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  Subtlety in derivation of Noether's theorem by Di Francesco

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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.

This post imported from StackExchange Physics at 2014-07-20 11:21 (UCT), posted by SE-user CAF
asked Jul 20, 2014 in Theoretical Physics by CAF (100 points) [ no revision ]

1 Answer

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The statement that the x-dependence is small means that not one $\omega$ is small, but the spatial derivatives are small. The justification for using low-infinitesimal orders is the same whether $\omega$ is a parameter or a slow-varying function, you can expand a differentiable function in orders. There is no subtlety.

answered Jul 25, 2014 by Ron Maimon (7,720 points) [ revision history ]

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