(The problem I have is really at the end, however, I have provided my whole argument in detail for clarity and completeness at the cost of perhaps making the thread very unappealing to read)

(c.f Di Francesco's book, P.41) We are given that the transformed action under an infinitesimal transformation is $$S' = \int d^d x \left(1 + \partial_{\mu}\left(w_a \frac{\delta x^{\mu}}{\delta w_a}\right)\right) L\left(\Phi + w_a \frac{\delta F}{\delta w_a}, [\delta^{\nu}_{\mu} - \partial_{\mu}\left(w_a\left(\frac{\delta x^{\nu}}{\delta w_a}\right)\right)](\partial_{\nu}\Phi + \partial_{\nu}[w_a\left(\frac{\delta F}{\delta w_a}\right)])\right)$$

To consider $\delta S = S' - S$, I am looking to expand the above result to first order.

I can multiply the brackets above to obtain $$S' = \int d^dx\, L(\Phi + w_a \frac{\delta F}{\delta w_a}, \partial_{\mu} \Phi + \partial_{\mu} (w_a \frac{\delta F}{\delta w_a}) - \partial_{\mu} (w_a \frac{\delta x^{\nu}}{\delta w_a})\partial_{\nu} \Phi) +$$ $$ \int d^d x\, \partial_{\mu} (w_a \frac{\delta x^{\mu}}{\delta w_a})L(\Phi + w_a \frac{\delta F}{\delta w_a}, \partial_{\mu}\Phi + \partial_{\mu}(w_a \frac{\delta F}{\delta w_a}) - \partial_{\mu} (w_a \frac{\delta x^{\nu}}{\delta w_a})\partial_{\nu} \Phi$$

Then I can taylor expand the expansion above to first order in the parameters. This gives $$S' = \int d^d x \left[L(\Phi, \partial_{\mu}\Phi) + \omega_{a} \frac{\delta F}{\delta \omega_a} \frac{\partial L}{\partial \Phi} + \partial_{\mu} (\omega_{a} \frac{\delta F}{\delta \omega_a}) \frac{\partial L}{\partial(\partial_{\mu}\Phi)} - \partial_{\mu}(\omega_a \frac{\delta x^{\nu}}{\delta \omega_a})\partial_{\nu}\Phi \frac{\partial L}{\partial(\partial_{\mu}\Phi)}\right] + \int d^d x \partial_{\mu}(\omega_a \frac{\delta x^{\mu}}{\delta \omega_a})[..]$$ where [..] is the terms in the bracket in the preceding integral. Most of these terms will be ignored (in fact all but the first) since they will of higher order in the parameters. The variation is then $$\delta S = S'-S = \int d^d x \omega_a \frac{\delta F}{\delta \omega_a}\frac{\partial L}{\partial \Phi} + \partial_{\mu} (\omega_a \frac{\delta F}{\delta \omega_a})\frac{\partial L}{\partial(\partial_{\mu}\Phi)} - \partial_{\mu}(\omega_a \frac{\delta x^{\nu}}{\delta \omega_a}) \partial_{\nu}\frac{\partial L}{\partial(\partial_{\mu}\Phi)}+\int d^d x \partial_{\mu}(\omega_a \frac{\delta x^{\mu}}{\delta \omega_a})L(\Phi, \partial_{\mu}\Phi)$$

Now perform the derivatives explicitly, grouping together terms in ##\partial_{\mu}\omega_a## and ##\omega_a## and imposing invariance of action: $$0 = \delta S = \int d^d x\,\partial_{\mu} \omega_a \left[\frac{\delta F}{\delta \omega_a} \frac{\partial L}{\partial(\partial_{\mu}\Phi)} - \frac{\delta x^{\nu}}{\delta \omega_a}\partial_{\nu}\Phi \frac{\partial L}{\partial(\partial_{\mu}\Phi)} + \frac{\delta x^{\mu}}{\delta \omega_a}L\right] +$$ $$ \omega_a\left[ \frac{\delta F}{\delta \omega_a}\frac{\partial L}{\partial \Phi} + (\partial_{\mu}\frac{\delta F}{\delta \omega_a})\frac{\partial L}{\partial(\partial_{\mu}\Phi)} - \partial_{\mu} (\frac{\delta x^{\nu}}{\delta \omega_a})\partial_{\nu}\Phi \frac{\partial L}{\partial(\partial_{\mu}\Phi)} + \partial_{\mu} (\frac{\delta x^{\mu}}{\delta \omega_a})L\right]$$

The answer in the book is that ##\int d^d x j^{\mu} \partial_{\mu}\omega_a= 0 ##.The terms multiplying ##\partial_{\mu}\omega_a## is exactly ##j^{\mu}## in the book. So I would have the right answer, provided all the terms multiplying ##\omega_a## would vanish. Indeed, the first two do as a result of applying the classical equations of motion for the field ##\Phi##. But I do not see how the final two terms vanish (or indeed if they do). I have tried again using the E.O.Ms, integration by parts to no success. I then thought I might be able to ignore these terms but I had no justification. Any thoughts would be great.

Many thanks.

**1. The problem statement, all variables and given/known data**(c.f Di Francesco's book, P.41) We are given that the transformed action under an infinitesimal transformation is $$S' = \int d^d x \left(1 + \partial_{\mu}\left(w_a \frac{\delta x^{\mu}}{\delta w_a}\right)\right) L\left(\Phi + w_a \frac{\delta F}{\delta w_a}, [\delta^{\nu}_{\mu} - \partial_{\mu}\left(w_a\left(\frac{\delta x^{\nu}}{\delta w_a}\right)\right)](\partial_{\nu}\Phi + \partial_{\nu}[w_a\left(\frac{\delta F}{\delta w_a}\right)])\right)$$

To consider $\delta S = S' - S$, I am looking to expand the above result to first order.

I can multiply the brackets above to obtain $$S' = \int d^dx\, L(\Phi + w_a \frac{\delta F}{\delta w_a}, \partial_{\mu} \Phi + \partial_{\mu} (w_a \frac{\delta F}{\delta w_a}) - \partial_{\mu} (w_a \frac{\delta x^{\nu}}{\delta w_a})\partial_{\nu} \Phi) +$$ $$ \int d^d x\, \partial_{\mu} (w_a \frac{\delta x^{\mu}}{\delta w_a})L(\Phi + w_a \frac{\delta F}{\delta w_a}, \partial_{\mu}\Phi + \partial_{\mu}(w_a \frac{\delta F}{\delta w_a}) - \partial_{\mu} (w_a \frac{\delta x^{\nu}}{\delta w_a})\partial_{\nu} \Phi$$

Then I can taylor expand the expansion above to first order in the parameters. This gives $$S' = \int d^d x \left[L(\Phi, \partial_{\mu}\Phi) + \omega_{a} \frac{\delta F}{\delta \omega_a} \frac{\partial L}{\partial \Phi} + \partial_{\mu} (\omega_{a} \frac{\delta F}{\delta \omega_a}) \frac{\partial L}{\partial(\partial_{\mu}\Phi)} - \partial_{\mu}(\omega_a \frac{\delta x^{\nu}}{\delta \omega_a})\partial_{\nu}\Phi \frac{\partial L}{\partial(\partial_{\mu}\Phi)}\right] + \int d^d x \partial_{\mu}(\omega_a \frac{\delta x^{\mu}}{\delta \omega_a})[..]$$ where [..] is the terms in the bracket in the preceding integral. Most of these terms will be ignored (in fact all but the first) since they will of higher order in the parameters. The variation is then $$\delta S = S'-S = \int d^d x \omega_a \frac{\delta F}{\delta \omega_a}\frac{\partial L}{\partial \Phi} + \partial_{\mu} (\omega_a \frac{\delta F}{\delta \omega_a})\frac{\partial L}{\partial(\partial_{\mu}\Phi)} - \partial_{\mu}(\omega_a \frac{\delta x^{\nu}}{\delta \omega_a}) \partial_{\nu}\frac{\partial L}{\partial(\partial_{\mu}\Phi)}+\int d^d x \partial_{\mu}(\omega_a \frac{\delta x^{\mu}}{\delta \omega_a})L(\Phi, \partial_{\mu}\Phi)$$

Now perform the derivatives explicitly, grouping together terms in ##\partial_{\mu}\omega_a## and ##\omega_a## and imposing invariance of action: $$0 = \delta S = \int d^d x\,\partial_{\mu} \omega_a \left[\frac{\delta F}{\delta \omega_a} \frac{\partial L}{\partial(\partial_{\mu}\Phi)} - \frac{\delta x^{\nu}}{\delta \omega_a}\partial_{\nu}\Phi \frac{\partial L}{\partial(\partial_{\mu}\Phi)} + \frac{\delta x^{\mu}}{\delta \omega_a}L\right] +$$ $$ \omega_a\left[ \frac{\delta F}{\delta \omega_a}\frac{\partial L}{\partial \Phi} + (\partial_{\mu}\frac{\delta F}{\delta \omega_a})\frac{\partial L}{\partial(\partial_{\mu}\Phi)} - \partial_{\mu} (\frac{\delta x^{\nu}}{\delta \omega_a})\partial_{\nu}\Phi \frac{\partial L}{\partial(\partial_{\mu}\Phi)} + \partial_{\mu} (\frac{\delta x^{\mu}}{\delta \omega_a})L\right]$$

The answer in the book is that ##\int d^d x j^{\mu} \partial_{\mu}\omega_a= 0 ##.The terms multiplying ##\partial_{\mu}\omega_a## is exactly ##j^{\mu}## in the book. So I would have the right answer, provided all the terms multiplying ##\omega_a## would vanish. Indeed, the first two do as a result of applying the classical equations of motion for the field ##\Phi##. But I do not see how the final two terms vanish (or indeed if they do). I have tried again using the E.O.Ms, integration by parts to no success. I then thought I might be able to ignore these terms but I had no justification. Any thoughts would be great.

Many thanks.

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