If we consider a transformation of a field ##\Phi \rightarrow \Phi + \alpha \frac{\partial \Phi}{\partial \alpha}## which is not a symmetry of a lagrangian then one can show that the Noether current is not conserved but that instead ##\partial_{\mu}J^{\mu} = \frac{\partial L}{\partial \alpha}##.(adsbygoogle = window.adsbygoogle || []).push({});

I think the way this is derived is as follows $$\delta S = \int d^4 x \left( \left( \frac{\partial L}{\partial \Phi} \delta \Phi - \partial_{\mu} \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \right) \delta \Phi + \partial_{\mu} \left(\frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) \right)$$ Then the first term is zero due to the equations of motion and so we are left with the second term.

Writing out ##\delta L = \frac{\partial L}{\partial \Phi}\delta \Phi + \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta (\partial_{\mu} \Phi)##, inserting what ##\delta \Phi## is we see that ##\delta L = \alpha \frac{\partial L}{\partial \alpha}## Then we can compare with the above and deduce the result more or less.

My questions are:

What permits the use of the equations of motion here? If the equations of motion hold then ##\delta S = 0 ## identically. Using the equations of motion gives me ##\int \partial_{\mu} J^{\mu} d^4 x = \delta S## but since I used the equations of motion isn't this just equal to zero?

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# I Technicality with Noether's theorem

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