comments: 1. A (quasi)symmetry transformation preserves the action $S$ (up to boundary terms). 2. When discussing a symmetry of the action $S$ under a group $G$, the variables/fields themselves do not transform in (possibly multiple copies of) the trivial irrep $G$. If they did, the corresponding conservation law would be a triviality a la $0=0$. (Note that horizontal transformations $\delta x=\ldots$ may induce different terminology of a rep: What appears as a trivial rep from one point of view (pov) may be non-trivial from another pov.) – Qmechanic Jul 21 '14 at 09:13
  • Given $$S = \int_{\mathcal D},\text{d}^d x \mathcal L(\phi, \partial \phi)$$ we have that $$S' = \int_{\mathcal D},\text{d}^d x \left|\frac{\partial x'}{\partial x}\right| \mathcal L(F(\phi(x)), \partial_{\mu}' F(\phi(x))),$$ so this would suggest to me that provided, a) the Jacobian factor $|\partial x'/\partial x|$ is unity, b)$F(\phi(x)) = \phi(x)$, c)$\partial_{\mu}' F(\phi(x)) = \partial_{\mu}\phi$

    then the transformation is deemed to be a symmetry. Would that be right?

     – CAF Jul 21 '14 at 10:45
  • I suppose those conditions a)- c) would always make $S'=S$, however, other symmetries could arise depending on the form and structure of the lagrangian. To illustrate what I mean, under the transformation $\phi'(x) = e^{i\theta}\phi(x)$ we have an invariant action for a lagrangian given here, bottom of first page, http://www.itp.phys.ethz.ch/research/qftstrings/archive/12HSQFT1/Chapter04.pdf – CAF Jul 21 '14 at 10:54